trigonometry
greatest common divisor
ratio
geometry
geometric transformation
rotation
complex numbers

[color=darkblue][b]Method I (complex numbers).[/b] Denote $ w = \cos\frac {\pi}{3} + i \cdot\sin\frac {\pi}{3}$. Observe that $ \|\begin{array}{c} w = \frac 12 + \frac {\sqrt 3}{2}\cdot i \\
 \\
w^3 = - 1\ ,\ w^2 = w - 1 \\
 \\
\overline w = \frac 1w = - w^2\end{array}$.

Consider the complex plane of the points $ X(x)$ with the affix $ x\in \mathcal C$. Thus,

$ \|\begin{array}{c} A(a)\ ,\ B(b)\ ,\ C(c)\ ,\ D(d) \\
 \\
X(x)\ ,\ Y(y)\ ,\ Z(z)\ ,\ T(t) \\
 \\
E(e)\ ,\ F(f)\ ,\ G(g)\end{array}$,

where the triangles $ ABX$, $ BCY$, $ CDZ$, $ DAT$ are equilaterally

(in the exterior of $ ABCD$),  the points $ E$, $ F$ are the midpoints

of $ [AT]$, $ [BY]$ respectrively and $ G$ is the center of $ \triangle CDZ$.

Ascertain $ \|\begin{array}{c} x = (1 - w)a + wb \\
 \\
y = (1 - w)b + wc \\
 \\
z = (1 - w)c + wd \\
 \\
t = (1 - w)d + wa\end{array}$ and $ \|\begin{array}{c} 2e = (1 + w)a + (1 - w)d \\
 \\
2f = (2 - w)b + wc \\
 \\
3g = (2 - w)c + (1 + w)d\end{array}$.

Calculate $ \|\begin{array}{c} 3(x - g) = 3(1 - w)a + 3wb - (2 - w)c - (1 + w)d \\
 \\
2(e - f) = (1 + w)a + (w - 2)b - wc + (1 - w)d\end{array}$.

Observe that $ \frac {3(1 - w)}{1 + w} = \frac {3w}{w - 2} = \frac {2 - w}{w} = \frac {1 + w}{w - 1} = 1 - 2w = - \sqrt 3\cdot i$ .

Thus, $ 3(x - g) = - 2i\sqrt 3(e - f)$, i.e. $ XG\perp EF$ and $ \sqrt 3\cdot XG = 2\cdot EF\ .$

[b]Method II (metrical).[/b] Denote $ \|\begin{array}{c} AB = a\ ,\ BC = b\ ,\ CD = c \\
 \\
DA = d\ ,\ AC = e\ ,\ BD = f\end{array}$.

Using the [b]generalized Pythagoras' theorem[/b] prove easily that

$ \|\begin{array}{c} \triangle GDE\ : \ GE^2 = \frac {3d^2}{4} + \frac {c^2}{3} - cd\cdot\cos(D + 60) = \frac {3d^2}{4} + \frac {c^2}{3} - \frac 12cd\cdot\cos D + \frac {\sqrt 3}{2}cd\cdot\sin D \\
 \\
\triangle GCF\ : \ GF^2 = \frac {3b^2}{4} + \frac {c^2}{3} - bc\cdot\cos(C + 60) = \frac {3b^2}{4} + \frac {c^2}{3} - \frac 12bc\cdot\cos C + \frac {\sqrt 3}{2}bc\cdot\sin C\end{array}\|\implies$

$ \|\begin{array}{c} GE^2 = \frac {3d^2}{4} + \frac {c^2}{3} - \frac 14\cdot (c^2 + d^2 - e^2) + \sqrt 3\cdot[ACD] \\
 \\
GF^2 = \frac {3b^2}{4} + \frac {c^2}{3} - \frac 14\cdot (b^2 + c^2 - f^2) + \sqrt 3\cdot [BCD]\end{array}\|$ $ \implies$

$ \boxed {GE^2 - GF^2 = \frac 14\cdot [e^2 - f^2 + 2(d^2 - b^2)] + \sqrt 3\cdot([ACD] - [BCD])}\ \ (1)\ .$

$ \|\begin{array}{c} \triangle XAE\ : \ XE^2 = a^2 + \frac {d^2}{4} + ad\cdot\cos (A - 60) = a^2 + \frac {d^2}{4} + \frac 12\cdot ad\cdot\cos A + \frac {\sqrt 3}{2}\cdot ad\cdot\sin A \\
 \\
\triangle XBF\ : \ XF^2 = a^2 + \frac {b^2}{4} + ab\cdot\cos (B - 60) = a^2 + \frac {b^2}{4} + \frac 12\cdot ab\cdot\cos B + \frac {\sqrt 3}{2}\cdot ab\cdot\sin B\end{array}\|\implies$

$ \|\begin{array}{c} XE^2 = a^2 + \frac {d^2}{4} + \frac 14\cdot (a^2 + d^2 - f^2) + \sqrt 3\cdot [ABD] \\
 \\
XF^2 = a^2 + \frac {b^2}{4} + \frac 14\cdot (a^2 + b^2 - e^2) + \sqrt 3\cdot [ABC]\end{array}\|$ $ \implies$

$ \boxed {XE^2 - XF^2 = \frac 14\cdot [e^2 - f^2 + 2(d^2 - b^2)] + \sqrt 3\cdot([ABD] - [ABC])}\ \ (2)\ .$

$ [ACD] - [BCD] = [ABD] - [ABC]$ $ \implies$ $ GE^2 - GF^2 = XE^2 - XF^2$ $ \implies$  $ \boxed {\ XG\perp EF\ }\ .$[/color]

[quote="chien than"]Let $ M;N$ be midpoint of $ AD;BC$,
Prove that $ MIN$ is equilateral triangle.[/quote]

May be it's wrong! Let $ A,D,C$ are fix point and $ B$ mutate then triangle $ MIN$ have $ M,I$ are fix point but $ N$ mutate! Now it's not a equilateral triangle for all case!

[color=darkblue][b]Remark.[/b] The enunciation of this problem becomes more simply if we renounce at the points $ Y$, $ Z$, $ T$ ![/color]

[quote="Chien Than"][color=darkred][b][u]An equivalent enunciation.[/u][/b] Let $ ABCD$ be a convex equilateral. Construct in the its exterior

the [b]equilateral triangle[/b] $ AXB$, the [b]isosceles triangle[/b] $ CGD$ for which $ \|\begin{array}{c} GC = GD \\
\ m(\widehat {CGD}) = 120^{\circ}\end{array}$

and the [b]right triangles[/b] $ AED$, $ BFC$ for which $ \|\begin{array}{c} EA\perp ED\ ,\ FB\perp FC \\
\ m(\widehat {ADE}) = m(\widehat {BCF}) = 30^{\circ}\end{array}\ .$

Prove that the lines $ XG$ and $ EF$ are perpendicularly.[/color][/quote]

[quote="chien than"]Let $ M;N$ be midpoint of $ AD;BC$,
Prove that $ MIN$ is equilateral triangle.[/quote]

I agree with [b][size=100]nkpt[/size][/b], that the triangle $ \bigtriangleup GMN$ ( [b][size=100]Virgil Nicula’s[/size][/b] wording ) is not equilateral, as I can see by the drawing.

As for the proposed problem, I see also by the drawing ( I have not in mind any proof ), that the perpendicularity $ XG\perp EF,$ is a part of a family of perpendicularities, relative to the configuration of a given quadrilateral $ ABCD.$

So, we define the points $ E',$ $ E'',$ on the line segment $ AT,$ such that $ E'A \equal{} E'E$ and $ E''A \equal{} E''E'$ and similarly the points $ F',$ $ F'',$ on the line segment $ BY,$ such that $ F'B \equal{} F'F$ and $ F''B \equal{} F''F'.$

We define the points $ T',$ $ Y',$ on the extensions of the line segments $ AT,$ $ BY,$ towards $ T,$ $ Y$ respectively, such that  $ TA \equal{} TT'$ and $ YB \equal{} YY'.$

Finally, we define the points $ X',$ $ X'',$ on the extension of the midperpendicular $ KX,$ of the side segment $ AB$ towards $ X,$ such that $ X'X \equal{} 2(XG')$ and $ X''X' \equal{} 2(X'X),$ where $ K,$ is the midpoint of the segment $ AB$ and $ G',$ is the center of the equilateral triangle $ \bigtriangleup XAB.$

In the above configuration, the line segments $ GK,$ $ GG',$ $ GX,$ $ GX',$ $ GX'',$ are perpndicular to $ T'Y',$ $ TY,$ $ EF,$ $ E'F',$ $ E''F'',$ respectively.

$ \bullet$ In more general configuration, on two arbitrary lines through the edges $ A,$ $ B,$ of a given segment $ AB,$ we define the points $ E'',$ $ E',$ $ E,$ $ T,$ such that $ AE'' \equal{} x,$ $ AE' \equal{} 2x,$ $ AE \equal{} 4x$ and $ AT \equal{} 8x$ and the points $ F'',$ $ F',$ $ F,$ $ Y,$ such that $ BF'' \equal{} y,$ $ BF' \equal{} 2y,$ $ BF \equal{} 4y$ and $ BY \equal{} 8y.$

On the midperpendicular of the segment $ AB,$ we define the points $ G',$ $ X,$ $ X',$ $ X'',$ such that $ KG \equal{} z,$ $ G'X \equal{} 2z,$ $ XX' \equal{} 4z,$ and $ X'X'' \equal{} 8z,$ where $ K,$ is the midpoint of the given segment $ AB.$

So, the lines through the points $ G',$ $ X,$ $ X',$ $ X''$ and perpendicular to $ TY,$ $ EF,$ $ E'F',$ $ E''F''$ respectively, are concurrent at one point so be it $ G,$ lies on the line through $ K$ and perpendicular to the line segment $ T'Y',$ where $ T',$ $ Y',$ are the points on the extensions of $ AT,$ $ BY,$ towards $ T,$ $ Y$ respectively, such that $ TT' \equal{} TA$ and $ YY' \equal{} YB.$

Kostas Vittas.

PS. I will post here later the relative schemas.

I"ll present soon an extension of this problem !

[quote="chien than"]Given a picture:
[img]http://img151.imageshack.us/img151/9356/200878823943564d0401288fk9.gif[/img]
where $ ABX$, $ BCY$, $ CDZ$, $ DAT$ are equilateral triangles. $ E$, $ F$ are the midpoints the segments $ AT$, $ BY$ and $ I$ the center of triangle $ CDZ$. Prove that $ XI \perp EF$[/quote]
Calculate the ratio $ \frac {XI}{EF}$.  :wink:

[quote="Virgil Nicula"][color=darkred][b][u]An extension.[/u][/b] Let $ ABCD$ be a convex equilateral. Construct in the its exterior

the triangles $ AXB$, $ CGD$ and $ BCY$, $ ADT$ so that $ \|\begin{array}{c} XA = XB\ ,\ GC = GD \\
\ CB = CY\ ,\ DA = DT\end{array}$

and $ m(\widehat {AXB}) = m(\widehat {ADT}) = m(\widehat {BCY}) = 2\phi$ , $ m(\widehat {CDG}) = \phi$.

Denote the midpoints $ E$, $ F$ of the segments $ [AT]$, $ [BY]$. Prove that $ \boxed {\begin{array}{c} EF\perp XG \\
\ EF = XG\cdot\sin 2\phi\end{array}}\ .$[/color][/quote]
[color=darkblue][b]Proof I[/b] (metric). $ 1\blacktriangleright$ Observe $ \{\begin{array}{c} AE = d\sin\phi\ ,\ BF = b\sin\phi \\
 \\
AX = BX = \frac {a}{2\sin\phi} \\
 \\
m(\widehat {XAE}) = 180^{\circ} + 2\phi - A \\
 \\
m(\widehat {XBF}) = 180^{\circ} + 2\phi - B\end{array}\|$ Apply the [b]Phytagoras' theorem[/b] :

$ \{\begin{array}{c} XE^2 = AX^2 + AE^2 - 2AE\cdot AX\cos\widehat {XAE} = \frac {a^2}{4\sin^2\phi} + d^2\sin^2\phi + ad\cdot\cos (A - 2\phi ) \\
 \\
XF^2 = BX^2 + BF^2 - 2BX\cdot BF\cos\widehat {XBF} = \frac {a^2}{4\sin^2\phi} + b^2\sin^2\phi + ab\cdot\cos (B - 2\phi )\end{array}\|$ $ \implies$

$ XE^2 - XF^2 =$ $ (d^2 - b^2)\sin^2\phi +$ $ (ad\cdot\cos A - ab\cdot\cos B)\cos 2\phi +$ $ (ad\cdot\sin A - ab\sin B)\sin 2\phi\ .$

But $ \{\begin{array}{c} ad\cdot\cos A - ab\cdot\cos B = \frac 12[(a^2 + d^2 - f^2) - (a^2 + b^2 - e^2)] = \frac 12[(d^2 - b^2) + (e^2 - f^2)] \\
 \\
ad\cdot\sin A - ab\cdot\sin B = 2([ABD] - [ABC])\end{array}$

$ \boxed {XE^2 - XF^2 = (d^2 - b^2)(\sin^2\phi + \frac 12\cos 2\phi) + \frac 12(e^2 - f^2)\cos 2\phi + 2([ABD] - [ABC])\sin 2\phi}\ \ \ (1)$

$ 2\blacktriangleright$ Observe that $ \{\begin{array}{c} DE = d\cdot\cos\phi\ ,\ CF = b\cdot\cos\phi \\
 \\
CG = DG = \frac {c}{2\cdot\cos\phi} \\
 \\
m(\widehat {EDG}) = D + 2\phi \\
 \\
m(\widehat {FCG}) = C + 2\phi\end{array}$ Apply again the [b]Phytagoras' theorem[/b] :

$ \{\begin{array}{c} GE^2 = DG^2 + DE^2 - 2\cdot DG\cdot DE\cdot \cos (D + 2\phi ) = \frac {c^2}{4\cdot\cos^2\phi} + d^2\cdot \cos^2\phi - cd\cdot\cos (D + 2\phi ) \\
 \\
GF^2 = CG^2 + CF^2 - 2\cdot CG\cdot CF\cdot \cos (C + 2\phi ) = \frac {c^2}{4\cdot\cos^2\phi} + b^2\cdot\cos^2\phi - bc\cdot\cos (C + 2\phi )\end{array}\|$ $ \implies$

$ GE^2 - GF^2 =$ $ (d^2 - b^2)\cos^2\phi +$ $ (bc\cdot\cos C - cd\cdot\cos D)\cos 2\phi - (bc\cdot\sin C - cd\cdot\sin D)\sin 2\phi\ .$

But $ \{\begin{array}{c} bc\cdot\cos C - cd\cdot\cos D = \frac 12[(b^2 + c^2 - f^2) - (c^2 + d^2 - e^2)] = \frac 12[(b^2 - d^2) + (e^2 - f^2)] \\
 \\
bc\cdot\sin C - cd\cdot\sin D = 2([BCD] - [ACD])\end{array}$

$ \boxed {GE^2 - GF^2 = (d^2 - b^2)(\cos^2\phi - \frac 12\cos 2\phi) + \frac 12(e^2 - f^2)\cos 2\phi + 2([ACD] - [BCD])\sin 2\phi}\ \ (2)$

Since $ \sin^2 \phi + \frac 12\cos 2\phi = \cos^2 \phi - \frac 12\cos 2\phi = \frac 12$ and

$ [ABD] - [ABC] = [ACD] - [BCD] = [AID] - [BIC]$, where $ I\in AC\cap BD$ obtain :

$ \boxed {XE^2 - XF^2 = GE^2 - GF^2} =$ $ \frac 12[ (d^2 - b^2) + (e^2 - f^2)\cos 2\phi] + 2\sin 2\phi([AID] - [BIC])$,

i.e. $ \boxed {\ EF\perp XG\ }\ .$

[b]Proof II[/b] (complex numbers). I"ll use the notations from the my upper proof. Denote $ w = \cos \phi + i\cdot\sin \phi\ .$

$ \{\begin{array}{ccccccc} b - x & = & (a - x)\cdot w^2 & \implies & x\cdot (1 - w^2) & = & b - a\cdot w^2 \\
\ d - g & = & (g - c)\cdot\frac {1}{w^2} & \implies & g\cdot (1 + w^2) & = & c + d\cdot w^2\end{array}\|$ $ \implies$

$ (g - x)\cdot (1 - w^4) = (c + d\cdot w^2)(1 - w^2) + (a\cdot w^2 + b)(1 + w^2)\implies$

$ \boxed {(g - x)\cdot(1 - w^4) = a\cdot w^2(1 + w^2) - b\cdot (1 + w^2) + c\cdot (1 - w^2) + d\cdot w^2(1 - w^2)}\ \ (3)\ .$

$ \{\begin{array}{ccccc} t - d = (a - d)\cdot w^2 & \implies & t = a\cdot w^2 + d\cdot (1 - w^2) & \implies & 2e = a + t \\
\ b - c = (y - c)\cdot w^2 & \implies & y\cdot w^2 = b + c\cdot (w^2 - 1) & \implies & 2f = b + y\end{array}\|$ $ \implies$

$ \{\begin{array}{ccc} 2e\cdot w^2 & = & a\cdot w^2(1 + w^2) + d\cdot w^2(1 - w^2) \\
\ 2f\cdot w^2 & = & b\cdot (1 + w^2) + c\cdot w^2(w^2 - 1)\end{array}\|$ $ \implies$

$ \boxed {2(e - f)\cdot w^2 = a\cdot w^2(1 + w^2) - b\cdot (1 + w^2) + c(1 - w^2) + d\cdot w^2(1 - w^2)}\ \ (4)\ .$

From the relations $ (3)$, $ (4)$ obtain $ 2(e - f)\cdot w^2 = (x - g)\cdot(w^4 - 1)\ .$ Observe that

$ \frac {w^4 - 1}{2w^2} =$ $ \frac { - 1 + \cos 4\phi + i\cdot\sin 4\phi}{2(\cos 2\phi + i\cdot \sin 2\phi )} =$ $ \frac {2i\cdot\sin 2\phi (\cos 2\phi + i\cdot\sin 2\phi )}{2(\cos 2\phi + i\cdot\sin 2\phi )} =$ $ i\ \cdot\ \sin 2\phi\ .$

In conclusion, $ e - f = i\cdot (x - g)\sin 2\phi$, i.e. $ \boxed {\begin{array}{c}
EF\perp XG\\\
EF = XG\cdot\sin2\phi\end{array}}\ .$

[b]Remark.[/b] Particular cases.

$ 1\blacktriangleright$ If $ \phi = 30^{\circ}$, then obtain the proposed problem.
$ 2\blacktriangleright$ If $ \phi = 45^{\circ}$, then obtain a [b]remarkable problem[/b] :[/color]

[quote="Virgil Nicula"][color=purple]Let $ ABCD$ be a convex equilateral. Construct in the its exterior the right isosceles triangles
$ AXB$, $ CGD$ and $ BCY$, $ ADT$ so that $ \|\begin{array}{c} XA \perp XB\ ,\ GC \perp GD \\
\ CB \perp CY\ ,\ DA \perp DT\end{array}$ .
Denote the midpoints $ E$, $ F$ of the segments $ [AT]$, $ [BY]$. Prove that $ XG\perp EF$.[/color][/quote]

[quote="Virgil Nicula"][color=purple]Let $ ABCD$ be a convex quadrilateral. Construct in the its exterior the right isosceles triangles
$ AXB$, $ CGD$ and $ BCY$, $ ADT$ so that $ \|\begin{array}{c} XA \perp XB\ ,\ GC \perp GD \\
\ CB \perp CY\ ,\ DA \perp DT\end{array}$ .
Denote the midpoints $ E$, $ F$ of the segments $ [AT]$, $ [BY]$. Prove that $ XG\perp EF$.[/color][/quote]

I think this is the well known [b][size=100]Van Aubel's theorem[/size][/b], which also states that $ XG = EF.$

http://www.mathworld.wolfram.com/vanAubelsTheorem.html

Kostas Vittas.

[quote="vittasko"][quote="Virgil Nicula"][color=purple]Let $ ABCD$ be a convex quadrilateral. Construct in the its exterior the right isosceles triangles
$ AXB$, $ CGD$ and $ BCY$, $ ADT$ so that $ \|\begin{array}{c} XA \perp XB\ ,\ GC \perp GD \\
\ CB \perp CY\ ,\ DA \perp DT\end{array}$ .
Denote the midpoints $ E$, $ F$ of the segments $ [AT]$, $ [BY]$. Prove that $ XG\perp EF$.[/color][/quote][color=darkred]I think this is the well known [b]Van Aubel[/b]'s theorem, which also states that $ XG = EF$.[/color][/quote]
[color=darkblue][b]Thank you, Kostas Vitas ![/b] You are right ! Sincerly, I didn't observe this. Thus, the my extension becomes more interestingly. I deleted the discordant big word "new". The [b]van Aubel[/b]'s theorem states that $ XG\perp EF\ ,\ XG=EF$ and the midpoints of the segments $ [XG]$. $ [EF]$, $ [AC]$, $ [BD]$ form a square. Come on, I and you, 'll try to find what become the another properties ($ XG=EF$ and the square) in the my extension. [/color]

[quote="Virgil Nicula"][quote="vittasko"][quote="Virgil Nicula"][color=purple]Let $ ABCD$ be a convex quadrilateral. Construct in the its exterior the right isosceles triangles
$ AXB$, $ CGD$ and $ BCY$, $ ADT$ so that $ \|\begin{array}{c} XA \perp XB\ ,\ GC \perp GD \\
\ CB \perp CY\ ,\ DA \perp DT\end{array}$ .
Denote the midpoints $ E$, $ F$ of the segments $ [AT]$, $ [BY]$. Prove that $ XG\perp EF$.[/color][/quote][color=darkred]I think this is the well known [b]Van Aubel[/b]'s theorem, which also states that $ XG = EF$.[/color][/quote]
[color=darkblue][b]Thank you, Kostas Vitas ![/b] You are right ! Sincerly, I didn't observe this. Thus, the my extension becomes more interestingly. I deleted the discordant big word "new". The [b]van Aubel[/b]'s theorem states that $ XG\perp EF\ ,\ XG = EF$ and the midpoints of the segments $ [XG]$. $ [EF]$, $ [AC]$, $ [BD]$ form a square. [b]Come on, I and you, 'll try to find what become the another properties ($ XG = EF$ and the square) in the my extension. [/b][/color][/quote]
[color=darkred][size=150][b]I found, Vittasko ![/b][/size]
In general case (my extension) $ \boxed {EF = XG\cdot\sin 2\phi }\ !$
(with the complex numbers, $ e - f = i\cdot (x - g)\sin 2\phi$).
I added in the my last reply this conclusion and its proof.
Remain only the extension of the ... square.[/color]

[quote="chien than"]Given a picture:
[img]http://img151.imageshack.us/img151/9356/200878823943564d0401288fk9.gif[/img]
$ ABX;BCY: CDZ;DAT$ are equilateral triangles.$ E;F$ are midpoint of $ AT;BY$
$ I$ be centre of triangle $ CDZ$
Prove that $ XI \perp EF$[/quote]
Another solution:
Let $ M,N,P,Q$ are the midpoints of $ AB,BC,CD,DA$
$ f$ is the rotation with angle $ \frac {\pi}{2}$. We have:
$ f(\overrightarrow{XI}) \equal{} f(\overrightarrow{XM}) \plus{} f(\overrightarrow{MP}) \plus{} f(\overrightarrow{PI}) \equal{} \frac {\sqrt {3} }{2} \overrightarrow{BA} \plus{} \frac {1}{2} f(\overrightarrow{AD}) \plus{} \frac {1}{2} f( \overrightarrow{BC}) \plus{} \frac {\sqrt {3}}{4} \overrightarrow{CD} \\
\equal{} \frac {\sqrt {3} }{2} \overrightarrow{BA} \plus{} \frac {1}{\sqrt {3}} \overrightarrow{QT} \plus{} \frac {1}{\sqrt {3} } \overrightarrow {YM} \plus{} \frac {1}{2\sqrt {3}} \overrightarrow{CD}$
$ \Rightarrow 2\sqrt {3} f(\overrightarrow{XI}) \equal{} 3\overrightarrow{BA} \plus{} 2\overrightarrow{QT} \plus{} 2\overrightarrow{YM} \plus{} \overrightarrow{CD} \equal{} 2\overrightarrow{BA} \plus{} 2\overrightarrow{YT} \equal{} 4\overrightarrow{FE}$
So $ XI \perp FE$ and $ \sqrt {3} XI \equal{} 2FE$

[quote="Virgil Nicula"][quote="PDatK40SP"] ... So $ XI \perp FE$ and $ \sqrt {3} XI \equal{} FE$[/quote]
I think that $ \sqrt 3\cdot XI \equal{} 2\cdot EF$. See the my last reply ![/quote]
Ah, I'm sorry  :oops:
[quote]$ 3\overrightarrow{BA} \plus{} 2\overrightarrow{QT} \plus{} 2\overrightarrow{YM} \plus{} \overrightarrow{CD} \equal{} \overrightarrow{BA} \plus{} \overrightarrow{YT} \equal{} 2\overrightarrow{FE}$[/quote]
This should be:
$ 3\overrightarrow{BA} \plus{} 2\overrightarrow{QT} \plus{} 2\overrightarrow{YM} \plus{} \overrightarrow{CD} \equal{} 2\overrightarrow{BA} \plus{} 2\overrightarrow{YT} \equal{} 4\overrightarrow{FE}$
So we have $ \sqrt{3} XI \equal{} 2 EF$
Thank you for your remark  :wink:  :)

[color=darkblue]Very nice your proof, [b]PDatK40SP[/b] ! Thank you for the your interest.
Maybe you can prove similarly and the my upper extension.[/color]

Similar and easier problem:
Given a picture:
[img]http://img81.imageshack.us/img81/184/vuminhthangsc6.jpg[/img]
$ BCX$ is equilateral triangle.$ E;F$ are midpoint of $ AB;AC$,respectively.
$ M$ be centre of triangle $ BCX$
Prove that $ AM \perp EF$

[quote="chien than"]Similar and easier problem. Given a picture ... $ BCX$ is equilateral triangle. $ E$, $ F$ are midpoint of $ AB$, $ AC$ respectively. $ M$ be centre of triangle $ BCX$. Prove that $ AM \perp EF$.[/quote]
[color=darkred]The upper your problem is a drivelling nonsense ![/color]

[color=darkblue]$ 3\blacktriangleright$ If $ \phi \equal{} 60^{\circ}$, then obtain the proposed problem in the "mirror" :[/color]
[quote][color=purple]Construct in the exterior of the convex quadrilateral $ ABCD$ the equilateral triangles $ AMB$, $ BNC$, $ CPD$ and $ DRA$.  Denote the midpoints $ E$, $ F$ of the segments $ [DR]$, $ [CN]$ and the center $ S$ of the triangle $ AMB$. Prove that $ SP\perp EF$ and $ EF \equal{} \frac {\sqrt 3}{2}\cdot SP$.[/color][/quote]


recurrence relation
algebra
Sequence
Inequality
IMO Shortlist

Define the numbers $a_0, a_1, \ldots, a_n$ in the following way:

\[ a_0 = \frac{1}{2}, \quad a_{k+1} = a_k + \frac{a^2_k}{n} \quad (n > 1, k = 0,1, \ldots, n-1).  \]

Prove that \[ 1 - \frac{1}{n} < a_n < 1.\]

We have $\frac {1}{a_k}-\frac{1}{a_{k+1}}=\frac{1}{n+a_k}$. Sum upp these and find that $ 2-\frac{1}{a_n}=\frac{1}{n+a_0}+...+\frac{1}{n+a_{n-1}}$/ Thus, $ 2-\frac{1}{a_n}<1$ , so $ a_n<1$. So, $ a_k<1$ for all k and so
 $ 2-\frac{1}{a_n}>\frac{n}{n+1} $ from where it's easy to find that $ a_n>1-\frac{1}{n} $

By the property of telescoping sum, we have quite easily that $\dfrac{1}{a_n}=2-\sum_{k=0}^{n-1}\dfrac{1}{n+a_k}$.

First we show that $a_n<1$ which is analogous to proving that $\sum_{k=0}^{n-1}\dfrac{1}{n+a_k}<1$. Now, we note that $\{a_k\}$ is an increasing sequence, hence $a_k\geq a_0$ for all $k\geq0$. This gives $\sum_{k=0}^{n-1}\dfrac{1}{n+a_k}\leq\dfrac{n}{n+a_0}=\dfrac{2n}{2n+1}<1$ and thus we are done.

Now we prove the other part of the inequality. We note that $\sum_{k=0}^{n-1}\dfrac{1}{n+a_k}\geq\dfrac{n}{n+a_n}$ because of increasing property of the sequence $a_n$. Now using the fact that $\dfrac{1}{a_n}=2-\sum_{k=0}^{n-1}\dfrac{1}{n+a_k}$ we have, after some algebra that $2a_n^2+(n-1)a_n-n\geq0$ implying, and keeping in mind that $\{a_k\}$ is a positive sequence, $a_n\geq\dfrac{-(n-1)+\sqrt{(n-1)^2+8n}}{4}$. It remains to show that this huge quantity is greater than $1-\dfrac{1}{n}$. By squaring both sides, and cancelling out terms, we come to $3n>2$ which is true for any $n$. Hence $a_n>1-\dfrac{1}{n}$.


I just have 2 questions that i cant seam to understand how to solve. If you could help me by showing me how you solve it that would be heaps of help ;)

[b]Question 1[/b]

A school invests $ \$5000$ at the end of each year at $ 6 \%$ p.a. towards a new library. How much will the school have after 10 years?

[b]Question 2[/b]

An employee contributes $ \$200$ into a superannuation fund at the end of each year. If the interest rate on this fund is $ 11.5 \%$ p.a., how much will the employee have at the end of 20 years?

I extremely appreciate every ones help =)

You should learn the formulas for simple interest:

$ I=Prt$
Where $ I$ is the amount of interest, $ P$ is the principal amount, $ r$ is the rate of interest, and $ t$ is the time.

And compound interest:

$ I=P{(}1+\frac{r}{n}{)}^{nt}$
Where $ I$ is the amount of interest, $ P$ is the principal amount, $ r$ is the rate of interest, $ t$ is the time, and $ n$ is the number of times interest is compounded per year (or whatever time period you have).

In this case, you need to use compound interest. I'll leave you to do the work.  :wink:


Have you ever tried drawing a triangle, only to notice that every attempt makes it look like an isosceles triangle?  I decided to make a problem out of it: how not isosceles can a triangle get?

More specifically, let a triangle have sides a,b,c, with a>= b>=c.  As the triangle becomes more isosceles, a/b or b/c is closer to 1 (but note that both are always greater than or equal to 1).  Therefore, we want to consider their minimum.  The problem becomes, what is sup(min{a/b,b/c})?

Just one last note, for those who don't know what a sup is.  A sup is the least upper bound of a set.  It's like the maximum, but doesn't have to be in the set.  The sup of the set {1<x<2} is 2, because it is the least upper bound.  Therefore, if you ended up with something like 1<=min{a/b,b/c}<100, the answer to the problem would be 100.

I like the result.

$\min{\left( \frac a b, \frac b c \right)} = \frac 1 2 \left( \frac a b + \frac b c \right) - \frac 1 2 \left| \frac a b - \frac b c \right |$

Maximum value is achieved when

$\frac a b - \frac b c = 0$

and the value of the minimum is then

$\min{\left( \frac a b, \frac b c \right)} = \frac 1 2 \left( \frac a b + \frac a b \right) = \frac a b$

Since the triangle inequality has to be satisfied,

$a \le b + c$

$\frac a b \le 1 + \frac c b = 1 + \frac b a$

Hence, the value of the minimum is bounded by

$\min{\left( \frac a b, \frac b c \right)} = \frac a b \le 1 + \frac b a$

and it is maximized at equality, i.e., when the triangle degenerates into a line segment:

$\frac a b = 1 + \frac b a$

$\left( \frac a b \right)^2 - \frac a b - 1 = 0$

$\frac a b = \frac{1 \pm \sqrt 5}{2}$

Since $a \ge b \ge 0$, the negative root is not acceptable and $\frac a b$ (the upper limit of the given minimum) is equal to the golden ratio $\frac{1 + \sqrt 5}{2}$.

[quote="blahblahblah"]I guess the only thing is that if he specifies the $\sup$, it's pretty clear that the answer will occur when the triangle becomes degenerate.[/quote]

Yeah, but you still need the derivation :)


function
number theory unsolved
number theory

Prove:
$\sum_{i=1}^r (-1)^{i-1} (_{i-1}^{r-1}) (x+r-i)^r=r!x+c_r$

that $c_r$ is an integer number not function of x.

amir, what is $x$? is it $\in$ the $\mathbb{R}$?

and put the$x=0$ then we will have $c_r$ as a function by $r$ so
what is $c_r$?
you said a integer independent from $x$ ,ok let it be but then what is the relation between the $r$ and $c_r$ :?

sorry i forgot , $x\in \mathbb{N}$

[quote="arman"]you said a integer independent from $x$ [/quote]
no I said [b]an[/b] integer not depending on $x$

Of course my dear friend,Of course , As you wish  :) .

But i didnt get it again.

What is $c_r$?(Surely it is not an arbitary integer ,is it ?)


Decide whether or not there exist primes p1, p2, ..., p2003 satisfying
(a) p1 < p2 < ... < p2003 \leq 1234567890;
(b) p1, p2, ..., p2003 form an arithmetic progression.

[i]Mexico 2000[/i] (I think, I'm not sure)

Well, (a) would be hard to *prove* on an olympiad, since the only known function for the number of primes less than some positive integer is only an approximation.

Actually they have to satisfy both conditions (a) and (b) ... :)
You'll see that this changes things.
It is not hard to solve it but I like this simple and nice problem.

Ah, my mistake, misread your problem.

2^2 posts for me :D

gogogo tycoon ... make some posts for the USA :D.

[quote]gogogo tycoon ... make some posts for the USA[/quote]

Yes..well, you see, I'm just an inexperienced youngster who is trying to work his way up to olympiad-level problems, so don't place too much hope in my being the next grobber in terms of post count  :) --I'll probably be just lurking around devouring the easier problems as I learn more (from the solutions of others).

Suppose that p_1=a, p_2=a+b,...,p_2003=a+2002b. Of course, a>=2 and b>=2. Also, 2002b<1234567890, so b has at most 7 digits. Now let q any prime smaller than 32. Suppose that q does not divide b. Then among the numbers {a+30b,a+31b,...,a+100b} at least one is divisible by q.So there exists i belonging to {30,31,...,100} such that a+ib=q. But a+ib>30*2=60, false. So b is divisible by any prime smaller than 31. In particular, it is greater than 2*5*11*13*17*19*23*29*31, which has at least 8 digits. So, the answer is no.


I made up this inequality.

Prove that for real x and y,

$ 2x^2 \plus{} y^2 \geq (4/5)((7/2)xy \plus{} y \minus{} x \minus{} 1).$

Without using calculus.

For all $x$, $y$ $\in \mathbb{R}$ we have $(3x-2y)^2+(x-y+2)^2\geq 0$

$\Longrightarrow 10x^2+5y^2-14xy+4+4x-4y\geq0$

$\Longrightarrow 2x^2+y^2\geq \dfrac{4}{5} \left[\dfrac{7xy}{2}+y-x-1 \right]$


Prove that: With each $ t > 0$ existent uniformization $ a,b,c \in Z$suct that:
$ 0 \ge a \plus{} b\sqrt {2} \plus{} c\sqrt {3} \ge t$

you want write $ \forall t>0,\exists a,b,c\in\mathbb{Z}: \ 0\le a\plus{}b\sqrt{2}\plus{}c\sqrt{3}\le t$.

let $ t>0$
$ \exists (p_n,q_n)\in\mathbb{N}^2$ such that $ p_n\plus{}q_n\sqrt{2}\to 0$ and  $ p_n\plus{}q_n\sqrt{2}>0$ (because $ \mathbb{R}\equal{}\overline{\mathbb{Z}\plus{}\sqrt{2}\mathbb{Z}}$)
then $ \exists N\in\mathbb{N}: \  0\le p_N\plus{}q_N\sqrt{2}\plus{}0\sqrt{3}\le t$


What is the greatest two-digit multiple of 13?

91 (13*7), as 104 (13*8) is too high.

You should probably memorize multiples of 13 up to 130, to improve time.


geometry
geometric transformation
analytic geometry
vector
calculus
integration
trigonometry

I'm very confused about this:

A triangle has latticial points as its vertices, and we know that no matter how we translate it s.t. it's still latticial, then the two triangles have no interior points in common.

They first ask if it's possible for the triangle to have area greater than $\frac 12$, and then they ask what the maximal area is. I got that it's $\frac 12$, and that sort of makes the problem sound really weird. What do you think? Did I make a mistake?

well, it is easy to see that the triangle cannot have interior lattice points, or boundary latice points(different from the vertices) by picks theorem one gets that its area is 1/2, the xtrange thing is that 1/2 would be the only area posible, because by pick's theorem also one gets area>=1/2. Another question is: does every latice triangle with area 1/2 satisfies the property?. So grobber i agree with you: the only possible area is 1/2 wich sounds weird.

I did exactly what you did, but then I thought that maybe the weird formulation is just a strange way of asking students not to use Pick's theorem :). I think all triangles of area $\frac12$ satisfy this, but I haven't tried to actually prove it.

In fact, it can be shown that all triangles of area $\frac 12$ satisfy the condition. If one such triangle doesn't, then there are two points $m,n$ inside it (I'm working with points as complex numbers) s.t. $m-n\in\mathbb Z_i$.

Try to show (it's fairly easy) that if we have a segment inside a triangle, then it can be translated s.t. one of its endpoints becomes a triangle vertex and the segment is still inside or on the boundary of the triangle. We do this for our segment and we get a lattice point inside or on the borders of the triangle (not a vertex), thus getting a contradiction.

I just checked my (official) archive of  and saw what I vaguely remembered when read this topic: the triangle in question WAS NOT supposed to have lattice points as its vertices. You may easily see that this changes the problem a bit :-))
   The problem indeed is problem No. 7 for senior students from the Autumn round ot
ToT 2001. The part a) was worth 3 points and b) was worth 6 points (and together
they were worth 8 points) :-))

So, can you traslate it because i dont understand the problem with the new condition.....Thank you :lol:

(Unofficial) translation from the official text:
   A triangle is placed on the coordinate plane so that its translations
by vectors with integral coordinates have no common interior points.
   a) Can the area of such a triangle be greater than 1/2 ?  (3 points)
   b) Find the maximum possible area of such a triangle. (6 points)

Can't we suppose WLOG that the coordinates of the triangle vertices are lattice too?? :? seems quite normal to me.

The only way to modify it a little in that sence would be to replace the text by "Vectors with an integer length" or something, then the result would be actually modified.

Yes I think we should replace the vectors with $''lattice vectors''$.
In that case I think the answer is 1.
For sure you can reach more than $\frac{1}{2}$

[quote="Peter VDD"]More? No iura, then the best we can do is an equilateral triangle with sidelength 1, so for a surface of only $\frac{\sqrt3}4$. :?[/quote]

No, Peter, it said "translations by vectors with integral coordinates", not translations by "Vectors with an integer length".

Consider an equilateral triangle of side length $\frac{2 \sin \frac{5 \pi}{12}}{\sqrt{3}}$, placed such that it is symmetrical about the line $y = x$.  It's easy to check that this satisfies the given conditions and has area greater than $\frac{1}{2}$ (although not that much greater...).


If x, y, z are positive real numbers, prove that
$ \frac{1}{y^{2}\plus{}yz\plus{}z^{2}}\plus{}\frac{1}{z^{2}\plus{}zx\plus{}x^{2}}\plus{}\frac{1}{x^{2}\plus{}xy\plus{}y^{2}}\ge\frac{9}{{(x\plus{}y\plus{}z)}^2}$

[quote="Luba1123"]If x, y, z are positive real numbers, prove that
$ \frac {1}{y^{2} \plus{} yz \plus{} z^{2}} \plus{} \frac {1}{z^{2} \plus{} zx \plus{} x^{2}} \plus{} \frac {1}{x^{2} \plus{} xy \plus{} y^{2}}\ge\frac {9}{{(x \plus{} y \plus{} z)}^2}$[/quote]
Let $ x\plus{}y\plus{}z\equal{}3u,$ $ xy\plus{}xz\plus{}yz\equal{}3v^2$ and $ xyz\equal{}w^3.$
Hence, your inequality is equivalent to $ A(u,v^2)w^3\plus{}B(u,v^2)\geq0.$
Id est, for the proof enough to check two cases:
1) $ z\equal{}0,$ $ y\equal{}1$;
2) $ y\equal{}z\equal{}1.$

if this problem was IMO's problem, can we solve like this?
can we use UVW method and ABC theorem  in the IMO?

[quote="Luba1123"]If x, y, z are positive real numbers, prove that
$ \frac {1}{y^{2} \plus{} yz \plus{} z^{2}} \plus{} \frac {1}{z^{2} \plus{} zx \plus{} x^{2}} \plus{} \frac {1}{x^{2} \plus{} xy \plus{} y^{2}}\ge\frac {9}{{(x \plus{} y \plus{} z)}^2}$[/quote]
This inequality is [b]Problem 2.19 - page 57[/b] in new book (where it is presented with a very simple solution by Cauchy-Schwarz):

Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, [url=http://can-hang2007.blogspot.com/2009/12/new-book-inequalities-with-beautiful_20.html][b][i]Inequalities with Beautiful Solutions[/i][/b][/url], GIL publishing house, 2009.

[quote="YESMAths"]
Obvious from Titu's Lemma or Engel's Lemma. :)[/quote]
Show your proof please. :wink:

This one is also true.
Let $x,y,z>0$ Prove that :
\[ \frac{x}{y^2+yz+z^2}+\frac{y}{z^2+zx+x^2}+\frac{z}{x^2+xy+y^2} \ge (x+y+z)/(xy+xz+yz)\]

[quote="elmrini"]This one is also true.
Let $x,y,z>0$ Prove that :
\[ \frac{x}{y^2+yz+z^2}+\frac{y}{z^2+zx+x^2}+\frac{z}{x^2+xy+y^2} \ge \frac{x+y+z}{xy+xz+yz}\][/quote]
$\sum_{cyc} \frac{x}{y^2+yz+z^2}-\frac{x+y+z}{xy+xz+yz}=\frac{(x+y+z)\sum\limits_{sym}(x^5y-x^4y^2)}{(xy+xz+yz)\prod\limits_{cyc}(x^2+xy+y^2)}\geq0$. :lol:

The following inequality is also true,
Let $x$, $y$ and $z$ be positive numbers. Prove that:
\[ \frac{x}{y^2+yz+x^2}+\frac{y}{z^2+zx+y^2}+\frac{z}{x^2+xy+z^2}\le\frac{x+y+z}{xy+xz+yz} \]


Jean reads 50 pages of a book in one hour. The book has 330 pages. At this rate, how many minutes will it take her to complete the book?

$ \frac{330}{50} \times 60 \equal{} \boxed{396 \ \text{minutes}}$.


limit
logarithms
calculus
calculus computations

$ \lim_{x \rightarrow 0} x^{x^x}$

$ \lim_{x \rightarrow 0} ({\frac {1}{x})^{\frac {1}{x}}}*x^{(x^x)}$

Both limits are from the right of zero.

Im not sure, but i think it would be 0. lim(x^x)=1, because as x gets infinitely close to 0, x^x becomes 1. thus, 0^1=0.

See also [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=225365[/url]

$ ({\frac{1}{x})^{\frac{1}{x}}}*x^{(x^{x})} \equal{} \exp({(\frac{\minus{}1}{x}\plus{}x^x)\ln(x)})$
$ \lim_{x\downarrow  0}{(\frac{\minus{}1}{x}\plus{}x^x)\ln(x)} \equal{} \minus{}\infty$
So, the second limit is 0.

1)

$ \lim_{x \to 0^{ \plus{} }} x^{x^{x}} \equal{} \lim_{x \to 0^{ \plus{} }} e^{ x^{x} \, \ln( x ) }$

since $ e$ is continuous, we can use the composition law of limits to focus entirely on the limit of the exponent given by:

$ \lim_{x \to 0^{ \plus{} }} x^{x} \, \ln (x )$

since both $ x^x$ and $ \ln(x)$ have existing limits when $ x \to 0^{ \plus{} }$ , we can then use the product rule of limits to get:

$ \lim_{x \to 0^{ \plus{} }} x^{x} \, \lim_{x \to 0^{ \plus{} }} \ln(x) \equal{} 1 \cdot \minus{} \infty \equal{} \lim_{t \to \minus{} \infty} t$

And so our answer is $ \lim_{t \to \minus{} \infty} e^{t} \equal{} 0$


Eight lamps, each with an on-off switch, are arranged in a cricle. A lamplighter can flip (in other words, change the state of) the switches, but he is not allowed to flip just one at a time. When he switches lamps on or off, he is required to flip the switches of three consecutive lamps simultaneously. (For example, he can flip the switches of lamps 7,8 and 1 at the same time) Prove that no matter what set of lamps was turned on at the start, the lamplighter can turn all the lamps on.

[hide="Solution"] This is very simple.  Let $F(k)$ denote the flipping of switches $k-1, k, k+1 \bmod 8$.

$S(k) = F(k), F(k+1)$ flips switches $k+2, k-1 \bmod 8$.

$R(k) = S(k), S(k+3)$ flips switches $k+5, k-1 \bmod 8$.

And so forth.  We can flip any two switches $k+(2+3m), k-1 \bmod 8$, and since $gcd(3, 8) = 1$ this is equivalent to being able to flip [b]any two[/b] switches.

If the number of lamps off is even, turn them on in pairs.

If the number of lamps off is odd, turn them on in pairs, then flip the last remaining lamp and two of its neighbors, then turn on the two neighbors again. [/hide]


inequalities
trigonometry
triangle inequality

prove the following inequality for a,b,c are three positive real :
$ \sqrt{a^2\plus{}b^2\minus{}ab}\plus{}\sqrt{b^2\plus{}c^2\minus{}bc} \geq \sqrt{a^2\plus{}c^2\plus{}ac}$

[hide="Hint"]$ \cos 60^\circ \equal{} \frac{1}{2}$

Also, an Engel Problem ;)[/hide]

For this inequality I have only
[hide="an algebraic proof"]\[ a^2\minus{}ab\plus{}b^2\equal{}(a\minus{}\frac{b}{2})^2\plus{}(\frac{\sqrt{3}}{2}b)^2\]
\[ b^2\minus{}bc\plus{}c^2\equal{}(b\minus{}\frac{c}{2})^2\plus{}(\frac{\sqrt{3}}{2}c)^2\]
Observe that $ (a\minus{}\frac{b}{2}\plus{}\frac{c}{2}\minus{}b)^2\plus{}(\frac{\sqrt{3}}{2}b\plus{}\frac{\sqrt{3}}{2}c)^2\equal{}a^2\plus{}ac\plus{}c^2$
Aplying Minkovski will prove the inequality and the equality case will be when $ (a\minus{}b/2,\sqrt{3}/2 \cdot b)$ and $ (c/2\minus{}a,\sqrt{3}/2 \cdot c)$ are proportional.That is equivalent to $ a(b\plus{}c)\equal{}bc$[/hide]
The geometric proof will have to wait a bit  :oops:

[hide="Geometric Strategy"]

Draw $ O$, $ A$, $ B$, $ C$ with $ \angle AOC \equal{} 120^\circ$ and $ OB$ the bisector of $ \angle AOC$. Let $ OA\equal{}a$, $ OB\equal{}b$, and $ OC\equal{}c$. Then apply triangle inequality to $ \triangle ABC$.[/hide]

[hide="something that looks like it yields a good solution"]
If $ \omega \equal{} e^{\frac {2i\pi}{3}}$, then $ |a \plus{} b\omega| \equal{} \sqrt {a^2 \plus{} b^2 \minus{} ab}$
[/hide]
[hide="completion of solution"]
Let $ \omega \equal{} e^{\frac {2i\pi}{3}}$.
$ |a \plus{} b\omega| \plus{} |b \plus{} c\omega| \equal{} |a \plus{} b\omega| \plus{} | \minus{} (b \plus{} c\omega)| \geq |a \minus{} c\omega|$ by the triangle inequality.

EDIT: Sorry, the above line is wrong. It should read:

$ |a\plus{}b\omega| \plus{} |b\plus{}c\omega| \equal{} |a\plus{}b\omega| \plus{} |\minus{}b\omega \minus{} c\omega^2| \geq |a\minus{}c\omega^2|$.
[/hide]


Find the number of permutations $( p_1, p_2, p_3 , p_4 , p_5 , p_6)$ of $1, 2 ,3,4,5,6$ such that for any $k, 1 \leq k \leq 5$, $(p_1, \ldots, p_k)$ does not form a permutation of $1 , 2, \ldots, k$.

I have 481 as my result calculating compliment cases

I have got it right now, it is 461 by PIE

i THINK WE CAN USE A RECURSION IDEA TOO
If we write T(n) as the number of permutations of 1,2,...,n such that (p(1),...,p(n-1)) is not a permutation of 1 to n-1 , then we can write T(n) in terms of T(1) to T(n-1) from where the rest of the computation shd be fairly simple ...........

These are called [i]indecomposable permutations[/i]; one can solve for the number of them of a given length using  generating functions, or use a recursion.  They are given by sequence [url=https://oeis.org/A003319]A003319[/url] in the [url=https://oeis.org/]OEIS[/url].

I am not sorry for reviving this topic  i think this is required.
I have also solved this problem in somewhat the same "ugly" way mentioned above just now and was(is) looking for an elegant one. i have not much knowledge of generating functions so [i]what is the recursion formed????[/i]

Maybe the recursion is this way: Take a permutation which is permissible for $n$, append $n+1$ to its end, and then just swap the $n+1$ and any other number. I think it works, but maybe we are missing out many cases  :(

[quote=AdithyaBhaskar]Maybe the recursion is this way: Take a permutation which is permissible for $n$, append $n+1$ to its end, and then just swap the $n+1$ and any other number. I think it works, but [color=#00f][b]maybe we are missing out many cases[/b][/color]  :([/quote]
Ya obviously you are missing out a hell lot of other cases. For the simplest example see that it you place $n+1$ in front then any permutation of {$1,2,...,n$} after that is a vaild permuntation(which brings in a good $n!$ new ways) :blush:

Can anyone generalize my recursion above???. U see that the main thing is that
the [b]number of caseworks to be done[/b] for $n^{th}$ recursion statement also itself follows a recursion of $C_n=1+C_{n-1}$( :P :rotfl: :P ).
I may sound inebriated but really that is the main [i]itch[/i] in the problem now.

[quote=XT]Hello everyone!  :D

Here I post my (ugly) idea to proceed which is simply to "Count 'em All!"  :?  :lol: It's not a complete solution, I'm just explaining how I tried to solve it. I've probably done some mistake so any correction or alternate solution is welcome!

Well, let's start!
We see that it can't be $p_6=6$ or it would fall taking $k=5$. How many of these permutations with $p_6=6$ are there? Easy! They are $6!\cdot\frac{1}{6}=5!=120$.
Going on it's clear that it can't be in any order that $5$ and $6$ take the two last positions or it won't work with $k=4$. This may happen only taking $p_5=6$ and $p_6=5$ (cause we have excluded $p_6=6$ here above!). So in how many ways may it happen? In $4!=24$ ways.
Again, $4$,$5$ and $6$ can't (in any order) be the last three $p_i$. How do we count the number of permutations where they are put in such way? The options are $3!\cdot3!$ but we must subtract the ones we have counted before (that are $2\cdot3!$ with $p_6=6$ and $1\cdot3!$ with $p_5=6,p_6=5$) so we have $3\cdot3!=18$ new "illegal" permutations.
...
On this way you check the number of illegal permutations with $k=2$ and $k=1$ and you're done. My result is $461$ possible permutations.

Has anybody solved it? Please post!  :lol:

Bye people![/quote]

The ugly method is the method so far

@Chandrachur
In case 3 $p_1$ cannot be equal to $1$. It will violate the condition for $k=1$ constraint,

[quote=Chandrachur]I have got a way which i describe as [b][i]QUASI-RECURSION[/i][/b] as it is actually a recursion but i found solving it more difficult than to go with it in a way that makes it somewhat like the solution @XT has provided above.
[Hide=.....]Take a breadth b4 u delve into something semi-cool semi-ugly.[/hide]
Let $S_n$ denote the set of required arrangements whose number of elements is required when u replace $6$ by $n$ in the question.Let $|S_n|=t_n$(Looks ambitious :huh: it isn't  :rotfl: )
We count $t_6$ by dividing that into 5 mutually disjoint cases namely when $p_i=6$ for $i=1,2,3,4,5$(assuming u already figured out why $p_6$ can't be $6$ :P )
[Hide=Case1_p_1=6]
The rem places can be filled in any way so $5!$ permutations.[/hide]
[Hide=Case2_p_2=6]
$p_1$ can be filled in $4$ ways ($2,3,4,5$) and others in any way(as $p_2=6$ has done all the disturbance that we wanted to create :P ). So $4*4!$ ways here.[/hide]
[Hide=Case3_p_3=6]
Only $(p_1,p_2)=(1,anything) or (2,1)$ are the cases to be thrown out. So $5!-(4!+3!)$ ways.[/hide]
[Hide=Case4_p_4=6]
In addition to the "to-be-thrown-out" cases mentioned in Case 3, $3$ more case $(p_1,p_2,p_3)=(3,2,1) or (3,1,2) or (2,3,1)$ gets added to the "to-be-thrown-out" list. Each has market value of $2!$(can be done in $2!$ ways :-D ). So total count here =$5!-(4!+3!+3*2!)$.[/hide]
[Hide=Finally_a_recursive_case_p_5=6]
A bijection of this sub-case with $S_{n-1}$ can be easily formed by swapping the $5^{th}$ number of an element of $S_{n-1}$ with $6$ and sending that to the last.(to formalize the bijection is your business @can't pain my hand more :dry: ).So count here is,simply =$t_5$.[/hide]
[color=#00f][b]So finally the recursion is[/b][/color] $t_6=(5!)+(4*4!)+(5!-(4!+3!))+(5!-(4!+3!+3*2!))+t_5$. or $t_6=390+t_5$.
Sad part is to relate $t_5$ and $t_4$ you will get a number other than $390$ in the recursion and [b]good part[/b] is that $5$ is quite close to $1$(thanks to question setter :omighty:  :P ).
But surely you can find a logical pattern and get(by logic)
$t_5=4!+3*3!+(4!-(3!+2!))+t_4$ or , $t_5=58+t_4$ and
$t_4=3!+2*2!+t_3$ or , $t_4=10+t_3$ and
$t_3=2!+t_2$ or , $t_3=2+t_2$. and $t_2$ is trivially $1$(check it!!!)
 :trampoline: So $t_6=390+58+10+2+1=461$. :D[/quote]
In case 3 $p_1$ cannot be equal to $1$. It will violate the condition for $k=1$ constraint,

Let's denote the set of permutations of $\{1, \dots, n\}$ as $P_n$, where $|P_n|=n!$ is the number of permutations. For a permutation $P=(\alpha_1 \dots \alpha_n)$, let's define the degree of the permutation, denoted as $\text{der} P$ (Degree of a permutation can be a different concept in the literature. Here we use our definition), for $1 \leq k \leq n$, $(\alpha_1\dots \alpha_k)$ represents a permutation of the set $\{1, \dots, k\}$, and the degree is the smallest $k$ for which this holds.

Let $Q_n$ denote the set of permutations of $\{1,\dots, n\}$ such that, for every $1\leq k \leq n-1$, $(\alpha_1\dots \alpha_k)$ is not a permutation of $\{1,\dots, k\}$. For example, $Q_1 = \{(1)\}$ and $Q_2 = \{(21)\}$. The problem asks for $|Q_6|$.

For $1\leq i \leq n$, let $D_i = \{P \in P_n \mid \text {der} P = i\}$. Note that $D_i$ and $D_j$ are disjoint for $i \neq j$, and $P_n = D_1 \cup \dots \cup D_n$. First, observe that $D_n= \{P \in P_n \mid \text {der} P = n\} = Q_n$.

If we append a permutation of $\{i+1, \dots, n\}$ to $Q_i$, we obtain a permutation in $D_i$. Therefore, we have the equation:

\[n! = |P_n| = |Q_n| + |Q_{n-1}|\cdot 1! + |Q_{n-2}|\cdot 2! + \dots + |Q_{1}|\cdot (n-1)! = \displaystyle \sum_{i=1}^{n} |Q_i|\cdot (n-i)!\]

Let's calculate for smaller values:

$|P_3| = 3! = |Q_3| + |Q_2|\cdot 1! + |Q_1|\cdot 2! \Longrightarrow |Q_3| = 3! - 1 - 2 = 3$.
$|P_4| = 4! = |Q_4| + |Q_3|\cdot 1! + |Q_2|\cdot 2! + |Q_1| \cdot 3! \Longrightarrow |Q_4| = 4! - 3 - 2 - 6 = 13$.
$|P_5| = 5! = |Q_5| + |Q_4|\cdot 1! + |Q_3|\cdot 2! + |Q_2| \cdot 3! + |Q_1| \cdot 4! \Longrightarrow |Q_5| = 5! - 13 - 6 - 6 - 24 = 71$.
$|P_6| = 6! = |Q_6| + |Q_5|\cdot 1! + |Q_4|\cdot 2! + |Q_3| \cdot 3! + |Q_2| \cdot 4! + |Q_1| \cdot 5! \Longrightarrow |Q_6| = 6! - 71 - 26 - 18 - 24 - 120 = 461$.


What is the value of $ 19^2 \plus{} 2(19) \plus{} 1$?

A more elaborate explanation would be...

Notice the trinomial square.  We can rewrite $ 19^{2}+2(19)+1 $ as $(19+1)^2=(20)^2=\boxed{400}$.


ratio
geometry
pigeonhole principle
calculus
integration
combinatorics solved
combinatorics

Given  $n^2$ points $A_1$, $A_2$, ..., $A_{n^2}$ in a circle of radius $10n$. ("In a circle" means "inside or on the boundary of the circle".)
 Prove that if $n$ is large enough, then for some distinct $i$ and $j$ the distance between $A_i$ and $A_j$ is less than $\sqrt{\left|i-j\right|}$.

Maybe I solved the problem but I'm not sure..
In my solution I get $10$ can be replaced by any constant.
Is it correct?
If so, I will post my soln.

Of course, 10 may be replaced to any constant. It is a problem of Sergei Ivanov, proposed to 239 lyceum olympiad in '03. I myself consider it very hard, and I did not succeed with it at my time. So, take my congratulations, Iura.

There is an official solution below.

We call points of the sequence
$\{A_i\}$ {\it marked points\/}.
%~  $i$~---  .
Assume that points are situated such that
the statement of problem fails, i.e.
$|A_iA_j|\ge\sqrt{|i-j|}$ for any $i$ and $j$.
In other words, difference of indices of any
two points does not exceed the square of a
distance between them.
We have to prove that $n$ is not too large.

For any square on a plane, define its {\it density\/}
as the ratio of the number of marked points in this square
to the area of this square.
Note that the density of a square with side
$20n$, which contains all our points, equals
$\frac1{400}$.

Denote $N=10000$.

Lemma. Let $K$ be a square with side $a>30$
and a density $\varepsilon\ge\frac1{400}$.
Then there is a square with side
$\frac aN$, whose density is not less then
$(1+\frac1{N^2})\cdot\varepsilon$.

Proof of the lemma.

Partition $K$ into $N^2$ squares with a side
$\frac aN$. Let's prove that one of them does
not contain marked points. Assume the contrary.
Let $p$ and $q$ be the maximal and minimal indices of marked
points in $K$. Consider a chain of $m\le 2N$
squares (any two consecutive squares in a chain have common side),
say, $K_1$, $K_2$, \dots, $K_m$,
in which $K_1$ contains $A_p$, and $K_m$
contains $A_q$.
Choose points $A_{p_i}$ in squares $K_i$ such that
$p_1=p,\,p_m=q$. We have $A_{p_i}A_{p_{i+1}}\le \frac{a\sqrt5}{N}$.
 So,
$|p_i-p_{i+1}|\le \frac{5a^2}{N^2}$.
Hence
\[
 q-p\le m\cdot\frac{5a^2}{N^2}\le 2N\cdot \frac{5a^2}{N^2}
 =\frac{10a^2}N. %=\frac{a^2}{1000} .
\]
The number of points in $K$ does not exceed $q-p+1$,
on the other hand, it equals
$a^2\varepsilon$.
So,
$a^2\varepsilon\le q-p+1\le \frac{10a^2}{N}+1$,
and $\varepsilon\le\frac{10}N+\frac1{a^2}$.
Substitute $N=10000$ and use $a>30$, we get
$\varepsilon<\frac1{1000}+\frac1{900}<\frac1{400}$,
a contradiction.

Hence, there is a square without marked points
and all marked points are covered by at most
$N^2-1$ squares. By pigeonhole principle, the density
of one of them is at least
$
\varepsilon \cdot \frac{N^2}{N^2-1}
> \varepsilon \cdot (1+\frac 1{N^2}) .
$

The lemma is proved.

The problem easily follows from the lemma.
If $n$ is large, we find squares with
more and more density, and finally we find a
square with density at least 100.
Clearly, the distance between some two points
in such square is less than 1 --- a contradiction.

Hello Everyone.
                       Thanks a lot Fedor Petrov for posting the solution to this beautiful problem. You said it was given at Lyceum Olympiad 03 , was it the elimination round problem. Btw, did anybody got success with this problem during the contest.

Well iura please post your solution also , I am interested and hope others also would like to see your solution.  You are a very good problem solver iura . Keep it up.  :lol:

                                                                                Bye

[quote="Euler_India"]Hello Everyone.
                       Thanks a lot Fedor Petrov for posting the solution to this beautiful problem. You said it was given at Lyceum Olympiad 03 , was it the elimination round problem. Btw, did anybody got success with this problem during the contest.
[/quote]

In which sense "elimination round"? 239 olympiad is unofficial at all, and the only  round is enough for making participants happy. Nobody solved it during the contest, and, as I know, after the contest, too.

I'm sorry but I must disappont you..(and myself of course)
My proof was wrong..
At least it's not so painful for me that so many wonderful problem solvers couldn't solve it too..

Hello ,
          Its a very beautiful problem and no doubt hard too.

     I am a great fan of such problems.

  Its a request to our Russian friends like Fedor Petrov, Myth , andrew and    others  to please post the problems of elimination rounds of St.PBMO which were'nt solved during the contest.
                                                   Thanks .

PS: To Fedor : Is Sergei Ivanov the same person who also won a gold medal at IMO 88 in Australia?

I couldn't resist giving a minor remark on a 8-year old proof (a beautiful proof, I must say!):

[quote="Fedor Petrov"]If $n$ is large, we find squares with
more and more density, and finally we find a
square with density at least 100.
Clearly, the distance between some two points
in such square is less than 1 --- a contradiction.[/quote]

It is not true that a square of density at least $100$ must always contain at least two marked points (in fact, just think of a sufficiently small square around one marked point). Thus, it is even less true that a square of density at least $100$ must always contain two marked points with distance less than $1$. However, this [b]is[/b] true when the square has integral sidelength (because we can then subdivide this square into small squares of sidelength $\frac12$, and at least one of these small squares must also have density $\geq 100$, and thus contain at least $\frac{100}{4}=25$ marked points, hence contain at least $2$ marked points, and the distance between these two points is obviously $\leq \sqrt2\cdot\frac{1}{2} < 1$). Now, since we have absolute freedom to choose the square that we are starting with, we can choose this square in such a way that after applying the Lemma to it sufficiently many times, its density becomes $\geq 100$ [b]and[/b] its sidelength becomes integral. Thus, we are done.

I just improved the LaTeX and wording of Fedor's solution.

[rule]
Of course, $10$ may be replaced to any constant. It is a problem of Sergei Ivanov, proposed to $239$ Lyceum Olympiad in $'03$. I myself consider it very hard, and I did not succeed with it at my time. So, take my congratulations, Iura. There is an official solution below.

We call the points the sequence $\{A_i\}$ [i]marked points[/i].

Assume that points are situated such that the statement of problem fails, i.e.
$$\forall i\neq j, \ |A_iA_j|\ge\sqrt{|i-j|}.$$
For any square on a plane, define its [i]density[/i] as the ratio of the number of marked points in this square and the area of this square. Note that the density of a square with a side length of $20n$, which contains all our points, equals $1/400.$ Denote $N=10000.$

[color=#c00][b]Lemma:[/b][/color] Let $K$ be a square with a side length of $a>30$ and a density of $\varepsilon\ge1/400.$ Then, there is a square with side $a/N$, whose density is not less then
$$\bigg(1+\frac1{N^2}\bigg)\cdot\varepsilon.$$

[rule]
[color=#3D85C6][b]Proof:[/b][/color] Partition $K$ into $N^2$ squares with sides of length $a/N$. Let's prove that one of them does not contain any marked points. Assume the contrary.

Let $p$ and $q$ be the maximal and minimal indices of the marked points in $K$. Consider a chain of $m\le 2N$ squares (any two consecutive squares in a chain have common side), say, $K_1, K_2, \dots, K_m$ such that $K_1$ contains $A_p$ and $K_m$ contains $A_q.$ Choose points $A_{p_i}$ in squares $K_i$ respectively, such that $p_1=p$ and $p_m=q$.

Therefore, we have $$A_{p_i}A_{p_{i+1}}\le \frac{a\sqrt5}{N}\Longrightarrow |p_i-p_{i+1}|\leq\frac{5a^2}{N^2}.$$
This furthermore implies
\[ q-p\le m\cdot\frac{5a^2}{N^2}\le 2N\cdot \frac{5a^2}{N^2} =\frac{10a^2}N =\frac{a^2}{1000} .\]
The number of points in $K$ does not exceed $q-p+1$ but on the other hand, it equals $a^2\cdot\varepsilon.$ Thus
$$a^2\cdot\varepsilon\le q-p+1\le \frac{10a^2}{N}+1\Longrightarrow\varepsilon\le\frac{10}N+\frac1{a^2}.$$
Substituting $N=10000$ and using $a>30$, we get
$$\varepsilon<\frac1{1000}+\frac1{900}<\frac1{400}$$
which is a contradiction.

Hence, there is a square without marked points, so all marked points are covered by at most  $N^2-1$ squares. By the pigeonhole principle, the density of one of them is at least
$$\varepsilon \cdot \frac{N^2}{N^2-1}
> \varepsilon \cdot \bigg(1+\frac 1{N^2}\bigg).$$
The lemma is proved.

[rule]
The problem easily follows from the lemma. If $n$ is large (enough), we can find squares with arbitrarily large density, so eventually, we can find a square with a density of at least 100. Clearly, the distance between some two points in such a square is less than $1$ which is yet again a contradiction.


geometry
circumcircle
geometric transformation
reflection
geometry proposed

Let $ ABC$ be an nonisosceles acute triangle.
Denote $ l_A,l_B,l_C$ the tangent line at $ A,B,C$ with respect to the circumcircle of $ ABC$.
It is known that line $ l$ passes through $ H$(othocenter) and $ l\perp OH$
$ A_1\equal{}l\cap l_A$
$ A_2$ is the reflection point of $ A_1$ wrt $ A$,
  $ B_2,C_2$ respectively
Prove that $ A_2,B_2,C_2$ collinear

[quote="livetolove212"]We know that the reflections of a line $ l$ which passes through $ H$ and perpendicular to $ OH$ in the sidelines $ BC, CA, AB$ concur at a point $ L$ lies on $ (O)$. $ L$ is called [b]Anti-Steinter[/b] point of $ l$ wrt triangle $ ABC$. Denote $ H'$ the reflection of $ H$ in the sideline $ AB$.
We will show that $ C_2L$ is a tangent of $ (O)$.
$ \Leftrightarrow OC_2\perp LC$
$ \Leftrightarrow \angle COC_2 \equal{} \frac {1}{2}\angle COL \equal{} \angle CH'L (1)$
Since $ C_1HOC$ is cyclic we get $ \angle COC_2 \equal{} \angle C_1OC \equal{} \angle C_1HC \equal{} \angle H'HA_1 \equal{} \angle LH'C$
Therefore $ (1)$ is true. Similarly we have $ A_2, B_2, C_2$ lie on a tangent through $ L$ of $ (O)$.[/quote]

Very nice solution,thank you so much!  :lol:
I only had an algebraic solution,I will post it next week.


Let $ a,b,c$ positive real numbers. Prove or disprove that

$ \sum\frac{a}{\sqrt{a\plus{}b}} \geq \frac{4(\sqrt{2a} \plus{} \sqrt{2b} \plus{} \sqrt{2c}) \minus{} 5\sqrt{a\plus{}b\plus{}c}}{4}$

______
 :!:  :!:

[quote="nguoivn"]It's obiviously trues because by Am-Gm, we have:
$ \frac {a}{\sqrt {a \plus{} b}} \plus{} \frac {\sqrt {a \plus{} b}}{2} \geq\ \sqrt {2a}$
Dear my friend, I think if you know exactly your inequality is trues, please write that "Prove".
Because with some ugly problems, if you write "prove or disprove", I think the readers won't interested it  :)[/quote]

Dear friend NGUOIVN, I write "prove or disprove" to avoid errors of calculation or digitization. Thanks for your suggestion :wink:


For $a,b,c$ positive reals such that $a+b+c=3$ prove that

$\frac{a^3b}{ab^2+1}+\frac{b^3c}{bc^2+1}+\frac{c^3a}{ca^2+1} \leq \frac{3abc}{abc+1}$

Sorry, for my first posted ineq. I copied erroneously a Vasc ineq, but this is the correct ineq.

Sorry, again.

yes,manlio,it is known,created by vasc.See topic"A nice cyclic inequality".
And along time ago,vasc posted a nice inequality,very similar to yours:
If a+b+c=3 then:
[tex]\frac{a}{b^2+2}+\frac{b}{c^2+2}+\frac{c}{a^2+2}\geq1[/tex]

Indeed,you last inequality is very nice.

Manlio, for $a=0$ your proposed inequality is not true. :P


A house with a square floor plan is to have each of its outer four walls finished with brick, stone, stucco, or wood. If no two adjacent outer walls are to have the same finish, in how many ways can the house’s walls be constructed? (One such way is brick on the front and back and stone on the left and right.)

This is from the 2006 National Sprint.

No solutions, please, just hints. The answer, for reference, is 84.

This was my (faulty) solution:

For the first wall, there are 4 different choices for finish we have. Then, for the one adjacent to it (doesn't matter which), there are 3 choices. For the other adjacent wall, there are 3 choices as well. Finally, for the last wall, there are two cases:

Case 1 - The walls adjacent to it have the same finish.

Then we have 3 choices.

Case 2 - The walls adjacent to it have different finishes.

Then we have 2 choices.

Thus, our final answer would be $ 4 \cdot 3 \cdot 3 \cdot 3 \plus{} 4 \cdot 3 \cdot 3 \cdot 2 \equal{} 180$, which is a lot higher than the real answer.

[b]Again, hints only please (or pointing out what was wrong with my solution, what I overlooked, etc.)[/b]

[hide="Hint"]Are you sure there are 3 choices for the last wall for ALL THREE choices of the adjacent wall?[/hide]

I think I have it. Can someone please verify if this is the correct solution?

In my second solution, I used complementary counting. If we have exactly two adjacent walls that are the same color, there are $ \left(\binom{4}{2} \minus{} 2\right) \times 30 \equal{} 120$ ways to do so. If we have exactly three walls all the same, there are $ \binom{4}{3} \times 12 \equal{} 48$ ways to do so. Finally, if all four walls are the same, there are 4 ways to do that. Thus, the answer is $ 4^4 \minus{} 172 \equal{} \boxed{84}$.

Assuming that is, thanks PowerOfPi.

Another problem I'm having difficulty with:

A 3x3x3 wooden cube is painted on all six faces and then cut into 27 unit cubes. One unit cube is randomly selected and rolled. What is the probability that exactly one of the five visible faces is painted?

From the 2004 National Team. The answer is 1/3, but I got 11/54.  :oops:

Again, hints only, please.

edit: Actually, got it. Never mind.

I think PowerOfPi was making the point that your first solution to the first problem also works, if you fix a little detail you overlooked (you did see that, right?). Your second solution is fine, though.

For the second problem:

[hide="Hint"]
You can either
get a cube with only one face painted and have the painted face not down or
get a cube with two faces painted and have a painted face down
[/hide]

EDIT: ninja'd.


I noticed that some resource pages have a solution button and some don't.
Sometimes when I copy&paste a solutionless page to microsoft word it shows up as a 3 column table with the problem number on the left, the problem in the center, and a solution button (with a valid link) on the right.
In the webpage there is no solution button.
Where did the solution button come from?
This happened on Internet explorer 7, firefox 2, and opera 9.

There's no bug involved. Those pages are not designed to be copy pasted.


Let $k$ be a constant number.Assumed that the equation $(\log_{2}x)^{2}-4k(\log_{2}x)+4k+3=0$ has real roots $\alpha ,\ \beta .$
Find the range of value for which $\log_{\alpha}{\beta}+\log_{\beta}{\alpha}$ can be valued.

what's it mean by range? The value is dependent upon k:

\[\frac{16k^{2}-8k-6}{4k+3}\](by newton sums).


conics hyperbola ellipse trigonometry algebra polynomial inequalities

Let $a,b,c,d\in \mathbb{R}$ such that:
  \[ab=c^2+4\cdot d^2=4\]
Prove that:
\[(a-c)^2+(b-d)^2\geq \frac{8}{5}\]

@Myth: Completely agree.

@problem: In [url=http://www.ihes.fr/PREPRINTS/M00/Resu/resu-M00-06.html]Ilan Vardi, [i]Mekh-Mat Entrance Examinations Problems[/i], 2000[/url], this is problem 8.

  Darij

Most of Mekh-Mat Entrance problems have a nice solution.
But in this case I can propose only the following idea:
Due to problem we are to estimate distance between hyperbola $xy=4$ and ellipse $x^2+4y^2=4$. We will consider only positive quadrant.
It is quite easy to calculate min value of $x\cos t+y\sin t$ on the hyperbola (denote $h_1(t)$) and max value of $x\cos t+y\sin t$ on the ellips (denote $h_2(t)$).
Then required distance [b](!)[/b] is at least $h_1(t)-h_2(t)=2\sqrt{2\sin 2t}-\sqrt{1+3\cos^2 t}$. Now just note $\max_{t}(h_1-h_2)(t)\geq(h_1-h_2)(\pi/3)\geq \sqrt{8/5}$.

That's all.

[quote="pigfly"][quote="Myth"]"Prove" is not a good title for thread.[/quote]
 :D  :D  :D Ok,I will correct it ;)[/quote]
 :(  :(  :(
I think, it is a bad style to state that problem is easy from the beginning. It can be easy for you, but many people may encounter difficulties solving it. And, of course, "nice" is an ultimately subjective opinion.
Try to use your imagination while naming problem!  ;)  :D

When I want to find a particular problem on MathLinks, I usually get tons of threads named "Easy", "Nice", "Easy and nice" or "Easy but nice". As a result, I have to read all of them to see which of them is the one I am looking for. Such way of giving names to problems is absolute nonsense. Why not just call it "(a-c)^2 + (b-d)^2 >= 8/5"? This has the advantage that every time I want to find it again, I know what to search for. And that the problem is "easy" you can tell in the post itself. (And "nice" is a completely subjective characteristic, as Myth said.)

@Myth: Это не случайная задача со вступительного экзамена, это "убойная задача". Тут не обязательно должно быть красивое решение.

  Darij

I think, we should organize "Topic's Title Naming Rules" thread.

[quote="darij grinberg"]
@Myth: Это не случайная задача со вступительного экзамена, это "убойная задача". Тут не обязательно должно быть красивое решение.

  Darij[/quote]

Why not writing within hide tags next time ?  :D

[hide="Myth and Darij"] What is so special about stating that some lethal problems are not supposed to have nice answers.  :P  [/hide]

[quote="orl"][quote="darij grinberg"]
@Myth: Это не случайная задача со вступительного экзамена, это "убойная задача". Тут не обязательно должно быть красивое решение.
Darij[/quote]
Why not writing within hide tags next time ?  :D[/quote]
Не хочу... Вот и всё.

[quote="mashimaro"]find the maxximum value of expression
     $x^2 - 4x + \frac{2}{x-2} $[/quote]
Why did you post it here? [b]Create separate topic next time![/b]

BTW. This expression doesn't have neither max neither min.

[quote=williamxiao]ummmmmm
that was posted 17 years ago[/quote]

lol, wonder if op will see this and respond

the last time they were on was 10 years ago though...


analytic geometry
combinatorics unsolved
combinatorics

Let $n \geq2$ be an integer number and $D_n$ the set of all the points $(x,y)$ in the plane such that its coordinates are integer numbers with: $-n \le x \le n$ and $-n \le y \le n$.

(a) There are three possible colors in which the points of $D_n$ are painted with (each point has a unique color). Show that with
any distribution of the colors, there are always two points of $D_n$ with the same color such that the line that contains them does not go through any other point of $D_n$.

(b) Find a way to paint the points of $D_n$ with 4 colors such that if a line contains  exactly two points of $D_n$, then, this points have different colors.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=14354 .

  darij


Utilizing Pascal's Identity, which states that $ \dbinom{n \minus{} 1}{k \minus{} 1} \plus{} \dbinom{n \minus{} 1}k \equal{} \dbinom nk$, we have

$ A \equal{} \left(\dbinom90 \plus{} \dbinom91\right) \plus{} \left( \dbinom92 \plus{} \dbinom93\right)$

$ \plus{} \cdots\plus{} \left(\dbinom98 \plus{} \dbinom99\right) \equal{} \dbinom{10}1 \plus{} \dbinom{10}3 \plus{} \cdots \plus{} \dbinom{10}9$, and

$ A \plus{} B \equal{} \dbinom{10}0 \plus{} \dbinom{10}1 \plus{} \cdots \plus{} \dbinom{10}{10} \equal{} 2^{10} \equal{} \boxed{1024}$.


Let $P$ and $Q$ be two horizontal neighbouring squares on a $n \times n$ chess board, $P$ on the left and $Q$ on the right. On the left square $P$ there is a stone that shall be moved around the board. The following moves are allowed:
1) move it one square upwards
2) move it one square to the right
3) move it one square down and one square to the left (diagonal movement)
Example: you can get from $e5$ to $f5$, $e6$ and $d4$.
Show that for no $n$ there is tour visting every square exactly once and ending in $Q$.

[hide="maybe this can help"]
it is basically a graph with $n^2$ vertices and direct edges, and it asks to prove that no Hamiltonian paths exist that start with $P$ and end with $Q$.  So maybe consider the parity/number of edges, especially those on the side and the corner (they're "special" in a sense).
[/hide]
I'll try to solve it later...

[hide]Assume there is such a tour.  Suppose we make $k$ diagonal moves in this tour.  Then we must make $k$ moves up and $k+1$ moves to the right to reach $Q$.  Therefore, we make $3k+1$ total moves.  To visit every square, we must make $n^{2}-1$ moves.  Therefore, $3k+2=n^{2}$, which is impossible because $2$ is not a quadratic residue $\pmod{3}$.  Therefore, no such tour exists.[/hide]


How many elements does the following set contain?
$ S\equal{}\{x,3x\minus{}1\}$,     $ x\in \mathbb R$

[quote="someperson01"]I believe 2, actually. x and 3x-1.[/quote]
Is that true if $ x\equal{}\frac{1}{2}$?

Good point! I completely missed that. Yes, in this case I believe there is one element. Will someone please look at this?  :blush:

[quote="Ognjen09"]How many elements does the following set contain?[/quote]

I think it's important to note that $ S$ here is really a "family" of sets, depending on the value of $ x$; if you have a different value of $ x$, you have a different set $ S$.

There are only two cases:

If $ x \equal{} \frac {1}{2}$, then $ x \equal{} 3x \minus{} 1$ and $ |S| \equal{} 1$.

If $ x\neq \frac {1}{2}$, then $ x\neq 3x \minus{} 1$ and $ |S| \equal{} 2$.


Show that this sequence contains infinite multiples of 7:

\[ a(1)=10, a(n)=a(n-1)+p(n) \]

(p(n) is the largest prime factor of n).

Does P(n)=n when n is prime itself??? Just 4 clarification sake..


Well, we can get a simple recurrence. Consider how many $ 2^0$ terms there can be.

We get that f(25)=f(12)+f(11)+...+f(1)+1 since once we use all of the 2^0 terms, we can divide by 2 (for example, the number of ways to make 24 with no 2^0 is the same as the number of ways to make 12).

Then just find the general formula, to the recurrence f(2n)=f(2n+1)=f(n)+f(n-1)+...+f(1)+1

With f(1)=1.


A large cube is formed by stacking $27$ unit cubes. A plane is perpendicular to one of the internal diagonals of the large cube and bisects that diagonal. The number of unit cubes that the plane intersects is
[asy]
size(120); defaultpen(linewidth(0.7)); pair slant = (2,1);
for(int i = 0; i < 4; ++i)
draw((0,i)--(3,i)^^(i,0)--(i,3)^^(3,i)--(3,i)+slant^^(i,3)--(i,3)+slant);
for(int i = 1; i < 4; ++i)
draw((0,3)+slant*i/3--(3,3)+slant*i/3^^(3,0)+slant*i/3--(3,3)+slant*i/3);[/asy]
$\textbf{(A)}\ 16\qquad
\textbf{(B)}\ 17 \qquad
\textbf{(C)}\ 18 \qquad
\textbf{(D)}\ 19 \qquad
\textbf{(E)}\ 20$

[hide]it's kinda hard to visualize at first, but the plane cuts through the midpoint of the 6 edges of the cube that are not connected to either of the two opposing vertices, forming a regular hexagon.  this plane will intersect all but the 4 cubes closest to each of the two vertices. $27-2(4) = \boxed{19 (D)}$

for a "diagram", i think there was an aime #15 from the 1980's that involved the shape i am referring to.

also, the choices make it so guessable...[/hide]

Hmm..

I wish I was just able to find it simply by visualizing.  :( I couldn't even find the number of cubes that the internal diagonal touch.. How would you find that? I don't know if this will help to find the number of the cubes that plane touch but I'm just curious.

Refer to AIME 1985/15 for a similar way of dissecting a cube (fold-up style).

[hide="Question on AHSME Solution"]
http://grab.by/oeB0 is the AHSME Solution I am referring to.

I have one question - how do you get that the equation of the plane is $x+y+z=9/2?$

Otherwise wow this is a really nice solution :o [/hide]

The vector that is normal/perpendicular to the plane $Ax + By + Cz = D$ is $(A, B, C)$.  You can probably prove this with dot products and/or cross products.

Thus, the plane $3x + 3y + 3z = k$ should be perpendicular to the vector from the origin to $(3, 3, 3)$.  The plane must also bisect the diagonal from $(0, 0, 0)$ to $(3, 3, 3)$, so it must hit $(1.5, 1.5, 1.5)$.  Plugging that in gives $3 \cdot 1.5 + 3 \cdot 1.5 + 3 \cdot 1.5 = k$, which gives $k = 13.5$.  Plugging back in gives $3x + 3y + 3z = 13.5$, or $x + y + z = 4.5$.


probability
integration
calculus
expected value
probability and stats

let $ X_1,...,X_n$ be independent random variables, all with the same distribution - uniform on $ [0,1]$.
Find $ E ( X_1 | \mbox{max}(X_1,...,X_n))$

For $ n=1$ everything is obvious, since $ \max\{X_1,\ldots,X_n\}=X_1$ holds. So assume $ n>1$ and let $ Y$ denote the maximum random variable. Then the joint probability law is given by $ \mathbb{P}(X_1\leq a,Y\leq b)=\left\{\begin{array}{ll}ab^{n-1}& \mbox{for }\;0\leq a \leq b\leq 1\\[0.2cm]0 & \mbox{otherwise}\end{array}\right.$.

This gives $ \mathbb{E}\left[X_1\cdot 1_{\{Y\leq y\}}\right]=\int_0^1\int_0^b\underbrace{X_1(a)1_{\{Y\leq y\}}(b)}_{\ldots = a \mbox{ if }b\leq y}(n-1)b^{n-2}\,da\,db=\frac{n-1}{2(n+1)}y^{n+1}$. Therefore, we can conclude $ \mathbb{E}[X_1|Y]=\frac{n-1}{2n}Y$, since

(i) $ \frac{n-1}{2n}Y$ is $ Y$-measurable

(ii) $ \mathbb{E}\left[\frac{n-1}{2n}Y\cdot 1_{\{Y\leq y\}}\right]=\int_0^1\frac{n-1}{2n}\underbrace{Y(b)1_{\{Y\leq y\}}(b)}_{\ldots =b \mbox{ if }b\leq y}nb^{n-1}\,db=\frac{n-1}{2(n+1)}y^{n+1}$

(iii) the sets $ \{Y\leq y\}$ generate the $ \sigma$-algebra of $ Y$

hjbrasch, you've done something wrong (probably something trivial?) because the answer is $ \frac{n \plus{} 1}{2n}Y$.  (Naively, we can see that the answer must be at least $ \frac{Y}{2}$: if $ X_1$ is not the largest of the $ X_i$, it is uniformly distributed in the interval $ [0, \max\{X_i\}]$.)

I would have computed as follows: if $ \max\{X_i\} \equal{} k$, we have $ \frac{1}{n}$ of the time that $ X_1 \equal{} k$ and $ \frac{n \minus{} 1}{n}$ of the time that $ X_1$ is uniformly distributed in $ [0, k]$, so it has an expected value of $ \frac{k}{n} \plus{} \frac{n \minus{} 1}{n}\cdot \frac{k}{2} \equal{} \frac{n \plus{} 1}{2n} \cdot k$.

thanks for help, but JB can you show it in more formal way, please?

thx, [b]JBL[/b], the calculation of $ \mathbb{E}[X_1\cdot 1_{\{Y\leq y\}}]\equal{}\int_0^yx_1dx_1\cdot\prod_{i\equal{}2}^n\int_0^ydx_i\equal{}\frac{1}{2}y^{n\plus{}1}$ instead of the incorrect previous "result" $ \frac{n\minus{}1}{2(n\plus{}1)}y^{n\plus{}1}$, therefore the correct conditional expected value is $ \frac{n\plus{}1}{2n}Y$ using a similar version of (i), (ii) and (iii) as before

P.S. I wandered about the counter-intuitive result in my first response, but was too lazy to cross-check my calcs  :ninja:

[quote="JBL"]I would have computed as follows: if $ \max\{X_i\} \equal{} k$, we have $ \frac {1}{n}$ of the time that $ X_1 \equal{} k$ and $ \frac {n \minus{} 1}{n}$ of the time that $ X_1$ is uniformly distributed in $ [0, k]$, so it has an expected value of $ \frac {k}{n} \plus{} \frac {n \minus{} 1}{n}\cdot \frac {k}{2} \equal{} \frac {n \plus{} 1}{2n} \cdot k$.[/quote]

Hi JBL, can you explain this heuristics in a bit more detail? I couldn't really understand ...if $ \max\{X_i\} \equal{} k$, we have $ \frac {1}{n}$ of the time that $ X_1 \equal{} k$ and $ \frac {n \minus{} 1}{n}$ of the time that $ X_1$ is uniformly distributed in $ [0, k]$,...etc

Given a bunch of i.i.d. variables $ X_1, \ldots, X_n$ with continuous distribution, the probability $ X_1$ is the largest is precisely $ \frac {1}{n}$.  Otherwise, some other $ X_i$ is the maximum.  Then we have $ \mathbf{E}(X_1 \mid X_1 \leq X_i \equal{} \max(X_1, \ldots, X_n) \equal{} k) \equal{} \frac {k}{2}$ because the conditional distribution of $ X_1$ under the restriction that $ X_1 < k$ is uniform in $ [0, k]$.

but if we assume that the joined density of $(X_1,Y)$ is $f(x,y)=(n-1)y^{n-2} \cdot 1_{\{x\leq y\}}$ and we know that the density of $Y$ is $f_Y(y)=ny^{n}$ and using the fact that $ E ( X_{1}| Y=y)=\frac{\int_0^1 x f(x,y)dx}{f_Y(y)}$  we will get the same result as hjbrasch found at the first topic : $ \mathbb{E}[X_{1}|Y]=\frac{n-1}{2n}Y $ !! I'm wondering where is the mistake because I'm convinced that the argument of JBL is correct ?!

Well, since neither of your densities integrates to $1$, it is clear that something in your computations went astray. As a matter of fact, the first density represents only the absolutely continuous part of the joint distribution while there is also a non-trivial measure sitting on the diagonal $X_1=Y$.

Yes, that should be the mistake ! but I couldn't see how the get the densities since we must count the contribution of  $X_1=Y$ ! [size=150]Fedja[/size], could you please develop the calculus or just the idea to get the density of Y and the joined ones ??  thanks in advance

As I noted in my previous post, given i.i.d. variables $X_1, \ldots, X_n$ with continuous distribution, we have for all $i$ that $\Pr(X_i = \max(X_1, \ldots, X_n)) = \frac{1}{n}$.


Find the largest $t$ so that

 $(1-t)(\frac{b}{a}+\frac{c}{b}+\frac{a}{c})+t(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\geq \frac{1}{3}(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$
inequality holds for all $a,b,c$ s.t $a\geq b\geq c>0$

This problem is not hard! :lol:
You are can using SOS method  :) If I'm not wrong then $t= \frac{2}{3}$

$\text{Yeah!!}\\ \text{ USE SOS we can find t\leq  \frac{2}{3}}$

[quote="mxy020426"]what's SOS?[/quote]
See [url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=sos&t=80127[/url].


Given that Joyce paints $ \frac14$ of her bedroom one day and her sister paints $ \frac23$ of the unpainted part of the room the next day. what fraction of the bedroom is yet to be painted?

After the first day, there are $ \frac44 \minus{} \frac14 \equal{} \frac 34$ left to be painted.

So now you have to subtract $ \frac34 \minus{} \frac23$

Use a common denominator, which is 12 $ \frac {9}{12} \minus{} \frac {8}{12} \equal{} \frac {1}{12}$ left

[color=blue]No need to quote the post directly before you. :)[/color]

[hide="solution"]
There is $ \frac{4}{4}\minus{}\frac{1}{4}\equal{}\frac{3}{4}$ of the room left after Joyce paints. Her sister paints $ \frac{2}{3}\times\frac{3}{4}\equal{}\frac{1}{2}$ of the room. That means there is now $ \frac{3}{4}\minus{}\frac{1}{2}\equal{}\boxed{\frac{1}{4}}$ of the room left.[/hide]

The mistake the previous poster made was that (s)he subtracted $ \frac{2}{3},$ but it doesn't say Joyce's sister painted $ \frac{2}{3}$ of the [u]whole[/u] room, she painted $ \frac{2}{3}$ of the [u]unpainted[/u] part.


algorithm trigonometry induction MATHCOUNTS AMC USA(J)MO USAMO

I can calculate really fast in my head(i.e. I can do 342x252 in 30 seconds), but it takes me about several hours to solve an decent hard olympiad math question.   I think I'm more of a calculation type, but I want you guys' honest opinions.

No one cares about calculation past middle school. General olympiads will do a very good job of improving your [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Prob_Solv.php]problem solving skills[/url].

Maybe he means olympiad-caliber questions that involve calculating a length rather than, say, proving a concurrency. These sorts of questions can be just as challenging.

Well, what I mean is for example...
Calculation: Expand (a-b+3c)^6. Do it in your head.
General: (Any IMO math problem).

Maybe I'm too subjective, as a past IMO contestant, but what's the point of high-leveled mental mathematics/arithmetics? Now, I know that most people won't find point in IMO mathematics either, but I guess that this sort of mental exercises don't give something valuable in the long-run. For instance, I do use mental calculations very often, but it is more of a control of the work being made on the paper (very often I don't try to get exact values-approximations (using power series and so) do the trick). On the other hand, IMO problem solving is a practice for forming the mind of a young mathematician/engineer/scientist and making him analyze, produce algorithms, find new ideas.

My two cents...

[quote]Expand (a-b+3c)^6[/quote]

Computers can do 200000000 such problems per second so it makes that sort of skill unvaluable.

Computers(and many people) can not solve IMO problems.

I think my choice is obvious :)

Wait, do "calculation olympiads" actually exist?

If you have to choose between the two, definitely go with an IMO-type olympiad; there's nothing rewarding about expanding trinomials.  Although, if you want an unbiased opinion then here is not really a good place to ask because most people here participate in problem-solving math competitions.

practicing calculations in your head help you a lot to open up your brain..
i have been practicing these calculations from a very young age and my this ability has helped me a lot..
just as i do calculations in my head fast..at the same rate i can try different approaches to a problem(IMO level ones)...
i keep on trying and trying at such problems and dont get tired before 7-8 hours..
the reason is i practice some really hard multiplications..and additions and other general stuff in my head..
i started of by multiplying two two-digit numbers when i was small..
and now i am able to multiply 2 100-digit numbers in my head..
ofcourse it takes time..about 6 hours..
its an equivalent of meditation..because you cut yourself off from the whole world..
such exercises help you a lot to build the muscles of your brain..and increase blood flow and oxygen supply..

Well, I don't consider myself to be very good at calculation, but that multiplication took about 20 seconds... Also, most olympiad problems are supposed to take a few hours to solve. (unless you are Alex Zhai, in which case it takes about 20 minutes  :wink: )

Doing calculations mindlessly is not that helpful. It could be good to sharpen your mind, but what you want from math contests is the critical thinking skills, creativity, and problem-solving, not the pure speed... It's like playing piano. You could practice Hanon studies (technical stuff) for years, or actually enter into the music, feeling it, living it. I think the latter is more enjoyable and more useful.

Also, as others have said, computation is mechanical - something machines can do. Humans have the capability for greater things. Of course, computation is not [i]bad[/i]... it's just that a "calculation olympiad" (I used to do some math contests that were just calculation, but that was elementary school) does not bring you to your full potential. Maybe you look cool, and other people can be like "HEY HE MULTIPLIES FAST" - but computers can multiply faster! If you want a challenge and a more rewarding experience, go for the IMO. I personally like calculation (except in Olympiad problems :mad: ), but I like Olympiads better for the reasons above.

I think there's a false dichotomy here.  It's obviously bad to continue doing calculations mindlessly forever, but I think being able to do a calculation sometimes helps give you some intuition about what's going on.  This, at least, seems valuable for helping in future mathematical pursuits.

Example: Expanding (a - b + 3c)^6 is not too useful by itself, but if you do this enough, you see how the symmetric sums work themselves out, you get some intuition for how symmetric sums and power sums interact, etc.  While this intuition doesn't lead directly to proofs, it can give you a really concrete grasp of the small cases that is useful later.

[quote]Calculation olympiad[/quote]
[quote]I can do 342x252 in 30 seconds[/quote]
[quote]To be honest, you're good at math, but you're a super genius on calculation stuff[/quote]
[quote]You're pretty good, I guess[/quote]

I think you guys got trolled.

hey i want to direct this primarily to not_trig about
his comment on calculation vs. creative/problem
solving,( also possibly others if they are interested in
answering my question), but it seems to me that calculation,
the "brute speed" part of math involves spatial manipulations,
but sometimes on things like math competitions mainly
logical intelligence is used, whether it be deduction or induction,
in other words, how do many of the people who perform well
at math contests "think", do they use logical skills, pattern recognition,
or does visualizing (seeing the problem in terms of pictures) help even more,
is it better to be a good visualizer (spatial-IQ), or just a logical thinker (logic
-IQ), it know it may sound a bit stupid cuz one may think that math is complete
logic, but for me I visualize things strongly, idk..... can someone explain?

My own opinion:

[u]"Calculation" olympiads[/u]

To be honest, I have never heard of these. From your example, it seems like you are referring to just a computation, rather than problems which just involve some computation.

Computations by themselves are not too useful, and doing them over and over is just useless. Obviously, in some problems, it's good to do computations for intuition or to develop a feel for the problem, but I'm pretty sure that's not what you are referring too.

[hide="Also"]I've already seen too many examples of people thinking that math is all about computation. Back in elementary school, math skill was judged (for me) based on how well we took timed tests (addition, subtraction, multiplication, division). Now, it's judged on how fast you can count something. Obviously a little extreme, but yeah. And people would be really impressed by seeing something multiplied really fast or something with unfamiliar symbols or Greek letters on it, so
the problem "Copy down the following symbols: $\alpha \beta \theta \phi \pi \Sigma \iota \gamma \kappa \mu \nu \Delta \epsilon \delta \zeta \omega || \triangle \angle \cong$" would at first glance look harder than "prove that there are no solutions to the equation $13^x+37^x=0$.[/hide]

I like the general Olympiads a lot :P

oh yeah, this kind of brings me to another question I've
always been wondering about, people say IQ and creativity
are uncorrelate (IQ being a more innate ability to solve problems
and learn), and I've heard you need a lot of creativity to
do well at math competitions, personally I've always thought that
people who win the USAMO or make it to IMO have IQs 140+? idk,
you guys probably heard about Richard Feynman- famous physics/math dude
, a genius in my opinion, he was tested at 125, which is high
on the bell-curve but not genius (certified genius being 140), so yeah is it
the creativity or the IQ that makes these kids do so well
at national, statewide, USAMO, IMO, you-name-it competitions, the reason i ask this
because I've always believed math is highly associated with the way some IQ tests work,
i know IQ is a pretty personal subject for many, but if someone can
enlighten me or give me insight on IQ relationship with doing at math contests :)

yeah guys, please forgive all these questions i keep asking
if it annoys you, but i'm new to this site and also new to
math competitions so i have a lot on my mind, but i have another
general question, i've noticed that a lot of these math people on this site
write their posts/comments in a very peculiar way I can't really explain well,
it's sort of like a mall-talk way if you will, "lol....", "I pwned this math competition...
" and especially the use of "darn" or "darjn" or whatever
 which i don't find as frequent in any other groups
as with the people on the art of problem solving site, even more interesting
is I here the terms "bash" like "bashy math"
which i'm not really sure what it means (maybe it means
hard or trivial...idk) and also if anyone knows what I'm talking about:
people here use "not [something]", like it makes the dialogue logical
in away, idk any insights... if you kinda know what I'm saying

IQs....
I don't think that the IQ test is an accurate test on how well people can learn and solve problems. Actually a test that I believe is more accurate (but not fool-proof) is the [url=http://www.naclo.cs.cmu.edu/]North American Computational Linguistics Olympiad[/url]. However, people with awesome IQs can be very smart, it's just that I haven't really seen them apply their smarts in any fashion that changes the world, or at least how it thinks. I may just live under a rock, and there are many such people, but I'm sure that Einstein and Feynman were not in that group of people.

"Mall speak"...
Well a possible explanation for the existence of this phenomenon is the fact that many students that frequent the AMC forums are either in Middle School or High School, and they've learned quite a few acronyms from just chatting online ("lol" meaning "laughing out loud", "pwned" is a word that essentially means "doing well at something", etc.) I'm not sure how the word "darn" became popular. However, when the USAJMO was first introduced, the AoPS user zephyredx used the word "dar(j)n" instead of "darn", and this just stuck with everybody (people went so far as to make a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=139&t=352709]"Stick j's in sentences" topic[/url] in the Fun Factory).

[quote="SetPoint"]Is MathCounts considered a calculation contest?[/quote]
I'm not sure about the definition of a "calculation contest". If it means what I think it means (straight-forward calculating), then that isn't a very good (or practical) contest to run. What would be the use of 228 people that can calculate 26 times 74 in 5 seconds? MathCounts is designed for Middle School students that think outside the box of the standard math curriculum (a very small box!). These outside-of-box-thinkers are invaluable to the field of scientific discovery, since they are never inside-box-discoveries (if it were inside the box, then we would have already discovered it).

If you're stuck on an inequality, using bruteforce calculation quickly and accurately can prove to be a useful skill to have.

I think general math olympiads really are harder. Calculation doesn't count for much nowadays, except to prove that you can understand how calculations work and invent ways to store things efficiently in your head.

[quote="AwesomeToad"]I've already seen too many examples of people thinking that math is all about computation. Back in elementary school, math skill was judged (for me) based on how well we took timed tests (addition, subtraction, multiplication, division). Now, it's judged on how fast you can count something. Obviously a little extreme, but yeah.[/quote]

No kidding. My 8-year-old brother and his classmates also do this a lot of the time, and I cringe every time they challenge each other to multiply "nine thousand billion trillion quadrillion" by "seventeen billion five thousand and ten."

[quote]And people would be really impressed by seeing something multiplied really fast or something with unfamiliar symbols or Greek letters on it, so
the problem "Copy down the following symbols: $\alpha \beta \theta \phi \pi \Sigma \iota \gamma \kappa \mu \nu \Delta \epsilon \delta \zeta \omega || \triangle \angle \cong$" would at first glance look harder than "prove that there are no solutions to the equation $13^x+37^x=0$.[/quote]

Of course, in this case, it [b]is[/b] harder because you're not given any context for the variables. :P

[quote="NuncChaos"][quote="sdkudrgn88"]No kidding. My 8-year-old brother and his classmates also do this a lot of the time, and I cringe every time they challenge each other to multiply "nine thousand billion trillion quadrillion" by "seventeen billion five thousand and ten."
[/quote]

Admittedly, though, for 8-year olds this might not be so bad; I remember when I was around that age and had no idea what algebra was, I still "discovered" (out of trying to do mental math) various factoring formulae. Little kids who do a lot of mental math develop (since they are young) strong concentration powers and "discover" neat patterns... This sort of discovery ability can then extend when they are older to pretty much anything.[/quote]

It's not just for 8-year olds; computations actually give highly useful intuition about numbers. Example: I ask hamster about a problem, he's like "Go do all 1024 cases for n = 5 and you'll have enough intuition to solve the problem."

(He was right, though it only took around 200 cases before i solved it)

[quote="SetPoint"]Is MathCounts considered a calculation contest? Cause I didn't like it that much.[/quote]

MathCounts is not a calculation contest. It involves calculations, as do most other contests, but it requires some degree of ingenuity to solve some of the problems.

Yes, but it is a very small degree of ingenuity, and many contestants, especially at the higher levels, have more then enough the necessary problem-solving skills, so it just becomes calculation.

It really doesn't matter about which one is actually better; do the competition you're better at. However, without regard to which one you're better at, you should definitely do the general math olympiad. Calculation Olympiads can be quite... boring, and it all comes down to who spends more hours a day working on simple problems and memorizing all the squares and cubes and all that stuff.

[quote="matt276eagles"]Wait, do "calculation olympiads" actually exist?[/quote]
[quote="AwesomeToad"][u]"Calculation" olympiads[/u]

To be honest, I have never heard of these.[/quote]
There are contests of this type, though I wouldn't call them "calculation olympiads".  National Assessment and Testing competitions are completely calculation based - either you know a problem or you don't, and you have to solve it as fast as possible.
[url]http://natassessment.com/[/url]
[url]http://en.wikipedia.org/wiki/National_Assessment_%26_Testing[/url]
The Fall Startup Event, for example, is a 100-question test which must be solved in 30 minutes - 18 seconds per problem.  The problems include simple addition, systems of two linear equations, simple applications of quadratic formula or power of point, etc.  If you know how to do them, they're easy, but it's so fast-paced that the whole thing is actually a very fun contest - it's like a race, where you're moving as fast as you possibly can.  It gets really exciting, for me anyway  :)

But while I find these contests fun, I think that proof-based math such as is found in the USAMO and IMO is much more interesting and valuable, and being able to do this type of math is much more impressive than being able to calculate simple things quickly.

Oh Thats what you mean by calculation olympiads. Well, they're not really olympiads, more like tests. Those problems are, well, not that fun, and one does not really need to actually do problems to get better at those.

[quote="filetmignon821"]Those problems are, well, not that fun[/quote]

That's subjective. I find computation tests fun, probably because I'm better at them compared with a standard Olympiad. It's not like all computation problems have to be ugly; for example, a telescoping sum is usually considered "nice" and computational.

[quote="filetmignon821"]one does not really need to actually do problems to get better at those.[/quote]

That's not true at all. There are number sense tricks, arithmetic strategies, and other things to learn. MATHCOUNTS is often viewed as a calculation-based competition. Yet there is never a winner who hasn't spent countless hours solving computational problems.

Thats true. However, many problems on the national assessment and testing require pure memorization to solve within the time limit, and it is often that one either gets it in a few seconds or doesn't get it at all. Its not that easy to telescope a series in half a minute.

[quote="filetmignon821"]Those problems are, well, not that fun[/quote]
I echo AIME15... in fact, I find these contests very, very fun, if in a different way than normal contests.  And I'm not even that good at them.  I just find them intense and exciting.
[quote="chinesebarefeet"]lol, just wondering, does anybody know how to find who the
person's real name is, because idk all the usernames that well
here yet?[/quote]
It's not necessary to know someone's true name.  In fact, I think one of the most beautiful things about online forums is that you don't know anyone's name.  But you can generally figure it out from the information on their profile if you really want to know.

another question... math contests... are they more about
logical thinking or abstract thinking, for example you can be
bad at "straight math" and good at logic (does anybody agree with that)
right? Like, i personally know people who are for example really
good at calculus, yes i would definitely need higher thinking for it, but
it's not really logic is it?, it's more abstract, also does anybody
else agree that for calculus it's more about visualization, like you
picture stuff in your head and see patterns instead
of lust logical reasoning.  What do you need most
for math competitions? Like i know this may seem a bit weird but
is it true that if people lack common sense, they do worse on math
competitions, especially with word problems, if you are bad at
reading comprehension isn't it quite possibly to misunderstand
the problem, and not see through the logic?

also just curious does anybody know if Calvin made MOP this year?

He made mop, but not black. He was a HM this year.

so who made black mop, and whoes going
to on the IMO


$ x,y\in Z$
find
$ xy\minus{}13x\plus{}16y\plus{}2\equal{}0$

[quote="arkgnael"]$ x,y\in Z$
find
$ xy\minus{}13x\plus{}16y\plus{}2 \equal{} 0$[/quote]

$ (x\plus{}16)(y\minus{}13)\equal{}\minus{}210$

This is going to have a lot of solutions. 32, to be precise. Basically you want $ (x\plus{}16)$ and $ (y\minus{}13)$ to be two quantities that when multiplied result in $ \minus{}210$.