Denote by $ X$, $ Y$, $ Z$ the intersections $ BB^{\prime} \cap CC^{\prime}$, $ CC^{\prime} \cap AA^{\prime}$ and $ AA^{\prime} \cap BB^{\prime}$, respectively. According to Pascal's theorem applied to the hexagon $ ABB^{\prime}PC^{\prime}C$ (or for example, from Pappus' theorem) we get that the points $ B_{1}$, $ X$, $ C_{1}$ are collinear. Similarly, the points $ C_{1}$, $ Y$, $ A_{1}$ are collinear and so are $ A_{1}$, $ Z$, $ B_{1}$. In this case, since the lines $ AY$, $ XB$, $ B_{1}A_{1}$ are concurrent at $ Z$, the triangles $ AXB_{1}$ and $ YBA_{1}$ are perspective, and therefore, by Desargues' theorem, the intersection points $ AX \cap YB$, $ XB_{1} \cap BA_{1} \equal{} \infty$, $ AB_{1} \cap YA \equal{} \infty$ are collinear. Thus, $ AX \| BY$. Similarly, we will get that $ BY \| CZ$ and so $ AX \| BY \| CZ$.
Denote by $ \left[\mathcal{P}\right]$ the area of the polygon $ \mathcal{P}$. In this case $ \left[XYZ\right]\equal{}\left[ABX\right]\plus{}\left[ACX\right]$, and since $ X$, $ B_{1}$, $ C_{1}$ are collinear it follows that $ \left[XYZ\right]\equal{}2\left[AC_{1}X\right]\plus{}2\left[AB_{1}X\right]\equal{}2\left[AB_{1}C_{1}\right]\equal{}\left[ABC\right]/2$, which is independent of the position of $ P$.

Here are another two (more) interesting results related to this configuration:

[b]Extension[/b]. Let $ I_{a}$, $ I_{b}$, $ I_{c}$ be the corresponding excenters of triangle $ ABC$. Denote by $ \alpha$, $ \beta$, $ \gamma$ the reflections of $ AX$, $ BY$, $ CZ$ into the lines $ I_{b}I_{c}$, $ I_{c}I_{a}$ and $ I_{a}I_{b}$, respectively. Prove that $ \alpha$, $ \beta$, $ \gamma$ concur at $ P$.

[b]Extension 2[/b]. Let $ MNQ$ be the anticomplementary triangle of $ ABC$. Denote by $ \alpha^{\prime}$, $ \beta^{\prime}$, $ \gamma^{\prime}$ the reflections of $ AX$, $ BY$, $ CZ$ into the sideliens $ NQ$, $ QM$ and $ MN$, respectively. Prove that $ \alpha^{\prime}$, $ \beta^{\prime}$, $ \gamma^{\prime}$ concur at a point $ P^{\prime}$ lying on the circumcircle of triangle $ ABC$ such that $ PP^{\prime} \| AX \| BY \| CZ$.

Let $ AA' \cap BC \equal{} A_2, BB' \cap CA \equal{} B_2, CC' \cap AB \equal{} C_2.$ Suppose $ P$ and $ B$ are on the opposite side of $ AC,$ so that $ A_2$ and $ C_2$ lie on the sides $ BC$ and $ AB,$ but $ B_2$ lies on the extension of side $ CA.$ We have \[ \tfrac{|BA_2|}{|A_2C|} \equal{} \tfrac{|BA|}{|AC|} \tfrac{\sin \angle BAA'}{\sin \angle A'AC} \equal{} \tfrac{c}{b} \tfrac{\sin \angle BPA'}{ \sin \angle A'PC} \equal{} \tfrac{c}{b} \tfrac{|PC|}{|PB|} \tfrac{|BA_1|}{|A_1 C|} \equal{} \tfrac{c}{b} \tfrac{|PC|}{|PB|} \equal{} \tfrac{\mathcal C}{\mathcal B}\] where $ \mathcal A \equal{} a \cdot |PA|, \mathcal B \equal{} b \cdot |PB|, \mathcal C \equal{} c \cdot |PC|.$ Similarly, $ \tfrac{|CB_2|}{|B_2A|} \equal{} \tfrac{\mathcal A}{\mathcal C}$ and $ \tfrac{|AC_2|}{|C_2B|} \equal{} \tfrac{\mathcal B}{\mathcal A}.$ Further, by [url=http://mathworld.wolfram.com/PtolemysTheorem.html]Ptolemy's theorem,[/url] we have $ \mathcal B \equal{} \mathcal A \plus{} \mathcal C.$ According to [url=http://mathworld.wolfram.com/RouthsTheorem.html]Routh's theorem[/url] and the fact that $ \lambda \equal{} \tfrac{\mathcal C}{\mathcal B}$ and $ \mu \equal{} \minus{} \tfrac{\mathcal A}{\mathcal C}$ and $ \nu \equal{} \tfrac{\mathcal B}{\mathcal A},$ we have that the area of the triangle is \[ |ABC|  \left|\frac{4\mathcal A\mathcal B\mathcal C}{(\mathcal A \plus{} \mathcal B \plus{} \mathcal C) (\minus{}\mathcal A \plus{} \mathcal B \plus{} \mathcal C) (\minus{}\mathcal A \minus{} \mathcal B \plus{} \mathcal C)} \right| \equal{} |ABC| \left| \frac{4\mathcal A \mathcal B \mathcal C}{(2 \mathcal B) (2 \mathcal C) (\minus{}2 \mathcal A)} \right|\]\[ \equal{} \frac{|ABC|}{2}\]

Let $X = BB'\cap CC'$, $Y = CC'\cap AA'$, $Z = AA'\cap BB'$. By Pascal's theorem on $BB'PC'CA$, $X$ lies on line $B_1C_1$; similarly, $Y\in C_1A_1$ and $Z\in A_1B_1$. Since $B_1C_1\parallel BC$, $\measuredangle{AB_1X} = \measuredangle{ACB} = \measuredangle{AB'X}$, so $AB'B_1X$ is cyclic. Since $B' = BX\cap PB_1$ and $AB'PB$ is cyclic as well, we thus obtain the spiral similarity $\triangle{AB_1X}\sim\triangle{APB}$.

Now we work in the complex plane with unit circle $(ABCP)$. Using $b_1 = \frac{a+c}{2}$, the aforementioned spiral similarity translates directly as
\[ \frac{b_1-a}{x-a} = \frac{p-a}{b-a} \Longleftrightarrow x = a + \frac{(a-b)(a-c)}{2(p-a)}. \](In fact, it is only slightly more annoying to compute $x$ without noticing $X\in B_1C_1$ or the spiral similarity.)

From here, one can directly compute $[XYZ]$ using a complex determinant. However, there is a nicer approach: in order to compute $\frac{x-b_1}{x-c_1}$, we use the analogous $\triangle{AC_1X}\sim\triangle{APC}$ to deduce
\[ \frac{x-b_1}{x-c_1} = \frac{(a-b_1)\frac{b-p}{a-p}}{(a-c_1)\frac{c-p}{a-p}} = \frac{a-c}{a-b}\frac{p-b}{p-c}. \]Now define $\alpha = (b-c)(p-a)$, $\beta = (c-a)(p-b)$, and $\gamma = (a-b)(p-c)$, and observe that $\alpha+\beta+\gamma=0$ (Ptolemey's theorem with complex numbers). Then $\frac{x-b_1}{x-c_1} = -\frac\beta\gamma$, so $X$ has barycentric coordinates $(0,-\frac\gamma\alpha,-\frac\beta\alpha)$ with respect to $\triangle{A_1B_1C_1}$. Hence the ratio between the signed areas of triangles $XYZ$ and $A_1B_1C_1$ is
\[
\det\begin{pmatrix} 0 & -\frac\gamma\alpha & -\frac\beta\alpha \\ -\frac\gamma\beta & 0 & -\frac\alpha\beta \\ -\frac\beta\gamma & -\frac\alpha\gamma & 0 \\ \end{pmatrix}
= -\frac1{\alpha\beta\gamma} \det\begin{pmatrix} 0 & \gamma & \beta \\ \gamma & 0 & \alpha \\ \beta & \alpha & 0 \\ \end{pmatrix}
= -\frac1{\alpha\beta\gamma} (-\gamma(-\alpha\beta) + \beta(\gamma\alpha)).
\](This is the matrix, in barycentric coordinates, corresponding to the affine transformation sending $\triangle{A_1B_1C_1}$ to $\triangle{XYZ}$.) Hence $[XYZ] = 2[A_1B_1C_1] = \frac12[ABC]$ is independent of $P$, as desired.

Could any one explain me this ???
area(XYZ) = area(ABX) +area(ACX) ???
Is true??? Sir

Yes pohatza is wrong in that step but still the equality is valid.Because

$[XYZ]=[XYB]+[BYZ]=[AYB]+[BYC]=[BA_1C_1]+[AC_1Y]+[CA_1Y]=[BA_1C_1]+[C_1B_1Y]+[A_1B_1Y]=[BA_1C_1]+[B_1A_1C_1]=\frac{[ABC]}{2}$

I think the proof is really nice.Well done pohatza!!

The vertices lie on the medial triangle of ABC by Pascal and by Thales' theorem we easily see the area is (ABC)/2

My solution:

Let $ X=BB' \cap CC', Y=CC' \cap AA', Z=AA' \cap BB' $ .

From Pascal theorem ( for $ BB'PC'CA $ ) we get $ X \in B_1C_1 $ .
Similarly, we can prove $ Y \in C_1A_1 $ and $ Z \in A_1B_1 $ .

From Pappus theorem ( for $ B-A_1-C $ and $ Z-A-Y $ ) we get $ BY \parallel CZ $ .
Similarly, we can prove $ CZ \parallel AX $ and $ AX \parallel BY $ .

Now we get $ [XYZ]=[XYA]+[XAZ]=[XBA]+[XAC]=\tfrac{1}{2} [ABC]=\text{Const} $ .

Q.E.D
____________________________________________________________
[quote=pohoatza]
[b]Extension[/b]. Let $ I_{a}$, $ I_{b}$, $ I_{c}$ be the corresponding excenters of triangle $ ABC$. Denote by $ \alpha$, $ \beta$, $ \gamma$ the reflections of $ AX$, $ BY$, $ CZ$ into the lines $ I_{b}I_{c}$, $ I_{c}I_{a}$ and $ I_{a}I_{b}$, respectively. Prove that $ \alpha$, $ \beta$, $ \gamma$ concur at $ P$.[/quote]
Let $ D=YZ \cap BC, E=ZX \cap CA, F=XY \cap XY $ .
Let $ P^* $ be the isogonal conjugate of $ P $ WRT $ \triangle ABC $ (at infinity) .

Since $ AX \parallel BY \parallel CZ $ ,
so from Desargue theorem ( for $ \triangle ABC $ and $ \triangle XYZ $ ) we get $ D, E, F $ are collinear ,
hence $ P(AX \cap \odot (ABC), A_1; B, C)=A(X, A'; B, C)=A(X,D;B,C)=-1 $ ,
so combine with $ A_1B=A_1C \Longrightarrow P \cup AX \cap \odot (ABC) $ is parallel to $ BC \Longrightarrow P^* \in AX  $ .

Similarly, we get $ P^* \in BY, P^* \in CZ $ (or follows from $ AX \parallel BY \parallel CZ $ ) $ \Longrightarrow \alpha, \beta, \gamma $ are concurrent at $ P $ .
____________________________________________________________
[quote=pohoatza]

[b]Extension 2[/b]. Let $ MNQ$ be the anticomplementary triangle of $ ABC$. Denote by $ \alpha^{\prime}$, $ \beta^{\prime}$, $ \gamma^{\prime}$ the reflections of $ AX$, $ BY$, $ CZ$ into the sideliens $ NQ$, $ QM$ and $ MN$, respectively. Prove that $ \alpha^{\prime}$, $ \beta^{\prime}$, $ \gamma^{\prime}$ concur at a point $ P^{\prime}$ lying on the circumcircle of triangle $ ABC$ such that $ PP^{\prime} \| AX \| BY \| CZ$.[/quote]
Seems not true  :maybe:

[hide=Solution]
Denote $X=BB'\cap CC'$, $Y=CC'\cap AA'$, and $Z=AA'\cap BB'$.

By Pascal on $BB'PC'CA$, $X\in B_1C_1$; similarly, $Y\in C_1A_1$ and $Z\in A_1B_1$. Now, we have
\begin{eqnarray*}
[XYZ] & = & [XYA] + [XAZ]\\
& = & ([XCA] + [CYA]) + ([XAB] + [BAZ])\\
& = & [C_1CA]\cdot\frac{B_1X}{B_1C_1} + [CC_1A] + [B_1AB]\cdot\frac{XC_1}{B_1C_1} + [BAB_1]\\
& = & \frac{[BCA]}{2}\cdot\frac{B_1X}{B_1C_1} + \frac{[CBA]}{2} + \frac{[CAB]}{2}\cdot\frac{XC_1}{B_1C_1} + \frac{[BAC]}{2}\\
& = & \frac{[ABC]}{2}\cdot\left(\frac{B_1X}{B_1C_1}-1+\frac{XC_1}{B_1C_1}-1\right)\\
& = & -\frac{[ABC]}{2},
\end{eqnarray*}
where areas are signed. Thus, $[XYZ]$ is independent of $P$.
[/hide]

[quote=April]Let $ ABC$ be a fixed triangle, and let $ A_1$, $ B_1$, $ C_1$ be the midpoints of sides $ BC$, $ CA$, $ AB$, respectively. Let $ P$ be a variable point on the circumcircle. Let lines $ PA_1$, $ PB_1$, $ PC_1$ meet the circumcircle again at $ A'$, $ B'$, $ C'$, respectively. Assume that the points $ A$, $ B$, $ C$, $ A'$, $ B'$, $ C'$ are distinct, and lines $ AA'$, $ BB'$, $ CC'$ form a triangle. Prove that the area of this triangle does not depend on $ P$.

[i]Author: Christopher Bradley, United Kingdom [/i][/quote]

By Pascal's Theorem, we see that $X=\overline{BB'} \cap \overline{CC'}$ lies on line $\overline{B_1C_1}$. Define $Y, Z$ similarly. Observe the

[b]Lemma.[/b] Let $A, B, C, D$ lie on a circle $\omega$ and $E$ be a point inside it. Lines $\overline{EA}, \overline{EB}, \overline{EC}, \overline{ED}$ meet $\omega$ again at $A', B', C', D'$ respectively. Then $(A,B,C,D)=(A',B',C',D')$.

(Proof) Consider a projective transformation which fixes $\omega$ and takes $E$ to its center. The result is obvious now. $\blacksquare$

Let $\overline{BB_1}$ meet $(ABC)$ again at $B_0$. Apply the previous lemma, to get $$(B_1C_1, X\infty) \overset{B}{=} (B_0A, B'C) \overset{B_1}{=} (BC, PA).$$ Let $P_1 \in (A_1B_1C_1)$ such that $A_1B_1C_1P_1$ and $ABCP$ are similar quadrilaterals.

Introduce barycentrics, $A_1=(1,0,0), B_1=(0,1,0)$ and $C_1=(0,0,1)$. Set $P_1=(u, v, w)$ and $a'=\tfrac{a^2}{u}, b'=\tfrac{b^2}{v}, c'=\tfrac{c^2}{w}$; by equation of circumcircle, we have $$a'+b'+c'=0.$$ By the cross-ratio condition, we have $X=(0:b':c')$. Likewise, $Y=(a':0:b')$ and $Z=(a':b':0)$. By area formula, $$\frac{[XYZ]}{[A_1B_1C_1]}=\frac{1}{(b'+c')(c'+a')(a'+b')}\det \begin{pmatrix} 0 & b' & c' \\ a' & 0 & c' \\ a' & b' & 0 \\ \end{pmatrix}=-2,$$ where area is signed. Thus, $[XYZ]$ doesn't depend on $P$, as desired. $\blacksquare$

[quote=gold1]can't  we solve it without using strong theorems like pascal's theorem?…[/quote]

Of course it can. The following thoughtless barybash was done by me without drawing any pictures, making zero synthetic observations, and not knowing Pascal's theorem. And it could be done in steady hands in <15 minutes (But for some reasons took me >5 hours because I was complex bashing first).

[hide=Vey trivial] We  use barycentric w.r.t $\Delta ABC$. Define $C* := AA' \cap BB'$ and so on. Let $P = (p : q: r)$ such that $$ \sum {a^2qr} = 0 $$. Assume that $p \neq q \neq r \neq p$ in the following solution (it can be trivially adjusted to handle that case). We have, $A_1 = (0:1:1)$, and the line through $P$ and $A_1$ is parametrized by $$ (p: pt+q :pt+r)$$. Let $A' = (p:pt'+q: pt'+r)$. As $A' \in \omega_{ABC}$, we have $$a^2(pt'+q)(pt'+r)+b^2p(pt'+r)+c^2p(pt'+q) = 0 $$ $$\Leftrightarrow t'^2 + t' \frac{a^2q+a^2r+b^2p+c^2p}{a^2p} = 0$$. Now $t = 0$ is one root of the equation, as $P \in \omega_{ABC}$, so we get the other root by Viete:  $$t' = -\frac{a^2(q+r)+p(b^2+c^2)}{a^2p}$$. Plugging that in, we get $$ A' = (\text{-----} : pt'+q :pt'+r) = (\text{-----} : \frac{a^2r+p(b^2+c^2)}{a^2} : \frac{a^2q+p(b^2+c^2)}{a^2})  = (\text{-----} : {a^2r+p(b^2+c^2)} :{a^2q+p(b^2+c^2)})  $$.

Now observe $$ {a^2r+p(b^2+c^2)}  = \frac{a^2rq+b^2pq+c^pq}{q} = \frac{b^2pq-b^2pr}{q} = -\frac{b^2p(r-q)}{q}$$.
Analogously, we get $$ :{a^2q+p(b^2+c^2)} = \frac{c^2p(r-q)}{r} $$.
Plugging that back in to the formula of $A'$, we get finally get $$A' = (\text{-----} : -\frac{b^2p(r-q)}{q} : \frac{c^2 p (r-q)}{r} )= (\text{-----} : - b^2r: c^2q)$$.

So, $$C* = (\text{-----} : -b^2r: c^2q) $$ and similarly we find $B' = (-a^2rq: \text{-----} : c^2 pq)$, so $$ C* = (-a^2rq:\text{-----} :c^2pq)$$.  Intersecting the cevians $AA'$ and $BB'$, we finally get $$C* = AA' \cap BB' = (a^2rq: b^2pr: -c^2pq)$$. Letting $X = a^2rq, Y = b^2pr, Z = c^2pq$, and noting $X+Y+Z = 0$, we get the normalized coordinate of $C*$ as $$C* = (-\frac{X}{2Z}, -\frac{Y}{2Z}, \frac{1}{2})$$.

Analogously, and finally using an well known formula, we get $$ \frac{ Area[ \Delta A* B* C* ]}{Area[ \Delta ABC]} =
\det \begin{pmatrix} \frac{1}{2} & \frac{-Y}{X} & \frac{-Z}{X} \\ \frac{-X}{Y} & \frac{1}{2} & \frac{-Z}{Y} \\ \frac{-X}{Z} & \frac{-Y}{Z} & \frac{1}{2} \\ \end{pmatrix}= \frac{1}{8XYZ} \det \begin{pmatrix} X &-Y & -Z \\ -X & Y & -Z\\ -X & -Y & Z \\ \end{pmatrix}= \frac{-1}{2} $$, so the area is independent of $P$, as desired. [/hide]

[quote=Konigsberg][quote=CantonMathGuy]
\begin{eqnarray*}
[XYZ] & = & [XYA] + [XAZ]\\
& = & ([XCA] + [CYA]) + ([XAB] + [BAZ])\\
& = & [C_1CA]\cdot\frac{B_1X}{B_1C_1} + [CC_1A] + [B_1AB]\cdot\frac{XC_1}{B_1C_1} + [BAB_1]\\
& = & \frac{[BCA]}{2}\cdot\frac{B_1X}{B_1C_1} + \frac{[CBA]}{2} + \frac{[CAB]}{2}\cdot\frac{XC_1}{B_1C_1} + \frac{[BAC]}{2}\\
& = & \frac{[ABC]}{2}\cdot\left(\frac{B_1X}{B_1C_1}-1+\frac{XC_1}{B_1C_1}-1\right)\\
& = & -\frac{[ABC]}{2},
\end{eqnarray*}
where areas are signed. Thus, $[XYZ]$ is independent of $P$.
[/quote]

Please explain what's the motivation for the steps in this manipulation. Thanks.[/quote]

Surprisingly I have the same sol. as cantonmathguy but for the sake of the motivation I say -
Note there are midpoints along with a preety diagram with area bisectors [medians] and so parallel lines  for the collinearity by pascals th. So a good area chasing ! :D  Still for a motivation you can call up the scenario of the proof[by trivial area chasing] of the th- The area of the  parallelogram formed by joining  the midpoints of a quadrilateral is half of the area of the original quadrilateral ! :)

[hide=Same as everybody else] WLOG assume $P \in \widehat{AC}$. Define $X = AA' \cap CC', Y = BB' \cap AA', Z = CC' \cap BB'$. Apply Pascal to $APCA'BC'$ to get $AB \cap PC', BC \cap PA', AA' \cap CC'$ to be colinear, i.e $X \in A_1C_1$. Similarly applying Pascal to $BPAB'CA'$ gives $Y \in B_1A_1$ and applying Pascal to $CPBC'AB'$ gives $Z \in C_1B_1$.

Since $B_1C_1 || BC$, we get $\Delta XC_1Z \sim \Delta XA_1C$, we get $\frac{XC}{XZ} = \frac{XA_1}{XC_1}$. Similarly, since $B_1A_1 || BA$, we get $\Delta XA_1Y \equiv \Delta XC_1A$, hence $\frac{XA}{XY} = \frac{XC_1}{XA_1}$. Multiplying these two, we get $\frac{XC \times XA}{XZ \times XY} = 1 (\star)$. Also let $P_A$ and $P_Y$ be the perpendicular from $A$ and $Y$ to $\bar{CZ}$. Clearly $\Delta AP_AX \sim \Delta YP_YX$, hence $\frac{AP_A}{YP_Y} = \frac{XA}{XY} (\blacklozenge)$.

So denoting $[MNP]$ as the absolute (not directed) value of the area of the triangle $MNP$, we see $$\frac{[AXC]}{[XYZ]} = \frac{XC \times AP_A}{XZ \times YP_Y} \overset{\blacklozenge}{=} \frac{XC \times XA}{XZ \times XY} \overset{\star}{=} 1$$. Hence, $[XYZ] = [AXC] = [AB_1C] = \frac{[ABC]}{2}$, a constant w.r.t $P$ as desired $\blacksquare$[/hide]

Define $X=BB'\cap CC'$, and similarly define $Y,Z$. Then, by Pascal on $PA'ABCC'$, we get that $Y$ lies on $A_1C_1$. Similarly, $X$ lies on $B_1C_1$ and $Z$ lies on $X_1Y_1$. Now, note that $\frac{XY}{XC}=\frac{XC_1}{a/2-XC_1}$, $\frac{XZ}{XB}=\frac{XB_1}{a/2-XB_1}=\frac{a/2-XC_1}{XC_1}$. So, $$\frac{[XYZ]}{[XBC]}=\frac{XY}{XC}\cdot\frac{XZ}{XB}\cdot\frac{\sin YXZ}{\sin BXC}=1\implies [XYZ]=[XBC]=\frac{1}{2}[ABC]$$ which is a constant, as desired.

In what follows, \([\bullet]\) denotes area. Let the triangle formed by the three lines be \(XYZ\). I claim \([XYZ]=\frac12[ABC]\), which will suffice. In what follows, the two claims allow the lemma to solve the problem.

[asy]     size(5cm); defaultpen(fontsize(10pt));     pair O,A,B,C,A1,B1,C1,P,Ap,Bp,Cp,X;     O=(0,0);     A=dir(110);     B=dir(210);     C=dir(330);     P=dir(50);     A1=(B+C)/2;     B1=(C+A)/2;     C1=(A+B)/2;     Ap=2*foot(O,P,A1)-P;     Bp=2*foot(O,P,B1)-P;     Cp=2*foot(O,P,C1)-P;     X=extension(B,Bp,C,Cp);

draw(P--Bp,gray);     draw(P--Cp,gray);     draw(X--B1,linewidth(0.4));     draw(Bp--X--C,dashed);     draw(circle(O,1));     draw(A--B--C--cycle);

dot("\(A\)",A,A);    dot("\(B\)",B,SW);     dot("\(C\)",C,C);     dot("\(B_1\)",B1,E);     dot("\(C_1\)",C1,NW);     dot("\(P\)",P,P);     dot("\(B'\)",Bp,Bp);     dot("\(C'\)",Cp,SW);     dot("\(X\)",X,W); [/asy]

[color=red][b]Claim:[/b][/color]      \(X\) lies on \(\overline{B_1C_1}\), and similarly \(Y\in\overline{C_1A_1}\), \(Z\in\overline{A_1B_1}\).

[i]Proof.[/i]      Apply Pascal theorem to \(BB'PC'CA\). \(\blacksquare\)

[color=red][b]Claim:[/b][/color]      \(\overline{AX}\parallel\overline{BY}\parallel\overline{CZ}\).

[i]Proof.[/i]      Apply Pappus theorem to \(BACZA_1Y\) to show \(\overline{BY}\parallel\overline{CZ}\). \(\blacksquare\)

[color=red][b]Lemma:[/b][/color]  [Fixed area on medial triangle]     Let \(ABC\) be a triangle with medial triangle \(A_1B_1C_1\). If \(X\), \(Y\), \(Z\) lie on \(\overline{B_1C_1}\), \(\overline{C_1A_1}\), \(\overline{A_1B_1}\) with \(\overline{AX}\parallel\overline{BY}\parallel\overline{CZ}\), then \([XYZ]=\frac12[ABC]\).

[i]Proof.[/i]      Using directed areas, we have \[ [XYZ]=[XYA]+[XAZ]=[XBA]+[XAC]=\tfrac12[ABC],\]     as needed. \(\blacksquare\)

[color=red][b]Remark:[/b][/color]      In general, for any triangle \(ABC\) with medial triangle \(A_1B_1C_1\) and points \(X\), \(Y\), \(Z\) on \(\overline{B_1C_1}\), \(\overline{C_1A_1}\), \(\overline{A_1B_1}\), the following two conditions are equivalent:     [list]         [*]\(\overline{AX}\parallel\overline{BY}\parallel\overline{CZ}\);         [*]\(A\in\overline{YZ}\), \(B\in\overline{ZX}\), \(C\in\overline{XY}\).     [/list]     After noting the first two claims, the problem statement is equivalent to the lemma.

[asy]
unitsize(1.25inch);
pair A, B, C, P, D, E, F, G, H, I, X, Y, Z;
A = dir(110); B = dir(200); C = dir(340);
draw(A--B--C--cycle); P = dir(240);
draw(circumcircle(A, B, C), dotted);
D = .5B+.5C; E = .5C+.5A; F = .5A+.5B;
G = 2*foot(0, P, D) - P;
H = 2*foot(0, P, E) - P;
I = 2*foot(0, P, F) - P;
Z = extension(A, G, B, H);
X = extension(B, H, C, I);
Y = extension(C, I, A, G);
draw(A--X, lightblue); draw(B--Y, lightblue); draw(C--Z, lightblue);
draw(D--E--F--cycle, dashed); draw(F--Y, dashed); draw(E--Z, dashed);
draw(Y--G, lightred); draw(B--H, lightred); draw(C--Y, lightred);
draw(P--G, dotted); draw(P--H, dotted); draw(P--I, dotted);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$P$", P, dir(P)); dot("$A'$", G, dir(G)); dot("$B'$", H, dir(H)); dot("$C'$", I, dir(90)); dot("$A_1$", D, dir(D)); dot("$B_1$", E, dir(0)); dot("$C_1$", F, dir(180)); dot("$X$", X, dir(265)); dot("$Y$", Y, dir(Y)); dot("$Z$", Z, dir(90));
[/asy]
Let $BB'\cap CC' = X$, $CC'\cap AA' = Y$, and $AA'\cap BB' = Z$. By Pascal's on $PB'BACC'$, $B_1, X, C_1$ are collinear. Similarly, $Y$ lies on $C_1A_1$ and $Z$ lies on $A_1B_1$. By similar triangles from parallel lines, $\frac{XC}{XY} = \frac{XB_1}{XC_1} = \frac{XZ}{XB}$, so $XC\cdot XB = XY\cdot XZ$ and $[XYZ] = [XBC] = \frac{1}{2}[ABC]$.

[asy]
size(10cm);

pair A=dir(110); pair B=dir(210); pair C=dir(330); pair P=dir(50); pair A_1=(B+C)/2; pair B_1=(A+C)/2; pair C_1=(A+B)/2; pair D=dir(256.1); pair E=dir(240.3); pair F=dir(177.3); pair X=extension(C,F,B,E); pair Y=extension(A,D,C,F); pair Z=extension(A,D,B,E);

dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$P$",P,dir(P)); dot("$A_1$",A_1,SE); dot("$B_1$",B_1,N*3); dot("$C_1$",C_1,NW); dot("$A'$",D,NE); dot("$B'$",E,dir(E)); dot("$C'$",F,NW); dot("$X$",X,dir(X)); dot("$Y$",Y,NE); dot("$Z$",Z,dir(Z));

draw(A--B--C--cycle); draw(unitcircle); draw(D--P);draw(E--P); draw(F--P); draw(X--Y--Z--cycle,blue+linewidth(3)); draw(X--B_1--Z,gray+dotted); draw(A_1--C_1,gray+dotted); draw(A--X,purple+linewidth(2)); draw(B--Y,purple+linewidth(2)); draw(C--Z,purple+linewidth(2));

[/asy]
Three steps - First, Pascal on $PC'CABB'$ gives $B_1,C_1, X$ collinear (similarly, $Y\in C_1A_1$ and $Z\in A_1B_1$). Then, Pappus on $BACZA_1Y$ shows $\infty_{AB},\infty_{AC}$, and $BY\cap CZ$ collinear (similar applications show that $AX||BY||CZ$). Last we have \[[XYZ]=[XYA]+[XAZ]=[XYA]+[XAC]=[YAC]=\frac12[ABC]. \]

Define the intersection of $BC$ and $B'C'$ to be $D$. Let $E,F$ be defined similarly

By Pappus on $\{F,A,E\}$ and $\{B,A_1,C\}$, we obtain $BE \parallel CF \parallel AD$. By Pappus on $\{E,A,F\}$ and $\{B,A_1,C\}$, we obtain $C_1,B_1,D$ are collinear. Hence we have  :

$$ [DEF]= [DEA]+[DAF]=[DBA]+[DAC]= 2[DC_1A]+2[DAB_1]= 2[AB_1C_1] = \underbrace {\tfrac {[ABC]}{2}}_{const}$$

Let $X=BB'\cap CC',Y=AA'\cap CC',Z=AA'\cap BB'.$

By the Pascal theorem for $ABB'PC'C$ we have $X\in B_1C_1,$ and analogously $Y\in A_1C_1,Z\in A_1B_1.$
Next triangles $AB_1X,YA_1B$ are homothetic, since $AY,XB,B_1A_1$ concur at $Z.$
Thus $AX\parallel BY,$ and also $AX\parallel CZ.$ Finally we see (directed areas) $$\text{area}(XYZ)=\text{area}(AXZ)-\text{area}(AXY)=\text{area}(AXZ)-\text{area}(AXB)=$$
$$=\text{area}(ABZ)=\text{area}(ABA_1)=\frac{1}{2}\text{area}(ABC).$$

Let $\overline{AA'},\overline{BB'},$ and $\overline{CC'}$ form triangle $DEF.$ By Pascal on $ABB'PC'C,$ we see $D$ lies on $\overline{B_1C_1}.$ Similarly, $E$ and $F$ lie on $\overline{A_1C_1}$ and $\overline{A_1B_1},$ respectively. By Pappus on $A,E,F$ and $B,A_1,C,$ we see $\overline{BE}\parallel\overline{CF}.$ Similarly, $\overline{AD}\parallel\overline{BE}\parallel\overline{CF}.$ Notice $[DEF]=[CBD]$ and $$[DEF]=[ADE]+[ADF]=[ABD]+[ACD]=[ABC]-[CBD]$$ so $[DEF]=\tfrac{1}{2}[ABC].$ $\square$

Let $AA', BB', CC'$ form a triangle $XYZ$. Using Pascal's Theorem, we have $D$ on $B_1C_1$, $E$ on $A_1C_1$, $F$ on $A_1B_1$. Now, Pappus' Theorem yields $AD \parallel CF \parallel BE$. Use barycentric coordinates. Note that we have $X=(0.5, m, 0.5-m), Y=(n, 0.5, 0.5-n), Z=(q,0.5-q,0.5)$. Thus the ratio area is $$\frac{-2mn+2mq-2nq+n}{2},$$ which comes out as $\frac{1}{2}$ upon applying the parallel condition. $\blacksquare$

Let $P_A$ be the intersection of $BB'$ and $CC'$, and define $P_B$ and $P_C$ similarly.

By Pascal on $PC'CABB'$, $C_1,B_1,P_A$ are collinear. Similarly, $A_1,B_1,P_C$ are collinear and $A_1,C_1,P_B$ are collinear.

Then, by Pappus, $AP_A,BP_B,CP_C$ are pairwise parallel. Then, $$[P_AP_BP_C]=[AP_AP_C]-[AP_AP_B].$$ Since $AP_A\parallel BP_B$, we have $[AP_AP_B]=[AP_AB]$, so $$[AP_AP_C]-[AP_AP_B]=[AP_AP_C]-[AP_AB]=[ABP_C].$$ However, from $A_1B_1$ passes through $P_C$ as shown earlier, and $A_1B_1\parallel AB$, we have $$[ABP_C]=[ABA_1]=\frac{1}{2}[ABC],$$ so we are done.

Let $AA'$ and $BB'$ intersect at $Z$ and similarly the other two points are $X$ and $Y$. By Pascal's on $B'PC'CAB$, $X$ is on $B_1C_1$, and similarly $Y$ is on $A_1C_1$, and $Z$ is on $A_1B_1$. Since $BA$ and $ZA_1$ are parallel, $AC$ and $A_1Y$ are parallel, by Pappus's theorem $BY$ and $ZC$ are parallel. Similarly, $AX$ is parallel to them both. Thus, we have
\begin{align*}
[XYZ] &= [XBY] + [ZBY] \\
&= [ABY] + [CBY] \\
&= [BAYC] \\
&= \frac12 [ABC]
\end{align*}
which is independent of $P$.

Pascal bash FTW
wlog p is on arc AB not containing C, and let D, E, F be the cyclic permutations of intersections of AA’, BB’, and CC’, according by BB’ with CC’ first. Now, on degenerate hexagons we find DYZ is a line, EZX is one, and FXY is another. Since XE//AC and similarly we have by sim triangles FE/FA=FX/FY=FB/FD, and consequently [DEF]=FE*FDsinDEF/2=FA*FBsinAFB/2=[AFB]=[ABC]/2, where the last step is because trigially they have same base but half the height (altitude dropped from C to XY is half to AB by sim triangles), which is constant. $\blacksquare$


Ok, let's have a group G and relation "conjugacy": x is in relation with y, iff there exists z from G so, that z.x.z^-1 = y. I'm looking for a proof, that conjugacy is equivalence....

Not sure it's the right sub-sub-sub-forum.

You just need to write it :

$e \cdot x \cdot e^{-1} = x$
$z \cdot x \cdot z^{-1} = y \Rightarrow z^{-1} \cdot y \cdot z = x$
$z_1 \cdot x \cdot z_1^{-1} = y$ and $z_2 \cdot y \cdot z_2^{-1} = z \Rightarrow z_3 \cdot x \cdot z_3^{-1} = z$ for a $z_3$ i'm sure you can find by your-seft  ;)


[quote="236factorial"]What is the total surface area of a hemisphere with radius 6? Express your answer in terms of $\pi$[/quote]

I don't remember what a hemisphere is. Is it $\frac{1}{4}$ of the entire sphere of just $\frac{1}{2}$ of it? Well, here are the answers for both!

[hide="One Fourth"]
The surface area of $\frac{1}{4}$ of a sphere consists of 3 quarter great circles, and $\frac{1}{4}$ the surface area of the entire sphere. That would be;

$\frac{4*\pi*6^2}{4}+\frac{3*6^2*\pi}{4*4}$

$\frac{4*4*\pi*36+3*36*\pi}{16}$

$\frac{576\pi+108\pi}{16}$

$\frac{\pi(684)}{16}$

$\frac{126\pi}{4}$

$\frac{63\pi}{2}$
[/hide]

[hide="One Half"]
Half of a sphere is composed of a great circle and half the surface area of the entire sphere so it would be:

$\frac{4*\pi*36}{2}+36\pi$

$72\pi+36\pi$

$108\pi$
[/hide]

It's 1/2 a sphere. Your second answer is correct (they never asked for a fraction, as with the first one).


graph theory
combinatorics proposed
combinatorics

$ n>2$ KGB secret agents are selected to do a mission. To increase the security, for any two agents, exactly one of them is spying on the other one. It turned out that for some integer $ 2<k\leq n$, one cannot select $ k$ distinct agents $ A_1,A_2,\cdots,A_k$ with $ A_i$ spying on $ A_{i\plus{}1}$ for $ i\equal{}1,2,\cdots,k\minus{}1$ and $ A_k$ spying on $ A_1$. Prove that it is possible to divide the agents into two non empty groups $ X$ and $ Y$ such that, any agent in $ X$ is spying on any agent in $ Y$.

I take it an agent can be spying on many other agents?

Yes.  Putting this in standard terminology: a tournament is a directed graph with exactly one edge between each pair of vertices.  Show that in every tournament, either there is a (directed) Hamiltonian cycle or there is some way to bipartition the vertices so that all edges between the parts are directed in the same direction.

Nice and classical result in the theory of directed graphs. Poland proposed it (in an ugly algebraic transcription) on the Long List of IMO 1989. We used it in some Romanian contest (Stars of Mathematics 2007 - find it in RMC2007).

[hide="Hint"]Look at a longest cycle.[/hide]


[size=150]$ t\leq2009,t\in N^{*},\forall n\in N^{*},(\sum^{n}_{k\equal{}0}C^{2k\plus{}1}_{2n\plus{}1}t^{k},2009)\equal{}1$
find the all possible of $ n$[/size]


For every positive numbers $a, b, c, x, y, z$  prove the inequality

$\sqrt[3]{a(b+1)yz} +\sqrt[3]{b(c+1)zx} +\sqrt[3]{c(a+1)xy} \leq \sqrt[3]{(a+1) (b+1)(c+1)(x+1)(y+1)(z+1)}$ .

I think holder or hugens ineq could help but i have not solution since now


induction
inequalities unsolved
inequalities

[quote="spanferkel"][quote="stancioiu sorin"]Prove that $a_1^2+3a_2^2+5a_3^2+...+(2n-1)a_n^2\le(a_1+a_2+a_3+...+a_n)^2$, with $a_1\geq a_2\geq a_3\geq ...$[/quote]follows imediately by induction[/quote]
   Thank you for idea.Now I think I can solve it.If not, please send me the demonstration.I'll try! ;)

[quote="spanferkel"][quote="stancioiu sorin"]Prove that $a_1^2+3a_2^2+5a_3^2+...+(2n-1)a_n^2\le(a_1+a_2+a_3+...+a_n)^2$, with $a_1\geq a_2\geq a_3\geq ...$[/quote]follows imediately by induction[/quote]
     Please explain me the demonstration. :) I can't solve it.

[quote="spanferkel"]this is so straightforward, man...
Start with proving
$(2n+1)a^2_{n+1}\le a^2_{n+1}+2a_{n+1}(a_1+...+a_n)$. Do you see how to proceed then? :)[/quote]
   I'm in the seventh grade, but I konow something about induction. :roll: In this case, could you post all the demonstration? :blush:


A man plants twelve rows of beans, each row twenty feet long. The plants in each row are spaced four inches apart. If the first and last plants in each row are four inches from the ends of the row, then the total number of plants is:

a. 732 b. 720 c. 718 d. 708 e. NG

[hide=" click to reveal hidden content"]

d. $ 708$;

$ 20$ feet $ \equal{} 120$ inches.

$ \frac{120}{4} \equal{} 60$

Therefore $ 59$ plants per row.

$ 59\times 12 \equal{} 708$

Therefore $ 708$ plants.

[/hide]

[hide]So, 20 feet is 120 inches.
Then you do 120.4=60 And then you know that there are 59 plants in one row.
59x12=708, so the answer is d[/hide]

Where are you getting these problems? Because you're reposting a lot of them, such as here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314452


The partial fraction decomposition of 1/[n(n+1)^2(n+2)^2] = A/n + 1/4n^2 + B/(n+1) + 1/(n+1)^2 + C/(n+2) + 1/4(n+2)^2 for some constants A, B, and C.
The sums Asum((3^.5-2)^n / n), Bsum((3^.5-2)^n / (n+1)), Csum((3^.5-2)^n / (n+2)) as they can all be evaluated through the ln n power series.  We consider the other summands separately.

sum(3^.5-2)^n [1/4n^2 + 1/(n+1)^2 + 1/4(n+2)^2], n=1..infinity)
=1/4sum(3^.5-2)^n / n^2, n=1..infinity) + (-2-3^.5)sum(3^.5-2)^(n+1) / (n+1)^2, n=1..infinity) + 1/4 (3^.5+2)^2 sum(3^.5-2)^(n+2) / (n+2)^2, n=1..infinity).

However the coefficients of the these sums are 0! (1/4+(-2-3^.5) + (3^.5+2)^2 = 1/4 -2 - 3^.5 + 1/4 (7+4*3^.5) = 0).  Therefore, the sum telescopes.

I leave the knitty gritties to whoever wants to carry through on these details.


function
LaTeX
number theory unsolved
number theory

Let $r_1$, $r_2$, $\ldots$, $r_m$ be a given set of $m$ positive rational numbers such that $\sum_{k=1}^m r_k = 1$.  Define the function $f$ by $f(n)= n-\sum_{k=1}^m \: [r_k n]$ for each positive integer $n$.  Determine the minimum and maximum values of $f(n)$.  Here ${\ [ x ]}$ denotes the greatest integer less than or equal to $x$.

It is obviosly $r_kn-1<[r_kn] \le r_kn.$ Therefore $0\le f(n)<m$. Therefore min=0. If n is integer, then f(n) is integer, and max=m-1.

How can we say that there always exists $n$ sunch that $f(n)=m-1$?

Ohhh my dear and first Math friend Mr.Shobber i think you are studying in Canada  ;)
Thanks for you

Davron

this is not directly related to the problem
just how are you supposed to type in f(n) and the Sigma sign anyway?

 madatmath

If you type this:

$\$$\sum_{n=1}^{m}$\$$

you'll get

$\sum_{n=1}^{m}$

(LaTeX works automatically, just include whatever you want LaTeX'ed between the dollar signs.)

and for $f(n)$ just type $\$$f(n)$\$$

farenhajt
Thanks for the $Tip$
madatmath


The surface area of a rectangular prism is 254. The volume is 252. The space diagonal has length $ \sqrt{146}$. What is the sum of the length, width, and height of the cube?

[quote="program4"]The surface area of a rectangular prism is 254. The volume is 252. The space diagonal has length $ \sqrt {146}$. What is the sum of the length, width, and height of the cube?[/quote]
With "cube", you meant "the rectangular prism"??. If this is what you meant, then the volume=252 is inessential.
We have: $ \sqrt{a^2\plus{}b^2\plus{}c^2}$=$ \sqrt{146}$ and $ ab\plus{}bc\plus{}ac\equal{}127$
From these 2 equations, $ (a\plus{}b\plus{}c)^2\equal{}146\plus{}2(ab\plus{}bc\plus{}ac)\equal{}400$. Therefore, $ a\plus{}b\plus{}c\equal{}20$.
I'm not sure what you meant by using the word "cube" though

[quote="program4"]What is the sum of the length, width, and height of the cube?[/quote]

I think it must be the length,width and height of the PRISM right??

[hide="my solution"]

I'm not sure with my solution since I didn't use the volume of the prism
Let the length$ =x$, width$ =y$, height$ =z$
$ 2(xy+xz+yz)=254$
$ x^2+y^2+z^2=146$

Summing two equations above, we get
$ (x+y+z)^2=400  x+y+z=20$

So the sum of the length,width,height of the prism is $ 20$

But maybe the value of the volume is given for finding the length,width,and the height of the prism

If we wants to find the value of $ x,y,z$, consider that $ x,y,z$ are the roots of
$ p^3-20p^2+127p-252=0$
$ (p-9)(p-4)(p-7)=0$
So the value of $ (x,y,z)=(9,4,7)$ and its permutation

[/hide]

I don't think this belongs in Intermidiate topics.


prove that : for any integer n,there exists k s.t:$ k2^n\plus{}1$ is a prime

[quote="bye"]prove that : for any integer n,there exists k s.t:$ k2^n \plus{} 1$ is a prime[/quote]

It's immediate with Dirichlet theorem.

if you want to proof that for some $ k$ there is infinitly many $ n$ st ; $ k2^{n}\plus{}1$ is prime ....so how you can do it ?

[quote="othman"]if you want to proof that for some $ k$ there is infinitly many $ n$ st ; $ k2^{n} \plus{} 1$ is prime ....so how you can do it ?[/quote]

This is a quite different problem.

Would you mean : "give a $ k$ such that there is infinitly many $ n$ st ; $ k2^{n} \plus{} 1$ is prime"
Or "For any given $ k$, show that there is infinitly many $ n$ st ; $ k2^{n} \plus{} 1$ is prime"

?

I think both of those questions are [url=http://mathworld.wolfram.com/ProthPrime.html]currently open[/url], although I can't find an explicit statement.

To pco : i mean the first one.
but how about this question : for some $ k$ all the numbers $ k2^n\plus{}1$ are composite this  is not an open problem ..have a look at :http://www.prothsearch.net/sierp.html


LaTeX
modular arithmetic
number theory
relatively prime
greatest common divisor
number theory unsolved

[hide="Hint"]
Sum and subtract $ 2^{p\plus{}1/2}$ in the left side and factorize.
[/hide]

Well before starting, I should remark that the identity which i'll propose later is very useful. ( I think it is also known as Sophie-Germain's Identity.). First we start with b=0. In this case we get 4^p + 1 =5 , p=1. Now we can assume that b is a positive integer. Now p is an odd prime. Let's put p=2k+1. Then start: 2^(2p) + 1. 2^(4k+2) + 2^(2k+2) + 1 - 2^(2k+2). (addiction and extraction.) Now (2^(2k+1) + 1)^2 - (2^(k+1))^2 . Now write it as follows: (2^(2k+1) + 2^(k+1) + 1) (2^(2k+1) - 2^(k+1) +1 ). Now if p is an odd prime then (p-1)/2 is at least 1 which is obviously positive. So k is a positive integer. Now we will try to determine the greatest common divisor of these two. Define it as d. d/(2^(2k+1) - 2^(k+1) + 1 and d/(2^(2k+1) + 2^(k+1) + 1. Here d/x means d divides x. Now it divides the sum of these two. d/2^(2k+2) + 2. And d/ 2. (2^(2k+1) - 2^(k+1) + 1) = 2^(2k+2) - 2^(k+2) + 2. So d divides the difference 2^(k+2). But the divisors of 2^(k+2) can be expressed in the form 2^m (m is an nonnegative integer.) But both 2^(2k+1) + 2^(k+1) + 1 and 2^(2k+1) - 2^(k+1) + 1 are odd numbers hence m=0. (i.e d=1). Now we also know that 2^(2k+1) + 2^(k+1) + 1 is greater than 2^(2k+1) - 2^(k+1) + 1. Now start with q=5. (We also know that b is a positive integer.) They are relatively prime, so 2^(2k+1) - 2^(k+1) + 1 should be 1. This is a contradiction. Now if q is different from 5 then exactly one of them should be 5 and the other should be q^b.
i) 2^(2k+1) - 2^(k+1) + 1 =5. Then 2^(2k+1)-2^(k+1) = 4. We know that k is a positive integer so both 2^(2k+1) and 2^(k+1) are divisible by 4. 2^(2k-1) - 2^(k-1) = 1. We know that 2k-1 greater than k-1. And if k-1 is different from zero, LHS should be an even number however the RHS is an odd number. Hence k=1. This gives p=3. q=13 and b=1.
ii) 2^(2k+1) + 2^(k+1) + 1 =5. But this is impossible because 2k+1 is greater than 2. So LHS is greater than 4.  :P

[quote="shoki"]find all (p,q,b) such that p and q are odd prime and b a natural number:
4^p + 1 = 5*(q^b)[/quote]
It's very easy if you note that: $ p \equal{} 2n\plus{}1$ then $ 4^p \plus{} 1 \equal{} (2^{2n\plus{}1} \minus{} 2^{n\plus{}1} \plus{} 1)(2^{2n\plus{}1} \plus{} 2^{n\plus{}1} \plus{} 1)$.
I got this idea because i know the Sophie Germain identity: $ a^4 \plus{} 4b^4 \equal{} (a^2\minus{}2ab\plus{}2b^2)(a^2\plus{}2ab\plus{}2b^2)$
Since $ (2^{2n\plus{}1} \minus{} 2^{n\plus{}1}\plus{}1, 2^{2n\plus{}1} \plus{} 2^{n\plus{}1} \plus{} 1) \equal{} 1$ it get's even easier.
Note that $ 2^{2n\plus{}1} \minus{} 2^{n\plus{}1}\plus{}1 > 2^{2n\plus{}1} \plus{} 2^{n\plus{}1} \plus{} 1$. If $ q \equal{} 5$, then $ 2^{2n\plus{}1} \minus{} 2^{n\plus{}1} \plus{} 1 \equal{} 1$, but it doesn't have any solutions, so $ q \neq 5$. If $ q^b > 5$ then $ 2^{2n\plus{}1} \minus{} 2^{n\plus{}1} \plus{} 1 \equal{} 5 \iff (2^n \minus{} 2)(2^n \plus{} 1) \equal{} 0 \iff n\equal{}1$. Hence $ p\equal{}3$. And $ 4^3 \plus{} 1 \equal{} 65 \equal{} 5 \cdot 13^1$, so $ (p,q,b) \equal{} (3,13,1)$ is a solution. Consider the case $ q^b < 5$. Since $ q \ge 3, b \ge 1$ we see that $ q \equal{} 3, b \equal{} 1$. So $ 4^p \plus{} 1 \equal{} 15 \iff 4^p \equal{} 14$, which doesn'thave any solutions.
So the only solution is $ (p,q,b) \equal{} (3,13,1)$.

nice problem shoki..
now find all prime p such that equations $ p \plus{} 1 \equal{} 2{x}^{2}$ and $ p^{2} \plus{} 1 \equal{} 2{y}^{2}$ have answer in integers.

$ p\plus{}1\equal{}2x^{2},p^{2}\plus{}1\equal{}2y^{2}$
$ x<y<p$ $ \implies$
$ x^{2}\equiv y^{2}\pmod{p}$ $ \implies$
$ i)x\equiv y\pmod{p}$  but $ x<y<p$ contradiction!
$ ii)x\equiv \minus{}y\pmod{p}$   $ p|x\plus{}y$ $ \implies$
$ x\plus{}y\equal{}p$ $ \implies$
$ y\equal{}p\minus{}x$ $ \implies$  $ p\plus{}1\equal{}2x^{2},p^{2}\plus{}1\equal{}2.(p^{2}\minus{}2px\plus{}x^{2})$
we get $ p\equal{}7,x\equal{}2,y\equal{}5  (7,2,5)$ :wink:

this is the generalization of that problem: find all non-negative integers m such that the number 4^(2m+1) + 1 has at most two distinct prime divisors(balticway02). it's easy to prove that 2m+1 (if m isn't 0) is prime if 5 doesn't divide it.and if 5 divides it,it's easy to show that the only solution is m=2.


integration
inequalities
trigonometry
function
real analysis
real analysis unsolved

[color=darkblue]let $ f$ be a continuous  $ \mathbb d{C^1}({[0,1],\mathbb{R})$   such that $ f(0)= f(1) = 0$.

Prove that: $ \int_{0}^{1}(f')^{2}\le \pi^2\int_{0}^{1}f^2$[/color]

Maybe we could use Cauchy Schwarz inequality.

Inequality is false. Example: if $ n\in\mathbb{Z}$ and $ |n|\geq 2$ then $ f(x)\equal{}\sin (n\pi x).$ This gives:

$ \int_{0}^{1}f'(x)^2 dx \equal{} n^2\pi^2\int_{0}^{1}\cos^2 (n\pi x) dx \equal{} \frac{n^2\pi^2}{2}.$

$ \pi^2\int_{0}^{1}f(x)^2 dx \equal{} \pi^2\int_{0}^{1}\sin^2 (n\pi x) dx \equal{} \frac{\pi^2}{2}.$

[color=red]oh I'm so sorry,I correct the statement :blush: [/color]

[color=darkblue]let $ f$ be a continuous $ \mathbb d{C^1}({[0,1],\mathbb{R})$ function such that $ f(0)=f(1)=0$.
prove that: $ \int_{0}^{1}(f')^{2}\ge \pi^2\int_{0}^{1}f^2$[/color]
[color=darkblue]what about the case of equality?[/color]

Keywords : [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=103717]Wirtinger's inequality.[/url]


[quote="mathemagician1729"]Two different circles are each tangent to both the x -axis and the
y -axis, as shown in the figure. One of the points of intersection of
the circles is (1, 5). What is the product of the lengths of their
radii?[/quote]

I think you are missing the diagram.

[quote="math154"]Err. Come on. What was like really close to this when you posted?[/quote]

What are you talking about... ?

As I [b]clearly[/b] stated in the other thread, [b]with justification[/b]...

Sorry guys I didn't know it was required to solve

No, it's [b]not[/b] required to solve it, as shown in the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1414928#1414928]other thread[/url].


Consider the system

$ \left\{\begin{array}{ll}x^2 \plus{} (y \minus{} 1)^2 \equal{} a^2 \\
y \equal{} |x| \end{array}\right.$,

where $ a$ is a positive real number.

Find the number of different solutions of this system in function of $ a$.

[hide="Answer"]

zero solutions for $ 0<a<\dfrac{\sqrt2}{2}$
two solutions for $ a\equal{}\dfrac{\sqrt2}{2}$ or $ a>1$
three solutions for $ a\equal{}1$
four solutions for $ \dfrac{\sqrt2}{2}<a<1$[/hide]


[quote="xp"]Εστω $ R$ σταθερο εωτερικο σημειο ενος τριγωνου $ ABC$ .Ευθεια $ (c)$ περιστρεφεται περι του $ R$ και οταν χωριζει το προηγουμενο τριγωνο σε τετραπλευρο και τριγωνο να βρεθει η μεγιστη τιμη του εμβαδου του τετραπλευρου.[/quote]

Ισοδύμναμα, θέλουμε τη μικρότερη δυνατή τιμή του τριγώνου. Το πρόβλημα είναι παλιό.

Προεκτείνουμε την AR και πάνω της παίρνουμε RD = AR. Από το D φέρνουμε παράλληλες στις AB, AC. Σχηματίζεται έτσι ένα παραλληλόγραμμο με πλευρές
ΑΕ (όπου Ε πάνω στην ΑΒ), ΕD, DF, FA.
H διαγώνιος ΕF είναι η ζητούμενη.
Η απόδειξη είναι εύκολη: Αν φέρουμε τυχαία διατέμνουσα από το R, σχηματίζονται δύο τριγωνάκια δεξιά και αριστερά του R, που είναι εύκολο να τα συγκρίνουμε.

Στην βιβλιογραφία η EF ονομάζεται "γραμμή του Φίλωνος", αλλά υπάρχουν και άλλες γραμμές με το ίδιο όνομα.

Τάκης.

Tακη η κατασκευη που αναφερεις ειναι οντως γνωστη και ωραια αλλα με μια μικρη συμπληρωση ελεγχουμε πιο απ τα $ AR,BR,CR$ ειναι το $ min$ για να προκυψει ...το ζητουμενο $ max$
Η δικη μου λυση καταληγει μετα απο λιγη γεωμετρια στο $ E\equal{}(ABC)\minus{}(E_1/2)[2\plus{}(x/a)\plus{}(a/x)] \leq{(ABC)\minus{}2E_1}$ οπου $ E_1$  ειναι το πιο μικρο εμβαδο απ τα εμβαδα των σχηματιζόμενων  παρ/μμων που κατασκευαζονται αν απ' το R φερουμε τις παράλληλες στις πλευρες του τριγωνου....

Χρηστος

[quote="xp"]με μια μικρη συμπληρωση ελεγχουμε πιο απ τα $ AR,BR,CR$ ειναι το $ min$ για να προκυψει ...το ζητουμενο $ max$
[/quote]

Σωστά. Είχα ξεχάσει να ελέγξω τις άλλες δύο κορυφές!
Ενδιαφέρον ο τύπος που γράφεις για το Εμβαδόν.

Τ.


algebra
polynomial
real analysis
real analysis solved

hello,

find a polynomial P(x,y) so that P is
a surjection from R^2 onto R*+ .

I must have been blinded for 10 minutes, since the example is really simple. Just take $ f(x,y)=y^2+(xy+1)^2 $. It verifies all conditions.


How many integers N less than 1000 can be written as the sum of j consecutive positive odd integers from exactly 5 values of  j>=1

i got 191, but there wasn't a solution

[quote="t0rajir0u"]I solved this awhile ago, which is why the solution is less than spectacular.  :|

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=465003#p465003[/quote]
... :huh: wow! :o


There are n boxes, each having a numbered lock. Now each box has a numbered key, ie box i has key j. Now u have to open all the boxes. Initially u hv no keys, so u hv to break open one box. Now, once u open d box, u get they key inside it, nd then go on opening boxes as long as possible. Then u break open another box. In how many ways can u open all d n boxes without breaking more than k locks.

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=243318[/url]


[quote="Mark1"]Hello.  I have a set of three problems that I am using to place into a PowerPoint presentation that I use to compare and associate to investment models.

#1: A regular octagon is inscribed in a circle of radius 15.8 cm.  Find the perimeter of the octagon.

Please show all work in detail.

Thank you.[/quote]

[hide]I drew two radii from two vertexes in the octagon to the center of the circle to form a triangle of angles 135/2, 135/2, and 45. So $\frac{\sin 135}{15.8}=\frac{\sin 45}{x}$, and $x\approx 12.09279646$ and that multiplied by 8 is the perimeter, $\boxed{96.7423717}$.[/hide]

[quote="DanK"]I got the same answer, but for some reason I decided to do it using right triangles and the half-angle trig identity...  ;)[/quote]

Dan, I have to say that, that is beautiful work, indeed.

Thank you  :)


Ring Theory
induction
superior algebra
superior algebra unsolved

Let $R$ be a commutative ring and $P_{1}, P_{2}, ..., P_{k}$ be distinct prime ideals in $R$. Suppose that $I$ is a subring of $R$ such that $I\subset \bigcup_{i=1}^{n}P_{i}$. Prove that $\exists j, \; 1\leq j\leq n,\; I\subset P_{j}$
PS The case $I$ considered as an ideal is quite well known.

[quote="-oo-"]but the usual proof can be copied (see for example Atiyah,McDonald: Commutative Algebra, Prop 1.11)[/quote]
Thank you for the reference you mentioned. I haven't seen this proof in my text ( but not problem) books. Any idea for the second part ? (I mean Extensions of a famous problem part ||)

It's sufficient to use elements ${x}_{i}$ which for every $i$ ,${x}_{i}$ be on the ${p}_{i}-{p}_{1}$ in the proof of that famous theorem.moreover in this case one can use less than usual usage of some conditions (like : the induction is needed one time, oppsite to the way of proving the theorem.)
But, in any manner,while
{
    going this way
    all the conditions is necessary
    as you know;
}

[mod.: DO NOT COMPLETELY WRITE IN CAPS!]


see the fig
4 children a,b,c,d r standing on stairs(3) and 1 behind

a man has 4 caps-2 black and 2 white
he puts them randomly on the 4
which one (of abcd) wud be able to tell which color cap is on his head?

a can see b & c
b can see c
c and d can't see ne1
 :rotfl:

B

[hide="Strategy"]If A yells out "Yes I can tell"(children can't keep silent :P ), then he is confident bcos B,C has both caps of same color. Hence B understands he is with the same color of C.
If A remains silent, B would realize he and C are wearing different caps.[/hide]

Nobody can see D, so he keeps his cap in his hands to see its color  :rotfl:


Let be $ a,b,c\in \mathbb{N}$ odd numbers . Prove that the number $ N\equal{}(a\plus{}ab\plus{}b)(b\plus{}bc\plus{}c)(c\plus{}ca\plus{}a)$ can't be a perfect square .

[hide]
Note that $ 2|(a\plus{}1)$ and $ 2|(b\plus{}1)$

Thus, $ 4|(a\plus{}1)(b\plus{}1) \equal{} a\plus{}ab\plus{}b\plus{}1$

Therefore, $ a \plus{} ab \plus{} b \equiv \minus{} 1 \pmod{4}$

Similarly, $ b \plus{} bc \plus{} c \equiv c \plus{} ca \plus{} a \equiv \minus{} 1 \pmod{4}$

Thus, $ N \equiv ( \minus{} 1)^3 \equiv \minus{} 1 \pmod{4}$, and thus, can not be a perfect square.
[/hide]


For a N-player symmetric game with S possible strategies, how many distinct strategy profiles are there?

There's some ambiguity for me regarding your definitions for symmetric and distinct.

If we care only about how many players use any particular strategy, then there are $ \binom{N\plus{}S\minus{}1}{N}$ different strategy profiles, by the balls-and-urns method.

If we distinguish strategy profiles based on what each player plays, then there are $ N^S$ different strategy profiles, since each player can choose from one of $ S$ strategies.

That is pretty much what I got, also involving the same logic. However, I wasn't sure of the distinction between symmetric and distinct in this context.

If you were to solve for NE for such a large set of strategy profiles, how would you go about doing it (keeping in mind there'll be a really large number of symmetric profiles)?  I had come across this problem a long time ago, with actual payoffs for each strategy profile given. I don't remember the source now, and so cannot post the file with the payoffs.

The only many-playered case I've studied is the simple one in which the payoff functions are linear and symmetric, ie $ u_i(s_1,s_2,\ldots,s_N) \equal{} a(s_1 \plus{} s_2 \plus{}\cdots \plus{} s_N) \minus{} bs_i$, for constants $ a$ and $ b$. (I hope that my thinking of the players as competing firms seeking maximal profits doesn't diverge from the original or general model too much.) Maximizing this is quite easy because of the linearity, and extrapolating one player's best strategy to form a Nash equilibrium is easy because of the symmetry. I think the resulting Nash equilibrium has everyone doing the same thing.

For higher-ordered payoffs, I'd assume there would just be more Nash equilibria, in addition to the obvious one where everyone does the same thing.

For less symmetric payoffs, there would be more work in doing the maximizations.

The point is that one would not solve this by checking every strategy profile. It's easier to write and solve several equations that a strategy profile must satisfy in order to be a Nash equilibrium.


geometry
3D geometry
sphere
integration
calculus
calculus computations

Consider a uniform distribution of mass on the sphere of radius $ r$ about the origin.  Show that

a)  Inside the sphere; the potential is constant and the gravitational field vanishes.

b)  Outside the sphere; the potential and field are the same as if the entire mass were located at the origin.

Use gauss's law, except for gravity.
[hide]
it's spherically symmetric. let R be distance from you to the origin.
we construct an imaginary sphere of radius R, center origin.
if R<r, then there is no mass inside the sphere and therefore no net field outwards.
if R>r, then there is the mass of the sphere inside. But this mass does not change as r<R<\infty, so we can replace it with a point mass.
so the fields are proven. Just integrate the field or whatever to get the potential.
\int{0}=constant, and the field == point => the potential == point.
[/hide]


Most of the time, multiplicity is used to tell how many times a Polynomial has a given root.  It is also used in prime factorizations to tell how many times a prime occurs.

Basically, multiplicity means how many times something occurs.


How do you put in formulas, so you don't need to say n^2, but just have n superscript 2? etc...
Thanks

Read all about it [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]here.[/url]

No, [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:About]here[/url], since that other page isn't updated or anything. :P

But my page had a link to you page, so mine is double more helpful.  (Why are we arguing about better links??)

[quote="AIME15"]Or, you could just type

[code]
$n^2$
[/code]

which shows up as

$ n^2$

:P[/quote]
That too. :D

Basically, enclose your math stuff in dollar signs. For some operations, you use the not-so-obvious notations inside them. and lots of people (including me) thought at first, while reading through the tutorial, that you had to download stuff to post $ \text{\LaTeX}$ in the forums. this isn't true; as you can see, the reply messages have it built into them.


I read that only 200 are being allowed in the competition. How many of those 200 spots are still available? Also, can we participate as just individuals if we do not have a team?

About 150 people have registered, so we can fit around 5 more full teams. Note that this is just a rough estimate (I don't have the list) and it is also as of October 6. Remember that if you register after the cap is reached, you will be placed on the waiting list and you still have a chance of participating. Please don't hesitate to register! Also, feel free to email me, pm me, or post here if you have any questions.

If you don't have a full team, you can still register. We'd prefer that anyone who can't field a full team register as a group of 5, but if you can't get that many (or want between 5 and 10), we can still accommodate that. You will be able to indicate on your registration form whether or not you would like to be combined with other partial teams.

Hey I sent you a PM regarding a problem. I'm freaking out! Can you respond soon please. Odd as it sounds, I'm going to be calling my teacher in a few moments.

We'll be announcing this on our website soon, but we've already filled up all 200 seats plus we have 3 teams on the waitlist right now. We still encourage teams to register in case other teams drop out or if we expand the cap.

We'll notify the teams that are definitely "in" (that is, either because they registered in time or if they got in via the waitlist) for the contest before Thanksgiving.


Solve for $ x \in [0,2\pi]$ :

$ \sin x < \cos x < \tan x < \cot x$

[hide="Progress"]
Argh, messed up again.

Too late... worthawholebean solved it already. :|
[/hide]

[hide="Solution outline"]
Eliminate intervals by contradiction until you're down to $ \left(0,\frac {\pi}{4}\right)$. Over this interval, $ \sin x < \cos x$ and $ \tan x < \cot x$ so all we need to find is when $ \cos$ and $ \tan$ intersect, which they will only do once because $ \tan$ is increasing and $ \cos$ is decreasing. In between this intersection, $ a$ and $ \frac {\pi}{4}$, the inequality will hold. So we solve and find $ a \equal{} \sin^{ \minus{} 1}\frac {\sqrt5 \minus{} 1}{2}$. So the desired interval is $ \left(\sin^{ \minus{} 1}\frac {\sqrt5 \minus{} 1}{2},\frac {\pi}{4}\right)$.

I actually drew out a massive graph for this with all 4 functions...[/hide]


geometry
ratio
trigonometry
geometry unsolved

$ABC$ is an acute triangle with altitude $CD$ and orthocenter $H$. We construct the square $BCMN$ external to the triangle. Let $P = AN \cap CD$. Prove that: $\frac{HP}{PD}=\frac{AB}{CD}$.

I shall use area ratios to prove this problem.

First we notice that $\frac{DP}{HP}=\frac{\triangle{ADN}}{\triangle{AHN}}$. Since $AH || BN$, so $\triangle{AHN}=\triangle{AHB}$, thus $\triangle{AHN}=\frac{1}{2} \cdot DH \cdot AB$.

Construct $NN' \perp AB$ with $N' \in AB$, thus $\triangle{ADN}=\frac{1}{2} \cdot AD \cdot NN'= \frac{1}{2} \cdot AD \cdot NB \sin{\angle{NBN'}}$. Since $\angle{NBN'}+\angle{CBA}=90^o$ and $NB=CB$,  so $\triangle{ADN}= \frac{1}{2} \cdot AD \cdot BC \cos{\angle{ABC}}$. Therefore:
\[ \frac{DP}{HP}=\frac{\triangle{ADN}}{\triangle{AHN}}=\frac{BC \cdot \cos{B} \cdot AD}{DH \cdot AB} \]
Since $BC \cdot \cos{B}=BD$, so plug it in the previous expression and simplify, we get:
\[ \frac{BD \cdot AD}{DH}=BD \cdot \frac{AD}{HD}=BD \cdot \frac{CD}{BD}=CD \]
Thus:
\[ \frac{DP}{HP}=\frac{\triangle{ADN}}{\triangle{AHN}}=\frac{BD \cdot AD}{DH} \cdot \frac{1}{AB}= \frac{CD}{AB} \]

[b]Proof.[/b] Denote $x=m(\widehat {DAP})$ and $y=m(\widehat {HAP})$; $BN\parallel AE\Longrightarrow m(\widehat {BNA})=y\ .$
From the [u]Sinus' theorem[/u] in the triangle $ABN$, i.e. $\frac{AB}{\sin y}=\frac{BN}{\sin x}$ results $\frac{\sin y}{\sin x}=\frac ca\ .$
Therefore, $\frac{PH}{PD}=\frac{AH}{AD}\cdot\frac{\sin y}{\sin x}=\frac{1}{\sin B}\cdot \frac ca\Longrightarrow\boxed {\frac{PH}{PD}=\frac{AB}{CD}}\ .$
If denote $Q\in AM\cap CD$, then similarly prove that $\frac{QH}{QC}=\cot A$.

[b]Remarks.[/b]
If $C=45^{\circ}$, then $\boxed {\frac{PH}{PD}=\frac{CH}{CD}}$, i.e. the points $P$ and  $C$ are conjugate harmonically w.r.t. the points $D$ and $H\ !$ Indeed, $\frac{CH}{CD}=$ $\frac{AH}{AD}\cdot \frac{\cos C}{\sin A}=$ $\frac{1}{\sin B}\cdot \frac{\cos C}{\sin A}=$ $\frac{c\cdot \cos C}{c\cdot \sin B\cdot \sin A}=$ $\frac{c\cdot \cos C}{a\cdot \sin C\cdot \sin B}=$ $\frac{c}{a\cdot \sin B}=$ $\frac{PH}{PD}\ .$

If $A=45^{\circ}$, then $\boxed {QH=QC}\ .$

[quote="shobber"]... Since $AH || BN$, so $\triangle{AHN}=\triangle{AHB}$...
[/quote]

I don't see how the first implies the other and I don't think it's true judging by your picture, but maybe I'm wrong. Could you please explain this part more in detail?

To [b]Virgil[/b]: nice solution, thank you

Since $AH||BN$, so the distance from $B$ to $AH$ is equal to the distance from $N$ to $AH$, which means, the height of $\triangle{ANH}$ and $\triangle{ABH}$ are equal, since they have the same base $AH$, so by $S= \frac{1}{2} \cdot h \cdot AH$, we get that the areas are equal.

Thank you - I misunderstood your notion at first because I thought you mean congruent triangles  :blush:


inequalities
calculus
vector
AMC
AIME
geometry
number theory

I'm studying my olympiad math primarily out of two books, AoPS2 and Problem solving strategies, with numerous supplements for euclidian geometry. My first question is if this knowledge alone, combined with massive problem solving, will be enough to make, say, IMO? I understand that reading these books won't make me an IMOer, and I probably never will be, but if a person knew only what he could glean from these books and a collection of olympiad problems, would he be equiped with all the tools he needed to make IMO? I would guess at the level of national competitor that training is much more practice than it is 'learning', per se, but what bits of knowledge are missing from those two books that one would need at the IMO?

I can only think of some advanced inequalities, but those are summarized in Mildorf's paper. I'd also wager that some more advanced number theory and graph theory knowledge is needed, but that is why I am asking, I'd like to know :blush:

My second question concerns higher mathematics. How important is it for an olympiad mathematician to know anything above very basic calculus and vectors? I have heard that some solutions are expidited by the use of group theory and advanced calculus, but would one be at a serious disadvantage if he or she did not know any advanced mathematics?

Thank you for your time,
PenguinIntegral

1.  In theory, once you have a decent enough Olympiad background with Olympiad Problem books, the most important tool to have is a mathematical mind, rather than any particular tips and tricks.  :)

2.  Engel comments that a knowledge of convexity, graph theory, and game theory are useful for olympiad problems, but again, your mind is your strongest asset.  I doubt any advanced calculus is necessary unless they gave an inequality easy enough that it could be brute-forced by Lagrange Multipliers.  :P

[quote="t0rajir0u"]1.  In theory, once you have a decent enough Olympiad background with Olympiad Problem books, the most important tool to have is a mathematical mind, rather than any particular tips and tricks.  :)
[/quote]
1) What is an "Olympiad Background"? How can I get one?
2) What Problem books do you reccomend?
3) Not all of us are Bartons :(  Some of us need tips and tricks, and I'm asking what knowledge (ie book knowledge, textbook, standard stuff) and tricks (sophie german indentity, demoive, veita, etc) are required/helpful that aren't found in the above books, or aren't common knowledge. I know advanced inequalities and graph theory are some, and I was wondering if anything more than what is taught in AoPS2 for NT is needed for IMO level competition.

[quote="t0rajir0u"]
2.  Engel comments that a knowledge of convexity, graph theory, and game theory are useful for olympiad problems, but again, your mind is your strongest asset.  I doubt any advanced calculus is necessary unless they gave an inequality easy enough that it could be brute-forced by Lagrange Multipliers.  :P[/quote]
I've heard from some german IMOers that group theory really helps. I'm wondering how much it helps...

4) How do I develop my mind? I'm not a very bright/sharp individual....

Thank you for your reply!

[quote="PenguinIntegral"]
1) What is an "Olympiad Background"? How can I get one?
2) What Problem books do you reccomend?
...
I've heard from some german IMOers that group theory really helps. I'm wondering how much it helps...
...
4) How do I develop my mind? I'm not a very bright/sharp individual....[/quote]

1)  Well, by "Olympiad background" I meant "studied Olympiad problems," which, of course, you have done :)
2)  http://www.amazon.com/gp/product/0817643052/104-7072709-9235156?v=glance&n=283155 comes very highly recommended from a friend of mine.  I personally like Problem Solving Strategies - if you can do most of those problems then I think you're in better shape than me because I can't :P
3)  I don't think it would be that helpful directly.  It's a good way to add rigor to a proof, but I don't know if knowing group theory necessarily means you'd be able to solve a problem that you wouldn't be able to solve without it.  Can your German IMO friends give some examples?
4)  Do more problems :P

[quote="t0rajir0u"]
4)  Do more problems :P[/quote]

I'm having trouble with this one. I can understand/solve most AIME level or lower problems just fine, because there is really only one way to approach most of them. If you see logs, you just use log identities and substitute. If you see coin flip problems, look for recursion, etc.

Olympiad problems generally don't fit into any nice catagories like AIME problems do, and are very hard. I have trouble with them, and I don't think just doing more is going to help :(

[quote="PenguinIntegral"]Olympiad problems generally don't fit into any nice catagories like AIME problems do, and are very hard. I have trouble with them, and I don't think just doing more is going to help :([/quote]

Well maybe I'm not the right person who could give advices but I believe that problem solving really helps, regardless for what kind of competition you are preparing. I think about it this way: the more problems you solve, the bigger chance is to see a similar problem during the competition. :wink:

[quote="PenguinIntegral"][quote="t0rajir0u"]
4)  Do more problems :P[/quote]
...
Olympiad problems generally don't fit into any nice catagories like AIME problems do, and are very hard. I have trouble with them, and I don't think just doing more is going to help :([/quote]

Well...there must be something that you remember from each question after you solved them (or even if you gave up and look at the solution).  So something like "when I see vectors I should try to convert it to a geometry problem", or like "if I have bunch of squares in a diophatine equation I should try complete the squares to factor", or something like that.  Those "tricks" (I rather call it insights) that you mention are merely stuff you get used to as you solve a lot of problems.

As for books-wise (personally I think the more books you read the better), I would suggest Art and Craft of Problem Solving, which is more advanced than AoPS II but easier than PSS.  However, he did go over how to think strategically and stuff like that, so I think it's quite good :)

another thing I do is that if I know I'm weak on some area (like counting :o ), I'll get books on that area of math and study it in addition to doing problems.  I think that's all. (But since you have PSS I think you should aim to solve EVERYTHING in that book :P )

Oh yeah for higher math stuff, let me just tell you I'm a stupid junior that's still taking precalc...(this past school year). Technically I don't even know "calculus"

Once you get to a certain point in terms of knowledge, nothing is more useful in getting better than simply doing problems. This point is usually once you have gotten through a good bit of AOPS2 (I dunno how much exactly, I never got that far in it).

Also, aren't you in like, seventh grade? Don't worry so much about not being able to do problems, you have four or five years to become decent.


real analysis
function
calculus
integration
topology
real analysis unsolved

What is the Lebesgue measure of Real line $ \mathbb{R}$ in $ \mathbb{R}^2$ ?
In my esteem, it may be $ 0$.
But I can't find open set $ O$ with $ m(O\minus{}\mathbb{R})\le \epsilon$.

For all integers $ n\geq 0$ and every $ \alpha\geq 0$, let $ C_{n,\alpha}$ be the open box $ ( \minus{} 1, \minus{} 1)\times ( \minus{} \alpha 2^{ \minus{} n},\alpha 2^{ \minus{} n})$

For all integers $ n$, let $ B_{n,\alpha} \equal{} (2n,0) \plus{} C_{|n|,\alpha}$

For all integers $ n\geq 0$, let $ D_{n,\alpha} \equal{} \{ \minus{} 2n \minus{} 1,2n \plus{} 1\}\times ( \minus{} \alpha 2^{ \minus{} n},\alpha 2^{ \minus{} n})$

(draw a picture)

Let $ A_\alpha \equal{} \bigcup^\infty_{i \equal{} \minus{} \infty} B_{i,\alpha} \cup \bigcup^\infty_{j \equal{} 0} D_{j,\alpha}$

(intuitively, $ A_\alpha$ is just a countable union of disjoint open boxes of constant width and exponentially decaying heights)

1)

$ A_\alpha$ is open and contains $ \mathbb{R}\times\{0\}$

2)

$ A$ is the disjoint union of the $ B_{i,\alpha}$'s and $ D_{j,\alpha}$'s

3)

each $ D_{j,\alpha}$ has measure $ 0$

4)

Each $ B_{i,\alpha}$ has measure $ \alpha\cdot 2\cdot 2^{1 \minus{} |i|}$

5)

Using countable additivity, computing $ m(A_\alpha)$ reduces to summing a geometric series.

Alternatively: You can use the countable additivity of Lebesgue measure to say that a countable union of sets of measure zero has measure zero. It's fairly easy to show that a finite line segment has measure zero.

blahblahblah's argument is in fact particularized down from an argument for the countable additivity, or at least for a general argument for the countable union of sets of measure zero.

For a cheap argument, try the set $ \{(x,y): |y|<\delta e^{\minus{}|x|}\}.$ This is open, contains the horizontal axis, and has measure $ 4\delta,$ which can be made arbitrarily small.

Again, this is not markedly different from blahblahblah's argument - it's just that he used a step function rather than a single-formula function like $ e^{\minus{}|x|}$ to the same end. The only thing that matters is the finite integral. If you'd rather use something like $ \frac1{1\plus{}x^2},$ that would work, too.

actually i originally posted the exact same thing you did, but i figured that any sort of argument which required some sort of version of the fundamental theorem of calculus for lebesgue integrals was maybe a little too advanced.

unfortunately writing it out in the way i did kind of obfuscates what is really going on.


group theory
abstract algebra
superior algebra
superior algebra solved

Let H and K be the subgroups of a finite group G.
Prove that $|HaK|=\frac{|H|\cdot |K|}{|H \cap aKa^{-1}|}$ for any a$\in G$.
[hide="what i did"]Let $L=aKa^{-1}$. Its easy to see that L is a subgroup of G and the order of L is equal to the order of K. there is a formula $|HL|=\frac{|H| \cdot | L|}{|H \cap L|}$. So what i have to prove is $|HaK|=|HL|$ :maybe: [/hide]

[quote="rustam"] So what i have to prove is $|HaK|=|HL|$ [/quote]

Well?

Just write out :

$|H L|=| H a K a^{-1 }| =| H a K|$


function
linear algebra
matrix
LaTeX
geometric sequence
algebra
system of equations

Source: GA Varsity Math Tournament 2001

46. Evaluate the determinant of the following matrix:

2 2 0 1
-1 x 2 1
-3 0 y -2
4 1 2 3

(is there a better way than messy expansion by minors? and how do you do matrices with latex?)

48. If $f(x)=x-[x]$, what is the exact value of the only positive real number x for which $f(x)$, [x], and $x$ form a geometric progression?

49. Find the exact values of all ordered pairs of numbers $(x,y)$ which satisfy the following system of equations:

$x^2 - xy + y^2 = 7$
$x - xy + y = -1$


Inequality
algebra
inequalities proposed

I also think so.
$ LHS \geq\ \frac{(a\plus{}b\plus{}c)^2\plus{}4}{a\plus{}b\plus{}c} \geq\ 4$ (Am-Gm)

 :)

Sorry but maybe my ineq is more difficult than this. Solutions for mine?

Yes, I think it's harder.
Using pqr, it's becomes:
$ p^3\minus{}2p^2\plus{}2pr\plus{}r \geq\ 0$
With $ p \geq\ 2$, ineq obviously trues.
With $ p \leq\ 2$, using Schur, it's becomes: $ (p\minus{}1)^2(p\minus{}2) \leq\ 0$ (trues)
Sorry because this proof is not njce  :oops:

[quote=sqing]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab + ac + bc = 1.$ Prove that
$$\frac {1}{b + c} + \frac {1}{c + a} + \frac {1}{a + b} + \frac {3}{a + b + c}\geq  \frac {(a \plus{} b \plus{} c)^2 \plus{} 4}{a \plus{} b \plus{} c}  \geq 4.$$
Nice.[/quote]

Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab + ac + bc = 1.$ Prove that
$$\frac {1}{b + c} + \frac {1}{c + a} + \frac {1}{a + b} + \frac {3}{a + b + c}\geq  \frac {(a \plus{} b \plus{} c)^2 \plus{} 4}{a \plus{} b \plus{} c}+\frac{9}{2}\,{a}^{2}{b}^{2}{c}^{2}\sqrt {3}.$$

[quote=thaithuan_GC]Let positive real numbers $ a,b,c$ satisfying $ ab\plus{}bc\plus{}ca\equal{}1$. Prove that $ \frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a}\plus{}\frac{1}{a\plus{}b}\minus{}\frac{1}{a\plus{}b\plus{}c}\ge 2$[/quote]

$$LHS=\frac{1}{a+b+c}\left (3+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \right)-\frac{1}{a+b+c}$$
$$LHS\ge \frac{1}{a+b+c} \left[3+\frac{(a+b+c)^2}{2(ab+bc+ca)} \right]-\frac{1}{a+b+c}$$
$$LHS\ge \frac{a+b+c}{2}+\frac{2}{a+b+c}\ge 2$$

[quote=arqady]For the following inequality there is a nice proof.
Let $ a,$ $ b$ and $ c$ are non-negative numbers  such that $ ab \plus{} ac \plus{} bc \equal{} 1.$ Prove that
\[ \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \plus{} \frac {1}{a \plus{} b} \plus{} \frac {3}{a \plus{} b \plus{} c}\ge 4
\]
[hide="remark"] The maximal $ k$ for which the inequality
\[ \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \plus{} \frac {1}{a \plus{} b} \plus{} \frac {k}{a \plus{} b \plus{} c}\ge \frac{k\plus{}5}{2}
\]
is true for all non-negative $ a,$ $ b$ and $ c$ such that $ ab\plus{}ac\plus{}bc\equal{}1$ is $ k\equal{}3.$ :wink:
[/hide][/quote]

Let $3u=a+b+c$, $3v^2=ab+bc+ca$ and $w^3=abc$, then
$\frac{9u^2+3v^2}{9uv^2-w^3}+\frac{1}{u}\geq 4$, since it's increasing function of $w^3$, enough to check when $w^3$ reaches minimum.
Case $a=c$
$(1-a) (24 a^4 - 15 a^3 + 17 a^2 - 7 a + 1)\geq 0$, which is obvious
Case $a=0$
$\frac{b+c}{bc}+\frac{4}{b+c}\geq 2\sqrt{\frac{4}{bc}}=4$

[quote=Mathskidd][quote=arqady][quote="nguoivn"]I also think so.
$ LHS \geq\ \frac {(a \plus{} b \plus{} c)^2 \plus{} 4}{a \plus{} b \plus{} c} \geq\ 4$ (Am-Gm)

 :)[/quote]
Yes! :lol:  It's also my proof.[/quote]

Dear Arqady,
how to prove the left inequality?[/quote]

<=> $(a+b+c)^2+4\geq4(a+b+c)$ <=> $(a+b+c-2)^2\geq 0$

[quote=arqady]For the following inequality there is a nice proof.
Let $ a,$ $ b$ and $ c$ are non-negative numbers  such that $ ab \plus{} ac \plus{} bc \equal{} 1.$ Prove that
\[ \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \plus{} \frac {1}{a \plus{} b} \plus{} \frac {3}{a \plus{} b \plus{} c}\ge 4
\]
[hide="remark"] The maximal $ k$ for which the inequality
\[ \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \plus{} \frac {1}{a \plus{} b} \plus{} \frac {k}{a \plus{} b \plus{} c}\ge \frac{k\plus{}5}{2}
\]
is true for all non-negative $ a,$ $ b$ and $ c$ such that $ ab\plus{}ac\plus{}bc\equal{}1$ is $ k\equal{}3.$ :wink:
[/hide][/quote]
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab + ac + bc = 1.$ Prove that
\[ \frac {1}{b^2 + c^2} + \frac {1}{c^2 + a^2} + \frac {1}{a^2 + b^2} + \frac {3}{a^2 + b^2 + c^2}\ge 4
\]

[quote=Mathskidd]<=> $(a+b+c)^2+4\geq4(a+b+c)$ <=> $(a+b+c-2)^2\geq 0$[/quote]

sorry,
it is the right, which I can get.[/quote]

Using a+b+c=p;ab+bc+ca=q;abc=r. The left becomes: $p^3-2p^2+2pr+r\geq0$. With $p\geq2$, inequality obviously true.
With $p\leq2$, using Schur, it's becomes $(p-1)^2(p-2)\leq0$ (true)

[quote=Mathskidd][quote=arqady][quote="nguoivn"]I also think so.
$ LHS \geq\ \frac {(a \plus{} b \plus{} c)^2 \plus{} 4}{a \plus{} b \plus{} c} \geq\ 4$ (Am-Gm)

 :)[/quote]
Yes! :lol:  It's also my proof.[/quote]

Dear Arqady,
how to prove the left inequality?[/quote]
Because $$\sum_{cyc}\frac{1}{b+c}=\sum_{cyc}\frac{ab+ac+bc}{b+c}=a+b+c+\sum_{cyc}\frac{bc}{b+c}\geq a+b+c+\sum_{cyc}\frac{bc}{a+b+c}=a+b+c+\frac{1}{a+b+c}.$$

[hide=Very nice. Thanks.][quote=TheEpicCarrot7]$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{a^2+b^2+c^2+3}{a+b+c-abc} \ge \frac{ab+bc+ca+3}{a+b+c-abc}=\frac{4}{a+b+c-abc}$$
Equality if $a=b=c=\frac{\sqrt{3}}{3}$[/quote]
[/hide]

Let $ a,b,c\ge 0$ and $ ab+bc+ca=1$. Prove that
$$\dfrac{1}{a+b}+ \dfrac{1}{b+c} + \dfrac{1}{c+a}-\dfrac{1}{a+b+c-abc}\ge \frac{9\sqrt 3}{8}$$


trigonometry
complex numbers
geometry unsolved
geometry

The points $ A, B$ and $ C$ in the Argand diagram are the vertices of an equilateral triangle described anticlockwise. Show that the complex numbers $ a, b$ and $ c$ representing $ A, B$ and $ C$ satisfy

$ 2c\equal{}(a\plus{}b) \plus{}i\sqrt{3}(b\minus{}a)$.

Find a similar relation in the case that $ A, B$ and $ C$ are the vertices of an equilateral triangle described clockwise.

(i) The quadrilateral $ DEFG$ lies in the Argand diagram. Show that the points $ P, Q, R$ and $ S$ can be chosen so that $ PDE, QEF, RFG$ and $ SGD$ are equilateral triangles and $ PQRS$ is a parallelogram.

(ii) The triangle $ LMN$ lies in the Argand diagram. Show that the centroids $ U, V$ and $ W$ of the equilateral triangles drawn externally on the sides of $ LMN$ are the vertices of an equilateral triangle.

Let $ O$ be center of equilateral $ \triangle ABC$ and $ \varrho$ its circumradius. Let $ c \equal{} o \plus{} \varrho \exp (i\phi).$ Then $ a \equal{} o \plus{} \varrho\exp i(\phi \plus{} 120^\circ),$ $ b \equal{} o \plus{} \varrho \exp i(\phi \minus{} 120^\circ).$

$ a \plus{} b \equal{} 2o \plus{} 2\varrho \exp (i \phi) \cos 120^\circ \equal{} 2o \minus{} (c \minus{} o) \equal{} 3o \minus{} c,$
$ b \minus{} a \equal{} \minus{} 2i\varrho \exp (i \phi) \sin 120^\circ \equal{} \minus{} i \sqrt 3 (c \minus{} o)$
$ (a \plus{} b) \plus{} i \sqrt 3 (b \minus{} a) \equal{} 3o \minus{} c \plus{} 3(c \minus{} o) \equal{} 2c.$

$ 2o \equal{} \frac {_2}{^3}(a \plus{} b \plus{} c) \equal{} (a \plus{} b) \plus{} i \frac {_{\sqrt 3}}{^3} (b \minus{} a)$

If equilateral  $ \triangle ABC$ is labeled clockwise, $ a, b$ get exchanged and $ (a \plus{} b) \plus{} i\sqrt 3 (a \minus{} b) \equal{} 2c.$

[b](i)[/b] Assume $ DEFG$ is labeled counter-clockwise.
Let $ \triangle PDE, \triangle RFG$ be drawn inward of $ DEFG$ $ \Longrightarrow$ labeled counter-clockwise.
Let $ \triangle QEF, \triangle SGD$ be drawn outward of $ DEFG$ $ \Longrightarrow$ labeled clockwise.

$ 2p \equal{} (d \plus{} e) \plus{} i \sqrt 3 (e \minus{} d)$
$ 2r \equal{} (f \plus{} g) \plus{} i \sqrt 3 (g \minus{} f)$

$ 2q \equal{} (e \plus{} f) \plus{} i \sqrt 3 (e \minus{} f)$
$ 2s \equal{} (g \plus{} d) \plus{} i \sqrt 3 (g \minus{} d)$

$ 2\overrightarrow{PQ} \equal{} 2(q \minus{} p) \equal{} (f \minus{} d) \plus{} i \sqrt 3 (d \minus{} f) \equal{} 2(r \minus{} s) \equal{} 2\overrightarrow{SR}$

$ \Longrightarrow$ $ PQRS$ is parallologram.

[b](ii)[/b] Assume $ \triangle LMN$ is labeled counter-clockwise $ \Longrightarrow$ equilateral triangles outward on the sides of $ \triangle LMN$ are labeled clockwise. Their centers are

$ 2u \equal{} (l \plus{} m) \plus{} i \frac {_{\sqrt 3}}{^3} (l \minus{} m)$
$ 2v \equal{} (m \plus{} n) \plus{} i \frac {_{\sqrt 3}}{^3} (m \minus{} n)$
$ 2w \equal{} (n \plus{} l) \plus{} i \frac {_{\sqrt 3}}{^3} (n \minus{} l)$

Triangle $ \triangle UVW$ is labeled counter-clockwise.

$ 2(u \plus{} v) \plus{} 2i \sqrt 3 (v \minus{} u) \equal{}$
$ \equal{} \left[(2m \plus{} l \plus{} n) \plus{} i \frac {_{\sqrt 3}}{^3} (l \minus{} n)\right] \plus{} i \sqrt 3 \left[(n \minus{} l) \plus{} i \frac {_{\sqrt 3}}{^3} (2m \minus{} l \minus{} n)\right] \equal{}$
$ \equal{} 2(n \plus{} l) \plus{} 2i \frac {_{\sqrt 3}}{^3}(n \minus{} l) \equal{} 4w$

$ \Longrightarrow$ $ \triangle UVW$ is equilateral.


geometry
circumcircle
incenter
geometry proposed

(1) Given a triangle $ ABC$. $ D$ is point on arc $ AC$ (do not contain $ B$) $ \omega$ is circle such that it [color=red]internally[/color] touches circumcircle of $ ABC$  and touches $ AC, BD$ at $ P,Q$ such that $ \omega$ and point $ B$ are exist in one side of $ AC$. Prove that [color=blue]Incenter[/color] of $ ABC$ , $ P, Q$ are collinear.

(2) Given a triangle $ ABC$. $ D$ is point on arc $ AC$ (do not contain $ B$) $ \omega$ is circle such that it [color=red]externally[/color] touches circumcircle of $ ABC$  and touches $ AC, BD$ at $ P,Q$ such that $ \omega$ and point $ B$ are exist in two side of $ AC$. Prove that [color=blue]Excenter [/color] ,which respect to $ B$,of $ ABC$ , $ P, Q$ are collinear.

I think (1) is well-known theorem... see [url]http://forumgeom.fau.edu/FG2003volume3/FG200325index.html[/url]
even this problem is in imo compendium!!

I can't understand for what point $ D$ is given?
For a proof see here:
http://www.mathlinks.ro/viewtopic.php?t=167283
and also here:
http://www.mathlinks.ro/Forum/viewtopic.php?p=367377#p367377


Παιδιά μπορείτε να υπολογίσετε το εξής ολοκλήρωμα?

$ {\int{\sqrt{1+9x^4}dx}}$

Νομίζω ότι καταλήγει σε ελλειπτικό ολοκλήρωμα (σύμφωνα με το Mathematica)


HELLO. this is my first blog
but i need help reallllll bad
hehheheh. okay:

Ricardo is wrapping a gift. He wants to cover an 8in x 10 in x5 in box with wrapping paper. How many square inches of paper will he need if the paper fits perfectly over the box?
If you can.. please explain  :)
Thank You!!

[hide]To find how much wrapping paper he needs to cover it he needs to find the surface area. To find the surface area of a box you add the are of each face so it would be $2(8*5+8*10+5*10)$ which is equal to $340$[/hide]

that is because
[hide]

[img]http://akimatsu.mypicgallery.com/mathmath/cube_large.jpg[/img]

because there are a total of 6 sides in a box. two of them will have the same area.

*ABCD and EFGH have the same area
*ABFE and CGHD have the same area
*AEHD and BFGC have the same area

so instead of multiplying all of the sides, we do $2(8\cdot10+10\cdot5+5\cdot8)=340$

[/hide]

[quote="helpmex3"]thanks. but i have another question
why is it [b]2[/b](8*5.....)[/quote]
Did you find the answer ?


Let $ a,b,c>0$ such that $ abc \le 1$. Prove that: \[ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c} \ge a\plus{}b\plus{}c\]

Take $ a \equal{} 9$, $ b \equal{} c \equal{} \frac{1}{3}$.

Thanks a lot "Soarer"  :oops:  I don't know but I get this smaller inequality from a bigger one I'm almost sure I didn't make mistakes in the steps until I came to this but   :huh: thanks again

You could have come to this inequality but I bet you had used another inequalities before obtaining it but not equivalent transformations  :wink:


Three students, with different names, line up single file. What is
the probability that they are in alphabetical order from
front-to-back?  Express your answer as a common fraction.

There are 6 ways for the students to stand in line without restrictions. There's only 1 formation for them to stand alphabetically. Probability=$ \boxed{\frac{1}{6}}$.

[hide=Smartgoodness' submission]There are $3\times 2\times 1=6$ formations that the $3$ students can line up in. $1$ of those $6$ formations is in alphabetical order. That means the probability of them lining up in an alphabetical manner is $\boxed{\frac16}$.[/hide]

[url=https://artofproblemsolving.com/community/user/32890]@gauss1181[/url]. I like your profile pic :D. It's kinda surprising to know that your accounts like, $15$ years old by now.


$ \int \dfrac{dx}{\sqrt {x}\sqrt {x \plus{} 1}} \equal{} ?$
Thank you.

(1) There is a post at the top that clearly says "do not post calculus problems in this forum!".

(2) Do not double post.  Find the right forum, and post once in that forum.


group theory
abstract algebra
number theory
least common multiple
superior algebra
superior algebra unsolved

[quote="sylow_theory"]If G is a finite group and it has subgroups of order n and m, prove that it has a subgroup of order lcm (n.m)[/quote]

This is not true. For example $A_{4}$, the alternating group of order 12, contains subgroups of orders 2 and 3, but it does not contain a subgroup of order 6.

If the group is not supposed to be abelian or something else, it's wrong in general (e.g. take $A_{5}$).

I think i made a mistake. yes you are right. i never looked for a counterexample, because i saw it as an excerise problem in some book. maybe i missed the word "abelian", then the result is trivial (if one is allowed to use the fundamental theorem).


The diameter of the cylinder is 3 inches and the height is 8 inches. A gallon = 231 cubic inches or 128 ounces. How many ounces of water does the cylinder hold? Use 22/7 as pi, round answer down.

A.18
B.36
C.31
D.16
E.NOTA

$ V\equal{}(1.5)(8)\frac{22}{7}\equal{}\frac{264}{7}$ so this is the volume of the cylinder in cubic inches.  Then $ \frac{264}{7}\cdot\frac{128}{231}\equal{}\frac{264}{1617}\equal{}20.9\mbox{ ounces}$.  That rounds down so I guess the answer is NoTA.

The correct answer is this though:

[hide]C. 31[/hide]

I just caught my mistake: $ V\equal{}\pi r^{2}h$.  I forgot to square 1.5.  Then we'll get the right answer.


topology LaTeX geometry MATHCOUNTS AMC AIME probability

Well if there is a topic about what type of math do you like then there should be the opposite!

Sorry if I left out a couple of topics  :lol:

I don't like topology very much. Maybe because lectures were quite poor for such abstract matter.
Generally speaking, I like analysis and dislike algebra.

oops i left out my dislike....i don't like statistics very much...u go through all that work just to find the mean of a 20  number set...

[quote="neelnal"]linear algebra is the worst. i like it but definitely nowhere as much as other subjects. all of the theorems,etc. seem so repetitive and simple and obviouis[/quote]
Did you take a college course in linear algebra yet? I've heard it's much better(harder too) than what you may think.

I personally like all contest math problems. But high-school math, bleh. Maybe BC calc can change my mind about that.
(disclaimer: I still don't know any college math kinda things, so I can't really say I like ALL math....and yes, I don't like statistics.)

Linear algebra begins to get interesting when it gets superior algebra ;)

I dislike any purely olympiad stuff, same for real analysis, numerics etc..
(I like most other advanced stuff)

[quote="blahblahblah"]If I never had to hear about graph theory again I would be pretty happy![/quote]

lol i hate it how you have to graph quintic functions...such a waste of time...id rather put it online, take a pic, and type the rest of my homework in LaTeX

[quote="blahblahblah"]Graph theory has nothing to do with graphing functions.[/quote]

I never said it does  :lol:  but I hate graphing all the same

I never liked geometry. It's too damn complicated! All those lines... if you're unlucky, you get face to face with a triangle or, God forbid, a circle; only then the real torment begins.

Geometry is what's held me back in Mathcounts, the AMC 10 and 12, and the AIME, and every other math competition.

[quote="jnApp66"]Geometry is what's held me back in Mathcounts [/quote]

yeah same here...in fact I was kidna hoping we'd learn Geometry in 8th grade instead of Alg 2...i might have had a better chance in MC

non-abstract math, in general.

Abstract math makes you wonder when you ever gonna reach your limit! I managed to do vectorspaces -> category theory, in one year, and nothing could stop me so far :D Most applied math can be done with a computer, so there's not really something challenging about that, or is there? :)

You really should have put Counting/Probability and Statistics in the poll. I loathe statistics and the questions you get in Counting/Probability are always annoyingly ambiguous. However, apart from those, I don't particularly like geometry, because the diagrams you draw always become a mess of symbols and you don't know what is what after a while. My favorite is number theory because it's the 'universal truth' (i.e. I prefer its axioms over geometry's) and it's fun.

[quote="7h3.D3m0n.117"]Haha so far, after this year in geo, I don't like it either.[/quote]

After what you told me about Geo, I'm thinking about taking the course of Pre-Calculus instead of Geometry  :P

I don't really like Geometry, it is far too complicated. :mad:

[quote="tjhance"][quote="7h3.D3m0n.117"][quote="tjhance"]What? geo is awesome[/quote]

Not the way my teacher teaches it.[/quote]

Well it's awesome outside of the curriculum[/quote]

So true, I hate having to explain every line of why two angles are equal, why certain points are collinear etc.

I don't really like Algebra,not at all,i like functional equaltions.My favourite type is geometry,I don't understand when someone said that he don't like geometry,geo is the most beaustiful part of math.Also i very like inequalities,combinatorics,number theory(not much,but  sometimes i find very beautiful problems from n.t).I think that Number theory is the most important and hardest part of math. :wink:

Actually I just realised I don't like geometry because I'm too lazy to draw stuff. In fact I can't remember when I did my last geometry problem, must have been about a month ago...

Algebra sucks. I just had to graph like 22 functions yesterday.

I think number theory is hard to me. Nevertheless, I will try my best to improve it.

Geometry. I apparently don't like drawing all those diagrams, and stuff.
Well, some parts of geometry are fun, but nothing about school goemetry.

Geometry is... okay. I actually used to think two-column proofs were awesome.

If there's any part of math I hate, it's arithmetic. Everybody in my class, when confronted with some complex calculations, visits me. I tell them, "Arithmetic is not math." That's what I feel. Sure, it's important and I'm not actually bad at it, but it's all brute force in my opinion.


geometry
analytic geometry
integration
trigonometry
calculus
calculus computations

Sketch the curve y^3=ax(x+y) and prove that the area of the loop is (1/60)a^2.

[hide="hint"]Use polar coordinates to reduce the problem to finding $ \frac {a^2}{2}\int_{\frac {\pi}{2}}^{\frac {3\pi}{4}}\left [\frac {\cos \theta (\cos \theta \plus{} \sin \theta)}{\sin ^3 \theta }\right]^2d\theta$[/hide]

Actually the integral $ \int_0^{\frac 14}y\sqrt{1\minus{}4y}\,dy$ you get using Cartesian coordinates is easier to find (especially if you write $ y$ as $ \frac 14[1\minus{}(1\minus{}4y)]$)


geometry
analytic geometry
graphing lines
slope
vector

So, I was trying to solve this problem, but I couldn't because I used cyclic reasoning.

Anyways here it is.

In triangle ABC, we have $AB=AC$.  Furthermore, D is the midpoint of BC, E is the foot of the perpendicular drawn from D to AC and F is the midpoint of DE.  Prove that $AF \perp BE$.

Thanks
And please show work.

I forgot to mention, please move if appropriate, cause I think this is an IMO shortlist.

[hide="Ugh, here is a really disgusting but valid proof:"]
Coordinate brute-forcing!  Lol.

Put $B(-1,0), \ A(0,\sqrt{x^{2}-1}), \ C(1,0) \implies D(0,0) \wedge \overline{AB}=\overline{AC}=x$.

Then, we have:
\[\triangle ABD \sim \triangle DCE \implies \overline{CE}=\frac{1}{x}\implies E\left( 1-\frac{1}{x^{2}}, \ \frac{ \sqrt{x^{2}-1}}{x^{2}}\right) \implies F \left( \frac{x^{2}-1}{2x^{2}}, \ \frac{ \sqrt{x^{2}-1}}{2x^{2}}\right) \]

Let $m \left( \overline{BE}\right)$ be the slope of segment $\overline{ BE}$. We have that
\[m \left( \overline{BE}\right) = \frac{ \left( \frac{\sqrt{x^{2}-1}}{x^{2}}\right) }{2-\frac{1}{x^{2}}}=\frac{\sqrt{x^{2}-1}}{2x^{2}-1}\]
and
\[m \left( \overline{AF}\right) = \frac{ \left( \frac{\sqrt{x^{2}-1}}{2x^{2}}-\sqrt{x^{2}-1}\right)}{ \left( \frac{x^{2}-1}{2x^{2}}\right) }= \frac{1-2x^{2}}{\sqrt{x^{2}-1}}\]
But we have that $m \left( \overline{BE}\right) = \frac{-1}{m \left( \overline{AF}\right) }$, so $\boxed{ AF \perp BE \text{ !}}$.
[/hide]
Please, [i]someone[/i] come up with a better proof than this. That was horrible.

I prove this problem by using inner product of vectors:
$2\vert{AF}.\vec{BH}=(\vec{AD}+\vec{AE})(\vec{BD}+\vec{DE})$
$=\vec{AD}.\vec{DE}+\vec{AE}.\vec{BD}=\vec{AD}.\vec{DE}+(\vec{AD}+\vec{DE}).\vec{BD}=\vec{AD}.\vec{DE}+\vec{DE}.\vec{DC}$
$=\vec{ED}.\vec{DE}+\vec{DE}.\vec{DC}=-DE^{2}+DE^{2}=0$

Denote: $K$ is the mid piont of $[EC]$.
We have: $\left \{\begin{array}{l}FK\| CD\\ AD\perp CD\end{array}\right\|\Longrightarrow\left\{\begin{array}{l}FK\perp AD\\ DE\perp AK\end{array}\right\|\Longrightarrow\left\{\begin{array}{l}AF\perp DK\\ DK\| BE\end{array}\right\|\Longrightarrow AF\perp BE$


Let n be an integer , $ n\geq 2 $ . Prove that : with every $ k \in Z $ : $ 2n-3 \leq k \leq \frac{n(n-1)}{2} $ , there exists n real numbers (different from together) such that : Among the numbers : $ a_i+a_j $ , $ 1 \leq i \leq j \leq n $ , there has exactly k numbers (differnet from together)

Unless I'm missing something, I think this can be done rather easily by induction.

First of all, if we show it for $n-1$, then when we go to $n$, it suffices than we can achieve everything in the interval $[2n-3,3n-6)$, because if we want $k\ge 3n-6$, we can form $k-(n-1)\ge 2(n-1)-3$ sums with $n-1$ numbers, and then take the $n$'th number "far away" from our $n-1$ numbers, so that it generates $n-1$ sums which have not been encountered before.

In order to get anything from $2n-3$ to $3n-7$, take first $1,2,\ldots,n$ for $2n-3$, then $1,2,\ldots,n-1,n+1$ for $2n-2$, and so on. It's easy to see that each time we increase the greatest number with one unit, the number of sums increases with one unit as well.

Hope there are no mistakes. If there are, sorry :).


function
number theory
greatest common divisor
least common multiple

I saw on the forum: " Note that 100=25*4 ;

phi(25)=20 and phi(4)=2 and LCM or 20 and 2 is 20
this means if you take any number, which is co-prime to 2 and 5, raised to power 20; its last 2 digit would be 01. "
what is the general theorem? proof? thanks.
and how do I prove that gcd(a + m·b, b) = gcd(a, b)?

[quote="srulikbd"]and how do I prove that gcd(a + m·b, b) = gcd(a, b)?[/quote]
[hide="Well..."]The sets of all common divisors of $a+m\cdot b,\; b$ and $a,\; b$ are the same: $d\mid a,b\implies d\mid a+m\cdot b$ and $d\mid b,\; a+m\cdot b\implies d\mid a+m\cdot b-m\cdot b= a$, so their maximums $\gcd(a+m\cdot b,\; b)$ and $\gcd(a,\; b)$ are equal.[/hide]


algebra
polynomial
function
calculus
integration
algebra proposed

Find all polynomial functions $f: \mathbb{R} \rightarrow \mathbb{R}$ with integer coefficients and all of these are bijective and satisfying the condition:
\[ \forall x\in\mathbb{R} , f^2(x)=f(x^2)-2f(x)+1  \]

hay Guys plz solve it  ,it isn't very hard  ;)

[quote="harazi"]Ehsan, do you read the proofs received for the problems you post? For example, I perfectly remember that a more general problem was posted by you and I solved it, so indeed it is not very hard, but I really don't think that posting the same problem 5 times is a good idea.[/quote]

I cann't understand what you said  ??????  :huh:  :huh:  :huh:

Or, you don't want to understand what Harazi wrote? :? i told you once about a problem with integrals, but you didn't hear me out. :roll:


We consider a cube with sides equal to $1$. Let $\mathcal A$ be a set of points which satisfies the following properties:
(a) $\mathcal A$ is in the interior of the cube (that means that the faces of the cube are also excluded);
(b) $\mathcal A$ has at least three points;
(c) The distance between any two points is at most $d \in \left( 0, 1 \right)$.

For all points $M \in \mathcal A$, let $x(M)$ be the smallest distance from $M$ to one of the others points from $\mathcal A$.

Prove that \[\sum_{M \in \mathcal A}\left( x(M) \right)^{3}< \left( d+\sqrt 3 \right)^{3}.\]

Any progress on it? It seems terribly hard :(

If moderators agree, then maybe we could put it in the Olympiad section...


abstract algebra
function
linear algebra
linear algebra unsolved

Let $(,)$ be a bilinear form on $V$, such that for $x,y\in V$ we know that \[(x,y)=0\Longleftrightarrow (y,x)=0\]Prove that $(,)$ is symmetric or skew-symmetric. (i.e. $(x,y)=(y,x)$ or $(x,y)=-(y,x)$ for each $x,y$.)

For any fixed $x$, $(x,y)$ and $(y,x)$ are both linear functionals in $x$; they have the same kernel by hypothesis. This defines a 1-dimensional subspace of functionals, so $(y,x)=\lambda(x)(x,y)$ for some function $\lambda(x)$. If $(x,y)\neq0$, $(x,y)=\lambda(y)(y,x)=\lambda(y)\lambda(x)(x,y)$, and $\lambda(x)\lambda(y)=1$.
If $(x,y)=0$ for all $y$, ignore $\lambda(x)$; any value works. We want to show that $\lambda$ is constant outside this subspace. Given $x$ and $y$, let $S_{x}=\{u: (x,u)=0\}$ and $S_{y}=\{u: (y,u)=0\}$. The union of these two proper subspaces cannot be all of $V$ (you need at least three, even in $F_{2}$), so there is some $z$ outside both $S_{x}$ and $S_{y}$. By the preceding calculation, $\lambda(x)\lambda(z)=1=\lambda(y)\lambda(z)$, and $\lambda(x)=\lambda(y)$. $\lambda$ is thus constant outside the degenerate part; its value must be $\pm1$, and we can extend this to the degenerate part trivially, obtaining either $(x,y)=(y,x)$ for all $x,y$ or $(x,y)=-(y,x)$ for all $x,y$.

Yes, it's been done before. My proof does cover the case grobber omitted, and is a completely different style.


function
logarithms
limit
real analysis
real analysis unsolved

Let $ x_n$ be the sequence defined by $ f_n(x_n)\equal{}2^n, \forall n \in N$ where
$ f_n(x)\equal{}x^n\plus{}lnx$.
Prove that $ x_n$ is convergent to 2.

Note that $ f_n(2)>2^n$ and $ f_n$ is an increasing function, so $ x_n<2.$ (And we know $ x_n$ exists by the Intermediate Value Theorem.)

By the Mean Value Theorem, $ \frac{f_n(2)\minus{}f_n(x_n)}{2\minus{}x_n}\equal{}f_n'(\xi)$ for some $ \xi\in(x_n,2).$

Hence, $ \frac{\ln 2}{2\minus{}x_n}\equal{}n\xi^{n\minus{}1}\plus{}\frac1{\xi}$ or

$ 2\minus{}x_n\equal{}\frac{\ln 2}{n\xi^{n\minus{}1}\plus{}\xi^{\minus{}1}}$

Since we also have that $ x_n>1$ (because $ f_n(x)\equal{}1<2^n$) we have that

$ n\xi^{n\minus{}1}\plus{}\xi^{\minus{}1}>n\xi^{n\minus{}1}>n,$ so that

$ 2\minus{}x_n<\frac{\ln 2}{n},$ proving our result.

Of course, the convergence is much faster than that. $ \xi>1$ is a gross underestimate since $ \xi$ is nearly $ 2$ for large $ n.$

$ x_n\in (1,2)$
and we have $ 1 \plus{} \frac {ln x_n}{(x_n)^n} \equal{} (\frac {2}{x_n})^n$
because $ x_n\in(1,2)\Rightarrow \lim_{n\rightarrow \infty} \frac {ln(x_n)}{(x_n)^n} \equal{} 0$ It follows from here that $ \lim_{n\rightarrow \infty} (\frac {2}{x_n})^n \equal{} 1 \Rightarrow x_n\rightarrow 2$


A metal cone of height 12 cm and radius 4 cm just fits into a cylinder of radius 4 cm and height 100 cm which is filled with water to a depth of 50cm. The cone is lowered at a constant rate of 4 cm per second (with respect to the walls of the cylinder). At what rate is the water in the cylinder rising when the vertex of the cone is immersed to a depth of 6 cm?

[hide="is this correct?"]
The radius $r$ of the cone at any point in time is $\frac{h}{3}$ So, Volume any point in time is $\frac{{\pi}h^3}{27}$. We basically want the rate of change of the volume when $h=6$
We also know that $4t=h$
So,
Replacing it with 1st equation gives:
$V=\frac{\pi(64t^3)}{27}$
We want the differential of this..
$\frac{dV}{dt}=\frac{\pi(64*3t^2)}{27}$
We want this when $t=1.5$ So replacing that should give us the rate?
Is this solution correct?
[/hide]

is this too easy to be answered in the pre-olympiad section? Nobody's replied for days. :?


calculus
integration
trigonometry
calculus computations

Calculate $I=\int_{0}^{\frac{\pi}{2}}\left(\sqrt{\sin x}+\sqrt{\cos x}\right)dx$

I think this should be posted in the college-playground section :D

[color=red][Done. -mathmanman][/color]

\begin{eqnarray*}\int_{0}^{\frac{\pi}{2}}\left(\sqrt{\sin x}+\sqrt{\cos x}\right) \, dx & = & 4 E\left(\left.\frac{\pi }{4}\right|2\right)\\ & = & \sqrt{\pi }\frac{ \Gamma \left(\frac{3}{4}\right)}{\Gamma \left(\frac{5}{4}\right)}\\ & \approx & 2.396280469471184414879845\cdots \end{eqnarray*}

I think there is no way to express the value of it as a nice closed form.

[quote="sos440"]I think there is no way to express the value of it as a nice closed form.[/quote]
That's correct

i am just in high-school and i have not understoud this :|
what does sos440 mean by $E(\frac{\pi}{4}|2)$.

It's not a notation I've ever seen before, but it's probably some variant on a binomial coefficient.


algorithm
combinatorics unsolved
combinatorics

Let $ a<b<c$ be three positive integer numbers .Prove that there exist a partition of $ N$ into three parts $ (X,Y,Z)$ such that $ a\plus{}X\equal{}b\plus{}Y\equal{}c\plus{}Z$ is equivalent to $ a\plus{}c\equal{}2b$ (Note that $ a\plus{}X$ denote the set $ {a\plus{}x | x\in X }$ )

Note that the condition $ a\plus{}X\equal{}b\plus{}Y\equal{}c\plus{}Z$ is equivalent to $ X \equal{} c\minus{}a\plus{}Z$ and $ Y\equal{}c\minus{}b\plus{}Z$.
If this condition is satisfied then, since $ a<b<c$ it follows that for each $ t \in X \cup Y$ there exists $ z \in Z$ such that $ z<t$.

- If $ a\plus{}c\equal{}2b$ :
We want to construct a partition with the desired property But the remark above shows that we have to use the following algorithm :
i) $ 0 \in Z,c\minus{}a \in X,c\minus{}b \in Y$.
ii) If $ t$ is the least positive integer not already used by the algorithm then $ t\in Z,t\plus{}c\minus{}a \in X,t\plus{}c\minus{} b\in Y$.

Clearly, from ii), $ X\cup Y \cup Z$ contains each non-negative integer and $ X\equal{} c\minus{}a\plus{}Z$ and $ Y\equal{}c\minus{}b\plus{}Z$.
Moreover, at each step, we first put in $ Z$ an integer $ t$ which is not in $ X$ neither in $ Y$, and then each number which is put in $ X \cup Y \cup Z$ is greater than $ t$. It follows that no element of $ Z$ can belong to $ X \cup Y$.

Now, assume, for a contradiction, that there exists $ t \in X \cap Y$.
Then, there exist $ z,z' \in Z$ such that $ t \equal{} c\minus{}a \plus{}z$ and $ t\equal{}z'\plus{}c\minus{}b$.
Using that $ a\plus{}c\equal{}2b$, we deduce easily that $ z' \equal{} z\plus{}c\minus{}b$. But, from the construction, $ z\plus{}c\minus{}b \in Y$, so that $ z' \in Y \cap Z$, a contradiction.

Thus, $ X,Y,Z$ are pairwise disjoint, and we are done.

- Conversly, assume that $ X,Y,Z$ form a partition of $ \mathbb{N}$ such that $ a\plus{}X\equal{}b\plus{}Y\equal{}c\plus{}Z$.
From our initial remark, we deduce that we must have i) and ii), which means that the algorithm is valid.
Thus, $ 0, \cdots, c\minus{}b\minus{}1 \in Z$ and $ c\minus{}b, \cdots ,2(c\minus{}b)\minus{}1 \in Y$ and $ c\minus{}a, \cdots, 2c\minus{}a\minus{}b\minus{}1 \in X$.
Moreover, the least element of $ X$ is $ c\minus{}a$ otherwise $ Z$ would have to contain a negative integer.

If $ a\plus{}c > 2b$ :
Then $ 0< b\minus{}a <c\minus{}b$ so that $ b\minus{}a \in Z$, which forces $ (b\minus{}a)\plus{}c\minus{}b \equal{} c\minus{}a$ to be in $ Y$. But, $ c\minus{}a \in X$ and $ X \cap Y \equal{} \emptyset$, a contradiction.

If $ a\plus{}c < 2b$ :
Since the least element of $ X$ is $ c\minus{}a$, and $ c\minus{}a > 2(c\minus{}b)$, it follows that $ 2(c\minus{}b)$ does not belong to $ X$. It does not belong to $ Y$ too, otherwise we would have $ c\minus{}b \in Y \cap Z$, a contradiction. Thus, $ 2(c\minus{}b) \in Z$.
Moreover, $ 2c\minus{}a\minus{}b$ cannot be in $ X$ (otherwise $ c\minus{}b \in Y \cap Z$) or in $ Y$ (otherwise $ c\minus{}a \in X \cap Z$). Therefore, $ 2c\minus{}a\minus{}b \in Z$.
It follows that $ 3c\minus{}2b\minus{}a \equal{} (2c\minus{}a\minus{}b) \plus{} (c\minus{}b) \in Y$.
But, $ 3c\minus{}2b\minus{}a \equal{} 2(b\minus{}c) \plus{} (c\minus{}a) \in X$, a contradiction.

Thus, $ a\plus{}c \equal{} 2b$, and we are done.

Pierre.


function
calculus
derivative
real analysis
real analysis unsolved

Doesn't this belong in the calculus/real analysis forum?

It does... but silouan, what is f? which domain/range? other properties? even if it's just the standard R->R it's easier if you write that...

I want only to say that the function is two times derivative

Well $x^3$ is a factor anyway... (as long as $f'''(0) \neq 0$)

well, a trivial answer would be that $f(x)=0$

Id imagine you would integrate f"(x) and f'(x) and use the other fact to determine the constant, then compare it to f^3(x).

Well, I'll suppose $f: \mathbb{R}\to\mathbb{R}$ with $f$ twice differentiable; now, we have
\[y''=-2y^3\]
so
\[y''y'=-2y^3y'\]
that is
\[\frac{d}{dt}(y'^2)=-\frac{d}{dt}(y^4)\]
or
\[y'^2=k-y^4\]
but $y'(0)=y(0)=0$, so $k=0$ and $y'^2=-y^4$ but this is possible only if $y'(x)=y(x)=0$ for all $x\in\mathbb{R}$.
So, $f\equiv0$.

Actually, we can simply invoke the existence-uniqueness theorem: 0 is a solution and there are no singular points of the differential equation, so any solution which agrees with zero up to the first derivative must be zero.