inequalities
geometry
incenter
circumcircle
analytic geometry
trigonometry
inradius

[color=darkred]Prove that in an [b]acute or right[/b] $ \triangle ABC$ there is the relation $ \boxed {\ IH\le IO\sqrt 2\ }$ ,

where $ I$ , $ H$ , $ O$ are incenter, orthocenter and circumcenter respectively.[/color]

The inequality is equivalent to proving that $ 3(R^2 + 3Rr + r^2)\leq s^2$ for acute of right triangles. This is true if my calculations [i]were[/i] correct.

EDIT: Here is what I did.  The inequality I found is based on your identity.  Let the circumcenter of the triangle be the origin of the complex plane.  Let $ \alpha, \beta, \gamma$ be the side lengths of the triangle, and let $ a,b,c$ be the complex coordinates of the vertices of $ A,B,C$.  In the complex plane, $ z_I = \frac {\alpha a + \beta b + \gamma c}{\alpha + \beta + \gamma}$ and $ z_H = a + b + c$.
Assuming that your inequality is true, I have
\begin{eqnarray*}IH^2 & \leq & 2OI^2\iff (z_I - z_H)\cdot (z_I - z_H)\leq 2 z_I\cdot z_I \\
\left( \frac {\alpha a + \beta b + \gamma c}{\alpha + \beta + \gamma} - (a + b + c)\right)\cdot\left( \frac {\alpha a + \beta b + \gamma c}{\alpha + \beta + \gamma} - (a + b + c)\right) & \leq & 2OI^2 \\
OI^2 - 2\left( \frac {\alpha a + \beta b + \gamma c}{\alpha + \beta + \gamma}\right)\cdot (a + b + c) + OH^2 & \leq & 2OI^2 \\
R^2 - 2Rr & \geq & 9R^2 - \sum\alpha^2 - 2\left( \frac {\alpha a + \beta b + \gamma c}{\alpha + \beta + \gamma}\right)\cdot (a + b + c)\end{eqnarray*}
Now,
\begin{eqnarray*}2\left( \frac {\alpha a + \beta b + \gamma c}{\alpha + \beta + \gamma}\right)\cdot (a + b + c) & = & \frac {(\alpha + \beta + \gamma)R^2 + \sum (\alpha + \beta) a\cdot b}{s} \\
& = & \frac {(\alpha + \beta + \gamma)R^2 + \sum (\alpha + \beta) \left(R^2 - \frac {\gamma^2}{2}\right)}{s} \\
& = & \frac {3(\alpha + \beta + \gamma)R^2 - \frac {1}{2}\sum (\alpha + \beta)\gamma^2}{s} \\
& = & \frac {1}{s}\left(3(\alpha + \beta + \gamma)R^2 - \frac {1}{6}\left( (\alpha + \beta + \gamma)^3 - \sum\alpha^3 - 6\alpha\beta\gamma\right)\right) \\
& = & \frac {1}{s}\left(6sR^2 - \frac {1}{6}\left( 8s^3 - 2s(s^2 - 3r^2 - 6Rr) - 24sRr\right)\right) \\
& = & 6R^2 - s^2 - r^2 + 2Rr\end{eqnarray*}
Substituting this back in gives us $ s^2\geq 2R^2 + 8Rr + 3r^2$.

Apparently, I had fogotten to substract $ \sum\alpha^3$.

Now, we have that your relation is equivalent to $ 8R\prod\cos^2\frac {A}{2}\geq R + 16\prod\sin\frac {A}{2} + 24R\prod\sin^2\frac {A}{2}$ for right and acute triangles.

As you can infer, I am not good at synthetic geometry. :( :blush:

EDIT 2: Here, $ a\cdot b = \frac {1}{2}(\bar{a}b + a\bar{b})$ is called the real product of $ a,b\in\mathbb{C}$.

[quote="boxedexe"]The inequality is equivalent to proving that $ 3(R^2 \plus{} 3Rr \plus{} r^2)\leq s^2$ for acute of right triangles.  Of course, this is true if my calculations were correct.[/quote]
[color=darkblue][b]Sorry, your inequality $ 3(R^2 \plus{} 3Rr \plus{} r^2)\leq s^2$ is falsely ![/b]

For example, the right triangle $ (3,4,5)\implies\left\|\begin{array}{c} R \equal{} \frac 52 \\
 \\
s \equal{} 6 \\
 \\
r \equal{} 1\end{array}\right\|$ .

Maybe $ 2R^2 \plus{} 8Rr \plus{} 3r^2\le s^2$ , where $ 2s \equal{} a \plus{} b \plus{} c$ ...

In [b]my opinion[/b], the proposed inequality isn't simply.[/color]

first of all by squaring both sides,it's sufficient to show that:

$ IH^2\leq 2IO^2$

now let $ R,r,p$ denote the circumradius,inradius and the semiperimeter of $ \triangle ABC$ respectively,now using the relations:

$ OI^2 \equal{} R^2 \minus{} 2Rr$

$ IH^2 \equal{} 4R^2 \plus{} 4Rr \plus{} 3r^2 \minus{} p^2$

we have to show that:

$ 4R^2 \plus{} 4Rr \plus{} 3r^2 \minus{} p^2\leq 2R^2 \minus{} 4Rr$

$ \iff 2R^2 \plus{} 8Rr \plus{} 3r^2\leq p^2$

which is a well-known inequality...

[quote="BaBaK Ghalebi"] [color=darkred]... $ \boxed {\ p^2\ge 2R^2 \plus{} 8Rr \plus{} 3r^2\ }$ which is a well-known inequality.[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b]  $ \frac {1}{4S}\cdot\sum a^2 \equal{} \frac {1}{4S}\cdot\sum \left(b^2 \plus{} c^2 \minus{} a^2\right) \equal{} \sum\frac {\cos A}{\sin A} \equal{}$ $ 2\cdot\sum \frac {\cos^2A}{\sin 2A}\stackrel {C.B.S.}{\ \ \ge\ \ }$ $ 2\cdot\frac {\left(\sum\cos A\right)^2}{\sum \sin 2A} \equal{} \frac {\left(1 \plus{} \frac rR\right)^2}{2\sin A\sin B\sin C} \equal{} \frac {\left(1 \plus{} \frac rR\right)^2}{\frac {S}{R^2}} \equal{} \frac {(R \plus{} r)^2}{S}$ .

Thus, $ \frac {1}{4S}\cdot\sum a^2\ge \frac {(R \plus{} r)^2}{S}$ , i.e. $ \boxed {\ \sum a^2\ge 4(R \plus{} r)^2\ }\ \Longleftrightarrow$ $ p^2 \minus{} r^2 \minus{} 4Rr\ge 2(R \plus{} r)^2\ \Longleftrightarrow\ \boxed {\ p^2\ge 2R^2 \plus{} 8Rr \plus{} 3r^2\ }$ .

[b]I don't knew that it is well-known. I proved it now. Has it one name ?! Know you a link or another proof of this "well-known" inequality ? Thanks ![/b]

And now [b]I"ll present the continuation of my proof.[/b]  From the well-known relations $ IO^2 \equal{} R^2 \minus{} 2Rr$ and $ IH^2 \equal{} 4R^2 \plus{} 4Rr \plus{} 3r^2 \minus{} p^2$ obtain :

$ \boxed {\ p^2\ge 2R^2 \plus{} 8Rr \plus{} 3r^2\ }$ $ \Longleftrightarrow$ $ 4R^2 \plus{} 4Rr \plus{} 3r^2 \minus{} IH^2\ge 2R^2 \plus{} 8Rr \plus{} 3r^2$ $ \Longleftrightarrow$ $ IH^2\le 2R(R \minus{} 2r)\ \Longleftrightarrow\ IH^2\le 2\cdot IO^2\ \Longleftrightarrow\ \boxed {\ IH\le IO\sqrt 2\ }$ .

In conclusion, for an [b]acute or right[/b] $ \triangle ABC$ , we have $ \boxed {\boxed {\ IH\le \sqrt 2\cdot IO\ }}$ and $ \boxed {\ \sum a^2\ge 4(R \plus{} r)^2\ \Longleftrightarrow\ p^2\ge 2R^2 \plus{} 8Rr \plus{} 3r^2\ \Longleftrightarrow\ IH\le \sqrt 2\cdot IO\ }$ .

[b][u]Remark.[/u][/b] [b]Generally[/b], in any $ \triangle ABC$ we have $ \boxed {\boxed {\ IH\le 2\cdot IO\ }}$ and $ \boxed {\ p^2\ge 3r(4R\plus{}r)\ \Longleftrightarrow\ IH\le 2\cdot IO\ }$ . Indeed, $ 3r(4R\plus{}r)\equal{}3\sum (p\minus{}b)(p\minus{}c)\le p^2$

and $ IH\le 2\cdot IO\ \Longleftrightarrow\ IH^2\le 4\cdot IO^2\ \Longleftrightarrow\ 4R^2\plus{}4Rr\plus{}3r^2\minus{}p^2\le 4R^2\minus{}8Rr$ $ \Longleftrightarrow$ $ p^2\ge 3r(4R\plus{}r)$ .[/color]

[quote="Virgil Nicula"][quote="BaBaK Ghalebi"] [color=darkred]... $ \boxed {\ p^2\ge 2R^2 \plus{} 8Rr \plus{} 3r^2\ }$ which is a well-known inequality.[/color] [/quote][color=darkblue]
[b]I don't knew that it is well-known. I proved it now. Has it one name ?! Know you a link or another proof of this "well-known" inequality ? Thanks ![/b]
[/color][/quote]
it's equivalent to the following inequalities:

[b]let $ x,y,z$ be positive reals such that $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz \equal{} 1$ then:

$ xy \plus{} yz \plus{} zx\leq\frac 12 \plus{} 2xyz$[/b]

[url]http://www.mathlinks.ro/viewtopic.php?p=876948&search_id=1962601750#876948[/url]

and

[b]in $ \triangle ABC$ we have:

$ 1 \plus{} 4\cos A\cos B\cos C\geq 2\sum\cos A\cos B$[/b]

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=718883#718883[/url]

Thank you, I found it and ... some my replies. I forgot even I worked with these.

I think that the above my proof of $ p^2\ge 2R^2 \plus{} 8Rr \plus{} 3r^2$ is new !

[quote="Virgil Nicula"]I think that the above my proof of $ p^2\ge 2R^2 \plus{} 8Rr \plus{} 3r^2$ is new ![/quote]
and nice! :)

Thank you.  Sometimes is good to forget somethings ...


conics
analytic geometry
geometry
circumcircle
trigonometry

Prove that pentagon $ ABCDE$ is cyclic if and only if
\[\mathrm{d(}E,AB\mathrm{)}\cdot \mathrm{d(}E,CD\mathrm{)} \equal{} \mathrm{d(}E,AC\mathrm{)}\cdot \mathrm{d(}E,BD\mathrm{)} \equal{} \mathrm{d(}E,AD\mathrm{)}\cdot \mathrm{d(}E,BC\mathrm{)}\]
where $ \mathrm{d(}X,YZ\mathrm{)}$ denotes the distance from point $ X$ ot the line $ YZ$.

I'll restate the problem with different notations.

[b]Problem:[/b] $ ABCDE$ is a pentagon and $X,Y,Z,W,P,Q$ are the orthogonal projections of $ E$ onto $ AB,$ $CD,$ $AC,$ $BD,$ $AD,$ $BC.$ Then $EX \cdot EY \equal{} EZ \cdot EW \equal{} EP \cdot EQ$ $\Longleftrightarrow$ $ ABCDE$ is cyclic.

We assume that $ ABCD$ is cyclic. If $ EX \cdot EY \equal{} EP \cdot EQ$ $ \Longrightarrow$ $ \triangle EXP \sim \triangle EQY$ since $ \angle XEP \equal{} \angle QEY.$ Then $ E$ is the center of the spiral similarity that takes $ \triangle EXP$ into $ \triangle EQY$ $\Longrightarrow$ circles $\odot (EXP)$ and $\odot (EQY)$ meet on $QY$ and angle chase yields $ \angle EBA \equal{} \angle ADE$ $ \Longrightarrow$ $ E \in \odot(ABCD).$ It remains to prove  $ EZ \cdot EW \equal{} EP \cdot EQ,$ which folllows from the fact that $ Y,W, Q$ are collinear on the Simson line with pole $ E$ WRT $ \triangle BCD$ and $Y,P,Z$ are collinear on the Simson line with pole $ E$ WRT $ \triangle ACD.$ Then $\odot (YWP)$ and $ \odot (YZQ)$ meet at $ E$ $\Longrightarrow$ $ E$ is the center of the spiral similarity taking $ \triangle EZP$ into $\triangle EQW,$ hence $ EZ \cdot EW \equal{} EP \cdot EQ.$

Besides this natural synthetic solution one could (not easily!) find the following one, which uses techniques of algebraic geometry. It involves some well-known properties of conics as well as some elements of complex geometry. I will write down the solution in different posts, for a better understanding.

At first I will reformulate the problem:

[i]Prove that pentagon $ A_0A_1A_2A_3A_4$ is cyclic iff the product $ \mathrm{dist}\(A_4,A_iA_j)\cdot\mathrm{dist}(A_4,A_kA_l)$ is the same for all choices of indices $ \{i,j,k,l\}=\{0,1,2,3\}.$ [/i]

[b]Solution.[/b] In what follows, we denote by lowercases the complex coordinates of points denoted by uppercases.
The equation of the oriented line $ A_jA_k$ is given by
$ f_{jk}(z,\bar{z})\equal{}\left|\begin{array}{ccc}
a_j & \bar a_j & 1 \\
a_k & \bar a_k & 1 \\
z & \bar z & 1 \\
\end{array} \right|\equal{}0$
or $ f_{jk}\equal{}(\bar a_j\minus{}\bar a_k)z\minus{}(a_j\minus{}a_k)\bar z\plus{}a_j\bar a_k\minus{}\bar a_ja_k\equal{}0.$
The distance from $ A_4$ to $ A_jA_k$ is $ \minus{}\frac{f_{jk}(a_4,\bar a_4)}{2\mathrm{i}|a_j\minus{}a_k|}.$
Next, it is not hard to verify that $ f_{jk}((1\minus{}t)z\plus{}tz',(1\minus{}t)\bar{z}\plus{}t\bar z'\equal{}(1\minus{}t)f_{jk}(z,\bar z)\plus{}tf_{jk}(z',\bar z'), \forall t\in\mathbb{R}$ and $ z,z'\in\mathbb{C}$.

The condition in the statement is equivalent to the fact that $ A_4$ lie on the following 2 conics passing through $ A_0,A_1,A_2,A_3$:
$ |a_0\minus{}a_3|\cdot|a_1\minus{}a_2|f_{01}f_{23}\minus{}|a_0\minus{}a_1|\cdot|a_2\minus{}a_3|f_{03}f_{12}\equal{}0\ \ (1)$
and
$ |a_0\minus{}a_2|\cdot|a_1\minus{}a_3|f_{01}f_{23}\minus{}|a_0\minus{}a_1|\cdot|a_2\minus{}a_3|f_{02}f_{13}\equal{}0\ \ (2)$
This is possible iff the 2 conics coincide.
Both of the conics belong to the pencil of conics determined by $ A_0,A_1,A_2,A_3.$
It is renowned that any two distinct conics of this pencil form a basis over $ \mathbb{R}$ for the pencil; in particular, any two of the three (degenerate) conics: $ f_{01}f_{23}\equal{}0,f_{02}f_{13}\equal{}0,f_{03}f_{12}\equal{}0.$
These three are therefore linearly dependent over $ \mathbb{R}$. Moreover, they satisfy, as we will show, the following Ptolemy-like relation:
$ f_{01}f_{23}\minus{}f_{02}f_{13}\plus{}f_{03}f_{12}\equal{}0\ \ (\star)$

Assume $ (\star)$ for the time being to rewrite (2) as:
$ (|a_0\minus{}a_2|\cdot|a_1\minus{}a_3|\minus{}|a_0\minus{}a_1|\cdot|a_2\minus{}a_3|)f_{01}f_{23}\minus{}|a_0\minus{}a_1|\cdot|a_2\minus{}a_3|f_{03}f_{12}\equal{}0\ \ (2')$
Now the two conics coincide iff:
$ |a_0\minus{}a_1|\cdot|a_2\minus{}a_3|\minus{}|a_0\minus{}a_2|\cdot|a_1\minus{}a_3|\plus{}|a_0\minus{}a_3|\cdot|a_1\minus{}a_2|\equal{}0.$
By [b]Ptolemy Theorem[/b] the points $ A_0,A_1,A_2,A_3$ are concyclic.

It only remains to establish $ (\star)$ for the proof to be complete.
Notice that $ f\equal{}f_{01}f_{23}\plus{}f_{03}f_{12}$ vanishes at any point $ z$ of the conic $ f_{02}f_{13}\equal{}0$: for example, if $ z$ lies on the line $ f_{02}\equal{}0$ then $ z\equal{}(1\minus{}t)a_0\plus{}ta_2,$ for some $ t\in\mathbb{R}$ so that
$ f(z,\bar{z})\equal{}t(1\minus{}t)(f_{01}(a_2,\bar a_2)f_{23}(a_0,\bar a_0)\plus{}f_{03}(a_2,\bar a_2)f_{12}(a_0,\bar a_0)\equal{}0.$
We used above the relation stated at the first step of the solution along with $ (f_{01}(a_2,\bar a_2)\equal{}f_{12}(a_0,\bar a_0)$ and $ f_{03}(a_2,\bar a_2)\equal{}\minus{}f_{23}(a_0,\bar a_0)$ which can be seen using determinants.
Consequently, one can find some $ \alpha\in\mathbb{R}$ such that $ f_{01}f_{23}\plus{}f_{03}f_{12}\equal{}\alpha f_{02}f_{13}$ and in a totally similar manner, one can find some $ \beta\in\mathbb{R}$ such that $ f_{01}f_{23}\minus{}f_{02}f_{13}\equal{}\beta f_{03}f_{12}$.
Combine the two to get $ (\alpha \minus{}1)f_{02}f_{13}\minus{}(\beta \plus{}1)f_{03}f_{12}\equal{}0$ and, furthermore, $ \alpha\equal{}1,\beta\equal{}\minus{}1$. This ends the proof.

I must say that this beautiful and surprising solution belongs to [b]Calin Popescu[/b].
_____________________________________________________________________________
[color=red]Moderator EDIT:[/color] [color=blue] Keep "Disable BBCode in this post" [b]unchecked[/b], in order to compile $ \color{blue}{\text{\LaTeX}}$ code.[/color]

One can easily solve the problem using the following [b][u]Lemma:[/u]

Let AB be a chord and P a variable point in the plane of the circle (O). If M, K and L are the feet of the perpendiculars from P onto AB and onto the tangents at A and B to the circle, then the following relation holds:

PM^2 = PK*PL ( 1 )

iff P belongs to the circle (O).[/b]

The proof comes directly from the similarity of the triangles PKM and PML: if P belongs to the circle, then the triangles are similar, from their similarity getting the relation (1).
If the relation (1) holds, then the triangles PKM and PML are similar (s.a.s., as if KA and LB meet at T, then PKTL is cyclic and, the bisectors of the angles <ATB and <KPL are parallel; PM being perpendicular onto AB is hence the angle bisector of <KPL).
But a simple angle chasing shows that these two triangles PKM and PML are similar to the triangle PAM and this compulsory brings P on the circumference ( BL is tangent to the circle (ABP), hence its center is the same with the one of (O) ).

Further, apply the lemma to (ABCD), supposed cyclic, and get (ABCDE).

Best regards,
sunken rock

I haven't read your posts carefully, but I'm afraid [b]Luis[/b] and [b]sunken rock[/b] have only proved the weaker problem: Given a cyclic quadrilateral $ ABCD$ and a point $ E$ in plane, then this point $ E$ lies on the circumcircle of $ ABCDE$ if and only if $ \mathrm{d(}E,AB\mathrm{)}\cdot\mathrm{d(}E,CD\mathrm{)} \equal{} \mathrm{d(}E,AC\mathrm{)}\cdot\mathrm{d(}E,BD\mathrm{)} \equal{} \mathrm{d(}E,AD\mathrm{)}\cdot\mathrm{d(}E,BC\mathrm{)}$.

The problem from the TST stated (and I don't see why this isn't clear from [b]Ahiles[/b]' post): Let $ A$, $ B$, $ C$, $ D$, $ E$ be five given points in plane (in general position if you like). Then they are [i]all[/i] concyclic if and only if $ \mathrm{d(}E,AB\mathrm{)}\cdot\mathrm{d(}E,CD\mathrm{)} \equal{} \mathrm{d(}E,AC\mathrm{)}\cdot\mathrm{d(}E,BD\mathrm{)} \equal{} \mathrm{d(}E,AD\mathrm{)}\cdot\mathrm{d(}E,BC\mathrm{)}$.

Actually, the part you missed (the identity implies the concyclicity of $ A$, $ B$, $ C$, $ D$) is the hardest part here, in my opinion.

I proved it with (the converse of) Ptolemy's theorem. I hope it will pass as synthetic.

Let $ k = \mathrm{d(}E,AB\mathrm{)}\cdot\mathrm{d(}E,CD\mathrm{)} = \mathrm{d(}E,AC\mathrm{)}\cdot\mathrm{d(}E,BD\mathrm{)} = \mathrm{d(}E,AD\mathrm{)}\cdot\mathrm{d(}E,BC\mathrm{)}$. The concyclicity of $ A$, $ B$, $ C$, $ D$ is equivalent with $ AC \cdot BD = AB \cdot CD + BC \cdot DA$. Multiplying this by $ k$, this rewrites as $ |EAC| \cdot |EBD| = |EAB| \cdot |ECD| + |EBC|\cdot |EDA|$, where $ |\mathcal{P}|$ is the unsigned area of the convex polygon $ \mathcal{P}$.
Expressing these areas differently, we get the new equivalent relation: $ \sin AEC \sin BED = \sin AEB \sin CED + \sin BEC \sin DEA$ (we have deleted the term $ EA \cdot EB \cdot EC \cdot ED$).
Denote $ \measuredangle AEB = x$, $ \measuredangle BEC = y$, $ \measuredangle CED = z$. In this case, the last relation can be rewritten as $ \sin (x + y) \sin (y + z) = \sin x \sin z + \sin y \sin (x + y + z)$, which can be easily verified (being true for any angles with magnitudes $ x$, $ y$, $ z$). Hence $ ABCD$ is cyclic.

Throughout a review of the proof, I've noticed that one (essential  :oops: ) thing misses, that is, the fact that our conic is indeed a circle.
I have reached that relation
$ |a_0 \minus{} a_1|\cdot|a_2 \minus{} a_3| \minus{} |a_0 \minus{} a_2|\cdot|a_1 \minus{} a_3| \plus{} |a_0 \minus{} a_3|\cdot|a_1 \minus{} a_2| \equal{} 0$
which is equivalently to the cross ratio
$ \frac {a_0 \minus{} a_1}{a_0 \minus{} a_3} : \frac {a_2 \minus{} a_1}{a_2 \minus{} a_3}$ being real and negative.
But this clearly holds. It is equivalent to the vanishing of the coefficient of $ z^2$ (and $ \bar{z}^2$ as well) in (1).
Hope everything's right now.

Use inversion with center E!! Let some invertion with center E be I.
H1,H2,H3,H4,H5,H6 are feet of perpendicular from E to AB,CD,AC,BD,AD,BC respectively.So E,A,H1,H3,H5 E,B,H1,H4,H6 E,C,H2,H3,H6 E,D,H2,H4,H5 are concyclic. Let circle whose diameter is EA EB EC ED be C1,C2,C3,C4 respectively.
Now Ci(i=1,2,3,4) move to line. We call that line li. So l1 meets l2 l3 l4 at H1' H3' H5' respectivrey.
And l2 meets l3 l4 at H6' H4', l3 meets l4 at H2'.By EH1*EH2=EH3*EH4=EH5*EH6 we can get EH1'*EH2'=EH3'*EH4'=EH5'*EH6'.
If E is miquel point for 4 lines, we can prove that equation by similarity. If 4 lines are given, we can decide only one E, so E is miquel point. Now we can prove that A', B', C', D' are coliner by simson's theorem(A' B' C' D' are feet of perpendicular from E to l1,l2, l3, l4 because EA EB EC ED are diameters).
Therefore A,B,C,D,E are concyclic.

[hide = Solution]
First, suppose that $A$, $B$, $C$, $D$, and $E$ lie on a circle with radius $R$. Then
\[\text{d}(E, AB) = \frac{2[EAB]}{AB} = \frac{EA \cdot EB \cdot \sin \angle AEB}{AB} = \frac{EA \cdot EB}{2R}\]
and similar equalities, so
\[\text{d}(E, AB)\text{d}(E, CD) = \text{d}(E, AC)\text{d}(E, BD) = \text{d}(E, AD)\text{d}(E, BC) = \frac{EA \cdot EB \cdot EC \cdot ED}{4R^2}\]
as desired.
-----
Now, suppose that $\text{d}(E, AB)\text{d}(E, CD) = \text{d}(E, AC)\text{d}(E, BD) = \text{d}(E, AD)\text{d}(E, BC)$ holds, and suppose that they are all equal to $x$. Then, we see that
\begin{eqnarray*}
4[EAB][ECD] & = & AB \cdot CD \cdot x,\\
4[EAC][EDB] & = & AC \cdot DB \cdot x,\\
4[EAD][EBC] & = & AD \cdot BC \cdot x.
\end{eqnarray*}

[b]Claim.[/b] $[EAB][ECD] + [EAC][EDB] + [EAD][EBC] = 0$, where areas are signed.

[i]Proof.[/i] We use Cartesian coordinates with $E = (0, 0)$, $A = (x_a, y_a)$, and similarly for $B$, $C$, and $D$. Without loss of generality, assume that none of the $y$-coordinates of $A$, $B$, $C$, or $D$ are zero. Then,
\[\frac{4[PAB][PCD]}{y_ay_by_cy_d} = \frac{4 \cdot \frac{x_ay_b - x_by_a}{2} \cdot \frac{x_cy_d - x_dy_c}{2}}{y_ay_by_cy_d} = (r_a - r_b)(r_c - r_d),\]
where $r_a = \tfrac{x_a}{y_a}$, and $r_b$, $r_c$, and $r_d$ are similarly defined. From similar equations, we see that
\begin{eqnarray*}
& & \frac{4}{y_ay_by_cy_d}([PAB][PCD] + [PAC][PDB] + [PAD][PBC])\\
& = & (r_a - r_b)(r_c - r_d) + (r_a - r_c)(r_d - r_b) + (r_a - r_d)(r_b - r_c)\\
& = & 0
\end{eqnarray*}
as desired.

From the claim, we see that $\pm AB \cdot CD \pm AC \cdot DB \pm AD \cdot BC = 0$ for some combination of signs; by the converse of Ptolemy's theorem, we see that $A$, $B$, $C$, and $D$ lie on a circle $\omega$.

Call a point $P$ [i]good[/i] if $\text{d}(P, AB)\text{d}(P, CD) = \text{d}(P, AD)\text{d}(P, BC)$. Observe that by the first part of the proof, all points on $\omega$ are good, and that any point $P$ on line $AD$ different from $A$ or $D$ is not good.

For a contradiction, suppose that $E$ does not lie on $\omega$. Then, $E$ must be good.

Define $f(P) = \text{d}(P, AB)\text{d}(P, CD) - \text{d}(P, AD)\text{d}(P, BC)$ for all points $P$, where distances are signed. It is easy to see that $f$ is quadratic on every line $\ell$, so if $\ell$ contains three good points, then all points on $\ell$ are good.

Let $\ell$ be a line through $E$ which intersects $\omega$ twice. Then, by the above paragraph, since three points on $\ell$ are good, all points on $\ell$ are good.

If $E$ lies inside $\omega$, then from the above paragraph we get that all points are good, a contradiction. Suppose that $E$ lies outside $\omega$; then taking the line through $E$ and the center of $\omega$, we see that $E'$, the inverse of $E$ wrt $\omega$, is good and lies inside $\omega$, a contradiction.

Thus, $E$ must lie on $\omega$ as well, so $ABCDE$ is cyclic, as desired.
[/hide]


How many more diagonals are in a octagon than in a hexagon?

[hide]The answer is 11.

 To find the number of diagonals in a figure, you need to know the number of sides. When you draw a diagonal, you connect a set of 2 points. To find the number of diagonals in an octagon, do 8 take 2. 8 * 7/ 2 * 1. This gives you 28, but there are 8 lines that you can draw that are not diagonals, so you subtract 8. This works for any amount of sides. You have the # of sides take 2, and subtract the # of sides. A hexagon has (6 take 2) - 6 diagonals, or 9 diagonals.[/hide]

[hide="to 4everwise"]
you posted the answer twice in two different ways[/hide]


rotation
LaTeX
geometry unsolved
geometry

I have the solution but it's a lot of random stuff.

1. From point A, draw an arc with radius AC, and from point B, draw an arc with radius BC. Find the other point of intersection (other than C), and label it E.
2. From point A, draw an arc with radius AD, and from point B, draw an arc with radius BD. Find the other point of intersection (other than D), and label it F.
3. Draw an arc from point F with radius CE.
4. Draw an arc from point C with radius EF.
5. Where the arcs from steps 3 and 4 intersect, label it G.
6. Find DF*CD/DG. This will be x.
7. Draw two circles with radius x centered on D and F.

The point of intersection will be one of the intersections of these circles, or, if they intersect only once, it will be that point only.

This is just what I think... too messy though.

"6. Find DF*CD/DG. This will be x."

how can i draw x segment only with compass ?  :(

Oh, does this mean that you cannot use a ruler to measure the lengths? I thought that you could calculate the answer by measuring the lengths and set it equal to x. Correct me if I'm wrong, but if you cannot measure the lengths, I believe that there is virtually no way to do this... :huh:

As for drawing x, you just need to perform the calculation to determine x and use it to draw two circles with radius x (explained in the next step). What I noted was not a segment, but a value for x. You will use this value in drawing the next circles. You are not drawing a segment of length x at any time. Understood? :)

Find DF*CD/DG. This will be x.

i only know the length  nDF or nCD or nDG on a circle,  $ n \in N$

 :(

Ok, how about this:

On a separate sheet, create a point O. From point O, create two different circles of radii DG and DF. Pick a point on the outer circle, label it N. Draw an arc with radius CD from point N, and pick one of the intersection points with the outer circle to be M. Now, with M and N as centers, draw two circles with arbitrary radius y > |DG - DF|. The two points of intersection with the [i]inner circle [/i] are points R and S (make sure that you make the arcs in the same direction and use only the first intersection point; for example, only rotate the arc clockwise and use the first point of intersection for each arc). The length RS is the value for x.

I hope that this is clear, because it looks terribly messy... Sorry :oops: Is this what you where asking for? In this case, you will not need a ruler to measure length. Hope this works for you!

On another note, where is this problem from? It looks like it would take very long to solve it with only a compass...  :D

thanks  :lol:

http://www.cut-the-knot.org/geometry.shtml

it's 198 item; the problem is solved there but i didn't understand that resolution for 3 years because x, so you make me understand construct x length only with compass.

thanks  :D

$ \mathbb{N}$

sorry i wrote N for odd numbers symbol but i see now in latex is \mathbb{N}  = |N  ? no.

i try $ \mathbb{N}$          \mathds{N} = |N ? no. Oh i don't know this comand to |N.

 :D


If $F_{n}$ is the $n$th Fibonacci number ($1, 1, 2, 3, 5, 8, 13, 21,...$):

a.) Prove that the series

$\sum_{n = 1}^{\infty}{\frac{1}{F_{n}}}$

converges, and

b.) Find the value of the above series.

[hide="a"] Trivial by comparison to, say, $\varphi^{n}$. [/hide]

Edit:  According to Wikipedia (and an old post I remember from ZetaX), the sum does not have a known closed form.

indeed this sum is known to be irrational

have anyone seen the proof ?
I searched for it but I didn't find :(

This question is probably really basic but doesn't every series converge i mean if you have the sum $ \sum_{n \equal{} 1}^{\infty}{x}$
then $ \displaystyle\lim_{n\to\infty}{x} \equal{} \infty$ then say instead of $ x$ you have $ \frac{1}{x}$ or whatever then $ \sum_{n \equal{} 1}^{\infty}{\frac {1}{x}}$ then \displaystyle\lim_{n\to\infty}{\frac{1}{x}} = \2 but what i mean to say is i can't think of a series that doesn't converge please help

Example: $ \sum_{n\equal{}1}^{\infty}n$ and $ \sum_{n\equal{}1}^{\infty}\frac{1}{n}$, both approach $ \infty$.  :wink:
(Or better yet, $ \sum_{n\equal{}1}^{\infty}1$ and $ \sum_{n\equal{}1}^{\infty}\frac{1}{1}$.)

Infinity is not an actual boundary, therefore saying $ \lim_{x\to\infty} \sum^{\infty}_{k \equal{} m} a_{k} \equal{} \infty$ is equivalent to saying that the series is limitless.

Of course, that depends on your notion of "convergence."  The standard definition takes $ \lim_{n \to \infty} \sum_{k\equal{}1}^{n} a_k \equal{} \infty$ to mean that a sum diverges.  Even if we include this in our notion of convergence, we can easily construct a sum whose partial sums don't have a limit, say, $ \sum (\minus{}1)^k$, or sums whose partial sums neither have a limit nor a bound, say, $ \sum (\minus{}2)^k$.  See the discussion [url=http://en.wikipedia.org/wiki/Divergent_series]on the Wikipedia page[/url].


rotation
integration
calculus
geometry
geometric transformation
calculus computations

Hello everyone,

I could not arrive at the correct answer for the following question and I have provided a self-created diagram below.

Thank you for your help.

---

18. Find the volume of the solid obtained by rotating the region bounded by $ y \equal{} x, y \equal{} 0, x \equal{} 2, x \equal{} 4$; about $ x \equal{} 1$

---

[img]http://i41.tinypic.com/35ivqx4.png[/img]

---

I separated the bounded region into two parts.

For the top portion, my Washer formula is $ \pi\int_2^4 (9 \minus{} y^2) dy \equal{} \minus{} \frac {2}{3}$.

For the bottom, it is $ \pi\int_0^2 (3^2 \minus{} 1^2) dy \equal{} 16$.

However, these two definite integrals added  do not give the correct volume, which is

[hide]$ \frac {76\pi}{3}$[/hide].

The problem is in your integral for the upper portion. The outer radius is given by $ x \equal{} 4,$ which is $ 3$ units away from the axis of rotation. You have that much correct. But the inner radius is given by the distance from $ x \equal{} y$ to $ x \equal{} 1,$ which is $ y \minus{} 1.$ (Not $ y,$ which is what you have.)

So that portion of the integral should be $ \pi\int_2^4\left(9 \minus{} (y \minus{} 1)^2\right)\,dy.$

Try that.

Thanks for your prompt reply.

I computed $ \pi\int_2^4\left(9 \minus{} (y \minus{} 1)^2\right)\,dy \equal{} \pi \frac {80}{3}$.

If my second part is correct, then $ \pi(\frac {80}{3} \plus{} 16) \equal{} \pi\frac {128}{3}$ which unfortunately is not the right answer.

Would there be a problem with my second integral?

Really? Calculate that again.

$ \int_2^4\left(9\minus{}(y\minus{}1)^2\right)\,dy\equal{}\int_1^3(9\minus{}u^2)\,du\equal{}\frac{28}3.$

Definitely not $ \frac{80}3.$

Thank you for your reply.

I fixed my calculation error.

However, why would the inner radius be $ y \minus{} 1$?

Is this because the rotational axis is $ x \equal{} 1$, so one unit of the graph $ x \equal{} y$ is "lost"?

The distance to the axis of rotation is $ x \minus{} 1.$ On this particular curve, $ x \equal{} y,$ so we make that substitution.

The volume can also be calculated by the "shell" method. May I assume that you've already done that? (As in perhaps that's your independent source for the answer?)


Given the system of equations

0.4sin(x) + cos(y) = -0.2
0.8sin(x)  - cos(y) = 1.1

Which value of y can be the solution?

If you combine the two, you get
$ 1.2\sin{x}\equal{}.9\Rightarrow x\equal{}\sin^{\minus{}1}(.75)$

Plug this into the first equation to get
$ .4(.75)\plus{}\cos{y}\equal{}\minus{}.2\Rightarrow\cos{y}\equal{}\minus{}.5\Rightarrow y\equal{}\cos^{\minus{}1}(.5)\equal{}\frac{2\pi}{3},\frac{4\pi}{3}$

The second equation will yield the same thing.

$ .4sin(x)\plus{}cos(y)\equal{}\minus{}.2$

$ .8sin(x)\minus{}cos(y)\equal{}1.1$

Add the two equations to get

$ 1.2sin(x)\equal{}.9$

$ sin(x)\equal{}\frac{.9}{1.2}$

$ x\equal{}sin^{\minus{}1}(\frac{.9}{1.2})$

$ x\approx 48.59$

Substitute

$ .4sin(sin^{\minus{}1}(\frac{.9}{1.2})\plus{}cos(y)\equal{}\minus{}.2$

$ .3\plus{}cos(y)\equal{}\minus{}.2$

$ cos(y)\equal{}\minus{}.5$

$ y\equal{}cos^{\minus{}1}(.5)$

$ y\equal{}120$

$ x\equal{}sin^{\minus{}1}(\frac{.9}{1.2})$

[quote="JRav"]First, note that I already solved the problem.  Second, your answer is not entirely correct since you are neglecting the second solution of $ y$.[/quote]

Sorry, don't mind my response JRav has everything you need.

JRav ketvoman's solution was different.And it's supposed to be appreciated.

[quote]JRav ketvoman's solution was different.[/quote]

Actually I'm pretty sure it's exactly the same.

JRavs right its the same but I initially didn't mean to copy his answer.  But
I still don't understand how you got two answers.

Actually there aren't just two there are infinitely many, there are two in the interval $ [0, 2\pi )$

This is because $ \cos$ is negative Quadrant I and IV

So the values would actually be $ y \equal{} \frac {2\pi}{3}, \frac {4\pi}{3} \plus{} 2k\pi$

[quote]Yeah...for these trig problems, people should really set a range, or the number of solutions is just going to be infinite =\
[/quote]
Actually, you'd want to specify a domain. :lol:

I am sorry 4 blaming u Jrav :blush: . didn't notice that much and the answers were different(u got an ans. more).anyway nice solution.

[quote="SnipedYou"]This is because $ \cos$ is negative Quadrant I and IV[/quote]

[url=http://mathworld.wolfram.com/Quadrant.html]Quadrants[/url] I and IV contain the [i]positive[/i] values of $ \cos \theta$, when you graph the unit circle (in a ‘real’ plane); I believe you mean Quadrants II and III.


Do you do it? Do you have videos on it? If you do, are the instructors Rodney Yee or Patricia Walden? They're great teachers.

Nope. The only "inner peace" I find is by doing AoPS math.

I HATE YOGA. I DO DO IT THOUGH(DRY LAND.) Hmmm....why yoga when you could run in circles, eh?  :blush:


trigonometry
limit
calculus
calculus computations

\[ \begin{gathered} {\text{if: }} \hfill \\
f\left( x \right) \equal{} \left\{ {\begin{array}{*{20}c} {x^2 \sin \left( {\frac {1} {x}} \right)} \\
{0{\text{ if }}x \equal{} 0} \\
\end{array} {\text{ if }}x \ne 0} \right. \hfill \\
{\text{Compute }}f'\left( 0 \right) \hfill \\
\end{gathered}\]
halp

The definition of differential coefficient : $ f'(0)\equal{}\lim_{x\rightarrow 0} \frac{f(x)\minus{}f(0)}{x\minus{}0}$.

SOLUTION:

 $ f'(0) \equal{} \lim_{x\rightarrow 0} \frac {f(x) \minus{} f(0)}{x \minus{} 0}$

As per the defination,

 $ f'(0) \equal{} \lim_{x\rightarrow 0} \frac {x^2sin(\frac{1}{x})\minus{}0}{x}$

$ f'(0) \equal{} \lim_{x\rightarrow 0} xsin(\frac{1}{x})$

$ f'(0) \equal{}0$

[quote="Kent Merryfield"]What is this a classic example (counterexample) of?

[hide="Answer"]If $ f$ is differentiable everywhere, then $ f'$ doesn't have to be continuous.[/hide][/quote]

Could you elaborate a little more on that sir?

What is there to elaborate on? The only detail to check is that $ f'(x)$ isn't continuous, and you should be able to do that yourself.


number theory
prime numbers
relatively prime

[quote="DaReaper"]Well, the question was "HOw many postitive prime numbers are less than $75$ ?"  , so...i just counted and got the answer, 21.
And the answer was right. But that counting took some time.

I was just wondering if there was an easier way to do this..like a formula or something..

* Also, why do i have "Hodge Conjecture" under my name?*[/quote]
I don't know of any easier way than counting. And you have Hodge Conjecture because you have reached the 50 post mark. At 100 posts you get Poincare Cinjecure. Then at 500 you get Yangmills Theory, or something like that.

[quote="DaReaper"]ohh...and the stars?[/quote]
That just tells you what kind of ranking you are. I don't know what else to call it. You have one star for being a Hodge Conjecture and so forth.

[quote="DaReaper"]So do you think there is an easy way to count those?[/quote]
Well, I don't see another way to count them.  Even though primes have the pattern of $6n\pm1$ (except for 2 and 3), primes are rather unpredictable.  I would probably count them.

[quote="DaReaper"]Well, the question was "HOw many postitive prime numbers are less than $75$ ?"  , so...i just counted and got the answer, 21.
And the answer was right. But that counting took some time.

I was just wondering if there was an easier way to do this..like a formula or something..

* Also, why do i have "Hodge Conjecture" under my name?*[/quote]

Now, to actually answer the question. Yes, you do have to count them. There is no "easy way" to do this. You could find all positive integers smaller than $75$ that are [b]relatively prime[/b] to $75$, which is $\phi(75)=40$ (I think).


Given a triangle ABC, let $ D$ and $ E$ be the points on $ BC$ such that $ BD \equal{} DE \equal{} EC$. A line $ p$ intersects $ AB,AD,AE,AC$ at $ K,L,M,N$, respectively. Show that $ KN \ge 3LM$

Denote $ x\equal{}KL, y\equal{}LM, z\equal{}MN$. We have to prove that $ x\plus{}z\geq 2y$
Since $ (K,N,L,M)\equal{}(B,C,D,E)$ we have:
$ \frac{x}{y\plus{}z}: \frac{x\plus{}y}{z}\equal{}\frac{\overline{BD}}{\overline{DC}}: \frac{\overline{BE}}{\overline{EC}}\equal{}\frac{1}{4}$
$ \Rightarrow 4xz\equal{}(x\plus{}y)(y\plus{}z)$
If $ y>\frac{x\plus{}z}{2}$ then $ x\plus{}y>\frac{3x}{2}\plus{}\frac{z}{2}\geq 2\sqrt[4]{x^3z}$
Similarly, $ y\plus{}z>2\sqrt[4]{xz^3}$ then $ (x\plus{}y)(y\plus{}z)>4xz$ which is a contradiction. So $ y\leq \frac{x\plus{}z}{2}$


For $ a,b,c$ positive reals prove that

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

$ < \equal{} > a^4 \plus{} b^4 \plus{} c^4 \minus{} b^2c^2 \minus{} c^2a^2 \minus{} a^2b^2 \geq 0$
BQ

Use

$ \sum \frac{ab(a\plus{}b)}{c^2\plus{}ab} \plus{} \sum \frac{c^2(a\plus{}b)}{c^2\plus{}ab} \equal{} 2\sum a$

and our inequality becomes

$ \sum \frac{c^2(a\plus{}b)}{(c^2\plus{}ab)} \leq \sum a$

but

$ \sum \frac{c^2(a\plus{}b)}{(c^2\plus{}ab)}\equal{}\sum \frac{c^2(a\plus{}b)^2}{(c^2\plus{}ab)(a\plus{}b)}\equal{} \sum \frac{(ca\plus{}cb)^2}{a(b^2\plus{}c^2) \plus{}b(a^2\plus{}c^2)} \leq \sum \frac{c^2a^2}{a(b^2\plus{}c^2)} \plus{} \sum \frac{c^2b^2}{b(a^2\plus{}c^2)} \equal{} \sum a$ :)


LaTeX
trigonometry
inequalities unsolved
inequalities

we have x,y<60° and x+y<90°
prove that:

1/sin(60°-x)+1/sin(60°-y)>=(4sinx+4siny)/[sin(60°+x)*sin(60°+y)]

i would appreciate if someone switched this in LaTeX

thanks... :)

and now someone a solution :?:

can anyone please help? arqady, moderators? :(


modular arithmetic
AoPSwiki
number theory
AoPS Books

In aops 1, it says 12 is congruent to 7 mod 5.

But I think 12 = 2 mod 5. What is the difference between = and congruent in this situation?

x = y ( mod z) where y is remainder, I think is for equals.

Like 98 = 3 mod 5

What is the difference between that and congruent?

I just know that congruent has 3 lines instead of two for equals..

I need serious help.  :o

okay

$12 \equiv 2 \pmod 5$ and $12 \equiv 7 \pmod 5$ are both correct answers

because if for the second one, $12-7$ is divisible by $5$.

but why? If both signs are the same I don't see how 12 is 7 mod 5.

12 divided by 5 has remainder 2.

One thing to note: 12 divided by 5 has remainder 2, but 7 divided by 5 also has remainder 2 ;)

To check if a congruence is true in the form
$a \equiv b \pmod{c}$ what you do is subtract b from a, and then see if it is divisible by c.

So to check $12 \equiv 7 \pmod{5}$ you subtract $7$ from $12$ which leaves us with $5$, this divides by $5$ evenly so it is true that $12 \equiv 7 \pmod{5}$.

Sadly this is about as far as my modular arithmetic knowledge goes.. the guys in the intermediate forum really know their stuff when it comes to mods eguegheg. I really need to get those AoPS books cause I cant find anywhere on the net to learn about mods really (used AoPSWiKi already :P)

so the three lines which means congruent is a lot different.

woah I'm going to have to look more into chapter 5. IT's giving me the heebie jeebies . * shudders *

[quote="now a ranger"]woah I'm going to have to look more into chapter 5. IT's giving me the heebie jeebies . * shudders *[/quote]

i don't think modular arithmetic is part of algebra II...so you can skip that and look into something else for  your competition  :)

[quote="i_like_pie"][quote="7h3.D3m0n.117"][quote="buzzer11"]isnt he asking the difference between $\equiv$ and $=$?
At least, that's what I thought...[/quote]

you can do all operations to both sides wit both types...its just one is used for modular arithmetic and one isn't  :wink:[/quote]
You can't divide in mods, unless the divisor has no common factors with the remainder.[/quote]

$ac\equiv bc\pmod{n}$

IMPLIES

$a\equiv b\pmod{\frac{n}{(n,c)}}$

[quote="vishalarul"]
 \[ac\equiv bc\pmod{n}\implies a\equiv b\pmod{\frac{n}{(n,c)}}\][/quote]

Right. For those who don't know the notation, $(n,c) = \gcd( n,c)$.

[b]Exercise[/b]: Prove this.

[b]Examples[/b]:

Solve these congruences:

$(1): \ \ 15a \equiv 5 \bmod 20$

$(2): \ \ 17b \equiv 55 \bmod 100$

$(3): \ \ 4^{c}\equiv 64 \bmod 77$

[hide="Answer to Original Question"]$\equiv$ is used in moduluses while $=$ is used in normal stuff.[/hide]


linear algebra
matrix
linear algebra unsolved

Prove that any real matrix $A$ of order $n$ and determinant 1 or -1 can be written as a product of $2n+9$ complex involutions. Can we find a bound independent of $n$?

just a remark:

suppose that one knows that in $M_2(\mathbb{R})$, every diagonal matrix with determinant $1$ is the product of $C$ symetries. Then for every $2 \times 2$ invertible diagonal matrix $D=(a,b)$, one can find $C$ symetries $S_1,S_2,..,S_C$ such that $S_1S_2..S_CD=diag(1,ab)$.

Now, if $M \M_n(\mathbb{R})$ is a matrix with determinant $+1$. Then $M=OS$ where $O$ is orthogonal direct and $S$ positive definite. Then $O$ is the product of $2$ symetries, and one can suppose $S$ diagonal. Now, one just have to decompose $S=diag(a_1,..,a_n)$, $a_i>0$ in a product of symetries. With the above assumption, I think on can show that $S$ is a product of at most $C([Log_2(n)]+1)$ symetries. ($[x]$ is integer part of $x$)

Hence one only need $C([Log_2(n)]+1) + 2$ symetries for $det(M)=1$. But what is this $C$ ? ($C=2$ ?)

back to work guys ! I have just read that Halmos & Co had proved that $4$ involutions suffices ..


Hello friends,

The event IOPC 2006 is now over. So, now I hope u can answer me. Though there were many interesting problems in the event (and i will post em all, stay assured) this was the one I found interesting in particular. The problem bears the name [b]Fermat and Gauss [/b]and here it is word by word (copied and pasted)...

[color=black][b]Problem

In the mansion of Mathematicus Supremus in heaven live all the great mathematicians of yesteryear's. However, even after losing their mortal self, these mathematicians have not stopped doing mathematics. They always come up with puzzles and riddles and baffle their fellow theoreticians with mind boggling questions. Gauss and Fermat happen to be among the most enthusiastic of these problem posers. Both are always trying to outsmart the other to establish their supremacy in the group.

The mathematicians of this mansion dine on a large round table. A number of chairs are arranged around the table. One night, Gauss and Fermat happened to reach for the dinner quite early. Losing no opportunity to show his supremacy, Fermat said: ``Hey Gauss, there is some time left. Why not play a game?''. Gauss, as always ready to outsmart Fermat, readily agreed. Fermat then said: ``Okay Gauss, the game will go like this. You will first sit on a chair. Then you will call out a number  between 1 to , where  is the number of chairs. I will then call out either clockwise or anticlockwise. If I call out clockwise, then you sit on the th chair in the clockwise sense from the chair you are presently sitting on, and if I call out anticlockwise you sit on the th chair in the anticlockwise sense. You will again choose a number, and I will again call out whether to move clockwise or anticlockwise, and so on. In this way, you have to sit on the maximum possible number of chairs''. Gauss replied: ``Hey Fermat, this is damn easy. I am sure that even any school kid will be able to play this game quite easily!''.

Input
The input consists of several lines. Each line has a number , the number of chairs around the dining table. The input is terminated by the case , which is not to be processed.  will always fit in a signed 64 bit integer representation.

Output
For each input , print the maximum number of chairs that Gauss can sit on. It should be assumed that Gauss and Fermat use their intellectual ability to the maximum possible extent.

Sample Input
4
6
12
0

Sample Output
4
4
8[/b][/color]

Please post ur solutions. I am relly dying to see them. Bye... :)

Post ur solutions fast (plzzzzz).

Well a few corrections. Number of chairs is between [b]1[/b] to [b]n[/b]. Also, the number which Gauss calls out is [b]k[/b] and [b]k[/b] lies b/w [b]1[/b] and [b]n[/b].

Well these are a few things which were dropped accidentally during the process of copy and paste. A few more mistakes may have crept in. Please bring them to my notice.

Good bye and Have a Good day...


integration
logarithms
symmetry
real analysis
real analysis solved

That's exactly what you need, and enough to solve the problem with the fact that $\frac{x^k}{q(x)}>0.$ $A(k)$ is defined for $-1<k<7$, and $A(k)=A(6-k)$ by the substitution $y=\frac1x,$ Also, $A''(k)=\int_0^\infty\frac{x^k\ln^2x}{q(x)}\, dx>0,$ so $A$ is convex. By this convexity and the symmetry of $A$, $A$ takes its minimum at $k=3.$


Hi.

I need help solving this one. I had be glad for a nice organized solution if possible, since this one is kinda hard.

Find conditions on c1...c2...cn so the following matrix will be diagnoizable.

[url=http://img515.imageshack.us/my.php?image=helpgi2.jpg][img]http://img515.imageshack.us/img515/9591/helpgi2.th.jpg[/img][/url]

I know that if $A$ is diagnoizable so is $A^{2}$.

But it doesnt work the other way around, so how would it help me to "look" at $A^{2}$ ?

And I know that $A^{2}$ is a diagonal matrix.

there is a kind of converse ! for example the converse is true if $A$ is invertible and we are working over $\mathbb{C}$ ..

I'd rearrange rows and columns to make $A$ block diagonal with about $\frac{n}{2}$ $2\times 2$ blocks. The problem reduces to whether each of those blocks is diagonalizable.
The condition is that $c_{k}c_{n+1-k}$ is a square for each $k$, and that $c_{n-k}$ is zero whenever $c_{k}$ is. To say anything more, you need to specify the field.


In a triangle abc , a point O exists such that angle OAB = OBC = OCA = x
Find the the value of cot(x) in the trig. ratios of angles A,B and C.

hello, or see here:
http://mathworld.wolfram.com/BrocardAngle.html
Sonnhard.

you could have just proved it you know.
Though i didnt know it was a standard thingy, thanks for the info.


ordinate

The tangent at pt P on the ellipse cuts the major axis at Q and directrix at R. If pt is the ORDINATE OF p ??

abscissa

The normal at an end of P of the latus rectum of the ellipse \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 cuts the y-axis at M, and PN is the ABSCISSA OF P

ordinate is the y coordinate

abscissa is the x coordinate


the number of parcels $a_ia_{i+1}a_{i+2}a_{i+3}$ with value $1$ is the same number of parcels with value $-1$. so, this implies that $n$ is even (the number of parcels is $n$).
so, we have $n/2$ parcels with value $1$ and $n/2$ with value $-1$.
see the product S of all parcels.
$1 = (a_1a_2..a_n)^4 = S = 1^{n/2} . (-1)^{n/2} = (-1)^{n/2}$.
so, $n/2$ is even, and the problem is solved.

[quote="riemann abdellah"]this is great solution[/quote]

thanks  :)


probability
calculus
integration
calculus computations

For fixed $\lambda>0,$ evaluate:

$e^{-\lambda}\sum_{k=0}^{\infty}\frac{k\lambda^k}{k!}$

and

$e^{-\lambda}\sum_{k=0}^{\infty}\frac{k^2\lambda^k}{k!}$

[quote="Kent Merryfield"]For fixed $\lambda>0,$ evaluate:

$e^{-\lambda}\sum_{k=0}^{\infty}\frac{k\lambda^k}{k!}$
[/quote]

[hide] Just differentiate the expansion of exp(x) to get the sum as $\lambda e^\lambda$, hence the whole expression is just $\lambda$[/hide]
[quote]
and

$e^{-\lambda}\sum_{k=0}^{\infty}\frac{k^2\lambda^k}{k!}$[/quote]

[hide]
Similar idea. The sum evaluates to $\lambda e^\lambda (\lambda+1)$ hence the expression is $\lambda  (\lambda+1)$ I have definatly seen these in a first year probabilty course, although the name of the distribution I cannot remember. My guess is we have just calculated the mean and variance, howver![/hide]

[quote="Kent Merryfield"][quote="dheardie"]I have definatly seen these in a first year probabilty course, although the name of the distribution I cannot remember. My guess is we have just calculated the mean and variance, howver![/quote]
It's called the Poisson distribution, and it's a very important one. It's the right distribution when the following is true:

We are counting the number of times something happens, so that our random variable takes on nonnegative integral values. We know a [i]rate[/i] of occurrance: so many cosmic rays pass through the counter per minute; so many raindrops during the sprinkle fall per square centimeter; so many customers per hour enter the store. We observe for a set time or set space. We know the mean, and that's all we know.

As for what we just calcululated - we found $E(Y)$ and $E(Y^2).$ That's the mean and enough information to determine the variance, but the variance itself is $E(Y^2)-E(Y)^2 = \lambda^2+\lambda-\lambda^2=\lambda.$[/quote]

Oh I rememeber it now! Yes I had almost the exact questions on the exam aswell, then the problem was something about the probabilty of cars passing on a highway...

What fun!


solve in $ \mathbb{R}$ the system :

$ 1 \minus{} \frac {12}{y \plus{} 3x} \equal{} \frac {2}{\sqrt {x}}$
$ 1 \plus{} \frac {12}{y \plus{} 3x} \equal{} \frac {6}{\sqrt {y}}$


I think the following is true, but I cannot find a proof (any proofs/counterexamples are welcome):

Let $a_1,...,a_n,b_1,...,b_n,c_1,...,c_n,d_1,...,d_n$ be nonnegative reals which satisfy the following:

$\sum_{i=1} ^n a_i = \sum_{i=1} ^n c_i$
$\sum_{i=1} ^n b_i = \sum_{i=1} ^n d_i$

$a_1b_1<a_2b_2<...<a_nb_n$
$c_1d_1<c_2d_2<...<c_nd_n$

and for all $i \in [1,n]$:

$a_ib_i=c_id_i$

Prove/disprove that it implies $a_i=c_i$ and $b_i=d_i$ for all $i$.


We divide a circle with three straight cuts into 7 regions. Is it possible that these 7 regions have equal areas?

Your source says it, this topic is too difficult for Intermediate Topics.

Please move this topic to at least Olympiad Section.

If there is a problem, please read this subforum's announcement.

Well I have a really ugly solution, I don't want to go through the computations here because well they are discusting, but heres a general outline:

[hide]

Note that the only way to get 7 regions from three lines is to have them make a triangle in the middle and have 6 regions eminating from the edges and vertices.

And because of this, on either side of each line there will be 3/7 of the area on one side and 4/7 on the other.   However also note that if 3/7 of the area is on one side of a chord that fixes the distance the chord is from the center.

Ok now consider the region that eminates from the a vertex of the inner triangle (I don't know how else to help visualize it).  Anyway, if we hold one chord in place and rotate one of the others from intersecting it on the circle to coinciding with it the area of this region increases from 0 to 3/7 continuously and increasing, so it hits 1/7 at one place, corresponding to one angle.

So only one angle between the chords will produce a region of the proper area.  But by symmetry this angle must be the same for the other vertices, so they must all be 60 degrees.

Ok now the center triangle is equilateral with area pi/7 and from here you can find measurements and voila you get a contradiction.... Just after some ugly computations.[/hide]


Given positive numbers $a_1$ and $b_1$, consider the sequences defined by
\[a_{n+1}=a_n+\frac{1}{b_n},\quad b_{n+1}=b_n+\frac{1}{a_n}\quad (n \ge 1)\]
Prove that $a_{25}+b_{25} \geq 10\sqrt{2}$.

[hide]We have $ a_2b_2 \equal{} 2 \plus{} a_1b_1 \plus{} \dfrac{1}{a_1b_1}\geq2 \plus{} 2 \equal{} 4$ by AM-GM and $ a_{n \plus{} 1}b_{n \plus{} 1} \equal{} 2 \plus{} a_nb_n \plus{} \dfrac{1}{a_nb_n}>2 \plus{} a_nb_n$.  So $ a_ib_i> 2i$ by induction and thus $ a_{25}b_{25}> 50$.  So by AM-GM
\[ a_{25} \plus{} b_{25}\geq 2\sqrt {a_{25}b_{25}}>2\sqrt {50} \equal{} 10\sqrt {2}
\]
as desired.[/hide]


find  all  positive integers     s >o,t>0   such that
$ 2^{2t \plus{} 3} \minus{} 3^{2s \plus{} 1} \equal{} 5$

if positive integer   s>1,  then $ 3^{2s\plus{}1} \plus{}5$  has a  odd  prime divisor?

the problem has been solved:
http://www.aoshoo.com/bbs1/dispbbs.asp?boardid=59&Id=13589
in chinese


Put down 12  identical circular coins on the table.

The coins may touch each other but they cannot

cover one another even partially.

Prove that the number of mutual points is less than 36.

[hide="Each coin..."]
...can touch at most 6 others. $6\times12=72$, but we have double counted, so $\leq36$ points where they touch. Clearly it's also strictly less than 36, since the northernmost (or equal northernmost) coin cannot be touching 6 others.

[/hide]


Excuse me, does anyone have the problems from the TST in USA 2004? Thanks a lot.

Why don't anyone reply me?Do you have???

[quote="nhat"]Katrina, if you want USA TST 2004 ,you should search the name 'Raducici' (i don't remember exactly) (the Mathlinker) and search all posts by him.You can find 6 problems of USA TST 2004 in his post :)  :)[/quote]

I can't find it.Can you help me?

[quote="Lovasz"]You should find "USA TST 2004" in http://www.unl.edu/amc/index.html. Good luck :)[/quote]

It hasn't got TST2004  :(


I was looking at the CFA website and I was wondering if an accounting major can take it after graduating? Is the CFA exam relatively similar to the CPA, is it required for the Wall Street Firms (not sure I will be there but curious), how difficult is it? So in general I was curious if it is people who do Wall Street type jobs that do the test. I also see there are actually 3 exams. So I guess what I am asking is, can anyone explain this whole test in more detail and specifically, what it can do for you?

As far as I know, the Chartered Financial Analyst (CFA) designation is open to just about anyone who can pass the test (you may need a bachelor's degree and possibly some experience as well - check their website for these details.)  The exam is fairly tough - I've known some Finance professors at good schools who've taken the exam more than once - after they became professors.

As far as "certifications" go, the CFA is about as good as it gets in the world of financial analysis.  Some university finance programs are designed to prepare students for this credential.  You'll need a solid foundation of mathematics, economics, accounting and finance (especially investments) coursework to prepare for the exam.

Don't confuse the CFA with the Certified Financial Planner (CFP) designation.   The CFA is much more difficult to attain and requires higher-level mathematical thinking.

Based on some conversations I've had with one of the question writers for the CFA, the organization puts a lot of work into creating and grading the exam.  The credential is not earned by very many folks and gets a fair amount of respect in many financial and investment circles.

The CFA and CFP are different from the Certified Professional Accountant (CPA) in at least one aspect - the CPA is a legal requirement to audit publicly traded companies while the CFA and/or CFP are not legally required for investment analysis or planning.  CPA Accountants are not typically required to have anywhere near the depth of investments knowledge required of a CFA.   Likewise, CFA's aren't required to have the depth of auditing knowledge possessed by CPA's.  Both should have a good grasp of managerial/cost and financial accounting.

Oh, so the CFA is not a requirement, its more like a bonus, and not too many people take it.

Not that it matters, but CPA is certified public accountant instead of certified professional accountant.

So approximatly how much % of finance people take the test and is just people that do investing, like other finance people, do they also take the test? For example, a CFO and others?

CFAs may be required by some employers but there are no legal requirements to use CFAs in the investment world.

You're right about the "public" in CPA (I must be getting too old or too careless.)

Public accounting usually refers to the external auditors of publicly traded companies.

There are probably less that 50,000 active CFA's worldwide at this time.  In the last few years approximately 7,000 new CFA's are minted annually.

For more information: [url=http://www.cfainstitute.org/index.html]CFA Institute[/url]


calculus
integration
function
real analysis
real analysis unsolved

Let $f: [0,1]\to [0,1]$ be a convex and increasing function such that $f(0)=0$ and $f(1)=1$. Prove that

                               \[ \int_{0}^{1}f^{2}(x)dx\geq\frac{3}{2}\int_{0}^{1}f(x)dx. \]

[quote="Ramanujan"]There is a mistake i think.... :)[/quote]
I think the problem is not correct

[quote=methylbenzene]I guess the coefficient should be reciprocal
\[\int_{0}^{1}f^{2}(x)dx\geq\frac{2}{3}\int_{0}^{1}f(x)dx\][/quote]

14 year revival :D

Even if the coefficient is $\frac23$ this is wrong also, I think. Think about a piecewise linear function which is defined as follows:
\(     f(x)=\left\{
\begin{aligned}
&\frac{2t}{3x}  & x\in[0,t] \\
&\frac{x-t}{3(1-t)}+\frac23  &  x\in[t,1]
\end{aligned}
\right.                                               \)
The function $f$ is of course convex and strictly increasing as long as $t$ is closed enough to $1$, and as $t\to1$, we know $f(x)\to \frac{2}{3}x$ in the $\ell_1$ sense. But this inequality can not holds for $f(x)=\frac{2}{3}x$.


superior algebra
superior algebra solved

Let a and b be elements of a group G. We say that a and b form a commutative pair if ab=ba. We know that for a certin group G in any set of 4 elements there are at least 2 commutative pairs. Show that G is commutative.

[i]Nicolae Paun (Romanian contest), 2003, 12'th grade[/i]

Suppose that a,b are in G such that ab =/= ba. (1)
Note that a =/= e and b =/= e from which we deduce that a,b,ab,ba are pairwise distincts.
Then, among thses 4 elements there are at least two commutative pairs.
But :
- If {a,ab} is a commutative pair, then  aab = aba. Multiplying at left by a <sup>-1</sup> , it leads to ab=ba, which contradicts (1).
- In the same way, we prove that each of {a,ba},{b,ab},{b,ba} is not acommutative pair.
It follows that there is at most one commutative pair in {a,b,ab,ba}. A contradiction. Thus ab=ba, for each a,b in G, and G is commutative.

Pierre.

Told you it was easy :D

This also holds: if in a group G there is at least a commutative pair in any subset of 3 elements then G is commutative. This is even easier. Maybe someone can find a generalisation of the type:

Find all pairs (t,n) s.t. if in a group there are at least t commutative pairs for each subset of n elements. Prove that G is abelian.

I have no idea if this problem is tough or easy, I haven't put any thought into it.

indeed an interesting generalisation grobber :D
by the way... u went to the contest?

I did, and at some point I thought I saw you there (I only know you from your avatar). Am I right? (I mean did you go there?)

About that generalisation: I don't know if it's Ok to waste lots of time with it, since it might be incredibly hard (I didn't see it anywhere, it's just something I asked myself...).

yes i was there :D :D
but why didn't u tell me anything???
next time u'll see me  stop me to talk :D

Remember some woman brought you to sit at a table in the first night there? I was at that table, but I wasn't sure it was you and I was afraid I'd be ridiculous if I ran after you shouting "Hey! Are you Kuba from Mathlinks?" :D


You have 14 points in a three dimensional space.

What is maximum number of  two dimensional intersecting flat planes possible interconnecting all the points ?

[hide]you take 3 points and that defines your first plane.  the second plane is created with one of those points, plus 2 other ones not already in a designated plane.  then you make the third plane with one of those points and 2 others, etc... so the number of points connected with $n$ planes is $3+2(n-1)=2n+1$ because the first plane uses 3 points, and all the other ones after that add 2 points to the intersecting plane thing.  so if $14=2n+1$, $n=6.5$ so we'll call it 7.  but that's only if there has to be a point at the intersection of each pair of planes.  otherwise, i think you can just put them in groups of 3 and the planes created will intersect.  as long as the planes aren't parallel.  but surely it is possible to put them in groups of 3 such that at least one plane is not parallel to all the others.  so in that case the answer would be $\frac{14}{3}$ which is rounded up to 5.  [/hide]


Μετά από πολύ καιρό απουσίας λόγο εξεταστικής και άλλων πραγμάτων θα βάλω μία άσκηση που με ταλαιπωρεί εδώ και καιρό και λύση δεν έχω βρει.

Να βρεθούν όλοι οι αριθμοί που $ x,y$ που ικανοποιούν την παρακάτω ισότητα:
$ 1!\plus{}2!\plus{}...\plus{}x!\equal{}y^{2}$.
Σίγουρα κάτι δεν βλέπω και δεν μπορώ να την λύσω....

[hide]Για $ x \leq 4$ χειρωνακτικά βλέπουμε ότι δεν ισχύει το ζητούμενο.
Για $ x \geq5$ ισχύει ότι  $ 1! \plus{} 2! \plus{} 3! \plus{} 4! \plus{} 5! \plus{} ... \plus{} x! \equiv 33 \plus{} 0 \equiv 3 mod 5$, άρα το ζητούμενο δε γίνεται να ισχύει (αφού $ n^2 \equiv 0, 1, 4 mod 5$).
Άρα η εξίσωση αυτή δεν έχει λύση στους ακεραίους.[/hide]

Αν δεν κάνω λάθος υπάρχει και μια γενίκευση αυτής της άσκησης που λέει ότι ο αριθμός $ 1! \plus{} 2! \plus{} 3! \plus{} 4! \plus{} ... \plus{} x!$ δεν γίνεται ποτέ τέλεια δύναμη, η οποία αποδεικνύεται με παρόμοιο τρόπο.

Για δοκίμασε x=1, y=1.

[quote="Spribo"]Για δοκίμασε x=1, y=1.[/quote]

Έχεις δίκιο Spribo, απροσεξία μου.  :blush:

Καλά το είχα πει ότι κάτι δε βλέπω και το χάνω...
Ευχαριστώ για τη λύση.... :lol:

Οπα.
Λυση υπαρχει και για$ x\equal{}3, y\equal{}3$ στην ασκηση του bouzoukman.
Γενικα παρατηρησε οτι με $ x \geq 3$ , $ LHS \equiv 0 (mod3)$

Νομίζω ότι η άσκηση αυτή είναι για το forum των ολυμπιάδων;

Anyway, για σκέψου τι γίνεται με το τελευταίο ψηφίο του LHS. :)

[hide]Μιας που δεν εχει δοθει πληρη απαντηση στο θεμα αυτο (για την τελεια δυναμη) βαζω μια συνοπτικη λυση.
Λοιπον,
ευκολα για $ x\geq 5, LHS \equiv 3 (mod10)$ αρα δεν ειναι σιγουρα δυναμη με εκθετη αρτιο.
για εκθετη περιττο με $ x\geq 9$
ειναι $ 1!\plus{}..\plus{}8! \equiv 9 (mod3^3)$ ενω $ 9!\plus{}..x!\equiv 0 (mod3^3)$
αρα δεν μπορει να ειναι ουτε περιττη δυναμη.
αρα ελεγχουμε τι γινεται στις υπολοιπες περιπτωσεις και βρισκουμε,$ x\equal{}3,x\equal{}1$[/hide]


For a theorem or an exercise, I have two or more proofs. How I can get  the following output:

Proof 1

Proof 2

Thanks!

[code]\usepackage{amsthm}
...
\begin{proof}[Proof 1]
...
\end{proof}
\begin{proof}[Proof 2]
...
\end{proof}[/code]

Thank you very much!

 1  is not upstanding

How to change \itshape or \slshape to \upshape?
upstanding

sorry, my english isnot good


How many chapters of a textbook can you go through? And please list concrete examples and the name of the textbook (even if you only did it once). I'm just curious.

I can do 1 chapter of a college textbook per day (without doing exercises).

But it's difficult for my attention span. I've done it by reading the entire chapter and highlighting a lot of phrases (1 chapter per day for Zumdahl) but was too darn tired to really do any exercises (though I probably could have done them if I had more self-efficacy). Then my attention span expires and I go forum or something. Still, at least it was good enough for a 790 on the SAT II Chemistry.

I'm doing it again for another textbook now. Ugh, same process once again...

Depends on the subject

Math: probably 2-4 chapters (with e
History: Maybe the whole book :D
Science: 1 chapter  :rotfl:
English: 4 chapters

Standarized (sp?) Test Preparation books: 5 Chapters
Contest Preparation books: 3 Chapters

Bleh... well I read about 200 pages of a typical intro level number theory textbook one time...

Math: 1.
History: 1.
Science: 1.
English: 0.

if i tried
i could finish my math text
finish my science text
get half way through history text

Well what do you mean "to go through"? IF its a math book does it contain doing the problems?

Well my calc book has long chapters and I need to review the concepts a couple times so maybe 5 chapters an entire day, then Id master those =).

Well, I can do about $1\frac{1}{2}$ chapters in 2 hours (complete with messing on my calc for half an hour) on my number theory book, so I'd say 7,8 chapters in a day.

for anything other than math, i can read for like 40 min. and go through like two-three chapters
for something like aops book 2 or other interesting math (acops etc.), i can go through like... half the book?  :D

Mathematics, about 2/3 of Volume 1, and 1/2 of volume 2(one or the other)

History, 3 Units(a combination of chapters)

Science, 4 chapters.

Vocabulary builder, 20 chapters

Not much! I will be fed up with studying...


Hello,

Find all positive integers n such that a_n= 3^(2n-1) - 2^(n-1) is a perfect square.

I've found perfect square for n=2, but I guess somehow it needs to be proven that it can be perfect square only if n=2.

Taking it mod 4 yields n<3 (otherwise the perfect square is $ (\minus{}1)^{2n\plus{}1}\plus{}0\equiv\minus{}1\pmod{4}$, which is impossible).

Can you please clarify that abit more to me, I cant seem to understand it, sorry  :blush:

EDIT: Got it now, thanks :)


[img]http://i212.photobucket.com/albums/cc112/le0thanh/dfsfs.jpg[/img]

Two rods of equal lengths and cross-sectional areas but of different materials are placed in thermal contact as shown in Figure 2. The thermal conductivity of Q			Figure 2
is half that of P. The outer end of P
is at 0o C and that of Q is at 100oC.
What is the temperature of the interface at steady state?

A	23.88 o C
B	33.33 oC
C	66.66 o C
D	80.39 o C

At steady state the rate of heat flow will be the same through both materials. It is
$ kA \frac{\Delta T}{L}$, where
$ k\equal{}$ thermal conductivity
$ A\equal{}$ area of cross section
$ L\equal{}$ length heat travels
$ \Delta T\equal{}$ temperature difference across the material
Since $ A$ and $ L$ are the same for both materials, $ Q$ having half the conductivity of $ P$ means that the temperature difference across $ Q$ will be double the temperature difference across $ P$. The total temperature difference $ (100$  degrees$ )$ will then be triple the temperature difference across $ P$, which will be
$ \frac {100}{3}\equal{}33.3$  degrees.


Those of you who have went to nationals, how much did this website help you to get there

I joined the website right after I won states, but it really helped me study before nationals.

I didn't join the website until the year after I went to nationals, so I didn't vote. However, it has helped me extremely since then.

I didn't find out about the site until after nationals, but it has helped me a lot since them, and I could still go again.  I took three of the classes (MathCounts Problem Series, Intro to Counting and Probability, and Intro to Number Theory), which helped a lot.  You learn a lot more in the classes than is on the "Do You Need This?" test.


Let $ABC$ be a triangle . Erect outside of triangle 2 equilateral triangles $BCA'$ , $ACB'$ . $A_0$ , $B_0$ respectively are the centroids of triangle $BCA' , ACB'$ . Let $P$ be the midpoints of $AB$ . Prove that : $B'PA_0 , A'PB_0$ are halfs of an equilateral triangle

He means the triangles have angles of 30°, 60°, 90°.

I see that the problem can be nicely solved using complex numbers.

Let's prove the result for triangle $A_0B'P$. Let $Q$, $R$ be the midpoints of $BC$ and $CA$, respectively. We claim that triangles $B'PR$ and $A_0QP$ are similar. In fact, since $PQ$ and $PR$ are respectively parallel to $CA$ and $CB$, we have $\angle ARP=\angle BQP$. In addition  angles $ARB'$ and $A_0QB$ are right, so it follows that $\angle A_0QP=\angle B'RP$. Now we have that $PR=a/2$, $PQ=b/2$, and we easily deduce from Pithagoras that $B'R=b\sqrt{3}/2$ and $A_0Q=a/(2\sqrt{3})$. Therefore $B'R/PQ=RP/A_0Q=\sqrt{3}$, and our assertion is proved. Hence $B'P/PA_0=\sqrt{3}$, and since the triangles clearly have two pairs or corresponding sides perpendicular to each other, it follows that $B'P$ and $PA_0$ must be perpendicular, too. And we're done. (By symmetry, we have the result for triangle $A'B_0P$ too.)

I think that, at least in this problem, an euclidean solution is far more elegant that a complex approach ;)


Suppose that the sequence {an} converges to a. Prove that the sequence {abs(an)} converges to abs{a}.

I am not exactly sure where to start on this problem.  Thanks

Either you say $|.|$ is continuous, or (more likely) you are expected to use the $\epsilon$/$\delta$ definition of a limit.

Start with  $|\ |x|-|y|\ | \leq |\ x-y\ |$ and so on ..
 :cool:


solving the equation:
$ [\frac{\sqrt{5x\minus{}2}\plus{}3}{5}]\equal{}[\frac{\sqrt{5x\plus{}3} \minus{}2}{5}]$

i don't think this equation has any solutions.

Let $ [\frac{sqrt{5x\minus{}2}\plus{}3}{5}]\equal{}k\equal{}[\frac{\sqrt{5x\plus{}3}\minus{}2}{5}].$
Then $ 25k^2\minus{}30k\plus{}9\le 5x\minus{}2<25k^2\plus{}20k\plus{}4$ and $ 25k^2\plus{}20k\plus{}4\le 5x\plus{}3<25k^2\plus{}70k\plus{}49$ and $ k\in Z_\plus{}$.
It give $ max(5k^2\minus{}6k\plus{}\frac{11}{5},5k^2\plus{}4k\plus{}\frac{1}{5})\le x<min(5k^2\plus{}4k\plus{}\frac{6}{5},5k^2\plus{}14k\plus{}\frac{46}{5})$.
Obviosly $ k\equal{}0$ had not solution and for $ k>0$ we get $ 5k^2\plus{}4k\plus{}\frac 15 \le x<5k^2\plus{}4k\plus{}\frac 65, \ k\in N$.

[quote="Rust"]Obviosly $ k \equal{} 0$ had not solution and for $ k > 0$ we get $ 5k^2 \plus{} 4k \plus{} \frac 15 \le x < 5k^2 \plus{} 4k \plus{} \frac 65, \ k\in N$.[/quote]

You're wrong on the case $ k\equal{}0$. It works with $ x\in[\frac{2}{5},\frac{6}{5})$
You're right on the other points.

And so the complete solution is :

$ x\in[\max(5k^2\plus{}4k\plus{}\frac{1}{5},\frac{2}{5}),5k^2\plus{}4k\plus{}\frac{6}{5})$, $ \forall k\in\mathbb{N}\cup\{0\}$


I'm not exactly sure if this question falls into this category but here goes...

If you have the book Mathematical Olympiad Challenges, is there a problem with the solution for 2.7 number 10?  I'm having difficulty understanding part of the solution, though I think that it may have an error.

why opening a new topic ?
You can easily find a topic about that book and you can ask your question there :)

I think that the solution is correct. It is a bit confusing.

Here is the idea: Take a sequence of $ n$ positive integers, $ x_1\le x_2\le...\le x_n$ such that $ \frac {1}{x_1} \plus{} \frac {1}{x_2} \plus{} ... \plus{} \frac {1}{x_n} < 1$. Let $ X_i \equal{} \frac {1}{x_1} \plus{} \frac {1}{x_2} \plus{} ... \plus{} \frac {1}{x_i}$. Then $ X_i\le U_i$ for $ i \equal{} 1,2,...,n$.

The reason that this is strange is because it is not an induction on $ n$, it is an induction on the index $ i$ or whatever. $ n$ is just a given integer the way that this induction goes...

Of course, normally, the hypothesis would be: $ X_n\le U_n$, but in this case, the solver uses a stronger hypothesis.

I'm not quite sure how we get: $ x_1x_2...x_nX_n\le x_1x_2...x_n \minus{} 1$ but I don't think it is that hard, I'm missing something.

The use of the abel summation is awkward because the solver introduces the $ x_1\le x_2 \le...\le x_n$ after it uses that assumption but that doesn't make it wrong.

I didn't know that there's a topic on that book...should check it out...

Anyhow, I didn't see how the Abel summation formula came out to be:

Sigma x_i/u_i < Sigma X_i * (x_i - x_i+1) ,

because the fraction is x_i/u_i, so I figured the inequality should be:

Sigma x_i/u_i = Sigma U_i * (x_i - x_i+1).

It confused me even more because X_i <= U_i for i < n-1, so it seemed like the inequality should be:

Sigma x_i/u_i = Sigma U_i * (x_i - x_i+1) >= Sigma X_i * (x_i - x_i+1).

Perhaps I'm not reading the solution properly, but if you can help clarify, thanks!

P.S. I'm sorry if I made the math hard to read.

Using the abel-summation formula, $ \sum_{i\equal{}1}^n\frac{x_i}{u_i}\equal{}\sum_{i\equal{}1}^{n\minus{}1}(x_i\minus{}x_{i\plus{}1})\cdot U_i\plus{} x_nU_n$

Now $ x_i\le x_{i\plus{}1}$ so $ (x_i\minus{}x_{i\plus{}1})$ is negative. Thus $ X_i\le U_i\implies U_i (x_i\minus{}x_{i\plus{}1})\le X_i(x_i\minus{}x_{i\plus{}1})$.

The $ x_n\ge 0$ and $ X_n\ge U_n\implies x_n U_n\le x_n X_n$.

I do have to say that I find this solution remarkable. All the terms in the abel summation with $ x_i$ increasing are negative except for the last one, and the way that the assumptions work, we can find an upperbound on all the terms.

EUREKA
I believe that I do understand it now!

Thanks for the help!


Pat wants to buy four donuts from an ample supply of three types of donuts: glazed, chocolate, and powdered. How many different selections are possible?

[quote="jhredsox"]isn't this supposed to be multiple choice???
-jorian[/quote]
They don't make much of a difference, besides your answer being a letter instead of the actual answer.

[hide]$15$ selections are possible.[/hide]

[color=blue]kyyuanmathcount:  Another AWS (answer w/out solution), I guess it had other content, but please, if you are going to answer a problem without giving a solution, don't bother to post.  [/color]


it appears that the application site has been stable (touch wood)  since we disabled file uploading three hours ago.  we'll have someone awake and monitoring the site until midnight pacific time (3 am eastern); after that, we'll all be asleep, so if the site goes down after that time it probably isn't coming back up until morning.

Thank you for your continued input. It's very much appreciated!

I'm going to sleep as well.  :)

[quote="xxreddevilzxx"]wow..... 209... with me solving only half the app quiz, is there still a chance of being that 40%?????[/quote]

I have no idea.

[quote="ztbb"][quote="MithsApprentice"]Can we get an estimate of how many on-time applications there are this year? (Which should tell us our chances of making it...)[/quote]

Right now we have 209 applications, so the acctepance rate will probably end up at around 40%.[/quote]

:( Well, I can always hope.

Actually this is good, because it means MathCamp is becoming more popular.

[quote="sky9073"][quote="dangvh"]any idea on when in may we will know if we get accepted or not?[/quote]
And another related question: How will we be notified of the decisions?
Since the postal service in Mexico isn't that reliable, I really hope it's not snail mail....[/quote]

Email, if you gave an email address in your application and indicated that you check it twice a week.


[quote="Metsjetsfan"]I hope this is the right forum...

Find b if x4 + ax3 + bx2 + cx + d has 4 non-real roots, two with sum 3 + 4i and the other two with product 13 + i.

Please show work.

Thanks a lot

edit: also, i am not really familiar with this topic, so I may require extra information to understand the answers[/quote]
[hide="partial solution"]
Call the roots w, x, y, z.
We know that $wx+wy+wz+xy+xz+yz=b$
$w+x=3+4i$
$yz=13+i$
$wx+wy+wz+xy+xz+13+i=b$
$wx+w(y+z)+x(y+z)+13+i=b$
$wx+(w+x)(y+z)+13+i=b$
$wx+(3+4i)(y+z)+13+i=b$
[/hide]

Erm, are we given anything else?

Like...a,b,c,d have to be real numbers?

[quote="diophantient"]What year is this problem from (just wondering)?[/quote]

This is a 1995 year question....and by the way, I am having many problems with this test....so this question may be the first of several more

and this is all the information the question provided

[quote="Art of Owna"][hide="hint"]For imaginaary roots, they come in complex conjugates

so lable
w=a+bi
x=c+di
y=a-bi
z=c-di..

then try multiplying w and x, and add y with z..you'll see that it works out nicely, then plug it into what diophantient has already done and you can get the answer.[/hide][/quote]

Sorry...I understood what diophant did, and I can see where you are going, but I still do not get how to w and x, and add y with z, and fitting it into the equation.

Could you give me a little more specific direction?

[hide]So we know that w+x=3+4i and yz=13+i

so w=a+bi
x=c+di
y=a-bi
z=c-di
so whats wx?  Well, w=a+bi, x=c+di, so wx=ac+adi+bci-ad

ok, whats y+z?  Well, y=a-bi z=c-di, so y+z=a+c-bi-di

But we know that w+x=a+bi+c+di=3+4i, so bi+di=4i, so -bi-di=-4i and a+c=3

So we can plug that into y+z, as a+c-bi-di=[b]3-4i[/b]

Also, yz=(a-bi)(c-di), ac+bd-adi-bci=13+i, so ac+bd=13, adi+bci=-i

so wx=[b]13-i[/b],

so plugging that in 13-i+(3+4i)(3-4i)+13+i=26+25=51[/hide]

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=394478#p394478
Also, if you click on the problem numbers [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1995]here[/url], you get linked to a place where the problem usually has a good solution or two :)

[quote="Art of Owna"][hide="hint"]For imaginaary roots, they come in complex conjugates

so lable
w=a+bi
x=c+di
y=a-bi
z=c-di..

then try multiplying w and x, and add y with z..you'll see that it works out nicely, then plug it into what diophantient has already done and you can get the answer.[/hide][/quote]

BTW, this is only true if the coefficients of the polynomial are real, which is why I asked.

It did specify in the problem that the coefficients of the polynomial were real.

Yeah, I tried looking for this AIME question; I looked in all the years except for 1985-1990, so I just gave up, and decided to ask on the forum, since Metsjetsfan didn't specify in the problem posted here.

Ok thanks a lot guys

I understand it now, and nice links diophant


ARML
AMC
AIME
probability
expected value

1. For how many pairs of consecutive integers in the set {1000, 1001, 1002,..., 2000} is no carrying required when the two integers are added?

2. Suppose that 7 boys and 13 girls line up in a row. Let S be the number of places in the row where a boy and a girl are standing next to each other. For example, for the row GBBGGGBGBGGGBGBGGBGG we have S=12. Find the average of S (if all possible orders of these 20 people are considered).

[hide="2"]
There are $20$ slots, and each choice of $7$ slots for the locations of the boys will correspond to one possible arrangement. So the total number of possible arrangements is $\binom{20}{7}$.

In the first and second slot, we can play a boy and girl next to each other in two ways,
\[bg, gb \]
There are then $18$ remaining slots, and $\binom{18}{6}$ ways to arrange the rest. The fraction of arrangements in which the first and second slot will have a boy and girl next to each other is equal to
\[\frac{2\binom{18}{6}}{\binom{20}{7}}= \frac{91}{190}\]
Similarly, for the second and third slots, third and fourth slots, and remaining pairs, the fraction of arrangements in which those pairs will have a boy and girl next to each other is $\frac{91}{190}$.

There are a total of $19$ pairs, so the expected value of $S$ is equal to $\frac{91}{190}\times 19 = \frac{91}{10}$.
[/hide]

[hide="1"]
Let's ignore the 1 in the thousands place and consider each of the remaining digits separately.

If we have a number $n\in[0,999]$, then if adding 1 to $n$ does not require any carrying, then we see that the 3 digits of $n$ range from 0 to 4, so this gives $5^{3}$ numbers.

If adding 1 to $n$ requires the units place to be carried, then the units digit must be 9, and the other 2 digits range from 0-4, so this gives $5^{2}$ numbers.

If adding 1 to $n$ requires the units place and the tens place to be carried, then the units and the tens digit must be 9, and the other digit ranges from 0-4, so this gives $5$ numbers.

If $n=999$, we need to add 1.

So our total is $125+25+5+1=156$.
[/hide]

[quote="mysmartmouth"]Just curious, where did these problems come from?  Because Joe "presented" #2 at the SC ARML lecture last year, and didn't provide a solution... :rotfl:[/quote]

The first one is from AIME 1992.

The second one is from AHSME 1989.


find all n that $ n^10\plus{}n^5\plus{}1$ is a prime number

Please do not double post. http://www.mathlinks.ro/viewtopic.php?t=239349
[hide="Solution"]Since the exponents are all different $ \mod 3$, the complex cubic root of unity is a root of that polynomial.
Or $ (n^2 \plus{} n \plus{} 1)$ divides it.
By synthetic division:
$ n^{10} \plus{} n^5 \plus{} 1 \equal{} (n^2 \plus{} n \plus{} 1) (n^8 \minus{} n^7 \plus{} n^6 \minus{} n^5 \plus{} n^4 \minus{} n^3 \plus{} n^2 \minus{} n \plus{} 1)$
If this is a prime then one of the following must hold:

1)  $ |n^2 \plus{} n \plus{} 1| \equal{} 1\implies n \equal{} 0\text{ or } \minus{} 1$, none giving prime in the main eqn..

2)  $ |n^8 \minus{} n^7 \plus{} n^6 \minus{} n^5 \plus{} n^4 \minus{} n^3 \plus{} n^2 \minus{} n \plus{} 1| \equal{} 1\implies$
$ |n(n \minus{} 1)(n^6 \plus{} n^4 \plus{} n^2 \plus{} 1) \plus{} 1| \equal{} 1$. $ n \equal{} 0,1$ are clearly solutions. If $ n > 1$ or $ < 0$ but $ \in \mathbb Z$ then
$ |n|\ge 1,|n \minus{} 1|\ge 1,|n^6 \plus{} n^4 \plus{} n^2 \plus{} 1|\ge 4\implies |n(n \minus{} 1)(n^6 \plus{} n^4 \plus{} n^2 \plus{} 1)|\ge 4\implies$
$ |n(n \minus{} 1)(n^6 \plus{} n^4 \plus{} n^2 \plus{} 1) \plus{} 1|\ge 3$ Hence the only integers $ n$ satisfying
$ |n(n \minus{} 1)(n^6 \plus{} n^4 \plus{} n^2 \plus{} 1) \plus{} 1| \equal{} 1$ are $ n \equal{} 0,1$, with only $ \boxed{n \equal{} 1}$ giving prime for the main polynomial.[/hide]


Αποδείξτε ότι η τομή $ I_1 \cap I_2$ των ιδεωδών $ I_1\equal{}<x^{7}y^{2} \plus{}x\plus{}y >$ και $ I_2\equal{}<x^{4}y^{8} \plus{}x\plus{}y >$ είναι ιδεώδες

lalas 2 ερωτήσεις!

1)Να δείξουμε ότι το $ I_1 \cap I_2$ είναι ιδεώδες τίνος δακτυλίου? Εννοείς του δακτυλίου των πολυωνύμων με ακέραιους συντελεστές και μεταβλητές τα $ x,y$ δηλαδή του $ \mathbb{Z}[x,y]$ ?

2) Εαν πράγματι εννοείς τον παραπάνω δακτύλιο, τότε η τομή δύο ιδεωδών είναι και πάλι ιδεώδες του δακτυλίου (πάρε ως δεδομένο ότι τα $ I_1,I_2$ είναι ιδεώδη του $ \mathbb{Z}[x,y]$ και με τον ορισμό δείξε ότι το $ I_1\cap I_2$ είναι ιδεώδες του $ \mathbb{Z}[x,y]$).

(Η πρόταση ισχύει για την τομή οσονδήποτε ιδεωδών $ I_i$ ενός δακτυλίου $ R$ ότι είναι και πάλι ιδεώδες του $ R$.)

Αλέξανδρος


Prove that -

sin 1 + sin 3 +sin 5 +....+sin 90 =  ((sin 50)^2)/sin 1

Are you sure it shouldn't be $\sin{1} + \sin{3} + \sin {5} +....+ \sin {89} = \frac{\sin^2{50}}{\sin{1}}$ instead of $\sin {90}$ ?

yeh sorri.....i meant up to sin 99 actrually

does anyone have a pure trignometric method....such that i can express the subject of the summation in the form of F(n+1) - F (n) or F (n) - F (n+1)???.........so that it cancels itself all the way leaving only ht efirst and last terms....:)


geometry
3D geometry
sphere
calculus
trigonometry
integration
frustum

Hi, I am trying to calculate the volume of a sphere of radius $a$ by the means of calculus but I keep getting the wrong answer. Here is what I did.

I divided the sphere into small vertical discs (i.e. a cylinder with an infinitesimal height). If you look at a single disc, then its base is a circle with radius $a\sin(\theta)$ (where $\theta$ is an angle measured from the center of the sphere).  Thus the area of the base of that disc is just the area of the circle i.e. $\pi a^{2}\sin^{2}(\theta)$. Also, the second edge of the cylinder is situated at an angle $d\theta$ from $\theta$. Thus the height of the cylinder would be approximately $a d\theta$. Thus the volume $dV$ of that disc would be base*height i.e. $(a d\theta) \cdot (\pi a^{2}\sin^{2}) = \pi a^{3}sin^{2}(\theta) d\theta$.

So,

$dV = \pi a^{3}sin^{2}(\theta) d\theta$
$\Rightarrow V = \pi a^{3}\int_{0}^{\pi}sin^{2}(\theta) d\theta$
$\Rightarrow V = \pi a^{3}(\frac{\pi}{2})$
$\Rightarrow V = \frac{{\pi}^{2}a^{3}}{2}$

which is obviously wrong. What did I miss??

The thickness of the disc (what you're calling the height of the cylinder) is not $a\,d\theta.$ It is $a\sin\theta\,d\theta.$

Make that correction, and your integral would become

$\pi a^{2}\int_{0}^{\pi}\sin^{3}\theta\,d\theta$

which would be correct.

[quote="Kent Merryfield"]The thickness of the disc (what you're calling the height of the cylinder) is not $a\,d\theta.$ It is $a\sin\theta\,d\theta.$[/quote]
How? I thought that the thickness of the disc was just the arclength which is subtended by the angle $d\theta$ and the radius is just $a$, the radius of the sphere?

You know what's really weird? The area of the side surface of each such "cylinder" depends only on its height.  :o This is true only for spheres in a $3$-dimensional space.

The result mlok is claiming was known to Archimedes.

Let it be known that I rediscovered this result independently of Archimedes, only some 22 centuries later.  :)

[quote="Kent Merryfield"]Remember that you're approximating a shape which isn't exactly a cylinder by a cylinder. The arclength is a measurement from that slanted edge for which you're disregarding the slant. What you want is the actual thickness of the disk, which is (approximately) the projection of that arclength into the direction perpendicular to the faces of the disk. Try drawing a little right triangle. That measurement of $a\,d\theta$ is the hypotenuse of the triangle, but you don't want the hypotenuse, you want one of the legs.[/quote]
I see what you mean. I realize now that the thicknes is not exactly $ad\theta$ but I am still confused about the $\sin(\theta)$ factor.  I drew my my little disc and tried to find the projection of arclength  but I can't figure out how you got $a\sin(\theta)d\theta$ ?!

[quote="mlok"]Draw a circle of radius $a$ in the plane. Let the cross-sections be perpendicular to the $x$-axis. Since $x=a\cos\theta$ on the circle, it follows that $dx=-a\sin\theta\,d\theta$. The thickness of a disk is $|dx|$.[/quote]
Got it! I get the right answer now. Thanks mlok and Kent.  :)

But one thing I still don't get is that why should it matter. The arclength and its leg should be close enough to get the right answer; they are both just approximations, right?

Also, even though after using $a\sin(\theta)d\theta$, I get a shape which is more like a cylinder, I am still using another approximation. I am assuming that $\pi (a\sin(\theta))^{2}$ is the area of both of the bases of the cylinder. But, it is only the area of one of the bases. The area of the other base is actually slightly larger--its $\pi ( a\sin(\theta+d\theta) )^{2}$.

[quote="EUCLID 3.14"]The arclength and its leg should be close enough to get the right answer[/quote]
They are "close" only in the sense that both are small. The [b]relative[/b] error in replacing the projection of arclength onto $x$-axis with arclength itself can be large. For example, let $a=1$ and consider the arc between $\theta=\pi/6$ and $\theta=\pi/4$. Its length is $\pi/12\approx 0.26$, and its projection onto the $x$-axis is $(\sqrt{3}-\sqrt{2})/2\approx 0.16$.

[quote="EUCLID 3.14"]I am assuming that $\pi (a\sin(\theta))^{2}$ is the area of both of the bases of the cylinder. But, it is only the area of one of the bases. The area of the other base is actually slightly larger--its $\pi ( a\sin(\theta+d\theta) )^{2}$.[/quote]
In this case, the relative error of approximation is $(\sin(\theta+d\theta)/\sin\theta)^{2}-1$, which approaches $0$ as $d\theta\to 0$.

[quote="mlok"]They are "close" only in the sense that both are small. The [b]relative[/b] error in replacing the projection of arclength onto $x$-axis with arclength itself can be large. For example, let $a=1$ and consider the arc between $\theta=\pi/6$ and $\theta=\pi/4$. Its length is $\pi/12\approx 0.26$, and its projection onto the $x$-axis is $(\sqrt{3}-\sqrt{2})/2\approx 0.16$.[/quote]
I see your point. But in your example you are comparing a *large* arclength, not an infinitesimal one, with its projection. How can you find the relative error between an infinitesimal arclength and its projection?

[quote="mlok"]In this case, the relative error of approximation is $(\sin(\theta+d\theta)/\sin\theta)^{2}-1$, which approaches $0$ as $d\theta\to 0$.[/quote]
How did you find that relative error?

Let's say that relative error in approximating $a$ with $b$ is something like $\frac{|a-b|}{|a|}$, or maybe $\frac{|a-b|}{|b|}$.

For infinitesimal arcs you are comparing the quantities $a d\theta$ and $a\sin\theta d\theta$. Regardless of what $d\theta$ stands for, the first quantity is twice as large as the second one when $\theta=\pi/6$. Do you agree that this is a significant difference?

Yes. I agree. How did you find the relative error of approximation for the area to be $(\sin(\theta+d\theta)/\sin\theta)^{2}-1$ by the way?

I divided $\pi ( a\sin(\theta+d\theta) )^{2}$ by $\pi (a\sin(\theta))^{2}$ and subtracted $1$ from the result.

Thank you for your kind help mlok. This all makes a lot of sense now.  :lol:

[quote="Kalle"]The truth is that the various 'approximations' are only lies to cover up the machinery that is integration. To prove, and to really understand, the formulae you need a definition of triple integrals and some variant of Fubini's theorem. The "this infinitesimal thickness is blah blah"-arguments would be worthless if the people teaching it were not sure that the resulting formulas were actually true. I would say you should just consider these "volume integrals" as unproven formulas.[/quote]I am not sure what you are trying to say here...Can you please expand?

Actually, I think I understand what Kalle was trying to say. After finding the volume by the disc method, I set to find the surface area of the sphere by the same method.
Thus, since the SA of the cylinder is $2\pi r h$, the SA of the small disc should be $2\pi a^{2}\sin^{2}(\theta) d\theta$ if $r = a\sin(\theta)$ and $h = a\sin(\theta)d\theta$
$\Rightarrow dS = 2\pi a^{2}\sin^{2}(\theta) d\theta$
$\Rightarrow SA = 2\pi a^{2}\int_{0}^\pi \sin^{2}(\theta) d\theta$
$\Rightarrow SA = 2\pi a^{2}(\frac{\pi}{2}) ={\pi}^{2}a^{2}$
which is not the right answer. However, if I set $h = a d\theta$, then I get the SA of the small disc to be $2\pi a^{2}sin(\theta) d\theta$ and
$dS = 2\pi a^{2}\sin(\theta) d\theta$
$\Rightarrow SA = 2\pi a^{2}\int_{0}^\pi \sin(\theta) d\theta$
$\Rightarrow SA = 2\pi a^{2}(2) = 4\pi a^{2}$
which is the exactly the right answer! However, I used two different values of $h$, $a\sin(\theta)d\theta$ and $a d\theta$ to get the right answer for the volume and the SA of the sphere, respectively !!!

The correct approximation for the surface area of such a "cylinder" (which is actually a [url=http://mathworld.wolfram.com/Frustum.html]frustum[/url]) is not $2\pi r h$, but $2\pi r s$. Here $s$ is the [url=http://mathworld.wolfram.com/SlantHeight.html]slant height[/url]. You have $h=a\sin\theta\,d\theta$ and $s=a\,d\theta$.

Draw a circular right cylinder of radius $1$ and height $0.1$. Then draw a frustum of a cone with the same radius and height, but with the side inclined at $45$ degrees. You will plainly see that the volumes of both solids are almost the same, but surface areas are quite different: the frustum has $40\%$ more of surface area.

I looked up on the web and I found out that the lateral surface area of the frustum of a cone is actually $\pi (r+R)s$, not $2\pi rs$.

That's why I wrote "approximation". For the frustums in this problem the difference between $R$ and $r$ is small enough to be irrelevant.

But then if $R$ is approximately equal to $r$, then $s$ should also approximately equal to $h$?

Anyways, looking back at this example, it seems to me that Kalle made a good point--you can't really justify these *approximations* unless you already know the answer beforehand. Can we really take an [b]objective[/b] approach in evaluating surface areas and volumes using the disc method (or the washer and the shell method for that matter)?

[quote="EUCLID 3.14"]But then if $R$ is approximately equal to $r$, then $s$ should also approximately equal to $h$? [/quote]
No. Please spend some time contemplating this point, with a drawing if it helps.

A rigorous approach to computations of volumes is as follows: (a) slice the solid; (b) find two right cylinders for each slice: one containing the slice and another contained in it; (c) write a formula for the sum of the volumes of all "big" cylinders and recognize it as a Riemann sum for a certain function $f$; (d) write a formula for the sum of the volumes of all "small" cylinders and recognize it as a Riemann sum for the same function $f$; (e) conclude that both sums converge to the integral of $f$.

[quote="mlok"][quote="EUCLID 3.14"]But then if $R$ is approximately equal to $r$, then $s$ should also approximately equal to $h$? [/quote]
No. Please spend some time contemplating this point, with a drawing if it helps. [/quote]

[img]http://www.inspection.gc.ca/english/fssa/labeti/nutrikit/images/secte20e.jpg[/img]

I drew the frustum out and stared at it for a while, but I still can't figure out why is it wrong to conclude that as $R_{2}\to R_{1}$, then $s \to h$?

If $R_{2}\to R_{1}$ and $h$ is fixed, then indeed $s\to h$. However, both $h$ and $s$ go to zero as you make slices thinner. Here's an example: take a cone of height $1$ with the base of radius $1$. Slice it by horizontal planes at equal distance $\frac{1}{n}$ from each other. The height of each slice is $\frac{1}{n}$, yet its slanted height is $\frac{\sqrt{2}}{n}$. This difference may look insignificant since $n\to\infty$, but remember that we must add the areas (or volumes) of $n$ slices together. Thus, an error of the size $\frac{1}{n}$ is going to mess up the result, but an error of size $\frac{1}{n^{2}}$ is not. This is the reason why some approximations for slices are legitimate but others are not.

Let's look at the volumes: numbering the slices $1,2,\dots,n$ starting from the top, we see that the slice numbered $i$ has bases with radii $R_{1}=\frac{i-1}{n}$ and $R_{2}=\frac{i}{n}$. Its volume is between $\pi R_{1}^{2}h$ and $\pi R_{2}^{2}h$, that is, between $\pi\frac{(i-1)^{2}}{n^{3}}$ and $\pi\frac{i^{2}}{n^{3}}$. If we use either of these numbers to approximate the volume, the error will be at most $\pi\frac{i^{2}-(i-1)^{2}}{n^{3}}=\pi\frac{2i-1}{n^{3}}\le \pi\frac{2n-1}{n^{3}}\le\frac{2\pi}{n^{2}}$ -- [u]with $n^{2}$ in the denominator[/u]. After summation over $i$ the total error is at most $\frac{2\pi}{n}$, which is harmless as $n\to\infty$.


Compute the value of $(1\times 2)+(3\times 4)+(5\times 6)+...+(99\times100)$.

errr... well here's an observation... i dont know how to finish the problem tho:

letting $s_n = \text{sum of the integers from 1 to n}$, we can turn that expression into

$2(s_1 + s_3 + s_5 + ... + s_{99})$ i think

i tried to simplify the stuff inside the parenthesis... but yea that got kinda hard?

eerrr actually i think there was a post like this except it went to (2004)(2005) posted just under this... you might want to look there

if you put in summation form, it is pretty clear
$\sum_{k=1}^{50} (2k-1)(2k)$
$\sum_{k=1}^{50} (4k^2 - 2k)$
$4\sum_{k=1}^{50} k^2 - 2\sum_{k=1}^{50}k$
then obviously sum of squares, etc, you got it from here

Yup, Altheman's got it  :D
That was the way I thought of.

Just curious, can someone else think of another way?


The diameter of a circle is increased so that the circumference increases by $ 20\%$. By what percent does the area increase?

However, the problem asks for the percent the area [b]increases[/b], so the answer would be 44%.

Another way is to substitute easy numbers. Let's say the original circle had a radius of 5. That means the circumference was $ 10\pi$. The problem stated that the diameter was increased so that the circumference would increase 20%, so the new circle had a circumference of $ 12\pi$ and a radius of $ 6$. So now we find the percent increase in area:

$ \frac{36\pi \minus{} 25\pi}{25\pi} \equal{} \frac{11\pi}{25\pi} \equal{} 44\%$

It goes from 1 to 1.2. Squaring (because it is 2D) gets 1.44 which is a 44% increase.

EDIT:Yay 1400 posts!


superior algebra
superior algebra unsolved

Suppose $ H \leq G$ is a group, then is there any simple way to check whether $ N( H ) \equal{} H$? $ N(H)$ is the normalizer of $ H$. Moreover, is there any special property once $ N(H) \equal{} H$ is satisfied?

(Edited from dumb comment). In general the normalizer grows $ N(H) > H$.


I'm not trying to refer to any specific problem, but I am asking if something like a 6-page solution for one problem is fine.
I know that rrusczyk answered my question about inelegant solutions being OK, but inelegant and long are two different words.

Thanks.

If it is a correct proof, then it will still get a 5/5. However, in the interests of the graders, you may want to consider trying to shorten said proof.  :P

An inelegant solution that is nonetheless well-written and correct will receive 5/5.
A lengthy solution that is well-written and correct will receive 5/5.

A poorly-written solution (of any length) may not necessarily receive 5/5.

yes, long solutions can get points.  I can't say how many times in year 18 (and even a few in year 19) when I wrote exceeding long solutions (about 3-5 page geometry/trig-bashing) and then saw the official solution was less than a page.  I still got 5's.  Once, I wrote a whole page induction proof and got a commendation when the official solution was only a paragraph.  So longer solutions are not necessarily a bad idea - just try to see if everything you write is necessary and well-explained or wordy/irrelevant.


algebra
polynomial
calculus
integration
modular arithmetic
algebra unsolved

Prove that $f(x)=x^{5}-x+a$ cannot be factored as the product of two nonconstant integral polynomials if the integer $a$ is not divisible by $5$.

I have done part of it, but now I am stuck.

let's consider the polynomial $\pmod 5$., or in $\mathbb{GF}_{5}$.

we have
$(x^{5^{4}}+x^{5^{3}}+x^{5^{2}}+x^{5^{1}}+x^{5}+1)(x^{5}-x+a)\equiv x^{5^{5}}-x$
so the splitting field of $f(x)$ over $\mathbb{GF}_{5}$ is $\mathbb{GF}_{5^{5}}$ or $\mathbb{GF}_{5}$, but the latter is impossible, because it would mean that f(x) has five prime factors of degree 1,or five roots in $\mathbb{GF}_{5}$: but for each $k$ $f(k)\equiv a \pmod 5$, and $a\neq 0 \pmod 5$. So, the splitting field is $\mathbb{GF}_{5^{5}}$,and that means that is irriducible in $\mathbb{GF}_{5}$, and so in $\mathbb{Z}$

bye bye

ah,the university.... :D

The method from http://www.mathlinks.ro/Forum/viewtopic.php?t=68010 works for all polynomials of type $x^{p}-x+a$, $p \nmid a$, $p$ prime.

I take it you mean this method:

[quote="ZetaX"]It's enough to prove that it is irreducible $\mod n$. So lets work $\mod n$ from now.

Faktor $f(x)=x^{n}-x-1$ totally as product of polynomials. Since $f(x+1)=f(x)$, the shift $x \to x+1$ permutes the factors of $f$.
Take any factor $g(x)$.
When $g(x+k)=g(x)$ for $k \neq 0$ holds for all different $x$, then $g$ is constant or has at least degree $n$ (the polynomial $g(x)-g(0)$ has $n$ roots).
So every nontrivial factor fulfills $g(x+k) \neq g(x)$ for the $n-1$ different $k$, so all we can reach out of $g(x)$ by shifting are different. Since every factor has at least degree $1$, we get $n$ factors of degree $1$, thus all of them. But a factor of degree $1$ implies that there is a root, which is impossible by Fermat's Little Theorem.[/quote]

Well i am sorry if this seems stupid but i just dont get it. First of all how is $f(x+1)=f(x)$? And also why does $g(x)-g(0)$ have $n$ roots? I just dont understand the solution at all, if somebody could find the time to explain it thoroughly to me i would appreciate it very much.


Let $S$ be a symmetric unitary $n\times n$ complex matrix. Prove that there is a unitary $n\times n$ matrix $A$ such that $S=A^tA$.

Use the polar decomposition; $S$ is invertible since it is unitary so $S = HU$ with $H$ hermitian positive definite and $U$ unitary.

$S$ being symmetric, then....


conics
ellipse
rotation
integration
calculus
calculus computations

A solid is formed by joining semicircular cross sections (whose bases lie in the $ xy$-plane and are perpendicular to the $ x$-axis) across the ellipse given by the equation $ 9x^2\plus{}4y^2\equal{}36$. What is the volume of this solid?

Let $ R$ be the region bounded by $ 6x^2\plus{}3y^2\minus{}12x\plus{}30y\plus{}63\equal{}0$. What is the resulting volume when $ R$ is rotated about the line given by $ \minus{}4x\plus{}3y\equal{}16$?

Does anyone have any hints on how to approach these types of problems?

The 2nd one is annoying. I've forgotten so much calc. it's a little sad. But What I did was find out that we have this ellipse:

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

And the line isn't going to pass through it, so we have an elliptical donut. Great. Why can't it be a circle? Well, I could not remember how to do this so I looked it up in a calc. book:

$ \frac {1}{1 \plus{} m^2} \int_p^q [f(x) \minus{} mx \minus{} b][1 \plus{} mf'(x)]dx$

Where the line has the equation y-mx +b.

see: http://www.stewartcalculus.com/data/ESSENTIAL%20CALCULUS%20Early%20Transcendentals/upfiles/projects/ess_wp_0702_stu.pdf