Let $ ABC$ be a triangle, $ G$ the centroid, $ O$ the circumcenter, $ K$ the symmedian point, and $ E$ the Euler reflection point. Prove that $ G$ and $ E$ are inverse wrt Brocard circle, i.e. circle with diameter $ OK$.

$ E$ is kimberling center $ X_{110}$, the intersection of reflection of Euler line in the sides of triangle $ ABC$.

The collinearity of $G , X_{110}$ and $X_{182}$ can be proved easily using barycentric coordinates

$\left[\begin{array}{ccc} 1 & 1 & 1\\  a^2(c^2-a^2)(a^2-b^2) & b^2(b^2-c^2)(a^2-b^2) & c^2(c^2-a^2)(b^2-c^2)\\ \sin A \cos (A- \omega) & \sin B \cos (B-\omega) & \sin C \cos (C-\omega)\end{array}\right]=0,$

where $\omega$ is the Brocard angle of $\triangle ABC.$

I found this problem by solving http://xtec.net/~qcastell/ttw/ttweng/resultats/r432.html. There are only problems with no solution in http://xtec.net/~qcastell/ttw/ttweng/portada.html. Isn't it? Then what is meaning of "not solved"?

Just a start, proving that inversion of $ E$ in Brocard circle is on Euler line.

Let $ (J)$ be the Brocard circle with diameter $ OK$ and center $ J \equiv X_{182}.$ Let $ (X), (Y), (Z)$ be the A-, B-, C-Apollonius circles of $ \triangle ABC$ centered at $ X \in BC, Y \in CA, Z \in AB.$ The A-, B-, C-symmedians are radical axes of $ (X), (Y), (Z)$ and $ (O)$ and these circles are all perpendicular to $ (O)$ $ \Longrightarrow$ $ X, Y, Z$ are poles of the symmedians $ AK, BK, CK$ and $ XYZ \perp OK$ is polar of the symmedian point $ K.$ Consequently, the line $ j \equiv XYZ$ is inversion of the Brocard circle $ (J)$ in the circumcircle $ (O).$ Denote $ \mathbf R_{(O)}, \mathbf R_{(J)}, \mathbf R_j$ reflections in $ (O), (J), j.$ Then $ \mathbf R_{(J)} \equal{} \mathbf R_{(O)} \circ \mathbf R_j \circ \mathbf R_{(O)}^{ \minus{} 1} \equal{} \mathbf R_{(O)} \circ \mathbf R_j \circ \mathbf R_{(O)}.$ Complete quadrilateral $ \mathcal Q$ formed by the lines $ (BC, CA, AB, XYZ)$ has 3 diagonals are $ AX, BY, CZ,$ let $ L, M, N$ be their midpoints. Circles $ (L), (M), (N)$ with diameters $ AX, BY, CZ$ are all tangent to $ OA, OB, OC$ at $ A, B, C,$ therefore powers of $ O$ to $ (L), (M), (N)$ are equal. They pass through vertices of $ \triangle ABC$ and are centered on its midlines, therefore they pass through its altitude feet, and powers of the orthocenter $ H$ to $ (L), (M), (N)$ are all equal to half the power of $ H$ to $ (O)$ $ \Longrightarrow$ Euler line $ OH$ is their common radical axis. Their centers form Newton line $ LMN$ of this complete quadrilateral and $ OH \perp LMN$ is its orthocentric (Steiner) line. It follows that 4 circumcircles $ (O) \equiv \odot(ABC), \odot (AYZ) \odot(BZX), \odot(CXY)$ of the 4 triangles of this complete quadrilateral concur at $ E,$ pole of the Euler line $ OH$ of $ \triangle ABC$ taken as Steiner line. Consequently, reflection $ E_j$ of $ E$ in $ XYZ$ is on $ OH.$ Initial reflection $ \mathbf R_{(O)}$ keeps $ E \in (O)$ in place, subsequent reflection $ \mathbf R_j$ takes it into $ E_j \in OH,$ and final reflection $ \mathbf R_{(O)}$ keeps it on $ OH.$ This means the reflection $ \mathbf R_{(J)}$ puts $ E$ on $ OH.$

It remains to show that $ EG$ cuts $ OK$ at its midpoint $ J.$ :maybe:

Let $ EJ$ cuts $ OH$ at $ G'$ Let $ F_a, F_b, F_c, F_j$ be feet of perpendiculars from $ E$ to $ BC, CA, AB, XYZ$ and $ E_a, E_b, E_c, E_j \in OH$ reflections of $ E$ in $ BC, CA, AB, XYZ.$ $ E_j$ is reflection of $ G'$ in $ (O).$ $ F_aF_bF_cF_j$ is Simson line with pole $ E$ WRT all 4 triangles of the complete quadrilateral $ (BC, CA, AB, XYZ).$ From this Simson line and cyclic $ EF_jXF_a, EF_jYF_b, EF_jZF_c,$ the angles $ \angle EXF_a \equal{} \angle EYF_b \equal{} \angle EZF_c$ are equal and isosceles $ \triangle EE_aX \sim \triangle EE_bY \sim \triangle EE_cZ$ similar $ \Longrightarrow$ spiral similarity $ \mathbf S_E$ with center $ E,$ rotation angle $ \phi \equal{} \angle EE_aX \equal{} \angle EE_bY \equal{} \angle EE_cZ$ and coefficient $ k \equal{} \frac{EX}{EE_a} \equal{} \frac{EY}{EE_b} \equal{} \frac{EZ}{EE_c}$ takes the Euler line $ OH \equiv E_aE_bE_c$ into the line $ XYZ.$ Consequently, perpendicular bisectors $ g, o$ of $ EG', EO$ cut $ XYZ$ at the images $ P \equal{} \mathbf S_E(G'), Q \equal{} \mathbf S_E(O).$ As $ XYZ$ is perpendicular bisector of $ EE_j,$ circumcircle $ (Q)$ of $ \triangle EOE_j$ is centered at $ Q.$ (This is Miquel circle of the complete quadrilateral $ (BC, CA, AB, XYZ),$ in this case centered on one of its lines.) From the reflection $ \mathbf R_{(O)},$ $ \angle (o, g) \equal{} \angle OEG' \equal{} \angle OE_jE \equal{} \frac{_1}{^2}\angle EQO \equal{} \frac{_1}{^2} \angle EPG'$ $ \Longrightarrow$ $ EP \parallel o.$ Since $ o \perp EO,$ $ EP$ is tangent of $ (O)$ at $ E.$ Circle $ (P)$ with center $ P$ and radius $ PE \equal{} PG'$ is centered on $ XYZ$ and perpendicular to $ (O),$ just like the 3 Apollonius circles $ (X), (Y), (Z)$ of $ \triangle ABC,$ therefore $ (P)$ is coaxal with the Apollonius circles and passes through their 2 intersection $ D_1, D_2,$ the 2 isodynamic points of $ \triangle ABC.$ This means that $ (P)$ is Parry circle of $ \triangle ABC$ and $ G' \equiv G$ its centroid.

[hide]However, I do not know a synthetic proof of the known fact that Parry circle, defined at $ \odot (GD_1D_2),$ goes through $ E.$ This may be it, if there was some other synthetic way to show that $ EJ$ cuts $ OH$ at the centroid $ G$ of $ \triangle ABC.$ [/hide]

Synthetic proof without using Parry circle $ (P)$:

$ O_a, O_b, O_c$ are circumcenters of $ \triangle AYZ, \triangle BZX, \triangle CXY$ intersecting at $ E \in (O).$ $ \angle EO_aA \equal{} 2 \angle EF_jA \equal{} \angle EXE_a$ $ \Longrightarrow$ isosceles $ \triangle EO_aA \sim \triangle EXE_a$ are similar, $ O_a \equal{} \mathbf S_E(A)$ and similarly, $ O_b \equal{} \mathbf S_E(B), O_c \equal{} \mathbf S_E(C)$ $ \Longrightarrow$ spiral similarity $ \mathbf S_E$ takes $ \triangle ABC$ with circumcircle $ (O)$ into $ \triangle O_aO_bO_c$ with circumcircle $ (Q).$ Circumcenter $ O$ of the remaining $ \triangle ABC$ of the complete quadrilateral $ (BC, CA, AB, XYZ)$ is also on $ (Q).$ Since $ O \in E_aE_bE_c,$ $ Q \in XYZ.$

A-symmedian cuts $ BC, OX, XYZ$ at $ K_a, X_a, L_a.$ Cross ratios $ (B, C, X, K_a) \equal{} (Y, Z, X, L_a) \equal{} \minus{} 1$ are harmonic $ \Longrightarrow$ if $ M_a$ is midpoint of $ XL_a,$ then $ \overline {M_aY} \cdot \overline {M_aZ} \equal{} M_aX^2.$ Circle $ (M_a)$ with diameter $ XL_a$ goes through $ X, X_a$ $ \Longrightarrow$ $ (M_a) \perp (O)$ and power of $ M_a$ to $ (O)$ is equal to $ M_aX^2.$ It follows that $ M_a \in XYZ$ is on radical axis $ EA$ of $ (O), (O_a).$

Let $ \triangle A_jB_jC_j$ with circumcircle $ (O_j)$ be reflection of $ \triangle ABC$ with circumcircle $ (O)$ in $ j \equiv XYZ.$  Let $ EX$ cut $ (O)$ again at $ N_a.$ $ N_a$ is reflection of $ E$ is perpendicular to $ EX$ through $ O,$ which is parallel to the perpendicular bisector $ O_bO_c$ of $ EX.$ From $ \overline{M_aE} \cdot \overline{M_aA} \equal{} M_aX^2,$ $ \angle AOO_j \equal{} \angle XAM_a \equal{} \angle M_aEX \equal{} \angle AEN_a \equal{} \frac {_1}{^2} \angle AON_a,$ $ N_a$ is also reflection of $ A$ in $ OO_j.$ Reflection of the line $ EP$ in the perpendicular bisector $ O_bO_c$ of $ EX$ goes through $ X$ and is perpendicular to the reflection of the segment $ OE$ in $ O_bO_c,$ i.e., also perpendicular to the segment $ ON_a.$ But reflection $ O_jA_j$ of $ OA$ in $ XYZ$ is parallel to $ ON_a$ and so the reflection of $ EP$ in $ O_bO_c,$ going through $ X,$ is perpendicular to $ O_jA_j,$ hence it is identical with reflection $ A_jX$ of $ AX$ in $ XYZ.$ Similarly, reflections of $ EP$ in $ O_cO_a, O_aO_b$ are identical with reflections $ B_jY, C_jZ$ of $ BY, CZ$ in $ XYZ.$ These reflections are all tangents of $ (O_j).$

Let $ R$ be Parry reflection point of $ \triangle ABC,$ i.e., reflection of $ O$ in $ E.$ Since reflections of $ EP$ in $ O_bO_c, O_cO_a, O_aO_b$ are tangents of $ (O_j)$, reflections of a parallel $ s \parallel EP$ through $ R$ in $ O_bO_c, O_cO_a, O_aO_b$ concur at $ O_j \in (Q).$ This means that $ s$ is Steiner line of $ \triangle O_aO_bO_c$ with the pole $ O_j$ and it cuts its Euler line $ XYZ$ at its orthocenter $ S.$ If $ M$ is midpoint of $ OE,$ then $ MQ \parallel EP \parallel RS$ and $ \frac {\overline{QP}}{\overline{PS}} \equal{} \frac {\overline{ME}}{\overline{ER}} \equal{} \frac {1}{2}.$ Since $ Q, S$ are circumcenter and orthocenter of $ \triangle O_aO_bO_c,$ $ P$ is its centroid. But $ P$ is image of $ G'$ is spiral similarity $ \mathbf S_E$ taking $ \triangle ABC$ into $ \triangle O_aO_bO_c,$ therefore $ G' \equiv G$ is centroid of $ \triangle ABC.$

I come back with a more concrete proof. In order to get that $E \equiv X_{110}$ is the inverse of $G$ in the Brocard circle, we need to show that $ EG$ passes through the midpoint $X_{182}$ of $\overline{OK}$ (already proved) and that the polar of $G$ WRT $\mathcal{B}$ passes through $E$ (to be proved).

Equation of Brocard circle $\mathcal{B}$ is given by

$\mathcal{B} \equiv a^2yz+b^2xz+c^2xy-\frac{a^2b^2c^2}{a^2+b^2+c^2}{(x+y+z) \left (\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z} \right)=0}$

Hence, polar $\tau$ of $G \ (1:1:1)$ WRT $\mathcal{B}$ is given by:

$\tau \equiv (b^2 - c^2)^2x+(c^2-a^2)^2y+(a^2-b^2)^2z = 0$

Indeed, coordinates of $X_{110}$ satisfy the equation of $\tau$ and the conslusion follows.

My solution:

Let $ \triangle A^*B^*C^* $ be the tangential triangle of $ \triangle ABC $ .

Since $ O, K, E $ is the incenter, Gergonne point, Feuerbach point of $ \triangle A^*B^*C^* $, respectively ,
so we get $ E $ is the center of the Feuerbach hyperbola  $ \mathcal{F} $ of $ \triangle A^*B^*C^* $ and $ O, K \in \mathcal{F} $ .

Since $ G $ is the cross point of $ \{ O, K \} $ WRT $ \triangle A^*B^*C^* $ ,
so from the generalization discussed in the topic [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=621222]Schwatt's lines[/url] we get $ OK $ is the polar of $ G $ WRT $ \mathcal{F} $ ,
hence from the theorem mentioned in the topic [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=50&t=623408]Rectangular circumhyperbola and circle[/url] (posts # 2 and # 4) $ \Longrightarrow G $ and $ E $ are inverse WRT $ \odot (OK) $ (Brocard circle) .

Q.E.D

Let $ \triangle A_1B_1C_1 $ be the first Brocard triangle of $ \triangle ABC $ and let $ T $ be the Tarry point of $ \triangle ABC. $ It's well-known that $ \triangle ABC $ $ \stackrel{-}{\sim} $ $ \triangle A_1B_1C_1 $ and $ G, $ $ O $ is the Centroid, Tarry point of $ \triangle A_1B_1C_1, $ respectively, so if $ M $ is the midpoint of $ OK, $ then we get $$ \triangle ABC \cup O \cup G \cup T \stackrel{-}{\sim} \triangle A_1B_1C_1 \cup M\cup G\cup O \Longrightarrow \triangle OGT \stackrel{-}{\sim} \triangle MGO, $$ hence $ \measuredangle OGT $ $ = $ $ \measuredangle OGM $ $ \Longrightarrow $ $ G, $ $ M, $ $ T $ are collinear. Notice that the steiner line of $ T $ WRT $ \triangle ABC $ is perpendicular to $ OK, $ so from $ \measuredangle GTO $ $ = $ $ \measuredangle MOG $ $ = $ $ \measuredangle (OK,OG) $ we get the second intersection of $ TG $ with the circumcircle of $ \triangle ABC $ is the anti-steiner point $ E $ of $ OG $ WRT $ \triangle ABC. $ Finally, from $ \measuredangle OEM $ $ = $ $ \measuredangle GTO $ $ = $ $ \measuredangle MOG $ we get $ {MO}^2 $ $ = $ $ MG $ $ \cdot $ $ ME. $


How many positive integers from 1 to 2008 have a cubic divisor other than 1?

[hide]
[b]Rephrase:[/b] How many positive integers not exceeding 2008 are divisible by the cube of a prime?

[b]Solution:[/b] The largest prime whose cube does not exceed 2008 is 11, so we are dealing with the set of primes {2, 3, 5, 7, 11}, whose cubes belong to { 8, 27, 125, 343, 1331}.

There are $ \lfloor 2008/8 \rfloor \equal{} 251$ multiples of 8, and similarly 74 multiples of 27, 16 multiples of 125, 5 multiples of 343, and just 1 multiple of 1331. Adding gives 347.

However, we must remove the double-counted numbers, which are the multiples of $ 2^3 \cdot 3^3 \equal{} 216$, of which there are 9, and the multiples of $ 2^3 \cdot 5^3 \equal{} 1000$, of which there are 2, giving a total of 11 overcounts.

The final answer is therefore [b][color=blue]336[/color][/b].[/hide]

Whoa, how did you change your name, TZF? (Sorry for spam.)


find all integer solution of the equation  $x^{2}=y^{7}+7$

[quote="Agrippina"]Clearly it's someone else's homework as well!

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=73519
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=67251[/quote]
   I see this topic. But can you prove this statement
                [quote="ZetaX"]  But either $y+2$ or $(y^6-2y^5+-...+2^6)$ has an odd number of prime factors $\equiv 3 \mod 4$, thus the product can't be the sum of two squares. [/quote]
      the sum of two squares can not has an odd number of prime factors  $\equiv 3 \mod 4$

[quote="ZetaX"]Let's prove a bit more: $n=a^2+b^2$ contains every prime factor $\equiv 3 \mod 4$ an even number of times in it's factorization.

Let $p \equiv 3 \mod 4$.
Lemma:
$a^2 + b^2 \equiv 0 \mod p$ has only the solution $a \equiv b \equiv 0 \mod p$.
proof:
Assume WLOG that $a \not\equiv 0 \mod p$. Then we get $(ba^{-1})^2 \equiv -1 \mod p$. Thus it suffices to show that $x^2 \equiv -1 \mod p$ has no solution. This is impossible since $p-1=4k+2$ and thus we would (by Fermat's little theorem) have $1 \equiv x^{p-1} \equiv x^{4k+2} \equiv x^2 \equiv -1 \mod p$, which is impossible.

So now:
Assume that $p^s$ exactly divides $n=a^2+b^2$.
Then $a\equiv b \equiv 0 \mod p$, thus $a=pa' , b= pb' \implies n= a^2+b^2= p^2 (a'^2+b'^2)$.
Now forget that factor $p$ since also we have a representation of $\frac{n}{p^2}$ as sum of two squares, and we can inductivly continue this till we get an $n'$ that cannot be divisible by $p$ since then it would also be divisible by $p^2$ and we could continue. Thus $s$ is even as we wanted to show.[/quote]
    thanks , ZETAX


Let $ a,b,c$ be non-negative real numbers. Prove that
$ \frac{ab^2\plus{}bc^2\plus{}ca^2}{a^2\plus{}b^2\plus{}c^2}\plus{}\frac{a^3b\plus{}b^3c\plus{}c^3a}{a^3\plus{}b^3\plus{}c^3}\le2\frac{ab\plus{}bc\plus{}ca}{a\plus{}b\plus{}c}$

[quote="dduclam"]Let $ a,b,c$ be non-negative real numbers. Prove that
$ \frac {ab^2 \plus{} bc^2 \plus{} ca^2}{a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac {a^3b \plus{} b^3c \plus{} c^3a}{a^3 \plus{} b^3 \plus{} c^3}\le2\frac {ab \plus{} bc \plus{} ca}{a \plus{} b \plus{} c}$[/quote]
Try $ a\equal{}3,$ $ b\equal{}2$ and $ c\equal{}0.$ :wink:


If a,b,c,d are positive reals, show that  at least one of the inequalities is not true :
1) a+b<c+d
2)(a+b)cd<ab(c+d)
3)(a+b)(c+d)<ab+cd

Suppose that the three inequalities are true.
$ cd(a\plus{}b)<ab(c\plus{}d)$
$ \implies (a\plus{}b)(ab\plus{}cd)<ab(a\plus{}b\plus{}c\plus{}d)<2ab(c\plus{}d)$
$ \implies (a\plus{}b)^2(c\plus{}d)<(a\plus{}b)(ab\plus{}cd)<2ab(c\plus{}d)$
$ \implies (a\plus{}b)^2<2ab$
$ \implies a^2\plus{}b^2<0$
And it is false, as claimed.


Could someone help me with this one:

A department store employs 63 men and 104 women. The total wages bill for one week is £10 672. If the mean weekly wage for men is £72, calculate the mean weekly wage for the women.

We know the total amount of money spent in wages on the men was $ 72 \times 63 \equal{} 4536$. Since the total amount was 10672, $ 10672 \minus{} 4536 \equal{} 6136$ was spent on women. There are 104 women, so the average wage for each woman is $ 6136 \div 104 \equal{} 59$.


power of 2
algebra
floor function
IMO
recurrence relation
Sequence
imo 1976

A sequence $(u_{n})$ is defined by \[ u_{0}=2 \quad u_{1}=\frac{5}{2}, u_{n+1}=u_{n}(u_{n-1}^{2}-2)-u_{1} \quad \textnormal{for  } n=1,\ldots \] Prove that for any positive integer $n$ we have \[ [u_{n}]=2^{\frac{(2^{n}-(-1)^{n})}{3}} \](where $[x]$ denotes the smallest integer $\leq x)$

we can prove by induction that u_n= 2^p(n) +1/2^p(n) for a funtion p:N->N_0 ( N union {0}) which satisfies the equations:

(1)  p(n+1) = p(n) +2p(n-1)
(2)  (-1)^(n-1)=p(n)-2p(n-1)
(3)  p(0)=0 , p(1)=1
and then solve that system to get the result.
sorry for non-latex text : /

My solution:
The first few values are easily verified to be $2^{r_{n}}+2^{-r_{n}}$, where  $r^{0}=0,r_{1}=r_{2}=1,r_{3}=3,r_{4}=5,r_{5}=11$,... Let us put $u_{n}=2^{r_{n}}+2^{-r_{n}}$ (we will show that
$r_{n}$ exists and is integer for each $n$). A simple calculation gives us $u_{n}(u^{2}_{n-1}-2)=2^{r_{n}+2r_{n-1}}+2^{-r_{n}-2r_{n-1}}+2^{r_{n}-2r_{n-1}}+2^{-r_{n}+2r_{n-1}}$. If an array $q_{n}$, with $q_{0}=0 and q_{1}=1$, is set so as to satisfy the linear recurrence $q_{n+1}=q_{n}+2q_{n-1}$, then it also satisfies $q_{n}-2q_{n-1}=-(q_{n-1}-2q_{n-2})=...=(-1)^{n-1}(q_{1}-2q_{0})$. Assuming inductively up to $n r_{i}=q_{i}$, the expression for $u_{n}(u^{2}_{n-1}-2) =u_{n+1}+u_{1}$ reduces to $2^{q_{n}+1}+2^{-q_{n}+1}+u_{1}$. Therefore, $r_{n+1}=q_{n+1}$. The solution to this linear recurrence with $r_{0}=0, r_{1}=1 is r_{n}=q_{n}=\frac{2^{n}-(-1)^{n}}{3}$, and since $[u_{n}]=2^{r_{n}}$ for $n\geq 0$, the result follow.


Two positive integers differ by $60.$ The sum of their square roots is the square root of an integer that is not a perfect square. What is the maximum possible sum of the two integers?

[hide=" Unsure..."]
If the answer is $100$, then I will post my solution.[/hide]

I have no idea if this is right or not..

[hide]Let the two numbers be $a$ and $b.$

We have $a-b=60$, or $a=b+60$.

If we consider the fact about the square roots, it becomes obvious that

$b+60=n^{2}p$

and

$b=m^{2}p$ for some integer $m$ and some non-square containing p. Subtracting one equation from the other, we get

$60=p(n+m)(n-m)$. Knowing that $n, m, p$ are all integers, we maximize $a+b$ by maximizing $n$ and $m$. Thus we get $p=3, n=6, m = 4.$ $a+b=156$, which is our answer.[/hide]

[quote="pianoforte"][hide]Our numbers are $x$ and $x-60$. We know that $\sqrt{x}+\sqrt{x-60}=\sqrt{m}$, for some integer $m$ not a perfect square. Squaring,
$2x-60+2\sqrt{x^{2}-60x}=m$. Since $m$ is an integer, $\sqrt{x^{2}-60x}$ must be an integer, so $x^{2}-60x$ is a perfect square $n^{2}$.
From the Quadratic Equation, $x=\frac{60\pm \sqrt{3600+4n^{2}}}{2}=30\pm\sqrt{900+n^{2}}$. Since $x$ is an integer, $900+n^{2}$ must be a perfect square $p^{2}$.
$900=(p+n)(p-n)$, the largest solution is $p=226, n=224$.
$x=30\pm226=256$. Answer is $\boxed{452}$?[/hide][/quote]

Remember, $m$ is not a perfect square...

[quote="The Zuton Force"]Cinco de Mayo, I got what you got

[hide="My Solution"]
Let these numbers be $a$, $b$, and $c$, and assume wlog that $b$ is not larger than $a$.

$\sqrt{a}+\sqrt{b}=\sqrt{c}$
Squaring, $a+b+2\sqrt{ab}=c$
So clearly, $ab$ is a perfect square. So, $a=n/k$ and $b=nk$ where $k|n$. Clearly $k$ is not 1, because that would make $\sqrt{c}$ and integer.

If we want to maximize the numbers, we should minimize the factor by which they differ. The logical value of $k$ is therefore $2$.

We want $2n-n/2=60\implies n=40$
Thus, $a=40/2=20$ and $b=2(40)=80$, and $a+b=100$.
[/hide][/quote]

[hide="Comments"]
Yup, that's basically what my solution was. But the reason that $k \neq 1$ isn't because that would make $\sqrt c \in \mathbb{Z}$ (it wouldn't necessarily, e.g. $\sqrt 2+\sqrt 2 = \sqrt 8$), its because it would make the two numbers the same (and so there couldn't be a difference of 60 between them).[/hide]

Haha Southpaw I got what you got.
Label the two numbers $\ a$ and $\ b$ so we write $\ a-b=60$ and that $\ \sqrt{a}+\sqrt{b}=\sqrt{c}$ squaring the second equation gives $\ a+b+2\sqrt{ab}=c$ and then using the first equation we can write this as $\ 2b+60+\sqrt{b(b+60)}= c$ since $\ 2b+60$ and $\ c$ are integers so must the $\ \sqrt{b^{2}+60b}$.  So $\ b^{2}+60b = k^{2}$ or $\ (b+30-k)(b+30+k)=900$.  Now since we want to make $\ b$ as large as possible it makes to sense to make these two factors as far apart as possible.  However, setting $\ b+30-k = 1$ and $\ b+30+k=900$ gives a nonintegral value of $\ b$.  Then trying $\ b+30-k=2$ and $\ b+30+k=450$ gives perfect square values for $\ a$ and $\ b$.  Finally $\ b+30-k = 6$ and $\ b+30+k =150$ gives $\ b=48$ and $\ a=108$ neither of which are perfect squares and making the final answer $\ \boxed{156}$

You're right, I was wrong in assuming $k$ had to be an integer

I think I found
[hide="a possibly nicer solution"]

Let our numbers be $ax^{2}$ and $ay^{2}$ ($x>y$), then their square roots are $x\sqrt{a}$ and $y\sqrt{a}$, whose sum is the square root of a non-square integer if $a$ is not a square.

We know $ax^{2}-ay^{2}=60$
$x^{2}-y^{2}=\frac{60}{a}$
$(x+y)(x-y)=\frac{60}{a}$
so $a$ is a non-square factor of 60 to make $\frac{60}{a}$ an integer, and for $x$ and $y$ to be integers, they must both be of the same parity, so $\frac{60}{a}$ either contains a factor of 4 or doesn't contain a factor of 2.

This means all possible $a$ are in the set $\left\{3,5,12,20\right\}$

Quickly checking, we get
$a=3\rightarrow(x+y)(x-y)=20\rightarrow(x,y)=(6,4)$
$3\times6^{2}+3\times4^{2}=156$

$a=5\rightarrow(x+y)(x-y)=12\rightarrow(x,y)=(4,2)$
$5\times4^{2}+5\times2^{2}=100$

$a=12\rightarrow(x+y)(x-y)=5\rightarrow(x,y)=(3,2)$
$12\times3^{2}+12\times2^{2}=156$

$a=20\rightarrow(x+y)(x-y)=3\rightarrow(x,y)=(2,1)$
$20\times2^{2}+20\times1^{2}=100$

So our maximum value is $\boxed{156}$
[/hide]

Mm, I got 452, like pianoforte. So there's no consensus on this problem as-is... can we make one?

I will post my solution later but right now I am making a PBJ for myself for dinner O_O

Edit: Oh, a and b can't be perfect squares, woops.

Edit2: Yeah, 156.

156 is larger than 100, so 100 is wrong.

$ a\minus{}b\equal{}60$. $ \sqrt{a}\plus{}\sqrt{b}\equal{}c\sqrt{d}$ where $ d$ is not divisible by the square of a prime (and is not equal to $ 1$). $ a\equal{}c^2d\plus{}b\minus{}2c\sqrt{bd}$. Hence $ bd$ and by symmetry of argument, $ ad$ are both perfect squares. Since $ d$ is a product of primes, $ a\equal{}dx^2$ and $ b\equal{}dy^2$ where $ x,y\in\mathbb{Z}^\plus{}$. Back in our equations above, $ x^2\minus{}y^2\equal{}\frac{60}{d}$, and $ x\plus{}y\equal{}c$ and we want to maximize $ (d)(x^2\plus{}y^2)\equal{}x^2d\plus{}x^2d\minus{}60\equal{}2xd^2\minus{}60$. So it suffices to maximize $ x^2d$.

$ (x\minus{}y)(x\plus{}y)\equal{}\frac{60}{d}\equal{}z$. Note that those two factors have the same parity mod $ 2$, so $ z$ must have an even number of factors of $ 2$. $ 60$ has two factors of $ 2$, so those have to go into either $ z$ or $ d$. However, $ d$ is not divisible by a square of a prime, the $ 4$ goes into $ z$ (and $ d$ is odd). So $ d\in\{3,5,15\}$ (note $ d\ne 1$). So $ (z,d)\equal{}(20,3),(12,5),(4,15)$. The max of $ x$ occurs when $ x\minus{}y\equal{}2$, $ x\plus{}y\equal{}\frac{z}{2}$, or $ x\equal{}1\plus{}\frac{z}{4}$,

So $ (z,d,x,dx^2)\equal{}(20,3,6,108),(12,5,4,80),(4,15,2,60)$. Then our number is $ 2dx^2\minus{}60\equal{}2*108\equal{}\boxed{156}$.

How do you get from "$ ad$ is a perfect square to $ a \equal{} dx^2$?

$ ad$ is a perfect square, and $ d$ is not divisible by a square of a prime.

So $ ad \equal{} q^2$. Let $ p$ be a prime that divides $ d$. Then $ p|ad \equal{} q^2$, so $ p|q$, so $ p^2|q^2 \equal{} ad$. So there are two factors of $ p$ among $ a$ and $ d$. But $ d$ has at most one of any prime factor. So $ p|a$. Hence all prime factors of $ d$ divide $ a$ (and $ d$ has at most one of any prime), so $ d|a$, so $ a \equal{} da'$ where $ a'$ is an integer. So $ (a'd)(d) \equal{} q^2$, so $ a' \equal{} \left(\frac {q}{d}\right)^2$. Since $ a'$ is an integer, it is the square of an integer. So $ a' \equal{} x^2$ for some integer $ x$, so $ a \equal{} dx^2$...

Its not really that hard a statement, but I suppose that the proof is a bit long.

[hide="Solution"]

$x-y=60$ and $\sqrt{x}+\sqrt{y}=\sqrt{z}$.  We therefore get $\sqrt{y+60}+\sqrt{y}=\sqrt{z}$.  Let $\sqrt{y+60}=m\sqrt{n}$ and $\sqrt{y}=a\sqrt{b}$.  We therefore get $b=n$ when $n$ and $b$ are perfect squares.

Therefore, $y+60=m^2*n$ and $y=a^2*n$.  Therefore, $60=(m^2-a^2)*n$.

We want to maximize $m^2*n$.  We have either $n$ is a multiple of $4$, or odd (theorem on perfect squares).  In the first case, $n=4, 12, 20, 60$ or $n=1,3,5, 15$.

However, $n=60$ gives $m^2-a^2=1$ which gives $a=0$ which definitely isn't maximal and same with $n=15$.  Also, $n=1,4$ gives $y$ being a perfect square and $y+60$ so that's absurd.

Therefore, $n=3, 5, 12, 20$.  The first case gives us $m^2-a^2=20$ which gives $m-a=2, m+a=10$ for maximum, or $m=6$.  Therefore, $m^2*n=108$ in this case.

$n=5$ gives $m^2-a^2=12$ or $m=4$ as maximum therefore $m^2*a=80$.

$n=12$ gives $m^2-a^2=5$ or $m=3$ as maximum therefore $m^2*a=108$

$n=20$ gives $m^2-a^2=3$ or $m=2$ as maximum therefore $m^2*a=80$

Therefore, maximum value of $m^2*a$ is $108$ in all cases.  Therefore, we get $y=48$ therefore the sum is $\boxed{156}$. [/hide]

*This is not a question, just an addition.

My solution was very similar to Binomial-theorem's. Essentially I assumed that there were no solutions to the expression:

$\sqrt{x}+\sqrt{x+60} = \sqrt{n}$

where x and x+60 are integers that have no perfect square divisors besides 1. I then used this expression to say that then the answer has to be in a format where I have $x \sqrt{n}, y \sqrt{n}$ where then they add together to give $\sqrt{n}(x+y)$. This expression was solvable beacuse now you just increment values of n (sqrt 2, sqrt 3) and then you get equations like $x^{2}-y^{2} = 60/n$. This was maximized when n was a small number (which was 3) and this gave us $x^{2}-y^{2}=20$, which was then solved by looking at the DOTS factorization and values that would work (6 and 4).

Pretty neat problem, although I would be curious as to how you can show that there are no solutions to the first statement for this problem, since I think they might exist in general.


number theory
relatively prime
combinatorics unsolved
combinatorics

Let $n$ and $k$ be relatively prime positive integers with $k<n$. Each number in the set $M={1, 2,..., n-1}$ is colored either blue or white. For each $i$ in $M$, both $i$ and $n-i$ have the same color. For each $i$ in $M$ not equal to $k$, both $i$ and $|i-k|$ have the same color. Prove that all numbers in $M$ must have the same color.


geometry
symmetry
articles
number theory
relatively prime
6th edition

Have you solved any of the problems? I didn't even came close to solving a problem  :(

If you did manage to solve one of them, please post the solution or the main ideas here

I solved question one.
[hide="extract from solution"]
Lemma
$\tau(kn-1)\equiv0\bmod k\,\forall n\in\mathbb{N}\,\iff m^{2}=1\bmod k\,\forall m\in\mathbb{Z}_{k}^{*}$,
the group of integers coprime to $k$ mod $k$.

Proof
Suppose $\tau(kn-1)\equiv0\bmod k\,\forall n\in\mathbb{N}$. Let $m$
be coprime to $k$. Since each element of $\mathbb{Z}_{k}^{*}$ has
a unique inverse, and $m^{2}=1\iff m=m^{-1}$, suppose $m\neq m^{-1}$.
By Dirichlet's theorem, we may choose distinct primes $p$ and $q$
such that $p=m\bmod k$, $q=-m^{-1}\bmod k$. Then $pq\equiv-1\bmod k$,
and $\tau(pq)=\tau(p)\tau(q)\equiv(m+1)(-m^{-1}+1)\equiv-1+m-m^{-1}+1\neq0\bmod k$.
Hence each element of $\mathbb{Z}_{k}^{*}$ is self-inverse.

Suppose now $m^{2}=1\bmod k\,\forall m\in\mathbb{Z}_{k}^{*}$. Let
$n$ be an arbitrary positive integer. Let $kn-1=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{r}^{a_{r}}$,
where $a_{1},\, a_{2},\ldots,\, a_{s}$ are all odd, $a_{s+1}\ldots a_{r}$
are all even. The $p_{i}$are clearly coprime to $k$. We have $\tau(kn-1)={\displaystyle \prod_{1}^{r}}(\sum_{j=0}^{a_{i}}p_{i}^{j})$.
It suffices to show that ${\displaystyle \prod_{1}^{s}}(\sum_{j=0}^{a_{i}}p_{i}^{j})\equiv0\bmod k$.
Since $p_{i}^{2}=1\bmod k$, we get $\sum_{j=0}^{a_{i}}p_{i}^{j}\equiv\frac{a_{i}+1}{2}(p_{i}+1)$.
Hence it suffices to show that $\prod_{1}^{s}(p_{i}+1)$. Since $p_{1}p_{2}\cdots p_{s}\equiv-1\times(p_{s+1}^{a_{s+1}}p_{s+2}^{a_{s+2}}\cdots p_{r}^{a_{r}})^{-1}\equiv-1$
(as $a_{i}$ is even for $i>r$), we get ${\displaystyle \prod_{i\in X}p_{i}}=-\left({\displaystyle \prod_{i\in S/X}p_{i}}\right)^{-1}=-{\displaystyle \prod_{i\in S/X}p_{i}}$
where $S=\{1,2,\ldots,s\}$ and $X\subset S$. This implies $\prod_{1}^{s}(p_{i}+1)=0\bmod k$
which implies the result; note that this argument fails for $k=2,$
as 1=-1 meaning that $kn-1$ does not necessarily have any prime factors
to an odd power.

Now, if $p\mid\phi(k)$, $p$ prime, then by Cauchy's theorem that
any group whose order is divisible by a prime has an element whose
order is that prime, we have an element of order $p$. Hence $p=2$
and $\phi(k)=2^{a}$ for some $a\in\mathbb{N}_{0}$. Hence $7\nmid k$,
as if it did then $6\mid\phi(k)$. So $k<50,$ as otherwise $49\neq1\bmod k$.
As it is, if $k>7,$then $k\mid48$.

The rest is a case analysis.

[/hide]

I didn't really look at q3, and made no progress on q2.

Jack

Well i solved problem 1 using dirichlet's theorem about the existence primes of the form $a+bn$ with $a,b$ given coprime integers - and I thought to have some argument avoiding this, but I am not really sure whether it will work.

For problem 2, we can get some conditions since the existence of the incircle is easily seen to be equivalent to some condition on the length of the sides; on the other hand, the incircle exists iff some circle exists teaching the sides from the outside, which gives some other condition. Those conditions can somehow be put together to give a solution of problem 2.

However, for problem 3 I didn't really have any idea - except for $n\leq 12$ ;).

Wow, Peter has just exactly summarized my solutions. (For Problem 1 anyways, I'm not so sure about Problem 2.) However, if anyone is interested in the details, here are my solutions. I didn't know where to start for problem 3.

[b]Problem 1:[/b]
For each positive integer $n$ let $\tau(n)$ be the sum of divisors of $n$. Find all positive integers $k$ for which $\tau(kn-1) \equiv 0$ (mod $k$) for all positive integers n.

[b]Solution:[/b] Clearly, $k=1$ works and $k=2$ doesn't work. For the proceeding discussion, we will only consider $k>2$.
	To start off, we show that the problem is equivalent to finding all $k>2$ such that $ab \equiv 1$ (mod $k$) $\Rightarrow a\equiv b$ (mod $k$)

	To prove this, assume that $ab \equiv 1$ (mod $k$) $\Rightarrow a\equiv b$ (mod $k$). For any $n\in \aleph$, $kn-1$ isn't a perfect square. This is because $t^2 = kn-1\equiv -1$ (mod $k$) $\Rightarrow t(-t) \equiv 1$ (mod $k$) $\Rightarrow t\equiv -t$ (mod $k$) $\Rightarrow 2t \equiv 0$ (mod $k$) $\Rightarrow$ either $0 \equiv 1$ (mod $k$) or $k|2$, either of which yields a contradiction with $k>2$. Therefore, the divisors of $kn-1$ can be partitioned into unordered pairs $(a_i, b_i)$, where $a_ib_i = kn-1$. Now, $a_ib_i \equiv -1$ (mod $k$) $\Rightarrow a_i(-b_i) \equiv 1$ (mod $k$) $\Rightarrow a_i \equiv -b_i$ (mod $k$) $\Rightarrow \tau(kn-1)=\sum(a_i+b_i) \equiv 0$ (mod $k$).

	Conversely, assume that $\tau(kn-1)\equiv 0$ (mod $k$) $\forall n\in \aleph$. Suppose that $ab\equiv 1$ (mod $k$) does not imply $a\equiv b$ (mod $k$). Then $\exists a_0$, $b_0$ such that $a_0 \neq b_0$ (mod $k$) and $a_0b_0 \equiv 1$ (mod $k$). Now, using Dirichlet's Theorem on primes in arithmetic progressions, $\exists$ primes $p$, $q$, such that $p \neq q$, $p \equiv a_0$ (mod $k$), $q\equiv -b_0$ (mod $k$). We have $pq=kn-1$ for some $n\in \aleph$, and $\tau(kn-1)=1+p+q+pq\equiv a_0-b_0 \neq 0$ (mod $k$), contradition. Therefore, $ab \equiv 1$ (mod $k$) $\Rightarrow a\equiv b$ (mod $k$).

	Thus, we have successfully converted the problem to finding all positive integer $k>2$ for which $ab \equiv 1$ (mod $k$) $\Rightarrow a\equiv b$ (mod $k$). It is straighforward to check that $k\in \{3,4,6,8,12,24 \}$ satisfy this condition (by checking the multiplication tables mod $k$). We proceed to show that there are no other solutions.

	Let $p$ be a prime power factor of $k$. $\forall a$ relatively prime to $p$, $\exists b$ such that $ab \equiv 1$ (mod $p$). Thus, we have, $\forall a$ relatively prime to $p$, $a^2 \equiv 1$ (mod $p$) and $ord_{p}(a) \leq 2$. This means that $p\leq 3$, since a maximum order of $p-1$ is always possible. Therefore, it suffices to consider $k=2^u3^v$ for non-negative integers $u$ and $v$. Now, we can repeat the first part of the above argument for prime power factor of $k$, $p^u$, and get that $ord_{p^u}(a) \leq 2$ $\forall a$ relatively prime to $p$. However, $3^2$ and $2^4$ don't satisfy this condition so neither $3^2$ nor $2^4$ divide $k$. This narrows down the solution set to $k\in \{3,4,6,8,12,24 \}$.

	Summarizing, the complete solution set is $k \in \{1,3,4,6,8,12,24 \}$.

[b]Problem 2:[/b]
Let $ABCD$ be a convex quadrilateral, and the points $A_1 \in (CD)$, $A_2 \in (BC)$, $C1 \in (AB)$, $C_2 \in (AD)$. Let $M=AA_2 \cap CC_1$, $N=AA_1 \cap CC_2$, prove that if three of the quadrilaterals $ABCD$, $A_2BC_1M$, $AMCN$, $A_1NC_2D$ are circumscriptive (i.e. there exists an incircle tangent to all the sides of the quadrilateral) then the fourth quadrilateral is also circumscriptive.

[b]Solution:[/b] We will use one well-known fact and two simple lemmas.

[b]Well-Known Fact:[/b] Convex quadrilateral $PQRS$ is circumscriptive iff $|PQ|-|RQ|=|PS|-|RS|$

[b]Lemma 1:[/b] Given ray $l$ with end point $X$, and point $Y$ not on the ray. Consider $P\in l$, then as $P$ moves away from $X$, $|PX|-|PY|$ is strictly increasing.

[b]Proof of Lemma 1:[/b] Consider $P$,$P' \in l$, such that $XP' > XP$, then $(|P'X|-|P'Y|)-(|PX|-|PY|)=|P'P|+|PY|-|P'Y|$ which is positive by the Triangle Inequality.

[b]Lemma 2:[/b] Let $PQRS$ be a convex quadrilateral with no parallel sides. Let $X=PQ \cap RS$, $Y=QR \cap SP$, and label the quadrilateral so that $R$ is within $PXY$. Then $PQRS$ is circumscriptive iff $|XP|-|YP|=|XR|-|YR|$.

[b]Proof of Lemma 2:[/b] Suppose first that $PQRS$ has incircle $O$. Let the points of tangency of $O$ with $PQ$, $QR$, $RS$ and $SP$ be $X_1$, $Y_2$, $X_2$, and $Y_1$ respectively. Then $|XP|-|YP| =(|XX_1|+|X_1P|)-(|YY_1|+|Y_1P|) =|XX_1|-|YY_1| =|XX_2|-|YY_2| =(|XX_2|-|X_2R|)-(|YY_2|-|Y_2R|) =|XR|-|YR|$.

Conversely, suppose that $|XP|-|YP|=|XR|-|YR|$. Let $O$ be the incircle of $PXS$, and let $R'$ be the point on $XS$ such that $R' \neq S$ and $YR'$(including its extensions) is tangent to $O$. By the previous paragraph, $|XP|-|YP|=|XR'|-|YR'|$, so $|XR'|-|YR'|=|XR|-|YR|$. By Lemma 1, $R=R'$. Thus, by our definition of $R'$, $PQRS$ is circumscriptive.

Returning to the problem, if $A_2BC_1M$ and $A_1NC_2D$ are both circumscriptive, then using Lemma 2 on $DC_2NA_1$ and $BC_1MA_2$, $|AD|-|CD|=|AN|-|CN|$ and $|AB|-|CB|=|AM|-|CM|$. This, in conjunction with our Well-Known Fact, implies that if one of $AMCN$ or $ABCD$ is circumscriptive, the other is also. Now, we have two cases left, and by symmetry, we just have to consider the case in which $AMCN$, $ABCD$ and $A_2BC_1M$ are circumscriptive. Using the Well-Known Fact and applying Lemma 2 on $BC_1MA_2$, we have $|AD|-|CD|=|AB|-|CB|=|AM|-|CM|=|AN|-|CN|$. By Lemma 2 on $DC_2NA_1$, $A_1NC_2D$ is circumscriptive. This completes the proof.

How do you prove the converse of the well-known fact?

i didn't know how to do it  :?

Think of the quadrilateral $PQRS$ as made up of 4 lines, call them $l$, $m$, $n$, $o$. Let $|PQ|$+$|RS|$=$|QR|$+$|SP|$, and call this equation 1. Suppose that $PQRS$ isn't circumscriptive, then consider the incircle $G$ of the triangle defined by $l$, $m$, $n$. Now, $o$ must not be tangent to $G$ by our assumption. We "slide" $o$ so that it is tangent to $G$ with the constraint that the orientation of $PQRS$ remains the same. We can show that there is one and only one way to do this. Moreover, this will always increase one side of equation 1 and decrease the other. However, we know that after sliding, equation 1 must be true, so we have a contradiction.

In fact, this was one of the problems in one of the earlier IMO's in the 1960's or 1970's, but I can't remember which. I always thought of this as well known.

Solved 1 and 2... in number one I used Dirichlet's theorem AND Bertrand's postulate, which was hardcore. My solution looks very similar to Jack's
For question 2, I used the incircle of one triangle and the excircle of the other, equated and got something really rather nice.

I have a small question: isn't problem 1 a problem that I posted on the forum some very long time ago and which was solved by Valentin? Or is it a variation, since I do not remember the exact statement of the problem I posted. For problem 3, again apologize, it was taken from a very nice article. Valentin can post the whole article, it is a very nice one.

Saul - which triangles!  Please betray to us your solution.  It seems that I have forgotten how to do geometry over the summer.
Jack

From memory, it was the A-excircle of $ANC_2$ and the incircle of $CC_2B$.

[quote="pavel kozlov"]2th problem is very easy, but I didn't solve 1th problem.[/quote]

don't worry. i solved neither.

harazi, do you still have that article?

[quote="harazi"]I have a small question: isn't problem 1 a problem that I posted on the forum some very long time ago and which was solved by Valentin?[/quote]That is correct. But it was a long time ago, and people have forgotten about it :D

I think  Problem1  can be solved  by a simple principle,not use Dirichlet's Theorem.

I think you should post this approach!!!  ;)

i think Dirichlet's Theorem should not be used because it's too difficult foe us to prove it.


Mathnerd314's album is available, without a description, on computers worldwide.

I saw this sign in the Boston subway, at Government Center.


11. Find the sum $ \frac{1}{(1)(3)}\plus{}\frac{1}{(2)(4)}\plus{}\frac{1}{(3)(5)}\plus{}\cdots\plus{}\frac{1}{(98)(100)}$

I thought I could do it, but couldn't.. :(

[hide]
First I changed it to $ \sum_{n\equal{}1}^{98}\frac{1}{n\cdot(n\plus{}2)}$ Solving for this gives $ \frac{14651}{19800}$.

I don't know if summation is appropriate here...
[/hide]

[quote="kunny"]Your answer is incorrerct, hunter34.[/quote]
Can you post the correct answer?

Sorry, [color=blue]hunter 34[/color]'s answer was correct. :oops:

Changing to summation notation does nothing more than save space, you should explain how to summed up the series.

[quote="kunny"]$ \boxed{\frac {1}{ab} \equal{} \frac {1}{b \minus{} a}\left(\frac {1}{a} \minus{} \frac {1}{b}\right)\ (0 < a < b)}$[/quote]
Wow.. I haven't thought of that way. But is that way faster than the summation?
Well, I don't know anything about summation, except that you use it to sum equations... :huh:

[quote="alanchou"]I believe that is used to evaluate the sum[/quote]
That's what I meant. I may have worded it wrong. Sorry.

I still don't get how to evaluate a summation. I mean it looks awesome, but...

[quote="hunter34"]View [url=http://www.artofproblemsolving.com/Wiki/index.php/Summation]this[/url][/quote]
Thanks for the link, but I still don't understand the precedure..Sorry.

Can you extend a little bit and show work process?

First we need to clarify what is meant by "summation."  Strictly speaking, all it means is that we are adding things together, which is basic arithmetic.

What you are asking about is more specific:  it is [i]series summation[/i], or how to add an unknown number of things together that have some pattern to them.  I think you are also using the word "summation" to mean "summation [b]notation[/b]," which is all it is:  notation.  Summation notation does not change the content of a problem.

kunny's hint suggests the following pattern:

$ \frac{1}{1 \cdot 3} \equal{} \frac{1}{2} \left( \frac{1}{1} \minus{} \frac{1}{3} \right)$
$ \frac{1}{2 \cdot 4} \equal{} \frac{1}{2} \left( \frac{1}{2} \minus{} \frac{1}{4} \right)$
$ \frac{1}{3 \cdot 5} \equal{} \frac{1}{2} \left( \frac{1}{3} \minus{} \frac{1}{5} \right)$
$ \frac{1}{4 \cdot 6} \equal{} \frac{1}{2} \left( \frac{1}{4} \minus{} \frac{1}{6} \right)$

These are all easily verifiable.  The general pattern (that is useful in this case) is

$ \frac{1}{k(k\plus{}2)} \equal{} \frac{1}{2} \left( \frac{1}{k} \minus{} \frac{1}{k\plus{}2} \right)$

and the reason it is useful is that stuff cancels.  In particular, the original summation can be rewritten (using the above) as

$ \frac{1}{2} \left( \frac{1}{1} \minus{} \frac{1}{3} \plus{} \frac{1}{2} \minus{} \frac{1}{4} \plus{} \frac{1}{3} \minus{} \frac{1}{5} \plus{} \frac{1}{4} \minus{} \frac{1}{6} \plus{} \frac{1}{5} \minus{} \frac{1}{7} \pm ... \plus{} \frac{1}{97} \minus{} \frac{1}{99} \plus{} \frac{1}{98} \minus{} \frac{1}{100} \right)$

Notice that there is a $ \plus{} \frac{1}{3}$ term and a $ \minus{} \frac{1}{3}$ term.  Similarly, there is a $ \plus{} \frac{1}{4}$ term and a $ \minus{} \frac{1}{4}$ term.  In fact, written this way, almost every term cancels, and the resulting sum is

$ \frac{1}{2} \left( \frac{1}{1} \plus{} \frac{1}{2} \minus{} \frac{1}{99} \minus{} \frac{1}{100} \right)$

This is called a [url=http://en.wikipedia.org/wiki/Telescoping_series]telescoping series[/url].

Oh, I have seen that pattern. I didn't know that it was telescoping series.

Thank You.

[quote="kunny"]$ \boxed{\frac {1}{ab} \equal{} \frac {1}{b \minus{} a}\left(\frac {1}{a} \minus{} \frac {1}{b}\right)\ (0 < a < b)}$[/quote]

in the summation, you can factor out the 1/(a-b), in this case 1/2, and throw it in at the end when your done with your canceling out and what not


Prove that if the points of the plane are covered red or blue, then there will be a red pair with distance 1, or there are 4 collinear points points with distance 1.

[quote="archimedes1"]Do you mean lattice points, like (x,y) for integral x,y?
Also, do you mean four collinear blue points with the disance between 2 neighboring points is 1?[/quote]

yes,yes


AMC AIME calculus integration geometry perimeter

I solved this AIME problem (#7 from 2003), but I want to check to see if my answer agrees with what other people got.

The points A, B, C lie on a line in that order with AB = 9, BC = 21. Let S be the set of all points D in the plane such that the distances AD, BD, CD are all integral and AD = CD. Find the sum of the perimeters of all triangles ACD with D in S.

[hide]Let BD = n, CD = m. So m:^2: - n:^2: = 152 - 6:^2: = 189.

(m+n)(m-n) = 189. They're both integers so just factor. I got m = 95, 33, 17, 15.

m = 15 produces a degenerate triangle. So the answer I get is (33+33+30) + (17+17+30) + (95+95+30) = 380 (assuming they mean non-congruent triangles). [/hide]

Although there's probably a better method.

Alright, I definitely did not follow that. Here's my answer:

[hide]

128

[/hide]

Explanation later.

FD does not have to be integral.

Edit: In fact, 33-33-30 gives FD =  :sqrt: 864 and BD = 30. So AD, BD, and CD are all integral as the problem states.

Sorry about that mistake Wyra.... BTW... Why did you do M^2-N^2 thing?

Heh, it's my fault. I know I didn't explain very thoroughly.

I'll put it in spoiler in case anyone's still working on it:

[hide]Let x = FD. (F is intersection of bisector and CD).

62 + x2 = n2

152 + x2 = m2

So m2 - n2 = 152 - 62.[/hide]

Nice job!  That's the same answer I got, but your method was better Wyra.


Is this just me or is something wrong on the latex? I typed up some of my solutions and tried to read them and it didn't give me anything but the first question.

Does anyone else see this problem?  Are you sure everything typesets correctly (i.e. are there any error messages)?

well there were errors but only the ones that were in the template and or questions. I checked many times.


calculus
integration
function
limit
calculus computations

New in here and I'm not sure this is a right forum to post
question like this one. My apologies if it's not a right forum! :)

Suppose $n \geq 1$ is an integer. Is it possible to have:
$\int_{0}^{1}x^{n}(1-x)^{n}\;dx = \frac{1}{4^{n}}$?

[quote="blahblahblah"]By the theory of beta functions, for $n\geq 1$, we have
\[\int^{1}_{0}x^{n}(1-x)^{n}=\frac{n!n!}{(2n+1)!}\]
So you want to solve
\[4^{n}(n!)^{2}= (2n+1)! \]
By Bertrand's postulate or whatever, you know that there will be no solutions for $n\geq 2$, since there is always a prime between $n$ and $2n$.  The cases $n=1$ and $n=0$ are easily seen not to work, so we're done.[/quote]
absolutely!

Bertrand seems like serious overkill here. Let's try something more elementary.

It does work for $n=0,$ of course; you just excluded that case in the setup.

Let $f(n)=\frac{4^{n}(n!)^{2}}{2n+1}.$  The question is whether we can ever have $f(n)=1$ for $n\ge 1.$ We do know that $f(0)=0.$

Then $\frac{f(n+1)}{f(n)}=\frac{2n+2}{2n+3}.$ Since this is positive and less than $1,\ f(n)$ is a strictly decreasing sequence of positive numbers. Since it is strictly decreasing, it will never equal $1$ again.

More detailed asymptotics for $f(n)$ would be a matter for Stirling's approximation.

e: ignore this post; I tried to show $(2n)!<4^{n}(n!)^{2}$, which actually isn't true.

In your last line, that should be $\prod_{j=1}^{n}\left(\frac{2j}{2j+1}\right).$

Erm... Isn't $\frac{1}{4}$ the maximum of $x(1-x)$ on $[0,1]$ attained at just one point?  From here it follows immediately that the integral is strictly less than $\frac 1{4^{n}}$.

thank you guys for replying. i probably should have posted the original
problem that i made it up and i was playing with, which is: are there any
natural numbers m, n, k for which we have: $ \frac{n!\;m!}{(n+m+1)!}=\frac{1}{k^{2}}$?
if m = n, then as [b]blahblah...[/b] showed, and was also what i got, that'd be
impossible. i'm aware that this problem probably doesn't have that much
to do with calculus anymore!  :)

Regards

There are certainly silly examples with $m=0$ and $n=k^{2}-1$ or the reverse. I don't think there are any others.

thanks jmerry. you're right. that's why i supposed
in the problem that m and n are natural numbers.

Oh, if we want it to, it does have to do with calculus, as fedja showed. Now let's take fedja's observation and push it a little harder.

Let $I_{n}=\int_{0}^{1}x^{n}(1-x)^{n}\,dx.$ Then:

$4^{n}I_{n}=\int_{0}^{1}(2x)^{n}(2-2x)^{n}\,dx.$

Make the substitution $u=2x-1,x=\frac{1+u}2.$

$4^{n}I_{n}=\frac12\int_{-1}^{1}(1+u)^{n}(1-u)^{n}\,du= \frac12\int_{-1}^{1}(1-u^{2})^{n}\,du$

Make another substitution: $v=\sqrt{n}u$ or $u=\frac{v}{\sqrt{n}}.$ The integral from the last line becomes

$\frac1{2\sqrt{n}}\int_{-\sqrt{n}}^{\sqrt{n}}\left(1-\frac{v^{2}}n\right)^{n}\,dv$

So that:

$\sqrt{n}\cdot4^{n}I_{n}=\frac12\int_{-\sqrt{n}}^{\sqrt{n}}\left(1-\frac{v^{2}}n\right)^{n}\,dv$

Take the limit as $n\to\infty$ and pass the limit inside the integral:

$\lim_{n\to\infty}\sqrt{n}\cdot4^{n}I_{n}= \frac12\int_{-\infty}^{\infty}e^{-v^{2}}\,dv= \frac{\sqrt{\pi}}2.$

Or, asymptotically, $I_{n}\approx\frac{\sqrt{\pi}}{2\sqrt{n}\cdot4^{n}}.$


a1,a2,.....,an ,b1,b2,......,bn  are reals

a1^2/b1+a2^2/b2+.......+an^2/bn>=(a1+a2+......+bn)^2/(b1+b2+...bn)

This is an inequality of Cauchy because i'm not sure.

It is Cauchy in Engel (or Andreescu) form and follows from application of Cauchy for numbers $\frac{a_{i}}{\sqrt{b_{i}}}$ and $\sqrt{b_{i}}$ (thus $b_{i}$ must be positive).


calculus
integration
number theory unsolved
number theory

Find all $ x,y$ such that they are all interger and $ 2x^3\plus{}1\equal{}y^3$

This equation is reduced to an elliptic curve:
\[ v^2 \equal{} u^3 \minus{} 27\]
with the following substitution: $ x\equal{}\frac{27 u \minus{} 3 u v}{u^3\minus{}108}$ and $ y\equal{}\frac{u^3\minus{}18v\plus{}54}{u^3\minus{}108}$.

[url=http://sagemath.org]SAGE[/url] tells that there is only one integral point $ (u,v)\equal{}(3,0)$ on this curve:
[code]sage: EllipticCurve([0,-27]).integral_points()
[(3 : 0 : 1)][/code]
which correspond to $ (x,y)\equal{}(\minus{}1,\minus{}1).$


150.01

Pwnt.

[hide=" "]actually 138[/hide]


induction
combinatorics unsolved
combinatorics

we have n point on a plane.
you need to give Each point a color:green or red.

if exist a straight line can separate two kinds of color,that is, green points on one side, red points on another side

ex：
     [geogebra]8b5fd3c7b56675be5521cc61c89bab78fb7ba305[/geogebra]

then this situation we call "good situation"。
how many Maximum number of "good situation"?

thanks~

I believe the answer is $ n^2 \minus{} n \plus{} 2$ , which can be attained when all the points lie on a circle.  Change the problem to counting the number of lines you can draw which seperate the points into distinct partitions, and prove by induction  (you have to multiply by 2 at the end to get the two possible colourings for each line).

I could be wrong though, haven't checked the details properly..


Let $ f$ be differentiable at $ x\equal{}0$, and also satisfy $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$. Prove that $ f$ is differentiable everywhere.

My idea is that you just need to show that $ f$ is continuous at every point, and thus differentiable since it would then be the case that $ f(x)\equal{}\lambda{x}$. But I am not sure how to go about doing this. Any suggestions?

Thanks. Man now I feel somewhat ashamed that I couldn't think of that.  :oops:

well to show the func to be a differentiable func
let x=a
f'(a)=lim(h->0){f(a+h)-f(a)}/h
      =lim(h->0){f(a)+f(h)-f(a)}/h
      =lim(h->0)f(h)/h
      =f'(0)  which is defined
Thus it is differentiable.


Yes, very hard...

http://www.mathlinks.ro/Forum/viewtopic.php?t=6026

  Darij


MIT
college
group theory
abstract algebra
ratio
puzzles

You have 12 marbles. All twelve of them are identical in looks and shape, but 1 of them is the slightest bit lighter or heavier then the other eleven. You have a scale that is old fashioned so that it only tells whether something is heavier or lighter than something else put on it. Each pan is large enough to hold all 12 marbles. Unfortunately your scale can only be used 3 times before it brakes.
Figure out a strategy that will tell you which one is the different one and whether it is heavier or lighter. It must be full proof with detailed explanations of why it works.

First one to figure it out gets a cookie point! This is also a game. The first one to get 5 cookie points becomes cookie king. I will post more logic problems as I get/solve them.

If there are and problems or not enough information post here and I will change it if need be.

Sooner or later you will get the fiendishly difficult question of this:

There is a closet that is pitch dark. There are 5 hats: 3 black, 2 red. 3 blind men walk into the closet choose a hat, put it on and walk out. How many red hats are there that were chosen by the blind men?

lol "brakes"?

I've heard of this one before...and in a similar problem, you only needed to weigh once...let's see if I could figure out.

Call them 1, 2, 3, 4, etc

1st weighing:

2, 3, 5, 12 vs 7, 8, 10, 11

2nd weighing:

3, 5, 6, 11 vs 1, 4, 8, 12

3rd weighing:

1, 2, 10, 11 vs 4, 5, 8, 9

and other solutions

Let L be the left side is heavier, R mean the right side is heavier, and N mean that they are balanced.

[hide="possibilities"]
[hide="1st: L"]
[hide="2nd: L"]
3rd: L
8 is light

3rd: R
5 is heavy

3rd: N
3 is heavy
[/hide]
[hide="2nd: R"]
3rd: L
Not possible

3rd: R
11 is light

3rd: N
12 is heavy
[/hide]
[hide="2nd: N"]
3rd: L
2 is heavy

3rd: R
10 is light

3rd: N
7 is light
[/hide]
[/hide]
[hide="1st: R"]
[hide="2nd: L"]
3rd: L
11 is heavy

3rd: R
Not possible

3rd: N
12 is light
[/hide]
[hide="2nd: R"]
3rd: L
5 is light

3rd: R
8 is heavy

3rd: N
3 is light
[/hide]
[hide="2nd: N"]
3rd: L
10 is heavy

3rd: R
2 is light

3rd: N
7 is heavy
[/hide]
[/hide]
[hide="1st: N"]
[hide="2nd: L"]
3rd: L
4 is light

3rd: R
1 is light

3rd: N
6 is heavy
[/hide]
[hide="2nd: R"]
3rd: L
1 is heavy

3rd: R
4 is heavy

3rd: N
6 is light
[/hide]
[hide="2nd: N"]
3rd: L
9 is light

3rd: R
9 is heavy

3rd: N
Not possible
[/hide]
[/hide]
[/hide]

word proven answers and you werent supposed to use anything besides your head. also anything is possible until you prove it can't. and if you can't your solution is not complete. Mit is hard when it comes to puzzles

[quote="Nandubetax"][quote="27r"]Marbles:

Weigh 6 marbles on one side and 6 on the other.
Take the heavier side and weigh 3 on one side and 3 on the other.
Take 2 marble from the heavier side and put 1 on one side, and 1 on the other.
If the two marbles balance, the one left out is the heavier one.
If one side is heavier than the other, then obviously, that's the heavier one.[/quote]

:D  Hehe, I already thought of that... then  I realized that it could be heavier[b]or lighter so...[/b][/quote]

Wait, I'm not seeing how it could be heavier or lighter.

[quote="Nandubetax"][quote="mathblitz"]You have 12 marbles. All twelve of them are identical in looks and shape,[b] but 1 of them is the slightest bit lighter or heavier then the other eleven.[/b] You have a scale that is old fashioned so that it only tells whether something is heavier or lighter than something else put on it. \[/quote]

It was given in the problem, since you don't no if its heavier or lighter, which you put all of the marbles on, the one that goes up could contain the lighter marble (if the marble was lighter) and the other side that went down down contain all regular marbles. But on the other hand, the one that went down could contain the abnormal heavier one (if the marble was heavier) and the side that went up could contain all regular marbles.[/quote]

Oh. I see now.

[hide="clue"]step one: seperate the 12 marbles into 3 groups of 4. than in 2 subgroups of 3 and 1.[/hide]

this question has the same value because I did not give the first weighing. the next one in 2-3 days will degrade the cookie point to 1/2. so hurry!

You sure its three and not four steps before the thing breaks

yeah

the answer is as simple as the problem so good luck!

I have a solution, although you didn't say why aleph0's one doesn't work. I believe it's correct, unless one of his cases is wrong.

if 9 or 6 are the ones I doubt he has sufficient information to prove it. especially 6

6 occurs when 1st and 3rd times are balanced but 2nd is not. 6 is the only one that only appears in the 2nd one.
Similarly for 9.

well could you explain it in words because im not that great with number answers... for puzzles.

Is this supposed to be some kind of grammar based question?  Looking at the second question in the original post, it seems like the meaning/answer could change depending on your grammatical interpretation.

after overlooking aleph0's work for a LONGGG TIME I deemed ti a good answer, but it was not the answer MIT was looking for. REMEMBER THIS IS A LOGIC PROBLEM. NOT a problem where there is a solid 3 steps. Aleph0 gets a cookie point for figuring out a different answer first, but I am still holding this problem here until someone figures out the simpler answer which someone would use if he knew nothing about that answer or did not have the brain capacity to hold all that information. LIKE ME!! :D

aleph0's solution has the advantage that the marbles to put in the scale are fixed in advance. With this, you need to figure out as you go:

First, I'll explain how to finish when you're down to 3 marbles (we'll need it later on):
[hide="Three marble technique"]
If you have 3 marbles left (A, B and C), and A and B have been weighed in the same pan of the scale against C earlier, weigh A vs B. If it balances, you can use the earlier result to determine whether C is heavy or light. If it doesn't balance, then you can use the earlier result to determine whether the odd marble is heavy or light, and select one of A and B with that information.
[/hide]

[hide="Solution"]
First weighing: 1, 2, 3, 4 vs 5, 6, 7, 8

If this comes up even, then the odd marble is among 9, 10, 11, 12. Then it's easy, [hide="but for completeness"]
weigh 9, 10 vs 1, 11 (1 is known to be ordinary)

If it balances, the odd marble is 12, and you still have one weighing to determine heavier/lighter.
If it doesn't balance, then use the 3 marble technique on 9, 10, 11 (weigh 9 vs 10).
[/hide]

If it doesn't balance, then one of 1-8 is odd, and 9 is ordinary.

Weigh 1, 2, 5 vs 3, 6, 9

If it balances, then the odd marble is either 4, 7 or 8, at which point we can use the 3 marble technique (weigh 7 vs 8).

If it still doesn't balance, then the odd marble is either 1, 2, 3, 5 or 6 (this is the hardest case).
However, by the previous weighings, we can narrow it down to either 1, 2, 6 or 5, 3. [hide="(Why?)"]

If 1, 2 or 6 is the odd marble, then the heavier pan will be the same in both weighings.
If 3 or 5 is the odd marble, the heavier pan will be different in the two weighings.
[/hide]

At which point use the 3-marble technique to finish it (if you're down to 5, 3 then add a random marble from 1-9 to complete the three. Doesn't really matter...)
[/hide]

VERY good job! the score is now

aleph0: 1 cookie point
ffao:1 cookie point

though ffao' explanation was good I find it better to label them into 3 groups:

group a, group b, and group c.

in each of these groups are 2 subgroups:

A: a1 and a 2-4 B: b1 and b 2-4 and the same with c.

Now the first thing to do is to weigh all of group a with all of group b. if they equal then we know that a marble in c is the odd one out. simply weigh 2-4with any other 3 marbles. besides c1. if they balance measure c1 with any marble. if they do not use the three marble method where the lighter/heavier is found depending on already known knowledge.

now lets say if the first weighing didn't balance:

on the scale currently are a1 a2-4 on the left and b1 and b2-4 on the right.

interchange the scale so that a1 and b2-4 are on the left, b1 and c2-4 are on the right and a2-4 and c1 are on the ground. id the scale balances than whichever way the a group tilted in the first weighing tells us whether it is heavy or lighter. use the 3 marble step method. if they tilt opposite directions than (lets say group b raised at the beginning) than when switched it still rose than group b has a light one. use the 3 marble method. if when you switch the marbles the scale is tilted the same as before than it is a1 or b1. another 3 marble method finishes this up.

new question tomorrow

speed question: pm me answer first three answers that are complete get a cookie poimt. first place gets 2.

there are 2 beakers of different liquids with the same amount in each one. you pour some of the liquid in beaker 1 into beaker 2. with a spoon you then mix it together. then you pour the same amount of liquid back into beaker 1 so that each beaker has the same amount of liquid in it. stir the liquids in beaker one together. compare the liquids in each beaker.

this answer should have a few things:

the answer (duh)
an example that explains the answer
ratio of the liquid of beaker 1 and the liquid of beaker 2 in each beaker.

the ratio may be your answer if your lazy.

Are the cookies still up for grabs? :p
Also is Mit that special society of people with high mental ability? Whats their site and how to register? Please send a PM if you know.
Thanks!

[quote="nuclear_alchemist"][quote="mathblitz"]MIT is a college Massachusetts institute of technology[/quote]
That I know. But then why is this topic titled Mit logic problems? I thought Mit was also an acronym for another organisation I had heard about which dealt with mental ability and stuff.[/quote]
You might be thinking of Mensa.

Oh yes! Thanks MCTP! I wish I could apply for Mensa, but unfortunately there is no centre within say 500 kms from where I live in Lucknow, India.


Let's all say congratulations to Sameer Mishra on his victory in the Scripps National Spelling Bee!(and a chance to be featured on [url=http://www.yahoo.com/]Yahoo![/url] for the day :) )

It [b][i]really[/i][/b] is amazing to have someone from our school win it like this, and that too, someone that many of us know this well.

Wow... this is amazing!

Just wait 'til Anirudh and I make IMO our senior years.

I still can't believe he did this... and to think that he almost lost in school due to a really easy word.

Wherever he is, I congratulate him.

Sameer is starting to become a YouTube sensation! lol

O - "NUMNAH" - O - ".....sorry...Numnah:" - N - U - M - N - A - H "...numnah" = Oonumnah

hehe

aww dang one of my friend's brother got 17th nationally (also from indiana)

but hes only a 7th grader so he should be one of the favorites for next year.

who was your friend? and who is his/her brother?

the guy i was talking about that was on espn.

too bad he lost now i cant rub it in his brothers face.

hes on the spelling bee website, he got out loike 6th or 7th round semifinalist stages.

[quote="MathAndKnowledge"]
Just wait 'til Anirudh and I make IMO our senior years.[/quote]
RIGHT...

_________________
Procrastination is the FASTEST way to get things done.
Talent=Apparent Skill/Work Ethic
If you like chess, go to http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193031 and post if interested in the 5 minute BLITZ TOURNAMENT!

Seconded, maybe Blue MOP, but Black? Not gonna happen.

what

_________________
Procrastination is the FASTEST way to get things done.
Talent=Apparent Skill/Work Ethic
If you like chess, go to http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193031 and post if interested in the 5 minute BLITZ TOURNAMENT!


rotation vector trigonometry integration parameterization calculus geometry

A small circular coil of N turns, radius a and total resistance R is rotated with uniform angular velocity $\omega$ about its vertical diameter in a horizontal magnetic field $\vec B_0 = B_0 \vec i$.

[color=blue][b](1)[/b][/color] Calculate the electromotive force $\epsilon$ induced in the coil and the mean power <P> required to keep the coil in motion. (Neglect the coil self-inductance.)

Let $\vec n$ be a unit vector perpendicular to the coil area. The electromotive force $\epsilon$ is equal to

$\epsilon = -\frac{\text d\Phi}{\text dt}$

where $\phi = \vec B_0 \cdot \vec S = N \pi a^2 B_0 \vec i \cdot \vec n$ is the magnetic flux through the coil. Since the coil is rotating with the uniform angular velocity $\omega$ about the vertical z-axis, $\vec n = \vec i \cos \omega t + \j \sin \omega t$ and

$\epsilon = -\frac{\text d\Phi}{\text dt} = -N \pi a^2 B_0 \frac{\text d(\vec i \cdot \vec n)}{\text dt} = -N \pi a^2 B_0 \frac{\text d(\cos \omega t)}{\text dt} = N \pi a^2 B_0 \omega \sin \omega t$

By Ohm's law, the induced current I through the coil and the dissipated power P are equal to

$I = \frac{\epsilon}{R} = \frac{N \pi a^2 B_0 \omega}{R}\ \sin \omega t$

$P = \epsilon I = \frac{\epsilon^2}{R} = \frac{N^2 \pi^2 a^4 B_0^2 \omega^2}{R}\ \sin^2 \omega t$

The mean dissipated power <P>, equal to the mean power required to keep the coil in motion, is

$\left<P\right> = \frac 1 T \int_0^T P\ \text dt = \frac{N^2 \pi^2 a^4 B_0^2 \omega^2}{R} \cdot \frac{1}{T} \int_0^T \sin^2 \omega t\ \text dt =$

$= \frac{N^2 \pi^2 a^4 B_0^2 \omega^2}{R} \cdot \frac{1}{T} \int_0^T \frac{1 - \cos 2\omega t}{2}\ \text dt =$

$= \frac{N^2 \pi^2 a^4 B_0^2 \omega^2}{R} \cdot \frac 1 T \cdot \left[\frac t 2 - \frac{\sin 2\omega t}{4\omega}\right]_0^T = \frac{n^2 \pi^2 a^4 B_0^2 \omega^2}{2R}$

where $T = \frac{2 \pi}{\omega}$ is the period of the coil rotation. Dimension check:

$[B_0] = \text T = \text{kg} \cdot \text{sec}^{-2} \cdot \text A^{-1}$

$[R] = \Omega = \text{kg} \cdot \text m^2 \cdot \text{sec}^{-3} \cdot \text A^{-2}$

$\left[\frac{n^2 \pi^2 a^4 B_0^2 \omega^2}{2R}\right] = \frac{\text m^4 \cdot \text T^2 \cdot \text{sec}^{-2}}{\Omega} = \frac{\text m^4 \cdot \text{kg}^2 \cdot \text{sec}^{-4} \cdot \text A^{-2} \cdot \text{sec}^{-2}}{\text{kg} \cdot \text m^2 \cdot \text{sec}^{-3} \cdot \text A^{-2}} =$

$= \text{kg} \cdot \text m^2 \cdot \text{sec}^{-3} = \text W = [\left<P\right>]$

A small magnetic needle is placed at the center of the coil, free to turn around the vertical z-axis in the horizontal plane, but unable to follow the rapid motion of the coil.

[color=blue][b](2)[/b][/color] Once the stationary regime is reached, the needle will set in a direction making a small angle $\theta$ with $\vec B_0$. Calculate the resistance R of the coil in terms of this angle and the other parameters of the system.

By Biot-Savart's law, the magnetic field $\vec B$ caused by the current I flowing through the coil is given by

$\vec B = \frac{\mu_0}{4 \pi} \cdot NI \oint \frac{\text d\vec l \times \vec r}{r^3}$

For $\vec B$ at the coil center, the path integral around the coil is easy to calculate, because r = a and

$\frac{d\vec l \times \vec r}{r^3} = \frac{\vec S\ \text d\phi}{r^3} = \frac{\pi a^2 \vec n\ \text d\phi}{a^3} = \frac{\pi \vec n\ \text d\phi}{a}$

$\vec B = \frac{\mu_0}{4 \pi} \cdot NI \int_0^{2\pi} \frac{\pi \vec n\ \text d\phi}{a} = \frac{\mu_0 \pi NI}{2a}\ \vec n = \frac{\mu_0 \pi^2 a^2 N^2 B_0 \omega}{2aR}\ \vec n \sin \omega t =$

$= \frac{\mu_0 \pi^2 a N^2 B_0 \omega}{2R}\ (\vec i \cos \omega t \sin \omega t + \vec j \sin^2 \omega t) = B_x \vec i + B_y \vec j$

Calculating separately the mean values $\left<B_x\right>,\ \left<B_y\right>$ of this magnetic field over the period T of the coil rotation, we need the following integrals:

$\frac 1 T \int_0^T \cos \omega t \sin \omega t = \frac 1 T \int_0^T \frac{\sin 2\omega t}{2} = \frac 1 T \left[-\frac{\cos 2 \omega t}{4\omega}\right]_0^T = 0$

$\frac 1 T \int_0^T \sin^2 \omega t = \frac 1 T \int_0^T \frac{1 - \cos 2\omega t}{2} = \frac 1 T \left[\frac t 2 - \frac{\sin 2 \omega t}{4\omega}\right]_0^T = \frac 1 2$

As a result, $\left<B_x\right> = 0$ and $\left<\vec B\right> = \left<B_y\right> \vec j$, where $\left<B_y\right> = \frac{\mu_0 \pi^2 a^2 N^2 B_0 \omega}{4R}$. The magnetic needle follows the superposition of the external magnetic field $\vec B_0$ with the mean magnetic field $\left<\vec B \right>$ generated by the rotating coil. Hence,

$\tan \theta = \frac{\left<B_y\right>}{B_0} = \frac{\mu_0 \pi^2 a N^2 \omega}{4R}$

$R = \frac{\mu_0 \pi^2 a N^2 \omega}{4\tan \theta}$

Dimension check:

$[\mu_0] = \text H \cdot \text m^{-1} = \text{kg} \cdot \text m \cdot \text{sec}^{-2} \cdot \text A^{-2}$

$\left[\frac{\mu_0 \pi^2 a N^2 \omega}{4\tan \theta}\right] = \text H \cdot \text m^{-1} \cdot \text m \cdot \text{sec}^{-1} = \text{kg} \cdot \text m \cdot \text{sec}^{-2} \cdot \text A^{-2} \cdot \text m \cdot \text{sec}^{-1} =$

$= \text{kg} \cdot \text m^2 \cdot \text{sec}^{-3} \cdot \text A^{-2} = \Omega = [R]$

The current balance consists of 6 identical single-turn coils $C_1, C_2, ..., C_6$ of radius a connected in series. The fixed coils $C_1, C_4$ and $C_3, C_6$ lie in 2 horizontal planes separated by a small distance 2h. The coils $C_2, C_5$ are carried on balance arms of a length d and in equilibrium, they are equidistant from the 2 horizontal planes. The coils $C_3, C_2, C_1$ are located directly above each other and the coils $C_6, C_5, C_4$ are also located directly above each other. The current I flows through the coils in such a direction that the magnetic force on $C_2$ from both $C_1, C_3$ is directed upwards, while the magnetic force on $C_5$ from both $C_4, C_6$ is directed downwards. A small mass m at a distance x from the fulcrum O is required to restore the equilibrium of the balance, when the current flows through the coils.

[color=blue][b](5)[/b][/color] Calculate the force F on $C_2$ due to the magnetic interaction with $C_1$. For simplicity, assume that the force per unit length is the same as the force between 2 long straight wires carrying parallel currents.

The magnetic field $\vec B_1$ generated by an infinite straight wire $L_1$ carrying the current I in the direction $\vec l$ at the radius vector $\vec h$ from the wire is given by Biot-Savart's law:

$\vec B_1(h) = \frac{\mu_0 I}{4 \pi} \int_L \frac{\text d\vec l \times \vec r}{r^3} = \frac{\mu_0 I}{4 \pi}\ (\vec l_0 \times \vec h_0) \int_{-\infty}^{+\infty} \frac{h\ \text dl}{(l^2 + h^2)^{3/2}} =$

$= \frac{\mu_0 I}{4 \pi h}\ \vec k_0 \int_{-\infty}^{+\infty} \frac{\text d\lambda}{(\lambda^2 + 1)^{3/2}} = \frac{\mu_0 I}{4 \pi h}\ \vec k_0 \left[\frac{\lambda}{\sqrt{\lambda^2 + 1}}\right]_{-\infty}^{+\infty} = \frac{\mu_0 I}{2 \pi h}\ \vec k_0$

where $\vec l_0, \vec h_0$ are unit vectors in the directions $\vec l, \vec h$ and $\vec k_0 = \vec l_0 \times \vec h_0$. There are other ways to calculate this magnetic field. From Biot-Savart's law and from symmetry, the magnetic field lines of force are circles concentric with the wire $L_1$. Using one of these circles of radius h for the closed path in Ampere's law

$\oint \vec B_1 \cdot \text d \vec k = \mu_0 I,\ \ 2\pi h \vec B_1(h) = \mu_0 I \vec k_0,\ \ \vec B_1(h) = \frac{\mu_0 I}{2 \pi h}\ \vec k_0$

because the magnetic field magnitude $B_1(h)$ is constant on this circle and $\vec B_1(h) \parallel \text d\vec k \parallel \vec k_0$. Let $L_2$ be another infinite straight wire parallel to $L_1$ at the distance h, carrying the same current I in the same direction. The Lorentz force $\text d\vec F$ on a charge $dQ = I\ \text dt$ moving in the wire $L_2$ at the velocity $\vec v = \frac{\text d\vec l}{\text dt}$ is

$\text d\vec F = \text dQ\ \vec v \times \vec B_1(h) = I\ \text dt\ \frac{\text d\vec l}{\text dt} \times \frac{\mu_0 I}{2 \pi h}\ \vec k_0 = \frac{\mu_0 I^2\ \text dl}{2 \pi h}\ \vec l_0 \times (\vec l_0 \times \vec h_0) = -\frac{\mu_0 I^2\ \text dl}{2 \pi h}\ \vec h_0$

$\frac{\text d\vec F}{\text dl} = -\frac{\mu_0 I^2}{2 \pi h}\ \vec h_0 = -\frac{\mu_0 I^2}{2 \pi h^2}\ \vec h$

Thus we see that the force F per unit length on the wire $L_2$ is directed oppositely to the radius vector $\vec h$ from the wire $L_1$ to the wire $L_2$, in other words, this force is attractive. In an antirely similar way, we could find that if the currents in the 2 wires were flowing in the opposite directions, this force would be repulsive with the same magnitude. As long as the distance h between the coils $C_1, C_2, C_3$ and $C_4, C_5, C_6$ is substantially smaller than their radius a, the result obtained for the infinite parallel wires can be applied to these coils. Thus the magnetic force on $C_2$ due to $C_1$ is

$\vec F_{12} = \oint_{C_1} \frac{\text d\vec F}{\text dl}\ \text dl = \int_0^{2\pi} \frac{\text d\vec F}{\text dl}\ a\ \text d\phi = -\frac{\mu_0 I^2 a}{h}\ \vec h_0 = -\frac{\mu_0 I^2 a}{h^2}\ \vec h$

[color=blue][b](6)[/b][/color] The current I is measured when the balance is in equilibrium. Give the value of I in terms of the physical parameters of the system. The dimensions of the apparatus are such that we can neglect the mutual effects of the coils on the left and on the right.

The torques $\vec T_2,\ \vec T_5$ of the magnetic forces $\vec F_{12} + \vec F_{32},\ \vec F_{45} + \vec F_{65}$ on the coils $C_2, C_5$ with respect to the fulcrum O are

$\vec T_2 = -\vec d \times (\vec F_{12} + \vec F_{32})  = -\vec d \times 2 \vec F_{12},\ \ \vec T_5 = \vec d \times (\vec F_{45} + \vec F_{65})  = \vec d \times 2 \vec F_{45}$

and the total torque of the magnetic forces is

$\vec T_2 + \vec T_5 = \vec d \times 2(\vec F_{45} - \vec F_{12}) = -\vec d \times  4 \vec F_{12} = \frac{4 \mu_0 I^2 a}{h^2}\ \vec d \times \vec h$

In equilibrium, this torque is balanced by the torque $\vec T$ of the additional mass m at the distance x from the fulcrum on the left:

$\vec T = -\vec x \times \vec F = \vec x \times m\vec g$

$\vec T_2 + \vec T_5 + \vec T = 0$

$\frac{4 \mu_0 I^2 a}{h^2}\ \vec d \times \vec h + \vec x \times m\vec g = 0$

If we put the positive x-axis from the left  to the right parallel to the balance arms in equilibrium and the positive x-axis upwards, then $\vec h = -h \vec k,\ \vec g = -g \vec k$ and in the equilibrium, $\vec d = d \vec i,\ \vec x = -x \vec i$.

$-\frac{4 \mu_0 I^2 ad}{h}\ \vec i \times \vec k + mgx\ \vec i \times \vec k = 0$

$-\frac{4 \mu_0 I^2 ad}{h} + mgx = 0$

$I = \sqrt{\frac{mgxh}{4 \mu_0 ad}}$

Dimension check:

$[\mu_0] = \text H \cdot \text m^{-1} = \text{kg} \cdot \text m \cdot \text{sec}^{-2} \cdot \text A^{-2}$

$\left[\sqrt{\frac{mgxh}{4 \mu_0 ad}}\ \right] = \sqrt{\frac{\text{kg} \cdot \text m \cdot \text{sec}^{-2} \cdot \text m^2}{\text{kg} \cdot \text m \cdot \text{sec}^{-2} \cdot \text A^{-2} \cdot \text m^2}} = \sqrt{\text A^2} = \text A = [I]$

yetti you're a beast :\.

I definitely enjoy reading your sokutions :).

[quote="Spoon"]yetti you're a beast :|.[/quote]
Himalayas, Himalayas ...

[quote="Spoon"]I definitely enjoy reading your solutions :).[/quote]
Thank you Spoon, it is nice to know that somebody reads them. I am trying to see, if I still can do these things, because after years of digging in specialized areas, I forgot a lot of the general stuff.

Yetti

Let M be the mass of the balance (except for m and the hanging parts), G its center of mass and l the distance $\overline{OG}$.

(7) The balance is stable agains deviations producing small changes $\delta z$ in the height of $C_2$ and $-\delta z$ in the height of $C_5$. Calculate the maximum value $\delta z_{\text{max}}$ so that the balance still returns to the equilibrium position when released.

Let $\theta$ be the oriented angle of the balance arms from the horizontal (equilibrium) position. If $\delta x > 0$, then $\theta < 0$ (the right arm is lower then the left arm). Consider the balance arms as a physical pendulum with the mass M and reduced length $l = \overline {OG}$. When using the reduced length (the distance from the fulcrum O to the center of mass G), the equations of motion of a physical pendulum are the same as those of a mathematical pendulum of the length l. Thus the pendulum potential energy is

$\delta V_P = Mgl(1 - \cos \theta) \approx \frac 1 2 Mgl \theta^2$

the force returnig the pendulum into the equilibrium position is

$\delta \vec F_P = -\frac 1 l\ \frac{\text d(\delta V_p)}{\text d\theta}\ \vec t_0 \approx -Mg \theta\ \vec t_0$

where $\vec t_0$ is a unit vector tangent to the circle centered at the fulcrum O and described by the center of mass G. Let $\vec l_0$ be a unit vector in the direction of $\vec l = \overline{OG}$. Since both the unit vectors $\vec l_0, \vec t_0$ lie in the zx-plane and since $\vec l_0 \perp \vec t_0$, their cross product is equal to $\vec l_0 \times \vec t_0 = -\vec j$. The torque of the force $\vec F_p$ with respect to the fulcrum O is then

$\delta \vec T_P = \vec l \times \vec F_P \approx -Mgl \theta\ \vec l_0 \times \vec t_0 = Mgl \theta\ \vec j$

When the balance was in equilibrium, the torque $\vec T_2 + \vec T_5$ of the magnetic forces $\vec F_{12},\ \vec F_{32},\ \vec F_{45},\ \vec F_{65}$ on the coils $C_2, C_5$ and the torque of the gravitational force $\vec F$ on the additional mass m were

$\vec T_{2,\text{eq}} + T_{5,\text{eq}} = -\vec d \times \vec F_{12} - \vec d \times \vec F_{32} + \vec d \times \vec F_{45} + \vec d \times \vec F_{65} =$

$= -\vec d \times 4\vec F_{12} = \frac{4 \mu_0 I^2 a}{h^2}\ \vec d \times \vec h = -\frac{4 \mu_0 I^2 ad}{h}\ \vec i \times \vec k = \frac{4 \mu_0 I^2 ad}{h}\ \vec j$

$\vec T_{\text{eq}} = -\vec x \times \vec F = \vec x \times m\vec g = mgx\ \vec i \times \vec k = -mgx\ \vec j$

(compare with the previous post and use $\vec d = d \vec i,\ \vec h = -h \vec k,\ \vec x = -x \vec i,\ \vec g = g \vec k$ and $\vec i \times \vec k = -\vec j$ for the correct signs). If the balance is not in equilibrium by the angle $\theta$, the torque of the additional mass m becomes

$\vec T = \vec x \times m\vec g = mgx \sin(90^\circ + \theta)\ \vec j  = mgx \cos \theta\ \vec j  \approx mgx (1 - \frac{\theta^2}{2}) \vec j $

$\delta \vec T = \vec T - \vec T_{\text{eq}} \approx -\frac{mgx \theta^2}{2} \vec j$

and the torque of the magnetic forces becomes

$\vec T_2 + T_5 = \vec d \times (\frac{\mu_0 I^2 a}{h - \delta z} + \frac{\mu_0 I^2 a}{h + \delta z} + \frac{\mu_0 I^2 a}{h - \delta z} + \frac{\mu_0 I^2 a}{h - \delta z}) \vec h_0 =$

$= -\vec d \times (\frac{2\mu_0 I^2 a}{h - \delta z} + \frac{2\mu_0 I^2 a}{h + \delta z}) \vec k = -d \sin (90^\circ + \theta) (\frac{2\mu_0 I^2 a}{h - \delta z} + \frac{2\mu_0 I^2 a}{h + \delta z}) \vec i \times \vec k =$

$= d \cos \theta\ \frac{2\mu_0 I^2 a (h - \delta z + h + \delta z)}{h^2 - (\delta z)^2}\ \vec j = \frac{4\mu_0 I^2 ahd \cos \theta}{h^2 - (\delta z)^2}\ \vec j$

For small angles $\theta$, we can substitute $\cos \theta \approx 1 - \frac{\theta^2 }{2}$ as before. In addition, $\delta z = -d \sin \theta \approx -d \theta$ and

$\frac{hd \cos \theta}{h^2 - (\delta z)^2} = \frac{d \cos \theta}{h [1 - (\frac{\delta z}{h})^2]} \approx \frac{d \cos \theta}{h} [1 + (\frac{\delta z}{h})^2] = \frac{d \cos \theta}{h} [1 + (\frac{d \sin \theta}{h})^2] \approx$

$\approx \frac d h (1 - \frac{\theta^2}{2})  (1 + \frac{d^2 \theta^2}{h^2}) \approx \frac d h [1 + (\frac{d^2}{h^2} - \frac 1 2 ) \theta^2]$

$\vec T_2 + \vec T_5 = \frac{4\mu_0 I^2 ad}{h} [1 + (\frac{d^2}{h^2} - \frac 1 2 ) \theta^2]\ \vec j$

$\delta \vec T_2 + \delta \vec T_5 = (\vec T_2 + \vec T_5) - (\vec T_{2,\text{eq}} + \vec T_{5,\text{eq}}) \approx \frac{4\mu_0 I^2 ad}{h} (\frac{d^2}{h^2} - \frac 1 2 ) \theta^2\ \vec j$

up to the 2nd order terms $\theta^2$. The total additional torque torque acting on the balance out of the equilibrium is

$\delta \vec T_2 + \delta \vec T_5 + \delta \vec T + \delta \vec T_P = \frac{4\mu_0 I^2 ad}{h} (\frac{d^2}{h^2} - \frac 1 2 ) \theta^2\ \vec j + \frac{mgx \theta^2}{2}) \vec j + Mgl \theta\ \vec j$

Recall that for $\delta z > 0$, the angle $\theta < 0$ (the right arm is lower then the left arm). For the balance to return to the equilibrium position, this additional torque must be in the direction opposite to the unit vector $\vec j$, which means that it must be

$\frac{4\mu_0 I^2 ad}{h} (\frac{d^2}{h^2} - \frac 1 2 ) \theta^2 + \frac{mgx \theta^2}{2}) + Mgl \theta < 0$

$\frac{4\mu_0 I^2 ad}{h} (\frac{d^2}{h^2} - \frac 1 2 ) \theta + \frac{mgx \theta}{2}) + Mgl > 0$ [color=white].[/color] (because $\theta < 0$)

Using the equilibrium condition $\frac{4\mu_0 I^2 ad}{h} - mgx = 0$, this simplifies to

$\frac{4\mu_0 I^2 ad^3}{h^3} \theta + Mgl > 0$

and using the same condition again,

$\frac{mgxd^2}{h^2} \theta + Mgl > 0$

$0 > \theta > -\frac{Mlh^2}{mxd^2}$

If we considered the angle of the balance arms from the horizontal (equilibrium) position $\theta > 0$ (the right arm is higher than the left arm), we could similarly deduce

$0 < \theta < +\frac{Mlh^2}{mxd^2}$

Combining,

$|\theta| < |\theta_{\text{max}}| \approx \frac{Mlh^2}{mxd^2} = \frac{Mlh^2}{md^2} \cdot \frac{mgh}{4 \mu_0 I^2 ad} = \frac{Mglh^3}{4 \mu_0 I^2 ad^3}$

$|\delta z| < |\delta z_{\text{max}}| \approx |d \theta_{\text{max}}| \approx \frac{Mlh^2}{mxd} = \frac{Mglh^3}{4 \mu_0 I^2 ad^2}$

It is not surprising that for rather small balance deviations from the equilibrium, the relatively strong (electro)magnetic force between the coils overcomes the relatively weak gravitational force keeping the balance in the equilibrium. This is an effect similar to a little magnet lifting an iron pin against the gravitational pull of the entire Earth.

The experimental setup consists of 2 identical metal disks D, D' of radius b mounted on a conducting shaft S. A motor rotates the shaft S at an adjustable angular velocity $\omega$. 2 identical coils C, C' of radius a and with N turns each surround the disks The coils are connected in such a way that the current I flows in them in opposite directions.

[color=blue][b](3)[/b][/color] Assuming that the current flowing through the coils C, C' creates uniform magnetic fields around the diskd D, D', equal to the fields at the coil centers. Calculate the electromotive force between the rims of the disks, assuming that the distance between the disks is much larger than the radii of the coils anf that $a \gg b$.

Let R be the measured resistance and V the voltage applied to the circuit. If the coil resistance was negligible, the conducting path through the coil C', the coil C, the disk D, the shaft S and the disk D' would present a short circuit for the voltage source. Let $R_C$ be the resistance of each coil. The resistances $R_C, R, R_C$ are connected in series, hence, assuming that the disk brushes are disconnected, the current flowing through the coils C, C' is

$I = \frac{V}{R + 2R_C}$

and the voltage drop across the measured resistance R is

$V_R = IR = V\ \frac{R}{R + 2R_C}$

Let $\vec n$ be a unit vector in the direction of the shaft S from the disk D to the disk D' and $\vec \omega = \omega \vec n$ the angular velocity of the shaft. Assume that the current I flows in the coils C, C' in such directions that the magnetic field at the center of C has the same direction as $\vec n$, while the magnetic field at the center of C' has the opposite direction. According to the answer to the question (2), the magnetic fields $\vec B,\ \vec B'$ at the centers of the coils C, C' are given by

$\vec B = -\vec B' = \frac{\mu_0 \pi NI}{2a}\ \vec n$

A free charge q at the radius vector $\vec r$ from the center of the metal disk D is subject to the Lorentz force

$\vec F = q\ \vec v \times \vec B = q\ (\vec r \times \vec \omega) \times \vec B = q\ [(\vec r \cdot \vec B) \vec \omega - (\omega \cdot \vec B)]  = -q\omega B\ \vec r$

because $\vec r \perp \vec B,\ \vec r \cdot \vec B = 0$ and $\vec \omega \parallel \vec B,\ \omega \cdot \vec B = \omega B$. This is equivalent to an induced electric field

$\vec E(\vec r) = \frac{\vec F}{q} = -\omega B\ \vec r$

The electromotive force (the induced voltage) $\epsilon_{43}$ from the center 3 and the rim 4 of the disk D is given by the integral of this electric field from the center to the rim along any disk radius L (cannot use R, because that's the measured resistance):

$\epsilon_{43} = \int_L \vec E \cdot \text d\vec r = -\int_0^b \omega B r\ \text dr = -\frac{\omega Bb^2}{2}$

Similarly, the electromotive force (the induced voltage) $\epsilon_{12}$ from the center 2 to the rim 1 of the disk D' is

$\epsilon_{12} = -\epsilon_{43} = \frac{\omega Bb^2}{2}$

Consequently, the total electromotive force (the total induced voltage) $\epsilon_{41}$ from the rim 1 of the disk D' to the rim 4 of the disk D is

$\epsilon_{41} = \epsilon_{43} + \epsilon_{32} + \epsilon_{21} = \epsilon_{43} - \epsilon_{12} = -2 \epsilon_{12} = -\omega B b^2$

because $\epsilon_{32} = 0$ (the shaft is conducting) and $\epsilon_{43} = -\epsilon_{12} = \epsilon_{21}$.

The disks D, D' are connected to the circuit through brush contacts at their rims (presumably when the disks are rotating, in order to avoid the short circuit in case the resistance of the coils C, C' is negligible).

[color=blue][b](4)[/b][/color] The resistance R is measured when the galvanometer G reads zero current. Give the resistance R in terms of the physical parameters of the system.

The current through the galvanometer is zero, when the induced voltage $\epsilon_{14}$ is exactly equal to the voltage drop $V_R$ across the resistor R developped by the circuit voltage source with the disk brushes disconnected.

$V_R = \epsilon_{14} = -\epsilon_{41}$

$\frac{VR}{R + 2R_C} = \omega B b^2 = \frac{\mu_0 \pi NI \omega b^2}{2a} = \frac{\mu_0 \pi N \omega b^2}{2a} \cdot \frac{V}{R + 2R_C}$

$R = \frac{\mu_0 \pi N \omega b^2}{2a}$

Thus we see that the measured value of the resistance R depends neither on the applied voltage V, nor on the coil resistance $R_C$, the only requirement is $V \neq 0$, so that it is possible to reduce the last but one equation by V.
Dimension check:

$[\mu_0] = \text H \cdot \text m^{-1} = \text{kg} \cdot \text m \cdot \text{sec}^{-2} \cdot \text A^{-2}$

$\left[\frac{\mu_0 \pi N \omega b^2}{2a}\right] = \frac{\text H \cdot \text m^{-1} \cdot \text{sec}^{-1}  \cdot \text m^2}{\text m} = \frac{\text{kg} \cdot \text m \cdot \text{sec}^{-2} \cdot \text A^{-2} \cdot \text{sec}^{-1}  \cdot \text m^2}{\text m} =$

$= \text{kg} \cdot \text m^2 \cdot \text{sec}^{-3} \cdot \text A^{-2} = \Omega = [R]$


function
inequalities
induction
algebra proposed
algebra

Find all functions $ f,g:R_+ \longrightarrow R_+ $ such that:
$f(g(x))=\frac{x}{xf(x)-2}$ and $g(f(x))=\frac{x}{xg(x)-2}$   for  $x>0$

Simple.
No such functions exist.
Note that since f,g>0 we have that f(x),g(x) > 2/x.
Now let the sequence an be defined as follows: a1=2 and 4an - an-1 - 7 =0 for all n>=2.
I shall prove that f(x),g(x) > an/x for all n>=1. Then since an has an explicit form an = A x1^n + B x2^n where both x1,x2>1
then the serie an diverges and so no two function exist.
To prove the inequality I shall employ induction. It is true for n=1.
Suppose true for n=k, then x/xf(x)-2=f(g(x))> ak/g(x) Now let ak=a.
So xaf(x) - xg(x) <2a and similarly xag(x)-xf(x) <2a
adding we get x(a-1)(f+g) <4a so that f+g<[4a/(a-1)]/x
Now this implies that f,g < b/x where b=4a/a-1
So x/xf(x)-2 = f(g(x)) < b/g(x) < (b/a)x and therefore  xf(x)-2 > (a/b) so that f(x) > [a/b +2]/x
but one can easily check that a/b +2 = ak+1
So that f(x),g(x) > ak+1/x the induction is complete and the result follows.
-Ali

There is a small problem in my solution, i need to work on it.
indeed f(x)=g(x)=3/x is a solution!
i will be back on this...

I don't remember where I saw this  :?
Anyway, I found the following relation, but I don't know if it will be useful
$g\left( x\right) -\frac{1}{g\left( f\left( x\right) \right) }=f\left( x\right) -\frac{1}{f\left( g\left( x\right) \right) }$ for all $x>0$

o.k it was much simpler than i expected:

Define F(x)=xf(X) and G(x)=xg(x)
then 1/F(g(x)) - 1/G(f(x)) = 2(1/F(x) - 1/G(x))
Now let s and d be supermum and infimum of {F(x),G(x)}. Note that s,d exist, see my first post.
Now 1/F(x)-1/G(x) =1/2( 1/F(g(x)) - 1/G(f(x)) ) = 1/4 (1/F(g(g(x)) - 1/G(f(f(x)) ) =1/8(...) =...
but note that 1/F - 1/G is a bounded expression and therefore F=G=constant
so f(x)=c/x and g(x)=d/x checking we get that c=d=3...
-Ali

[quote="ali"]Define F(x)=xf(X) and G(x)=xg(x)
then 1/F(g(x)) - 1/G(f(x)) = 2(1/F(x) - 1/G(x))[/quote]
How can you prove that $\frac{1}{F(g(x))} - \frac{1}{G(f(x))} = 2(\frac{1}{F(x)} - \frac{1}{G(x)})$ ??
If we suppose that this relation is true then :
$\frac{1}{F(g(x))} - \frac{1}{G(f(x))} = \frac{1}{g(x)f(g(x))} - \frac{1}{f(x)g(f(x))}$
$=\frac{1}{g(x)\frac{x}{xf(x)-2}}- \frac{1}{f(x)\frac{x}{xg(x)-2}}$
$=\frac{F(x)-2}{G(x)} - \frac{G(x)-2}{F(x)}$
$=\frac{F(x)}{G(x)}- \frac{G(x)}{F(x)} + 2(\frac{1}{F(x)}-\frac{1}{G(x)})$
$=2(\frac{1}{F(x)}-\frac{1}{G(x)})$ (as you said)
which means that : $\frac{F(x)}{G(x)}- \frac{G(x)}{F(x)} =0$
Hence : $F(x)=G(x)$ and $f(x)=g(x)$
but we do not have that  :?

But your proof shows that F=G but not F=cte

Method of step-by-step-estimation can be used here. First we know $\alpha_0 >x f\left(x\right),\  x g\left(x\right) >\beta_0$ where $\alpha_0=\infty,\ \beta_0=2$. Suppose we have $\alpha_n >x f\left(x\right),\   x g\left(x\right) >\beta_n$, by simple arithmatic we get $\alpha_{n+1} >x f\left(x\right),\   x g\left(x\right) >\beta_{n+1}$, where $\alpha_{n+1}=\frac{2\beta_n}{\beta_n-1},\ \beta_{n+1}=\frac{2\alpha_n}{\alpha_n-1}$, it's easy to know that $\alpha_n,\ \beta_n \to 3$, so $f\left(x\right)=g\left(x\right)=\frac{3}{x}$.


Hello,

Just in case someone wants to walk arround and watch (I don't know if it is still possible to register), but next Saturday February 26th is the Second Anual Georgia Math Competition. It is in the GATech campus, I suppose it is going to be in the Instructional Center, but for sure there will be some add telling exactly where it is going to be outside the Skiles Building (686 Cherry St.)

Best regards,

[quote="solafidefarms"]I'm going in JV division :-) Will you be there? :?

Billy[/quote]

Yes Billy, I will, but not as a competitor  :D , it has been a long time since the last time I compited in a math competition. I shope to see you there, my picture is in the Avatar!, and Good luck  :first:

Best regards,

[quote="lyf3xb3xsw33t"]Hey, I'm going to the math competition at GA Tech, too! But they chose a bad day: it conflicts with science olympiad and some orchestra things, so we only have 2 people on our team.  :(  We'll see how that goes. I'm going to be on the Varsity team for Parkview High School. Hope to see you there! :lol:

Linda[/quote]
Good!, I'll see you there Linda!

Best,

Hello,

Well, the GAHSMC was yesterday. I met two AoPSers, both of them made it to the semifinal, one in Varsity, and one in Junior Varsity, and one of them (even being an 8th grader) was one of the first 5 places in Junior Varsity (usually intended for 9th and 10th grader).

It was a wonderful day, even when it costed me 40 hours awake in a row (and I was not one of the organizers in charge!!), and I have to say that I had never seen that much food to disapear in that few time (1020 donouts, 25 galons of orange juice, 1200 soda cans, 300 bottles of water, 175 pizzas, 600 disserts, and other few things)

There were about 400 participants (students) plus, of course, some parents and several coaches.

Best regards,

Unfortunately, I did not see the other AoPSer, but did see a USAMTSer who was not a frequent AoPSer. I still like the fact I got 4th!

I didn't know that you were Linda Liu! You have done pretty well in the past, haven't you?

Billy

[quote="lyf3xb3xsw33t"]heh, yeah. I'm Linda Liu. And well, last year was my small claim to fame. 2nd in junior varsity. But I've been slacking off lately.  :P  Life's been hectic.[/quote]

But any way you did it quite good. Try hard and I am sure you may reach the first place in Varsity next year.

Best,


Let $ X$ be a compact metric space, and $ f: X \to X$ an isometry. Show that $ f$ is bijective.

You posted this already! :mad: ... and the solution was given. See [url=http://www.mathlinks.ro/viewtopic.php?t=228548]here[/url].


rotation
analytic geometry
geometry
geometric transformation

[quote="GameBot"]When the point $ (4, 1)$ is rotated 90 degrees counterclockwise about the point $ (1, 0)$, what are the coordinates of the point at which it lands?[/quote]

You move up 1, right 3 to get from (1, 0) to (4, 1). Counterclockwise, you instead move left 1, up 3. This gives $ \boxed{(0,3)}$

It seems right, but could someone please explain why that works?
Thanks :)

Since the two points are $ 3$ and $ 1$ apart...

You just do the opposite to get a correct rotation of $ 90$ degrees.


probability
geometry
probability and stats

Selecting a point P in random in the circle $ S\equal{}{(x,y) : x^2\plus{}y^2<\equal{}R^2}$.

Whats the probability that the interval of length R centered at P and parallel to the x-axis is contained in S.

How do I do that one?

thanks!

[quote="RealTime"]How do I do that one?[/quote]

Find the set of points for which the desired property holds.  Calculate its area.  Then divide by the area of the disk.  (This is purely a geometry problem.)


symmetry
number theory unsolved
number theory

let x,y and z positive integers
show that (xy+1)(xz+1)(yz+1) is a perfect square if and only if xy+1, yz+1, and zx+1 are perfect squares

Surfing on the web I found finally a solution to the problem, which I post here.

We proceed for contradiction. Suppose there are three non-negative integers x, y and z such that (xy+1)(yz+1)(zx+1) is a perfect square but at least one of xy+1, yz+1 or zx+1 isn't a perfect square. Suppose wlog that $x\leq y\leq z$. The "magic recipe" to solve the problem is to observe that defining the integer number w as \[w=2xyz+x+y+z-2\sqrt{(xy+1)(yz+1)(zx+1)}\] the four numbers x, y, z, w satisfy the symmetric equation \[2(x^{2}+y^{2}+z^{2}+w^{2})-(x+y+z+w)^{2}-4xyzw-4=0\] so that, for symmetry, we also have \[2xyw+x+y+w-z=\pm 2\sqrt{(xy+1)(yw+1)(wx+1)}\] and so (xy+1)(yw+1)(wx+1) is a perfect square.
Also, the equation linking the four numbers can be rewriten in the three forms \[(x+y-z-w)^{2}=4(xy+1)(zw+1)\]  \[(y+z-x-w)^{2}=4(yz+1)(xw+1)\]  \[(z+x-y-w)^{2}=4(zx+1)(yw+1)\] from where it is immediate to prove that w is non-negative and that at least one of the numbers xy+1, yw+1, wx+1 is not a perfect square.
And we finish by observing that \[(2xyz+x+y+z-2\sqrt{(xy+1)(yz+1)(zx+1)}) (2xyz+x+y+z+2\sqrt{(xy+1)(yz+1)(zx+1)}) = (z-x-y)^{2}-4(xy+1)<z^{2}\] implies $w<z$.

SUMMING UP: starting from a non-negative solution $(x,y,z)$ we obtain another non-negative solution $(x,y,w)$ such that $x+y+z>x+y+w$, and we can proceed inductively obtaining a third solution from $(x,y,w)$, and so ad infinitum.
CONTRADICTION


check out this boring question-

let x=1,y=1
x=y
=x^2=yx
=x^2-y^2=yx-y^2
=(x+y)(x-y)=y(x-y)
=x+y=y

Tell me whats wrong in this.its a pathetic question

[hide] See this

$3\times 0 = 5\times 0 $ If divide $0$ , $\frac{3\times 0}{0}=\frac{5\times 0}{0}$ so become $ 3=5 $ .  Try to figure out why this wrong and that will answer your question ....  :D  [/hide]

yep,same thing.the part where x-y is cancelled is wrong.told ya this sum is pathetic.

Surprisingly enough, I didn't think of that at first.

i thought it was that addition thingy....x+y=y
*feels stupid*

Wait, if you think this should be in mathcounts, why isn't it?


Did anyone take this competition?

scores (Green Level):

        -- case number --
                  1  2  3  4  5  6  7  8  9 10
navigate    *  *  *  *  *  x   t   t   t
marathon   *  *  *  *  *  *  *  *  t  t
dquery       *  *  *  *  *  *  t   t   t  t
dstats        *  *  *  *  *  *  t   t   t  t

and i have absolutely no idea what my number score is going to be from this test data :)

It's hard to predict the numerical score from the test data. Different problems and different test cases can be weighted very differently. But that certainly looks pretty good. Are you hoping to get to camp?

Hmmm... well that would certainly be nice.

Quite honestly I just started doing these in november, and I was recently forced into green, so this is my first green-level competition.

But you had a perfect score in Orange in December. Sounds like a great start to me!

navigate    *  *  *  *  *  *  t  s  t
marathon    *  *  *  *  *  *  *  t  t  t
dquery      *  *  *  *  *  *  *  t  t  t
dstats      *  *  *  *  *  *  *  t  t  t

yeah i think I made the same mistake you did...

I also know not one weighted tree algorithm, although they tried to trick people into not realizing these were all trees. (They gave you the number of farms AND the number of roads? Please. It should be pretty obvious roads = farms - 1.) So you must have done the same thing I did on marathon too. When looking for maximal distances between two farms, I only checked distances where one was an "endpoint" (i.e. only had one road coming off it). That seemed to get me those couple extra test cases correct.

So now I have to ask my teacher for weighted tree algorithms. There HAVE to be some, and this contest (with the exception of question 1, and kind of 4) were standard results, (Distance between two points, diameter of the graph) so I found them kind of annoying.

Im still wondering what I did wrong on 4... oh well. I should get answers by next week.

i am not very familiar with these tree algorithms... do you guys want to maybe illustrate how the optimum solution to solve these problems?

ps topcoder is nice... i like it...  texas137  what is your id on topcoder?

>"ps topcoder is nice... i like it... texas137 what is your id on topcoder?"

I don't actually program. My son does this stuff. (I guess that makes me a groupie.) I'm just passing along what he says about it. He did a few rounds about a year ago, but none recently. Many of the top names on Topcoder also made it to the camp level or better in USACO.

[quote="texas137"]do you guys know about Topcoder? The contests are not nearly as intelligent as USACO, but they occur every week and you might enjoy doing some of them for extra practice.
http://www.topcoder.com[/quote]
The contests on Topcoder are not easy. They're geared toward developing actual software.

[quote="Setofallsets"]Nukular, this is pretty funny - I got exactly the same results.  Did you get the 6th test case in "navigate" wrong because they asked about the distance from a node to itself?

I felt pretty dumb afterwards because there were nice (and not [i]too[/i] hard) solutions to a couple of them.  It rather bothers me that all of the problems on this contest were [i]tree[/i] algorithms, especially since I really didn't know in what direction to go with them.  I'd really like to see more problems like the "cowq" one on the December contest, which don't require much background but do require a good bit of ingenuity and some mathematics.[/quote]

I still can't find the tree algorithms, and <i>Kolstad won't give them to me!</i> Good job anyhow, it wasn't hard to figure out who you were since there are only two 437's.

Overall, I found the results... surprising. I know a lot of people that took green, but don't keep contact with many.

>"The contests on Topcoder are not easy. They're geared toward developing actual software."

No, Topcoder isn't "easy". When I said the problems are less intelligent than USACO, I meant that they were intellectually less elegant, not that they are easy (remember, I'm passing along someone else's opinion here). The contests are different. But many present or former USACO people show up on Topcoder. It's worth taking a look at.

its "easy" because it is expected that a coder is to complete 3 problems in 75 minutes.  talking about div 1, the easy problem (easy/medium/hard) everyone usually gets it...  the medium problem takes some work, which takes a little bit less work, which constitutes an hard USACO orange problem i'd say or easy USACO green... the catch is you must do it within the time limit, and 100% correct code that gets everything.  the hard one is usually tricky, and most ppl cant get it...  its usually DP or some graph problem for the majority part.

hahaha,  I luv all the DP problems that I can do, and hate the all the rest (which is a majority).  go check out TopCoder SRM 183 Div 1 hard (Ambigram) for a hard DP problem.

where on the site do they have past problems?

either go to the contest applet, and goto practice rooms and select the respective contest, or goto stats, and select the contest and then view the problem.  contest applet lets you submit solutions and test them also