algebra
polynomial
modular arithmetic
Divisibility
number theory
IMO
IMO 1984

Find one pair of positive integers $a,b$ such that $ab(a+b)$ is not divisible by $7$, but $(a+b)^7-a^7-b^7$ is divisible by $7^7$.

So we want $7 \nmid ab(a+b)$ and $7^7 | (a+b)^7-a^7-b^7 = 7ab(a+b)(a^2+ab+b^2)^2$, so we want $7^3 | a^2+ab+b^2$.
Now take e.g. $a=2,b=1$ and get $7|a^2+ab+b^2$. Now by some standard methods like Hensels Lemma (used to the polynomial $x^2+x+1$, so $b$ seen as constant from now) we get also some $\overline{a}$ with $7^3 | \overline{a}^2+\overline{a}b+b^2$ and $\overline{a} \equiv a \equiv 2 \mod 7$, so $7\nmid \overline{a}b(\overline{a}+b)$ and we are done. (in this case it gives $\overline{a}=325$)

I found in without Hensels Lemma (which I don't know :( ), just with few computations that putting $b=1$, $a\equiv_7 2$ and $a^3\equiv_{7^3}1$ gives $a=2^{\cdot 7^2}$ which works... so $a\equiv_{7^3}324$... so $(a,b)=(324,1)$ is a solution

can someone fully explain the solution to this problem?
i dont fully understand ZetaX's solution.
a=325 is not a solution :wink:

the smallest solution would be 18,1

is there a way to find all the solutions?

[quote="ZetaX"]$325 \to 324$ ;)

Hensel guarantates us to find all solutions, simply by it's way of working. You can take a look at it on the web.[/quote]

sorry, i dont understand,  could you explain better?!?

isnt $\overline{a}$ also a solution?

[quote="Kalva"]We find that $(a+b)^{7}-a^{7}-b^{7}= 7ab(a+b)(a^{2}+ab+b^{2})^{2}$. So we need find a, b such that $a^{2}+ab+b^{2}$ is divisible by $7^{3}$.

At this point I found $a = 18$, $b = 1$ by trial and error.

A more systematic argument turns on noticing that $a^{2}+ab+b^{2}= \frac{a^{3}-b^{3}}{a-b}$, so we are looking for $a$, $b$ with $a^{3}\equiv b^{3}\pmod{73}$. We now remember that $a^{\varphi(m)}\equiv 1 \pmod m$. But $\varphi(73) = 2 \cdot 3 \cdot 49$, so $a^{3}\equiv 1 \pmod{343}$ if $a = n^{98}$. We find $2^{98}\equiv 18 \pmod{343}$, which gives the solution 18, 1.

This approach does not give a flood of solutions. $n^{98}= 0, 1, 18,$ or $324$. So the only solutions we get are $(1, 18); (18, 324); (1, 324)$.[/quote]

[Moderator: Added Source]

[quote="chien than"]We find that $(a+b)^{7}-a^{7}-b^{7}= 7ab(a+b)(a^{2}+ab+b^{2})^{2}$. So we need find a, b such that $a^{2}+ab+b^{2}$ is divisible by $7^{3}$.

At this point I found $a = 18$, $b = 1$ by trial and error.

A more systematic argument turns on noticing that $a^{2}+ab+b^{2}= \frac{a^{3}-b^{3}}{a-b}$, so we are looking for $a$, $b$ with $a^{3}\equiv b^{3}\pmod{73}$. We now remember that $a^{\varphi(m)}\equiv 1 \pmod m$. But $\varphi(73) = 2 \cdot 3 \cdot 49$, so $a^{3}\equiv 1 \pmod{343}$ if $a = n^{98}$. We find $2^{98}\equiv 18 \pmod{343}$, which gives the solution 18, 1.

This approach does not give a flood of solutions. $n^{98}= 0, 1, 18,$ or $324$. So the only solutions we get are $(1, 18); (18, 324); (1, 324)$.[/quote]

Please dont copy and paste other solutions!!
this is a copy of the Kalva's solution:
http://www.kalva.demon.co.uk/imo/isoln/isoln842.html
at least post the source! :wink:

That's poor, indeed. And it explains the mass of really stupid and foreseeable $\LaTeX$ mistakes I had to remove.

[quote="ZetaX"]After heavily editing of the previous post, I've time to answer now  :D

I think the explaination at http://en.wikipedia.org/wiki/Hensel%27s_lemma is good enough, tell me if you have specific problems.[/quote]
sorry, im bad at NT :blush: ,
this is as far as i get it:
ok, so let $f(a)=a^{2}+a+1$ (b=1)
then $f(2)\equiv 0 \pmod{7}$
so by Hensels lemma there exists an integer $0\le \overline{a}\le 6$ s.t.
$f(2+7\overline{a})\equiv 0 \pmod{7^{2}}$
or $(2+7\overline{a})^{2}+2+7\overline{a}+1\equiv 0\pmod{7^{2}}$
now what?
i dont see how you get $\overline{a}=325$?

[quote="orl"]Find one pair of positive integers $ a,b$ such that $ ab(a \plus{} b)$ is not divisible by $ 7$, but $ (a \plus{} b)^7 \minus{} a^7 \minus{} b^7$ is divisible by $ 7^7$.[/quote]
By writing everything in terms of $ a\plus{}b$ and $ ab$ we easily get $ 7^3 \mid a^2\plus{}ab\plus{}b^2$. Setting $ z \equal{} ab^{\minus{}1}$, where $ b^{\minus{}1}$ is the inverse of $ b$ $ \bmod 7$. We get $ 7^3 \mid z^2\plus{}z\plus{}1 \mid 4z^2\plus{}4z\plus{}4 \equal{} (2z\plus{}1)^2\plus{}3$. Then first $ 7 \mid (2z\plus{}1)^2\plus{}3$, gives a solution $ z \equiv 2 \bmod 7$. Then substitute $ z \equal{} 7k\plus{}2$, and works $ \bmod 7^2$, and repeat this 'till you get a solution. I found $ z \equal{} 37$. Then $ (a,b) \equal{} (37,1)$ is a solution :)

How to get all solutions to this problem?

@3above, this isn't exactly an NT NT problem, at least the way I did it... it's more like expand everything, group together and cancel terms till you get $343 \mid a^2 +ab+b^2$, and since you just need to find $1$ solution, arbitrarily put $b=1$ and bash it out to hit $18$ as your first solution- actually, it's not even a proper bash , as it's pretty simple to estimate first solution should be kind of around of $18$ (as $324$ is pretty close to $343$)... btw the stats for this year is pretty... well, gold was a $40$ :)

At the IMO I did something like this to find a pair $(a,b)$ such that $7^3\big|a^2+ab+b^2$.
First realized that either $a\equiv 2b\pmod7$ or $b\equiv 2a\pmod7$; I chose the first one.
Then I took the square of $a-2b$:
$$ 0 \equiv (a-2b)^2 = a^2+ab+b^2 -b(5a-3b) \pmod{49} $$
so we need $49\big|5a-3b$.
Took squares again:
$$ 0 \equiv (5a-3b)^2 = 25(a^2+ab+b^2)-b(55a-34b) \pmod{7^4} $$
... and then I was the stranger who submitted $(34,55)$ instead $(1,18)$. :-)

It could be done in one step, by taking cubes:
$$ 0 \equiv (a-2b)^3 = (a-7b)(a^2+ab+b^2) + b^2(18a-b) \pmod{7^3} $$
so, under the assumption $a\equiv 2b\pmod 7$, we can see that $7^3\big|a^2+ab+b^2$ is equivalent with $18a\equiv b\pmod{7^3}$.

You can not always assume a Coefficient be 1 (0, -1, etc.) and find it works.
I solve "a^2+a*b+b^2" by the method of exhaustion (Code).
1 18
2 36
3 54
4 72
5 90
6 108
8 144
9 162
10 153
11 134
12 115
13 96


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be a group such that
$Z(G) \cong \mathbb{Z}/3\mathbb{Z}$
and
$G/Z(G) \cong S_{3}$
($S_{3}$ is symmetric group on 3 letters).

Show that $G$ must be isomorphic to $(\mathbb{Z}/3\mathbb{Z})\times S_{3}$.

The group $G/Z(G)$, being isomorphic to $S_{3}$, is generated by the subset $\Delta$ of its elements of order $3$. For each element $\tilde{g_{i}}\in \Delta$, let $g_{i}$ be a representent of $\tilde{g_{i}}$ in $G$. Then the subgroup $K=<\{g_{i}\}>$ has order $3!=6$ or $18$. Since all the $g_{i}$ have order $3$ or $9$, each of them is contained in a $3$-Sylow of $G$. But $G$ has only one $3$-Sylow $H$ (because $1$ is the only divisor of $|G|$ which is $1 \ mod \ 3$), whence $\forall i, g_{i}\in H$, $K \subseteq H$, and $|K|=6$. Thus $K$ is a set of representents of $G$ and is also a subgroup of $G$ (*); the mapping $\phi: G/Z(G) \rightarrow Z(G)$ defined by $\phi((\tilde{g_{i}},z_{i}))=g_{i}z_{i}$ is easily checked to be bijective (since $|G/Z(G)|.|Z(G)|=|G|$). It also defines an homomorphism, since  $\phi((\tilde{g_{i}g_{j}},z_{i}z_{j}))=g_{i}g_{j}z_{i}z_{j}=g_{i}z_{i}g_{j}z_{j}=\phi((\tilde{g_{i}},z_{i}))\phi((\tilde{g_{j}},z_{j}))$. Note that the first equality is made possible by (*), while the second one is made possible by the fact that $z_{i}\in Z(G)$.

Unfortunately I don't know what a Sylow is. This is from an exam in a first, basic, algebra course. Is it possible to translate the solution to newbie terms?

Well, in this case, I'm afraid there is a much easier solution I missed out ... a $3$-Sylow is a subgroup of $G$ with order $3^{u}$, where $u$ is maximal (in our case, $u=2$)


inequalities
LaTeX
trigonometry
geometry
angle bisector
triangle inequality
geometry unsolved

In an acute - angled triangle ABC, the following lines are concurrent :
(1) the angle bisector of angle $\angle A$ ;
(2) the median through B and the midpoint of [AC] ;
(3) the altitude through C.
Show that $\angle A > 45^{\circ}$.

[i]Alternative formulation.[/i] The bisector $AD$, the median $BM$ and the height $CH$ of the acute-angled triangle $ABC$ intersect in one point. Hereby, $D$ lies on $BC$, $M$ lies on $AC$, and $H$ lies on $AB$.
Prove that the angle $\measuredangle BAC$ is greater than $45$ degrees.

[quote="Arne"]In an acute - angled triangle ABC, the following lines are concurrent :
(1) the angle bisector of angle $\angle A$ ;
(2) the median through B and the midpoint of [AC] ;
(3) the altitude through C.
Show that $\angle A > 45^{\circ}$.[/quote]

Let D be the foot of the altitude from C, E the foot of the bisector of angle A, and X a point on the line AB s.t. CX perpendicular to AC.

From Menelaus' thm for triangle ADC and transversal line EB we get DE/EC*1*AB/DB=1, so AD/AC=ED/EC=DB/AB (*). Let's assume that A<=45 deg. This means AC>=CX>XD, so AC^2>AC*DX, so AD*AX>AC*DX iff AD/AC>DX/AX iff (from (*)) DB/AB>DX/AX (**). Since we know angle ACB<90=angle ACX, it means that AB=AX-k and DB=DX-k with k positive. We got (DX-k)/(AX-k)>DX/AX iff AX<DX, which is false, so we have found a contradiction => our assumption that A<=45 deg is false.

[quote="Arne"][i]Alternative formulation.[/i] The bisector $AD$, the median $BM$ and the height $CH$ of the acute-angled triangle $ABC$ intersect in one point. Hereby, $D$ lies on $BC$, $M$ lies on $AC$, and $H$ lies on $AB$.
Prove that the angle $\measuredangle BAC$ is greater than $45$ degrees.[/quote]

[hide]I don't know how to latex and my solution will probably have lots of holes in it.

Let the sides opposite of angles ABC be abc respectively.

Since all the cevians intersect at one point we can apply Angle Bisector theorem and Ceva's theorem.

b/c*bcosA=/(c-bcosA)=1
c*cosA/(c-bcosA)=1
ccosA=c-bcosA
cosA=c/(c+b)

If angle A is less than 45 degrees then its cosine is less than radical two over two

c/(c+b)>sqrt(2)/2
c>(sqrt(2)+1)b

now we can reenter this back

(b^2+c^2-a^2)/2/b/c>sqrt(2)/2
((4+2sqrt(2))b^2-a^2)/(3+2sqrt(2))/b^2>sqrt(2)
((4+2sqrt(2))b^2-a^2 )>(3sqrt(2)+4)b^2
-((sqrt(2))b^2>a^2

which is weird[/hide]

[quote="Arne"][i]Alternative formulation.[/i] The bisector $AD$, the median $BM$ and the height $CH$ of the acute-angled triangle $ABC$ intersect in one point. Hereby, $D$ lies on $BC$, $M$ lies on $AC$, and $H$ lies on $AB$.
Prove that the angle $\measuredangle BAC$ is greater than $45$ degrees.[/quote]

some progress-

By Ceva, $1=\frac{AH.CM.BP}{HB.AM.PC}=\frac{AH.BP}{HB.PC}=\frac{AH.AB}{HB.AC}=\frac{bc \cos A}{b(c-b\cos A)}\Rightarrow cos A=\frac{c}{c+b}$

Also, to prove $A>45^\circ$ it is sufficient to prove $\cos A < \frac{1}{\sqrt2}\Leftrightarrow \frac{b}{b+c}>\sqrt2-1$

Can somebody finish this?

EDIT: Sorry, ehe~ got in first. I'll post anyway... :)

The triangle inequality holds too.....

 :maybe:

vulgarfraction, i'm stuck there too.......

 :huh:

[Note that the problem is unsolved in the Pre-Olympiad section: http://www.mathlinks.ro/Forum/viewtopic.php?t=104180 ]

[quote="Arne"][i]Alternative formulation.[/i] The bisector $AD$, the median $BM$ and the height $CH$ of the acute-angled triangle $ABC$ intersect in one point. Hereby, $D$ lies on $BC$, $M$ lies on $AC$, and $H$ lies on $AB$.
Prove that the angle $\measuredangle BAC$ is greater than $45$ degrees.[/quote]

Assume contrary. Then suppose they cevians concur at P. Let X be the point on AB such that $\angle ACX = 90$, and let XP meet AC at Y. By Menelaus,

$\frac{AY}{CY}= \frac{AX}{XH}\cdot \frac{HP}{CP}= \frac{AX}{XH}\cdot \frac{AH}{AC}= \frac{AC}{XH}> 1$ (in fact 45 degrees is not optimal; the optimum is whatever satisfies $\sin^{2}\theta = \cos \theta$)

So that means B must lie farther from A than X, and so $\angle ACB > 90$, which is bad.


please how old the contestants in maldova TST are?.

It is junior TST. The contestants are under 15,5 years old.

[quote="Ahiles"]Find all integers  $ (x,y,z)$, satisfying equality:
$ x^2(y \minus{} z) \plus{} y^2(z \minus{} x) \plus{} z^2(x \minus{} y) \equal{} 2$[/quote]
[hide]
First assume either $ x\ge y\ge z$ or $ z\ge y\ge x$

Rearrange the given equation to \[ xz(z\minus{}x)\plus{} y(x^2\minus{}z^2)\equal{} 2\plus{}y^2(x\minus{}z)\] Since $ x\minus{}z\equal{}0$ leads to a contradiction, divide: \[ y(x\plus{}z)\minus{}xz\equal{}\frac{2}{x\minus{}z}\plus{}y^2\]
In case $ x\ge y\ge z$, get $ x\minus{}z\equal{}1,2$, and in case $ z\ge y\ge x$, get $ z\minus{}x\equal{}1,2$

------------------
First for $ x\minus{}z\equal{}1$, there are two cases:

[u]Case 1:[/u] $ x\minus{}y\equal{}1$ and $ y\minus{}z\equal{}0$
This leads to $ z^2\minus{}y^2\equal{}2$ which has no solutions

[u]Case 2:[/u] $ x\minus{}y\equal{}0$ and $ y\minus{}z\equal{}1$
This leads to $ x^2\minus{}y^2\equal{}2$ which has no solutions

$ z\minus{}x\equal{}1$ works a similar way, so no solutions for $ x\minus{}z\equal{}\pm1$

------------------
Second, for $ x\minus{}z\equal{}2$, there are three cases:

[u]Case 1:[/u] $ x\minus{}y\equal{}2$ and $ y\minus{}z\equal{}0$
This leads to $ 2z^2\minus{}2y^2\equal{}2$ or $ z^2\minus{}y^2\equal{}1$ which is a contradiction to $ y\minus{}z\equal{}0$

[u]Case 2:[/u] $ x\minus{}y\equal{}0$ and $ y\minus{}z\equal{}2$
This leads to a similar contradiction as Case 1

$ z\minus{}x\equal{}2$ works a similar way for these two cases

------------------
[u]Case 3:[/u] $ x\minus{}y\equal{}1$ and $ y\minus{}z\equal{}1$
First we get $ x^2\minus{}2y^2\plus{}z^2\equal{}2$
Here $ x,y,z$ are consecutive integers so plug it into the above equation:
\[ (y\plus{}1)^2\minus{}2y^2\plus{}(y\minus{}1)^2\equal{}2\]
which is always true. So $ (x,y,z)\equal{}\boxed{(k\plus{}1,k,k\minus{}1)}$ works for integer $ k$

Now for $ z\minus{}x\equal{}2$, the case is $ z\minus{}y\equal{}1$ and $ y\minus{}x\equal{}1$
First we get $ \minus{}x^2\plus{}2y^2\minus{}z^2\equal{}2$
Here $ z,y,x$ are consecutive integers so plug it into the above equation:
\[ \minus{}(y\minus{}1)^2\plus{}2y^2\minus{}(y\plus{}1)^2\equal{}2\]\[ (y\minus{}1)^2\minus{}2y^2\plus{}(y\plus{}1)^2\equal{}2\] which is a contradiction, so no solutions.

------------------
So $ (x,y,z)\equal{}(k\plus{}1,k,k\minus{}1)$ for integer $ k$.[/hide]

Note that the expression $ x^2(y \minus{} z) \plus{} y^2(z \minus{} x) \plus{} z^2(x \minus{} y)$ factors as $ (x \minus{} y)(x \minus{} z)(y \minus{} z)$.

@post 4 nice sol! The key idea of ur solution is just to use a lemma[hide=.]2/(x-z) is an integer, which we can prove by rearranging into terms in multiples of (x-z).[/hide] Then use a ton of casework to get sols of consecutive three numbers. @above that is a very nice factorization, and now we can try x-y=1, x-z=2, y-z=1, x-y=2, and it’s negative versions of casework, which I leave as an exercise to others, and then simply solve.


modular arithmetic
number theory unsolved
number theory

Prove that there exist infinite many narual numbers $n$ such that:
a) $n^2+n+1|(n^2)!$;
b) $n^2+n+1|n!$.

For a), take $n=3m+1$. Then $n^2+n+1=9m^2+6m+1+3m+1+1=3(3m^2+3m+1)$. Since $n^2=9m^2$, both $3$ and $3m^2+3m+1$ are factors of $(n^2)!$ and we're done (take large enough $m$).

For b), take $n$ a perfect square, say, $n=k^2$. Then $n^2+n+1=k^4+k^2+1=(k^2+k+1)(k^2-k+1)$. Now we want this to divide $(k^2)!$. Obviously now any large enough $k \equiv 1 \pmod{3}$ works.

Here is my solution:

[b]lemma:[/b].
There exists infinitely many pair (a;b) such that $(2a+1)^2+3=13b^2$ with   $(a_0=3;b_0=2)$
Use lemma, for any (a;b) such that b>13. we will show that

$(\frac{\sqrt{13b^2-3}-1}{2})! \vdots 13b^2.$

because
$\frac{\sqrt{13b^2-3}-1}{2}>2y>13 \Rightarrow (\frac{\sqrt{13b^2-3}-1}{2})! \vdots 13b^2=4$$(a^2+a+1).\Rightarrow a!\vdots a^2+a+1$ and we're done.
i'm sorry, my English is very bad.


It has been proven that every integers not of the form $ 9n\pm 4$ is expressible as the sum of four cubes.

http://mathworld.wolfram.com/CubicNumber.html

[quote="manjil"]Every multiple of $ 3$ can be expressed as the sum of four cubes.

Can we replace $ 3$ here by any even number?[/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?t=134862


calculus
integration
ratio
geometry
perimeter
number theory unsolved
number theory

Call a scalene triangle K [i]disguisable[/i] if there exists a triangle K′ similar to K with two shorter sides precisely as long as the two longer sides of K, respectively. Call a disguisable triangle [i]integral[/i] if the lengths of all its sides are integers.
(a) Find the side lengths of the integral disguisable triangle with the smallest possible perimeter.
(b) Let K be an arbitrary integral disguisable triangle for which no smaller integral
disguisable triangle similar to it exists. Prove that at least two side lengths of K are
perfect squares.

[hide="Solution, but not sure!"]
Well, two equal triangles are similiar, so we take the smallest positive integers: 1,1 and 1!!!     :ninja:           [/hide]

Bugi, triangles are scalene (raznostranicni), right?

Let $ K$ have side lengths $ a \ge b \ge c$; then $ K'$ has side lengths $ d \ge a \ge b$ and the similarity condition implies

$ \frac {a}{d} \equal{} \frac {b}{a} \equal{} \frac {c}{b} < 1$

which implies $ d, a, b, c$ is a geometric progression.  Write its common ratio as $ \frac {r}{s}, (r, s) \equal{} 1$; then we can write $ (a, b, c) \equal{} (ks^2, ksr, kr^2)$ for integers $ k, s, r$.  (Note that $ d \equal{} \frac {ks^3}{r}$ need not be an integer.)

a) The integral disguisable triangle of minimal perimeter occurs when $ k \equal{} 1, r \equal{} 2, s \equal{} 3$; this gives $ 4 \plus{} 6 \plus{} 9 \equal{} 19$.  (This is the valid common ratio of minimal height; $ r \equal{} 1, s \equal{} 2$ violates the triangle inequality.)

b) A primitive integral disguisable triangle must have $ k \equal{} 1$ and is therefore of the form $ (s^2, sr, r^2)$.

Call a scalene triangle K [i]disguisable[/i] if there exists a triangle $K'$ similar to $K$ with two shorter sides precisely as long as the two longer sides of $K$, respectively. Call a disguisable triangle [i]integral[/i] if the lengths of all its sides are integers.
(a) Find the side lengths of the integral disguisable triangle with the smallest possible perimeter.
(b) Let $K$ be an arbitrary integral disguisable triangle for which no smaller integral disguisable triangle similar to it exists. Prove that at least two side lengths of $K$ are perfect squares.


inequalities
inequalities proposed
algebra

This one is very nice. I may be wrong, but isn't there a more general inequality (which I don't know how to solve) due to Kiran Kedlaya?
  First of all, we observe that a*(a+b)/2*(a+b+c)/3>=1/27*(a+a+a)(a+sqrt(ab)+b)(a+b+c). From AM-Gm we deduce that (x=a+a+a, y=a+sqrt(ab)+b, z=a+b+c):
  a/x+a/y+a/x>=3a/(xyz)^1/3, a/x+sqrt(ab)/y+b/z>=3sqrt(ab)/(xyz)^1/3 and a/x+b/y+c/z>=3(abc)^1/3/ (xyz)^1/3. Sum up these inequalities and you will find that  1/27*(a+a+a)(a+sqrt(ab)+b)(a+b+c) is at least ((a+sqrt(ab)+(abc)^1/3)/3)^1/3 and the coclusion follows.

[quote="Mircea Lascu"]Let a, b, c be positive real numbers. Show that
$(*)$ $\frac{a + \sqrt{ab} + \sqrt[3]{ abc}}{3} \leq \sqrt[3]{a\cdot\frac{a+b}{2}\cdot\frac{a+b+c}{3}}.$
Establish when equality holds.[/quote]
Let $a=x^6,$ $b=y^6$ and $c=z^6,$ where $x,y,z$ are positive numbers.
Then $(*)\Leftrightarrow\sqrt[3]{\frac{x^6+y^6}{2}\cdot\frac{x^6+y^6+z^6}{3}}\geq\frac{x^4+xy^3+y^2z^2}{3}.$
But $\frac{x^6+y^6}{2}\geq\frac{x^6+x^3y^3+y^6}{3}.$ Hence, remain to prove that
$\sqrt[3]{\frac{x^6+x^3y^3+y^6}{3}\cdot\frac{x^6+y^6+z^6}{3}}\geq\frac{x^4+xy^3+y^2z^2}{3}\Leftrightarrow$
$\Leftrightarrow\sqrt[3]{3(x^6+x^3y^3+y^6)(x^6+y^6+z^6)}\geq x^4+xy^3+y^2z^2\Leftrightarrow$
$\Leftrightarrow\left(1^3+1^3+1^3\right)^{\frac{1}{3}}\left((x^2)^3+(xy)^3+(y^2)^3\right)^{\frac{1}{3}}\left((x^2)^3+(y^2)^3+(z^2)^3\right)^{\frac{1}{3}}\geq$
$\geq1\cdot x^2\cdot x^2+1\cdot xy\cdot y^2+1\cdot y^2\cdot z^2,$ which true by holder's inequality.

(a+a+a+a+a+a)(a+a+a+a+a+a)(a+a+a+b+b+b)(a+a+b+a+b+b)(a+a+b+b+c+c)(a+a+b+b+c+c) $\ge 8(a+\sqrt{ab}+\sqrt[3]{abc})^3$ by holder's inequality.

What about
$\frac{a+\sqrt{ab}+\sqrt[3]{abc}+\sqrt[4]{abcd}}{4}\leq\sqrt[4]{a\cdot\frac{a+b}{2}\cdot\frac{a+b+c}{3}\cdot\frac{a+b+c+d}{4}}$

?? :wink:

the general inequality is true, and it's exactly what kedlaya proved in his paper, isn't it? :maybe:

[quote="Mircea Lascu"]Let a, b, c be positive real numbers. Show that
(a + \sqrt ab + 3\sqrt abc)/3 \leq 3\sqrta(a+b)/2 (a+b+c)/3.
Establish when equality holds.[/quote]
Can anyone retype the problem? I can not read it.

[quote="sqing"]Let a, b, c be positive real numbers. Show that\[{a + \sqrt{ab} + \sqrt[3]{ abc}} \leq \frac{4}{3}(a+b+c).\][/quote]
${a + \sqrt{ab} + \sqrt[3]{ abc}} =a +\frac{ \sqrt{4ab}}{2} + \frac{\sqrt[3]{ 64abc}}{4}$

$\leq a +\frac{a+4b}{4} + \frac{a+4b+16c}{12}=\frac{4}{3}(a+b+c).$

[quote="Lovasz"]Applying Minkowski’s inequality, \begin{eqnarray*}\sqrt[3]{(a+a+a)\left(a+\frac{a+b}{2}+b\right)(a+b+c)}&\ge& \sqrt[3]{a^{3}}+\sqrt[3]{ab\frac{a+b}{2}}+\sqrt[3]{abc}\\ &\ge& a+\sqrt[3]{ab\sqrt{ab}}+\sqrt[3]{abc}\\ &\ge& a+\sqrt{ab}+\sqrt[3]{abc}\end{eqnarray*}[/quote]
________
Very very nice! Congratulations!
The problem is solved very elegantly in " Inequalities,theorems,technique and selected problems", author Z.CVETKOWSKI.(see pb 97,pag.272)

[quote="sqing"]Let $a, b $ be positive real numbers. Show that\[a + \sqrt{ab} \leq \frac{ \sqrt{2}+1}{2}(a+b).\]
For $a, b, c , d$  be positive real numbers , have \[{{a + \sqrt{ab} + \sqrt[3]{ abc}}+\sqrt[4]{ abcd}}  \leq \lambda (a+b+c+d),\]
then $\lambda=?$[/quote]
For $a, b, c , d$  be positive real numbers , have ${{a + \sqrt{ab} + \sqrt[3]{ abc}}+\sqrt[4]{ abcd}}  \leq \lambda (a+b+c+d),$ then $\lambda_{min}=?$

[quote="sqing"]Let $a, b $ be positive real numbers. Show that\[a + \sqrt{ab} \leq \frac{ \sqrt{2}+1}{2}(a+b).\][/quote]
$\frac{ \sqrt{2}+1}{2}(a+b)=a+(\frac{ \sqrt{2}-1}{2}a+\frac{ \sqrt{2}+1}{2}b)\ge a + \sqrt{ab}.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=549731&hilit=Slovenia+2012]Generalization of Slovenia 2012:[/url]
Let $a$, $b$, $c$ and $d$ are non-negative numbers. Prove that:
\[a+\sqrt{ab}+\sqrt[3]{abc}+\sqrt[4]{abcd}\leq\frac{10}{7}(a+b+c+d).\]
If $a,$ $b,$ $c$ are positive real numbers, [url=https://artofproblemsolving.com/community/c6h348191p1867758]then[/url]
$$3\left(a+\sqrt{ab}+\sqrt[3]{abc}\right) \le 2\left(4+\frac{\sqrt{ab}}{a+b}\right) \sqrt[3]{a\cdot \frac{a+b}{2}\cdot \frac{a+b+c}{3}}.$$
Let $a,b,c>0$ ,[url=https://artofproblemsolving.com/community/c6h216090p1200135] prove that[/url]
$$ 3(a + \sqrt {ab}+\sqrt [3]{abc})\leq (8 + \frac {a+b}{2\sqrt {ab}} ){\sqrt [3]{a\cdot \frac {a + b}{2}\cdot \frac {a + b + c}{3}}}$$
[url=https://artofproblemsolving.com/community/c6h230072p1273967]h[/url]

[quote="sqing"]For $a, b, c , d$  be positive real numbers , have ${{a + \sqrt{ab} + \sqrt[3]{ abc}}+\sqrt[4]{ abcd}}  \leq \lambda (a+b+c+d),$ then $\lambda_{min}=?$[/quote]
$\lambda_{min}=\frac{86+25\sqrt{13}}{108}.$
For $a, b, c , d,e$  be positive real numbers , have ${{{a + \sqrt{ab} + \sqrt[3]{ abc}}+\sqrt[4]{ abcd}}+\sqrt[5]{ abcde}}   \leq \lambda (a+b+c+d+e),$ then $\lambda_{min}=?$

The first inequality has been appeared here https://www.facebook.com/pages/B%E1%BA%A5t-%C4%90%E1%BA%B3ng-Th%E1%BB%A9c-Inequality/874561135893908 since august 2014 . There are two proofs : mine and mister Marian Dinca's proof.
Here is my proof ( The Lemma was invented by me and was already sent and accepted by AMM Magazine ) :


agha in soali ro ke mikham begam 100 sale ke daram roosh fek mikonam vali be hich natijei nemiresam:

ye graphe 2n+1 rasi darim.be ezaye har n raas yek raase digar vojood darad ke be in n raas vasl hast....sabet konid ke ke rasi vojood dare ke be hameye baghieie rous vasle..... :(

inam halesh saman joon hala mitooni beri be hadiano alipouro lotfio hame soalo bajavab begi :P
http://www.mathlinks.ro/Forum/viewtopic.php?t=51


vector
real analysis
real analysis unsolved

The temperature at the point (x,y) on a metal plate is $T(x,y)=50-x^{2}-2y^{2}$. If an ant is located at the point P(4,3), find its path if it continuously seeks the coldest points on the plate.

Since $\nabla T(4,3)$ pointing towards the maximum value at that point the answer would just be $-\nabla T(4,3)$, right?

Which is:
$\nabla T=\left< -2x, -4y\right>$ therefore $-\nabla T(4,3)=\left< -8, -12\right>$
My only problem is that this seems almost too simple.  :D

That's becuase it is too simple. As soon as the ant moves, it will be at a different point, and the gradient at that different point will point in a different direction. The path the and follows will be a curve, and that curve will be the solution of a differential equation.

Okay, I have spent the last hour searching for a section in my book that shows how to make a  curve that "will be the solution of a differential equation." Everything I have read points to getting a maximum/minimum or the gradient or the directional derivative.

Okay, I found this obscure reference in my class notes. Unfortunately, I don't understand what I wrote:
$\vec{r}(t)=\left<x(t),y(t)\right>$ = path of ant $\frac{dr}{dt}=\left<x'(t),y'(t)\right>$
$\nabla T=\left<-2x,-4y\right>$ motion towards coldest (direction of motion)
direction = $-\nabla T$ $-\nabla T \parallel \vec{r}'(t) \therefore \vec{r}'(t)=k\nabla T$
$\frac{dy}{dt}=kA$ and $\frac{dx}{dt}=kB$ $\therefore$ $\frac{dy}{dx}=\frac{A}{B}@(4,3)$

Unfortunately, that doesn't clear it up for me....heh. Maybe one of you guys can figure out what my notes mean...heh.


quadratics
inequalities
Cauchy Inequality

Reals $a,b,c>0$, Prove $\sum\sqrt{a^{2}+bc}\leq \frac{3}{2}(a+b+c)$

As I know a lot of ineqs. with $a^{2}+bc$ are very hard, if you have some other ineqs in this type, you can post it here and share with us.

please delete this post, i think i was wrong  :blush:

considering the following quadratic equations
$\frac{1}{4}ax^{2}+bx-c=0$
$\frac{1}{4}bx^{2}+cx-a=0$
$\frac{1}{4}cx^{2}+ax-b=0$

they all have some interesting discriminants
$\Delta_{1}=b^{2}+ac$
$\Delta_{2}=c^{2}+ab$
$\Delta_{3}=a^{2}+bc$

observe that we have not reduced generality of the problem,
hence we have brought no restriction so far.

add all 3 equations and one gets , after some calculations:
$x^{2}-4x-4=0$
wich has roots $x_{1,2}=2\cdot (-1 \pm \sqrt{2})$
(let $x_{1}$ be the biggest of the two roots)

this is the important step in the proof :[b]
but we can also find x as beeing a solution of the equalities
above that is[/b]

$x=\frac{-b+\sqrt{\Delta_{1}}}{\frac{a}{2}}$
$x=\frac{-b+\sqrt{\Delta_{2}}}{\frac{b}{2}}$
$x=\frac{-b+\sqrt{\Delta_{3}}}{\frac{c}{2}}\ (2)$

in the above 3 equations after some calculations we get to

$a\frac{x_{1}}{2}+b \geq \sqrt{\Delta_{1}}$
$b\frac{x_{1}}{2}+c \geq \sqrt{\Delta_{2}}$
$c\frac{x_{1}}{2}+a \geq \sqrt{\Delta_{3}}$

adding them we get

$(a+b+c)(1+\frac{x_{1}}{2})\geq \sum \sqrt{\Delta_{i}}$

but $(1+\frac{x_{1}}{2})=+\sqrt{2}$
so we shall have a better inequality than the first:

$\sum \sqrt{\Delta_{i}}\leq\sqrt{2}(a+b+c) <1.43(a+b+c)<\frac{3}{2}(a+b+c)$

please correct me if i am wrong

p.s. i have corrected some errors

[quote="spx2"]
$\sum \sqrt{\Delta_{i}}\leq\sqrt{2}(a+b+c)$
[/quote]
It's wrong for $a=b$ and $c\rightarrow0^{+}.$ :wink:
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63645

arqady you are  right,i'm sorry if i mislead anyone,
now i'm looking at what i wrote and i can't find the
mistake,please help me

[quote="spx2"]... i can't find the
mistake,please help me[/quote]
I think, the mistake here:
[quote="spx2"]... after some calculations we get to

$a\frac{x_{1}}{2}+b \geq \sqrt{\Delta_{1}}$
$b\frac{x_{1}}{2}+c \geq \sqrt{\Delta_{2}}$
$c\frac{x_{1}}{2}+a \geq \sqrt{\Delta_{3}}$
[/quote] :wink:

arqady yes,you are right,the explanation would be
that if $c=0$ or $c \to 0$ then there is a big error division by 0
at formula $(2)$.

does the proof hold for all other cases?
(by other cases i mean those that do not include conditions like
 $a\to 0$ , $b\to 0$ , $c\to 0$)

p.s. i have modified so that it can be seen what formula $(2)$ is

[quote=Ji Chen][quote="libra_gold"]If $ x,y,z$ be nonnegative real numbers, then

$ \sqrt {x^{2} \plus{} yz}\plus{}\sqrt {y^{2} \plus{} zx}\plus{}\sqrt {z^{2} \plus{} xy}\leq \frac {3}{2}(x \plus{} y \plus{} z).$[/quote]Using the Cauchy inequality,

$ \left(\sum{\sqrt{x^2\plus{}yz}}\right)^2\leq\sum{\left(3x^2\plus{}4yz\plus{}zx\plus{}xy\right)}\sum{\frac{x^2\plus{}yz}{3x^2\plus{}4yz\plus{}zx\plus{}xy}}$

$ \equal{}3(x\plus{}y\plus{}z)^2\sum{\frac{x^2\plus{}yz}{3x^2\plus{}4yz\plus{}zx\plus{}xy}},$

we only need to prove

$ \frac{x^2\plus{}yz}{3x^2\plus{}4yz\plus{}zx\plus{}xy}\plus{}\frac{y^2\plus{}zx}{3y^2\plus{}4zx\plus{}xy\plus{}yz}\plus{}\frac{z^2\plus{}xy}{3z^2\plus{}4xy\plus{}yz\plus{}zx}\leq \frac{3}{4},$

with equality if $ x\equal{}0,y\equal{}z$.

[hide="Proof."]$ \frac{3}{4}\minus{}\sum{\frac{x^2\plus{}yz}{3x^2\plus{}4yz\plus{}zx\plus{}xy}}\equiv\frac{F(x,y,z)}{4\left(3x^2\plus{}4yz\plus{}zx\plus{}xy\right)\left(3y^2\plus{}4zx\plus{}xy\plus{}yz\right)\left(3z^2\plus{}4xy\plus{}yz\plus{}zx\right)}$.

$ F(x,y,z)\equal{}F(x,x\plus{}s,x\plus{}t)$

$ \equal{}243x^6\plus{}486(s\plus{}t)x^5\plus{}3\left(104s^2\plus{}301st\plus{}104t^2\right)x^4$

$ \plus{}(s\plus{}t)\left(74s^2\plus{}439st\plus{}74t^2\right)x^3\plus{}\left(5s^4\plus{}101s^3t\plus{}228s^2t^2\plus{}101st^3\plus{}5t^4\right)x^2$

$ \plus{}st(s\plus{}t)\left(5s^2\plus{}22st\plus{}5t^2\right)x\plus{}8s^2t^2(s\minus{}t)^2\geq 0$,

which is clearly ture for $ x\equal{}\min \{x,y,z\}$.[/hide][/quote]

What makes you think of the factors $3x^2\plus{}4yz\plus{}zx\plus{}xy$?

i have much simpler solution than all, finally.
WLOG, a+b+c=3
(a*a+b*c+1)\2>=√a*a+bc
Similarly,applying in remaining three..
And, a*b+b*c+c*a>=a+b+c=3.
I hope,my solution is right.
Please, reply,if there is some mistake..

[quote=bel.jad5][quote=Loserfistte]And, a*b+b*c+c*a>=a+b+c
I'm sure of it. Try to prove it first yourself, it's not tough.[/quote]

take $a=b=0$[/quote]

Sorry to all, i made mistake.

[quote=libra_gold]Thanks arqady.

I also see this ineq unsolved in that post.

$\frac{1}{\sqrt{4a^{2}+bc}}+\frac{1}{\sqrt{4b^{2}+ac}}+\frac{1}{\sqrt{4c^{2}+ab}}\ge \frac{4}{a+b+c}$[/quote]

How about: $$\frac{1}{\sqrt{4a^2+bc}}+\frac{1}{\sqrt{4b^2+ca}}+\frac{1}{\sqrt{4c^2+ab}}\ge\frac{2}{\sqrt{ab+bc+ca}}$$

[quote=jokehim]
How about: $$\frac{1}{\sqrt{4a^2+bc}}+\frac{1}{\sqrt{4b^2+ca}}+\frac{1}{\sqrt{4c^2+ab}}\ge\frac{2}{\sqrt{ab+bc+ca}}$$[/quote]

see [url=https://artofproblemsolving.com/community/c6h51089p24385918]here[/url]

Assume $a\geq b\geq c$，then
$$\begin{aligned}
&\dfrac{3}{2}(a+b+c)\\
=&\dfrac{a+(a+c)+(a+b)+(2b+2c)}{2}\\
\geq&\sqrt{a(a+c)}+\sqrt{(a+b)(2b+2c)}\\
=&\sqrt{a^2+ac}+\sqrt{2(b^2+ca+bc+ab)}\\
\geq&\sqrt{a^2+ac}+\sqrt{b^2+ca}+\sqrt{bc+ab}\\
\geq&\sqrt{a^2+bc}+\sqrt{b^2+ca}+\sqrt{c^2+ab}
\end{aligned}$$

[quote=Zuyong][quote=libra_gold]Thanks arqady.

I also see this ineq unsolved in that post.

$\frac{1}{\sqrt{4a^{2}+bc}}+\frac{1}{\sqrt{4b^{2}+ac}}+\frac{1}{\sqrt{4c^{2}+ab}}\ge \frac{4}{a+b+c}$[/quote]

Any idea?[/quote]

\begin{align*}\frac1{\sqrt{4a^2+bc}}+\frac1{\sqrt{4b^2+ca}}+\frac1{\sqrt{4c^2+ab}}&=\sqrt{\sum_{\rm cyc}\frac1{4a^2+bc}+2\sum_{\rm cyc}\frac1{\sqrt{{\left(4b^2+ca\right)}{\left(4c^2+ab\right)}}}}\text,\\&\geqslant\sqrt{\sum_{\rm cyc}\frac1{4a^2+bc}+2\sum_{\rm cyc}\frac{4{\left[{\left(4b^2+ca\right)}+{\left(4c^2+ab\right)}\right]}}{{\left(4b^2+ca\right)}^2+6{\left(4b^2+ca\right)}{\left(4c^2+ab\right)}+{\left(4c^2+ab\right)}^2}}\text,\\&\overset?>\frac4{a+b+c}\textnormal.\end{align*} But I cannot see this ineq in that post.

[quote=Rhapsodies_pro][quote=Zuyong][quote=libra_gold]Thanks arqady.

I also see this ineq unsolved in that post.

$\frac{1}{\sqrt{4a^{2}+bc}}+\frac{1}{\sqrt{4b^{2}+ac}}+\frac{1}{\sqrt{4c^{2}+ab}}\ge \frac{4}{a+b+c}$[/quote]

Any idea?[/quote]

\begin{align*}\frac1{\sqrt{4a^2+bc}}+\frac1{\sqrt{4b^2+ca}}+\frac1{\sqrt{4c^2+ab}}&=\sqrt{\sum_{\rm cyc}\frac1{4a^2+bc}+2\sum_{\rm cyc}\frac1{\sqrt{{\left(4b^2+ca\right)}{\left(4c^2+ab\right)}}}}\text,\\&\geqslant\sqrt{\sum_{\rm cyc}\frac1{4a^2+bc}+2\sum_{\rm cyc}\frac{4{\left[{\left(4b^2+ca\right)}+{\left(4c^2+ab\right)}\right]}}{{\left(4b^2+ca\right)}^2+6{\left(4b^2+ca\right)}{\left(4c^2+ab\right)}+{\left(4c^2+ab\right)}^2}}\text,\\&\overset?\geq\frac4{a+b+c}\textnormal.\end{align*}[/quote]

It is wrong direction


The probability of getting 3 coins the same is a 50%X50%X50%, so the answer would be 1/8 chance, which is a 12.5% chance. Is that the mistake?

what if we look at it this way...the first two coins have the following possibilities (H=heads, T=Tails)

coin 1:coin 2
H:T
H:H
T:H
T:T

so couldn't one argue that the probability of two coins having the same side up is 50%, with the 50% you get when you toss the third coin, so couldn't one argue that the final probability is 25%?

[quote="seamusoboyle"]Three coins are flipped. At least two of them are the same, ie two heads or two tails. Lets say we get two heads. The other coin has to be either heads or tails, of which there is a 50 percent chance of either. Therefore, when tossing three coins, the probability of getting three the same is 50 percent. This is clearly absurd, where is the mistake?[/quote]
The mistake is that there is no error if at least two have to be the same then there would be a 50% chance of getting all three the same

The first coin is arbitrary. You get heads or tails and you base the next two coin flips on it. You could say that the first flip has a 100% chance. The next flip has a 1/2 chance of being the same and the flip after that also has a 1/2 chance of being the same. Therefore, you have a 25% chance of all three coins being the same. Isn't that beautiful?  :lol:

Yeah, I think my way of doing (1/2) :^3: is wrong, it is probably 25%.  :?

[quote="seamusoboyle"]Lets say we get two heads. [/quote]

That is the exact spot where the mistake lies.
Here you are computing
P(A|B) where A is the event where you get heads and B is the event that you have two heads already.

However when flipping three coins, you do not have the given condition. Therefore, you are computing the probability for a different question.

you don' t know if they're H or T
so if they (two) are H, the chance is 50%*50%
so you must add this probability too

Remember that showing what the answer is is not the same thing as showing what the mistake is in this particular argument.

One person so far has put their finger on the mistake, but they didn't articulate it very clearly.

[quote]Three coins are flipped. At least two of them are the same, ie two heads or two tails. Lets say we get two heads. The other coin has to be either heads or tails, of which there is a 50 percent chance of either. Therefore, when tossing three coins, the probability of getting three the same is 50 percent. This is clearly absurd, where is the mistake?[/quote]

There's nothing wrong.  You just left out part of the answer.  There is a 50% chance of 3 heads...If at least two of them are heads.[/quote]

[quote="gopherhole112"]
There's nothing wrong.  You just left out part of the answer.  There is a 50% chance of 3 heads...If at least two of them are heads.[/quote]

Are you sure about that?

[quote="mysmartmouth"][quote="seamusoboyle"]Three coins are flipped. [color=blue]At least two of them are the same, ie two heads or two tails.[/color] Lets say we get two heads. The other coin has to be either heads or tails, of which there is a 50 percent chance of either. Therefore, when tossing three coins, the probability of getting three the same is 50 percent. This is clearly absurd, where is the mistake?[/quote]

[color=blue]There is the mistake![/color][/quote]

What I meant was, the mistake is that you are saying "at least."

[quote="mysmartmouth"]What I meant was, the mistake is that you are saying "at least."[/quote]

But you are wrong.  If I flip three coins, there is a 100% probability that two of them will turn out the same, no matter what.  Then, (Seamus argues), the third coin can be either the same or different than these two.  Thus, there is a 50% probability that all 3 are the same.

Now, it's clear that this is wrong -- there is only a 2/8 = 25% chance that all three coins come out the same.  What you (all) have to do is figure out [i]why[/i] this argument is wrong.

[quote="seamusoboyle"]Three coins are flipped. At least two of them are the same, ie two heads or two tails. Lets say we get two heads. The other coin has to be either heads or tails, of which there is a 50 percent chance of either. Therefore, when tossing three coins, the probability of getting three the same is 50 percent. This is clearly absurd, where is the mistake?[/quote]

You didn't count the two heads in the beginning.

[quote="Priapus"]what i think:

two of them have to be the same, but that doesn't affect each of the coins. That is simply an outcome of the experiment. Each coin is independent, and the assumption that you get 2 heads will not always work. the probability is not 50%, because in your assumption, you are assuming [i]the first two come out heads, which is clearly not true because each coins chance of landing heads is 1/2, so 2 heads = 25% chance [/i]. I think thats the mistake.[/quote]

It dosnt matter if its two heads or two tails, there is still a 50 percent outcome for both.

[quote="probability1.01"]When there are "at least two coins the same," we didn't take into account the combinations possible (i.e. we have THH, HTH, HHT).  Also, we kinda forgot about two tails.[/quote]

I think this is more or less the point.

If [b]the first two[/b] coins would be the same, there is a 50% change of getting three times the same result.

But with [b]at least two[/b] coins the same, say H, we still have four possible [b]variations[/b]: HHH, THH, HTH, HHT. Only one of them has three times the same result, so the probability is 25%.

[quote="Stijn"][quote="probability1.01"]When there are "at least two coins the same," we didn't take into account the combinations possible (i.e. we have THH, HTH, HHT).  Also, we kinda forgot about two tails.[/quote]

I think this is more or less the point.

If [b]the first two[/b] coins would be the same, there is a 50% change of getting three times the same result.

But with [b]at least two[b] coins the same, say H, we still have four possible [b]variations[/b]: HHH, THH, HTH, HHT. Only one of them has three times the same result, so the probability is 25%.[/quote]Indeed stijn, that is the (only) correct answer to this problem :)

"2 coins at the same side" is the truth, not a possible chance! So we have to assume that:
1/ Get at the H: 50% to get same side of the 2 coins and 50% for the last coin, so we get 50%*50%
2/ As for the T.


geometry
LaTeX
ratio
symmetry
trigonometry
geometry unsolved

Can anyone help me with this geometry problem?

Consider a triangle ABC. Take points [b]A[/b][i]1[/i], [b]B[/b][i]1[/i], [b]C[/b][i]1[/i] such that
[b]A[/b][i]1[/i] belongs to BC and B[b]A[/b][i]1[/i] : [b]A[/b][i]1[/i]C = 2:1
[b]B[/b][i]1[/i] belongs to CA and C[b]B[/b][i]1[/i] : [b]B[/b][i]1[/i]A = 2:1
[b]C[/b][i]1[/i] belongs to AB and A[b]C[/b][i]1[/i] : [b]C[/b][i]1[/i]B = 2:1

Consider the smaller triangle which sides belong to the lines A[b]A[/b][i]1[/i], B[b]B[/b][i]1[/i], C[b]C[/b][i]1[/i]. Prove that its area is equal to 1/7 of the area of the triangle ABC by using centre of masses.

hint: First consider only a part ..A has mass 1, B has mass 2, C has mass 4....

$\text{\LaTeX}$ing the problem:

Consider a triangle $\triangle ABC$. Take points $A_1, B_1, C_1$ such that
$A_1$ belongs to $BC$ and $BA_1 : A_1C = 2:1$
$B_1$ belongs to $CA$ and $CB_1 : B_1A = 2:1$
$C_1$ belongs to $AB$ and $AC_1 : C_1B = 2:1$

Consider the smaller triangle which sides belong to the lines $AA_1, BB_1, CC_1$. Prove that its area is equal to $\frac 1 7$ of the area of the triangle $\triangle ABC$ by using centre of masses.

The triangle centroid $G$ (the intersection of medians) is the center of mass of mass points with equal mass (for example $m_A = m_B = m_C = 1$) placed at the vertices $A, B, C$. This configuration has the same center of mass as if the mass point $m_A$ stays at the vertex $A$ and the other two mass points $m_B, m_C$ both move from the vertices $B, C$ to the midpoint $D$ of the segment $BC$ (the foot of the median $AD$). But then the center of mass $G$ cuts the median $AD$ in the ratio $AG : GD = (m_B + m_C) : m_A = 2 : 1$ (note the reversed order of the masses).

Put the masses $m_A = 1, m_B = 2, m_C = 4$ at the vertices $A, B, C$. Since $m_A : m_B = 1 : 2$ and $m_B : m_C = 1 : 2$, the center of mass $G_1$ of this configuration is the intersection of the cevians $AA_1 \cap CC_1$, such that $AC_1 : BC_1 = m_B : m_A = 2 : 1$ and $BA_1 : CA_1 = m_C : m_B = 2 : 1$ (note the reversed order of the masses). Thus this center of mass is one vertex of the triangle of interest. This configuration has the same center of mass as if the mass points $m_A, m_B$ are moved to the point $C_1$. But then the center of mass $G_1$ cuts the cevian $CC_1$ in the ratio $CG_1 : C_1G_1 = (m_A + m_B) : m_C = 3 : 4$.

Put the masses $m_A = 2, m_B = 4, m_C = 1$ at the vertices $A, B, C$. Since $m_A : m_B = 1 : 2$ and $m_C : m_A = 1 : 2$, the center of mass $G_2$ of this configuration is the intersection of the cevians $CC_1 \cap BB_1$, such that $AC_1 : BC_1 = m_B : m_A = 2 : 1$ and $CB_1 : AB_1 = m_A : m_C = 2 : 1$ (again, note the reversed order of the masses). Thus this center of mass is another vertex of the triangle of interest. This configuration has the same center of mass as if the mass points $m_A, m_B$ are moved to the point $C_1$. But then the center of mass $G_2$ cuts the cevian $CC_1$ in the ratio $CG_2 : C_1G_2 = (m_A + m_B) : m_C = 6 : 1$.

Combining these 2 results:

[b](1)[/b] The cevian $CC_1$ is cut by the centers of mass $G_1, G_2$ in the ratios $CG_1 : G_1G_2 : G_2C_1 = 3 : 3 : 1$.

When cycling the masses 1, 2, 4 through the vertices $A, B, C$, in exactly the same way, it can be shown that:

[b](2)[/b]  The cevian $AA_1$ is cut by the centers of mass $G_3, G_1$ in the ratios $AG_3 : G_3G_1 : G_1A_1 = 3 : 3 : 1$.
[b](3)[/b]  The cevian $BB_1$ is cut by the centers of mass $G_2, G_3$ in the ratios $BG_2 : G_2G_3 : G_3B_1 = 3 : 3 : 1$.

Thus the sides of the triangle $\triangle G_1G_2G_3$ are equal to $G_1G_2 = \frac 3 7 CC_1$, $G_2G_3 = \frac 3 7 BB_1$, $G_3G_1 = \frac 3 7 AA_1$. Every triangle can be transformed by a parallel projection into an equilateral triangle. Parallel projection preserves all area ratios and all segments ratios on the same line. Project the triangle $\triangle ABC$ into an equilateral triangle $\triangle A'B'C'$ and let $a = A'B'$ be the side length of this equilateral triangle. Because of symmetry of this equilateral triangle, the projected cevians $A'A_1' = B'B_1' = C'C_1'$ are equal and the projected triangle $\triangle G_1'G_2'G_3'$ is also equilateral. Calculating length of the cevian $C'C_1'$ by cosine theorem:

$C'C_1'^2 = B'C_1'^2 + B'C'^2 - 2 B'C_1' \cdot B'C' \cos{\widehat{C_1'B'C'} =}$

$= \frac{a^2}{9} + a^2 - \frac {2a^2}{3} \cos{60^o} = \frac{a^2}{9} + a^2 - \frac {a^2}{3} = \frac 7 9 a^2$

$C'C_1' = \frac{\sqrt 7}{3} a$

Consequently, $G_1'G_2' = G_2'G_3' = G_3'G_1' = \frac 3 7 C'C_1' = \frac 3 7 \cdot \frac{\sqrt 7}{3} a = \frac{a}{\sqrt 7}$

The areas $\triangle_G, \triangle$ of the equilateral triangles $\triangle G_1'G_2'G_3', \triangle A'B'C'$ are in the ratio of squares of their sides:

$\frac{\triangle_G}{\triangle}  = \frac{G_1'G_2'^2}{A'B'^2} = \frac{a^2}{7 a^2} = \frac 1 7$

The more general result is known as Routh's theorem:
[url=http://mathworld.wolfram.com/RouthsTheorem.html]mathworld.wolfram.com/RouthsTheorem.html[/url]


Only post images that haven't been posted yet. Don't copy your avatar.
[img]http://content.pyzam.com/funnypics/9/pyzamdogtastelikechicken.jpg[/img]

Or, you could simply go to [url]http://www.failblog.org[/url]


The rules of chess are changed as follows. White and Black alternately make [b]two legal moves [/b]starting with White. I say that even if White does not win, White can always guarantee at least a tie against the opponent. Is my opinion correct?

Regards,
Shamik Banerjee

Is there an easy way to see checks in this variation?

A check would be considering the next two moves of a given player.

Suppose, the Black king is on A3. The White can declare a check after placing his Queen E7 provided the relevant paths justify the check on White's next 2 moves.

Hope, I could make it clear.

Shamik

[hide]Assume the second player has a winning strategy. The first player to move can move Nc3 and then Nb1, keeping the position the same. In effect, the first player has become the second player. So the first player has at least a drawing strategy.[/hide]

Could you please explain me a bit more?

"The first player to move can move Nc3 and then Nb1, keeping the position the same. In effect, the first player has become the second player."

I understand that the first player has become the second player. But, to enforce a draw the first player (White) should have been aware at the very first what is going to be the winning strategy of the second player (Black).

It means suppose I become the first player (White) I cannot so easily enforce a draw with, say, a grandmaster when both of us take 2 moves at a time. I can only force a draw if I was aware what is going to be the winning strategy of Black.

Would wait for your explanations.

Thanks,
Shamik

I only know a proof for the existance of a drawing strategy for White, which I posted above. The actual drawing strategy is probably much more complicated and unknown.

this is an actual variant of chess played, therefore I assume it is indeterminable (at least without a supercomputer)


1) A two headed coin and three normal coins are in a jar. A coin is selected at random and tossed three times with heads resulting each time. What is the probability that it was a two-headed coin? I have a final answer of 8/11, but I am not sure.

2)5 married couples sign up for a dance class. The instructor pairs the five men with the five women at random. What is the probability that all the five men will be dancing with their wives? What is the probability that at least one man will be dancing with his wife?

1> ur ans. (8/11 ) to the problem seems to be correct.

 The problem is an application of the Baye's theorem, the mutually exclusive and exhaustive events being "the 2 headed coin is chosen" and " anormal coin is chosen".

Let's denote these two events by E1 and E2 respectively. Note that P(E1) = 1/4 and P(E2) = 3/4

Let the actual event be denoted by E which is HHH

We have to find P(E1/E) which, by Baye's theorem is nothing but

P(E1)*P(E/E1) / {P(E1)*P(E/E1) + P(E2)*P(E/E2)}

= 1/4 * 1/{ 1/4 * 1 + 3/4 * (1/2)^3 }  = 8/11

The no. of ways the couples could be formed  up is 5!. Out of these 5! ways, there's only one way in which the couples could be correctly formed.
So the probability of the event = 1/5! = 1/120

Ans. for the second part = 8/15

Let's first try to calculate the no. of ways in which every man will dance with a [b]woman other than his wife[/b].

This is a very famous problem and we shall try to develop a generic recurrence relation for this.

Let the total no. of couples be n

Then the total no. of ways to form the couple correctly would be n!

Again , all these ways could be classified into the following categories

 No. of wrong couples       No. of correct couples

 0			n

 1			n-1

 2			n-2

...

...

n			0

Let Wr denote the no. of  ways in which r couples are wrongly formed. Corresponding to each of those ways, the correct couples could be formed in C(n,n-r) ways.

So, summing up we have  n! = W0*C(n,n) + W1*C(n,n-1) + .....+ Wn*C(n,0)

=> [b]Wn   = n!  -  { W0*C(n,n) + W1*C(n,n-1) + .....+ Wn-1*C(n,1) }[/b]

Now observe that W0 = 1, W1 = 0. With these we can evaluate W2, W3 etc.

We can calculate W5 as 56.

So the probability that at least one man will dance with his wife  = 1 - 56/120 = 64/120 = 8/15

[quote="allrighty"]5 married couples sign up for a dance class. The instructor pairs the five men with the five women at random.

Extension of the problem: Find the probability that exactly one man will be dancing with his wife.[/quote]
Even more interesting : Find the probability that exactly four men  will be dancing with theirwives .  :D
(Have seen this problem, more than once in contests...:)!)

Did one couple get a divorce some time between when the problem was originally posted and now? :D

[quote="Gyan"][quote]What is the probability that at least one man will be dancing with his wife?[/quote]

As  santosh said this is a classic problem. The probablity of this is around $(1-1/e)$ ( When the number of couples is large, it really does not depend  "that much"  on  that number - eg probablity would be almost same  even if there were 100 couples)

So answer is about $0.63 $  exact answer would be   $76/120$  or $19/30$

$p =  1 - 1/2! + 1/3! - 1/4! + 1/5! $[/quote]

Hello,

My soln. to the second problem is INCORRECT..Thanks Gyan , for the correction. Actually the summation won't result in n!. This is because Wr denotes the no. of ways in which not a single couple is formed correctly(in a set of r couples)

Actually the correct recurrence relation for  Wn would be

Wn+1 = n * ( Wn + Wn-1)

U can get an exhaustive proof of the same at

http://www.physics.harvard.edu/probweek/sol16.pdf


floor function
modular arithmetic
Sequence
algebra
recurrence relation
IMO Shortlist

Let the sequence $ a(n), n \equal{} 1,2,3, \ldots$ be generated as follows with $ a(1) \equal{} 0,$ and for $ n > 1:$

\[ a(n) \equal{} a\left( \left \lfloor \frac{n}{2} \right \rfloor \right) \plus{} (\minus{}1)^{\frac{n(n\plus{}1)}{2}}.\]

1.) Determine the maximum and minimum value of $ a(n)$ over $ n \leq 1996$ and find all $ n \leq 1996$ for which these extreme values are attained.

2.) How many terms $ a(n), n \leq 1996,$ are equal to 0?

Letting $ n\mapsto 2n$ and $ n\mapsto 2n + 1$ we can define the sequence as
\[ a(2n) = a(n) + ( - 1)^n,\quad a(2n + 1) = a(n) + ( - 1)^{n + 1}
\]
for all $ n$. Or equivalently,
\[ a(2n + \epsilon) = a(n) + ( - 1)^{n\bmod 2 + \epsilon}\qquad(*)
\]
where $ \epsilon\in\{0,1\}$.

Let $ n = \overline{a_ka_{k - 1}\dots a_1a_0} = 2^ka_k + 2^{k - 1}a_{k - 1} + \dots + 2a_1 + a_0$, where $ a_i\in\{0,1\}\forall i$ and $ a_k = 1$, be the binary representation of $ n$. We get using $ (*)$,
\begin{align*}a(n) & = a(\overline{a_ka_{k - 1}\dots a_1a_0}) \\
& = a(2^ka_k + 2^{k - 1}a_{k - 1} + \dots + 2a_1 + a_0) \\
& = a(2^{k - 1}a_k + 2^{k - 2}a_{k - 1} + \dots + 2a_2 + a_1) + ( - 1)^{a_1 + a_0} \\
& = a(\overline{a_ka_{k - 1}\dots a_1}) + ( - 1)^{a_1 + a_0}\end{align*}
And an inductive argument shows that
\[ a(\overline{a_ka_{k - 1}\dots a_1a_0}) = ( - 1)^{a_k + a_{k - 1}} + ( - 1)^{a_{k - 1} + a_{k - 2}} + \dots + ( - 1)^{a_1 + a_0}
\]
Now to find $ \max\{a(i)\}$, we need to have $ a_i + a_{i + 1}\equiv 0\pmod 2$ for $ i = 0,1,\dots,k - 1$. Since $ a_k = 1$, we conclude that $ a_0 = a_1 = \dots = a_{k - 1} = 1$, and thus $ a(n)$ is maximum for $ n = 1 + 2 + 2^2 + \dots + 2^k = 2^{k + 1} - 1$ in which case $ a(n) = a(2^{k + 1} - 1) = k$. Therefore for $ n\le 1996$ we conclude $ a(2^{10} - 1) = a(1023) = 9$ is maximum.

Similarly, for the minimum we must have $ a_i + a_{i + 1}\equiv 1\pmod 2$ for the same values of $ i$ as before, and with $ a_k = 1$ we get $ a_{k - 1} = 0,a_{k - 2} = 1,\dots$, with alternating $ 0$'s and $ 1$'s. Hence $ n = 2^{k} + 2^{k - 2} + \dots = \frac {2^{k + 2} - 1}{3}$ for even $ k$, and $ \frac {2^{k + 2} - 2}{3}$ for odd $ k$, and in these cases $ a(n) = - k$. Therefore $ n\le 1996$ implies for even $ k$, $ a(n) = a\left(\frac {2^{12} - 1}{3}\right) = a(1365) = - 10$, or for odd $ k$, $ a(n) = a\left(\frac {2^{11} - 2}{3}\right) = a(682) = - 9$.

Therefore, in conclusion, the intended maximum is $ a(1023) = 9$, and the minimum is $ a(1365) = - 10$.

For the second part of the problem, we need half of the elements of $ \{a_k + a_{k - 1},a_{k - 1} + a_{k - 2},\dots,a_1 + a_0\}$ to be even, and the remaining ones odd. In that case $ k$ must be even. Also, given such a sequence $ a_k,a_{k-1},\dots,a_{k-i}$ there is only one choice for $ a_{k-i-1}$. Hence we have a bijection with the number of ways to choose $ k/2$ arbitrary elements from a $ k$ element set, which is $ \binom{k}{k/2}$. Again, since $ 1996_2 = 11111001100$, we get $ k\le 10$. Hence the intended number of ways is
\[ \binom{10}{5} + \binom{8}{4} + \binom{6}{3} + \binom{4}{2} + \binom{2}{1} = 350
\]
and with $ a(1) = 0$, the total number of $ i$'s not exceeding $ 1996$ such that $ a(i) = 0$ is $ 351$.

Ps. Solution might contain calculation mistakes. :blush:


Remnok the Nomad has a brother, Damon the Nomad, who's fond of coming up with math puzzles. Anyway, he came up with one that uses his own name:

Assume each letter in the following formula represents a unique digit: (for example, ABCD could be the number 1234)

NOMAD x 4 = DAMON

What is the number represented by the digits DAMON?

Does anyone have an idea ?

we did the exact problem for four digits in our college entrance exam
shouldn't be tough

N must be even, and 4N must be less than 10. So N is 2. So D is 3 or 8. But it must be either 8 or 9, so it's 8. 2 numbers are already set now. 4O is less than 10, so O is 0 or 1, etc.


conics
hyperbola
symmetry
algebra
pen
Vieta
number theory

I use a delicate argument involving properties of hyperbolas. If anyone has a nicer argument, I would be very interested.

Let $H_q: x^2+y^2-qxy+1=0$, a hyperbola symmetric in $x,y$. I want to show that only $H_3$ contains lattice points. Since it is symmetric, we work only on the positive branch with $x \leq y$.

Observe that $T: (x,y) \rightarrow (qx-y, x)$ allows us to generate new points on the hyperbola, since $(qx-y)^2+x^2-qx(qx-y)+1=x^2+y^2-qxy+1$. If $x<y$, such a new point must be either further down the branch or on a different branch.

The other positive branch (which is the only possibility since $x$ is positive) contains the point $(y,x)$ by symmetry, so if our new point lies on this branch, $y=qx-y$, which is equivalent to $y= \frac{qx}{2}$. Intersecting this line with $H_q$, $x^2(q^2-4)=1$, but this is clearly not a lattice point, so this possibility doesn't concern us.

So by repeated application of $T$ I must at some point reach a lattice point at which $x=y$, since otherwise I have an infinitely decreasing sequence of positive integers, so here $(q-2)x^2=1$. Both factors are integers, and $x^2$ is positive, so $q=3$ must be the only hyperbola containing lattice points, which was to be shown.

This problem seems hard. I doubt I would have found it if I hadn't read Engel's Number Theory chapter.

i have one ;) vieta-jumping again.
\[ \frac {x^{2} \plus{} y^{2} \plus{} 1}{xy} \equal{} k
\]
->
\[ x^{2} \minus{} kxy \plus{} y^{2} \plus{} 1 \equal{} 0
\]
we assume that we have a solution, otherwise the problem would be trivial. if $ x \equal{} y$ we get $ x,y|1$ so $ \frac {x^{2} \plus{} y^{2} \plus{} 1}{xy} \equal{} 3$ and we have finished.
$ k \geq 2$ must hold because of AMGM. so if we assume $ x > y$ we get a new solution $ x_{2} < y$ with $ x_{2} \equal{} \frac {y^{2} \plus{} 1}{x_{1}}$ (by vieta). it must be a "really" $ <$ for $ y > 1$. So we can jump down by vieta and vieta again and get $ y \equal{} 1$ and $ x \equal{} 1$. because of the fact that we have already done this, the problem got PWN3D and the quotient is always $ 3$.

Naphthalin

Nice. I think that's kind of related to my method, but much better.

thx^^ the method of vieta-jumping works really often. too often if you ask me ;)

Naphthalin

I'm a beginner and I'm having some problem understanding your solution. What do you mean, using vieta's jumping up and down until, ..." ?? I couldn't get the last part of A3 either. Anyone's answer would be a great help. thnxs

I'll write Napthalin's solution a bit easier for you.

Assume there exists $ x,y,k > 0$ with $ x^{2} \minus{} kxy \plus{} y^{2} \plus{} 1 \equal{} 0$ (else there is nothing to prove), and take the pair $ (x,y)$ where $ x \plus{} y$ is minimal, wlog $ x\ge y$. If $ x \equal{} y$ or $ y \equal{} 1$ we easily get $ x \equal{} y \equal{} 1$ and $ k \equal{} 3$.
For $ k\not \equal{} 3$ (by AM-GM this is $ k > 3$) we have $ y > 1$, so solving for $ x$ we get two [i]positive[/i] roots $ x,x' \equal{} \frac {ky\pm\sqrt {k^2y^2 \minus{} 4y^2 \minus{} 4}}{2}$.
By the minimality of $ x \plus{} y$, we need $ x'\ge x > y$, while de Vieta tells us $ xx' \equal{} y^2 \plus{} 1$, contradiction.

OK,and more(I think it is very nice)
*Let $ p,q\in \mathbb Z^\plus{},p\le q$
Find $ a,b$ satisfying $ \frac{(a\minus{}b)^2}{pab\minus{}q}\equal{}k (k\in \mathbb N)$

*$ \frac{(a\minus{}b)^2\plus{}m}{pab\minus{}q}\equal{}k(m\mathbb N^*)$ so K is belong to a finite set
They are nice,do you think that

Here is my way to express the same ideas than those proposed by Naphtalin and Peter.
Let (m, n) be a solution of equation $ m^2 + n^2 - qmn + 1 = 0 $ with $0 < m \le n $ and $m + n$ minimal. We show that m = n.
As (qm - n, m) is another solution, we have, by minimality, $ m + n \le qm - n + m$ so $2n \le qm$ and $2n^2 \le qmn$; thus $ 0 = m^2 + n^2 - qmn + 1 \le m^2 - n^2 + 1 $ which is possible only when m = n.

I like andrejilievski's solution best.
Sometimes when I am doing vieta jumping, I get x2 = something, and then I go blank like "so what".

Also, we can manipulate to get sqrt {k^2b^2 - 4b^2 - 4} = (b^2+1-a^2)/a, and I get nothing out of it.

Nice proofs all. They are all about the same to me. But what is "Vieta Jumping"? This is the first time I hear.

See [url=https://projectpen.wordpress.com/]Project Pen[/url] for a nice Vieta Jumping explanation. Could someone read my solution and see if it is correct?

Consider the set  $$S(k) = \left\{ (x,y)\in\mathbb{Z}^+\times\mathbb{Z}^+ : \dfrac{x^2+y^2+1}{xy}=k.\right\}.$$ By AM-GM, we see that $k>2$. Now if $x=y$, then $2x^2+1=kx^2\implies \dfrac{1}{k-2} = x^2$, and hence $x=y=1, k=3$. Now let $(x,y)$ be the element of $S(k)$ with minimal $x+y$ and assume WLOG that $x>y$. Now consider the quadratic in $X$ $$X^2+y^2+1=kyX \iff X^2-kyX+y^2+1=0.$$ One solution is obviously $X=x$; let the other one be $X = x_1$. Then by Vieta's formulas, $$ x_1 = ky-x = \dfrac{y^2+1}{x}.$$ The first equality tells us that $x_1$ is integer, while the second one tells us that $x_1$ is positive; hence, $x_1\in\mathbb{Z}^+$ and $(x_1,y)\in S(k).$ However, note that $x^2-y^2=(x-y)(x+y)>1$, and thus $$x^2>y^2+1 \implies x>\dfrac{y^2+1}{x} = x_1,$$ a contradiction to the minimality of $(x,y)$ in $S(k)$. Therefore, $x=y$, from which our conclusion follows.

hamup1, your solution is correct :), but I don't know if you can take the minimal x+y, because don't you have to rigorously prove for all x,y? Please correct me if I am wrong, I am a novice at this  :)

Also found here, but they solve with infinite descent:

https://en.wikipedia.org/wiki/Vieta_jumping#Constant_descent_Vieta_jumping

[quote=Peter]Let $x$ and $y$ be positive integers such that $xy$ divides $x^{2}+y^{2}+1$. Show that \[\frac{x^{2}+y^{2}+1}{xy}=3.\][/quote]

Use Vieta jumping

Apart from the solution we can show, that there exist infinitely many such $(x,y)$ where $x\le y$.
Take $x_0=1\wedge y_0=2$. Then the equation holds. Define $(x_{n+1},y_{n+1})=(y_n,3y_n-x_n)$ for $n\ge 0$. Since $x_n\le y_n$ then by induction hypothesis $x_n,y_n\in Z_+\wedge x_n\le y_n\implies \left(x_{n+1}\ge x_n\wedge y_{n+1}\ge 2y_n>y_n\right)$ so we constructed infinitely many solutions.

[quote=mavropnevma][quote="andrejilievski"]Using the lemma $ x^2+y^2+z^2=kxyz $ has solutions only for $ k=1 $ or $ k=3 $, taking $ z=1 $ we have $ x^2+y^2+1=kxy $ so if $ k=1, xy=x^2+y^2+1\ge\ 2xy+1 $ which is not true so it must be $ k=3 $ or $ x^2+y^2+1=3xy $[/quote]
Quite a hefty lemma, using the theory of Markoff (and Hurwitz) equations ...[/quote]

Rip


Nebraska, the home of the AMC, changed its license plate scheme.  Each old license plate
consisted of a letter followed by four digits.  Each  new license
plate consists of  three letters followed by
three digits.  By how many times is the number of possible license plates
increased? Express your answer to nearest tenth decimal.

The old liscense plate scheme had $ 26\times{10^4}$ possibilities. The new scheme has $ 26^3\times{10^3}$ possibilities. Therefore, we have $ \frac{26^3\times{10^3}}{26\times{10^4}}\equal{}\frac{576}{10}\equal{}\boxed{57.6}$.

Ugh, dumb arithmetic mistake (for me, its really easy to confuse $ 24^2\equal{}576$ and $ 26^2\equal{}676$). Thanks.

Switching from 5 to 6 digits/letters? Ok. Something sounds off.

wow, you just bumped an 11-yr-old thread!

[quote=hansenhe]$K$?$I don't understand.$[/quote]

He means that if you don't have something important to say, please do not bump old threads, thanks! :D


LaTeX
Vieta
quadratics
algebra
quadratic formula

[hide="Solution"]
$ x^2 \plus{} x \plus{} 58 \equal{} a^2$
$ x^2 \plus{} x \plus{} 58 \minus{} a^2 \equal{} 0$

By Vieta's formulas, the coefficient of the x term is equal to -(p + q), where p and q are the roots of the equation.

$ 1 \equal{} \minus{} (p \plus{} q)$
$ p \plus{} q \equal{} \minus{} 1$

The sum of all values of x should be -1.[/hide]

I'm not sure I understand the problem.  Is (a positive integer)^2 fixed or can it be any positive integer? If it can be any, then I've found the values of x that give a square to be $ x\equal{}57, \minus{}58, 18, \minus{}19, 2, \minus{}3, 6, \minus{}7$.  The sum of these is $ \minus{}4$.

[quote="11235"][b](x^2)+x+58= (a positive integer)^2
Find the sum of all values of X.[/b]
(Source: This is from a ICTM 8 Person Fresh-Soph Competition.)

Sorry that I do not know LaTeX[/quote]

this is merely an applcation of vieta´s formula.i´ll show it for the quadratic equation :
$ x^2\plus{}bx\plus{}c\equal{}0$ .if the roots of this equation are $ y$ and $ z$, this equation can be factorised as following: $ (x\minus{}y)(x\minus{}z)$.easy to see that this implies that $ y\plus{}z\equal{}\minus{}b$.therefore we have that your searched answer is $ \minus{}1$.

obs: vieta´s formula can be applied to a polynomyal of degree $ n$, where $ n$ is a positive integer.The proof of this is analogous to the proof of degree $ 2$.

Sorry that I am bringing this back after such a long time but will vieta's formula give all the values for which the expression x^2+x+58 is any positive integer squared (in this case)? What you are doing is finding the sum of the roots to the equation but what we are trying to find is the [b]sum of values (not roots)[/b] that make the expression equal the square of a positive integer.

For example:
x^2+x+58 can equal a positive integar such as 8^2
x^2+x+58=64
x^2+x-6=0
(x+3)(x-2)=0
x= -3,2

This is what Machodog did but I do not understand how he knows that those are the only values that give a positive integer squared.

@11235

$ x^2 \plus{} x \plus{} 58 \minus{} a^2$ can be factorised into $ (x \minus{} a)(x \minus{} b)$ where $ a$ and $ b$ are the solutions.

$ (x \minus{} a)(x \minus{} b) \equal{} x^2 \minus{} (a \plus{} b)x \plus{} ab$ and you compare it with  $ x^2 \plus{} x \plus{} 58 \minus{} a^2$

the coefficient of $ x$ is $ 1$, so $ \minus{} (a \plus{} b) \equal{} 1$ and $ a \plus{} b \equal{} \minus{} 1$

You can also try the quadratic formula

$ a\equal{}1 b\equal{}1$ and the surd part would be canceled out upon summation, so $ c$, and hence the value of a, doesn't matter.

@10000th User
I think it's quite ridiculous that you should dismiss almost all the posts above as totally irrelevant, even when the question is to be interpreted this way.

And I think $ a$ is a constant. And we simply need to compute the answer, which is -1, for a certain $ a$. $ a$ can take an infinite number of values.

[quote="10000th User"]All the proposed hints and somewhat "solutions" are totally irrelevant for this problem, except Machodog's. We would like to see how you got those solutions, Machodog. 11235's last reply in this topic points in the right direction. The answer is not -1.[/quote]

The question asked:

[quote]Find the sum of all values of X.[/quote]

Perhaps he stated the original question wrong, but the sum of all values of X, not a, can be found via vieta's forumla.

$ x^2 + x + 58 = a^2$
Complete the square:
\begin{eqnarray*} \left(x + \frac {1}{2}\right)^2 + \frac {231}{4} & = & a^2 \\
\frac {(2x + 1)^2}{4} + \frac {231}{4} & = & a^2 \\
4a^2 - (2x + 1)^2 & = & 231 \\
(2a + (2x + 1))(2a - (2x + 1)) & = & 3\times7\times11 \\
\end{eqnarray*}
Now we just have to do casework.  I will do one, the rest are the same:

Suppose:
$ 2a + (2x + 1) = 231$, and $ 2a - (2x + 1) = 1$
Subtracting the first equation from the second, we get:
$ 2(2x + 1) = 230 \implies x = 57$
A quick check shows that $ x = 57$ does indeed give a square.
To find all the values of $ x$ that give a square, just do the same thing but with the other ordered pairs that $ (r,s)$ such that $ rs = 231$.  For instance $ 2a + (2x + 1) = 1$, $ 2a - (2x + 1) = 231$ gives $ x = - 58$.

The question does not actually ask us to find the values of $ x$ that give a square.  We can still find the sum of all the $ x$ values though by saying:
Let
$ 2a + 2x + 1 = r \\
2a - (2x + 1) = s$
where $ r,s$ are integers such that $ rs = 231$
Adding we get:
$ 4a = r + s \implies a = \frac {r + s}{4}$
How many values of $ a$ are there?  $ 231 = 1\times231 = 3\times77 = 11\times21 = 33\times7$, so there are $ 4$ values of $ a$.  As has already been pointed out, for a [b]given[/b] value of $ a$, the sum of the values of $ x$ that satisfy $ x^2 + x + 58 = a^2$ is $ - 1$, so for four values of $ a$ the sum is $ - 1\times4 = - 4$.

@ Machodog

$ a$ can take an infinite number of values.

For any $ a$, $ x \equal{} \frac{\minus{}1}{2} \minus{} \sqrt{a^2\minus{}57.75}$ or $ \frac{\minus{}1}{2} \plus{} \sqrt{a^2\minus{}57.75}$

Oh WOOPS for some reason I thought that $ x$ had to be an integer  :blush: .

If x was not an integer then how could the expression equal the square of a positive integer, which is obviously also an integer?

Just as a random example, let $ x\equal{}\minus{}\frac{1}{2} \plus{} \frac{5\sqrt{273}}{2}$.

Then $ x^2\plus{}x\plus{}58\equal{}1764\equal{}42^2$


$ p$ is a prime,as is $ 8p^{2}\plus{}1$. Find the value(s) of $ p$, and prove that your value(s) are the only ones possible.

[hide]
$ 1)\ p \equal{} 3 \implies 8p^2 \plus{} 1 \equal{} 73$ which is a prime

$ 2)\ p > 3 \implies p^2\equiv1\ (mod\ 3)$ (Fermat's little theorem) $ \implies 8p^2 \plus{} 1\equiv9\equiv0\ (mod 3)$
so $ 8p^2 \plus{} 1$ is not a prime for any prime $ p > 3$

$ p \equal{} 3$ is the only solution[/hide]


Let $ a,b,c$ be the sides of a triangle such that: $ a^3\plus{}b^3\plus{}c^3\equal{}3$.Prove that:
$ \frac{a^3}{a^2\plus{}ab\plus{}b^2}\plus{}\frac{b^3}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{c^3}{c^2\plus{}ca\plus{}a^2} \ge 1$

[quote=bararobertandrei][quote=Mathskidd]Dear luofangxiang,
$\sum_{cyc}\sqrt{(x+y)(x+z)}\geq \sum_{cyc}x+\sqrt{3\sum_{cyc}xy}$
How to get this inequality ?
Is it Cauchy?[/quote]

Let $\sqrt{y+z}=a$ and the other ones, where a,b,c are lenghts of sides of an acute triangle ABC. So we have to prove that:
$\sum_{cyc}ab \geq 1/2* \sum_{cyc}a^2+2S \sqrt{3}$ where S is area of triangle
So $\sum_{cyc}2ab - \sum_{cyc}a^2 \geq 4S \sqrt{3}$
And this is equivalent to $4R+r \geq p \sqrt{3} $ where $2p=a+b+c$, which is obvious[/quote]

Dear bararobertandrei, why is that obvious ?

[quote=Mathskidd]Dear luofangxiang,
$\sum_{cyc}\sqrt{(x+y)(x+z)}\geq \sum_{cyc}x+\sqrt{3\sum_{cyc}xy}$
How to get this inequality ?
Is it Cauchy?[/quote]

Any other idea of proving it?

$\sum_{cyc}\sqrt{(x+y)(x+z)}\geq \sum_{cyc}x+\sqrt{3\sum_{cyc}xy}\iff  a+b+c\ge \sqrt{3(ab+bc+ca)} $

where

$a=\sqrt{(x+y)(x+z)}-x$
$b=\sqrt{(x+y)(y+z)}-y$
$a=\sqrt{(z+y)(x+z)}-z$

because

$ab+bc+ca=xy+yz+zx$

[quote=luofangxiang][img]http://s16.sinaimg.cn/middle/0018zOAxgy71nkPoQbBcf&690[/img][/quote]
Nice solution, luofangxiang!  :-D. This inequality was proposed by me when I am 16 year old. I also saw that you had solutions for some inequalities of me (proposed in 2009) but I think that if you really like inequalities, you can try with some harder inequalities. Example, the following inequality also nice (and some other inequalities of Vasc and Can_hang2007).
 http://www.artofproblemsolving.com/community/c6h549060

I know you search for your inequality is mainly elaborate you.good luck, and you and other friends of ingenious solution to the problem, thus improve their problem solving ability, occasionally put forward their own ideas. You mention this inequality, in front of me read, think you for a long time haven't a clue, admire this problem mudok train of thought
我寻找您的不平等主要是学习您的精巧的出题思路及你和其他的朋友巧妙的解决问题的办法,从而提高解决问题的能力,偶尔也会提出自己的想法。你提到这个不等式,在我面前读,觉得你很长一段时间没有一个线索,欣赏这个问题mudok的思路

[quote=mudok][quote=quykhtn-qa1] This inequality was proposed by me when I am 16 year old.[/quote]

It was proposed 7 years ago, but it was unsolved for 7 years![/quote]
If I remember exactly, my proof is similar to luofangxiang, except I don't use the inequality
 $$\sum_{cyc}\sqrt{(x+y)(x+z)}\geq \sum_{cyc}x+\sqrt{3\sum_{cyc}xy}.$$
Indeed, we only need using the Cauchy-Schwarz inequality to obtain
$$\left(\sum\sqrt{a^3+b^3}\right)^2=2\sum a^3+2\sum \sqrt{(a^3+b^3)(a^3+c^3)} \geq 6+2\sum\left(a^3+b^{\frac{3}{2}}c^{\frac{3}{2}}\right)=12+2\sum b^{\frac{3}{2}}c^{\frac{3}{2}}.$$
It make the proof is simpler and nicer. Also, I inspired to create this inequality by a nice identity
$$\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}=\dfrac{b^2}{a+b}+\dfrac{c^2}{b+c}+\dfrac{a^2}{c+a}.$$

You are the Vietnamese，
Whether do you CAN？

[quote=smoothtorus]How did this post go unanswered for over 7 years?![/quote]
I agree with you! It's  very easy inequality.


P and Q are the point on the sidea AB and BC of a triangle ABC respectively such that BP = 3PA and QC = 2BQ. K is the midpoin of the segment PQ. Prove that the area of the traingle AKC is equal to 11S/24 where S is the area of the triangle ABC.

Please someone help me! I need this asap!! Someone said it has something to do with:

"triangles that have same base and common vertex have equal areas"

but thats all I know. Please help!

I write $ A(T)$ for the area of the triangle $ T$ .

$ A(APC)\equal{}\frac{1}{4}A(ABC)$
$ A(AQC)\equal{}\frac{2}{3}A(ABC)$

So $ A(AKC)\equal{}\frac{A(APC)\plus{}A(AQC)}{2}\equal{}\frac{11.A(ABC)}{24}$

How did you get those results? Or more so, how to prove them?

The area of triangles with the same base is proportionnal to the altitude and you can use Thales .

What is Thales, and what is the altitudes again :( Im so sorry! please bear with me

An exemple ( sorry my english is very poor ) :

$ A(ABC)\equal{}\frac{AC.BH}{2}$ with $ BH$ the altitude of $ ABC$
$ A(APC)\equal{}\frac{AC.PI}{2}$ with $ PI$ the altitude of $ APC$

By using Thales in the triangles $ API$ and $ ABH$ ( edges are proportionnal ) :

$ PI\equal{}\frac{1}{4}.BH$ and $ A(APC)\equal{}\frac{1}{4}A(ABC)$ .

Thank you very much! I got it!! yay!! THanks again!


Hey ppl,
           how are all of you? Ppl like varunarasu n all, long time no talk? Ppl, jus get back n do some ob on AOPS as well da..

in fact i d love it if ppl actually spare a few mins from their bz 24x7 studying time and do post here :P
it can be a grand reunion  :wink:

yes..but do tell the meganerds :rotfl:  :P

so i have company :!:  (at last)
awesome ,Soumyashant you of all people forgot? :rotfl:    :rotfl:

nice 2 see you here soumya  :)

@gaurav:y only soumya could give you company  :rotfl:  :rotfl:

I was also thinking on same lines. :rotfl:

Ritu and soumya(both girlish names and both brilliant ones)  :omighty:

[quote="gauravpatil"]
awesome ,Soumyashant you of all people forgot? [/quote]

Well its human to forget[hide]but till now i was thinking soum is a beast :rotfl: [/hide]

[quote="skand"]Ritu and soumya(both girlish names and both brilliant ones[/quote]

Thats like underestimating SOUMYASHANT NAYAK :o

[quote="rituraj007"]i tink there wasnt ne girl at inmotc last yr.. :rotfl:  :D  :P[/quote]
That's because poor Palak was mistaken for a girl :).  Curse all uni-sex names...


Euler s Theorem
number theory
Sequence
relatively prime
IMO
IMO 1970

Prove that we can find an infinite set of positive integers of the from $2^n-3$ (where $n$ is a positive integer) every pair of which are relatively prime.

We construct them inductively. If $a_1,\ldots,a_n$ have been constructed, then we take $a_{n+1}=2^{\varphi(\prod_{i=1}^na_i)}-3$.

I posted it some time ago, without knowing that it is from IMO, at http://www.mathlinks.ro/Forum/viewtopic.php?t=42703

We proceed inductively by constructing a sequence $\{n_k\}$ of numbers that work. Notice that for the base case, $2^2-3$ and $2^3-3$, $2^4-3$ are relatively prime.

Now, let $S$ be the set of all $p_i$ such that $p_i \mid 2^{n_k} - 3$ for some $k$. Then we can pick $n_{k+1} = \prod_{p_i \in S} (p_i - 1)$. This works because $2^{n_{k+1}} - 3 \equiv -2 \pmod {p_i}$ for any $p_i \mid n_k$, and thus is relatively prime to all previous elements.


Let $a,b,c,x,y,z$ are positive real such that $cy+bz=a; az+cx=b; bx+ay=c.$
find the minimun value of
\[ f(x,y,z)=\frac{x^2}{1+x}+\frac{y^2}{1+y}+\frac{z^2}{1+z}  \]

solving the system we get $x = \frac{b^2+c^2-a^2}{2bc}$, $y = \frac{a^2+c^2-b^2}{2ac}$
and $z = \frac{b^2+a^2-c^2}{2ab}$.

so, we can suppose $x = cosA$, $y = cosB$ and $z = cosC$.

in http://www.mathlinks.ro/Forum/viewtopic.php?t=59649

was proved that the minimum of $f(x,y,z) = 1/2$.  :)


ratio
geometry
trigonometry
trig identities
Law of Sines

This problem appeared on a local interschool test, and to my dismay, no one answered or even attempted it (along with nine others).  Wanting not to waste my math problems, I decided to start a weekly problem page.  Here's the first question: Triangle ABC with [tex]AC=BC=13[/tex] and [tex]AB=10[/tex] is inscribed in a circle with center [tex]\emph{O}[/tex].  Segment [tex]BO[/tex] is extended so that it divides [tex]\bigtriangleup ABC [/tex] into two regions.  Find the ratio of the area of the smaller of these parts to the area of the larger.  Express your answer as a common fraction.

[hide]The ratio is the same as AD/DC (with D = AC inter BO) since triangles have same height.

There are a lot of equals angles and equal lengths. Using law of sines in triangles AOD and ODC I  find L/(2L-l) as a general formula. Here L=13 and l=10 gives 328/169. Is it correct ??

L=l gives a ratio of 1 which is fortunate ![/hide]

[quote="msmswatson"]P.S. How do I activate html in my signature?[/quote]
I think you have to do [url=http://www.olemiss.edu/~sswatson]My Weekly UM Problem page[/url] instead of the standard <a href> tag...

Your general formula is indeed correct, though you substituted incorrectly.  Good job.


function
floor function
logarithms
probability
probability and stats

Let $ X$ be the Cantor distribution and let $ F$ be his distribution function, show that $ Y \equal{} 1 \plus{} \lfloor \log{F(X)}/ \log{q} \rfloor$ has a geometric distribution ($ 0<q<1$).

This result is well known.
Let $ X$ be a random variable with a continuous distribution function $ F$, then $ F(X)$~ uniform on $ (0,1)$:
if $ 0 \le y \le 1$, then $ P(F(X) \le y) \equal{} P(X \le F^{ \minus{} 1}(y)) \equal{} F(F^{ \minus{} 1}(y) \equal{} y$. There is a small difficulty if $ F$ is not strictly increasing, but this is overcome by defining $ F^{ \minus{} 1}(y) \equal{} sup(x: F(x) \equal{} y)$.

Cantor distribution is a r.v. with a continuous distribution function $ F$, so $ F(X)$~$ U(0,1)$.
Hence
$ P(Y \equal{} k) \equal{} P( \lfloor \log{U}/ \log{q}\rfloor \equal{} k \minus{} 1) \equal{} P(q^k < U \le q^{k \minus{} 1}) \equal{} q^{k \minus{} 1}(1 \minus{} q)$, $ k \ge1$.

This sort of problems is for people who have just started studying probability, i dont see any reason to post it on forum.

I don't see the problem.
In the forum there are students who have just started studying probability, researcher, and so on...
This exercises are for people who have just started studying probability.


[quote="jibonmrito"]let G be a group of order 15 and A, B are two subgroups of G of order 5 and 3 respectively. show that, G=AB.[/quote]
I assume that $HN=\{hn|h\in H,\, k\in K\}$

[b]Theorem:[/b] If $H$ and $K$ are subgroup of a finite group then,

$|HN|=\frac{|H\parallel K|}{|H\cap K|}=\frac{3\cdot 5}{|H\cap K|}$

Now here's the thing.
When elements in $H$ are raised to 3 they produce the identity (it is cyclic).
Similarly when elements in $G$ are raised to 5 they produce the identity.
(it is cyclic).

Then, that means that $H\cap K ={e}$

Thus,

$|HN|=\frac{15}{1}=15$

So,

$HN=G$

Q.e.d.

[quote="HilbertThm90"]In anticipation of the next question, I think I'll attempt to clear this up. Every $K$ should be an $N$ or every $N$ should be a $K$, and correspondingly every $k$ should be an $n$ or every $n$ should be a $k$. It should just stay consistent throughout.[/quote]
Thanks.

-----

Here is another way.
Since 15=5*3 and $5\not \equiv 1 (\mod 3)$ it must be that this group is abelian (in fact cyclic).
Hence all the subgroups are normal.

You can regard the 5 element subgroup as $N$ the normal subgroup and the 3 element subgroup as $K$ an arbitraty subgroup.
Then by the 2nd Isomorphism Theorem:
$(NK)\N \simeq K/(H\cap K)$
Since they are finite you can write the numbers in and use the fact that the intersection between them is 1 as mentioned above,
Then,
$|NK|/5=3/1$
So,
$|NK|=15 \to NK=G$
Q.E.D.


Eu acho que entendo melhor alíngual de vocês, por isso  escrevi isso aqui. Gostaria da discussão desse problema :

  Prove que [tex]\frac{(\sum_{i=1}^{n}a_i)^k}{(\sum_{i=1}^{n}b_i)^p}[/tex] <= [tex]\sum_{i=1}^{n}\frac{a_i^k}{b_i^p}[/tex], para [tex]a_i,b_i[/tex] inteiros positivos e [tex]k>p[/tex].

  P.S: Não sei se está certo, mas gostaria da paticipação de vocês para sabermos quendo essa desigualdade é verdadeira.

Meu Solução
[hide="Solucao"]No, não esta certo. Tomade $a_m=b_m=1$ para $m=1,2,\dots,n$, então $\frac{(\sum_{i=1}^{n}a_i)^k}{(\sum_{i=1}^{n}b_i)^p}=n^{k-p}\geq n = \sum_{i=1}^{n}\frac{a_i^k}{b_i^p}$. Isto funciona si $k-p>1$, mas possa ser modifique para quemquer valor de $k-p$. O outro direcção esta falso tambén, al menos em geral, não sei se a circunstância do quem $a_m$ son enteiros positivos cambia algo. Desculpa meu terrível portugese. [/hide]

Ahora en un idioma en el que me siento un poco más cómodo, el problema que Lucas propone es el de probar o desprobar que $\frac{(\sum_{i=1}^{n}a_i)^k}{(\sum_{i=1}^{n}b_i)^p}\leq\sum_{i=1}^{n}\frac{a_i^k}{b_i^p}$ donde $a_i$ y $b_i$ son enteros positivos, y $k>p$, no se si son enteros positivos o no!

[quote="Lucas Molina"][tex]k,p[/tex] são inteiros positivos.[/quote]Entonces quizás es $\frac{(\sum_{i=1}^{n}a_i)^k}{(\sum_{i=1}^{n}b_i)^p} \geq \sum_{i=1}^{n}\frac{a_i^k}{b_i^p}$ ?

[quote="Lucas Molina"]Você conhece a desigualdade de Radon? Ela parece ser parecida com essa, mas eu não tenho a prova dela... Você poderia demonstrá-la, por favor?[/quote]Buscando achade quem
\[\sum_{i=1}^{n}{a_i^{k+1} \over b_i^k}\geq{(\sum_{i=1}^{n}a_i)^k \over (\sum_{i=1}^{n}b_i)^{k+1}}\]
Não sei a prova, amanhã intento.

Lucas,  fala (lei) você espanhol? Meu portuguese es terrível, eu prefira escrever espanhol!


When I download AMC10  2008  pdf file, there are no diagram  in problems. (for example #15
), only some strange code. Can someone tell me how to view pictures in this file?

Many thanks!

I don't think you can. What I've always done is take the un-printable version and copy and paste it onto a word file, then edit it slightly so that no problems are cut-off.

Or what you could do is copy the asymptote code that is between the tags, including the tags:
[code]
[asy]
[/asy]
[/code]
and paste it in a new post. Then click 'Preview' and that will generate an image. You can save this image and then add it to the PDF if you have a PDF editor.

I'm not 100% sure on this matter but I think that it was due to copyright law. I think you can download the exams but not the images so that's why they are not available.

One other suggestion is to save the entire page as a cluster of file. If you have Firefox, you can click File, Save Page As, and just go from there. I'm sure it'll be same for other browsers too.


Prove that if $a,b,c$ are positive real numbers such that $x+y+z+xyz=4$, then $ x+y+z geq xy+yz+zx.$

The same substitution $x=2a/(b+c)$ and so on reduces the inequality after some computations to Schur.

Can you explain better your substitution? Cause I think that
$\frac{2a}{b+c}+\frac{2b}{a+c}+\frac{2c}{a+b}+\frac{8abc}{(a+b)(b+c)(c+a)}$
is not $4$ in general. Or it is?  :huh:

Oh, now I see, it's bigger or equal to 4. But equality occurs only when $a=b=c$ => $x=y=z$, but there ARE such $x,y,z$ that are not all equal and $x+y+z+xyz=4$ holds, or not?

my approach was rather different.
let $f(t)=(t-x)(t-y)(t-z)=t^3-at^2+bt-c$
we have $a+c=4$
with this and $f(t)=0$ should have three roots in$(0,+\infty)$
can we prove $a \geq b$
I haven't finished it yet :(

I wanted to tell, that approach that Killer used is just getting into account triple $1,1,1$ and no other triples of $x,y,z$. Or am I wrong?

And also if $x,y$ are nearly 0 and $z$ is nearly 4, then inequality is reversed.

Another solution: assume that $x+y+z<xy+yz+zx$. Schur inequality can also be written as $ 9xyz/(x+y+z)\geq 4(xy+yz+zx)-(x+y+z)^2$. You will deduce thus that $9xyz> (x+y+z)^2 (4-(x+y+z))=xyz (x+y+z)^2$. Thus $ x+y+z<3$, which is clearly falseL

[quote="zhaobin"][quote="harazi"]Thus $ x+y+z<3$, which is clearly falseL[/quote]
i don't see :(
but i think you are wrong ;)[/quote]
i think he's correct.

I know in the last period I did not practice inequalities, but I still hope I can solve this kind of problems correctly.  ;)

Assume $ x \ge y \ge z$.
Let $y=\frac{4-xz}{x+z+xz}$ and we have
$x+y+z \ge xy+yz+zx \Leftrightarrow (x+z-2)^2+xz(x-1)(1-z) \ge 0$


Are there any general extensions or related theorems to Ptolemy's Theorem? Or like, any kinds that could be used in a similar case like Ptolemy's Theorem?

http://en.wikipedia.org/wiki/Casey's_theorem
This is Casey's theorem which the generalised theorem of ptolemy

Hope it is helpful

Daniel

By generalized theorem, do you mean that Ptolemy's Theorem was derived from Casey's?

In fact Casey's theorem is recent while Ptolomey is old. But yes, Ptolomey theorem is a special case of Casey's theorem
Read the link I provided and you will found out more :D


Of the final five contestants in a television show, three are female
and two are male. If two are chosen randomly to be the final
contestants, what is the probability that both are female?
Express your answer as a common fraction.


Prove that in any triangle $ ABC$ the following equality is true :

$ \frac{\sin \ A}{\sin \ B\cdot \sin \ C}\plus{}\frac{\sin \ B}{\sin \ C\cdot \sin \ A}\plus{}\frac{\sin \ C}{\sin \ A\cdot \sin \ B}\equal{}2(\mathrm{ctg}\ A\plus{}\mathrm{ctg}\ B\plus{}\mathrm{ctg} \ C)$

hello, setting $ \sin(\alpha)=\frac{a}{2R}$, $ b=\sin(\beta)=\frac{b}{2R}$, $ \sin(\gamma)=\frac{c}{2R}$, and the theorem of cosine, we get
$ 2(\frac{Ra^2}{abc}+\frac{Rb^2}{abc}+\frac{Rc^2}{abc})=2(\frac{R(c^2+b^2-a^2)}{abc}+\frac{R(a^2+c^2-b^2)}{abc}+\frac{R(a^2+b^2-c^2)}{abc})$ and this is true.
Sonnhard.


calculus
derivative
trigonometry
real analysis
real analysis unsolved

I'm in an elementary PDE's class and I'm trying to do problems in my textbook. I'm having trouble starting. I was wondering if someone could show me how to do this example, so I can model similar solutions after it.

Find the solution and give a physical interpretation of the problem u[size=84]xx[/size] = 4u[size=84]t[/size] (i.e. second partial derivative of u with respect to x = 4 times the first partial derivative of u with respect to t)
for 0 < x < 2, t > 0,
with the conditions
u(0,t) = 0
u(2,t) = 0
u(x,0) = 2sin(pi*x/2) - sin(pi*x) + 4sin(2pi*x)

Thanks.

The only thing you need to know here is that $ u(x,t)\equal{}e^{\minus{}a\pi^2L^{\minus{}2}k^2t}\sin (k \pi L^{\minus{}1}x)$ solves $ u_t\equal{}au_{xx}$, $ u(0,t)\equal{}u(L,t)\equal{}0$, $ u(x,0)\equal{}\sin (k \pi L^{\minus{}1}x)$ (plus, of course, that the sum of solutions is again a solution, etc.).

OK, a physical interpretation. It's the wave equation on an interval, clamped at both ends. That's a string instrument; the main real-world difference is that there's a bit of damping.

So it is. How did I misread that? My alleged physical interpretation is irrelevant.


geometry
geometric transformation
rotation
reflection

How many ways are there to put five beads on a necklace if there are eight distinct beads to choose from, and rotations and reflections of the necklace are considered the same?

First, you have to choose which beads are put into the bracelet.  This is simply $ \binom{8}{5}$ or $ 56$.

Without accounting for rotations and reflections, we would have $ 120$ bracelets.  For rotations, we must divide by $ 5$, to get $ 24$.  For reflections, we must divide by $ 2$, to get $ 12$.

$ 56 \times 12 \equal{} \boxed{672}$

I used Ernie's method as well, came to post it and found it here already.


Find the remainder when $ 0!\plus{}1!\plus{}2!\plus{}3!\plus{}4!\plus{}5!\plus{}....\plus{}2006!$ is divided by $ 60$.

[hide="Answer"]
$ 5!\equal{}120$ is a multiple of $ 60$, and all the terms past $ 5!$ are multiples of $ 5!$, so they are all multiples of $ 60$.
Therefore, the remainder of that sum when divided by 60 is $ 0!\plus{}1!\plus{}2!\plus{}3!\plus{}4!\equal{}1\plus{}1\plus{}2\plus{}6\plus{}24\equal{}\boxed{34}$
[/hide]

Find the remainder of $ \displaystyle\left(\frac {2}{5!}\right)\displaystyle\sum_{i \equal{} 0}^{2006}i!$
[hide]Clearly, $ n!$ is a divisor of all factorials with a base $ b$ such that $ b\ge n$
So, the remainder $ r \equal{} \displaystyle\sum_{i \equal{} 0}^{4}i! \equal{} 34$[/hide]

 :rotfl: Wow, I thought that $ 5!\equal{}60$ and still got it right because I made the exact same elimination. The lesson? Always use variable forms of definition to justify your answer!


[quote="Ihatepie"]How do you do this?

In the figure, each of the circles has a radius of 6, and DE = 6. Find the area of the shaded region to the nearest square unit. Use 3.14 for pi.

It is basically two circles intersecting to make a venn-diagram like shape. Points D and E are the points at which the two circles intersect, and the shaded region is the intersection of the two circles.[/quote]
Umm this isn't possible without the picture, because you don't know how far in the points intersect.
Oh wait nevermind, I got it.
Umm....
I'll solve later.

[quote="Ihatepie"]How do you do this?

In the figure, each of the circles has a radius of 6, and DE = 6. Find the area of the shaded region to the nearest square unit. Use 3.14 for pi.

It is basically two circles intersecting to make a venn-diagram like shape. Points D and E are the points at which the two circles intersect, and the shaded region is the intersection of the two circles.[/quote]

[hide]Draw DE, and then draw the radii from the center of one of the circles to the points of intersection.  Note that an equilateral triangle is formed, and that the area that's in the sector cut off minus the area of the equilateral triangle is half the shaded region.  The area of the sector is $ \frac{1}{6}6^2\pi\equal{}6\pi$, and the area of the triangle is $ \frac{6^2\sqrt{3}}{4}\equal{}9\sqrt{3}$, so the area of the circular segment is $ 6\pi\minus{}9\sqrt{3}$ and the area of the whole shaded region is $ 12\pi\minus{}18\sqrt{3}$.  Do that on the calculator.[/hide]

[hide]
Let's call the centers of the two circles $ A$ and $ B$. We draw the line $ \overline{AD}$ and $ \overline{BD}$, and since those a radii also, their lengths are also 6. Let's also draw the line $ \overline{AB}$ which connects the centers of the two circles. Then, let's have the intersection of $ \overline{AB}$ and $ \overline{DE}$ be called $ C$. Therefore, $ \overline{DC} \equal{} \overline{EC} \equal{} \frac {1}{2}\overline{DE} \equal{} 3$. Finally, $ F$ and $ G$ are the intersections of the line $ \overline{AB}$ to the two circles.

We're looking at two right triangles, $ \triangle ACD$ and $ \triangle BCD$, both of whose right angles are $ C$.
These are in fact 30-60-90 triangles, since the length of the hypotenuse ($ \overline{AD}$), 6, is twice the length of leg $ \overline{DC}$, 3. Therefore, $ \angle DAC \equal{} \angle DBC \equal{} 30^\circ$. It follows that the lengths of $ \overline{CA}$ and $ \overline{CB}$ are both $ 3\sqrt {3}$.

From the diagram, finding the area of the region shared by both circles just boils down to finding the area of the rhombus in blue plus the areas of the four "pie crusts."

Finding the area of one of the pie crusts is trivial: calculate the area of sector $ DAG$ minus $ \triangle DAG$, which is $ \frac {30}{360}*\pi*6^2 \minus{} \frac {1}{2}*3*6 \equal{} 3\pi \minus{} 9$. Thus four of those equals $ 12\pi \minus{} 36$.

Calculating $ \overline{FG}$ is also trivial: $ 2(6 \minus{} 3\sqrt {3}) \equal{} 12 \minus{} 6\sqrt {3}$. Therefore, the area of the rhombus is half the product of its diagonals: $ \frac {1}{2}(12 \minus{} 6\sqrt {3})(6) \equal{} 36 \minus{} 18\sqrt {3}$.

Add them all up: $ (36 \minus{} 18\sqrt {3}) \plus{} (12\pi \minus{} 36) \equal{} 12\pi \minus{} 18\sqrt {3} \approx 6.522 \approx 7$.

So the answer is 7.

Whew (and a lot of LaTeX)!
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Edit: I think Alan's solution was more elegant... :blush: But I used more LaTeX! :D


Find the pairs (c,d) for which the equation
$x^{3}-8x^{2}+cx+d=0$
 has three (not necessarily) different positive integer roots.

Do you mean "not necessarily different"?

Either way the problem seems routine.

[hide="Solution"]Well, if we have roots $k, l, m$, then $k+l+m = 8$ which only has finitely many solutions in the positive integers.  Then $c = kl+lm+km, d =-klm$.

WLOG $k \le l \le m$:  then these solutions are

$k = l = 1, m = 6 : (c, d) = \boxed{(13,-6)}$
$k = 1, l = 2, m = 5 : (c, d) = \boxed{(17,-10)}$
$k = 1, l = 3, m = 4 : (c, d) = \boxed{(19,-12)}$
$k = 2, l = 2, m = 4 : (c, d) = \boxed{(20,-16)}$
$k = 2, l = 3, m = 3 : (c, d) = \boxed{(21,-18)}$ [/hide]


Hi! Let say that we have double sequence $ \{a_{mn}\}_{\mathbb{N}\times\mathbb{N}}:$

$ \begin{array}{l l l l l l}
a_{11} & a_{12} & \ldots & a_{1n} & \ldots & \to\infty\\
a_{21} & a_{22} & \ldots & a_{2n} & \ldots & \to\infty\\
\vdots & \vdots & \ddots & \vdots \\
a_{m1} & a_{m2} & \ldots & a_{mn} & \ldots & \to\infty\\
\vdots & \vdots &  & \vdots & \ddots\\
\downarrow & \downarrow & & \downarrow\\
0 & 0 &  & 0 & & \textbf{?}
\end{array}$

Where every $ a_{mn}\neq\infty$ and every $ a_{mn}\neq 0.$

What can we say about it when $ m\to\infty$ and $ n\to\infty$, ie. about the $ a_{\infty\infty}$? I was wondering if it is always $ \neq\infty$? Under what conditions it could be $ 0$ (if it could be)?

The answer is that under the circumstances you've set up, there's just no good way to define $ ``a_{\infty\infty}".$ You can do pretty much anything, including having it tend to zero or infinity.

Very simple example: let $ a_{mn}\equal{}\frac{n}{m}.$ That satisfies your framework.

Now consider various subsequences:

Consider $ a_{k,3k}\equal{}\frac{3k}{k}.$ That tends to $ 3.$

Consider $ a_{3k,k}\equal{}\frac{k}{3k}.$ That tends to $ \frac13.$

Consider $ a_{k,k^2}\equal{}\frac{k^2}{k}.$ That tends to $ \infty.$

Consider $ a_{k^2,k}\equal{}\frac{k}{k^2}.$ That tends to $ 0.$

Basically, we can make this do anything we want it to do.

Thanks! And what can be said about $ a_{kk}$ subsequence?

Nothing. Nothing at all. Obviously, for the sequence I named, $ a_{kk}$ tends to $ 1.$ But what if we had $ a_{mn}\equal{}\frac{n^2}{m}\ ?$ Or $ a_{mn}\equal{}\frac{n}{m^2}\ ?$

Thanks. So, can we always find subsequence $ a_{ij}$, where we pick just one element of a sequence from every coloumn and every row (we don't jump over some columns or rows), wich convergence to $ \infty$?

Another question would be like to first one, but convergence is to 0.

No: what if $ a_{m, n} \equal{} 1$ for all $ m, n$?  Really, there is no good general statement to be made without imposing some conditions.

I am afraid that I do no understand. We are not shure that such subsequence exists?

But how it could be proved? I don't have any counterexample in my mind. I am not wondering about radom subsequence wich could be formed in the way I said, but can we be shure that such subsequence of [b]our[/b] double subsequence exists?

I just gave you an example of a "double sequence" that has neither of the properties you asked about.  There is no general behavior to be found here: pretty much any statement of the form, "For every double sequence, we have ..." is either false or trivial.

[Edit: sorry, I've been misunderstanding what you wrote in the initial post this whole time.]

Maybe my english is realy bad (and it's very lete where I am :) ), but double sequence defined that $ a_{mn}\equal{}1$ for all $ m,n$ doesn't have properity that:
$ a_{1n}\to\infty$ when $ n\to\infty$
$ a_{2n}\to\infty$ when $ n\to\infty$
...

and properity that:
$ a_{m1}\to 0$ when $ m\to\infty$
...
(like I posted in first post). I had my first double sequence in this topic in my mind. Sorry if there was a misunderstanding.

I am not sure how to interpret your question in post #5.

If you mean: Can you always choose an order of columns $ n_1,n_2,...$ and an order of
rows $ m_1,m_2,...$ (so that the $ n_i$ form a permutation of the columns and the $ m_i$
a permutation of the rows) such that $ a_{m_in_i} \to \infty$ for $ i \to \infty$?
In this case the answer is no.
(Try to construct a counterexample, where the rows are of the form $ 0,...,0,1,2,3,...$ .)

But something weaker is true: You can always choose $ n_1<n_2<...$ such that
$ a_{in_i} \to \infty$ for $ i \to \infty$. (The proof of this is not too difficult.) But here you miss some columns.

[b]Question:[/b] Must there exist a subsequence $ a_{m_k, n_k}$, where $ \{ m_k \}$ and $ \{n_k \}$ are both increasing sequences of positive-integer indexes, such that $ \{ a_{m_k, n_k} \}_{k \equal{} 1}^{\infty}$ converges to a non-zero finite value?

An answer to that: Let $ a_{m,n}\equal{}\begin{cases}n&n\ge m\\ \frac1m&m>n\end{cases}$. All of the values are either large or small, so nothing can converge to a value strictly between zero and $ \infty$.

[b]solyaris[/b], you have interpreted my post #5 correctly, so thanks for that. But I don't quite follow what you said about constructing counterexample, because it said that $ a_{mn}\neq 0$ and $ a_{mn}\neq\infty$ for all $ m,n\in\mathbb{N}$ (and I mean not just diagonally, but for every element of the double sequence).

About the second thing (the weaker statement) - thanks for that, because it is something that I needed too.

I forgot this additional requirement, but it doesn't really matter anyway, just replace the 0s by $ \frac 1 m$s.


\[ \boxed{\text{Required to prove }\sin{2A} + \sin {2B} = 4\sin{A}  \sin {B} \Leftrightarrow \triangle ABC \text{ is right angled.} }\]
\begin{align*}
\sin{2A} + \sin{2B} & = 2\sin{A}\cos{A} + 2\sin{B}\cos{B} \\
\text{if }\triangle ABC \text{ is right angled, } &\Leftrightarrow \cos{A} = \sin{B}, \text{ and } \cos{B} = \sin{A} \\
\text{then }2\sin{A}\cos{A} + 2\sin{B}\cos{B} & = 2\sin{A}\sin{B} + 2\sin{B}\sin{A} \\
&= 4\sin{A}\sin{B} \\
\end{align*}

[quote="Johnny"]
\[ \boxed{\text{Required to prove }\sin{2A} + \sin {2B} = 4\sin{A} \sin {B} \Leftrightarrow \triangle ABC \text{ is right angled.} }
\]

\begin{align*} \sin{2A} + \sin{2B} & = 2\sin{A}\cos{A} + 2\sin{B}\cos{B} \\
\text{if }\triangle ABC \text{ is right angled, } & \Leftrightarrow \cos{A} = \sin{B}, \text{ and } \cos{B} = \sin{A} \\
\text{then }2\sin{A}\cos{A} + 2\sin{B}\cos{B} & = 2\sin{A}\sin{B} + 2\sin{B}\sin{A} \\
& = 4\sin{A}\sin{B} \\
\end{align*}
[/quote]

You have proved that if ABC is a right angle, then the statement is true.

However, proving the converse (or the reverse) of a statement does not prove the statement.

For example, since a dog has 4 legs, does this mean that a 4 legged animal is a dog?

[quote="thanhnam2902"]Let $ ABC$ is a triangle and let: $ \boxed{\sin{2A} \plus{} \sin {2B} \equal{} 4\sin{A} . \sin {B}}$.

Prove that $ ABC$ is a right triangle.[/quote]
Johnny,

as given, this is equivalent to an if-then statement.

It is not an if-and-only-if statement, with that double-
sided arrow that you typed out.

Although poorly written, the solution does work. All of those steps are reversible.

EDIT: I do agree with the next post, though. I did mean "poorly written".

[quote="MellowMelon"]Although poorly written, the solution does work. All of those steps are reversible.[/quote]

But then i dont think that considering that statement true and then reaching on the given condition is correct , like i had this question :
if z=x+iy and arg|(z-1)/(z+1)| = pi/2 , show that x^2 + y^2=1 , i assumed it to be true and got to the arg part but all i got in the examination by a Zero.
We can use contradiction but i dont think Going backwards is a correct way.

I agree with your posts I proved one direction only.  I tried going from the identity to getting ABC is right angled but caught in a vicious circle.  Anyone actually prove the converse?

[quote="ocha"][hide="solution"]
Notice that we have

$ \sin(2A) \plus{} \sin(2B) \equal{} 2\sin(A \plus{} B)\cos(A \minus{} B) \qquad(1)$
and
$ 4\sin(A)\sin(B) \equal{} 2\cos(A \minus{} B) \minus{} 2\cos(A \plus{} B) \qquad(2)$

Now combine $ (1)$ and $ (2)$, let $ A \plus{} B \equal{} x, A \minus{} B \equal{} y$ with $ 0 < x < \pi$ and $ 0\le y < \pi$ we have

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

This is clearly true when $ x \equal{} \frac {\pi}{2}$ (i.e. C is right-angled) we must show that this is the only case. So suppose $ x \neq \frac {\pi}{2}$ (which we must assume otherwise we will be dividing by zero)

$ \cos(y) \equal{} \frac {\cos(x)}{1 \minus{} \sin(x)}$

$ \Rightarrow \cos(y) \equal{} \frac {\cos(x)(1 \plus{} \sin(x))}{1 \minus{} \sin^2(x)}$

$ \Rightarrow \cos(x)\cos(y) \equal{} 1 \plus{} \sin(x)$

However, $ \cos(x)\cos(y) \le 1 \Longrightarrow \sin(x) \equal{} 0$ which is impossible because $ 0 < x < \pi$

Thus there are no solutions, Contradiction!

Thus it follows that $ ABC$ is  a right triagnle  :lol: [/hide][/quote]

Wow, it is quite nice!

I hate to be a fly in your ointment but
\[ A \plus{} B \equal{} x\text{ So } 0 < x \le 2\pi
\]

Sorry for the above I almost forgot that \[ A\plus{}B \equal{} \pi \minus{} C,\] So your proof is correct as is your ineqaulity.

My apologies.

Nice solution by the way! :blush:

It's equivalent to $ \frac{cos A}{sin B}\plus{}\frac{cosB}{sinA}\equal{}2$. Fix A, then as B increases, sin B increases and cos B decreases, thus the function is strictly decreasing, which completes the proof, as a right angle triangle works.