floor function
geometry
3D geometry
pyramid
symmetry
trigonometry
tetrahedron

A tripod has three legs each of length 5 feet.  When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is 4 feet from the ground.  In setting up the tripod, the lower 1 foot of one leg breaks off.  Let $h$ be the height in feet of the top of the tripod from the ground when the broken tripod is set up.  Then $h$ can be written in the form $\frac m{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime.  Find $\lfloor m+\sqrt{n}\rfloor.$  (The notation $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$.)

[hide="A straightforward but messy solution"]Let's first examine the non-broken tripod: if $P$ is the intersection of the altitude of the pyramid formed with the base, and we let the base triangle be denoted $\Delta ABC$, then we have $AP = BP = CP = 3$ by the Pythagorean theorem.  We also know that $\angle APB = \angle BPC = \angle CPA = 120^{\circ}$ by symmetry, so by the law of Cosines we have $AB = BC = AC = 3 \sqrt{3}$.  Then the cosine of the angle between any two tripod legs is given by
$\cos \theta = \frac{23}{50}$

Now we examine the broken tripod.  Let the base triangle be given by $\Delta ABC'$, where $BC'$ and $AC'$ are the base lengths of the tripod faces with side lengths 4 and 5.  Applying the Law of Cosines, we find that
$BC' = AC' = \sqrt{\frac{113}{5}}$
(I'm skipping over a lot of messy arithmetic in this problem, but you can verify it for yourself).  Now let's examine $\Delta ABC'$.  Let $P'$ be the intersection of the new height and the base.  Note that by the Pythagorean theorem we have $AP' = BP' = \sqrt{25 - h^2}$, and $C'P' = \sqrt{16 - h^2}$, where $h$ is the unknown altitude.  Note also that $\angle AC'P' = \angle BC'P' = 1/2 \angle AC'B$, as $\Delta AC'P'$ and $\Delta BC'P'$ are congruent.  Now we apply law of Cosines to $\Delta ABC'$ to find first that
$\cos AC'B = \cos \alpha = \frac{91}{226}$
Applying the half-angle identity for cosine, we have
$\cos \frac{\alpha}{2} = \frac{1}{2} \sqrt{\frac{317}{113}}$
Now we apply Law of Cosines to $\Delta AC'P$ (or $\Delta BC'P$) to find
$25 - h^2 = \frac{113}{5} + 16 - h^2 - 2 \sqrt{\frac{113}{5}} \sqrt{16 - h^2} \cos \frac{\alpha}{2}$
Now substituting in for $\cos \frac{\alpha}{2}$ and doing a ton of tedious arithmetic (thankfully all but one $h^2$ cancels, so we don't have to resort to anything insane), we find that
$h = \frac{144}{\sqrt{1585}}$
So we have
$\lfloor m + \sqrt{n} \rfloor = \lfloor 144 + \sqrt{1585} \rfloor$
and noting that $40^2 = 1600$, so that $\lfloor \sqrt{1585} \rfloor = 39$, we have
$\lfloor 144 + \sqrt{1585} \rfloor = 144 + 39 = \boxed{183}$[/hide]

[hide="A similarly messy solution, but without cosine law"]
This is really hard to explain without a diagram, but here goes:

First consider the non-broken tripod. You can think of this as a tetrahedron, with height $4$, sides $5$, and an equilateral triangle base of side $3 \sqrt{3}$

Now, consider cutting this tetrahedron in half (or a projection onto a 2D plane if you wish to think of it that way), along the plane which includes the height of the trapezoid and one of the legs. We now consider this cross-section, which is a triangle which we shall name $ABC$, with $AB = 5$, $BC = \frac{9}{2}$, and altitude $AD = 4$.

The beauty of this cross-section is that, if you consider the similar cross-section of the broken tripod, you can draw in its cross-section superimposed on the one of the normal tripod. You do this simply by drawing a line $CE$ which divides $AB$ (the leg that gets  broken), such that $EB = 1$ and $AE = 4$.

The new height of this broken tripod, is then the altitude of triangle $ACE$ dropped from $A$ to $CE$. Let's name this $AF$.
Let's also designate the intersection of $CE$ and $AD$ to be $G$, and draw a perpendicular from point $E$ to $BC$, and designate this line $AH$.

Now comes the messy part; i will skip through a lot of the calculations, but you can verify them if you want:

$EB = 1$, and by similar triangles to $EDB$, $EH = \frac{4}{5}$, and $HB = \frac{3}{5}$.

Now, triangle $CDG \sim CHE$ , and you know $CD = \frac{3}{2}$, $CH = CB - HB = \frac{39}{10}$, and $HE = \frac{4}{5}$, so you set up the ratio and find $GD = \frac{4}{13}$. This gives us $AG = 4-\frac{4}{13}=\frac{48}{13}$.

Now, triangle $CDG \sim AFG$, so we wish to set up a ratio to find $AF$. To do this, we find $CG$ by pythagoras, which comes out to be $\frac{\sqrt{1585}}{26}$. So now, we set up the ratio:
\begin{eqnarray*} \frac{CG}{AG} &=& \frac{CD}{AF}\\ \frac{\frac{\sqrt{1585}}{26}}{\frac{48}{13}} &=& \frac{\frac{3}{2}}{h}\\ h &=& \frac{114}{\sqrt{1585}} \end{eqnarray*}
I know, I know, i skipped a whole bunch of ugly math...
but after this, we have $m = 114$ and $n = 1585$, so $\left\lfloor m + \sqrt{n}\right\rfloor = 183$
[/hide]

OK, I am writing a little clean solution:

[hide]
We will use $[...]$ to denote volume (four letters), area (three letters) or length (two letters).

Let $T$ be the top of the tripod, $A,B,C$ are end points of three legs. Let $S$ be the point on $TA$ such that $[TS]=4$ and $[SA]=1$. Let $O$ be the center of the base equilateral triangle $ABC$. Let $M$ be the midpoint of segment $BC$. Let $h$ be the distance from $T$ to the triangle $SBC$ ($h$ is what we want to find).

We have the volume ratio: $\frac{[TSBC]}{[TABC]} = \frac{[TS]}{[TA]}=\frac{4}{5}$

So $\frac{h\cdot [SBC]}{[TO]\cdot [ABC]}=\frac{4}{5}$

We also have the area ratio: $\frac{[SBC]}{[ABC]}=\frac{[SM]}{[AM]}$

The triangle $TOA$is a 3-4-5 right triangle so $[AM] = \frac{3}{2}\cdot[AO]=\frac{9}{2}$ and $\cos{\angle{TAO}}=\frac{3}{5}$.

Applying Law of Cosines to the triangle $SAM$ with $[SA]=1$, $[AM]=\frac{9}{2}$ and $\cos{\angle{SAM}}=\frac{3}{5}$, we find:

$[SM]=\frac{\sqrt{5\cdot317}}{10}$.

Putting it all together, we find $h=\frac{144}{\sqrt{5\cdot317}}$.

[/hide]

I went for a much more complicated solution...and messed it up somewhere along the way.  Well my answer was about the same, less than a unit off...

[hide="A solution using coordinates"]
As in the previous solutions, it's not hard to get that the feet of the original tripod form an equilateral triangle of side length $3 \sqrt{3}$.

Place the feet of the tripod at $P(3, 0, 0)$, $Q(-3/2, 3 \sqrt{3}/2, 0)$, and $R(-3/2, -3 \sqrt{3}/2, 0)$, and the top at $V(0, 0, 4)$.

Let $VP$ be the leg that gets broken, so it is broken at the point
\[ S = \frac{1}{5} V + \frac{4}{5} P = \left( \frac{12}{5}, 0, \frac{4}{5} \right). \]

Let the equation of the plane passing through points $Q$, $R$, and $S$ be $Ax + By + Cz = 1$, so we get the system
\begin{eqnarray*} -\frac{3}{2} A + \frac{3 \sqrt{3}}{2} B &=& 1, \\ -\frac{3}{2} A - \frac{3 \sqrt{3}}{2} B &=& 1, \\ \frac{12}{5} A + \frac{4}{5} C &=& 1. \end{eqnarray*}

Solving, we get $A = -2/3$, $B = 0$, and $C = 13/4$, so the equation of the plane is
\[ -\frac{2}{3} x + \frac{13}{4} z = 1, \]
or $8x - 39z + 12 = 0$.

Recall that the distance from the point $(x_0, y_0, z_0)$ to the plane $Ax + By + Cz + D = 0$ is
\[ \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}. \]

Plugging these in, we get that the distance from point $V$ to plane $QRS$ is
\[ \frac{|-39 \cdot 4 + 12|}{\sqrt{8^2 + 39^2}} = \frac{144}{\sqrt{1585}}. \]

Then $\lfloor 144 + \sqrt{1585} \rfloor = 183$.
[/hide]

Normally, I wouldn't endorse coordinates, but the calculations in this case are quite reasonable.

[quote="joml88"]Wow, I feel brilliant.  I found $h$ but thought I was wrong since $m+n>1000$.  So I just randomly guessed.   :wacko:[/quote]

Same :(. I was like I can't have a $1585$ in the denominator that's too big...

[quote="zanttrang"]Note that by the Pythagorean theorem we have $AP' = BP' = \sqrt{25 - h^2}$, and $C'P' = \sqrt{16 - h^2}$, where $h$ is the unknown altitude.[/quote]

I got up to this point. Here, I used the circumradius area formula, $A = \dfrac{abc}{4R}$ and equated it with the easily calculated area to find the circumradius. Then, my fallacy was that the circumradius is the second leg of the right triangle that has a hypotenuse of one of the feet of the tripod and the other leg being the height. This is obviously false since that would mean the height is the leg of two non-congruent right triangles, both with other leg being the circumradius, but one with hypotenuse $4$ and other with hypotenuse $5$. However, I do not understand then where to proceed from here. I'm kind of lost by the other solutions. Where, wrt to the base, does the height "land"?

[quote="wilson97"]In the aopswiki solution: http://www.artofproblemsolving.com/Wiki/index.php/2006_AIME_I_Problems/Problem_14 , why is

$ \frac{[SBC]}{[ABC]}=\frac{[SM]}{[AM]} $  true?[/quote]

The ratio of the areas of two triangles equals the ratio of their respective altitudes when their bases are equal.

[hide=Fast 2D coordinates]
Let $G$ be the centroid of the equilateral triangle base $ABC$ and let $T$ be the other vertex of the original tripod. Let $5S = 4A+T$ and set up coordinates with $G(0,0),A(3,0),T(0,4) \implies S(\frac{12}{5},\frac{4}5), \frac{B+C}{2} = (-\frac{3}{2},0)$. Therefore the line through $S, \frac{B+C}{2}$ is
$$\ell: 8x - 39 y + 12 = 0.$$
Use distance to line formula  for $T, \ell$ to finish.
[/hide]

This is really easy with complex numbers.

Oops here's another slow synthetic solution.

First we want to figure out the dimensions of the new tripod. Check that the base is an equilateral triangle of side length $3\sqrt 3$, so by Stewart's on one of the lateral triangles, we obtain that the new base of the tripod is an isosceles triangle with side lengths $2\sqrt 7, 2\sqrt 7, 3\sqrt 3$.

Now, let $O$ be the foot of the perpendicular from the apex of the tripod. Observe that the distances from $O$ to the three vertices of the base are $x, x, \sqrt{x^2-9}$ by perpendicularity lemma. Using the altitude relation, we obtain $$\sqrt{x^2-9} + \sqrt{x^2-\frac{27}4} = \sqrt{\frac{317}{20}},$$ which solves to get $x^2 = \frac{4 \cdot 34^2}{5 \cdot 317} + 9$. Thus, the length of the altitude is $$\sqrt{25-y} = \sqrt{\frac{16 \cdot 5 \cdot 317 - 68^2}{5 \cdot 317}} = \sqrt{\frac{144^2}{5 \cdot 317}},$$ so the answer is $\lfloor 144 + \sqrt{5 \cdot 317} \rfloor = \boxed{183}$.

Let $A,B,C$ be the feet of the tripod in some order, and let $D$ be the highest point. Note that $\triangle ABC$ is equilateral with side length $3\sqrt{3}$. Hence, $[ABC] = \frac{(3\sqrt{3})^2\sqrt{3}}{4}$, so $\text{Vol}(ABCD) = \frac{1}{3}\cdot 4\cdot [ABC] = 9\sqrt{3}$.

Now, let $C'$ be the foot of the cut leg. From volume ratios, $\text{Vol}(ABC'D) = \frac{4}{5}\text{Vol}(ABCD) = \frac{36\sqrt{3}}{5}$. Now, we compute $[ABC']$. Note that $\cos\angle ADC =\frac{23}{50}$ from LoC, so
\[ AC' = \sqrt{5^2 + 4^2 - 40\cdot \frac{23}{50}} = \sqrt{\frac{113}{5}}, \]
from LoC in $\triangle ADC'$. Similarly, $BC' = \sqrt{\frac{113}{5}}$. Now, we have all three side lengths of $\triangle ABC'$, so we use pythag to find the height to $AB$ and use base*height/2 to get $[ABC'] = \frac{3}{4}\sqrt{\frac{951}{5}}$. Setting up an equation for the volume of $ABC'D$ gives
\[ \frac{1}{3}\cdot h\cdot \frac{3}{4}\sqrt{\frac{951}{5}} = \frac{36}{5}\sqrt{3}\implies h = \frac{144}{\sqrt{1585}}.\]
Hence, the answer is $144 + \lfloor \sqrt{1585}\rfloor = 144 + 39 = \boxed{183}$

Solved with [b]amuthup[/b], [b]js104[/b], [b]crazyeyemoody907[/b], [b]ETS1331[/b], [b]dottedcaculator[/b], [b]v4913[/b], [b]CT17[/b], [b]CyclicISLscelesTrapezoid[/b], [b]squareman[/b], and [b]bluelinfish[/b].

Let the feet be $A$, $B$, $C$ and the top of the tripod be $T$. Note that the distance between the projection of $T$ onto plane $ABC$ and any one of the feet is $\sqrt{ 5^{2}-4^{2} }=3$, so the side length of $\triangle ABC$ is $\frac{3}{\frac{\sqrt{3 }}{3}}=3\sqrt{ 3 }$.

Place the tripod on the coordinate plane, with the ground being $y=0$. Let the feet of the tripod be $A,B=\pm\left( \frac{3\sqrt{ 3 }}{2},0,0 \right)$ and $C=\left( 0,0, \frac{9}{2} \right)$. We then must have $T=\left( 0,4, \frac{3}{2} \right)$.

Suppose the lower foot of leg $TC$ breaks, so that the new end is $C'=\frac{4C+T}{5}=\left( 0, \frac{4}{5}, \frac{39}{10} \right)$.

We wish to find the equation of plane $ABC'$: $ax+by+cz=d$. Since the line with parametric equation $(t,0,0)$ (aka $AB$) is on this plane, $a=d=0$. We can then see that the plane's equation is therefore $39y-8z=0$ (or any scaling thereof).

Our desired $h$ is the distance from $T$ to this plane, which is:
$$
\frac{\left\lvert  39\cdot 4-8\cdot \frac{3}{2}  \right\rvert}{\sqrt{ 39^{2}+8^{2} }}=\frac{144}{\sqrt{ 1585} }.
$$

So our answer is $\lfloor 144 + \sqrt{ 1585 }\rfloor=144+39=\boxed{183}.$


For positive numbers $ x, y, z$ , prove that,

$ \displaystyle \frac {\left\{\prod(x \plus{} y)\right\}^{2}}{\prod x}\geq\frac {\left\{\frac {8}{9}\left(\sum x\right)\left(\sum xy\right)\right\}^{2}}{\prod x}\geq\frac {64}{27}\left(\sum x\right)^{3}.$
Thanks

[quote="Shanku"]For positive numbers $ x, y, z$ , prove that,

$ \displaystyle \frac {\left\{\prod(x \plus{} y)\right\}^{2}}{\prod x}\geq\frac {\left\{\frac {8}{9}\left(\sum x\right)\left(\sum xy\right)\right\}^{2}}{\prod x}\geq\frac {64}{27}\left(\sum x\right)^{3}.$

The source where I got it says it is obvious. However I am not finding the obvious links.[/quote]

If something not obvious is not out of my attention the left hand side of the 1st part of the inequality reduces to $ 9(x \plus{} y)(y \plus{} z)(z \plus{} x)\geq 8(x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx)$ which is obviously true, since it is of the form $ \sum_{cyc}x(y \minus{} z)^2$.
Now the second part of the inequality reduces to $ \left(\sum_{cyc}xy\right)^2\geq 3xyz\sum_{cyc}x$ which is also obviously true.
Sorry for interupting at this part of the forum.


Let  $ a, b, c$ be the sides and $ l_a, l_b, l_c$ the bisectors of an triangle $ ABC$. Prove that

$ (a \plus{} b \plus{} c)^2 > l_a(b \plus{} c) \plus{} l_b(c \plus{} a) \plus{} l_c(a \plus{} b) > \frac {(a \plus{} b \plus{} c)^2}{2}$ .

________
 :|

[quote="Ligouras"]Let  $ a, b, c$ be the sides and $ l_a, l_b, l_c$ the bisectors of an triangle $ ABC$. Prove that

$ (a + b + c)^2 > l_a(b + c) + l_b(c + a) + l_c(a + b) > \frac {(a + b + c)^2}{2}$ .

________
 :|[/quote]
we know #$ l_a=\frac{2bccos{\frac{A}{2}}}{b+c}$
SO $ l_a(b + c) + l_b(c + a) + l_c(a + b)=2(\sum bccos{\frac{A}{2})}$


inequalities
geometry
inequalities unsolved

Let $ a,b,c,x,y,z>0$. Prove that $ \frac{a(z\plus{}y)}{b\plus{}c}\plus{} \frac{b(z\plus{}x)}{a\plus{}c} \plus{}\frac{c(x\plus{}y)}{b\plus{}a} \geq \sqrt{3(xy\plus{}yz\plus{}zx)}$.

[quote="Spribo"]Let $ a,b,c,x,y,z > 0$. Prove that $ \frac {a(z \plus{} y)}{b \plus{} c} \plus{} \frac {b(z \plus{} x)}{a \plus{} c} \plus{} \frac {c(x \plus{} y)}{b \plus{} a} \geq \sqrt {3(xy \plus{} yz \plus{} zx)}$.[/quote]
See here:
http://reflections.awesomemath.org/2008_1/unexpected_ineq.pdf

[b]Following well-know-inequality:[/b]
$ \mbox{If a, b, c sides of triangle ABC, and: }m,n,p > 0 \\
\mbox{and F \equal{} area of triangle ABC, rezult: } \\
a^2m \plus{} b^2n \plus{} c^2p\ge\sqrt {mn \plus{} mp \plus{} np}.4F; \\
\mbox{ let: } a \equal{} \sqrt {x \plus{} y}, b \equal{} \sqrt {x \plus{} z}, c \equal{} \sqrt {y \plus{} z}\mbox{ and} \\
m \equal{} \frac {a}{b \plus{} c}, n \equal{} \frac {b}{a \plus{} c}, p \equal{} \frac {c}{a \plus{} b}\Rightarrow \\
\Rightarrow \sum_{cyclic}{\left(\sqrt {x \plus{} y}\right)^2.\frac {a}{b \plus{} c}}\ge\sqrt {\sum_{cyclic}\frac {a}{b \plus{} c}.\frac {b}{a \plus{} c}}.\sqrt {\sum_{cyclic}{xy}} \\
\mbox{and }\sum_{cyclic}{\frac {a}{b \plus{} c}.\frac {b}{a \plus{} c}}\ge\frac {3}{4} \mbox{ (well \minus{} know inequality)} \\
\mbox{and prowe the inequality.}$
view topics: http://www.mathlinks.ro/viewtopic.php?t=261034


Multiply each number by $ 2$ to get $ 3,11,19,\ldots,83,91$.

The number of times this sequence increases by $ 8$ is $ \frac{91\minus{}3}8\equal{}11$,

so the number of terms is $ 11\plus{}1\equal{}\boxed{12}$.

Or, you could subtract $ 0.5$ from each term.  Then we have $ 1,5,9,13,\ldots,41,45$.  This increases by $ 4$ every step, so we have $ \frac{45\minus{}1}{4}\plus{}1\equal{}\boxed{12}$.

I guess the fastest way is to add $ 2.5$ to each term to get

$ 4,8,\ldots,48$, and dividing through by $ 4$, we get

$ 1,2,\ldots,12$, which obviously has $ 12$ terms.

[quote=pog]I want one added to my post count too![/quote]
We subtract $0.5$ from each number to get:
$$1, 5, 9, 13, ...., 41 , 45$$
, then we subtract one from each number to get
$$ 0, 4, 8, 12, .... 40, 44$$
and then we divide each number by four to get
$$ 0, 1, 2, 3, ...., 10, 11$$,
so we have $\boxed{12}$ numbers.

[quote=ernie]The multiplying by $ 2$ is unnecessary.[/quote]

Ya

[quote=Smartgoodness][quote=jonathanprieto]El .5 en cada número no importa. avanza de 4 en 4, le resto 1 a la lista 1,5,9,...45 para conseguir 0, 4,...44, luego tenemos
44/4 + 1 (cero) = 11 + 1 = 12. ya estuvo.[/quote]

Translated (from Spanish into English) :

The .5 in each number does not matter. advances from 4 to 4, I subtract 1 from the list 1,5,9, ... 45 to get 0, 4, ... 44, then we have
44/4 + 1 (zero) = 11 + 1 = 12. It was already.

steps for mine:

$$1.~\text{Add }0.5.$$
$$2.~\text{Divide by }2.$$
$$3.~\text{Add }1.$$
$$4.~\text{Divide by }2.$$

You should have:

$$1,2,3,4,\ldots ,11,\boxed{12}.$$[/quote]

Thank you for the translation


algorithm
number theory
prime numbers
number theory open

A perfect number x is a number which  equals to the sum of all it's divisors
(including 1, but without x).
Euclid proved that if p -is prime and p+1=2^{k} then 2^{k-1}*p is perfect.
Eiler proved that every even perfect number is equal to 2^{k-1}*p, where p -is prime and p+1=2^{k}.
It is unknown are there odd perfect numbers.
The question which is interesting to me is: Is it true that every even perfect number
ends on 6 or 28? First five perfect numbers are 6,28,496,8128,33550336.
If someone knows sixth perfect number please tell it to me. Is it ends on 28?

The first perfect numbers are:
$6\\28\\496\\8128\\33550336\\8589869056\\137438691328\\2305843008139952128$
$\cdots$
It is well known that all [u]even[/u] numbers (except $6$) end in $16,28,36,56,76$ or $96$.
But it is not known if any odd perfect number exist.

You can easily show your statement when looking at all prime residue classes $\mod 100$.

On February 18, 2005, Martin Nowak from Germany, found the new largest known prime number (42 nd):
$2^{25964951}-1$.

Very easy problem: what are the two last digits of the corresponding perfect number?  :)

[quote="blang"]
But it is not known if any odd perfect number exist.[/quote]

Have a look here : http://arxiv.org/abs/hep-th/0401052  :)

Did you check the proof?  ;)
The problem seem still to be open [url=http://mathworld.wolfram.com/OddPerfectNumber.html][u]here[/u][/url] :(
Sept 2004, Hare wrote: "It is unclear if the recent paper on the Number Theory Archive claiming to prove that there are no odd perfect number, successfully solves this problem or not".
Do you have some new precisions?

in more general, i know that there is a theroy about numbers called friends (amiable-im not sure about the translate) wich have the following : S(m)-m=S(n)-n were S(x) means the sum of all divisors of x ,integer.the smallest pair is (220,284).there is not known  an algorithm wich finds  all this kind of pairs.EULER found a particular prove and discover aprox 50 new pairs. in his prove he takes forms for m and n. (it's very long).
220 and 284 were considerated for a long time two magic numbers with special power.if i remember well...Pitagora was the one who  found first 2 friends numbers.

the english word is "amicable", but the definition is different: $\sigma(m)=\sigma(n)=m+n$, so each is the sum of the "aliquot" parts of the other ($m=\sigma(n)-n$ and $n=\sigma(m)-m$).

[url]http://mathworld.wolfram.com/AmicablePair.html[/url]
[url]http://www.shyamsundergupta.com/amicable.htm[/url]

Igor did you read the proof ?
I don't follow the author just from the start  :P

yap,you're right  :blush:  :blush:  :blush: i'm sorry about that...i'm not sure what happened :oops:

Is there a simple test for primality of Merseene number?
thx
Bomb

the standard test is:

define $s_2:=4, s_{n+1} := s_n ^2 -2$. Then $2^p -1$ (with $p>2$ being prime) is a prime itself iff $2^p - 1 | s_{p}$. (as much as I know it's the best known test for prime numbers)

This is a special version of a more general test:
Let be $p$ a given odd interger.
Let be $a$ any integer that fulfils $\left( \frac{a}{p} \right) = -1$ ( $\left( \frac{a}{p} \right)$ means the jacobi symbol ).
Then $p$ is prime
iff
there is a $\epsilon \in (\mathbb{Z} / p \mathbb{Z})[\sqrt a]$ that fulfils:
a) $\epsilon \overline{\epsilon} =1 $ ($\overline{\epsilon}$ means the 'conjugate' of $\epsilon$)
b) $\epsilon ^ {p+1} =1$
c) $\epsilon ^\frac{p+1}{q} \neq 1$ for all prime divisors $q$ of $p+1$.


algebra
polynomial
calculus
integration
arithmetic sequence
number theory unsolved
number theory

Does anyone have a non very difficult and elementary proof of the following theorem?
  ' If f is a polynomial with integer coefficients such that in any infinite arithmetic progression there is an element x such that f(x) is a square then f is the square of a polynomial'?

This theorem is not due to Sierpinski, but to Davenport (this one is ok  ;)  ), Schinzel and Lewis.
Their paper appeared in Acta Arithmetica, IX (1964), p.107-116.
I would not say that the proof is completely elementary...and it refers to some results due to Hasse that I did not know (which does not prove anything).
It is a shame tyhat you didn't ask for it before yesterday because I would bring you the paper last sunday (I have no scan...). Maybe I could send it to you by classical mail. Send me your room adress by pm.

To be rigourous, they proved a more general result :

Let $f(x,y)$ be a polynomial with integral coefficients. Suppose that every arithmetical progression contains some integer $x$ such that the equation $f(x,y)=0$ has an integral solution in $y$. Then, there exists a polynomial $g(x)$ with rational coefficients such that $f(x,g(x)) = 0$ identically.

It is an immediate corollary to deduce that :

Let $k>1$ be an integer and let $f(x)$ be a polynomial with integer coefficients. Suppose that every arithmetical progression contains some integer $x$ such that $f(x)$ is a $k$-th power of an integer. Then $f(x)=(g(x))^k$ identically, where $g(x)$ is a polynomial with integral coefficients.

Pierre.

Pierre, if I found this problem earlier, be sure I would have asked you some information about you. Unfortunately, after the confererence at ENS I received a message from a friend in which he told me someone gave this theorem to him and he was asking for my help. Be sure I will always ask for some information. I would be very glad if you could send me the file by e_mail. I will give you the address in a pm. Thanks a lot.


[i]One day, a kid Michael was invited to visit a Math class in the home planet of E.T.

The teacher asked, "If there are 24 classes in a school, and exactly 30 students in each class, then how many students are there in the school?"

A student William quickly put up his hand,"Sir, it should be 720."

Teacher replied: "I am sorry, William, your calculation is incorrect. Think again."

Another girl Mary then put up her hand with uncertainty,"I am not sure if my answer is correct...."

Teacher: "Don't worry, just tell me your answer, Mary."

"I think...it should be 740..."

"Good girl ! "  the teacher praised the girl and wore a smile in his face.[/i]

"What?????"   Michael was so confused.

[b]Question: What's wrong with the class?[/b]

20 of the students are pregnant?

Also
[hide]
I think that they use the digit 4 instead of 5.
[/hide]

20 students got pregnant ? :ninja:
I have never thought about that before.
This answer is creative but it isn't the truth.
The movie clip in this link reveals the secret:
http://www.youtube.com/watch?v=1KpripUvLs8

Uh, isn't that just a music clip?

I'm afraid I don't understand how that explains it.

EDIT: Ah.  :)

Look at the hand of E.T. again.
The truth is out there!

[hide]
They count in base 8. Seven fingers, no thumbs, one flashlight.[/hide]

[hide]Let b equal the number base. Then
$ (2b\plus{}4)(3b)\equal{}7b^2\plus{}4b$
$ 6b^2\plus{}12b\equal{}7b^2\plus{}4b$
$ b^2\equal{}8b$
$ b\equal{}8$

In other words, ET has only four fingers on each hand~[/hide]


$K > L > M > N$ are positive integers such that $KM + LN = (K + L - M + N)(-K + L + M + N)$. Prove that $KL + MN $is composite.

This is from IMO 2001. Again, it has been posted before(even the geometric solution using Ptolemy). Search it yourself.

This is really too well-known.

IMO 2001 Problem 6. Solutions on at least 100 different places on the web and on this site.

How many time did you send this problem?Anyway,here is link to the [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=imo+2001+geometry+problem&t=32032]solution[/url].
(If you realy want the solution.)But it is very nice problem and darij solved it very very nice.   :love:

can u solve it without using batlamious?
i my self solved it by batlamious but iwant the other new solutions. :D

i think you are so cool ingeometry ashegh and you can solve it without using batlamius,so try it and sent your solution! ;)

your wellcome dear jensen.and i know that u can solve it without using bathlamious.
come on boy ,solve it. :D

[quote="ashegh"]can u solve it without using batlamious?[/quote]

Right. I was getting used to Pythagoras being Fishagoures already, but now I see that Ptolemaius (Ptolemy) became Batlamious! Guys, check your spelling! Please!


I have though of a puzzling part of refraction that has caused a bit of a problem. it is that if the light slows down then where does it get the energy from to speed back up once it has left the medium it was refracted into?

Light has no mass. Energy is only expended in accelerating an object with non-zero rest mass. Therefore no energy is gained or lost by light particles (or photons) when their velocity chenges. In fact the knietic energy of an object is given by:

$ E_{K}\equal{} E\minus{}E_{o}\equal{} mc^{2}\minus{}m_{0}c^{2}$

$ m$ and $ m_{o}$ are the relativistic and rest masses respectively. $ m$ is given by:

$ m \equal{}\frac{m_{0}}{\sqrt{1\minus{}\frac{v^{2}}{c^{2}}}}$

Therefore:

$ E_{K}\equal{} m_{0}c^{2}\left(\frac{1}{\sqrt{1\minus{}\frac{v^{2}}{c^{2}}}}\minus{}1\right)$

Hence an object with no rest mass, such as a photon will have no kinetic energy. The above equation shows why it is impossible to accelerate an object with non-zero rest mass to the speed of light. As $ v\to c$ the denominator of the fraction tends to zero, and $ E_{K}\to\infty$. So it would take infinite energy to accelerate an object with mass to the speed of light. Anyway, I have diverged slightly from the main problem.

The energy of a photon is dependent on its frequency alone: $ E \equal{} hf$ where $ h$ is Plank's constant. But the frequency is conserved when entering a different optical medium. Hence the light's energy is conserved, no energy is lost or gained by the photon during the transition between any two media. So nothing to worry about. (It is in fact due to the conservation of energy, that the frequency must remain constant).


algebra
polynomial
calculus
derivative
calculus computations

well, got the following results for $ \ x^3\plus{}4x^2\minus{}20x\minus{}81 \equal{}0$ :

$ \ x_1 \equal{} \sqrt[3]{\frac{2315}{54}\plus{}\sqrt{\frac{3371}{108}}}\plus{}\sqrt[3]{\frac{2315}{54}\minus{}\sqrt{\frac{3371}{108}}}$

$ \ x_2 \equal{} w\sqrt[3]{\frac{2315}{54}\plus{}\sqrt{\frac{3371}{108}}}\plus{}\overline{w}\sqrt[3]{\frac{2315}{54}\minus{}\sqrt{\frac{3371}{108}}}$

$ \ x_3 \equal{} \overline{w}\sqrt[3]{\frac{2315}{54}\plus{}\sqrt{\frac{3371}{108}}}\plus{}w\sqrt[3]{\frac{2315}{54}\minus{}\sqrt{\frac{3371}{108}}}$

But the original equation is $ \sqrt{x}\equal{}18\minus{}x^2$

The equation, after a change has become $ \ (x\minus{}4)(x^3\plus{}4x^2\minus{}20x\minus{}81)\equal{}0$

4 As one of the roots, and the fact the original equation is not polynomial, there is no guarantee that a root complex, or both, are solutions of this equation .... doubt ....

obviously:

$ \ w\equal{}\frac{\minus{}1}{2}\plus{}i\frac{\sqrt{3}}{2}$

This is continued from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=253802]here.[/url] I'm not entirely sure why you paid close attention to Sonnhard Graubner's post and ignored mine.

I suggested that you make the substitution $ y \equal{} \sqrt {x}.$ That gives us the equation

$ y^4 \plus{} y \minus{} 18 \equal{} 0$

$ (y \minus{} 2)(y^3 \plus{} 2y^2 \plus{} 4y \plus{} 9) \equal{} 0.$

The solution $ y \equal{} 2$ gives us $ x \equal{} 4.$

The equation $ y^3 \plus{} 2y^2 \plus{} 4y \plus{} 9$ has no nonnegative roots. (Call it Descartes' rule of signs, but it's the most obvious application of that: if $ y\ge0$ than that cubic polynomial is $ \ge 9.$) It does have at least one negative real root. Does it have any other real roots? The derivative of $ y^3 \plus{} 2y^2 \plus{} 4y \plus{} 9$ is $ 3y^2 \plus{} 6y \plus{} 4 \equal{} 3(y \plus{} 1)^2 \plus{} 1$ which has no real roots, so by Rolle's Theorem, the original cubic has exactly one negative real root.

If you want to use Cardano-Tartaglia to write a formula for that root, go ahead. I tend to find that answers written in raw Cardano-Tartaglia form (as in your first post on this topic) are seldom much help in actually thinking about problems. So let's just say that there is a real number $ y_0$ such that $ y_0 < 0$ and $ y_0^4 \plus{} y_0 \minus{} 18 \equal{} 0.$

Our difficulty now is the usual convention surrounding the symbol $ \sqrt {x}.$ If $ y \equal{} y_0 < 0$ and $ x \equal{} y^2,$ then $ x$ is a positive real number. But then, by convention, $ \sqrt {x} > 0,$ meaning that $ \sqrt {x}\ne y_0$ and we have not found a solution to our original problem. The original equation has $ x \equal{} 4$ as its only real root, as long as we use the usual convention concerning $ \sqrt {x}.$

It is true that $ y^4 \plus{} y \minus{} 18$ has four complex roots: one positive, one negative, and two as a conjugate pair. The squares of these four numbers are in some sense roots of the original equation $ x^2 \plus{} \sqrt {x} \equal{} 18$ but only if you adopt a "multi-valued" interpretation of what you mean by the symbol $ \sqrt {x}.$

So is correct: 4 is the only root.

Sorry for the new topic, it was a mistake.


algebra
polynomial
superior algebra
superior algebra open

let $k$ be an algebraically closed field. a morphism $\varphi$ of $n$-dimensional affine space $k^n$ into itself is given by $n$ polynomials $f_1, ..., f_n$ in $n$ variables: $\varphi(x_1, ..., x_n) = (f_1(x_1,...,x_n), ..., f_n(x_1, ..., x_n))$. it is called an automorphism if there exists an inverse morphism $\psi$. let $J=\det|\frac{\partial f_i}{\partial x_j}|$ be the jacobian.
a) if $\varphi$ is an automorphism, show that $J$ is a nonzero constant polynomial.
b) show the converse of a).

a) is easy, but b) is still unsolved.

Peter

b) is also open in the case n=2 !
This problem is called the Jacobian conjecture, many false proofs exists in the litterature.

A few new results were published after 2004 and I thought of updating them(Also it is the 80th anniversary of the conjecture!)

Dixmier implies Jacobian Conjecture Here
http://journals.math.ac.vn/acta/pdf/0702205.pdf and here https://arxiv.org/abs/math/0512171
Moh's paper https://eudml.org/doc/152524 was counter-argued by his students here https://arxiv.org/abs/1604.07683v2 and here
https://arxiv.org/abs/1708.09367. Also Essen proposed a generalized version of JC as(*)
A paper on this is https://www.sciencedirect.com/science/article/pii/S0022404903002561
For 2-D Jc $(x,y)\mapsto(f(x,y),g(x,y))$ is thought to be a counter example(I don't know about this).Also I think that the conjecture was proved to be true in the symmetric an cubic linear case which used  nilpotency etc.I couldn't find a reference for this,so can anyone provide?
(*)[Quote=Essen]...it suffices to prove JC for all $n \geq 2$ and for all $F's$ of the form: $F=(l_1,…,l_r,x_{r+1}+M_{r+1},…,x_n+M_n)$ where each $l_i$ is linear and each $M_j$ is a monomial.[/quote]


linear algebra
matrix
linear algebra unsolved

In the ring $\mathbb{Z}_{n}$ of residues modulo $n$ calculate $\det A_{n},$ $\det B_{n}$ where $A_{n}=(\overline{i}+\overline{j})_{\overline{i},\overline{j}=0,1,\ldots,n-1},$ $B_{n}=(\overline{i}\cdot\overline{j})_{\overline{i},\overline{j}=0,1,\ldots,n-1},$ $n\ge2.$
(V. Mazorchuk)

I hope I didn't misunderstand the notation.
For $A_{n}$, its rank is 2 obviously, thus the det = 0 for $n \ge 3$, for n = 2 the det is -1 since the matrix is $\begin{array}{cc}0 & 1 \\ 1 & 0 \\ \end{array}$
For $B_{n}$, the first row is 0, so det is 0.


limit
real analysis
real analysis solved

Here my proof with Landau technic
g_n(x)=1+tgx+...+tg(nx) i use tgu=u+u^3/3 + o(u^3) when u->0
g_n(x)=1+\sum_{k= 1 to n}(kx+k^3x^3/3 + o(x^3)) when x->0
=1+\sum_{k=1 to n}kx + o(x)
now g_n(x)^(1/(n^3.x)=exp(ln(g_n(x))/(n^3x)=
exp((n(n+1)(2n+1)/n^3 o(1)) -->exp(1/3) when first x->0, and second
when n->+oo


Ross Mathematics Program
induction
combinatorics proposed
combinatorics

On a circular road there are some gas stations. Together, they have exactly as much gas as a car would need in order to go around the circle once. Prove that there is a gas station where you can start with an empty tank s.t. you can go around the circle.

I found it [url=http://www.cut-the-knot.org/proofs/GasStations.shtml]here[/url].

BTW it's Beijing 64.
Assume that car begins travel with the quantity of gas enough to go around the circle without supply. Take the station (or one of the stations) on which the quantity of gas is minimal. If we start from this station the quantity of gas always will be not less than the initial one (0).

Thanks! On that website the exact same solution is given. I think I have another one, but it might turn out to be the same :).

A long time ago Moubinool posted a problem. I can't remember what it sounded like, but it was similar to this: given some numbers around the circle s.t. their sum is $0$, prove that we can choose one s.t. all the partial sums (in a clockwise direction) are $\ge 0$. We can apply this here, considering the gas stations as positive numbers (because the car can get more fuel) and the gaps between them as negative numbers. My solution to this problem was, however, a bit longer.

Yes Grobber, it was given in our 'Concours Gnral' in 1997 (I think), and it was, as you, the first thing which came to my mind while reading the statement of the problem.

Pierre.

I've tried to locate that problem recently, but failed :?. Do you happen to know where it was posted?

This problem (the original one Grobber posted) is attributed to the Hungarian mathematician Laslo Lovas (spelling?) in David A. Klarner, [i]The Mathematical Gardner[/i], essay 5.4 by Ross Honsberger, with the same solution as Sasha gave.

  Darij

I've found that :
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=537&highlight=concours+general

[color=red][[b]Moderator edit:[/b] This problem is [i]not exactly[/i] what grobber uses, because it requires the sum of the numbers on the circle to be $1$ and the partial sums to be positive, whereas the version grobber uses requires the sum of the numbers on the circle to be $0$ and the partial sums to be nonnegative. However, its solution is more or less the same as the solution of the version used by grobber.][/color]

Pierre.

For 1 or 2 gas stations, it's trivial.
Let's assume that it's true for some n gas stations. Let's look at n+1 stations: there must be at least one station that has enough gas to arrive at the next station. If we look at this station and the station after it as [b]one[/b] station, then we have n stations, and by the assumption, we're done.
So by induction, it's true for every n.


Let $ ABC$ be an equilateral triangle and $ P$ in its interior. The distances from $ P$ to the triangle's sides are denoted by $ a^2, b^2,c^2$ respectively, where $ a,b,c>0$. Find the locus of the points $ P$ for which $ a,b,c$ can be the sides of a non-degenerate triangle.

Using areal (or trilinear - both the same here) co-ordinates we are looking for the a region on which $ (\sqrt{x}\plus{}\sqrt{y}\minus{}\sqrt{z})(\sqrt{x}\minus{}\sqrt{y}\minus{}\sqrt{z})(\minus{}\sqrt{x}\plus{}\sqrt{y}\plus{}\sqrt{z}) > 0$ and on which $ \sqrt{x}\plus{}\sqrt{y}\plus{}\sqrt{z}$ is strictly positive.

Multiplying it all together and factorising we get $ 2(xy\plus{}yz\plus{}zx)\minus{}(x^2\plus{}y^2\plus{}z^2)>0$.

It's easy to show that this is the interior of a conic which is a circle and passes through the three midpoints of the sides, and that it is therefore the interior of the incircle.

So this is our locus.

For those who are interested, here are the official solutions of the problems.


inequalities
Euler
inequalities unsolved

With positive real numbers $ x, y, z$, prove that: $ \sum\limits_{cyc} {\left[ {\sqrt[3]{{4(x^3  \plus{} y^3 )}} \plus{} \frac{{2x}}{{y^2 }}} \right]}  \ge 12$

[quote="L_Euler"]With positive real numbers $ x, y, z$, prove that: $ \sum\limits_{cyc} {\left[ {\sqrt [3]{{4(x^3 \plus{} y^3 )}} \plus{} \frac {{2x}}{{y^2 }}} \right]} \ge 12$[/quote]
By AM-GM we have $ \sum {\left( {\sqrt[3]{{4\left( {x^3  \plus{} y^3 } \right)}} \plus{} \frac{{2x}}{{y^2 }}} \right)}  \ge \sum {\left( {x \plus{} y} \right) \plus{} \frac{{2x}}{{y^2 }}}  \ge 12$.


calculus
integration
trigonometry
calculus computations

Evaluate $\int_{0}^{\pi}\frac{\cos 4x-\cos 4\alpha}{\cos x-\cos \alpha}\ dx.$


limit
trigonometry
algebra
polynomial
calculus
calculus computations

1.  Lim             (sinx - cosxsinx) / x^2
    x->0

2.  Lim                (tanx) / (x+(pie/4))
   x->(pie/4)

3.  Lim             ((2x^3) + 1) / ((5x^3) + (2x^2) + 1)
    x->inf.

Thank you all in advance!

I will omit $x\to 0$ in limits.
$\lim \frac{\sin x - \cos x \sin x}{x^2} = \lim \frac{\sin x (1-\cos x)}{x^2} = \lim \frac{\sin x}{x} \frac{2\sin^2 \frac{x}{2} }{x} = \lim \frac{\sin x}{x} \frac{\sin \frac{x}{2}}{\frac{x}{2}} \sin \frac{x}{2} = 1\cdot 1 \cdot 0 = 0$

\[\lim_{x \rightarrow \frac{\pi}{4}} \frac{\tan{x}}{x + \frac{\pi}{4}} =  \frac{\tan{\frac{\pi}{4}}}{\frac{\pi}{4} + \frac{\pi}{4}} = \frac{2}{\pi}.\]

So, you just plug in $\frac{\pi}{4}$...

The limit of the quotient of two polynomials (of equal degrees) is just the quotient of the coefficients of their highest degree terms :huh:

[quote="Riastrad"][quote="Arne"]The last one: \[\lim_{x \rightarrow \infty} \frac{2x^3 + 1}{5x^3 + 2x^2 + 1} = \frac{2}{5}.\][/quote]

How did you cancle out the x's?[/quote]

basically divide numerator and denominator by $x^3$


Point $ P$ is inside a rectangular box.  The distances from point $ P$ to four of the vertices of the box are 1,2,3, and 4.  If the distance from $ P$ to another vertex is greater than 5, find that distance.

No one attempted this problem? I admit that it is a rather difficult problem compare to first five.

[hide="Solution"]
This problem can be solved with [url=http://www.artofproblemsolving.com/Wiki/index.php/British_Flag_Theorem]British Flag Theorem[/url]. Basically, this theorem states that with plane $ ABCD$ and point $ P$ that is on the plane or just on the space, the equation $ PA^2\plus{}PC^2 \equal{} PB^2\plus{}PD^2$ holds true.

Knowing that, we know what to do... Casework.

Construct the rectangle such that one face has points $ ABCD$.  With respect to the center of the rectangle, construct $ A',B',C',D'$ so that they are opposite of $ A,B,C,D.$

WLOG, let $ PA \equal{} 1$.  By considering rectangle $ ABCD$, we can say that either $ PB \equal{} 2$ or $ PC \equal{} 2$.  Also, the key point is to note that $ PA'$ has to be maximized because $ PA \plus{} PA'$ is the diagonal of the rectangle, and thus, more than 5 (so that the length from $ P$ to one vertice can reach beyond 5).

Assume $ PC \equal{} 2.$ Then $ 3 \neq PA, PA', PB, PD, PC'$ (use the above reasoning to see why these points do not work) so $ PD' \equal{} 3$ or $ PB' \equal{} 3$. Assume then $ PD' \equal{} 3$ then $ 4 \neq PA, PA', PB, PD, PC$ so $ PC' \equal{} 4$ or $ PB' \equal{} 4$. In the first case, the diagonal becomes $ \sqrt{2^2 \plus{} 4^2}$ and that's less than 5. In latter case, the diagonal becomes 5 so this is still bit short.  Thus, $ PB \equal{} 2$.

With $ PB \equal{} 2,$ then $ 3 \neq PA, PA', PB'$.  So:

$ PC \equal{} 3, PC' \equal{} 3, PD \equal{} 3,$ or $ PD' \equal{} 3$

We can consider casework by first two and the last two instead of considering all four.

If $ PC \equal{} 3,$ then $ PD \equal{} \sqrt{6}$ from the rectangle $ ABCD$.  Also, $ 4 \neq PA, PA', PB', PD', PC'$ so nope, this is not the right case.

If $ PD \equal{} 3,$ then $ PC \equal{} \sqrt{12}$.  Then, $ 4 \neq PA, PA', PB', PD'$ which leaves $ PC' \equal{} 4$. So, the diagonal (since we know that $ PC$ and $ PC'$), becomes $ \sqrt{12 \plus{} 4^2} \equal{} \sqrt{28}$.

Now that we know the diagonal, we can find the other values.  $ PD \equal{} \sqrt{19}, PB' \equal{} \sqrt{24},$ and most importantly, $ PA' \equal{} \sqrt{27}$, which is the answer since it's over 5. [/hide]


[hide]the answer is C. the equation is equal to [1/(x+y)]*[(x+y)/xy] which simplifies to 1/xy[/hide]

Can you post where these problems come from?

I'm just asking this for curiosity. Also, even though it is fine to make separate posts, if you are posting as challenges, I would be more grateful if you post all problems in one thread.

Thanks!


number theory
prime numbers
inequalities unsolved
inequalities

Let $ k$ be a positive integer, we write k as a product of prime numbers (not necessarily distinct, par example 18=2.3.3). Let $ T(k)$ be the sum of all factors in the factorization above. Find the largest constant $ C$ such that $ T(k) \geq C.lnk$ for all positive integer k.

I haven't any solution for this nice problem! But i think that it's a very difficult problem and it's not easy to try to solve it! I'm trying and i'll post my proof while i have it!

Before I post my solution, I'd like to know the source of the problem (It's strange that it is still an open problem, because the solution I have is pretty straightforward).

Is it (maybe) a problem from a (running) contest? Hope not.

Hi [b]hsiljak[/b]:
I'll say of the source of this problem. It's a problem which my friend gave in 2006. But he haven't any solution for it!
Previous month, i meet him and we said of it during a long times. So i decide post it in our forum! :D

It is an easy problem . Try to solve it if you want to send it to math magazine in Viet Nam . You must know that shouldn't post it here . If you want to find solution [b]try to solve it or wait for 3 months[/b].


Three circles of radii x, 2x, and 3x are inscribed inside a circle of radius 12 and tangent to each of the other two inscribed circles. What is the value of x?

Connecting the radii, we have a triangle with sides $ 3x$,$ 4x$, and $ 5x$, with area $ 6x^2$ (since it is a right triangle). The lines through the points of tangency to the outer circle and the centers of the circle meet at the center. From here we get that the center of the big circle is distances of $ 12 \minus{} x$, $ 12 \minus{} 2x$, and $ 12 \minus{} 3x$ from the vertices. If we assume the center is inside the triangle, we get no solution. Thus, we assume the center is outside. By Heron's,
$ \sqrt {12(x)(2x)(12 \minus{} 3x)} \plus{} \sqrt {12(x)(3x)(12 \minus{} 4x)} \minus{} \sqrt {12(2x)(3x)(12 \minus{} 5x)} \equal{} 6x^2$
Solving, $ x \equal{} 2$. 0 is actually the minimum of $ \sqrt {24(12 \minus{} 3x)} \plus{} \sqrt {36(12 \minus{} 4x)} \minus{} \sqrt {72(12 \minus{} 5x)} \minus{} 6x$, so I think there is probably a nicer solution either using inequalities or the fact it is a right triangle.

The answer can also be found by [url=http://mathworld.wolfram.com/DescartesCircleTheorem.html]Descarte's Circle Theorem[/url]: $ \left(\frac{1}{x}\plus{}\frac{1}{2x}\plus{}\frac{1}{3x}\minus{}\frac{1}{12} \right)^2 \equal{} 2\left(\frac{1}{x^2}\plus{}\frac{1}{4x^2}\plus{}\frac{1}{9x^2}\plus{}\frac{1}{144} \right)$ $ \Longrightarrow \left(\frac{\minus{}x \plus{} 22}{12x}\right)^2 \equal{} \frac{x^2 \plus{} 196}{72x^2}$ $ \Longrightarrow  x^2 \plus{} 44x \minus{} 92 \equal{} (x\minus{}2)(x\plus{}46) \equal{} 0$, and $ x \equal{} 2$.

Or you can connect the center of the circle to the centers of the circles with radius $ 2$ and $ 3$.  You then get a rectangle, with opposite sides having lengths $ 3x$ and $ 12\minus{}3x$, so $ \boxed{x\equal{}2}$.


The roots of the equation $ x^{3}\minus{}19x\equal{}\minus{}30$ are a, b, and c. Find the numerical value of the product:
(a+1)(b+1)(c+1).

This is how I tackled the problem:

[hide]First, I set my goal as factoring out an (x-r) where x is a root from $ x^3\minus{}19x\plus{}30$. Knowing r must be equal to (some factor of 30)/(some factor of 1.. haha), I ran through the options in my head, using synthetic division and got a 2. I factored the whole thing to get (x-2)(x+5)(x-3), thus the factors were 2, 3, and -5. Plugging them in for a, b, and c, I got -48. [/hide]

Does anyone have a helpful way to solve cubic equations other than the technique I used? Thanks.

Well in tougher cases, you wont be able to find the roots. That's where Vieta's Formulas and Newton's Sums come in. Vieta's are the relationships between the coeffiecients of a polynomial and its roots. Newton's is..well something you won't need in this problem but I suggest you learn for future problems.

P.S. I don't want to type all of Vieta's formulas so please look them up before looking at my solution. Otherwise, you won't understand.

[hide="Look up Vietas!"]First expand $ (a \plus{} 1)(b \plus{} 1)(c \plus{} 1) \equal{} (a \plus{} b \plus{} c) \plus{} (ab \plus{} ac \plus{} bc) \plus{} abc \plus{} 1$. By Vieta's formulas,

$ a \plus{} b \plus{} c \equal{} 0$

$ ab \plus{} ac \plus{} bc \equal{} \minus{} 19$

$ abc \equal{} \minus{} 30$

Plugging in, we have $ (a \plus{} 1)(b \plus{} 1)(c \plus{} 1) \equal{} 0 \plus{} \minus{} 19 \plus{} \minus{} 30 \plus{} 1 \equal{} \boxed{ \minus{} 48}$.[/hide]

I think modularmark understated how important knowing these relationships are.  Viete's formulas crop up EVERYWHERE and can be used in a variety of situations  - even in some unsuspecting ones.

http://www.artofproblemsolving.com/Wiki/index.php/Vieta%27s_Formulas

http://www.artofproblemsolving.com/Wiki/index.php/Newton_sums

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

That should be enough for a basic understanding.

There is also a nice solution involving the Factor Theorem.

[hide="My Solution"]
We have $ (a \plus{} 1)(b \plus{} 1)(c \plus{} 1) \equal{} \minus{}(\minus{}1 \minus{} a)(\minus{}1 \minus{} b)(\minus{}1 \minus{} c)$. Notice that $ x^3 \minus{} 19x \plus{} 30 \equal{} (x \minus{} a)(x \minus{} b)(x \minus{} c)$ for all $ x$. Substituting $ x \equal{} \minus{}1$, we obtain $ (\minus{}1 \minus{} a)(\minus{}1 \minus{} b)(\minus{}1 \minus{} c) \equal{} (\minus{}1)^3 \minus{} 19(\minus{}1) \plus{} 30 \equal{} \minus{}1 \plus{} 19 \plus{} 30 \equal{} 48$. Now we take the negative of that to get the desired result, $ \boxed{\minus{}48}$.
[/hide]


How many different arrangements of the six letters of the word "YELLOW" can be made if the first letter must be W and the last letter must be L?

There are $ 360$ total ways to arrange.  After that, there is a $ \frac{1}{6}$ chance the first letter is W and $ \frac{2}{5}$ chance that the last letter is L.  $ 360 \times \frac{1}{6} \times \frac{2}{5}\equal{}\boxed{24}$.

Or, after picking a W and an L, we still have to arrange YELO in the middle.  There are $ 4!\equal{}\boxed{24}$ ways to do so


geometry
circumcircle
trapezoid
symmetry
geometric transformation
reflection
power of a point

In triangle $ ABC$ with $ m(\angle{C})\equal{}2\cdot m(\angle{A}),$ $ (CD$ is the internal bisector of $ \angle{C}, \  (D\in [AB]).$ Let $ S$ be the center of the circle located on the same side
of line $ AC$ as $ B,$ which is tangent to $ AC$ and externally tangent to the circumcircles of $ \triangle{ACD}$ and $ \triangle{BCD},$ respectively. Prove that $ \boxed{AB\perp CS.}$

I've seen the problem in Crux Mathematicorum, but I don't remember exactly the reference. Can someone help?
[hide="Main idea of the solution I have in mind"]Consider an inversion with respect to the vertex $ C$. It becomes more or less straightforward after that.[/hide]

[quote="FelixD"]What does $ m( \angle C)$ mean?[/quote]

It means the measure of angle $ C.$

Let $(S)$ be the circle tangent to $AC,$ $ (K_1) \equiv{} \odot(DAC)$ and $ (K_2) \equiv{} \odot(DBC)$ externally. Since $ \angle DAC \equal{} \angle BCD,$ then ray $ CB$ is tangent to $ K_1$ at $ C.$ Inversion with center $ C$ and power equal to the power of $ C$ WRT $ (S)$ takes $ (K_1)$ into the line $ k_1$ passing through the inverse $ A'$ and tangent to $ (S).$ Similarly, $ (K_2)$ is taken into the line $ k_2$ passing through $ B'$ and tangent to $(S).$ Further, $ k_1 \equiv{} \overline{A'D'}$ is parallel to $ CB',$ since $ CB$ is tangent to $ (K_1).$  $ k_1,$ $ k_2$ meet at the inverse $ D'$ of $ D$ and since $ A,D,B$ are collinear, then $AB$ is taken into the circumcircle $\omega$ of the isosceles trapezoid $ A'D'B'C.$ By conformity, angle between $ AB$ and $ AC$ equals the angle between $\omega$ and $ A'C.$  Thus, $ \angle D'CA' \equal{} \angle A'D'C$ $\Longrightarrow$ $ \triangle A'D'C$ and $ \triangle D'B'A'$ are isosceles with apices $ A'$ and  $D'.$ Thus, $S$ is the the center of $\omega$ $\Longrightarrow$ $AB \perp CS.$

Very nice problem.
Consider the composition $P$ of symmetry wrt to line $CD$ and inversion with center $C$ and square radius $CA*CB$ now let $CD$ cut circle $ABC$ again at $E$. Then $P(D)=E$ $P(A)=B$ so circles $ADC$ and $BDC$ go to $BE$ and $AE$. and line $CA$ goes to $CB$ so the circle which is externally tangent to circles $ADC$ and $BDC$ and line $AC$(circle $k$) goes to circle $m$ which is tangent to segments $BE,EA,BC$. However $\angle BCE=\angle ECA=\angle BAC$ so $BC=BE=EA$ and the circumcenter of $ABC$ point $O$ is equidstant from $BE,EA,BC$. So the center of $m$ is $O$. So now $P(S)$ is a point on $CO$ and therefore $\angle SCA=\angle BCO=90-\angle CAB$ so $CS\perp AB$.

[size=150]Remark:[/size] [b]The circle $S$ touches $AC$ at the symmetrical of $C$ about $A$![/b]

Best regards,
sunken rock

Here is another approach that also proves sunken rock's remark.

Let $P,Q,R$ be the tangency points of $(S)$ with $AC,\odot(DAC),\odot(BDC),$ respectively. $D$ is clearly midpoint of the arc $AQC$ and if $\odot(BCD)$ cuts $AC$ again at $E,$ then $B$ is midpoint of the arc $CRE.$ Inversion WRT $(D,DC)$ swaps $AC$ and $\odot(DAC)$ leaving $(S)$ fixed and inversion WRT $(B,BC)$ swaps $AC$ and $\odot(BCE)$ leaving $(S)$ fixed $\Longrightarrow$ $P \in DQ,$ $P \in BR$ and $DC^2=DP \cdot DQ,$  $BC^2=BP \cdot BR$ $\Longrightarrow$ $B,D$ have equal powers WRT $C$ and $(S)$ $\Longrightarrow$ $BDA$ is radical axis of $C$ and $(S)$ $\Longrightarrow$ $CS \perp AB$ and $AB$ bisects the tangent from $C$ to $(S),$ i.e. $P$ is reflection of $C$ on $A.$

[quote=pohoatza][color=#f00]I've seen the problem in Crux Mathematicorum[/color], but I don't remember exactly the reference. Can someone help?
[hide="Main idea of the solution I have in mind"]Consider an inversion with respect to the vertex $ C$. It becomes more or less straightforward after that.[/hide][/quote]

v21 n01. Problem 2010.


The owner's manual for a string trimmer says that a 16 : 1 ratio of gasoline to 2-cycle oil MUST BE used for its fuel. How many gallons of gasoline would be needed to go with an 8 ounce packet of oil, if 1 gallon is 128 ounces?

[hide] Say the amount of gasoline needed is $x$ [b]ounces[/b].  Therefore, $\frac{x}{8}=\frac{16}{1}$.  Mulitplying 8 on both sides, we get that that $x=128$ ounces.  Therefore, we have that x is $\frac{128}{128}=1$ gallon.  Therefore, the answer is $1$.  [/hide]

Good job but I also like to see how the question was actually solved.

Thank you The QuattoMaster 6000!


search
superior algebra
superior algebra unsolved

sir

    I want the solution of the problem.. If G is a finite group.Show that the number of elements x of G with x^2 not equal to 1 is even.Conclude that if G is a finite group of even order then G has an element of order 2.....

Please post a new problem into a new thread every time.

[quote="coquitao"]Take a look at the following thread:

[url]http://www.mathlinks.ro/viewtopic.php?search_id=875870790&t=225332[/url]

I'm sure that it is going to prove useful to you.

Best of luck...  :wink:[/quote]

Given that G is finite
  Let no.of elements of G be 'm'
  Let A= no.of elements that x^2 is not equal to 1
       B= no.of elements that x^2 is equal to 1
Therefore G=A union B
Since B is even, then G is even if A is even otherwise G is odd if A is odd[/hide]

Instead of

A = no.of elements that x^2 is not equal to 1

write

A = elements that x^2 is not equal to 1.

Same for B.

Then, G = A union B would actually hold (the equality wasn't correct as you'd written it before.)

The conclusion of the argument is now as you said:

1- |A| is always even. No matter what the order of G is...
2- Now, if the order of G is even, then |B| would have to be even because |G| = |A| + |B|.
3- Since the identity of G belongs to B, we conclude that |B|>=2. The conclusion follows now from this inequality.

Hope it helps...

[quote="barathisid"] Since B is even, then G is even if A is even otherwise G is odd if A is odd[/quote]

Contrast this line with my previous post. We can't tell whether B is even or not... The one that is always even is A, not B.


Find all $ f: \mathbb{R}^{\plus{}}\rightarrow \mathbb{R}$ such that
(i) $ f(1)\equal{}2008$
(ii) $ |f(x)|\leq x^2\plus{}1004^2$ for all $ x>0$
(iii) $ f(x\plus{}y\plus{}\frac{1}{x}\plus{}\frac{1}{y})\equal{}f(x\plus{}\frac{1}{y})\plus{}f(y\plus{}\frac{1}{x})$

Let $ z_1\equal{}x\plus{}\frac 1y,z_2\equal{}y\plus{}\frac 1x$. If $ z_1z_2\ge 2$ exist real solution for x,y. Therefore
$ f(z_1\plus{}z_2)\equal{}f(z_1)\plus{}f(z_2) \ \forall z_1z_2\ge 2$.
from (ii) we get $ f(x)\equal{}ax$,from (i) a=2008.

[quote="outback"][quote="Rust"] If $ z_1z_2\ge 2$ exist real solution for x,y. [/quote]

Can you show?[/quote]
$ z_2 \equal{} y \plus{} \frac 1x \equal{} y \plus{} \frac {1}{z_1 \minus{} \frac {1}{y}} \equal{} \frac {z_1y^2}{z_1y \minus{} 1}\to y \equal{} \frac {z_1z_2\pm\sqrt {z_1z_2(z_1z_2 \minus{} 4)}}{2z_1}$.
It is true for $ z_1z_2\ge 4$.

And so you can conclude $ f(x)\equal{}ax$ for which $ x$? (Not for all positive real $ x$, I think...  :wink: )

It is easy, but long to write.
1)Let $ z_1>0,z_2>0$. We can chose $ z_3$, suth that $ z_1z_3\ge 4,z_2z_3\ge 4$.
Then $ f(z_2\plus{}z_3)\equal{}f(z_2)\plus{}f(z_3)$.
$ f(z_1\plus{}z_2)\equal{}f(z_1\plus{}z_2\plus{}z_3)\minus{}f(z_3)\equal{}f(z_1)\plus{}f(z_2\plus{}z_3)\minus{}f(z_3)\equal{}f(z_1)\plus{}f(z_2)\plus{}f(z_3)\minus{}f(z_3)\equal{}f(z_1)\plus{}f(z_2) \ \forall z_1,z_2>0$.
2)We can define $ f(\minus{}z)\equal{}\minus{}f(z),f(0)\equal{}0$, then $ f(x\plus{}y)\equal{}f(x)\plus{}f(y) \ \forall x,y\in R$.
3) If $ \frac{f(x)}{x}\not \equal{}const$, then exist $ f(x_1)\equal{}a_1x_1,f(x_2)\equal{}a_2x_2,x_1>0,x_2>0$, Let $ m_n\equal{}[\frac{nx_1}{x_2}]$, then
$ 0\le nx_1\minus{}m_nx_2<x_2, f(nx_1\minus{}m_nx_2)\equal{}na_1x_1\minus{}m_na_2x_2\equal{}n(a_1\minus{}a_2)x_1\plus{}a_2(nx_1\minus{}m_nx_2)\to \infty$, when $ n\to \infty$ contradition with ii).


[quote="thomaskurian89"]Prove that there are infinitely many positive integers $ n$ such that $ n(n \plus{} 1)$ can be expressed as a sum of two positive squares in [i]at least[/i] two different ways. (Here $ a^2 \plus{} b^2$ and $ b^2 \plus{} a^2$ are considered as the same representation.)[/quote]
Let $ n \equal{} x^2 \plus{} y^2 \equal{} k^2$ (There exists infinitly many pythagorean triples)
Then $ n(n\plus{}1) \equal{} (x^2\plus{}y^2)(k^2\plus{}1) \equal{} (xk \minus{} y)^2 \plus{} (x \plus{} yk)^2 \equal{} (xk \plus{} y)^2 \plus{} (x \minus{} yk)^2$ QED


[b]1.) On the picture provided, write $A^3$ as a function of $a$.
2.) On the picture provided, write $A^n$ as a function of $a$ and $n$, where $n$ is a positive integer.[/b]

If you're talking about matrix multiplication here's a solution.
Let $a_{ij}$denote the entry in the ith row and jth column
For $A^n$, $a_{11}=\frac{1}{2^n}$, $a_{12}=0$, $a_{21}=\frac{a}{2^{n-1}}+a^n$, and $a_{22}=a^n$.

drunner, are you sure?
AT n=1 you'd get $a_{11}=1/2$ $a_{12}=0$ $a_{21}=2a$ $a_{22}=a$  THe problem tells us that n=1 is $a_{11}=1/2$ $a_{12}=0$ $a_{21}=a$ $a_{22}=a$
[hide]The only square that's not correct on drunner's answer is $a_{21}$.  I think it's $a^n+\frac{a^{n-1}}{2}+...+\frac{a}{2^{n-1}}$.  THere's 2 rules to this function.  FIrst, only positive numbers may be taken to the $a$ power.  SEcond, the terms are grouped by power of a (from largest to smallest) and there's only one term per power of a.  If there is such a conflict, the second term becomes undefined.  Any value that becomes undefined because of these rules are then set at 0.
Now let's show this inductively.
BAse case: n=1 creates $a+1/2+a^{-1}/4+...+a$.  Using the rules, every term except the first a equals 0.
Now assume it true for a given n.  To prove it for n+1, use matrix multiplication.  Here, we are only going to multiply the relevant rows that will create $a_{21}$ of $A^{n+1}$
$1/2(a^n+\frac{a^{n-1}}{2}+...+\frac{a}{2^{n-1}})+a^{n+1}=a^{n+1}+a^n/2+\frac{a^{n-1}}{4}+...+\frac{a}{2^n}$.  If you plugged n+1 into the original equation you also get this.  [/hide]

I really trying to get the idea. Can you please tell me how to get the answer on letter a? On how to write $A^3$ as a function of $a$? I searching for this topic on the web. Thank you very much!

willydavidjr, if the formula for $A^n$ has been found, then the first part about finding $A^3$ is just substituting 3 for n.  That's $a_{11}=1/8$, $a_{12}=0$ $a_{21}=1/4a+1/2a^2+a^3$, $a_{22}=a^3$ for $A^3$.

Dengmi can you help me how to write the formula for the function $A^n$?
I only know how to write the function for $A^3$ by multiplying the matrix three times. But for the variable n, well its quite tough for my part. Please help how did you get the answer.

Delta the definition was quite tough on the site you have given. But I really appreciate that. What I am up to is how to get the value for $A^n$ on any matrices. I am looking for the technique. Please help, thanks a lot!


calculus
real analysis
real analysis unsolved

[quote="anik_andrew"]Prove $1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n} = \sum_{j=0}^{n-1} \frac{C_{n}^{j}}{n-j}(-1)^{n-j+1} =\sum\limits_{j=1}^{n}\frac{(-1)^{j+1}}{j}{n\choose j}=: (-1)^{n-1}L_n(2)$[/quote]
[color=blue][b]Nice ! Only few remarks : [/b][/color] for $p\in {\mathbb N}^*$ consider the numbers $\left(L_n(p)\right)_{n=1}^{\infty}$ where  $\begin{array}{|c|}\hline \displaystyle L_n(p)=\displaystyle \sum\limits_{k=1}^{n}\frac{(-1)^{n-k}}{k^{p-1}}{n\choose k}\\ \hline \end{array}\; .$
Further denote the $n-th$ harmonic number by   $H_n : =\displaystyle \sum\limits_{k=1}^{n}\frac{1}{k} .$
Please verify following equalities between these numbers :
\[ (1)\; \; \; \; \; \; \left\{\begin{array}{lcl} L_n(1)&=& \displaystyle (-1)^{n-1}\\ L_n(2)& =&\displaystyle (-1)^{n-1}H_n \\ L_n(3)&=& \displaystyle (-1)^{n-1}\sum\limits_{k=1}^{n}\frac{1}{k}H_k \\ & \vdots & \end{array}\right.  \]
\[ (2)\; \; \; \; \; \; \; \; L_n(p+1)=\sum\limits_{k=1}^{n}\frac{(-1)^{n-k}}{k}L_k(p)  \]
\[ (3)\; \; \; \; L_n(p+1)=\frac{L_n(p)-L_{n-1}(p+1)}{n}\; .  \]
[color=blue] [b] Methods  to prove  above formulas [/b][/color][size=75] (I hope that there are not errors)[/size]  :
[b]
---Calculus of Finite Differences,
---Interpolation,
---Divided differences,
---Theory of determinants (Vandermonde  type) . [/b]


calculus
integration
calculus computations


Let $ a,\ b,\ c$ be positive numbers such that $ \sqrt {\frac {a}{b}} \plus{} \sqrt {\frac {b}{c}} \plus{} \sqrt {\frac {c}{a}} \equal{} 3$.

Prove that $ \ \frac {b^6}{a^6} \plus{} \frac {c^6}{b^6} \plus{} \frac {a^6}{c^6}\geq 3$.

We do not need the condition, just direct 3-variable AM-GM.

Cheers,

Rofler

I do know that, I dare to post it.

I made the problem from $ x^3\geq 3x\minus{}2\ (x>0)$.

By  the way, the given condition is a little subtle...
$ \sqrt {\frac {a}{b}} \plus{} \sqrt {\frac {b}{c}} \plus{} \sqrt {\frac {c}{a}} \equal{} 3$.
Because from AM-GM we easily have $ \sqrt {\frac {a}{b}} \plus{} \sqrt {\frac {b}{c}} \plus{} \sqrt {\frac {c}{a}}\geq 3$
So the condition actually is the equality case of AM-GM. So all of its variables $ \frac ab; \frac bc;\frac ca$ are equal. (and, equal to $ 1$).
Hence $ \frac {b^{6}}{a^{6}} \plus{} \frac {c^{6}}{b^{6}} \plus{} \frac {a^{6}}{c^{6}} \color{red}{ \equal{} } 3$.
Please correct me if I am wrong. :)


If in a triangle $ a \equal{} 5, b \equal{} 4$ and cos $ (A\minus{}B) \equal{}\frac{31}{32}$, prove that cos $ C \equal{} \frac{1}{8}$ and that $ C \equal{} 6$.

Hi!  This is my first post!  I hope I use these buttons right.

Suggestion
[hide] Draw out the figure and tackle the problem using geometry.  You know angle A is bigger than angle B because length A is bigger than length B.  Draw a line from vertex A to line BC with one angle being A-B and the other angle being B.  Try to create two equations from this figure and solve for C. [/hide]

Thanks winternights. I got there after this inspired advice.


Let $a;b;c$ be positive real numbers satisfying $a^{2}+b^{2}+c^{2}=3$.Find $k$ max such that:
$ab^{2}+bc^{2}+ca^{2}\geq k(ab+bc+ca)^{2}$


function
algebra
rational function
algebra unsolved

Solve the following system:

$ a \plus{} b \minus{} c \minus{} d \equal{} 1$
$ a^2 \plus{} b^2 \minus{} c^2 \minus{} d^2 \equal{} 3$
$ a_3 \plus{} b^3 \minus{} c^3 \minus{} d^3 \equal{} \minus{}5$
$ a^4 \plus{} b^4 \minus{} c^4 \minus{} d^4 \equal{} 15$
$ a > 0$ (I don't know if this information is important)

With Maple.
In any commutative ring of characteristic $ 0$, the system is equivalent to this one:
$ b^2\plus{}b\minus{}2\equal{}0,d^2\plus{}2d\plus{}1\equal{}0,a\equal{}\minus{}1\minus{}b,c\equal{}\minus{}2\minus{}d$.

More generally, we have a look on the system:
$ (S): a\plus{}b\minus{}c\minus{}d\equal{}\alpha,a^2\plus{}b^2\minus{}c^2\minus{}d^2\equal{}\beta,a^3\plus{}b^3\minus{}c^3\minus{}d^3\equal{}\gamma,a^4\plus{}b^4\minus{}c^4\minus{}d^4\equal{}\delta$.
Let $ a\plus{}b\equal{}u_1,ab\equal{}u_2,c\plus{}d\equal{}v_1,cd\equal{}v_2$. The system $ (S)$ can be rewritten as a system in $ u_1,u_2,v_1,v_2$ because $ a^2\plus{}b^2\equal{}u_1^2\minus{}2u_2,a^3\plus{}b^3\equal{}u_1^3\minus{}3u_1u_2,a^4\plus{}b^4\equal{}u_1^4\minus{}4u_1^2u_2\plus{}2u_2^2$.
If $ \alpha^4\minus{}4\alpha\gamma\plus{}3\beta^2\not\equal{}0$ and if you are a good calculator then you obtain an unique value of $ u_1,u_2,v_1,v_2$ as rational function of $ \alpha,\beta,\gamma,\delta$.

hello, we have $ a\equal{}\minus{}2,b\equal{}1,c\equal{}\minus{}1,d\equal{}\minus{}1$ or $ a\equal{}1,b\equal{}\minus{}2,c\equal{}\minus{}1,d\equal{}\minus{}1$ as the solutions.
Sonnhard.


[quote="MCrawford"]How many square inches are in the area of the largest circle which can be drawn on a piece of paper measuring 8 inches by 11 inches?[/quote]

[hide="Hint"]
First draw the largest square you can in that rectangle, then draw the largest circle you can draw in the [b]square[/b]. That will be the largest circle you can draw in the rectangle.
[/hide]

[hide]8 is the diameter so the area is 16pi[/hide]

[hide]
The largest circle possible is a circle with a diameter of 8 inches.
$4^2\cdot\pi=\boxed{16\pi}$ square inches.[/hide]


limit
trigonometry
real analysis
real analysis unsolved

[quote="kunny"]$a_n={\frac{1}{\displaystyle\cos \left[\frac{\pi}{4}\left(1-\frac{1}{2^n}\right)\right]}}\ (n\geq 0),$ right?[/quote]
Your solution isn't correct. Note that:
$\lim_{n\to +\infty}\frac{1}{\displaystyle\cos \left[\frac{\pi}{4}\left(1-\frac{1}{2^n}\right)\right]}=\sqrt{2}$

This is right. Now find a sequence $\left\{\theta_n\right\}\subseteq (-\frac{\pi}{2},\frac{\pi}{2})$ such that $\sin(\theta_{n+1}-\theta_n)=\cos{\theta_{n+1}}$.

$\theta _{n+1}=\frac{1}{2}\theta _n+\frac{\pi}{4},$ right? :roll:

Right. For $\theta_0=0$ you find $\theta_n=\frac{\pi}{2}(1-\frac{1}{2^n})$, so
$a_n=\frac{1}{\displaystyle \cos \frac{\pi}{2} (1-\frac{1}{2^n})}$.
Now compute $\lim_{n\to +\infty} \frac{2^n}{a_n}$.

Oops! I was wrong in calculation of finding the root of the characteristic equation, $\alpha =\frac{1}{2}\alpha +\frac{\pi}{4} \Longleftrightarrow \alpha =\frac{\pi}{2}.$ anyway $\lim_{n\to\infty} \frac{2^n}{a_n}=\lim_{n\to\infty} \frac{\displaystyle \sin \frac{\pi}{2^{n+1}}}{\displaystyle \frac{\pi}{2^{n+1}}}\cdot \frac{\pi}{2}=\frac{\pi}{2}.$ :lol:

[quote="kunny"]Hum, regrettable!

Let $a_0+a_1+\cdots +a_{n-1}=\tan \theta _n\ (n\geq 1),$ we have $\sin (\theta _{n+1}-\theta _n)=\cos \theta _{n+1}.$ I wonder this appoach is wrong.[/quote]

kunny , I dont understand why can we make such substitution ? It seems so weird yet nice  :)

[quote="kunny"]We can imagine the formula $1+\tan ^ 2 \theta =\frac{1}{\cos ^ 2 \theta}.$[/quote]

ok thanks Kunny,but how you apply that to your question ?

i try to let $a_n=2\tan^2\theta_{n}$ but doesnt seem to work :(

For Tokyo University's problem, we should find another substitution.

[quote="kunny"]Here is the problem from Tokyo University entrance exam 1975. I have omitted hint.

Solve the following reccurence.

\[ a_1=\sqrt{2},\ a_{n+1}=\sqrt{2+a_n}\ (n=1,2,\cdots).  \][/quote]

Put $a_n=b_n+\frac{1}{b_n}$, then $a_n=a_{n+1}^2-2$ implies $b_n+\frac{1}{b_n}=b_{n+1}^2+\frac{1}{b_{n+1}^2}$ i.e. $b_{n+1}=\sqrt{b_n}$; if $b_1$ satisfied $\sqrt{2}=b_1+\frac{1}{b_1}$, then $b_n=b_1^{2^{-n}}$.


quadratics
Vieta
algebra
pen
ratio
number theory
relatively prime

Show that if $x, y, z$ are positive integers, then $(xy+1)(yz+1)(zx+1)$ is a perfect square if and only if $xy+1$, $yz+1$, $zx+1$ are all perfect squares.

Method of infinite descent (fall)-
assume $x\leqq\leq z$ are integers satisfying minimum sum such that any of $xy+1,yz+1,zx+1$ isn't perfect square.
Consideration regarding number $t=x+y+z+2xyz\pm
2\sqrt{(xy+1)(yz+1)(zx+1)}$ (integer solutions of quadratic equation $t^2+x^2+y^2+z^2-2(xy+yz+zt+tx+zx+ty)-4xyzt-4=0$):
We get the following  identities
 $(x+y-z-t)^2=4(xy+1)(zt+1)$,
$(x+z-y-t)^2=4(xz+1)(yt+1)$,
$(x+t-y-z)^2=4(xt+1)(yz+1)$.
From this system observe that not all $xt+1,yt+1,zt+1$ are the squares, while $(xt+1)(yt+1)(zt+1)$ is.
Now any of products $(xy+1)(yt+1)(tx+1)$ is square.
Let's prove that by this we got the least solution $(x,y,t)$:
Since from the system is $zt+1\geq 0$,$t\geq -\frac1z$,  for $z>1$ must be $t\geq 0$.
It is easy to  check that both cases $t=0$ and $t=1$ are impossible. Therefore all numbers $(x,y,t)$ are positive integers.
But,according to assumption about minimality it must be $t\leq
z$, and from Vieta's  formulae we get $tt_1<z^2$ ($t_1$ is second solution of the quadratics).Contradiction.
$\blacksquare$

[quote="Anonymous"]We get the following  identities
 $ (x \plus{} y \minus{} z \minus{} t)^2 \equal{} 4(xy \plus{} 1)(zt \plus{} 1)$,
$ (x \plus{} z \minus{} y \minus{} t)^2 \equal{} 4(xz \plus{} 1)(yt \plus{} 1)$,
$ (x \plus{} t \minus{} y \minus{} z)^2 \equal{} 4(xt \plus{} 1)(yz \plus{} 1)$.[/quote]

could someone explain how these are found from what is said above?

thank you for helping me to understand...

The identities follow from the assumption that $ t$ is a root of $ t^2\plus{}x^2\plus{}y^2\plus{}z^2\minus{}2(xy\plus{}yz\plus{}zt\plus{}tx\plus{}zx\plus{}ty)\minus{}4xyzt\minus{}4\equal{}0$

We shall prove that xy + 1, yz + 1, zy + 1 are pairwise relatively prime. Suppose there is a prime p1 such that p1 divides both xy+1 and yz+1. then p1 also divides z(xy+1) - y(zx+1) = z-x. Analogously if there are primes p2 and p3 such that p2 divides yz + 1, zy + 1 and p3 divides xy + 1,zy + 1 then p2 divides z-y and p3 divides x-y. Then there are numbers f1, f2, f3, such that p1f1= z-x; p2f2= z-y; p3f3=x-y. This gives us that p2f2 - p1f1= p3f3 and p3f3 - p2f2 = p1f1. Then p1f1 + p2f2 = p2f2 - p1f1 ==> p1f1=-p1f1 <==> p1f1=0. Contradiction. So these numbers are pairwise relatively prime hence result.

You have only shown that one pair is relatively prime.
You assumed that every pair has a common prime factor, but what you really should assume is: one pair has a common prime factor.

I all of a sudden don't understand my own solution anymore...

Can anyone tell me why $ z^2 \minus{} x(2z \minus{} x) \minus{} y(2z \minus{} y) < z^2$ in my last line?

Maybe I'm just tired right now...

Zetax: in fact I proved it for a single pair of terms, but since the expressions xy + 1, yz + 1, xz + 1, are symmetric i could assume for a single pair without lost of generality. Also, the primes p1, p2, p3, i didnt say they were equal.

maybe i cant see your point: can you explain?

You assumed that every pair out of $ xy\plus{}1,yz\plus{}1,zx\plus{}1$ has a common factor and got a contradiction. Thus you just proved that not _all_ pairs have a common factor, one or two pairs may still have.
By the way, you also made a sign-mistake, p3f3 - p2f2 = p1f1 should in fact be p3f3 - p2f2 = -p1f1. This destroys your contradiction.

[quote="Peter"]
Can anyone tell me why $ z^2 \minus{} x(2z \minus{} x) \minus{} y(2z \minus{} y) < z^2$ in my last line?

Maybe I'm just tired right now...[/quote]

No, it's not because of your tiredness. The above solution is just not correct.. yet:

You shouldn't suppose wlog that $ xy \plus{} 1$ is not a perfect square, but that $ z\ge x,y$
And you actually get $ (xt \plus{} 1)(yt \plus{} 1)(xy \plus{} 1)$ being a square by multiplying $ 4(xz \plus{} 1)(yt \plus{} 1)$, $ 4(xt \plus{} 1)(yz \plus{} 1)$ and $ (xy \plus{} 1)2$ and then dividing by $ 16(xy \plus{} 1)(xz \plus{} 1)(yz \plus{} 1)$, which is known to be a perfect square.

You also know that $ yt \plus{} 1$ is a perfect square iff $ xz \plus{} 1$ is, and the same with $ xt \plus{} 1$ and $ yz \plus{} 1$ (and that's why you don't need xy+1 to be a perfect square).
The $ t \equal{} 0$ case also implies $ xz \plus{} 1$ and $ yz \plus{} 1$ being a square, so you still have a contradiction there.
And with all this, the answer to your question also becomes trivial.

everyone knows that A.1 is the hardest of PEN

A non-empty set of positive integers is defined "good" whenever the product of any two distinct elements is smaller of some squares by $1$: with this convention $\{x,y,z\}$ is good. Let's start proving that if $\{x,y,z\}$ is good, then $\{x,y,z,w\}$ is good too, whenever
\[w=x+y+z+2xyz\pm 2\sqrt{(xy+1)(yz+1)(zx+1)}\]
is a positive integer. The two values of $w$ are roots of the following equivalent quadratic equations:
\[(x+y-z-w)^2=4(xy+1)(zw+1),\text{ }(1) \\
(x-y+z-w)^2=4(xz+1)(yw+1),\text{ }(2) \\
(x-y-z+w)^2=4(yz+1)(xw+1).\text{ }(3)\]
And that's clear that $\alpha w+1$ is a integer that can be expressed as ratio of two squares, i.e. it's a square too, for all $\alpha \in \{x,y,z\}$. Since the problem is symmetric in $x,y,z$ it can be assumed without loss of generality that $x\le y\le z$. Let's suppose for the sake of contradiction that not all sets $\{x,y,z\}$ are good, i.e. there exists a set such that at least one between $xy+1,yz+1,zx+1$ is not a perfect square. Choose now such a set $\{x,y,z\}$ such that $x+y+z$ is minimal, and define $w$ using the negative square root part. It's claimed that $0<w<z$ and $xy+1,yw+1,wx+1$ are not all perfect squares, but their product it is. This will contradict the minimality of $x+y+z$, showing that $\{x,y,z\}$ has to be good by force. Then, multiplying (1), (2) and (3) we obtain that $(xy+1)(yw+1)(wx+1)$ is a perfect square. Moreover, e.g. $yz+1$ is a perfect square if and only if $xw+1$ is a square too, so not all $xy+1,yw+1,wx+1$ can be perfect squares. We are missed to show only thavt $0<w<z$. Let's rewrite (1) as
\[zw+1=\frac{(x+y-z-w)^2}{4(xy+1)}\ge 0 \implies w \ge -z^{-1}.\]
If $z=1$ then we are forced to $x=y=z=1$, but $(xy+1)(yz+1)(zx+1)$ would not be a square. That means that $w\ge 0$, as far as it is a integer. But if $w=0$ then looking at (1), (2) and (3) each term in $xy+1,yz+1,zx+1$ would be a perfect square, against our assumptions. That means that $w\ge 1$. Hence, proving that $w<z$ will be enough to reach our contradiction. Define $w^\prime$ the other (greater) root of (1); then
\[w w^\prime=x^2+y^2+z^2-2(xy+yz+zx)-4<z^2-x(2y-x)-y(2z-y)<z^2,\]
implying that $w<z$, as far as $0<w<w^\prime$, and it completes the proof.

How does one think of a solution like this?? In particular, the value for $t$, the infinite descent, etc.
Thanks!

Proving "if (xy+1)(yz + 1)(zx + 1) is a perfect square then xy + 1, yz + 1, and zx + 1 are all perfect squares".

I am trying Proof by Contradiction, so...

Assume ~(xy + 1, yz + 1, and zx + 1 are all perfect squares).

     Question: is the negative of this statement equivalent to "at least one is not a perfect square" in which case I would have to go through cases, looking for a contradiction in each case; or, is it enough to assume that exactly one is not a perfect square and then find the contradiction?

Thanks for the help.

I have completed a possible proof of this theorem - it is attached. I do have one weak spot, possibly, it is clearly labeled in the proof.

Please check to see if my logic holds.

I look forward to your responses.

[quote=davisv]I've updated my proof. The new draft is attached.[/quote]
Where is the case when all are not perfect squares?

[quote=equationcrunchor][quote=Anonymous]
Consideration regarding number $t=x+y+z+2xyz\pm
2\sqrt{(xy+1)(yz+1)(zx+1)}$ (integer solutions of quadratic equation $t^2+x^2+y^2+z^2-2(xy+yz+zt+tx+zx+ty)-4xyzt-4=0$):
[/quote]

Can someone explain to me how this person created a quadratic equation with the correct roots to solve the problem?[/quote]

pell equality

[quote=Peter]Ok then, I'll rewrite it more detailed.

The if part is trivial. For the only if part, assume $ x,y,z$ are such integers with $ xy \plus{} 1, yz \plus{} 1, zx \plus{} 1$ not all perfect squares, moreover pick them such that $ x \plus{} y \plus{} z$ is minimal. Without loss of generality $ xy \plus{} 1$ isn't a perfect square.

Now let $ t$ be the smallest positive root of $ t^2 \plus{} x^2 \plus{} y^2 \plus{} z^2 \minus{} 2(xy \plus{} yz \plus{} zt \plus{} tx \plus{} zx \plus{} ty) \minus{} 4xyzt \minus{} 4 \equal{} 0$, an equation which rewrites equivalently as
\[ \\
(x \plus{} y \minus{} z \minus{} t)^2 \equal{} 4(xy \plus{} 1)(zt \plus{} 1), \\
(x \plus{} z \minus{} y \minus{} t)^2 \equal{} 4(xz \plus{} 1)(yt \plus{} 1), \\
(x \plus{} t \minus{} y \minus{} z)^2 \equal{} 4(xt \plus{} 1)(yz \plus{} 1).
\]
But our quadratic equation has roots $ t \equal{} x \plus{} y \plus{} z \plus{} 2xyz\pm 2\sqrt {(xy \plus{} 1)(yz \plus{} 1)(zx \plus{} 1)}$, which is given to be integer, so the right hand side is a square each time. But then the left hand side needs to be a square, which gives that $ (xt \plus{} 1)(yt \plus{} 1)(xy \plus{} 1)$ is a perfect square.
Now, since the centered equations above tell us that $ xt \plus{} 1\ge0,yt \plus{} 1\ge0,zt \plus{} 1\ge0$, we have $ t\ge \minus{} \frac1{\max(x,y,z)} > \minus{} 1$ (since $ x \equal{} y \equal{} z \equal{} 1$ isn't a solution).
If $ t \equal{} 0$, then working out the formula for $ t$ gives $ (x \plus{} y \plus{} z)^2 \equal{} 4(xy \plus{} yz \plus{} zx \plus{} 1)$ or $ (x \plus{} y \minus{} z)^2 \equal{} 4(xy \plus{} 1)$, which means $ xy \plus{} 1$ is a square, contradiction, so we have $ t > 0$ and by the minimality of $ x \plus{} y \plus{} z$ we must have $ t\ge z$.
But for the two roots $ t,t'$ of our quadratic equation we have by de Viete that $ tt' \equal{} x^2 \plus{} y^2 \plus{} z^2 \minus{} 2xy \minus{} 2yz \minus{} 2zx \minus{} 4 < z^2 \minus{} x(2z \minus{} x) \minus{} y(2z \minus{} y) < z^2$, contradiction.

keywords: vieta[/quote]

I did not understand why you checked $t=0$ case. You have already said $t$ is the smallest positive root of equation $ t^2 \plus{} x^2 \plus{} y^2 \plus{} z^2 \minus{} 2(xy \plus{} yz \plus{} zt \plus{} tx \plus{} zx \plus{} ty) \minus{} 4xyzt \minus{} 4 \equal{} 0$

Claim 1: If $\prod_{cyc} (xy+1) = (xyz+\frac{x+y+z}{2})^2$, I am done.

If I expand, I get $(\sum_{cyc} xy) + 1 = \frac 12 \sum_{cyc} xy + \frac 14 \sum_{cyc} x^2$

$\sum_{cyc} x^2 - 2\sum_{cyc} xy + 4=0$

As a quadratic in $x$: $x^2-(2y+2z)x+((y-z)^2-4)=0$. Its discriminant is $4((y+z)^2 - (y-z)^2 +4) =16(yz+1)$ is a perfect square, as needed.

Now suppose $\prod_{cyc} (xy+1) = (xyz+\frac{x+y+z-l}{2})^2$ where $l\in \mathbb{Z}$. It's clear $l\ge 0$.

$\prod_{cyc} (xy+1) - (xyz+\frac{x+y+z}{2})^2 = -\frac14 \sum_{cyc} x^2 +  \frac 12 \sum_{cyc} xy + 1$

$-\frac14 \sum_{cyc} x^2 +  \frac 12 \sum_{cyc} xy + 1 = -\frac l2 (2xyz+x+y+z-\frac l2)$

$\sum_{cyc} x^2 -2 \sum_{cyc} xy -4 = l (4xyz+2x+2y+2z-l)$

$x^2 - (2y+2z)x +(y-z)^2 - 4 - 4lxyz - 2lx - 2ly - 2lz + l^2=0$

$x^2 - (2y+2z+4lyz+2l)x + (y^2+z^2+l^2 - 4 - 2yz -2ly - 2lz)=0$

Note this expression is fully symmetric: $x^2 - 2yx-2zx-4lyzx-2lx + (y^2+z^2+l^2 - 4 - 2yz -2ly - 2lz)=0$

so $x^2+y^2+z^2+l^2 - 4lxyz - 2(xy+xz+xl+yz+yl+zl) - 4=0$

We know $x,y,z,l\in \mathbb{Z}_{\ge 0}$. If $xyz=0$ the statement is clear. Let $(x,y,z,l)$ be the solution with $x+y+z+l$ minimal. I claim $\min\{x,y,z,l\}=0$. Furthermore WLOG $x$ is the largest.

First we can assume $x<l+y+z+2xyz$ because otherwise $(2y+2z+4lyz+2l-x,y,z,l)$ gives a solution with smaller sum. Let $r=2y+2z+4lyz+2l-x$

We consider their product. Note $xr=y^2+z^2+l^2-2(yz+lz+yl) < 3x^2$, so it follows that $x\sqrt{3}>r$ so we can bound $(y^2+z^2+l^2 - 4 - 2yz -2ly - 2lz)=xr\ge \frac 29 (2y+2z+4lyz+2l)^2$

There is another case where $r<0$. In this case, $y^2+z^2+l^2 - 4 - 2yz -2ly - 2lz < 0$ and its absolute value is at least $2y+2z+4lyz+2l+1$, and we can get a contradiction.


superior algebra
superior algebra solved

Another problem if anyone has time. Sorry for the terrible notation/format. New to the forums and haven't learned how to use all its functionalities yet.

Suppose x and y are integers. Suppose p = x^2+y^2 where p is a prime integer. Prove p is irreducible in Z[i].

Yes, and this gives that $p$ is [u]not[/u] irreducible in $\mathbb{Z} [i]$

I was given this problem by my math professor to show that given x and y (which are integers) and p=x^2+y^2 (where p is a prime integer), p is irreducible in Z[i] and thus a Gaussian prime. But p is in fact not irreducible??? I am confused. Is this professor asking me to prove something that is not true?

If you heard him right, he made a mistake. You can show that $x+yi$ is a gaussian prime.


calculus
Princeton
college
algebra
partial fractions

Hi,i am a 10th grader and lately im kinda bored from the same old high school mathematics(AMC,AIME..........................) .

I would like to start studying Calculus(but i have no Calc background whatsoever).

Can someone help me find a book that  teaches you Calculus from the very beginning...(like limits....ehh intregals) but that also later on teaches u some more advanced topics( like for example giving a 7th grader the AOPS introductory books, they start off easy but then get kinda hard^^)....or maybe what i need is a set of books?

Also does anyone know if the book that David Patrick is working on is the type of book that i just described.....both nooby and advanced terms

Any help would be appreciated :)

PS:i had an account and my little brother erased it ..........
on purpose   :dry:  :mad:

Get an AP review book, like Princeton Review.  I used that to teach myself calculus from the beginning.  Or you could just use a textbook.  Of course, you'll have to know some pre-calculus topics first, like $ e$, basics of limits and series, partial fractions,...

hmm, i'm kinda in the same fix as you however i have had limits, though they are easy to learn, i learned them myself before we went over it in class. I would recommend just getting an old calculus textbook used, that's what i plan on doing. I'm not sure about gettting the AP book, it seems like it might be too superficial, idk.  I think calculus might be kinda easy to learn as long as you take the time to understand everything.

[quote]I think calculus might be kinda easy to learn as long as you take the time to understand everything.
[/quote]
That basically the attitude you need.  When I tought myself, I read through the chapters once just to get an overview, then I read through it again to digest it and work out the problems.  And then I read it again if I felt I had to.


geometry
inequalities
inequalities proposed

In every triangle

$\sum \frac{m_a^2}{bc} \leq \frac{5}{4} + \frac{R}{2r}$

with usual notation.

Note that the LHS equals [tex]\sum \frac{\frac{1}{4}(2b^2 + 2c^2 - a^2)}{bc} = \frac{\displaystyle \sum a(2b^2 + 2c^2 - a^2)}{4abc}[/tex].

Also, letting the area be [tex]K[/tex] and the semiperimeter be [tex]s[/tex],

[tex]\frac{R}{2r} = \frac{\frac{abc}{4K}}{\frac{4K}{a+b+c}} = \frac{abc(a+b+c)}{16K^2} = \frac{abc(a+b+c)}{16s(s-a)(s-b)(s-c)} \\ = \frac{abc(a+b+c)}{16\frac{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}{16}} = \frac{abc}{(a+b-c)(a-b+c)(-a+b+c)}[/tex]

So we want to prove the following inequality:

[tex]\frac{\displaystyle \sum a(2b^2 + 2c^2 - a^2)}{4abc} \leq \frac{5}{4} + \frac{abc}{(a+b-c)(a-b+c)(-a+b+c)}[/tex].

However, I'm stuck here...would someone mind providing a hint to help me finish the problem?

I posted a stronger inequality at:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16365

Can somebody connect the dots...?

Yeah, we jusy need to write $m_a^2=\frac{b^2+c^2}2-\frac{a^2}4$.


I disagree with the answer in my textbook. Can someone please tell me if I've missed something?

'A circle touches the [i]x[/i]-axis and also touches the circle [i]x[/i]^2 + [i]y[/i]^2 = 4. Find the equation of the locus of its centre.'

My book gives the answer as:
[i]x[/i]^2 = 4(1 + [i]y[/i])
but this doesn't allow for negative [i]y[/i] values
My answer is: ([i]x[/i]^2 - 4)^2 = 16[i]y[/i]^2

Thanks for your time.
jimi

EDIT: Just read the one below its much nicer, ya the book just simplifies your answer..

Let $C_{1}(O,2)$ be the first circle
and $C_{2}(P,r)$ the second circle

Let $M(x,0)$ be the projection of $P$ onto the x-axis.

Then $r$ is the distance $PM$

$OP-2=PM\Rightarrow$

$\sqrt{x^{2}+y^{2}}-2=r$

The book probably substitutes $r=y$ then we get

[hide="book s solution"]$\sqrt{x^{2}+y^{2}}-2=y \Rightarrow$

$\sqrt{x^{2}+y^{2}}=y+2 \Rightarrow$

$\left(\sqrt{x^{2}+y^{2}}\right)^{2}=(y+2)^{2}\Rightarrow$

$x^{2}+y^{2}= y^{2}+4y+4 \Rightarrow$

$x^{2}= 4y+4 = 4(y+1)$

$\hline$[/hide]

But if we substitute $r=|y|$ in $(1)$ we get

[hide="jimi s solution"]$\sqrt{x^{2}+y^{2}}-2 = |y| \Rightarrow$

$\left(\sqrt{x^{2}+y^{2}}-2\right)^{2}= |y|^{2}\Rightarrow$

$\left(\sqrt{x^{2}+y^{2}}\right)^{2}-4\sqrt{x^{2}+y^{2}}+4= y^{2}\Rightarrow$

$(x^{2}+y^{2})-4\sqrt{x^{2}+y^{2}}+4= y^{2}\Rightarrow$

$x^{2}-4\sqrt{x^{2}+y^{2}}+4 = 0\Rightarrow$

$x^{2}+4 = 4\sqrt{x^{2}+y^{2}}\Rightarrow$

$(x^{2}+4)^{2}= 16(x^{2}+y^{2})\Rightarrow$

$(x^{2}+4)^{2}-16x^{2}= 16y^{2}\Rightarrow$

$(x^{2}-4)^{2}= 16y^{2}$
[/hide]


geometry
3D geometry
combinatorics unsolved
combinatorics

Find the largest natural number $n$ such that any $2n$x$2n$ board covered with non-overlapping 1x2 tiles can be divided into two pieces with a straight line without splitting any tiles.

I think it's $n=3$. Every dividing line of a 6x6 board (there are 10 of them) has an even number of unit squares on either side. So it must as well hit an even number of tiles. So we would need at least 20 tiles to cut all the lines, but there are only 18 of them  :(
This means any tiling of a 6x6 board can be divided somewhere.
For 8x8 and more, it's easy to find tilings that can't.

Now the same question for a (2n)^3 cube and
(a) 2x1x1 tiles
(b) 2x2x1 tiles  :D

[quote="spanferkel"]
Now the same question for a (2n)^3 cube and
(a) 2x1x1 tiles
(b) 2x2x1 tiles  :D[/quote]

I believe it can be done the same way
(a) $n=1$
(b) $n=2$


Prove that there exist a quadrilateral where the three sides are equal and both the diagonals and the fourth side are equal.

Hint
[hide]
There exists an isosceles trapezoid satisfying the conditions. Can you find it? :)[/hide]

[hide]Let $AB=AC=BD=x$ and $BC=CD=AD=y$ from Ptolemy theoreme for any inscribed quadrilateral we have $AB*CD+AD*BC=AC*CD$.

In our case $xy+x^2=y^2$ or $z^2-z-1=0$ $z= \frac yx$ so $z=\frac xy=\frac{1+ \sqrt5}{2}$ :D .

Actually if u draw a regular pentagone, this is a part of the picture![/hide]

[quote="delta"]$AB*CD+AD*BC=AC*CD$.[/quote]

Typo :)

[quote="delta"]from Ptolemy theoreme for any inscribed quadrilateral[/quote]

You don't know that it's cyclic. I guess you're just constructing one in such a way that it's cyclic?

All isosceles trapezoids are cyclic...  :)

What's the Ptolemy Theorem though?

For cyclic quad $ABCD$,

$(AB)(CD) + (BC)(AD) = (AC)(BD)$

you could just say:
observe a regular pentagon ABCDE, ABCD is cyclic since regular polygons are cyclic, AB=BC=CD, and AC=BD=AD since they respectively subtend the same degree arcs, ABCD is our desired quadrilateral

Nice...

Is there some type of generalized formula you could derive from that?  Like there is an $n$-sided polygon where $n-1$ of the sides are congruent and the last side is congruent to all of the diagonals for all $n>3$, and that polygon can be found by taking a regular polygon with $n+1$ sides and eliminating one of its vertices...

I'm pretty sure that works...


[color=darkblue]Let $ \Gamma_A$ be the $ A$-mixtilinear incircle of $ ABC$, which is tangent to the circumcircle at $ A'$ and to the side $ AB$ at $ D$. Find the length of $ A'D$.[/color]

[quote="Ahiles"][color=darkblue]Find the length of $ A'D$.[/color][/quote]

What is the application of finding the lenght of A'D? , In terms of which elements of $ ABC$? , Could you show at leats the expression that we should get? thanks in advance

Best Regards


integration
calculus
calculus computations

Change variables according to $ x^{3}\equal{}\frac{1}{t^{3}\minus{}2}$,

[hide="and so"]$ P \equal{}\int(2x^{\minus{}6}\plus{}x^{\minus{}9})^{2/3}\,dx \equal{}\int x^{\minus{}4}(2\plus{}x^{\minus{}3})^{2/3}\,dx \equal{}\minus{}\int t^{4}\,dt$

$ \equal{}\minus{}\frac{1}{5}t^{5}\plus{}C \equal{}\minus{}\frac{1}{5}(2\plus{}x^{\minus{}3})^{5/3}\plus{}C$.
[/hide]


I have a Mastery of Counting Fundamentals (MCF), MCF level 1 and MCF level 2, and I only have problems for level 2 (7 problems).

Another question: What is "Basic Pigeonhole"? I have had about 300 problems, and none show up in that subject.

Both dragon96 and I have over a thousand points, and we still have received no Basic Pidgeonhole questions, which leads me to think it's one of those empty topics.

I've finished all the problems and I have had no "Basic Pidgeonhole" problems either.

[quote="r15s11z55y89w21"]I don't know about others, but upon viewing my report, I discover that "Mastery of Counting Fundamentals" and "Mastery of Counting Fundamentals Level 1" encompass exclusively the exact same problems. I just wish to know if there are others for which there is something different. Also, is this as supposed to be?[/quote]
I think they might be worded differently to test your consistency with the problems. That's happened to me with some of the problems.

[quote="r15s11z55y89w21"]I don't know about others, but upon viewing my report, I discover that "Mastery of Counting Fundamentals" and "Mastery of Counting Fundamentals Level 1" encompass exclusively the [b]exact[/b] same problems. I just wish to know if there are others for which there is something different. Also, is this as supposed to be?[/quote][hide= ][/hide]


[b]1.[/b]Sa se demonstreze ca pentru diferite alegeri ale semnelor + si - din expresiile \[ \pm 1 \pm 2 \pm 3 \pm ... \pm(4n\plus{}1)\] se obtin numere impare mai mici sau egale cu $ (2n\plus{}1)(4n\plus{}1)$

si inca o problema poate chiar mai interesanta decat prima
[b]2.[/b] Fie n un numar natural arbitrar si k un numar intreg.Sa se demonstreze ca daca k admite o reprezentare de forma \[ \pm 1^n \pm 2^n \pm ... \pm m^n\] , m natural, pentru o anumita alegere a semnelor + si -, atunci admite o infinitate de reprezentari de aceasta forma.

problema1
termenii expresiei din enunt sunt numerele naturale de la 1 la 4n+1.este usor de vazut ca de la 1 la 4n avem 2n numere pare si 2n numere impare, si cum 2n+1 e impar atunci suma are 2n termeni pari si 2n+1 termeni impari.atunci este evident ca numarul obtinut va fi impar(este o ,,suma'' de numere pare si de un numar impar de numere impare).rezultat maxim se obtine atunci cand se alege doar plus.deci numerele vor fi mai mici sau egale decat 1+2+3+...(4n+1)=(4n+1)(4n+2)/2=(2n+1)(4n+1) :)


let [tex]  x,  y  [/tex] and  [tex] z [/tex] be nonnegatif real numbers such that:
[tex] xyz(x+y+z) =1 [/tex]
Prove that: [tex]
\surd ( x^2 + \frac{1}{y^2} )( y^2+\frac{1}{z^2} )( z^2+\frac{1}{x^2} )=(x+y)(y+z)(z+x)[/tex]


Hi.

Given: $ x^5 \plus{} x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1 \equal{} 10$ and $ \minus{} 1$ isnt a root to the equation.

Find $ (x \plus{} 1)^4$.

Thanks.

"and $ \minus{}1$ isn't a root to the equation?" It's not anyways, is it? Or do you mean $ 0$ instead of $ 10$ on the RHS?

[hide="Assuming that"]
The solutions are the sixth roots of unity (except for $ 1$, of course) excluding $ \minus{}1$.
[/hide]

What makes you think there'll be a nice solution to this? The other 4 roots yield 4 different values for $ (x \plus{} 1)^4$.

Whoa; I just realized there was another part of the problem - I thought it was just find the roots. That's what I get for trying to do math at night...


trigonometry
algebra
polynomial
Vieta
function
algebra unsolved

Assuming that the roots of $x^3 + p \cdot x^2 + q \cdot x + r = 0$ are real and positive, find a relation between $p,q$ and $r$ which gives a necessary condition for the roots to be exactly the cosines of the three angles of a triangle.

[quote="Arne"]Let $a$, $b$ and $c$ be the roots of the equation. It is well-known that $a$, $b$ and $c$ are the cosines of the angles of a triangle if and only if $a^2 + b^2 + c^2 + 2abc = 1$. Now, using the Vieta relations, we get $(p^2 - 2q) -2r = 1$ or $p^2 = 2q + 2r + 1$.[/quote]

Actually, the condition is necessary, but generally not sufficient! For example, setting $a = b = 2,$ and $c = -1,$ we get: \[ a^2 + b^2 + c^2 + 2abc = 4 + 4 + 1 - 8 = 1, \] but obviously $a, b,$ and $c$ are not the cosines of the angles of a triangle.

If—as stated in the problem—all three quantities are positive, then $a^2 + b^2 + c^2 + 2abc = 1$ implies $0 < a, b, c < 1.$ The inverse cosines $\alpha = \arccos a$ and $\beta = \arccos b$ are thus necessarily each strictly between zero and $\frac\pi2$ and so there exists a triangle with $0 < \alpha, \beta < \frac\pi2$ as two of its angles. Furthermore, for $0 < \theta < \frac\pi2$ the square of the cosine is a strictly decreasing function of $\theta.$ Observing that $\cos^2(\theta) + \cos^2(\frac\pi2 - \theta) = \cos^2\theta + \sin^2\theta = 1$ we see that $a^2 + b^2 < 1 \implies \alpha + \beta > \frac\pi2,$ and thus $\gamma = \pi - \alpha - \beta < \frac\pi2.$

Therefore, in the triangle whose angles have positive cosines $a$ and $b$ with $a^2 + b^2 < 1,$ we have $x = \cos\gamma > 0.$ Now $x$ must satisfy the equation: \[
x^2 + 2abx + (a^2 + b^2 - 1) = 0, \] whose  roots have a product $(a^2 + b^2 - 1)$ and sum $-2ab$ that are both negative. Thus, the roots have opposite signs and so $x$ is the only positive solution. Since $c > 0$ also satisfies the same equation, $c$ is indeed the cosine of the third angle.

The positivity condition is essential, and cannot be ignored.


Let $ f: \mathbb R\to\mathbb R$ be continuous. $ \lim_{x\to\infty}f(x)\equal{}\infty$ and $ \lim_{x\to\minus{}\infty}f(x)\equal{}\minus{}\infty.$ Prove that $ f$ is surjective.

Bolzano-Cauchy intermediate value theorem.

Thanks, but I don't know how to apply it. :(


Let $a,b,c>0$ such that $a+b+c\geq\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$.
  Is this true: $abc\leq 1$ ? ;)

Yes,but the reverse inequality $abc\geq1$? ;)


function
number theory
prime factorization
superior algebra
superior algebra unsolved

Suppose that f is a multiplicative function with f(1)=1. Show that the summation over d|n of u(d)f(d) = (1-f(p1))*(1-f(p2))*.....*(1-f(pk))

where n=(p1^a1)*(p2^a2)*...*(pk^ak) is the prime factorization for n.  u(d) is the Mobius function.

I divided the divisors d of n, into p1, p2,..,pk so that I have k different sums of u(p_i)f(p_i) over each p_i.  I don't know where to go from here.

by the definition of $ \mu$, we have $ \mu(p_{j_1}\cdots p_{j_i})\equal{}(\minus{}1)^i$, so
\[ (1 \minus{} f(p_1))\cdots(1 \minus{} f(p_k)) \equal{} \sum_{1\leq j_1\leq \cdots \leq j_i \leq k} ( \minus{} 1)^if(p_{j_1})f(p_{j_2})\cdots f(p_{j_i}) \equal{}\]

\[ \sum_{i \equal{} 0}^k\mu(p_{j_1}\cdots p_{j_i})f(p_{j_1}\cdots p_{j_i}) \equal{}\]

\[ \sum_{d|n}\mu(d)f(d),\]
because whenever $ d$ is not squarefree, $ \mu(d) \equal{} 0$.


inequalities
function
trigonometry
inequalities proposed

This cant be new but I have not seen it cited anywhwere

In the acute triangle ABC prove that

tanA/2 + tanB/2 + tanC/2 <2

Since  the triangle is acute, $(\pi/4, \pi/4, 0)$ majorizes $(A/2, B/2, C/2)$.

Apply the Majorization (Karamata) Inequality to the convex function $f(x) = \tan{x}$ on ${}[0,  \pi/4]$, we have

$ 2 = f(\pi/4) + f(\pi/4) + f(0) > f(A/2) +  f(B/2) + f(C/2)$. The inequality is strict because none of the angles can be 0.

This inequality can be improved :
For any acute triangle ABC, we have

sum(tan A/2 ) + tan(A/2)*tan(B/2)*tan(C/2) < 2

And 2 is the best constant.

There is an easy solution for this inequality:

Because ABC is an acute triangle 0<tanA/2,tanB/2,tanC/2<=1
Hence
(1-tanA/2)(1-tanB/2)(1-tanC/2)>0
--> 1-tanA/2-tanB/2-tanC/2+tanA/2tanB/2+tanB/2tanC/2+tanC/2tanA/2-tanA/2tanB/2tanC/2>0
And so Because tanA/2tanB/2+tanB/2tanC/2+tanC/2tanA/2=1 then we have:

tanA/2+tanB/2+tanC/2<2-tanA/2tanB/2tanC/2<2 !


Each side of square $ ABCD$ is doubled in length to form square $ EFGH$.  The perimeter of square $ EFGH$ is 40 cm.  How many square centimeters are in the area of square $ ABCD$?

[asy]pair a,b,c,d,e,f,g,h;
a=(0,0);
b=(10,0);
c=(10,10);
d=(0,10);
e=(15,0);
f=(35,0);
g=(35,20);
h=(15,20);
draw((11,5)--(14,5),Arrow);
draw(a--b--c--d--cycle);
draw(e--f--g--h--cycle);
label("$A$",a,S);
label("$B$",b,S);
label("$C$",c,N);
label("$D$",d,N);
label("$E$",e,S);
label("$F$",f,S);
label("$G$",g,N);
label("$H$",h,N);[/asy]

Or, since the larger square has a perimeter of 40, we find that it's length is 10. Half of 10 is 5, so $ 5^2 \equal{} \boxed{25}$.

Or even, since the larger square is twice the smaller one, the perimeter of the smaller one is 40/2 = 20.  So one side is 5 and the area is 25.


Bueno ..
aki

la parte  2 del libro de geo

SeCCION: Problemas

cARLOS

[color=blue]
 Buenos dias,

les informo que por restricciones en mi
cuota de uploading...

quitare estos pdf,
por tanto, descarguenlos  antes del lunes en la noche,

Gracias[/color]


function
algebra
domain
integration
complex analysis

Given continuously differentiable functions $ u(x,y), v(x,y)$, let $ f\equal{}u\plus{}iv$.  Suppose for every $ z_0$ in the domain of $ f$, $ D$, there exists an $ r_0$ such that \[ \int_{|z\minus{}z_0|\equal{}r}fdz\equal{}0\] for all $ r<r_0$.  Show that $ f$ is analytic in $ D$.

this is just green's theorem, and you will get that for any point $ z_0$ in the domain, the cauchy reimann equations hold...


In a acute triangle $ABC$, $BD \perp AC$, $CE \perp AB$, $BD \cap CE = H$. Construct a circle with diameter $DE$. The circle meets $AB$ and $AC$ at $F$ and $G$ respectively, $FG$ meets $AH$ at $K$. If $BC=25$, $BD=20$, $BE=7$, find the length of $AK$.

[quote="stranger"]2004 China League[/quote]
Thanks for pointing it out. I shall edit.


Let point $K$ be on side $BC$ of $\triangle ABC$. The incircle of $\triangle ABC$ is tangent to side $BC$ at point $D$. Point $Q$ is the center of the incircle of $\triangle ABK$ and $R$ is the center of the incircle of $\triangle ACK$. Show that $\angle QDR$ measures 90 degrees.

Very nice problem!

  See :

 http://www.mathlinks.ro/Forum/viewtopic.php?t=59205

 [u]Babis[/u]


i was wondering if i could ask a teacher to look at my proof and say whether or not its good i dont mean like for her to tell me wut i did wrong just say good or bad because i have really bad grammar skills and i'm positive that i solved the Qs and i have the ans i just dont know if my style off writing will make the cut of 5 :winner_first:    or even 3 points :winner_third:  :blush: and it would make me feel better if i could get assurance from someone that my solution is good so if someone could just tell me whether or not that would be breaking the usamts honor code

You cannot do this.  You may not get any assistance from any other person.

You don't even need to do that. You just need to look over your proof for any holes, and if there aren't any it'll probably get a 5.

You probably won't get any points off for style as long as you explain everything. Remember the page how to write proofs (which you can find if you go to USAMTS and click on the link back to AoPS): it's better to overdo it than to underdo it.

What I posted does not go against honor code!

u are absolutely correct Heero except for the fact that i find my work flawless to my eyes i can never find my own mistakes because they are not mistakes in my sense if u no wut i mean

No.

(You can wait until *after* the round is over, when it is allowed to discuss the problems, many people post their solutions, and you can look over their proof writing, and also post your own solutions, and ask any particular questions you may have about writing.)

This isn't discussing the problem.

Okay, so we still have a month left. Then leave it alone for like, two weeks, and then you can look at it with fresh eyes. If you can't, then start practicing. It's an important skill in competition.


Hi everyone,

Could you please tell me if this is right so far and help me fin cn?

Find the Taylor series for f(x) centered at the given value of a. Also, find R

f(x)=x^-2, a=1

f(x)=x^-2        f(1)=1          c1=1
f'(x)=-2x^-3      f'(1)=-2        c2=-2
f''(x)=6x^-4        f''(1)=6       c3=6
f'''(x)=-24x^-5      f'''(x)=-24   c4=-24

f^(n)(x)=-1^(n)      f^(n)(1)=-1^n    cn=(-1)^n/n

R=?????????

Thank you very much

I believe you need to find the limit of $ \frac{a_{n\plus{}1}}{a_{n}}$ for which it is less than 1.  I can't remember for sure but I would guess wikipedia has something on it.