geometry
inradius
trigonometry
geometry unsolved

Construct a triangle given its three exradii.

Denote a = BC, b = CA, c = AB the sides, $\alpha = \angle A,\ \beta = \angle B,\ \gamma = \angle C$, the angles $r, r_a, r_b, r_c$ the inradius and exradii against the vertices A, B, C, s the semiperimeter and $\triangle$ the area of the triangle $\triangle ABC$.

$r = (s - a) \tan \frac \alpha 2 = (s - b) \tan \frac \beta 2 = (s - c) \tan \frac \gamma 2$

$r_a = s \tan \frac \alpha 2,\ r_b = s \tan \frac \beta 2,\ r_c = s \tan \frac \gamma 2$

$r_ar_b + r_br_c + r_cr_a = s^2 \left( \tan \frac \alpha 2 \tan \frac \beta 2 + \tan \frac \beta 2 \tan \frac \gamma 2 + \tan \frac \gamma 2 \tan \frac \alpha 2 \right) =$

$= s^2 \left( \frac{r^2}{(s - a)(s - b)} + \frac{r^2}{(s - b)(s - c)} + \frac{r^2}{(s - c)(s - a)} \right) =$

$= \triangle^2 \frac{3s - (a + b + c)}{(s - a)(s - b)(s - c)} = s^2$

Let F be the 1st Fermat point of a triangle $\triangle XYZ$ such that $FX = r_a,\ FY = r_b,\ FZ = r_c$. Since $\angle XFY = \angle YFZ = \angle ZFX = 120^\circ$, the triangle $\triangle XYZ$ is trivial to construct. The area of this triangle is

$|\triangle XYZ| = \frac 1 2 \sin 120^\circ \left(r_ar_b + r_br_c + r_cr_a\right) = \frac{s^2 \sqrt 3}{4}$

Construct an equilateral triangle with the same area - its side will be s, the semiperimeter of the triangle $\triangle ABC$. For example, denote x = YZ and let $h_x$ be the X-altitude of this triangle. Then

$|\triangle XYZ| = \frac 1 2 h_x x = \frac{s^2 \sqrt 3}{4},\ \ h_x = \frac{s^2 \sqrt 3}{2x}$

Draw an equilateral triangle with the altitude x = YZ. Its side is $p = \frac{2x}{\sqrt 3}$.  Construct the square root

$\sqrt{h_x p} = \sqrt{\frac{s^2 \sqrt 3}{2x} \cdot \frac{2x}{\sqrt 3}} = s$

Once the semiperimeter s of the triangle $\triangle ABC$ is constructed, all its angles can be constructed and then the triangle itself. For example, draw the circle $C_a(E_a, r_a)$, pick an arbitrary point P on this circle and draw a tangent to $C_a$ at P. Let A be a point on this tangent, such that AP = s. Construct the other tangent AQ from A to $C_a$ by reflecting P in $AE_a$. The line $AE_a$ bisects the angle $\angle PAQ$ and a normal to $AE_a$ at A bisects the complementary angle. Draw a circle $C_b(E_b, r_b)$ centered on this normal and tangent to the lines AP, AQ. One of the lines AP, AQ is the common internal tangent of the circles $C_a, C_b$, say the line AQ. Draw the other common internal tangent of these 2 circles by reflecting the common internal tangent AQ in the center line $E_aE_b$. The other common internal tangent meets the lines AP, AQ at points B, C, respectively, completing the construction of the triangle $\triangle ABC$.


If $ \pi\equal{}3.1415926...$, what is the exact value of $ |\pi\minus{}3.14|\plus{}|\pi\minus{}\frac{22}{7}|$? Express your answer as a common fraction.

$ |a\minus{}b|\equal{}a\minus{}b$ iff $ a\geq b$, and $ |a\minus{}b|\equal{}b\minus{}a$ iff $ a<b$.

[quote="Mewto55555"]I remember one guy on my team said he left it as 2/700  :rotfl:[/quote]

That guy happens to be me... :P

There isn't really a fast way to find $\dfrac{22}{7}-3.14$ other than noting their common denominator is $\gcd(100,7)=700.$

[quote=hamon][quote=priceyfalcon]How do you get $3.14$ + $22/7$ quickly?[/quote]

I don't think we need a 12 year bump here[/quote]

According to the sheriff, who gave me a report ban for reporting this, you're supposed to bump alcumus (and I guess FTW), as it's open forever.


According to their website ([url]http://www.usaco.org[/url]), they're starting the contest in October this year, instead of November.  That's a total of 7 contests this year.

jeese, i dunno if i'll be able to do all of those.  Rarely do I have 3-4 open hours on a weekend.

you don't have to do all of the rounds in order to participate. Just do what you can.

btw - People who are interested in USACO might want to take advantage of the new discussion forums there.

Just wanted to bump this, since the compeition is this weekend.

Are you talking about the online 'practice' contests or the actual USACO contests part of the selection for IOI???

I'm pretty sure that this is part of the selection for the US IOI team.

Nope, that's just assignment for the appopriate division. It's written in rules that it  doesn't count to IOI team selection.


The number of real values of $ x$ satisfying the equation $ 2^{2x^2 \minus{} 7x \plus{} 5} \equal{} 1$ is:

$ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ \text{more than 4}$

All zeroes of $ 2x^2\minus{}7x\plus{}5\equal{}0$ satisfy this (and they are the only solutions). Since the discriminant is $ 9>0$, there are $ \boxed{\textbf{(C)}\ 2}$ solutions.

Solvent
$2^\gamma=1\implies\gamma=0$
$2x^2-7x+5=\gamma$
$\gamma=0$
$\implies $2$x$^2$-7$x$+5$=0 (Transit law)
x identical to $1$, x idempotent to $-\frac52$ solutes
--> Double (C)

god is greta
-tony BARKER

Note that the only solutions to this equality occur when you choose the roots of the polynomial $f(x)=2x^2 - 7x + 5$.  From here, there are a couple of approaches (some displayed above), but the two I will use are Descartes Rule of Signs and the quadratic formula.

[b]Solution 1[/b]: Using DRoS, we see that $f(x)=2x^2 - 7x + 5$ has two changes in sign which means that it has two positive real roots.  If we take $f(-x)=2x^2 +7x + 5$, we see that there are $0$ changes in sign so it has no negative roots.  Hence, the answer is $ \boxed{\textbf{(C)}\ 2}$.

[b]Solution 2[/b]: Using the quadratic formula, we find that $f(x)=2x^2 - 7x + 5$ has two real roots, $\frac{5}{2}$ and $1$.  This not only confirms our previous solution, but also gives us the answer again $ \boxed{\textbf{(C)}\ 2}$.

[b]Disclaimer[/b]: I know Descartes only gives the maximum amount of roots a polynomial can have and that solution 1 is essentially cheese, but it typically works for second and third degree polynomials (there are exceptions, as always) in my experience and it is basically confirmed to be true by solution 2.  Just wanted to put that out there because you should [i]never[/i] use Descartes as viable proof for the roots of a polynomial.


calculus
integration
trigonometry
calculus computations

Calculate: $ \int\frac{\cos (x)}{\sqrt{\cos (2x)}}\, dx$.

$ \int \frac {\cos x}{\sqrt{\cos 2x}}\ dx \equal{} \int \frac {1}{\sqrt {1 \minus{} \sin ^ 2 x}} (\sin x)'\ dx \equal{} \sin ^{ \minus{} 1}(\sin x) \plus{} C$.

hello kunny, i think it's $ \cos(2x)\equal{}1\minus{}2\sin^2(x)$.
Sonnhard.

Oops!

$ \int \frac{\cos x}{\sqrt{\cos 2x}}\ dx\equal{}\frac{1}{\sqrt{2}}\int \frac{1}{\sqrt{1\minus{}(\sqrt{2}\sin x)^2}}(\sqrt{2}\sin x)'\ dx$

$ \equal{}\frac{1}{\sqrt{2}}\sin ^ {\minus{}1} (\sqrt{2}\sin x)\plus{}C$.

hello, here is my solution
by substituting $ t \equal{} \sqrt {2\cos^2(x) \minus{} 1}$ we get $ dt \equal{} \frac { \minus{} 2\cos(x)\sin(x)}{\sqrt {2\cos^2(x) \minus{} 1}}dx$ and from here $ dx \equal{} \frac {\sqrt {2\cos^2(x) \minus{} 1}}{ \minus{} 2\cos(x)\sin(x)}dt$ and $ \sin(x) \equal{} \pm\sqrt {\frac {1 \minus{} t^2}{2}}$ and our integral is
$ \pm\frac{\sqrt {2}}{2}\int \frac {dt}{\sqrt {1 \minus{} t^2}} \equal{} \pm\frac{\sqrt {2}}{2}\arcsin(t) \plus{} C$.
Sonnhard.


In this problem we have: $ F_0\equal{}1, F_1\equal{}1$ or $ F_1\equal{}1, F_2\equal{}1$?


Does anyone have any recommendations for undergraduate research opportunities during the summer? Also, could people share their experiences?

Hello,
I think you can try REUs, but you must be a US Citizen or a Permanent Citizen to participate in them. They are also quite competitive, but worth applying to. Of course, you can get in touch with a mathematics professor, ask him or her about some interesting mathematics problems to look into or books to read. When school starts again, you can talk to him or her about some new thoughts you've had, etc. You can also intern at various places as a trader, etc. I know JP Morgan and DE Shaw are always looking for interns, especially those who are more mathematically inclined. Best of luck in your endeavors!


geometry
geometric transformation
reflection
trapezoid
geometry unsolved

Does anyone know how to do this:

[color=green]Prove or disprove by giving a counter example.

"If two medians of a triangle are of the same length, thenthe triangle is isosceles"[/color]

Thanks very much!!!  :lol:

Denote the reflections $A'$, $B'$ of the points $A$, $B$ w.r.t. the midpoints of the sides $[BC]$, $[CA]$ respectively. Then $m_a=m_b$ $\Longleftrightarrow$  $AA'=BB'$ $\Longleftrightarrow$ the quadrilateral $ABA'B'$ is a isosceles trapezoid $\Longleftrightarrow$ $BA'=AB'$ $\Longleftrightarrow$ $AC=BC$ $\Longleftrightarrow$ $a=b\ .$


Prove that the coordinates of the centroid of a coordinatized triangle are $ (a,b)$ where $ a$ is the arithmetic mean of the x-coordinates of the vertices of the triangle and $ b$ is the arithmetic mean of the y-coordinates of the triangle.

[hide="Solution"]Let $ \vec{P}$ be the centroid. This is equivalent to proving that \[ \vec{P}=\frac{\vec{A}+\vec{B}+\vec{C}}{3}\]$ \vec{P}$ is $ 2/3$ of the way from $ A$ to the midpoint of $ BC$, so
\begin{eqnarray*}
\vec{P}&=&\vec{A}+\frac{2}{3}\left(\frac{\vec{B}+\vec{C}}{2}-\vec{A}\right)\\
&=&\frac{\vec{A}+\vec{B}+\vec{C}}{3}\end{eqnarray*}And we're done.[/hide]

How would we have proven it without the knowledge that the centroid divides its distance to the vertex as 2/3 of the distance from the vertex to the midpoint of the opposite side?

(BTW, is there an elegant way to prove it?  I know a way using vectors.)

Of course these proofs are wrong because they assume the existence of the centroid.

Defn: The centroid is the intersection of the medians of a triangle.

First, we have to check that our definition makes sense. That is to say, we need to show that the three medians are concurrent.

We can easily verify through vectors that $ G\equal{}\frac{A\plus{}B\plus{}C}{3}$ lies on the line that connects $ A$ and $ \frac{B\plus{}C}{2}$.

By symmetry of argument, $ G$ lies on all of the medians, hence $ G$ is the centroid (and our definition makes sense).

Can we prove it without using coordinates at all?  Also the existence and uniqueness of the centroid.

Of course.

Existence: use Ceva's theorem.

I don't think that uniqueness is really something to focus on here. The intersection of two lines is usually a point...


Find all $ x,y,z\in N$ such that
$ 4^x\plus{}4^y\plus{}4^z$ is a perfect square.

More
Find all $ (x,y,z)\in N$
$ a^{2x}\plus{}a^{2y}\plus{}a^{2z}$ is a perfect square.


Give a,b,c,d are positive numbers, and a+b+c+d=4, prove that
\[ 5(ab\plus{}bc\plus{}cd\plus{}da\plus{}ac\plus{}bd)\leq abc\plus{}bcd\plus{}cda\plus{}dab\plus{}16.\]
\[ 3(ab\plus{}bc\plus{}cd\plus{}da\plus{}ac\plus{}bd)\leq abc\plus{}bcd\plus{}cda\plus{}dab\plus{}14.\]

[quote="shijin93"]Give a,b,c,d are positive numbers, and a+b+c+d=4, prove that

\[ 3(ab \plus{} bc \plus{} cd \plus{} da \plus{} ac \plus{} bd)\leq abc \plus{} bcd \plus{} cda \plus{} dab \plus{} 14.
\]
[/quote]
$ \frac{22}{9}(ab \plus{} bc \plus{} cd \plus{} da \plus{} ac \plus{} bd)\leq abc \plus{} bcd \plus{} cda \plus{} dab \plus{} \frac{32}{3}$ is true too.

[quote="shijin93"]Give a,b,c,d are positive numbers, and a+b+c+d=4, prove that
\[ 5(ab \plus{} bc \plus{} cd \plus{} da \plus{} ac \plus{} bd)\leq abc \plus{} bcd \plus{} cda \plus{} dab \plus{} 16.
\]

\[ 3(ab \plus{} bc \plus{} cd \plus{} da \plus{} ac \plus{} bd)\leq abc \plus{} bcd \plus{} cda \plus{} dab \plus{} 14.
\]
[/quote]
1)$ a\equal{}b\equal{}c\equal{}d\equal{}1, 30<20$ :wink:
2) My solution very ugly(Murhed, Schur)

Sorry I mistyped it. It must be this
\[ 5(ab\plus{}bc\plus{}ca\plus{}da\plus{}db\plus{}cd)\leq abc\plus{}bcd\plus{}cda\plus{}dab\plus{}26.\]


quadratics
calculus
calculus computations

How do you check in general whether a differential is exact?
My teacher said something about "Euler's Criterion" but that's about Quadratic Reciprocity and has nothing to do with exact differentials (I think).

In two dimensions, $ dF \equal{} M(x,y)dx \plus{} N(x,y)dy$ is an exact differential iff $ \frac {\partial M}{\partial y} \equal{} \frac {\partial N}{\partial x}$.

The corresponding criterion for $ n$ dimensions involves $ \frac {n(n \minus{} 1)}{2}$ such relations, and can be similarly derived from the fact that $ \frac {\partial^2 F}{\partial x_i \partial x_j} \equal{} \frac {\partial ^2 F}{\partial x_j \partial x_i}$ for all $ 1\leq i,j\leq n$.  See [url=http://mathworld.wolfram.com/ExactDifferential.html]MathWorld[/url] for the two-dimensional case, or [url=http://en.wikipedia.org/wiki/Exact_differential]Wikipedia[/url] for a slightly more general discussion.


algebra
polynomial
least common multiple
linear algebra
linear algebra unsolved

Let $P, Q$ be polynomials of degree $m, n$ respectively. Prove that the necessary and sufficient condition for$P$ and $Q$have common root is that the following system is linear independent:
$(P, XP, X^{2}P,..., X^{n-1}P, Q, XQ, X^{2}Q, ..., X^{m-1}Q)$

$X \in R$ is the variable of the polynomials.

It's still trivially wrong; the condition should be linear dependence rather than linear independence. With that fix, it's routine (over an algebraically closed field). Any common factor gives a linear dependence relation, and any linear dependence relation gives that the least common multiple has degree less than $m+n$. Dividing into $PQ$, we get a nontrivial common factor.
It's also trivially false over any field that's not algebraically closed; $x^{2}+1$ and $x^{2}+1$ have no common roots over $\mathbb{R}$.


Where would I send an e-mail to ask a clarification question?

usamts@usamts.org, but be warned in advance that we almost never "clarify" the problem questions while a round is in progress.

So I should probably state that I am assuming a property at the beginning, right?

Say I stated that I assume that property #X is assumed for the problem because it seemed ambiguous. If the actual solution states that property #X is false, what would happen?


linear algebra
matrix
trigonometry
geometry
geometric transformation
rotation
trig identities

A transformation carries a triangle with vertices (0,0), (13,0), (3,2) to the triangle with vertices (0,0), (12,5), (2,3).

(a) Describe the transformation.
(b) Where does the transformation send the point (6,0)?

Thanks.

[hide="(a)"]Since the transformation sends $ (0,0)$ to itself, we know that it does not involve a translation. As such, we can express the transformation as follows, and solve for $ a,b,c,d$:
\[ \begin{bmatrix}
x\\
y\end{bmatrix}\begin{bmatrix}
a & b \\
c & d\end{bmatrix} \equal{} \begin{bmatrix} x'\\
y'\end{bmatrix}\]
When $ (x,y)\equal{}(13,0)$ and $ (x',y')\equal{}(12,5)$, we have $ 13a\equal{}12$ and $ 13c\equal{}5$, implying that $ a\equal{}\frac{12}{13}$ and $ c\equal{}\frac{5}{13}$.

Also, when $ (x,y)\equal{}(3,2)$ and $ (x',y')\equal{}(2,3)$, we have $ \frac{36}{13}\plus{}2b\equal{}2$ and $ \frac{15}{13}\plus{}2d\equal{}3$. Solving yields $ b\equal{}\minus{}\frac 5{13}$ and $ d\equal{}\frac{12}{13}$. So our transformation matrix is then\[ \begin{bmatrix}
12/13 & \minus{}5/13 \\
5/13 & 12/13\end{bmatrix}.\] Note that if we let $ \theta\equal{}\sin^{\minus{}1} \frac 5{13}$, then the matrix is in the form\[ \begin{bmatrix}
\cos\theta & \minus{}\sin\theta \\
\sin\theta & \cos\theta\end{bmatrix},\]so this transformation is simply a counterclockwise rotation by the angle $ \theta$.
[/hide]

[hide="(b)"]
Now just simply plug in $ (x,y)\equal{}(6,0)$ and we get:\[ \begin{bmatrix}
6\\
0\end{bmatrix}\begin{bmatrix}
12/13 & \minus{}5/13 \\
5/13 & 12/13\end{bmatrix} \equal{} \begin{bmatrix} 72/13\\
30/13\end{bmatrix},\]so the resulting point is $ \left(\frac{72}{13},\frac{30}{13}\right)$.[/hide]

Thank you.  That solution's beautiful.  Is there an alternative solution not involving matrices?

If you draw a diagram, it's hard not to notice that there are circles around the origin, and it's easy to show: $ \sqrt{2^2\plus{}3^2}\equal{}\sqrt{3^2\plus{}2^2}$ and $ \sqrt{5^2\plus{}12^2}\equal{}\sqrt{13^2}$. So now it seems like it is a rotation, so to prove it we just need to show that the the two given triangles are congruent, which is easily done with distance formula bashing. To find the angle of rotation, note that it is the vertex angle of the isosceles triangle $ (0,0)$, $ (13,0)$, and $ (12,5)$, which has sides $ 13,13,\sqrt{26}$, so by the Law of Cosines, $ 13^2\plus{}13^2\minus{}2\cdot13^2\cos\theta\equal{}26\implies\cos\theta\equal{}\frac{12}{13}\implies\theta\equal{}\cos^{\minus{}1}\frac{12}{13}\equal{}\sin^{\minus{}1}\frac5{13}$.

Without matrices for the second part, note that $ A\equal{}(0,0)$, $ B\equal{}(6,0)$, and $ C\equal{}(13,0)$ are on the same line, and that $ AB\equal{}\frac6{13}AC$. Since rotations of lines preserve the relative positions of the points on the lines, and $ C$ rotates to $ (12,5)$, $ B$ is just $ \left(\frac6{13}\cdot12,\frac6{13}\cdot5\right)\equal{}\boxed{\left(\frac{72}{13},\frac{30}{13}\right)}$.

Thanks to both of you.  Your solutions are crystal clear.

The reason you should suspect that rotation is the answer is that the double-angle formulas $ \sin 2 \theta \equal{} \frac {2 \tan \theta}{1 \plus{} \tan^2 \theta}$ and $ \cos 2 \theta \equal{} \frac {1 \minus{} \tan^2 \theta}{1 \plus{} \tan^2 \theta}$ tell you that twice one of the acute angles in a $ 2 \minus{} 3 \minus{} \sqrt {13}$ triangle is one of the acute angles in a $ 5 \minus{} 12 \minus{} 13$ triangle.  Equivalently this is because $ (3 \plus{} 2i)^2 \equal{} 5 \plus{} 12i$.


Let three equal circles $C_{1},\ C_{2},\ C_{3}$  externally tangent ($C_{1}$  with $C_{2}$,$C_{2}$ with $C_{3}$,$C_{3}$with $C_{1}$) and these circles are internally tangent to onother circle $C_{4}$ .
Let a point $S$ in $C_{4}$  . Let the tangents from $S$ to $C_{1}$,$C_{2}$,$C_{3}$ at the points $A,\ B,\ C$ .
Prove that $SA=SB+SC$  or $SB=SC+SA$ or $SC=SA+SB$ .

this easily follows from the generalised Ptolemy theorem!
let r be the length of the common tangent of two of the circles then we have (if for example we assume S to be in the arc AB) r*SA +r*SB=r*SC
giving the required result!

[color=darkblue]This problem is a particular case of the [b][u]Casey's theorem[/u][/b] which is similarly with the particular case of the [b][u]Ptolemeus' theorem[/u][/b] for an equilateral triangle.[/color]

What theorem called <<generalised Ptolemy>> ?

Generalized Ptolemy = Casey's theorem. See [url=http://www.google.ro/search?q=casey+theorem&start=0&ie=utf-8&oe=utf-8&client=firefox-a&rls=org.mozilla:en-US:official]here[/url].


geometry
rectangle
combinatorics unsolved
combinatorics

I want a general solution to problem such as these, the question is to find out how many paths of the shortest possible length there are between A and B  , how do I think and what is the fastest way to solve it?

Any help would be greatly appriciated.

It is the standard combinatorics problem with just one rectangle but with some subcases.

For any small situation more complicated than just a rectangle, the fastest way is often just to do it by hand, as if you were building Pascal's triangle.  In this case, that should take under a minute.

One can describe a general formula for the number of paths from one corner to another of a rectangular grid that doesn't pass through a given set of points.  It's just inclusion-exclusion taken over paths passing through those points in all of the possible orders.  It's not practical to implement in any but the most simple cases, unfortunately.

I don't think your solution is correct libra_gold or I made a mistake evuluating it :D   But I don't get the correct value.

Sorry, I found something wrong.

In fact we should delete some paths which are calculated twice.

I correct my solution as following:

Denote $A(0,0), B(5,6)$

Then the path can be

(1) $(0,0)\to (2,2)\to (5,6)$, the number of the path is $\binom{4}{2}\binom{7}{3}$

(2) $(0,0)\to (1,3)\to (2,3)\to (5,6)$, the number of the path is $\binom{4}{1}\binom{6}{3}$

(3) $(0,0)\to (1,4)\to (2,4)\to (5,6)$, the number of the path is $\binom{5}{1}\binom{5}{3}$

The correct answer should be $340$.


group theory
abstract algebra
search
superior algebra
superior algebra theorems

Someone knows the book/article  regards the theory of groups
 above all the subgroup promormal and the T-groups?:(
  many thanks!   :)
sastra

I usually use follow book to search defines:

http://www.amazon.com/Introduction-Theory-Groups-Joseph-Rotman/dp/0387942858

To hope it help you so  :wink:


geometry rectangle trigonometry parallelogram

Let $ABCD$ be a rectangle with $BC=3AB$. Show that if $P,Q$ are the points on side $BC$ with $BP = PQ = QC$, then \[\angle DBC+\angle DPC = \angle DQC.\]

[hide]Let $\angle DBC = \alpha$, $\angle DPC = \beta$, $\angle DQC = \gamma$. Then $\alpha+\beta = \gamma$ iff $\cos (\alpha+\beta) = \cos \gamma$ because $0 < \gamma < \frac{\pi}{2}$ and $0 < \alpha+\beta < \frac{3\pi}{2}$.
Hence iff $\frac{BC}{BD}\cdot\frac{PC}{PD}-\frac{AB}{BD}\cdot\frac{AB}{PD}= \frac{AB}{DQ}$ or after substitution $\frac{5AB^{2}}{BD\cdot PD}= \frac{AB}{DQ}$. After squaring becomes $25AB^{2}\cdot DQ^{2}= BD^{2}\cdot PD^{2}$ and by Pythagoras $50AB^{4}= 50AB^{4}$.[/hide]

[hide="Easy..."]
Same angle labeling as [b]Andreas[/b] ( $\angle DBC = \alpha, \ \angle DPC = \beta, \ \angle DQC = \gamma$).
Then, $\alpha = \arctan \left( \frac{1}{3}\right), \ \beta = \arctan \left( \frac{1}{2}\right), \ \gamma= \frac{\pi}{4}$.
Since $0 < \alpha, \beta, \gamma \leq \frac{\pi}{4}, \ \ \alpha+\beta = \gamma \iff \tan(\alpha+\beta) = \tan(\gamma) \iff \frac{ \frac{1}{3}+\frac{1}{2}}{1-\left(\frac{1}{2}\right) \left( \frac{1}{3}\right) }= 1$, which is true. [/hide]

This was an Olympiad problem??

[hide]Let $\angle DBC=\alpha, \angle DPC=\beta, \angle DQC=\gamma$.
$\sin \gamma=\frac{1}{\sqrt{2}}$.
$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha=(1/\sqrt{10})(2/\sqrt{5})+(1/\sqrt{5})(3\sqrt{10})=\frac{1}{\sqrt{2}}$.
Thus, $\sin\gamma=\sin(\alpha+\beta)$.
Taking arcsin of both sides, $\gamma+2k\pi=\alpha+\beta$, for integer k. Since $0<\gamma,\alpha,\beta<180$, $\gamma=\alpha+\beta$.[/hide]

yeah, with trig. is all easy.

is there a way to prove it without trig.?

[hide="ha! no trig!"]
Place points $X$ and $Y$ on $\overline{AD}$ such that $AX=XY=YD$, and let $O$ be the intersection of $\overline{BD}$ and $\overline{PY}$. Let $l=BP$; then $XY=l$ and simple isosceles right triangles give us $BX=PY=l\sqrt{2}$. Also, the diagonals of parallelogram $BPDY$ bisect each other, so that $OP=\frac{1}{2}PY=\frac{l}{\sqrt{2}}$. But this means that $\triangle BXY$ and $\triangle BPO$ are similar (by SAS similarity; also note that $\angle BPO=\angle BXY=135^\circ$). Now we get congruent angles all over the place:
\[\angle DBC=\angle OBP=\angle XBY=\angle BYP=\angle YPD\]
so that
\[\angle DBC+\angle DPC=\angle YPC=\angle DQC\]
as desired.
[/hide]

I don't get where is $P$....

give a image please

[color=darkred]Short and nice solution[/color][hide]
See attached diagram!$AE=DF=AB$ and in $\Delta EPD$ $EP=DP$ and $\angle EPD=90$ [/hide]

[hide="Natural first attempt for me; ugly compared to others after checking theirs"]
Let first angle be called 1, second called 2, third 3.
Then $\cos 1=\frac{1}{\sqrt{10}}~\cos 2+\frac{1}{\sqrt{5}}~\cos 3=\frac{1}{\sqrt{2}}$
$\cos (1+2)=\frac{1}{\sqrt{10}}\frac{1}{\sqrt{5}}-\frac{3}{\sqrt{10}}\frac{2}{\sqrt{5}}=-\frac{5}{50}=-\frac{1}{\sqrt{2}}$
considering angle bounds, this implies that 1+2=3[/hide]

[hide]excellent proof "delta" :idea: [/hide]


I already have aops vol. 1 and vol. 2 and introduction the counting and probability.
I want to qualify for the AIME (I'm taking the AMC 12 this year) and I am not sure if the books I already have are adequate enough. Is there anything in The first steps for math Olympians that I cannot find in the books I already have?

AOPS 1 and AOPS 2 will teach you in a much better way than first steps for math olympians.

Probably not. The books you already have cover more than the first steps book. If you do some practice AMC 10s as well, you should be fine.

FSMO does a good job preparing for the AMCs but thats about it.

it has a pretty good coverage of topics and you can finish the book pretty quickly, (one chapter wont take more than 1-1.5 hrs)

the problems are pretty nice i guess. the first few chapters are very elementary and the last 4 are geared towards topics on amc12.

 aops v1 and v2 definitely  have more material and they are much more involved than FSMO, but the only thing is, v1 and v2 will take more time to properly finish because you have to reallly think deeply about many of the problems.  while fsmo might only have 1 or 2 "deep" problems per section.

only get FSMO if you need to do really well on AMC but you dont have alot of time to study.

its more at V1 level but its problems come from AMC12, AMC10(some of them) and AHSME.

[quote="giratina150"]The Math Olympiad books will help you more on AMC 8, while the Aops Vol. 1 and 2 are great for all AMCs[/quote]

This is to some extent, false. FSFMO is quite helpful in preparing for the AMC10 & 12. Although the way the topics are covered in FSMO is somewhat not to my liking, it has great exercises and it can be helpful if you're running low on time.

I think that FSMO and the AoPS v1 and v2 are excellent. But I would start with AoPS v1 if you have time. For review, use FSMO.


trigonometry
trig identities
Law of Cosines
geometry
Law of Sines

Are we allowed to brute force this. :rotfl:

[hide="General solution outline"] We can write out $\frac{8\pi}{35}$ as $\frac{15\pi}{35}-\frac{7\pi}{35}$. We can use the trig identity $\cos(\pi-x)=-\cos x$ to rewrite the angle as a negative cosine of $\frac{30\pi}{35}-\frac{7\pi}{35}$. We can write the last one out also as a subtraction of $\frac{25\pi}{35}-\frac{7\pi}{35}$. We can simplify the fractions and use cosine subtraction identities to expand. We can group together the terms with $\cos{\frac{\pi}{5}}$ and $\sin{\frac{\pi}{5}}$. We notice that, after some usage of the cosine identity and its corresponding (not the same) identity for sine, the sum of the arguments of the cosine and sine fuctions is $\pi$ for each group. Therefore, these angles are parts of a triangle with special angles. We can use trig identities and the Law of Cosines and Law of Sines to find the cosines and sines of each of the angles. We can substitute in the values to see that the identity is true. (I haven't done the last part :oops: )[/hide]


I have to do a presentation about the conic section hyperbola. My schoolbook has just a little information since it is not in the school program, i. e., it is not officially taught, it is optional. So, although I am searching in google, I would like to see your sugestions not only about sites to seek information but also about methods of presentation to make it more interesting. Thank you and sorry my english.

How about trying Wikipedia?  Google, in my opinion, [i]is[/i] a pretty good place to search.

http://mathworld.wolfram.com/

and do a search.


The sums of N terms of two arithmetic progressions are in the ratio (7N+2)/(N+4). Find the ratio of their 5th term.

Thanks!

You've posted this problem in two forums, when it need only be placed in one.  It has been answered in the other forum so I'm locking this post, next time, however, please post problems in only one forum.


This problem has been driving me nuts for a couple of days.

[img]http://www.gamepsychos.com/squarecircle.gif[/img]

The Blue Lines are 24cm long and the red one is 16cm long. The square is in the exact center of the circle.

What is the radius of the circle?

Sorry for your inconvenience but please delete it as a substitute for somebody because I cannot delete this post from me.

I don't think the side of the square is $12\sqrt{2}$ because the point where the blue line meets the circle isn't collinear with two of the vertices of the square.

Solution:

If the blue lines are parallel to the red line, they all incline at 45 degrees.
Let $2a$ be the length of the diagonal. We get $r=a+16$.
Draw lines from the centre to the both ends of the blue line.
These form a right-angled triangle with sides of length $a,24,r$.
So $a^{2}+24^{2}=r^{2}\Rightarrow a^{2}+24^{2}=(a+16)^{2}\Rightarrow a=10$.
Therefore, $r=26$.

Sorry.
When a right-angled isosceles triangle was made, I misunderstood it.

I introduce a solution in a thing with much effort.

$\cos \angle FGB = \frac{GB}{GF}= \frac{16}{8\sqrt{13}}=\frac{2}{\sqrt{13}}$
$\cos \angle OGH = \frac{GH}{OG}= \frac{4\sqrt{13}}{OG}$

Thus, $\cos \angle FGB = \cos \angle OGH \iff \frac{2}{\sqrt{13}}= \frac{4\sqrt{13}}{OG}\iff OG=26$

The answer that I made a mistake in not to misunderstand it deleted it.


vector
analytic geometry
linear algebra
linear algebra unsolved

S={ x e R3 / x1 + 2x2 - x3 = 0}. Find a base B from R3 that contains a base of S and a base of S orthogonal, and that the vector (0,5,-2) hast coordinates (0,1,1) in the base B.

I have come till here:

S=<(1,0,1)(0,1,2)>
S orthogonal=(-1,-2,1)

Putting these two together i get a Base, not B, lets call it Z.

Z={(1,0,1)(0,1,2)(-1,-2,1)}
Coordinates of (0,5,-2) in Z are (-2,1,-2).

How can i go on? Please help.

The only choice for the part of the basis orthogonal to $S$ is size- since our first try gave a coefficient of $-2$, let's multiply that one by $-2$; the third vector in the basis is now $(2,4,-2)$. We can find the second basis vector by subtracting the third from $v=(0,5,-2)$, since we know the coefficients of $v$ with respect to this basis. That makes the second basis vector $(-2,1,0)$. We have plenty of choice now for the first basis vector; anything in $S$ and not a multiple of $(-2,1,0)$ will work.


Okay, obviously I know my rank, but I have no idea what my score was at states and chapter both. How do you  people know your scores? Is there a website you can find it at? Forgive me if I'm ignorant. :lol:

Don't they post the answers up at your state competition?
They should also mail the scores to your school.
Chapter, I don't know. Didn't get my scores there.

The scores should have been given to your coach for both state and chapter.  If you want to know your scores, just ask him/her.

[quote="236factorial"]The scores were told to us by our mathcounts coach, and also, we got our test back too![/quote]

Getting the test back was quite unusual.  They do not give the tests back.  It is not that I do not believe you 236factorial  :) !

Every coach gets a scoresheet back.  Your coach should be able to tell you your score on the sprint, target and team rounds.  You can also know which problems you missed from this sheet.

I agree. usually caoches only get their scoresheets back, and tests are basically never given back, and if they did, it wold probably be given back after all the other state competitions have been finished. Prevents cheating, see?

Okay, thanks everybody. I'll have to talk to my coach and see if he received anything.


function
integration
calculus
real analysis
real analysis unsolved

Let $ f: R\minus{}>R$ be a continuous function with $ \int_x^y f(t)dt<\equal{}f(x\plus{}y)$ for any real $ x,y,$. Prove $ f\equal{}0$.

$ 0 \equal{} \int_x^x f(t)\; dt \leq f(2x)$ for all real $ x$,
thus $ f(x) \geq 0$ for all $ x$.
Suppose $ f$ has a minimum in $ x$,
then $ \int_{\frac{x}{2}\minus{}1}^{\frac{x}{2}\plus{}1}f(t)\; dt \leq f(x)$,
but the integration interval has length 2, and the function values in that interval are at least $ f(x)$,
thus the inequalty can only be fullfilled when $ f(t) \equal{} 0$ for $ t \in [\frac{x}{2}\minus{}1,\frac{x}{2}\plus{}1]$.
Now it is easy to see that the function is 0 for all arguments, by moving the $ x$-value to the left and the right.
If $ f$ does not have a minimum, then it has an infinum  $ \alpha$. Choose a very small $ \epsilon>0$.
Then the reasoning remains the same, when you start with a $ x$-value for which $ |f(x)\minus{}\alpha| < \epsilon$.


calculus
derivative
calculus computations

A man 5 ft tall walks at the rate 4 ft/sec directly away from a street light which is 20 ft above the street. At what rate is the length of this shadow changing?  I don't understand like What's the difference between if the question asks you the rate of the tip of his shadow changing and the length changing?  Hope you can give me some hint. thanks.

It's based on related rates, which obviously requires derivatives.  You need to set up an equation that describes the scenario (looks like a triangle will do), and then take the derivative for each term with respect to time.  By plugging in what you know, you can solve for what you don't.


analytic geometry
geometry
3D geometry
sphere
number theory
relatively prime
complex numbers

The first statement is a little off.  First, we can't represent $(3, 4, 5)$ by that formula; we need to allow the first two coordinates to be switched.  Second, even with that fix, we can't represent $(9, 12, 15)$; we need to handle nonprimitive triples with an extra parameter.

Well, I think I have fixed it, but either way I am referring to primitive Pythagorean triples.

Hmm now that I think about it maybe it should be moved to the Pre-Olympiad forum. If a moderator could move it please... :)

[quote="t0rajir0u"][hide="Cheap"] We will instead find the integer solutions to $a^{2}+b^{2}+c^{2}+d^{2}= e^{2}$ and then consider those solutions where one of $a, b, c, d$ is zero.

We will be making use of the Lipschitz quaternions, which are the quaternions with integer components (there is another name for them, but I don't recall).

We desire to find a Lipschitz quaternion whose norm is the square of an integer.  Hence, we square an arbitrary Lipschitz quaternion (because the norm is multiplicative):

$(x+yi+zj+wk)^{2}= (x^{2}-y^{2}-z^{2}-w^{2})+(2xy) i+(2xz) j+(2xw) k$

(I will spare you the details; quaternion multiplication is just a hassle.)  The norm of this quaternion is the square of an integer, that is,

$(x^{2}-y^{2}-z^{2}-w^{2})^{2}+(2xy)^{2}+(2xz)^{2}+(2xw)^{2}= \left( x^{2}+y^{2}+z^{2}+w^{2}\right)^{2}$

We want to set each of these to zero in turn.

The first term is troublesome, and suggests that we may not be able to generate all solutions in this way, since setting it to zero is equivalent to solving our original problem...

The other terms are symmetric, so WLOG we let $w = 0$.  Then

$(x^{2}-y^{2}-z^{2})^{2}+(2xy)^{2}+(2xz)^{2}= \left( x^{2}+y^{2}+z^{2}\right)^{2}$

Which should generate (if not all, at least a lot of) primitive solutions.  [/hide][/quote]

lol i was thinking that too...i love using difficult results to trivialize problems...anyway, i don't think there is any one solution, see:
[url]http://mathworld.wolfram.com/PythagoreanQuadruple.html[/url]

In retrospect, [hide="this identity..."] $(r_{0}^{2}-(r_{1}^{2}+r_{2}^{2}+...+r_{n}^{2}))^{2}+\sum_{k=1}^{n}\left( 2r_{0}r_{k}\right)^{2}= (r_{0}+(r_{1}^{2}+r_{2}^{2}+...+r_{n}^{2}))^{2}$ [/hide] is somewhat obvious from the given, and I really didn't need to go into the quaternions and stuff.  :|

By the way, an idea:  if we instead use the Hurwitz quaternions, which allow for $x, y, z, w$ to be half-integers, and then multiply by $2$, we may be able to generate some more solutions.

Edit:  [hide="Hmm, it is troublesome..."] The solutions that the Lipschitz quaternions miss occur when we let $x = \frac{a}{2}, y = \frac{b}{2}, z = \frac{c}{2}, w = \frac{d}{2}$ where $a, b, c, d$ are odd integers.  This produces, after the appropriate multiplication,

$\left( \frac{a^{2}-b^{2}-c^{2}-d^{2}}{2}\right)^{2}+(ab)^{2}+(ac)^{2}+(ad)^{2}= \left( \frac{a^{2}+b^{2}+c^{2}+d^{2}}{2}\right)^{2}$

Verify that when $a, b, c, d$ are all odd every component is an integer.

However, we run into a problem:  because of the odd condition, we cannot set any of the terms to zero except the first.

This is quite unsatisfactory.  All it tells us is that given an entirely odd solution $a, b, c, d$ to

$a^{2}= b^{2}+c^{2}+d^{2}$

We can generate another solution where at least two of the squares being added are odd (this is a case not covered by the Lipschitz-only formulae).

Unfortunately, entirely odd solutions don't exist.  Odd squares are always congruent to $1 \bmod 4$, so where the LHS is congruent to $1 \bmod 4$, the RHS would be congruent to $3 \bmod 4$.    [/hide]

Just a small remark: in any dimension, you can get all solutions by solving $x_{1}^{2}+...+x_{n}^{2}=1$ in rational numbers, thus look for rational points on the unit sphere. One can parametrise all solutions as the intersections of the sphere with lines through $(1,0,0,0,...,0)$ and the points $(0,a_{2},...,a_{n})$ with all $a_{i}$ rational.

[quote="chess64"]All triples of relatively prime positive integers $(a,b,c)$ (with $a$ even) that are solutions to the equation $a^{2}+b^{2}=c^{2}$ are given by the formula $(2rs,r^{2}-s^{2},r^{2}+s^{2})$ where $r$ and $s$ are positive integers with $r>s$.

Find a similar formula for the integer solutions to the equation $a^{2}+b^{2}+c^{2}=d^{2}$.[/quote]
Equation--->
d^2-c^2=a^2+b^2 --->
(d-c)(d+c)=a^2+b^2
Since if (a+bi)(c+di)=e+fi, then (a^2+b^2)(c^2+d^2)=e^2+f^2 by some basic binomial expansion.
Since complex numbers are closed over multiplication, that means that if d-c and d+c are the sum of two squares, then its cool.
The only restrictions on d-c and d+c is that they have the same parity and sum to 2d.
But we know that 2d is the sum of four squares by four square theorem because 2d is an integer.
So if d is any number, therefore, it can be done unless 2d is the sum of two squares.

Um, as for a generalized solution..... that looks like messy/ugly quaternion expansions because for some a and b for d that can be represented as such, GCD(a,b)>sqrt(2d/3) which lends itself to massive ugliness....
Oh and there should be no explicit thing for d besides that it includes everything except 2d=x^2+y^2 which implies d not being x^2+y^2 because 2 is the sum of two squares for some x and y and I doubt there exists an explicit formula for such numbers.

[quote="Scrambled"]So if d is any number, therefore, it can be done.[/quote]

This is true (without the "unless...") because $d^{2}= 0^{2}+0^{2}+d^{2}$.

oops
well i'm trying to ignore the solutions with 0's in them

Different problem.  Also much harder.  :|

sometimes there are compounded quadruples

like 3^2+4^2=5^2

5^2+12^2=13^2

therefore 3^2+4^2+12^2=13^2 and the result is a quadruple.

[hide="a full solution"]My Diophanine equations book gives this solution:

At least two of $a,b,c$ must be even; otherwise $d^{2}\equiv 2, 3 \mod 4$, which is impossible.  So let $a = 2x$ and $b = 2y$ for positive integers $x$ and $y$.  Also let $d = c+z$.  Then we have

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

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

We conclude that $z = 2t$ for some positive integer $t$.  Plugging this result into the above equation and solving for $c$ gives

$c = \frac{x^{2}+y^{2}-t^{2}}{t}.$

Now we conclude from $d = c+z$ that

$d = \frac{x^{2}+y^{2}+t^{2}}{t}.$

with the condition that $t | x^{2}+y^{2}, \ t \leq \sqrt{x^{2}+y^{2}}$.  We now have our full solution to the equation:

$a = 2x, b = 2y, c = \frac{x^{2}+y^{2}-t^{2}}{t}, d = \frac{x^{2}+y^{2}+t^{2}}{t}$

with $t | x^{2}+y^{2}, \ t \leq \sqrt{x^{2}+y^{2}}.$

We know this is the full solution to the system because we made no specific assumptions about the form of $a,b,c,d$ to come to our answer.  The solution can be verified by plugging into the original equation.[/hide]

[quote="t0rajir0u"]Cool solution.

[hide="Observation"] This is just the solution I gave divided by $x$, which implies that the quaternion solutions give [i]multiples[/i] of each primitive solution, and that all there is left to do is to divide out by a greatest common factor. [/hide][/quote]
Yeah, I noticed that too when I looked back at yours and compared it to my book's - it's always fun to see two very different solution methods turn up the same answer!


This is basically a split-off of the thread about religion, because we were arguing about a whole bunch of different issues at the same time. :D

Happy debating! :)

I don't care :no:, because i just want to live my life without debate  :rotfl:
debate=pointless (just kidding)

Historians have never proved god true or not, how would you expect to come close?

[quote="236factorial"][quote="hello"]no, there is no evidence for a supernatural spirit, why need to believe in something for which there is no evidence? Thats a indirect idea of my thuoghts.[/quote]

that's what I think. If there is a god, we can only know it in silence. People say praying is "talking to god" sometimes, but he never replies.  :([/quote]

*sigh* yes he does. God doesn't talk in the way that we do. :noo: When you pray for something to happen, then he makes it happen. But you have to realize, God isn't stupid. If you pray for something that might not benefit you, God will do something that is better for you. For example, if you pray for a car, and you get a bike instead, that might have been better. Maybe you would have died in a car accident the next day.

Let me give you an example. In 5th grade I was the worst player in my chess club. In 6th grade, I hadn't improved at all. In 7th grade, I started praying a lot for success in chess. That year, I won 2nd in the championships.  In 8th grade, I continued praying and I got 1st place. Nobody would have thought that I could win 1st place, or even place in the championships before I started praying. :)

[quote="236factorial"][quote="chess64"]. For example, if you pray for a car, and you get a bike instead, that might have been better. Maybe you would have died in a car accident the next day.[/quote]

You could've crashed into a tree in your bike or get ran over by a car  :rotfl:[/quote]

*sigh* you obviously don't understand. Maybe if you would have gotten your car then you would die in a car accident; only God knows. Then He will give you a bike instead b/c He knows what will happen.

The religious state of AoPSers is in a hopeless state. :noo:

[quote="236factorial"][quote="stupidkid"][quote="chess64"]The religious state of AoPSers is in a hopeless state. :noo:[/quote]
Eh, if you look at the poll on religion, most people here are religious with the most number of Christians.[/quote]

stupidkid is right, but in america, there is that freedom of religion, right?[/quote]

From experience, I can say that most Christians know less about Christianity than I do. ;)

edit: And I'm Muslim. :)

[quote="236factorial"]I hope chess64 isn't offended by that  :D[/quote]

hehehe I hope not, I'm a christain too and I find it highly amusing.

[quote="Sunny"]Hehehe that reminds me of a joke I heard,

"When we talk to God, it's called praying.  When God talks to us, it's called crazy."[/quote]

I also remember someone saying something like, "praying is talking to God, and meditation is listening to God."

Personally I don't believe in the existence of a higher being, but who knows?  I'll acknowledge the possibility and wait to find out for sure when the time comes, lol.  I also tried telling someone once that there's no proof of a God, and therefore there's no reason to believe in one.  The reply I got was, "that's why it's called faith."

[quote=tacowizard][quote]How do you know nothing can be absolutely good?
[/quote]

I'm so tired of my arguments being ignored, so here you go

[quote=tacowizard]Nope. I didn't say that God exists but is 'morally messed up'. The requirements for being God are being absolutely good and perfect. Even if someone or something is strong enough to be as strong as God is described, if they aren't absolutely good, they can't be God.

And since decisions can't be absolutely good and perfect as they always benefit one side and hurt others, and so are only good or bad relative to one side(eg. losing a war is bad for one country, but good for another) for the universe as a whole, God can't exist even if there is someone with as much strength as him because that person wouldn't be morally perfect as that can't exist as good and evil are all relative to individuals(literally my entire moral relativism argument) and so they can't be God.

If absolute good can't exist by moral relativism, then one of the requirements of God simply can't be fulfilled.

Once again, selfishness as a whole isn't a very large component of human nature in politics, as those who care about themselves care not for power but a luxurious life. Therefore, those who actually care about power want to do something with it. What those people want to do they believe is good, and what they hate, they think is evil. Everyone thinks that they are good, and that the enemy is evil, or at the very least less good than them.

Henceforth, good and evil are subjective relative to the universe absolutely. They can only be applied to groups. For example, if a country won a war, then that was good for the country. But to the country that lost, that was bad. We also can't use absolute good or evil in this case as some groups are benefited whereas others are harmed, and multiple frameworks exist in this case. Some argue for the greatest majority, whereas others argue for the least well off. We have no way of determining which framework is better here, because we can't determine 'betterness'.

Thus, absolute good can't exist, and as being absolutely good is a requirement of being God, God can't exist.[/quote][/quote]

Relative to who? To sides of the universe? To material effects? Those are still material sides and material actions, which God is separate from. Yes, some things might be good or bad momentarily, but ultimately, God is the only enjoyer, so if it all goes to God, it is still considered good.

Also, it wasn't that I ignored you. It was that I wasn't participating in the debate back when you posted that.

So you're saying that because God benefits from these things, God is absolutely good?

Good is based on the outcomes of decisions on groups and individuals. As every decision harms some groups and benefits others, no decision can be absolutely good, only good to some groups and evil to others. Frameworks like utilitarianism, Rawls's Theory, etc. aren't accurate as we have no way of determining whether or not which system is better than other systems as 'better' isn't defined, as its relative to good and bad.

Essentially, every hero is only a hero to those that they benefit, and every evildoer is only evil to those that they harm. Harm isn't limited to physical harm. By harm, I mean to conflict with the ideals of the individual or state(eg. a man reading a newspaper is harmed morally when he reads of a murderer, because murder is against his morals and a woman who was physically hurt is harmed morally as being hurt is against her morals), and benefiting is what benefits those morals.

In order for something to be absolutely good, there can't be any ambiguity. As no action can be universally good due to groups being hurt or benefited by the action, and if we accept the inaction of this 'God' as actions, this 'God' 's actions simply can' t be absolutely good as whatever his decisions are, they would always benefit some groups and hurt others.

A decision absolutely good simply isn't possible due to morally exclusive moral goals. For example, let's say that in the start of the Spanish Civil War in 1936, Francisco Franco decides that the democratic government is ineffective, and is composed of poor leadership that creates immense wealth gaps and alienates Spain. He then starts a coup to overthrow this democratic government, which is legitamately democratic. Franco's goal is to increase the material wealth of Spain, whereas the loyalist government's goal was to preserve the ideals of democracy. There obviously can't be any compromise here due to the value of government legitamacy. By the loyalist government's moral code, Franco having part of Spain means that part of Spain would have an illegitamate ruler, which by their legal code is injust. By Franco's code, the monarchy ought to be restored as it is the legitamate government, and so any trace of the Republic, which is incredibly ineffective at ruling, must be abolished as the Spanish are one people and so can't be seperated. Compromises are only created when both sides are benefited although possibly to lesser amounts, not when both sides are harmed when one side could benefit. And even if compromise can occur, it would be injust by both groups as their views are mutually exclusive.

As being absolutely good is a requirement of being God, even if some godlike being existed, they wouldn't be God even if they made the universe and were omnipotent.

[quote]Relative to who? To sides of the universe? To material effects?[/quote]

Not sure what you mean by this, but good and evil, as I already said, are relative to the in and out groups, or who were benefited(the in) and who are harmed(the out). There isn't any absolute good unless it benefits everyone's moral framework, which is impossible as mutually exclusive groups exist.

[quote=Doodles250]@SilverLightning Then what caused the Big Bang? The universe just randomly started existing with nothing to start it?[/quote]

The Big Bang is the moment that space-time came into existence. Before the Big Bang there was no space or time. So, it is actually meaningless to ask what caused the Big Bang to happen – there was no Universe in which that cause could have existed.

Quantum physics has shown us that some events have no cause at all. Things can happen randomly for no particular reason. This unpredictable and nature of the Universe is experimentally verified but it is a fundamental property of the Universe.

[quote=pezzeepenguin69]

Okay... that's the thing... you're viewing it from the stance that something must be sustaining the universe right now... If it isn't God, it is nothing? Make that make sense. I don't believe that we can definitively say that it's God, but that's what separates theists from atheists. Theists believe that God holds the energy that sustains the universe while atheists believe that nothing holds the universe in place. It just sustains itself in a circle of life. Idk. Idc. It doesn't matter.[/quote]

No I am not. Nothing is sustaining the universe. The universe is going to rip itself apart in around a googol years. The universe is inherently chaotic.

We have also found no life forms outside of Earth so how does a circle of life hold the universe in place???

[quote=shadowking2001][quote=tacowizard]To be honest it doesn't even matter if something made the Big Bang as there's a million different possible fictional and real deities that represent a 'first god' in their corresponding religions, and God only refers to 3 of them.

And then the Big Bang could have just been made literally by itself because the Big Bang could have been 'above reality' and then part of it leaked into the material universe-who knows.[/quote]

I'm not really sure. If a deity does not exist, can morals exist?[/quote]

Morals are an entirely evolutionary construct.

[quote=shadowking2001][quote=nsd08]

Morals are an entirely evolutionary construct.[/quote]

How so? How could morals evolve into existence?[/quote]

For example, consider the moral that murder is wrong.  Some people think murder is wrong and some think it is justified.  However, those who think it is justified are at an evolutionary disadvantage, as people would not want to be murdered no matter whether they think they are justified in murder.  Thus, they have less kids and spread their ideas less.  Thus, evolution by natural selection explains why everyone thinks murder is wrong.

[quote=shadowking2001]But its just as easy to say that God designed the world that way, so the morals would be there anyway.

To counter, why would people value human life?[/quote]

That is indeed a possible theory.  However, it is not the only theory.

People value human life because they value their own life (due to evolutionary pressure).  Thus, it is advantageous for everyone if people value life.

[quote=shadowking2001]But its just as easy to say that God designed the world that way, so the morals would be there anyway.

To counter, why would people value human life?[/quote]

No, because people thinking that murder is good is ultimately disadvantageous to the survival as a human race as a whole.

[quote=shadowking2001][quote=tacowizard]To be honest it doesn't even matter if something made the Big Bang as there's a million different possible fictional and real deities that represent a 'first god' in their corresponding religions, and God only refers to 3 of them.

And then the Big Bang could have just been made literally by itself because the Big Bang could have been 'above reality' and then part of it leaked into the material universe-who knows.[/quote]

I'm not really sure. If a deity does not exist, can morals exist?[/quote]

Please read his arguments...he posted numerous arguments.

[quote=shadowking2001]But its just as easy to say that God designed the world that way, so the morals would be there anyway.

To counter, why would people value human life?[/quote]

God couldn't make the world that way because God has to be absolutely good and having human life valued over other things causes suffering relative to groups such as animals, the environment, and saving certain groups of humans over others isn't absolutely good as one thing(human lives) are benefited whereas other aspects(another amount of human lives) are harmed. Compromise might occur, and a middle ground can be established sometimes, but there are other times when two groups want mutually exclusive things, such as if one group wants to abolish a dictatorship, whereas others see a monarchy as a legitamate government. To have a 'middle ground' would be unjust by both sides as half the country would be ruled by an illegitamate government. In this situation, its impossible to make an absolutely good decision as whatever decision made would help one group and hurt the other. As these situations exist, and there can't be absolutely good decisions made during them, God can't make absolutely good decisions and can only make decisions good for some groups and harmful to others, and therefore isn't absolutely good.

If God isn't absolutely good, then God isn't the 'God' as a requirement of being God is being the embodiment of good, which means God has to be absolutely good.

And no, we can't decide to simply look at the majority. For example, it's obviously wrong for a doctor to kill 1 person, steal all of his organs, and use them to save 3 patients.

And we can't decide to simply benefit the worst well off first. For example, it's obviously wrong for all rich people to be forced to donate all of their wealth to the poor, as they worked for that wealth and if that happened then nobody would want to work hard to become rich anymore.


algebra
polynomial
number theory proposed
number theory

Let $ \tau (k)$ denote the number of positive divisors of $ k$ and let $ \tau (0) \equal{} \infty$.

Let two nonconstant polynomials with integer coefficients be $ f(x)$ and $ g(x)$.  If $ \tau (f(n)) \equal{} \tau (g(n))$ for all integers, $ n$, then prove that $ f \equal{} \pm g$.

EDIT: You're right - sorry.  Problem edited.

As stated, $ f \equal{} \minus{}g$ works, as does $ f, g$ constant and equal to different integers with the same number of positive divisors.

Sorry for the delay; here's the solution:

[hide="Solution"]
Assume that $ f(n) \ne cg(n)$ for a constant $ c$.  Let $ f(n) \equal{} \prod_{i \equal{} 1}^k f_i (n)^{d_i}$ and $ g(n) \equal{} \prod_{i \equal{} 1}^l g_i (n)^{e_i}$, where the $ f_i$ and $ g_i$ are irreducible polynomials.  Note that, for an irreducible polynomial $ p$ and an arbitrary polynomial $ q$ so that $ p(x) \not | q(x)$, there are polynomials $ r(x)$ and $ s(x)$ so that $ pr (x) \plus{} qs (x)$ is constant.  Therefore, the of primes, $ K \equal{} \{k_1, k_2, k_3, \cdots , k_x \}$, so that some member of $ K$, say $ k$, dividing both numbers in the following possibilities is finite:

a) $ f_i (m), f_i' (m)$
b) $ f_i (m), g_j (m)$
c) $ f_i (m), f_j (m)$
d) $ g_i (m), g_j (m)$
e) $ g_i (m), g_i'(m)$
f) $ g_i (n)$ for all $ n$
g) $ f_i (n)$ for all $ n$

for some $ i, j$ (the solutions to f) and g) are clearly finite since $ f(0)$ and $ g(0)$ are finite).

[b]Lemma:[/b] Take an irreducible polynomial $ t(n) | g(n)$; there exists a prime $ q$ so the number of $ 1\le n\le N$ with the property that there exists a prime not in $ K$ that divides $ t(n)$ at least $ q \minus{} 1$ times is less than $ \frac {N}{l}$.

[i]Proof: [/i]For any prime, $ r$ sufficiently large, that divides $ t(n)$, there are at most $ \deg g$ solutions to $ t(n) \equiv 0\bmod r$ since $ t (n)$ is irreducible.  By Hensel's Lemma, the "lifted" solution to $ t (rt \plus{} n) \equiv 0\bmod r^2$ has precisely one solution for each solution of $ t (n) \equiv 0\bmod r$.  Therefore, there are at most $ \deg g_i$ solutions to $ t (n) \equiv 0\bmod r^{q \minus{} 1}$, and so for an extremely large $ N$, there are at most $ \sum_{q} \frac {N\deg g_i}{r^{q \minus{} 1}}$ solutions for $ n$ to $ g_i (n) \equiv 0\bmod r^{q \minus{} 1}$.  With $ q$ large enough, this sum will be less than $ \frac {N}{l}$.

For any $ k_i$, increase $ q$ even more so that $ g (n) \not \equiv 0\bmod k_i^{q \minus{} 1}$ has at least a solution.  Use the Chinese Remainder Theorem to obtain an $ n$ so that $ g (n) \equiv 0\bmod \prod_{i \equal{} 1}^x k_i^{q \minus{} 1}$ has no solution.  For all $ i$, take $ t_i(n) \equal{} g_i(Mn \plus{} y)$ so that $ Mn \plus{} y \equiv n\bmod \prod_{i \equal{} 1}^x k_i^{q \minus{} 1}$ and $ Mn \plus{} y \equiv m\bmod p^{r \plus{} 1}$, where $ p$ is a prime to be chosen later (but $ p > \max {k_i}$ so that the Chinese Remainder Theorem works) and $ r$ is a prime greater than $ q\max \{d_i \}\max \{e_i \}$.  Let $ g (Mn \plus{} y) \equal{} t(n)$.

By the Lemma, for any $ N$, less than $ \frac {N}{l}$ numbers $ n$ between $ 1$ and $ N$ let a $ t_i (n)$ have a prime factor that divides it at least $ q \minus{} 1$ times.  Hence, there is an $ n$ so that $ g(Mn \plus{} y)$ has no prime factor dividing it more than $ r \minus{} 1$ times.

Since $ f(m)$ and $ g(m)$ are not off by a constant, there exists an $ f_m (n)$ with $ 1\le m\le k$ that divides $ f$ a different number of times than it divides $ g$.    Let $ f_m (n)$ be $ h(n)$; say that $ h(n)^a \parallel{} f(n)$ and $ h(n)^b \parallel{} g(n)$ with $ a\ne b$.  Since there are infinitely many primes dividing $ h$, we can pick a prime $ p$ be larger than any of the $ k_i$ so that $ p$ divides $ h(n)$, but is not a factor of $ h'(n)$ for some $ n$.  By Hensel's Lemma, for any large integer $ k$, we may pick an $ m$ so that let $ p^k \parallel{} h(m)$, and it follows that $ p^{ak} \parallel{} f(m)$ and $ p^{bk} \parallel{} g(m)$.  Pick $ k$ so that $ ak\equiv \minus{} 1\bmod r$.  This implies that $ bk\not \equiv \minus{} 1 \bmod r$ as long as $ r > a \minus{} b$ (which is true since $ r > \max \{d_i\}\max \{e_i\})$.  Note that since $ p^{ak} \parallel{} f(n)$, $ r | (ak \plus{} 1) | \tau (f(m \plus{} p^{r \plus{} 1}\cdot n))$ for all $ n$.  Therefore, $ r | \tau (g(m \plus{} p^{r \plus{} 1}\cdot n))$ for all $ n$, so $ r | \tau (t(n))$ for all $ n$.  However, this means that there is a prime that divides $ t(n)$ at least $ r \minus{} 1$ times for all $ n$, which is a contradiction, so $ f(n) \equal{} cg(n)$ for some constant $ c$ and all integers $ n$.

Without loss of generality, let $ |c| \ge 1$.  Then, $ g(n) | f(n)$ for all $ n$.  Hence, if $ |g(n)| < |f(n)|$ for some $ n$, we would get that $ \tau (g(n)) < \tau (f(n))$ (indeed, $ f(n)$ contains all factors of $ g(n)$ and $ f(n)$ itself), which is impossible.  Hence, $ |c| \equal{} 1\implies c \equal{} \pm 1\implies f \equal{} \pm g$ as long as $ f$ is nonconstant.

[/hide]


algebra
polynomial
function
calculus
integration
Functional Analysis
real analysis

Hi all

I have been working on this problem for a while now, but haven't made too much progress:

Consider the Hilbert space L^2([0,1]) with usual inner product norm. We wish to apply the G-S process to the set {1,x,x^2,...} to get polynomials {P_0, P_1, P_2,...}. Determine a recurrence relation between P_n, P_(n+1) and P_(n+2) and then determine a general formula for P_n which can be expressed in terms of elementary functions and derivatives.

Now I haven't even got past the first part. I calculated the first few P_n, and hypothesized a recurrence relation which seems to fit: P_(n+1)' / sqrt(2n+3) = ((n-1)*4 +2)*P_n / sqrt(2n+1) + P_(n-1)' / sqrt(2n-1).

But I have been unable to prove this, and I have no idea how to start on the 2nd part. I'm hoping that someone has seen this before and can give me an idea of how to proceed. Please do not give a full solution as I want to try this myself (Or at least hide your solution), but if you could give me an idea on how to prove the recurrence relation or how to find the "closed" formula for part 2 that would be greatly appreciated, as I am quite stuck!

Thanks very much.

That recursion looks right.

These are Legendre polynomials. With the nice formula, you can prove orthogonality using integration by parts. The formula will also help to prove the recursion.

Thanks very much. I knew I had seen these polys somewhere before.


A bounded sequence $ (x_n)_{n\ge 1}$ of real numbers satisfies $ x_n \plus{} x_{n \plus{} 1} \ge 2x_{n \plus{} 2}$ for all $ n \ge 1$. Prove that this sequence has a finite limit.

$ x_n \plus{} x_{n \plus{} 1} \ge 2x_n \plus{} 2\to x_{n \plus{} 1}\ge x_n \plus{} 2\to x_n\ge 2n \plus{} x_0$ is not bounded.

I think the condition should be $ x_{n}\plus{}x_{n\plus{}1}>\equal{}2x_{n\plus{}2}$.

Thank you for pointed out my mistake. Yes, it must be $ x_n\plus{}x_{n\plus{}1}\ge 2x_{n\plus{}2}$. Thanks

We can prove by induction that : $ x_{n\plus{}m}\leq \frac{x_m\plus{}2x_{m\plus{}1}}3\plus{}\frac{x_{m\plus{}1}\minus{}x_m}3 (\minus{}\frac 12)^{n\minus{}1}$.The sequence $ x_n$ is bounded . Let $ I\leq S$ be the [b]Inferior,Superior limits[/b] of $ x_n$
Therefore there  are increasing sequences $ p(n),q(n)$ of natural numbers such that $ x_{p(n)}\rightarrow I, x_{q(n)}\rightarrow S$.
The sequence $ x_{p(n)\plus{}1}$ is bounded , so there is $ u(n)$ such that $ x_{p(u(n))\plus{}1}$ is convergent to $ L \leq S$
We can write $ x_{q(2p(u(n)))}\equal{}x_{q(2p(u(n)))\minus{}p(u(n)) \plus{}p(u(n))}\leq$$ \frac{x_{p(u(n))}\plus{}2x_{p(u(n))\plus{}1}}3\plus{}\frac{x_{p(u(n))\plus{}1}\minus{}x_{p(u(n))}}3 (\minus{}\frac 12)^{q(2p(u(n)))\minus{}p(n)\minus{}1}$
It's easy to see that $ q(n) \geq n$ , so  $ q(2p(u(n)))\geq 2p(u(n))$, so for $ n\rightarrow \plus{}\infty$ we obtain $ S \leq \frac{I\plus{}2L}3 \leq \frac{I\plus{}2S}3\Longrightarrow S\leq I$
Finally $ I\equal{}S$, and the sequence $ x_n$ is convergent .
 :cool:


algebra
polynomial
number theory unsolved
number theory

I think it's nice problem please help me to solve it.

No one will open a doc file. Post it here.


let  x,y,z>0
 1\3<=yx+ y z +z x  <=3

determine the range of values for
x y z
and
x+z+y

We know that:
$x^{2}+y^{2}+z^{2}\ge xy+yz+zx \implies$
$(x+y+z)^{2}\ge 3(xy+yz+zx) \ge 3\cdot\frac{1}{3}= 1$
But i can't get the maximum. If you let $x,y,z\ge 0$ then this maximum doesn't exist. If we let $xy+yz+zx$ fixed. $x+y+z$ goes to a maximum when $x^{2}+y^{2}+z^{2}$is maximal.
$1= \left(\frac{3}{3}\right)^{\frac{3}{2}}\ge \left(\frac{xy+yz+zx}{3}\right)^{\frac{3}{2}}\ge xyz$
I can't find a minimum.

Assume the maximum value of $x+y+z$ is $k$, so $x+y+z \leq k$ for every $x,y,z$ that satisfy the property from the problem.

Set $x=k+z, y=z=\frac{\sqrt{k^{2}+9}-k}{3}$
$xy+yz+zx=3z^{2}+2kz=3$

Then,$x+y+z=3z+k>k$. Contradiction. So $x+y+z$ has no maximum value.

Assume the minimum value of $xyz$ is $k$, so $xyz \geq k$ for every $x,y,z$ that satisfy the property from the problem.

Set $x=\frac{k}{3}, y=1, z=\frac{3-xy}{x+y}=\frac{4}{x+1}-1 < 4-1=3$. [$xy+yz+zx=3$]

Then, $xyz<k$. Contradiction. So $xyz$ has no minimum value.

I was thinking about this for the maximum and the minimum.
Take $x\ge1$ and $y=z=\frac{1}{x}$
Then we have that $xy+yz+zx= 1+\frac{1}{x^{2}}+1 \le 3$ so it satisfies the condition.
Now make $x\rightarrow \infty$ and we have  $x+y+z>x$ so it has no bound.
And $xyz=x\cdot\frac{1}{x^{2}}=\frac{1}{x}$ so for a very big $x$ we have that $xyz \rightarrow 0$ and this has it has no bound.


In a standard 52 card deck, I draw two cards, withcout replacement.  What is the probability that I draw atleast one spade and one king (the king of spades accounts for both)?

[quote="ckck"][quote="Art of Owna"]In a standard 52 card deck, I draw two cards.  What is the probability that I draw atleast one spade and one king (the king of spades accounts for both)?[/quote]
[hide]There are 13 spade cards. There are 4 king cards. Thats a total of 17. But we counted the king of spades twice, so subtract one. Then we have 16 to get. Then the probability is $\frac{16}{52}\cdot\frac{15}{51}$. Simplifying $\frac{4}{13}\cdot\frac{5}{17}$. The final answer is $\frac{20}{221}$. [/hide]
Is there any replacements?[/quote]

No replacements.  Ill change that.  I actually dont know the answer, I was just wondering what some solutions were.

EDIT: But wait, the question said atleast one king and atleast one spade, isnt your way only counting the ways without both a kind and a spade and subtracing that from one (beucase if I get a kind of hearts and a nine of hearts, that wouldnt count towards the required)

[hide]
I will solve by finding the opposite probability

16 of the cards are spades or kings
36 are neither so the number of ways of drawing neither spades nor kings is

$36*35$

Ways to draw spades, but no kings
$12*11$

Ways to draw kings, but no spades
$3*2$

Probability of not getting at least 1 spade and 1 king:
$\frac{(36)(35)+(12)(11)+(3)(2)}{(52)(51)}$

Probability of getting at least 1 spade and 1 king
$1-\frac{(36)(35)+(12)(11)+(3)(2)}{(52)(51)}$

don't feel like simplifying

I probably made a mistake somewhere.
[/hide]


Express the positive difference between $ \frac{3}{4}$ of $ \frac{3}{5}$ and 20 percent of 30 percent as a decimal.


Έστω $ a, b, c$ θετικοί πραγματικοί ώστε $ a \plus{} b \plus{} c \equal{} 3$

Να δείξετε την παρακάτω ανισότητα:

$ \frac{a(a \plus{} 1)}{b \plus{} 1} \plus{} \frac{b(b \plus{} 1)}{c \plus{} 1} \plus{} \frac{c(c \plus{} 1)}{a \plus{} 1}\geq 3$

Θέλω να δω έξυπνες λύσεις... :wink:

[color=red][Moderator edit: Ρε αδερφέ μάθε latex καμιά φορά  :D ][/color]

Μια σχετικά ικανοποιητική λύση

Από τον αγαπημένο μας BCS έχουμε
$ \sum\frac{a^2}{b\plus{}1}\geq\frac{(a\plus{}b\plus{}c)^2}{3\plus{}a\plus{}b\plus{}c}\equal{}\frac{3}{2}$

και $ \sum\frac{a}{b\plus{}1}\equal{}\sum\frac{a^2}{ab\plus{}a}\geq\frac{(a\plus{}b\plus{}c)^2}{ab\plus{}bc\plus{}ca\plus{}a\plus{}b\plus{}c}\geq\frac{3}{2}$

Προσθέτοντας we get a big QED  :)

Andreescu δηλαδη  :P

Δεν μου λες, εισαι σιγουρος οτι $ a\plus{}b\plus{}c\geq ab\plus{}bc\plus{}ca$ ?

Ωραίος Σιλουανέ αυτό είχα στο μυαλό μου. Mailo ισχύει $ (a\plus{}b\plus{}c)^2>\equal{}3(ab\plus{}bc\plus{}ca)$ :wink:

Βγαίνει και με το δικό μου αγαπημένο τρόπο:

από αναδιάταξη (θέτοντας και ότι $ 0 \le a \le b \le c$) είναι

$ \frac {a(a \plus{} 1)}{b \plus{} 1} \plus{} \frac {b(b \plus{} 1)}{c \plus{} 1} \plus{} \frac {c(c \plus{} 1)}{a \plus{} 1} \geq$

$ \frac {a(a \plus{} 1)}{a \plus{} 1} \plus{} \frac {b(b \plus{} 1)}{b \plus{} 1} \plus{} \frac {c(c \plus{} 1)}{c \plus{} 1} \equal{}$

$ a \plus{} b \plus{} c \equal{} 3$ q.e.d.  :P

maniopa den vgainei me anisotita tis anadiataksis kathws oi triades pou exeis an ypotheseis oti $ a<\equal{}b<\equal{}c$ den einai omoia diatetagmenes.
diladi exeis men $ a(a\plus{}1)<\equal{} b(b\plus{}1) <\equal{}c(c\plus{}1)$
alla gia tin alli triada exeis
$ 1/(c\plus{}1) <\equal{} 1/(b\plus{}1) <\equal{} 1/(a\plus{}1)$
kai oxi
$ 1/(b\plus{}1) <\equal{}1/(c\plus{}1)<\equal{} 1/(a\plus{}1)$   pou xrisimopoieis.

[quote="Athinaios"]maniopa den vgainei me anisotita tis anadiataksis kathws oi triades pou exeis an ypotheseis oti $ a < \equal{} b < \equal{} c$ den einai omoia diatetagmenes.
diladi exeis men $ a(a \plus{} 1) < \equal{} b(b \plus{} 1) < \equal{} c(c \plus{} 1)$
alla gia tin alli triada exeis
$ 1/(c \plus{} 1) < \equal{} 1/(b \plus{} 1) < \equal{} 1/(a \plus{} 1)$
kai oxi
$ 1/(b \plus{} 1) < \equal{} 1/(c \plus{} 1) < \equal{} 1/(a \plus{} 1)$   pou xrisimopoieis.[/quote]

Χρησιμοποιώ μια όχι τόσο γνωστή μορφή της ανισότητας:

αν οι n-άδες $ a,b$ έχουν αντίθετη διάταξη και $ c$ αναδιάταξη του b είναι $ a_1b_1 \plus{} a_2b_2... \le a_1c_1 \plus{} a_2c_2...$

τώρα είναι $ \frac {1}{a \plus{} 1} \ge \frac {1}{b \plus{} 1} \ge \frac {1}{c \plus{} 1}$

και $ a(a \plus{} 1) \le b(b \plus{} 1) \le c(c \plus{} 1)$ άρα εφαρμόζοντας αναδιάταξη όπως ανέφερα έχουμε ότι:

$ \frac {a(a \plus{} 1)}{a \plus{} 1} \plus{} \frac {b(b \plus{} 1)}{b \plus{} 1} \plus{} \frac {c(c \plus{} 1)}{c \plus{} 1} \le \frac {a(a \plus{} 1)}{b \plus{} 1} \plus{} \frac {b(b \plus{} 1)}{c \plus{} 1} \plus{} \frac {c(c \plus{} 1)}{a \plus{} 1}$

Στο ίδιο πνεύμα κινείται και η παρακάτω άσκηση:

αν $ a,b,c,d > 0$ να δειχτεί ότι

$ (\frac {(a \plus{} b)(c \plus{} d)}{ab \plus{} cd})^2 \plus{} (\frac {(b \plus{} c)(d \plus{} a)}{bc \plus{} ad})^2 \plus{} (\frac {(a \plus{} c)(b \plus{} d)}{ac \plus{} bd})^2 \ge 4$

[color=red]Edit: έκανα ένα λάθος πριν.[/color]


logarithms
LaTeX
algebra unsolved
algebra

Please Help :blush:

evaluate and prove log(2)^log(2)^log(2).......

thanks :)

$ \left( \left( \left( \left( \log 2 \right)^\log 2 \right)^\log 2 \right)^\log 2 \right)^\log 2^{\cdot^{\cdot^\cdot}}$ ?
or
$ \log2^\left( \log2^\left( \log2^\left( \log2^\left( \log 2 \right) \right) \right) \right)^{\cdot^{\cdot^\cdot}}$ ?

it's the second one. by the way, how did you type that equation

[quote="seraphim"]how did you type that equation[/quote]
This site can use $ \text{\LaTeX}$  :)

$ \log2^\left( \log2^\left( \log2^\left( \log2^\left( \log 2 \right) \right) \right) \right)^{\cdot^{\cdot^\cdot}}$
$ \Rightarrow x = (\log 2)^x$

i got the x- log(2)^x = 0 part, just that how would you evaluate the equation algeraically and prove the solution other than by using a calculator?


What is the remainder when $5^{999,999}$ is divided by 7?

[hide]These kinds of problems often come up on mathcounts.

We want to find $5^{999,999} (\text{ mod }7)$

So, if we look at the cycles of $5^x$ (mod 7),

5 *5=25=
4 *5=20=
6 *5=30=
2 *5=10=
3 *5=15=
1 *5=5=
5

So there is a cycle of 6 remainders, 546231.

so: $5^{999,999}$ is the same as $5^{999,999\text{ (mod 6)}}$

and $999,999 \text{ (mod 6)} = 3$

so: the remainder is the same as that of $5^3$ which is $\boxed{6}$

[/hide]

EDIT: solution hidden

I did it that way too and found no pattern. must've messed up in my calculations  :?
this is #29 on the '92-'93 National sprint round.

What's Fermat?

i think he means fermat's little theroem

look it up on mathworld or...

$A^{p-1}=1\bmod p$  where $p$ is a prime and $A$ is a number not divisible by $p$

that helps but it doesn't solve the problem on its own. The only thing that that can tell us (as far as i can tell) is that there are 6 different remainders possible.

[hide]We need to find $5^{999,999} \pmod {7}$.

From Fermat's little theorem, we have $5^{6} \equiv 1 \pmod {7}$.

$999999 \equiv 3 \pmod {6}$, so $5^3 \equiv 6 \pmod {7}$[/hide]

I realized after posting it that it's basically the same thing you did. $5^3\equiv 6\bmod {7}$, so we have $5^{999,999}\equiv 6^{333,333}\bmod{7}$, and go from there. ;)

[quote="chess64"]That's nice too...
[size=67]but I still like my way better. :D [/size][/quote]Let's vote. I think JesusFreak's is little nicer, sorry. :P After all you just applied Fermat in the normal fashion.

guys...[hide="stop spamming"]STOP SPAMMING!!![/hide]

please dont get off topic or make useless posts like "interesting problem"


AMC AIME USA(J)MO USAMO geometry USAMTS Support

I'm sure a lot of people will disagree with me for saying this, or for being so blunt, but I've never been more embarrassed about the USA's top mathematics students.

First, it was "Boo hoo.  I should get to take the AIME twice, because I had a bad day."

Now, it's "Boo hoo.  I shouldn't have to take the AMC or the AIME.  I'm too good for them.  I already qualified for MOSP, once."

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=138675]Fifteen forum pages of whining[/url], and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=139901]growing[/url]!  Good grief!

It's not all MOSPers, but to those who have made such claims, you are an embarrassment.

You are given a free summer math program to learn advanced mathematics.  (MOSP)  Then you are given a free seven-month online program to learn more advanced mathematics.  (WOOT)  Then you think you are so much better than everyone else that you can sit on your lazy derrieres.

Excuse me.  But if you can't qualify for the AIME, you are NOT a well-rounded mathematician, and you don't deserve to be considered for one of the mathematicians that will receive free training or one of the six student-mathematicians that will represent the United States of America at an International Mathematical Olympiad.

Excuse me again.  But if you can't qualify for the USAMO, you are NOT a well-rounded mathematician, and you don't deserve to be considered for one of the mathematicians that will receive free training or one of the six mathematicians that will represent the USA at an IMO.

Don't get me wrong.  I understand the AIME and the USAMO are different kinds of tests.  I happen to think you should be able to do well on *both* in order to qualify for MOSP and to represent the USA at the IMO.  I also have been supportive of the USAMO expansion.  However, the main effect of the expansion seems to have been to increase the number of whiners on the bubble.  To those who are whining.  Stop whining.  You're an embarrassment.

To those who have gone to MOSP and are whining now, stop whining.  If you can't qualify for USAMO, even with all the advantages you've been given, step aside gracefully, and let the more talented and well-rounded students have a crack at it.

To all those involved with writing and administering and scoring the AMC contests, I say "Thank you!"  I am a coach who has had a few students qualify for the AIME, and one student qualify for MOSP.  I have known, and perhaps assisted in small ways, others who have qualified for AIME, USAMO, and MOSP.  Thank you for spending your time and your energy and your money to produce the AMC series of tests and beyond.  You don't deserve the unprecedented negative karma that's been directed toward your series of contests by some of the very students you seek to serve.

[quote="rcv"]
Excuse me.  But if you can't qualify for the AIME, you are NOT a well-rounded mathematician, and you don't deserve to be considered for one of the mathematicians that will receive free training or one of the six student-mathematicians that will represent the United States of America at an International Mathematical Olympiad.[/quote]

What about if you're a very deep thinker, but not so fast?

Is ANY of this based on actual reason/evidence, or is everything just an opinion, which you base on to insult a bunch of random people?

Cheers.

Yes, there is actual reason/evidence. This is a problem that I've pointed out to several of my friends and we all had a good laugh about it. There is WAY TOO MUCH whining on this forum nowadays.

I'll quote two people (not whiners) who described the situation the best, though it may be strange taking it out of context. Also, I believe the evidence takes the form of those 15 pages in that link.

[quote="white_horse_king88"]Aka: Compile a master list of all the ways people "ought" to be allowed to qualify for the USAMO.

I agree with the USAMTs thing.
Agreed with the Blue/Black Mop thing.
Also Red Mop. That shouldn't be a problem.
Also those that placed individually on the HMMT. Those problems are almost like AIME problems.
Also people who think the AIME screwed them over.
And like ARML top ten and Mu Alpha Theta and stuff.
There's probably more I missed.[/quote]

It case you missed it, that was heavy sarcasm. You know it's bad when Mu Alpha Theta is called as a selection factor for the USAMO  :lol:

[quote="Ignite168"]
I dislike the fact that when the AMC writers set an obstacle, even if it's plain annoying computation, that people try to change that obstacle, rather than working to overcome it by simple methods like checking your work extremely closely and such. It seems to go against the spirit of "Art of Problem Solving."[/quote]

Thus, if you're a deep thinker, but you're not fast, get fast. You don't need to be that fast for the AMC/AIME anyway. You just need to not be slow.

-end of quotes-

If you want to take the USAMO or go to MOSP, you should have to take the AMC/AIME because it is only fair for everyone. Also, the system has shown itself to work, with very few outliers, year after year. Like I mentioned earlier in another thread, the only major problems we've seen have only began occurring after the expansion, which was a noble effort but too many of you have taken it for granted quickly and are now expecting more.

[quote]Fifteen forum pages of whining, and growing! Good grief! [/quote]
Well half of it is people refuting the whining but yeah ok.

[quote="rcv"]Don't get me wrong.  I understand the AIME and the USAMO are different kinds of tests.  I happen to think you should be able to do well on *both* in order to qualify for MOSP and to represent the USA at the IMO. [/quote]

Since you seem to be attacking me personally in claiming that I wish to sit upon my "lazy derriere" (interesting accusation, considering my work ethic), I will take the time to respond once by pointing out the obvious flaw in your logic.

Olympiad mathematics is tested by the USAMO, not the AIME. The AIME is simply easier to grade. Your second statement above is wholly unjustified.

Eh you don't need to be a "well-rounded" mathematician to do well on IMO, since being able to solve problems very quickly and make no computation error is not tested on IMO.

Also, no MoPpers whined about the system because they didn't qualify for USAMO this year; in fact, most people who like this idea are non-MoPpers.

Dude, I agree with the "what if you're a deep thinker, but not fast" point, and I'd also like to add "what if you're bad at really computation heavy math, but can do the really deep stuff," because both describe me.

I'm not saying that I wouldv'e made MOSP or IMO this year, but I'd feel fairly confident in saying that if I had the opportunity to take USAMO this year, I'd at least come close (unless all 6 problems were geometry or something, which wouldn't happen anyways).

The point is not that they wanna be lazy, but that they wanna spend their time doing real math, and practicing for silly AMC and AIME stuff, which matters more whether you're good at taking them than whether you can do deep math.

Also, notice that they're not acting "better than everyone" by asking for a free pass to MOSP/IMO again.  Rather, they want to be guaranteed to have the chance to COMPETE FOR THEIR SPOT AGAIN, among all of the others trying out.

This feels like more of a rant, opinion, or personal thought post than that 15 page thread that you mentioned.

As Arnav said earlier, you seem to be directly attacking certain individuals. Might I mention, these individuals are discussing ways of possibly changing the AMC structure. This discussion is quite furtile, and I think it is much more substantial and meaningful than your personal embarrassment.

[quote="PenguinIntegral"]I personally think all these problems would have been avoided had the problems not sucked.[/quote]

Amen to that, my brother.

[quote="PenguinIntegral"]I personally think all these problems would have been avoided had the problems not sucked.[/quote]

Is it just me, or is saying you did badly because of bad problems just more complaining?  The problems were written to test if you could do them.  Therefore they cannot possibly be bad, if you got it right, you can do them, if you didn't, you can't, and that's what the test is for in the first place.

This is probably due to the fact that I didn't fail the AIME as badly a most people, but I'm getting very annoyed with the complaining as well. Thank you rcv for bringing this up!

Yeah, this completely ignores the fact that the person who brought up the possibility of automatically allowing MOPpers back into the USAMO in the first place (me) is not a MOPper and will almost certainly never be a MOPper. This is not about MOPpers being better than everyone else. As I mentioned before, if you don't like saying that they automatically "qualify" for the USAMO that one year, you can always have a separate list of MOPpers who didn't "qualify" but are "invited" to take the USAMO. Such a rule just very slightly increases the reward for people who score extremely high on the USAMO, and I think it's a justified increase. It's not like it would be "once in, always in". A MOPper would only be automatically allowed into the USAMO for one year--if they want to get in automatically again, they have to make MOP again. It would not hurt non-MOP-level contestants at all--we have as many USAMO spots to compete for as ever. The AMC people would only need to add maybe 5 spots at the most to deal with the very, very few people who have proven that they are great at the USAMO and weak on the AIME.

Also, I want to clarify that I, for one, do not blame the test, the writers, or the organizers for anything. I'm very, very thankful that they make the effort to give the AMC tests every year, and I don't think the issues this particular AIME raised could have been predicted in advance or prevented. But I also don't think that should stop me or anyone else from suggesting possible rule changes if we really feel that they would make the tests more fair.

Thank you thank you thank you rcv for taking a stand on this! But guys, as you seem to insist on complaining wherever you get a chance, please note what rcv pointed out at the top of this thread. There is already a 15-page haven of whining, so do it there.

I agree with the topic of this thread, but not rcv's reason for choosing that topic.  It is an embarassment.  I am not saying that the supposed "whining" is an embarassment, but that we have a SIX-PAGE THREAD THAT CONSISTS ALMOST ENTIRELY OF TWO FACTIONS SLINGING MUD AT EACH OTHER.  Art of Problem Solving should be a haven of young mathematicians, not of a mudslinging contest to rival recent elections.  Maybe this AIME was worse than normal, maybe not.  My personal inclination is toward the former, but that is not the point here.  The point is that we should be able to hold reasonable discussions, with out dividing into two factions bashing each other over six pages.  In less than 48 hours, I might add.  Please, everyone, grow up.  I expect better of America's best future mathematicians.

It seems to me that the central augument of the complaining is that AIME is for selecting USAMO, and since USAMO focuses on proof type problems, therefore any significant computation in AIME would make a bad AIME.  However, that does not seem to be what the AMC says.  Here is a copy of what AMC says on their website:

The American Mathematics Competitions (AMC) is dedicated to the goal of strengthening the mathematical capabilities of our nation’s youth. We believe that one way to meet this goal is to identify, recognize and reward excellence in mathematics through a serics of national contests called the:

    * American Mathematics Contest 8 (AMC 8),
    * American Mathematics Contest 10 (AMC 10),
    * American Mathematics Contest 12 (AMC 12),
    * American Invitational Mathematics Examination (AIME), and
    * United States of America Mathematical Olympiad (USAMO).

My understanding of this is that either AIME or USAMO is one in a series of exams that serve the same purpose: strengthing the mathematical capabilities of our nations's youth.  It goes without saying that certain amount of computational skill is very important.  So either intentionally or unintentionally, the relatively more significant computational skill this AIME requires might be just the right kind of wakeup call.  I've been a scientist and an engineer, and my own experience told me that the ability of dealing with numbers in various ways without much error is very very important even in our computer-dominated era, unless probably you'll be a pure theretical mathematician.  How many of you will be pure theretical mathematician?  I don't think there'll be many.

I hope I understand it correctly.

I think that merely acting considerately suffices.  We are entitled to express our opinions politely.

I also note that few of last year's MOPpers commented on the offending fifteen-page thread, fewer of whom were complaining.

I also think that it is difficult for those not accustomed to olympiad problems to realize the relation between olympiad and AIME problems, but I surmise that most such people have been to MOP.  However, I have been to MOP, so I might be biased.

[quote]There's no good response to: you are a conservative bigot who does not favor change.[/quote]
What...? I didn't say that. I thought your argument sounded like you thought the tweaks people are suggesting are bad partly because they're tweaks. If that's not what you meant, I apologize...but I didn't call you conservative, a bigot, or anything remotely similar to either of the two.
[quote]Alright... we have lots of people here to like change. That's cool. And by the way, slippery slope is when the outcome does not match the trend. In this case, the fact that the trend in opinion shows complete favoring of expansion of qualification criteria and reduction of competitiveness for the... uh... "ideologically correct purpose of mathematics"... is definitely a trend. Not slippery slope.[/quote]
Okay. Do you dislike the trend? Personally, I don't. Nobody's going to say, "Oh, it's not impressive that you made the USAMO anymore because 500ish other people in the entire nation did too!"

By the way, I loved #10. My reaction was basically, "I SOLVED A PROBLEM THAT THE 2-TIME BLUE MOPPER AT MY SCHOOL DIDN'T!" I wish the AIME tested nothing but combinatorics :lol:

[quote="solafidefarms"]All right, we've gone through the mudslinging. This is the [i]American[/i] Invitational Math Exam, so we should abide by the democratic principles of the United States. [Please don't argue that the US isn't a democracy :P.] Let's hold an election to determine whether we should ask for a change or not!

[That's only half in jest; we need some solution that everyone will abide by on here.][/quote]
What's a democracy if <1% of the people will know about the vote?

I'm sorry to burst everyone's bubbles but AoPS does not represent the American Math World.

I think the fact that this thread is now six pages long only adds to the embarrassment that rcv mentioned in the very first post. And now I'm contributing to it!

You know what would be REALLY funny... if the AMC people wrote this AIME specifically to weed out people who (1) don't read carefully and (2) can't compute quickly and accurately. Maybe that's what they wanted; it's their competition, so they get whatever they want. Sure you can say it wasn't good, but maybe they think it's a good test, so how can you argue with that?

Personally, I did reasonably well on the AIME. So maybe I'm just saying this because of that. But really, this AIME was not that bad of a test. A fair number of the questions had merit, and a couple of them didn't. It happens.

um. . . to penguin. . . "counting" is not the same as "computation". . . i thought the most confusing part of question 10 was finding a way to count every solution without overcounting anything, but that's not computation. after the test, my father (university math professor) went through the test, did all the problems, and he thought the test as a whole was [b]fine[/b].

Eh, I thought the test was actually pretty good. Perhaps it wasn't to your taste, but I'm sure that for the most part, it was about the same as the AIMEs in the past. Since it made it past the rigorous editing, I'm sure we can safely assume that the test was well-written.

This isn't Mu Alpha Theta guys, there aren't so many problems and these tests are as close as you will ever get to "good" math tests. Be thankful that you have it and don't take it for granted.

Past tests aren't always the same as the current year's tests.

It will be okay guys.

Just look at the difference from HMMT '05 to HMMT '06

[quote="chess64"]You guys are acting like this is the first AIME I've ever taken. I've taken the AIME before, and AIME 2007I was not like [i]any[/i] of the previous exams, all the way back to 1983.[/quote]

Well, what do you expect? They're not going to throw the same problems at you again (actually, #11 is nearly identical to AIME 1995 # 13).

[quote="Phelpedo"]Well, what do you expect? They're not going to throw the same problems at you again (actually, #11 is nearly identical to AIME 1995 # 13).[/quote]

Wow... #10 was also extremely similar to an old AIME question (except it was $4 \times 4$ before). I have to say, that's kind of lame.

mysticterminator, what a baller

he sums it up so well:

[quote]and it's really really hot in my room So take off all your clothes I am gettin so hot,(uh,uh,uh,OO) I wanna take my clothes off (oo)[/quote]

[/quote]


$ \frac {1}{x \minus{} y} \plus{} \frac {1}{y \minus{} z} \plus{} \frac {1}{z \minus{} x} \equal{} \frac {3}{2}$ is true for some real numbers $ x,y,z$. What value may expression $ \frac {1}{(x \minus{} y)^2} \plus{} \frac {1}{(y \minus{} z)^2} \plus{} \frac {1}{(z \minus{} x)^2}$ have?

$ \frac{1}{x\minus{}y}\plus{}\frac{1}{y\minus{}z}\plus{}\frac{1}{z\minus{}x}\equal{}A$
$ \frac{1}{{x\minus{}y}^2}\plus{}\frac{1}{{y\minus{}z}^2}\plus{}\frac{1}{{z\minus{}x}^2}\equal{}B$
$ x\minus{}y\equal{}a, y\minus{}z\equal{}b, z\minus{}x\equal{}c$
$ A^2\equal{}B\plus{}2(\frac{1}{ab}\plus{}\frac{1}{ac}\plus{}\frac{1}{bc})\equal{}B\plus{}\frac{2}{abc}(a\plus{}b\plus{}c)\equal{}B$ because $ a\plus{}b\plus{}c\equal{}0$.
So $ B\equal{}\frac{9}{4}$


geometry
algebra
convex polygon
perimeter
geometric inequality
IMO
IMO 1984

Let $ d$ be the sum of the lengths of all the diagonals of a plane convex polygon with $ n$ vertices (where $ n>3$). Let $ p$ be its perimeter. Prove that:
\[ n\minus{}3<{2d\over p}<\Bigl[{n\over2}\Bigr]\cdot\Bigl[{n\plus{}1\over 2}\Bigr]\minus{}2,\]
where $ [x]$ denotes the greatest integer not exceeding $ x$.

Consider all the pairs of non-adjacent sides in the polygon. Each pair uniquely determines two diagonals which intersect in an interior point. The sum of these two diagonals is greater than the sum of the two sides (triangle inequality). Each side will appear in a pair $ n\minus{}3$ times. And each time the side appears it also includes two diagonals. Thus, $ (n\minus{}3)p < 2d$.

Note that for any diagonal, its length is less than the sum of the sides on either side of it (triangle inequality). Consider all the diagonals from a fixed vertex and using the least number of sides in each application of the triangle inequality. We can sum everything up over all the diagonals considering the two cases (even or odd) to get the result.


[quote="duc 95"]A quite difficult inequality[/quote]

sorry

[quote="duc 95"]I cannot solve it now . can you help me?[/quote]

difficult inequality


induction
LaTeX
number theory proposed
number theory

Let $ p > 3$ be positive integer number.  $ x_1,x_2$ be roots of an equation $ x^2 \minus{} px \plus{} 1 \equal{} 0$.Denote $ a_n \equal{} x_1^n \plus{} x_2^n$

Prove that $ (p \minus{} 1)|a_n$ is not true

\[ x_1^2 \plus{} x_2^2 \equal{} (x_1 \plus{} x_2)^2 \minus{} 2x_1x_2 \equal{} p^2 \minus{} 2\]

\[ \implies p \minus{} 1\nmid p^2 \minus{} 2\]
which is true for ($ p\in N$)

nice solution shoki. i guess thats what victory.US meant. else its trivial.

[quote="shoki"]u mean $ a_n \equal{} x_1^n \plus{} x_2^n$,no?
we have $ a_n \equal{} a_{n \minus{} 1}.(x_1 \plus{} x_2) \minus{} x_1x_2(a_{n \minus{} 2}) \implies a_n \equal{} a_{n \minus{} 1}(p) \minus{} a_{n \minus{} 2},a_1 \equal{} p,a_0 \equal{} 2$
we prove it by induction,
let's suppose that
$ p \minus{} 1 | a_n \implies p \minus{} 1| a_{n \minus{} 1} \minus{} a_{n \minus{} 2} \equal{} (p \minus{} 1)a_{n \minus{} 2} \minus{} a_{n \minus{} 3} \implies p \minus{} 1 | a_{n \minus{} 3}$
but we know that $ p \minus{} 1 \nmid a_{n \minus{} 3}$.
so we have a contradiction.
(note that at first we must prove $ p \minus{} 1 \nmid a_1,a_2,a_3$. which is easy)[/quote]

yes, nice solution  :)  :) , it seem like my way. [hide]i had a mistake in latex and i edited  :oops:[/hide]


From numbers $ 1,2,3,...,37$ we randomly choose 10 numbers. Prove that among these exist four distinct numbers, such that sum of two of them equals to the sum of other two.

There are $ \binom{10}{2}\equal{}45$ pairs of elements from the drawn set (let's call it $ S$).
Since there are 36 possible non-negative differences of numbers between 1 and 37, there are 9 coincidences of the form $ a\minus{}b\equal{}c\minus{}d$ with $ a,b,c,d\in S$.
If all four numbers in such a coincidence are different, then (adding $ b$,$ d$ to both sides) we are done with the proof already.
So let us assume that all of these coincidences are of the form $ a\minus{}b\equal{}c\minus{}a$ for some $ a,b,c\in S$
Let $ M$ be the maximum and $ m$ be the minimum numbers from $ S$.
Notice that $ c\minus{}M$ and $ m\minus{}b$ cannot yield positive differences when $ c,b\in S$, hence numbers $ a$ in $ a\minus{}b\equal{}c\minus{}d$ must be all from the remaining 8 elements of $ S$.
Since there are 9 coincidences, for some two of them we have the same $ a$, i.e. $ a\minus{}b_1\equal{}c_1\minus{}a$ and $ a\minus{}b_2\equal{}c_2\minus{}a$.
Substracting both equalities we have $ b_2\plus{}c_2\equal{}c_1\plus{}b_1$.  It is east to notice that $ b_1,b_2,c_1,c_2$ must all be different.

best regards,
Wojtek


Let a trianlge  $ABC$ ,  the bisector $AD$ of its angle $A$ , the midpoint $M$  of side $BC$ and $E$ the intersection point of the circumcircles of triangles $ABC$ and $ADM$. Prove that $EA \perp AD$.

  [u] Babis [/u]

 [i]I think it deserves to have all solutions of this classical problem.I have seen this problem as a subquestion in a Mandelbrot Competition.[/i]

This is an easy, but cute problem. Call G the intersection of $AD$ with the circumcircle of $\Delta ABC$. Call $E'$ the intersection point of $GM$ with the same circumcircle.

[u][b]Solution 1:[/b][/u] Now, it's obvious that $\angle ACG = \angle ADB$, so $AE'MD$ is a cyclic quadrangle. So it follows that $E'=E$, and $EA \perp AD$

[u][b]Solution 2:[/b][/u] $\Delta DMG$ and $\Delta E'AG$ are similar, so $\left|GM\right|.\left|GE'\right| = \left|GD\right|.\left|GA\right|$ and $AE'MD$ is a cyclic quadrangle. Again, $E=E'$ and we are done.


algebra
polynomial
inequalities
triangle inequality

Given a quartic polynomial $ f(x)=x^{4}+ax^{3}+bx^{2}+cx+d$ with $ 4$ real roots and $ |f(i)|=1$, prove that $ a=b=c=d=0$.

[quote="t0rajir0u"][hide="Well..."] $ |f(i)| = |(i-p)(i-q)(i-r)(i-s)| \ge 1$ by Triangle Inequality, where $ p, q, r, s$ are the roots.  The minimum occurs when $ p = q = r = s = 0$. [/hide][/quote]

Very nice proof, but

[hide]is that really Triangle Inequality?  It looks like you're just using $ |ab|=|a\parallel b|$.[/hide]

Yeah, I thought triangle inequality is just |a+b|<=|a|+|b|

But the proof still works.

Hmm.  It's not actually triangle inequality, sorry.  $ |i-p| \ge |i| = 1$ follows because $ i-p$ is the hypotenuse of a right triangle with sides $ i, p$, so I suppose in some sense it would just be trivial inequality.


$ x\geq y\geq 1$ prove:

$ \frac{x\minus{}y}{\sqrt{x\plus{}y}}\plus{}\frac{y\minus{}1}{\sqrt{y\plus{}1}}\plus{}\frac{1\minus{}x}{\sqrt{x\plus{}1}}\geq 0$

$ \to (x \minus{} 1) \frac {\sqrt {x \plus{} 1} \minus{} \sqrt {x \plus{} y}}{\sqrt {(x \plus{} 1)(x \plus{} y)}} \plus{} (y \minus{} 1) \frac {\sqrt {x \plus{} y} \minus{} \sqrt {y \plus{} 1}}{\sqrt {(x \plus{} y)(y \plus{} 1)}} \geq 0$
or $ \frac {(x \minus{} 1)(1 \minus{} y)}{\sqrt {(x \plus{} 1)(x \plus{} y)}.(\sqrt {x \plus{} 1} \plus{} \sqrt {x \plus{} y})} \plus{} \frac {(x \minus{} 1)(y \minus{} 1)}{\sqrt {(y \plus{} 1)(x \plus{} y)}.(\sqrt {y \plus{} 1} \plus{} \sqrt {x \plus{} y})} \geq 0$
later
$ \frac {(x \minus{} 1)(y \minus{} 1)}{\sqrt {x \plus{} y}} . \frac {x \minus{} y \plus{} \sqrt {x \plus{} y}(\sqrt {x \plus{} 1} \minus{} \sqrt {y \plus{} 1})}{\sqrt {(x \plus{} 1)(y \plus{} 1)}.(\sqrt {x \plus{} y} \plus{} \sqrt {x \plus{} 1}).(\sqrt {x \plus{} y} \plus{} \sqrt {y \plus{} 1})} \geq 0$
obvious true

[quote="zaya_yc"]$ x\geq y\geq 1$ prove:

$ \frac {x \minus{} y}{\sqrt {x \plus{} y}} \plus{} \frac {y \minus{} 1}{\sqrt {y \plus{} 1}} \plus{} \frac {1 \minus{} x}{\sqrt {x \plus{} 1}}\geq 0$[/quote]
Setting $ 2a^2 \equal{} y \plus{} 1,2b^2 \equal{} x \plus{} 1,2c^2 \equal{} x \plus{} y$ $ (a,b,c > 0)$, then $ a \le b \le c$ and
\[ x \equal{} b^2 \plus{} c^2 \minus{} a^2,y \equal{} c^2 \plus{} a^2 \minus{} b^2,1 \equal{} a^2 \plus{} b^2 \minus{} c^2
\]
And the inequality becomes
\[ \sum \frac {(b^2 \plus{} c^2 \minus{} a^2) \minus{} (c^2 \plus{} a^2 \minus{} b^2)}{\sqrt {2c^2}} \ge 0
\]

\[ \Leftrightarrow \sum \frac {b^2 \minus{} a^2}{c} \ge 0
\]

\[ \Leftrightarrow \sum ab(b^2 \minus{} a^2) \ge 0
\]

\[ \Leftrightarrow (c \minus{} b)(c \minus{} a)(b \minus{} a) (a \plus{} b \plus{} c)\ge 0
\]
which is true. :)