calculus conics hyperbola MATHCOUNTS Asymptote algebra function

This is a countdown problem from this year: I believe they said on TV that it isn't hard if you have calculus lol.

If $y=\frac{3x}{x-2}$ and x is allowed to be any real number other than 2, what is the only value that y cannot be?

I came up with a solution, but if you guys would tell me if there is a better one I would appreciate it.

[hide="my solution"]
since the equation will be graphed as a hyperbola, we know that the value that it will not be will be "reached" when x = $\infty \text{ or }-\infty$ Plugging in one of those values would go as follows:

since $\infty - 2 = \infty$, we have $y=\frac{3\cdot\infty}{\infty}$ which equals $\boxed3$. This is the right answer, but as I said before, if there is a quicker way, I would like to know it.
[/hide]

[quote="chess64"][quote="math92"]i don't think there is a quicker way to do that problem[/quote]

???? didn't you just see nolachrymose's solution?  ;)[/quote]
all that showed up on nolachrymose's post was "clearly y can never be three".  That's all that showed up.  Something must be wrong with my computer.

so which would be faster, mine or nolachrymose's?

You have paper and pencils when you do the round, so this shouldn't be too hard. If you think fast, 45 seconds should be more than enough to solve this problem, especiallly for a national competitior.

Hamster's method solves the problem in less than 5 seconds, without a pencil.  Divide the coefficient of x in the numerator by the coefficient of x in the denominator, and you are done.  That's what I did while watching the broadcast.

Hamster:  You have discovered a calculus-based solution to the problem -- perhaps the very method to which Mr. Rusczyk referred.  I think this is the fastest method, but it requires you to either know some calculus or to use your instincts that as [i]x[/i] becomes infinite, [i]y[/i] will get very close to a forbidden value.

You also noted that this is a hyperbola.  I'm not sure all National MATHCOUNTS competitors would have recognized that.  But since you know [i]x[/i] and [i]y[/i] have forbidden values, you can assume the asymptotes for the hyperbola are parallel to the [i]x[/i]- and [i]y[/i]-axes.

If you recognize the hyperbola, you can rearrange the equation as follows:

\begin{eqnarray}
y&=&\frac{3x}{x-2} \\
xy-2y&=&3x \\
xy-2y-3x+6&=&6 \\
(x-2)(y-3)&=&6
\end{eqnarray}

[The solution by nolachrymose, which simply solves for [i]x[/i] works great, too.]

From equation (4), we see it would be absurd if $x=2$ or $y=3$.

If this were a Sprint question, I would definitely suggest you write out the Algebra.  But for a National Countdown question, you might be better off to go with your instincts, even if you don't know Calculus.

If the timing of the ESPN broadcast accurately represents the real conditions, the contestant buzzed in with the correct answer about 23 seconds after the moderator called for the question.

Hamster and Ravi B have the right idea.  It is a pre-calculus problem to do with horizontal and vertical asymptotes of rational functions.
The domain is all reals except 2, which means a vertical asymptote of x=2.  The problem is to find the range.  The range of the function is all reals except the y-value of the horizontal asymptote.  Since the degrees of numerator and denominator are the same, that y-value of the horizontal asymptote is the ratio of the leading coefficients of the numerator and denominator, which is 3/1 = 3.  This is the excluded value.  If you graph the function on a calculator you will see why it works.

To find the range of a rational function, you can find the inverse (by switching x and y and then solving for y) and then determine its domain.  This also is a pre-calculus approach, and is essentially the one taken by nolachrymose.   However, it is not quick enough for countdown and Mark (I think it was him) presumably just figured out that the horizontal asymptote is y = 3.  Very impressive.

Another way to see which $y$ value is missing is to split off a constant:
\[y = \frac{3x}{x-2} = 3 + \frac{6}{x-2} \ne 3.\]
But that's only if you want to know why Hamster's method works -- it's not what you should do during the Countdown round.

Right I see. I did graphs of hyperbolas generated by rational functions like these in algebra I so that's how I recognized it. Thank you for all your insights.

[quote="math92"]i still can't see the latex![/quote]

 :roll: too bad

I split and moved an advanced discussion of hyperbolas over to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40003]here[/url].


Let $ S\equal{}\{0,1,1,2,2,3,3,3,4\}$. How many different 4-digit numbers can be formed with all its digit chosen from $ S$?

[hide="casework"]
[u]1) all digits are distinct[/u]

for the first digit $ 4$ possibilities (can't have $ 0$) for the second again $ 4$, $ 3$ for third, $ 2$ for fourth - $ 96$ in total

[u]2) 2 non-distinct digits and 2 distinct[/u]

first, we can pick our non-distinct digits on $ 3$ ways and place them on $ \binom 4 2$ ways, for the remaining $ 2$ digits we have $ 4\cdot 3$ ways to pick them, but we have to subtract cases where $ 0$ is the first digit; there are again $ 3$ ways to pick and $ \binom 3 2$ ways to place non-distinct digits and $ 3$ ways to pick remaining digit, in total we have

$ 3\cdot 12 \cdot \binom 4 2 \minus{} 3\cdot 3 \cdot \binom 3 2 \equal{} 189$

[u]3) a pair of non-distinct digits[/u]

we can pick a pair on $ 3\cdot 2$ ways and place them on $ \binom 4 2$ ways - $ 36$ in total
[u]
4) 3-non distinct digits[/u]

we can only pick $ 3$'s and place them on $ 4$ different ways; there are $ 4$ ways to pick remaining digit, but we have to subtract case where $ 0$ is the first digit - $ 4\cdot 4 \minus{} 1 \equal{} 15$

We can form $ \boxed{336}$ different 4-digit numbers.[/hide]

[quote="Flame"]
[u]3) a pair of non-distinct digits[/u]

we can pick a pair on $ 3\cdot 2$ ways and place them on $ \binom 4 2$ ways - $ 36$ in total

[/quote]

Do you mean "[i]two pairs of non-distinct digits[/i]"? For example, 1212.
In this case we have $ \binom 3 2$ ways to pick the pairs and place them in $ \frac{4!}{2!2!}$ ways - that would be $ 18$ in total.

Yes, you are right :)
I can't edit anymore, so it will have to stay like this.


geometry
incenter
circumcircle
trigonometry
ratio
number theory
relatively prime

Triangle $T_1$ has sides of lengths 4, 5, and 6. A second triangle $T_2$ is formed by using the excenters of $T_1$ as vertices. If $[T_i]$ denotes the area of triangle $T_i$, then the value of $\frac{[T_1]}{[T_2]}$ can be written in the form $m/n$, where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.

(Note: An excenter is the center of an exscribed of a triangle. An exscribed circle is tangent to two extensions of the sides of a triangle and is also tangent to the third side.)

[color=blue][b]1. solution:[/b] [/color]Denote a = BC = 4, b = CA = 5, c = AB = 6 the sides of the triangle $\triangle ABC$ and X, Y, Z its excenters against the vertices A, B, C, respectively. The internal bicector of the angle $\angle A$ passes through the excenter X and the external angle bisector of this angle passes through the excenters Y, Z. Since the internal and external bisectors of an angle are perpendicular to each other, $XA \perp YZ$, which means that XA is the X-altitude of the triangle $\triangle XYZ$. Similarly, YB, XC are the Y-, Z-altitudes of this triangle. The intersection I of the lines XA, YB, ZC, the incenter of the triangle $\triangle ABC$, is then the orthocenter of the triangle $\triangle XYZ$. Hence, the triangle $\triangle ABC$ is the orthic triangle of the triangle $\triangle XYZ$. The first consequence is that the circumcircle $(O_1)$ with radius $R_1$ is the 9-point circle of the triangle $\triangle XYZ$, passing through its altitude feet. Hence it is also the circumcircle of the medial triangle $\triangle X' Y'Z'$ of the triangle $\triangle XYZ$, where X', Y', Z' are the midpoints of YZ, ZX, XY. If $(O_2)$ is the circumcircle of the triangle $\triangle XYZ$ with radius $R_2$, then $R_2 = 2 R_1$. The second consequence of the triangle $\triangle ABC$ being the orthic triangle of the triangle $\triangle XYZ$ is similarity of the triangles

$\triangle XYZ \sim \triangle ABZ \sim \triangle AYC \sim \triangle XBC$

For example, the triangles $\triangle XYZ, \triangle ABZ$ have the angle $\angle Z$ common and $\frac{ZA}{ZX} = \cos \widehat Z = \frac{BZ}{YZ},$ hence, they are similar by SAS. Since the triangle $\triangle ABC$ with the given sides a, b, c is acute ($a^2 + b^2 > c^2$) and the triangle $\triangle XYZ$ is also acute (its orthocenter identical with the incenter of $\triangle ABC$ is in its interior), it follows that

$180^\circ = \angle YAZ = \angle YAC + \angle CAB + \angle BAZ = 2 \angle X + \angle A$

$\angle X = 90^\circ - \frac{\angle A}{2}$ and similarly, $\angle Y = 90^\circ - \frac{\angle B}{2},$ $\angle Z = 90^\circ - \frac{\angle C}{2}$.

Denote x = YZ, y = ZX, z = XY the sides of the triangle $\triangle XYZ$. The areas $|T_1|, |T_2|$ of the triangles $T_1 \equiv \triangle ABC,\ T_2 \equiv \triangle XYZ$ are equal to

$|T_1| = \frac{abc}{4R_1} = \frac{8R_1^3 \sin \widehat A \sin \widehat B \sin \widehat C}{4R_1} = 2R_1^2 \sin \widehat A \sin \widehat B \sin \widehat C$

$|T_2| = \frac{xyz}{4R_2} = \frac{8R_2^3 \sin \widehat X \sin \widehat Y \sin \widehat Z}{4R_1} = 2R_2^2 \sin \widehat X \sin \widehat Y \sin \widehat Z$

Substituting $R_2 = 2R_1$ and $\sin \widehat X = \sin \left(90^\circ - \frac{\widehat A}{2}\right) = \cos \frac{\widehat A}{2},$ etc. into the formula for the area $|T_2|$,

$|T_2| = 8R_1^2 \cos \frac{\widehat A}{2} \cos \frac{\widehat B}{2} \cos \frac{\widehat C}{2}$

As a result, the area ratio is equal to

$\frac{|T_1|}{|T_2|} = \frac{\sin \widehat A \sin \widehat B \sin \widehat C}{4 \cos \frac{\widehat A}{2} \cos \frac{\widehat B}{2} \cos \frac{\widehat C}{2}} = 2 \sin \frac{\widehat A}{2} \sin \frac{\widehat B}{2} \sin \frac{\widehat C}{2} =$

$=2 \sqrt{\frac{1 - \cos \widehat A}{2} \cdot \frac{1 - \cos \widehat B}{2} \cdot \frac{1 - \cos \widehat C}{2}}$

Using the cosine theorem formula for $\cos \widehat A$, we have

$\frac{1 - \cos \widehat A}{2} = \frac{2bc - (b^2 + c^2 - a^2)}{4bc} = \frac{a^2 - (b - c)^2}{4bc} =$

$= \frac{(c + a - b)(a + b - c)}{4bc} = \frac{(s - b)(s - c)}{bc}$

and similarly, $\frac{1 - \cos \widehat B}{2} = \frac{(s - c)(s - a)}{ca},$ $\frac{1 - \cos \widehat C}{2} = \frac{(s - a)(s - b)}{ab}$, where $s = \frac{a + b + c}{2} = \frac{15}{2}$ is the semiperimeter of the triangle $\triangle ABC$. Substituting this to the last area ratio expression leads to

$\frac{|T_1|}{|T_2|} = 2 \sqrt{\frac{(s - b)(s - c)}{bc} \cdot \frac{(s - c)(s - a)}{ca} \cdot \frac{(s - a)(s - b)}{ab}} = \frac{2(s - a)(s - b)(s - c)}{abc}$

Plugging in $a = 4, b = 5, c = 6$ amd $s - a = \frac 7 2,\ s - b = \frac 5 2,\ s - c = \frac 3 2$, we get $\frac{|T_1|}{|T_2|} = \frac{7}{32},$ $\frac{|T_2|}{|T_1|} = \frac{32}{7}$.

[color=blue][b]2. solution:[/b] [/color] Having done this elementary straightforward calculation, we can play smart after the fact. Using Heron's formula for the area $|T_1|$,

$\frac{2(s - a)(s - b)(s - c)}{abc} = \frac{2 |T_1|^2}{s \cdot 4R_1 |T_1|} = \frac{|T_1|}{s R_2}$

In order for this to be equal to the area ratio $\frac{|T_1|}{|T_2|}$, we have to show that the area $|T_2|$ of the triangle $\triangle XYZ$ is equal to the product of its circumradius $R_2$ and the semiperimeter s of its orthic triangle $\triangle ABC$. But this is true for any acute triangle $\triangle XYZ$, because then both its circumcenter $O_2$ and its orthocenter I are in its interior and

$|T_2| = |\triangle O_2XY| + |\triangle O_2YZ| + |\triangle O_2ZX| =$

$= \frac{R_2^2}{2} (\sin \widehat {XO_2Y} +\sin \widehat {YO_2Z} + \sin \widehat {ZO_2X}) = \frac{R_2^2}{2} (\sin 2 \widehat Z +\sin 2 \widehat X + \sin 2\widehat Y) =$

$= R_2^2 (\sin \widehat Z \cos \widehat Z + \sin \widehat X \cos \widehat X + \sin \widehat Y \cos \widehat Y) =$

(using the extended sine theorem and similarity of the triangles $\triangle XYZ \sim \triangle ABZ \sim \triangle AYC \sim \triangle XBC$)

$= \frac{R_2}{2} (z \cos \widehat Z + x \cos \widehat X + y \cos \widehat Y) = \frac{R_2}{2} (AB + BC + CA) = s R_2$

I do not mind, if you guys hide the solution, I just do not like to do it myself. (This geometry problem was 8 pages out of scope with no reply).

Yetti


Let $ a,\ b,\ c$ be positive reals such that $ \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \plus{} \frac {a \plus{} b}{c} \equal{} 2\left(\frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ca}\right).$

Prove that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 3\geq 2(ab \plus{} bc \plus{} ca).$

$ \sum{bc(b\plus{}c)}\equal{}2(a\plus{}b\plus{}c)$

Multiplying both sides by $ a\plus{}b\plus{}c$
$ (a^2 \plus{} b^2 \plus{} c^2)(a\plus{}b\plus{}c) \plus{} 3(a\plus{}b\plus{}c)\geq 2(ab \plus{} bc \plus{} ca)(a\plus{}b\plus{}c)$
$ a^3\plus{}b^3\plus{}c^3\plus{}\sum{bc(b\plus{}c)}\plus{}3(a\plus{}b\plus{}c)\geq 2\sum{bc(b\plus{}c)}\plus{}6abc$
$ a^3\plus{}b^3\plus{}c^3\plus{}\frac{1}{2}\sum{bc(b\plus{}c)}\geq 6abc$

Which is true by $ AM\minus{}GM$.

Your solution is correct, cefer. :)

Here is my solution:

For $ a>0,\ b>0,$ we have $ (a\plus{}1)(b\minus{}1)^2\plus{}(a\minus{}1)^2(b\plus{}1)\equal{}ab(a\plus{}b)\minus{}4ab\plus{}a^2\plus{}b^2\minus{}(a\plus{}b)\plus{}2$

$ \equal{}(a\minus{}b)^2\plus{}(ab\minus{}1)(a\plus{}b\minus{}2)$

$ \equal{}(a\plus{}b)^2\plus{}(ab\minus{}1)(a\plus{}b)\minus{}2(3ab\minus{}1)\geq 0.$ Similarly we have $ (b\plus{}c)^2\plus{}(bc\minus{}1)(b\plus{}c)\minus{}2(3bc\minus{}1)\geq 0,\ (c\plus{}a)^2\plus{}(ca\minus{}1)(c\plus{}a)\minus{}2(3ca\minus{}1)\geq 0,$

Adding these inequalities gives $ 2(a^2\plus{}b^2\plus{}c^2)\plus{}(ab\minus{}1)(a\plus{}b)\plus{}(bc\minus{}1)(b\plus{}c)\plus{}(ca\minus{}1)(c\plus{}a)\geq 4(ab\plus{}bc\plus{}ca)\minus{}6.$

Here we have $ \frac{b\plus{}c}{a}\plus{}\frac{c\plus{}a}{b}\plus{}\frac{a\plus{}b}{c}\equal{}2\left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\right)$

$ \Longleftrightarrow ab(a\plus{}b)\plus{}bc(b\plus{}c)\plus{}ca(c\plus{}a)\equal{}2(ab\plus{}bc\plus{}ca)$

$ \Longleftrightarrow (ab\minus{}1)(a\plus{}b)\plus{}(bc\minus{}1)(b\plus{}c)\plus{}(ca\minus{}1)(c\plus{}a)\equal{}0$.

Thus we obtain $ 2(a^2\plus{}b^2\plus{}c^2)\geq 4(ab\plus{}bc\plus{}ca)\minus{}6$, yielding $ a^2\plus{}b^2\plus{}c^2\plus{}3\geq 2(ab\plus{}bc\plus{}ca).$

From the condition we have $ \sum bc(b + c) = 2(a + b + c)$. But from Chebyshev's inequality we have
\begin{eqnarray*}3\cdot 2(a + b + c)
& = & 3\sum bc(b + c) \\ & \ge & (bc + ca + ab)(b + c + c + a + a + b) \\
& \ge & 3(abc)^{\frac 23}\cdot 2(a + b + c)\end{eqnarray*}
Where the last inequality is AM-GM. Thus $ (abc)^{\frac 23}\le 1$.

So we'll be done if we can show that $ a^2 + b^2 + c^2 + 3(abc)^{\frac 23}\ge 2(ab + bc + ca)$. But this follows from Schur and AM-GM:
\begin{eqnarray*}(x^{\frac 23})^3 + (y^{\frac 23})^3 + (z^{\frac 23})^3 + 3(xyz)^{\frac 23} & \ge & \sum_{\text{sym}}\left(\left(x^{\frac 23}\right)^2y^{\frac 23} + x^{\frac 23}\left(y^{\frac 23}\right)^2\right) \\
\ & \ge & \sum_{\text{sym}}2xy \end{eqnarray*}


AMC
AIME
function
trigonometry
quadratics
modular arithmetic
geometry

There have some requests for an AIME problem solving marathon (basically it was all shawn). So here. Try to be original don't just use old AIME problems.

well i didnt technically request it...I was just suggesting but oh well, thanks for creating it anyways..

now we'll now rely on mathcrazed for the questions...

or lets see...

heres a play of an old AIME question...(year like 1985)

1. (like number 1) How many of the first 999 positive numbers can [x]+[2x]+[3x] equal if x is a real number and [y] denotes the floor function (greatest integer less than y)?

lol i think i made it too easy since gcf(1,2,3)=1...
oh well.

(notice i had to change the question from first 1000 postivie numbers to first 999 positive numbers...)

2. (like number 1) What is tan(arctan(x)+arctan(y)) if x and y are the solutions to the equation a^2+4a-10=0

[hide="#1"]
I CAN'T COUNT
[/hide]

[hide="#2.. I think I got this wrong"]
trig yay..

Using the quadratic equation, it's obvious that $ a \equal{} \minus{} 2 \pm \sqrt {14}$

trig identity:

$ \tan{\alpha \plus{} \beta} \equal{} \frac {\tan{\alpha} \plus{} \tan{\beta}}{1 \plus{} \tan{\alpha}\tan{\beta}}$

$ \alpha \equal{} \tan^{ \minus{} 1}{( \minus{} 2 \minus{} \sqrt {14})}$
$ \beta \equal{} \tan^{ \minus{} 1}{( \minus{} 2 \plus{} \sqrt {14})}$

since $ \tan{(\tan^{ \minus{} 1}{x})} \equal{} x$, plugging in:

$ \frac {( \minus{} 2 \minus{} \sqrt {14}) \plus{} ( \minus{} 2 \plus{} \sqrt {14})}{1 \minus{} ( \minus{} 2 \minus{} \sqrt {14})( \minus{} 2 \plus{} \sqrt {14})}$

$ \equal{} \frac { \minus{} 4}{1 \minus{} (4 \minus{} 14)}$

$ \equal{} \boxed{ \minus{} \frac {4}{11}}$

except that's not really AIME format.. did I do something wrong?

EDIT: formatting[/hide]

[hide="2"]for $ \tan{\arctan{x} \plus{} \arctan{y}}$, we would use the sum formula for tan so we get $ \frac{x \plus{} y}{1\minus{}xy}$. now you find the roots of the equation $ a^2 \plus{} 4a \minus{}10$ which are $ \minus{}2\pm \sqrt{14}$ so you just add them and get $ \minus{}4$ and if you multiply them you get $ \minus{}10$ so $ \minus{}\frac{4}{1\minus{}(\minus{}10)}$ = $ \frac{\minus{}4}{11}$, darn sooey stole my thunder.[/hide]

[quote="cognos599"][hide="2"]for $ \tan{\arctan{x} \plus{} \arctan{y}}$, we would use the sum formula for tan so we get $ \frac {x \plus{} y}{1 \minus{} xy}$. now you find the roots of the equation $ a^2 \plus{} 4a \minus{} 10$ which are $ \minus{} 2\pm \sqrt {14}$ so you just add them and get $ \minus{} 4$ and if you multiply them you get $ \minus{} 10$ so $ \minus{} \frac {4}{1 \minus{} ( \minus{} 10)}$ = $ \frac { \minus{} 4}{11}$, darn sooey stole my thunder.[/hide][/quote]

lol...or you could use the fact that the sum of the roots of a quadratic is $ \minus{}\frac{b}{a}$ and the product is $ \frac{c}{a}$
heh...

[hide="okay I got #1 :D "]
but again, I'm probably wrong

It was kinda brute force, but whatever:

when $ x = \frac{1}{3}$, the value of the sum, $ v$, equals 1.

similarly,

$ x = \frac{1}{2}$ then $ v = 2$

$ x = \frac{2}{3}$ then $ v = 3$

$ x = 1$ then $ v = 6$

since $ \lfloor{x} = \lfloor{x + n}$ where $ n$ is an integer, I assumed this pattern of four integers repeated for every six.

$ 999 \times \frac{4}{6} = \boxed{666}$[/hide]

[quote="MustItAlwaysBeMe"][quote="cognos599"][hide="2"]for $ \tan{\arctan{x} \plus{} \arctan{y}}$, we would use the sum formula for tan so we get $ \frac {x \plus{} y}{1 \minus{} xy}$. now you find the roots of the equation $ a^2 \plus{} 4a \minus{} 10$ which are $ \minus{} 2\pm \sqrt {14}$ so you just add them and get $ \minus{} 4$ and if you multiply them you get $ \minus{} 10$ so $ \minus{} \frac {4}{1 \minus{} ( \minus{} 10)}$ = $ \frac { \minus{} 4}{11}$, darn sooey stole my thunder.[/hide][/quote]

lol...or you could use the fact that the sum of the roots of a quadratic is $ \minus{} \frac {b}{a}$ and the product is $ \frac {c}{a}$
heh...[/quote]
whoa.. that makes it a whole lot more efficient..

then it'd simply be $ \frac{x \plus{} y}{1 \plus{} xy}$ instead of going through all my $ \tan{(\tan^{\minus{}1}{x})}$ crap and just plugging in.. blah I'm retarded.

[quote="MustItAlwaysBeMe"][quote="cognos599"][hide="2"]for $ \tan{\arctan{x} \plus{} \arctan{y}}$, we would use the sum formula for tan so we get $ \frac {x \plus{} y}{1 \minus{} xy}$. now you find the roots of the equation $ a^2 \plus{} 4a \minus{} 10$ which are $ \minus{} 2\pm \sqrt {14}$ so you just add them and get $ \minus{} 4$ and if you multiply them you get $ \minus{} 10$ so $ \minus{} \frac {4}{1 \minus{} ( \minus{} 10)}$ = $ \frac { \minus{} 4}{11}$, darn sooey stole my thunder.[/hide][/quote]

lol...or you could use the fact that the sum of the roots of a quadratic is $ \minus{} \frac {b}{a}$ and the product is $ \frac {c}{a}$
heh...[/quote]

NICE, nice catch subs that is a way better way of doing it.

btw sooey, the answer to your question is cursed :ninja: . lol 666

[color=darkblue][size=117]Problem #3 (based on 1983 AIME #10):[/size][/color]

(I lack the ability to create mathematically beautiful problems T_T)

The numbers $ 12245$, $ 10241$, and $ 14552$ have something in common. Each is a five-digit number beginning with 1 that has exactly two identical digits. If $ x$ represents the quantity of such numbers, what is $ x \pmod {1000}$?

[hide="Solution"]
of the form $ 11xyz$, $ 1x1yz$, $ 1xy1z$, and $ 1xyz1$, there are

$ 4 \times 9 \times 8 \times 7$ numbers.

of the form $ 1xxyz$ and similar rearrangements of the digits, there are

$ 10 \times \dbinom{4}{2} \times 8 \times 7$ numbers.

\[ 4 \times 9 \times 8 \times 7 + 10 \times \dbinom{4}{2} \times 8 \times 7\]

\[ = (36 + 60)(8 \times 7)\]

\[ = 5376 \equiv \boxed{376} \pmod{1000}\][/hide]

[color=darkblue][size=117]Problem #4 (copied straight from 1983 AIME #12 XD):[/size][/color]

In a given circle, the length of diameter $ \overline{AB}$ is a two digit integer. Reversing the digits of AB gives the length of a perpendicular chord $ \overline{CD}$. The distance from their intersection point H to the center of the circle is a positive rational number. Determine the length of $ \overline{AB}$.

[hide="Solution"]
simple pythagorean stuff..

Let the length of the diameter = $ 10a + b$, with $ a$ and $ b$ representing the two digits.

Then the chord's length = $ 10b + a$, since the digits are reversed.

Let's also have the distance from H to the circle's center = $ x$, which is a positive rational number, which also means it's an integer (somebody correct me if I'm wrong).

We're solving for the length of $ \overline{AB}$.

Drawing a diagram makes it obvious that

\[ (\frac}10a + b}{2})^2 = x^22 + (\frac{10b + a}{2})^2\]

\[ 100a^2 + 20ab + b^2 = 4x^2 + 100b^2 + 20ab + a^2\]

\[ 99a^2 - 99b^2 = 4x^2\]

\[ 3^2 \times 11 (a-b)(a+b) = (2x)^2\]

since $ x$ is an integer, $ a+b = 11$ and $ a-b$ must be a perfect square that differs from $ 11$ by an even number. Such squares are 9 and 1, but 9 would make $ a = 10$, which is not possible.

Solving for $ a$ and $ b$ we get $ a = 6$ and $ b = 5$.

\[ \overline{AB} = 10a + b = \boxed{065}\][/hide]

Just remember that 33^2+56^2=65^2

also note this question was altered to become a AMC 10 number 24 like a couple of years ago.

It was sorta like: there is an integer m sucht that m^2=x^2-y^2 where x and y have their digits reversed.  What is m+x+y...

and who says the AMC/AIME questions dont "reapeat"...

Infact, there were two AIME questions in the 80s/90s that used almost the exact same concept, just different numbers...

but i think now for the modern AIMEs, they always come up with pretty unique questions (like the one about the planets...)

yea number 3 is also a pretty easy problem for a 10 on AIME...these days I would expect it to be like number 3ish since its just two cases.

[size=150][u]PROBLEM 5 [/u] [/size] (this problem is inspired by mathcrazed-and his cool construction stuff he knows how to do)

Suppose we have a triangle $ ABC$ that is equilateral.  We have a point P inside the triangle such that the distance from P to A is $ 5$, P to B is $ 12$, and P to C is $ 13$.  $ AB \equal{} m \plus{} n\sqrt {p}$ where $ p$ is not divisible by the square of any integer.
Find $ m \plus{} n \plus{} p$.

Ill say this is a number 10ish on the AIME
(if you want some inspiration do number 23 on the AMC 2006 12B)

[hide="#5 GEOMETRY SAVES LIVES"]
omgggg I have a solution but can't draw the freakin image.. don't have a program. It's pretty easy, though..

Rotate $ \triangle{PCB}$ clockwise about $ B$ so that $ \overline{BC}$ matches up with $ \overline{AB}$, creating $ \triangle{P'AB}$.

Note that $ m\angle{PBC} \equal{} m\angle{P'BA}$.

It's easily seen that $ \angle{P'BP} \equal{} 60^\circ$, so $ \overline{P'P} \equal{} 12$.

We've got a right triangle $ APP'$ and an equilateral triangle $ BPP'$.

Using the Pythagorean Theorem we can find the length of $ \overline{AB} \equal{} \sqrt {169 \plus{} 60\sqrt {3}}$

(I think Shawn meant to say $ \overline{AB} \equal{} \sqrt {m \plus{} n\sqrt {p}}$, cause $ (\overline{AB})^2 \equal{} 169 \plus{} 60\sqrt {3}$ doesn't factor into anything. Says my TI-89 T :P )

so solving for $ m \plus{} n \plus{} p \equal{} 169 \plus{} 60 \plus{} 3 \equal{} \boxed{232}$.

btw I loved this problem :D geometry rocks.
[/hide]

[quote="MathAndKnowledge"][hide="Problem #6"]This one I invented myself.  If cognos can milk a cow in 3 hours, undefined can milk a cow in 8 hours, mathandknowledge can milk a cow in 1 hour, mustitalwaysbeme can milk a cow in 19 hours, unimpossible can milk a cow in 9 hours and 30 minutes, mathcrazed can milk a cow in 4 hours, and Nathan Benjamin can milk a cow in 100 hours (he's too busy dominating some math competition to actually milk the cow), then how many seconds will it take all seven of these people to milk a single cow (assuming that the cow has 7 utters)?[/hide]

Note that I can solve this problem in less than 30 seconds with a calculator, and a couple of minutes without a calculator.  Note that the previous sentence implies that there is an easy way to do this type of problem.[/quote]

one thing about AIME problems, they usually aren't hard-core arithmetic as much as clever reasoning and all answers in AIME are from 000-999. how come you are the fastest milker( practice makes perfect  :D , jk). i am too lazy to do it without a calculator.

[quote]one thing about AIME problems, they usually aren't hard-core arithmetic as much as clever reasoning and all answers in AIME are from 000-999. how come you are the fastest milker( practice makes perfect  , jk). i am too lazy to do it without a calculator.[/quote]

Yes, I suppose...  But do you BTW know of the method that I'm talking about?

harmonic mean. $ \frac{7}{1 \plus{} \frac{1}{3} \plus{} \frac{1}{4} \plus{} \frac{1}{8} \plus{} \frac{1}{9.5} \plus{} \frac{1}{19} \plus{} \frac{1}{100}}$. that is the number of hours at least.

oh whoops I made a typo on the second page.. the first line of math, at the end it should've been $ (n\minus{}1)^4$ and $ n^4$. bah.

*spazzing from lack of activity*

Basically same as AoPS vol 1 #109:

Show that the sum of five consecutive perfect squares equals a perfect square.

sleep.deprivation.*gasp*.low.bludsuger.*gassssp*.PIKACHU

*bump*

you use mods for my (super-easy) problem. :|

Okay guyz:

You have two tangent circles and two parallel lines tangent to one of each of these circles. Let the two circles be $ \phi$ and $ \lambda$ and let the lines be $ l_1$ and $ l_2$. Prove that the three tangential points are collinear. See picture.

 [geogebra]d9c7ed5ae9daedd73a1574da54de71f20cc73363[/geogebra]

uh do I have to be rigorous or something with this?

[asy]size(200);

D((-6,-6)--(-6,6),black,Arrows); D((10,-6)--(10,6),black,Arrows);
D((-6,0)--(10,0),black); D((0,-6)--D("P",(0,3),NW)--(0,6),black+dashed,Arrows);
D(CP(D("\phi",(-3,0)),D("L_1",(-6,0),W)));
D(CP(D("\lambda",(5,0)),D("L_2",(10,0),E)));
D("T",(0,0),SE);
MP("l_1",(-6,6),W); MP("l_2",(10,6),E);[/asy]

$ \overline{PT}$ passes through point $ T$ tangent to both circles. Thus $ m\angle{L_1TP} = m\angle{L_2TP} = \frac {\pi}{2}$ and $ m\angle{L_1TL_2} = \pi$. A 180-deg angle is a line, so $ L_1$, $ T$, and $ L_2$ are collinear.

And Asymptote > Geogebra:

If the volume of the innermost octahedron equals $ 1$, find the volume of the outermost tetrahedron. All polyhedra have edges, vertices, or faces that are tangent to edges, vertices, or faces of at least one other polyhedron.

[asy]import three;
size(400);
currentprojection=perspective((50,35,30));
defaultpen(linewidth(.8));

triple O=(0,0,0);
triple AA=(0,0,1), AB=(1,0,0), AC=(0,1,0), AD=(-1,0,0), AE=(0,-1,0), AF=(0,0,-1);
triple BA=(1,1,1), BB=(-1,1,1), BC=(-1,-1,1), BD=(1,-1,1), BE=(1,1,-1), BF=(-1,1,-1), BG=(-1,-1,-1), BH=(1,-1,-1);
triple DA=(0,0,2), DB=(2,0,0), DC=(0,2,0), DD=(-2,0,0), DE=(0,-2,0), DF=(0,0,-2);
triple EA=(0,0,3), EB=(3,0,0), EC=(0,3,0), ED=(-3,0,0), EE=(0,-3,0), EF=(0,0,-3);
triple FAA=(3,-3,3), FAB=(3,-3,0), FAC=(3,0,3), FAD=(0,-3,3), FBA=(-3,3,3), FBB=(-3,3,0), FBC=(0,3,3), FBD=(-3,0,3), FCA=(3,3,-3), FCB=(3,0,-3), FCC=(3,3,0), FCD=(0,3,-3), FDA=(-3,-3,-3), FDB=(0,-3,-3), FDC=(-3,-3,0), FDD=(-3,0,-3);

D(AA--AB--AF--AD--cycle--AC--AF--AE--AB--AC--AD--AE--cycle);
DPA(BA--BB--BC--BD--BA^^BE--BF--BG--BH--BE^^BA--BE^^BB--BF^^BC--BG^^BD--BH,red);
DPA(BA--BC--BH--cycle--BF--BC^^BF--BH,green);
D(EA--EB--EF--ED--cycle--EC--EF--EE--EB--EC--ED--EE--cycle,orange);
DPA(FAA--FBA--FCA--cycle--FDA--FCA^^FBA--FDA);[/asy]

It's awesome how simple the relationships are.

You gotta be evil to draw that in asymptote :o

[quote="cognos599"][hide="my solution"]well if x is a solution so the equation, then it must be a solution to

\begin{eqnarray*}x & = & \sqrt {73} + \frac {61}{\sqrt {73} + \frac {61}{\sqrt {73} + \cdots}}.\end{eqnarray*}.

so $ x = \sqrt {73} + \frac {61}{x}$.

now you solve the quadratic equation $ x^2 - \sqrt {73}x + 61 = 0$

and you get the two roots to be $ \frac {\sqrt {73} \pm \sqrt {317}}{2}$, so if you add the two together you get $ \sqrt {317}$ and if you square that, you get $ \fbox{317}$. voila! [/hide][/quote]
If x is a solution to the equation it doesn't imply it is a solution to \begin{eqnarray*}x & = & \sqrt {73} + \frac {61}{\sqrt {73} + \frac {61}{\sqrt {73} + \cdots}}.\end{eqnarray*} because there are two solution to the original equation with one positive and one negative, while the expression you typed is clearly nonnegative. In fact the expression you type only corresponds to the positive solution in the original equation (so the converse to your statement is the correct one).

Hi. I'd like to start this again. I didn't think it was necessary to create a new thread about this, so here is new problem:

The polynomial $ 1\minus{}x\plus{}x^2\minus{}x^3\plus{}\cdots\plus{}x^{16}\minus{}x^{17}$ (or $ \sum_{k\equal{}0}^{17}(\minus{}x)^k$ for those who know summation) can be written as $ a_0\plus{}a_1y\plus{}a_2y^2\plus{}\cdots\plus{}a_{16}y^{16}\plus{}a_{17}y^{17}$ where each $ a_i$ is a constant and $ y\equal{}x\plus{}1$ Determine the value of $ a_2$

[quote="mathwizarddude"][hide]$ x \equal{} y \minus{} 1\implies\binom22 \plus{} \binom32 \plus{} ... \plus{} \binom{17}2 \equal{} \binom{18}3$ from the hockey stick identity. [/hide]
I'll post the next problem if this is correct.
ABC is an equilateral triangle. A point S is chosen in the plane of the equilateral triangle, and AS=3, BS=5, CS=8. Find BC.[/quote]

You need to compute your final answer since that is the accepted form on AIME and you should practice doing that, it will help some.

You're right though. But $ \binom{18}{3} \equal{} \boxed{816}$

Also, try to be a little more rigorous. You used the binomial theorem I see. But you need to explain how the $ a_2$ value builds up from the $ y^2$ coefficients

For example:
Every coefficient of $ y^2$ in $ (y\minus{}1)^{m}$ is $ \binom{m}{2}$ could be added :)

Common guys aren't we going to continue with this?

I'll post a new problem:

Find the least positive integer such that when its leftmost digit is deleted the new integer is $ \frac{1}{29}$ of the original integer.

Let the number be $ N\equal{}\sum_{k\equal{}0}^n a_k*10^k$ ($ 0\le a_k\le9$). Then
$ N\equal{}29(\sum_{k\equal{}0}^{n\minus{}1}a_k*10^k$
$ a_n*10^n\equal{}28(\sum_{k\equal{}0}^{n\minus{}1}a_k*10^k$

$ 28|a_n*10^n\implies a_n\equal{}7,n\ge2$.
For $ n\equal{}2$ we get $ 5^2\equal{}25\equal{}10a_1\plus{}a_0\implies N\equal{}725$

For next problem, look up my last post in this thread (my problem is unsolved yet).


Need some help with this one please.

Quoted from an EC problem my teacher gave me:

"The ideal gas law is PV=nRT where P is pressure, V is volume, n is the number of moles of gas (mole is a big number of molecules), R is constant, and T is temperature in degrees Kelvin.

1. What happens when you heat a closed system? That is, what happens when temperature rises, gas cannot enter or escape, and the volume is fixed?
2. What happens when temperature drops and pressure remains the same?
3. Why does hot air rise?
4. Ever notice that it's usually colder at the top of a mountain that at lower elevations? In fact, it is 7 degrees C colder per vertical kilometer. This is known as the adiabatic lapse rate. Adiabatic expansion is the expansion of a gas at constant energy, resulting in a drop in temperature. Sketch out this situation using P1, V1, n1, T1, and P2, V2, n2, T2. (Meaning P to the 1st power etc.) Try taking a set volume and moving it upwards. How does pressure change at high altitudes?
5. Look up a value for R. Be sure to notice the units. Using appropriate units or conversions, figure out how many moles of gas (i.e. solve for n) are present at 1 atom pressure in 10 L of gas at 20 degrees C (be sure to convert to Kelvin.)"

1. Pressure increases(thats why gas cylinders explode on heating)
2. volume decreases
3. P remains constant(atm. pressure) temp increases-vol. increases. mass constant. density decreases.
4. question was not completely clear to me.  :| for the last part of question- pressure decreases
5 :arrow: this is for one who likes calculating

... Shouldn't this belong in the Chemistry section of the Forums?

not many people visit the chemistry forum, I think  :arrow:

Could anbody help with the math part of this one?


Let $1 \leq k \leq n.$ Consider all finite sequences of positive integers with sum $n.$ Find $T(n,k),$ the total number of terms of size $k$ in all of the sequences.


Consider the following game: at the beginning, $ n$ fair coins lie tails-up on a table.  At each step, all coins facing tails-up are picked up and tossed simultaneously.  This process continues until at least $ k$ coins are heads-up.  Let $ f(n, k)$ be the expected number of steps until this happens.

1) Prove that $ f(n \plus{} m, n)$ is a strictly increasing function of $ n$ (for any fixed $ m$).
2) Can you prove that $ f(n, k)$ is a strictly increasing function of (integer values of) $ k$ for fixed $ n$ and $ 0 \leq k \leq n$?  What is its behavior as a function of $ n$ for fixed $ k$?
3) Give a recursion for $ f(n \plus{} m, n)$.  Is it true that $ f(n, n) > n$ for all $ n$?
4) Find a closed form for $ f(n, 1)$.
5) Can you find a closed form for $ f(n, 2)$?

[Edited to correct misstatements.]

? I believe you have mixed up the "increasing" in 1) and the "decreasing" in 2.

[hide="4"]
Consider what happens after 1 turn. Two cases:

No coins are flipped heads-up: This occurs with probability $ \frac{1}{2^n}.$ The expected number of more turns needed is f(n,1).

At least one coin is flipped heads-up: Occurs with probability $ 1\minus{}\frac{1}{2^n}.$ Then, we need 0 more turns.

So f(n,1) = 1 (this turn) + $ \frac{1}{2^n}f(n,1)$ + $ 1\minus{}\frac{1}{2^n}(0)$ = 1 +  $ \frac{1}{2^n}f(n,1).$

Subtracting $ \frac{1}{2^n}f(n,1)$ from both sides, we get $ \frac{2^n\minus{}1}{2^n}f(n,1) \equal{} 1,$ and $ f(n,1) \equal{} \frac{2^n}{2^n\minus{}1}.$
[/hide]

[hide="5?"]
On the first turn, there are three cases: $ 0$ heads, $ 1$ head, and $ 2$ or more heads.

If there are $ 0$ heads, then we need $ f(n,2)$ more steps.  This happens with probability $ \frac{1}{2^n}$.

If there is $ 1$ head, we keep it and begin flipping the $ n\minus{}1$ tails, needing $ 1$ more head.  So we need $ f(n\minus{}1,1)\equal{}\frac{2^{n\minus{}1}}{2^{n\minus{}1}\minus{}1}$ more steps, as pythag011 showed.  This happens with probability $ \frac{n}{2^n}$.

If there are $ 2$ or more heads, we are done.  This happens with probability $ \frac{2^n\minus{}n\minus{}1}{2^n}$.

So

$ f(n,2)\equal{}\frac{1}{2^n}(f(n,2)\plus{}1)\plus{}\frac{n}{2^n}\left(\frac{2^{n\minus{}1}}{2^{n\minus{}1}\minus{}1}\plus{}1\right)\plus{}\frac{2^n\minus{}n\minus{}1}{2^n}$

$ (2^n\minus{}1)f(n,2)\equal{}\frac{n2^{n\minus{}1}}{2^{n\minus{}1}\minus{}1}\plus{}2^n\plus{}2n$

$ f(n,2)\equal{}\frac{3n2^{n\minus{}1}\plus{}2^{2n\minus{}1}\minus{}2^n\minus{}2n}{2^{2n\minus{}1}\minus{}2^n\minus{}2^{n\minus{}1}\plus{}1}$
[/hide]

That seems too ugly to be right . . .


Do anyone of you have the pdf  of excerpts from the AoPS vol 1 & 2?

http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=1

I think you can view an excerpt for Vol 1. and Vol. 2. That link is Vol. 1, I think.

Thank you, but I'm aware of that link.I had gotten hold of a 100 page pdf of the AoPS vol 1 from a discussion thread. Is it possible I get one like that of the 2nd Volume also?

the 100 page pdf was a special combination packet that shows off the Intro series.   It is not an excerpt, really.


Here is the signiature I was trying to use:

[code]Let $ (x\frac{d}{dx})^{k}f(x) = \underbrace{(x\cdot \frac{d}{dx}(x\cdot \frac{d}{dx}(\cdots)))}_{\text{k times}}f(x)$. Then
$ \sum_{i=1}^{n}i^{k}= \lim_{x\to1}\left[(x\frac{d}{dx})^{k}\left( \frac{1-x^{n+1}}{1-x}\right)\right]$
[/code]

If vim is right (and I trust it), there are only 226 (<255) characters, but the site is rejecting it for being too long. Have the limits been changed and the sig page is just misquoting them, or is this a bug?

Actually the way that is stored in the database will exceed 255 characters (it's not being stored exactly as you see it).

[quote="Hamster1800"]Darn. What, exactly, causes the expansion? Is it simple enough for you to elaborate so that I can avoid these problems in the future?[/quote]It depends on what you write. For example $ \$$ is a special character, so we must "escape" it, by replacing it with something else. The same stands for many other things.

Basically if you don't have BBCode (that includes $ \text{\LaTeX}$) you have exactly 255 characters available. Otherwise you will have slightly less.


quadratics
modular arithmetic
number theory proposed
number theory

Find all primes p with following property:

For any positive integers x,y, if p divides x^2 + y^2, then p divides both x and y.

[quote="Anthony2"][quote="jin thynj"]Own?  :rotfl:[/quote]
Well, I've honestly came to this question by myself, but I should have figured out that It's probably known. Anyway if you smile, you live longer, so you're welcome :).

[quote="cnyd"]if $ p|x^{2} \plus{} y^{2}$ $ \iff$  $ p|(x,y)$

$ \implies$  $ \minus{} 1,(modp)$  is non-quadratic residue

$ \implies$  $ p\equiv \minus{} 1\pmod{4}$[/quote]
I'm probably dumb, but I don't understand how you prove that if p is 1 mod 4, then it has quadratic residue -1.[/quote][/quote]
$ ((2n)!)^2 \equiv (4n)! \equiv \minus{}1 \bmod p$ where $ p\equal{}4n\plus{}1$.


quadratics
conics
LaTeX
geometry
vector
linear algebra
matrix

Hello

I'm not sure wether this problem is too "easy" to post here but I'm kinda desperate  :?

I have a system of two equations with two unknowns, which I should be able to solve for analytically. The problem is that the equations are nonlinear and a bit tricky (for me atleast), so I'm wondering if anyone has any idea what to do?

Both the equations are on the form:
Ax^2+By^2+Cx+Dy+Exy+F=0

where A-F are known constanst, x and y are the ones I wish to solve for.

Any help is much appreciated and if anyone knows how to use Matlab to solve this it would be a perfect bonus!

The best method I know requires guesswork:
Does some linear combination zero the quadratic terms?
Is some linear combination a perfect square?
Does some linear combination factor?

That last is actually very likely, but spotting a factorization may be difficult. Once we have one of these we can reduce to one or more systems of a quadratic and a linear equation. Solve the linear equation for one unknown in terms of the other, and substitute into the quadratic.

This is mostly out of place because it's not calculus.

Yes, you should [u]not[/u] try solving for [i]x[/i] and [i]y[/i] analytically in terms of the coefficients of the terms!
Solving for the intersection of two generic circles (which has no xy term and which has equal coefficients for the x and y terms) is just barely doable (with very capable with math software like MATHCAD).

You should either try another approach to arrive at your ultimate solution (there's probably more to the problem, right?) or follow JMERRY's advice.

Good luck (and very merry Christmas)!

When the form of the equations is as you stated, then the two equations represent conic sections (but we cannot be sure of what type unless we know the coefficients, of course). I know how to solve the system if someone tells me how to use Latex on the posts. Please tell me fast, I'm eager to post my first solution!

Yes, this would be (fairly) easily solvable if we knew the specific coefficients.

(I'm assuming that you know the numerical coefficients, ROADRUN, and they aren't just variables.)

PS: When I saw A-F, I thought that it was the difference between A and F, not A, B, C, D, E, and F (I now think the latter one is what you really meant.)

By the way, can we give direct help on your problem?  There might be a different, much simpler approach rather than brute algebra.  (I run into massive, complex equations all the time when I try to solve geometry problems because I do them the wrong way.)

@arjunlandes:
See the recommended links  in Valentin Vornicu's signature (here, for example, you can find his sig: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31922 ).
Also, you may put the cursor above someone else's formula and you'll see, for a couple of seconds, the formula's LaTeX codification.
Try this, for instance  :
$\sqrt{x_5^{n+3}}$
(you must begin and end with $\$$; these two $\$'s$ does not appear when you put the cursor above the formula, but believe me, they are there !:) )

Using the quadratic formula, we obtain
$y=\frac{-(Bx+D)\pm\sqrt{(Bx+D)^2-4(E)(Ax^2+Cx+F)}}{2(E)}$
Given a system of two equations of this form, we get $0,1,2,3,4,$ or infinite points of intersection.

First off, sorry for posting in the wrong section and thanks for moving it! :)

Second off, A-F meant all the constanst A thru F, not the difference, sorry for not clearing that up ;)

Ok, I'll start from the beginning.

The two equations look like this (i'll try some latex...bare with me ;) ):

$(r^2 v_1^t K v_1)(- u_2^t K u_2) - ( u_2^t K u_2)(r s v_1^t K v_2) = 0$

$(r^2 v_1^t K v_1)(u_1^t K u_1) - (u_2^t K u_2)(s^2 v_2^t K v_2) = 0$

where $r, s$ are known scalars, $u_1,u_2,v_1,v_2$ are known $3 \times 1$-vectors and $K$ is a $3 \times 3$-matrix.

$K = \left[ \begin{array}{ccc} \alpha_x^2+x_0^2 & x_0 y_0 & x_0\\ x_0 y_0 & \alpha_y^2+y_0^2 & y_0\\ x_0 & y_0 & 1 \end{array} \right]$

where $x_0,y_0$ are known and $\alpha_x, \alpha_y$ are the ones I want to get.

This system should have at most 4 solutions, but the ones for $K_{11}, K_{22}$ that causes the $K$-matrix to be nonreal and positive definite should be discarded.

As you said, maybe it was wrong of me to start expanding the equations in the first place...

@arjunlandes:

Not really sure I understand what $a7c71c1d24y$ is, I'm assuming its the $y$-variable but written in some way I don't follow. Hope you can post the next message to finish up the solution :)

*bump*

Did everyone stop replying all at once?  :?


In how many ways can 20 be expressed as a sum of powers of 2?

For example, $ 4\equal{}2^{2}\equal{}2^{1}\plus{}2^{1}\equal{}2^{1}\plus{}2^{0}\plus{}2^{0}\equal{}2^{0}\plus{}2^{0}\plus{}2^{0}\plus{}2^{0}$
So 4 can be expressed in 4 different ways.

[hide="Hint"]Let $ f(n,m)$ be the number of ways of writing $ n$ as the sum of exactly $ m$ powers of $ 2$.

Try to establish the following recursive properties:

$ \blacktriangleright f(2k,m)\equal{} f(k, m)\plus{}f(2k\minus{}1, m\minus{}1)$
$ \blacktriangleright f(2k\plus{}1, m) \equal{} f(2k, m\minus{}1)$[/hide]

@Tor

Just a check of how well I remember how to do those:
Is the solution simply to check the coefficient on $ x^{20}$? It counts the number of ways to use powers of two to reach 20, so that should be it, right?

Yep, that's what you do. My question is, how do you make practical use out of that -- without multiplying out? I can't find a significantly nicer form for the function than what you've put it in as.

[quote="t0rajir0u"][hide="Alternately"] The relevant generating function is $ (1 \plus{} x \plus{} x^2 \plus{} ...)(1 \plus{} x^2 \plus{} x^4 \plus{} ...)(1 \plus{} x^4 \plus{} x^8 \plus{} ...)...$. [/hide][/quote]

OK,, how does generating function work usually?? Can someone explain basics about generating function or provide links please??

ALso, how does this one work for the question above?

A generating function of this type will give you the number of solutions $ (m_1, m_2, \ldots m_n)$ to $ a_1 m_1 \plus{} a_2 m_2 \plus{} a_3 m_3 \plus{} \ldots \plus{} a_n m_n \equal{} k$, where $ a_i$ are given weights, $ k$ is a given value, and $ m_i$ are non-negative integers.

The general form of such a function is $ (1 \plus{} x^{a_1} \plus{} x^{2 a_1} \plus{} \ldots) (1 \plus{} x^{a_2} \plus{} x^{2 a_2} \plus{} \ldots) (1 \plus{} x^{a_3} \plus{} x^{2 a_3} \plus{} \ldots) \ldots$, and the answer is the coefficient of $ x^k$ in the expansion of the above.

When you expand a product of polynomials, you consider the product of each instance when you pick one term from each of the factors. So here, you sum over all possible ways of picking one term of the form $ x^{m_1 a_1}$, one term of the form $ x^{m_2 a_2}$, etc., where $ m_i$ are arbitrary non-negative integers. Their product is $ x^{m_1 a_1 \plus{} m_2 a_2 \plus{} \ldots \plus{} m_n a_n}$.

If we now collect all terms such the exponent $ m_1 a_1 \plus{} m_2 a_2 \plus{} \ldots \plus{} m_n a_n$ is equal to the desired exponent $ k$, then the coefficient on this collected term will give the number of different ways we could achieve $ k$ in the form $ m_1 a_1 \plus{} m_2 a_2 \plus{} \ldots \plus{} m_n a_n$.


LaTeX
real analysis
real analysis unsolved

i havent yet installed latex so please bear with me. here is the qn

Let (Pn) be a sequence and f and bijection from N onto N. define another sequence Qk=Pf(k). such a sequence is called a re-arrangement. are limits of a sequence unaffected by re arrangement? wat if f is an injection? a surjection?[/quote][/list]

You don't need to install $\LaTeX$ to use it here- just put the code in your post between dollar signs, and let the forum do the rest.

Yes, a rearrangement of a convergent sequence is convergent to the same limit. Any rearrangement of a Cauchy sequence is Cauchy as well. It's a nice exercise in understanding that "eventually" quantifier in the definition of limit.

An injection: How do you know that $Q$ is even defined everywhere? It can't be nice in general, because it can do anything at the points that didn't come from what we knew before.

A surjection: Those early values that aren't close to the limit can come back as often as we want (as long as we leave room for each value to show up at least once). If every other element of the new sequence is $f(1)$, it's unlikely our new sequence converges to the same limit.


Το παρακάτω θέμα τέθηκε σε πρόσφατο διαγωνισμό στην Βουλγαρία (πάλι θέλει την άτιμη μέθοδο  :mad: .)
Να λυθεἰ στους ακεραίους το σύστημα

$ 3a^{4} \plus{} 2b^{3} \equal{} c^{2}$

$ 3a^{6} \plus{} b^{5} \equal{} d^{2}$

Καλά Χριστούγεννα !

Και από μένα χρόνια πολλά σε όλους και καλές γιορτές!

Θα δείξουμε ότι οι αριθμοί $ a, b, c, d$ διαιρούνται από οσοδήποτε μεγάλες δυνάμεις του $ 3$ θέλουμε και άρα η μοναδική λύση του συστήματος είναι η $ a\equal{}b\equal{}c\equal{}d\equal{}0$. Θα ασχοληθούμε μόνο με τη δύναμη του 3 που διαιρεί κάθε αριθμό. Προς τούτο η εξίσωση $ 3^ka^4\plus{}2*3^lb^3\equal{}3^mc^2$ θα συμβολίζεται με $ k, l, m$ και αντίστοιχα για τη δεύτερη εξίσωση. Τότε η αρχική κατάσταση γράφεται $ 1, 0, 0 | 1, 0, 0$. Όμως $ x^2\equiv0,1\pmod{3}$ και άρα εύκολα παίρνουμε ότι $ 3/b$ και επομένως $ 3/c,d$ και $ 3/a$ και συνεπώς πάμε στην κατάσταση $ 4, 3, 2 | 6, 5, 2$ --> $ 4, 3, 4 | 6, 5, 6$ και ούτω καθεξής. Αρκεί λοιπόν να δείξουμε ότι αυτή η διαδικασία δεν τερματίζει. Πράγματι για να συμβεί αυτό πρέπει και στις δύο εξισώσεις τουλάχιστον $ 2$ εκθέτες του $ 3$ να είναι ίσοι. Όμως γενικά οι εξισώσεις θα είναι της μορφής $ 1\plus{}4x, 3y, 2z | 1\plus{}6x, 5y, 2t$. Άρα θα έχουμε $ 3y\equal{}2z, 5y\equal{}2t$ (οι άλλες περιπτώσεις εύκολα αποκλείονται). Τότε όμως παίρνουμε πάλι ότι $ 3/b',c'$. Άρα η διαδικασία δεν τερματίζει και το ζητούμενο αποδείχτηκε  :wink: .


advanced fields
advanced fields unsolved

Suppose that the topological space $Z$ has a universal covering. Show that for any coverings $p_{1}: X \to Y$ and $p_{2}: Y \to Z$, $p_{2}\circ p_{1}: X \to Z$ is a covering.

Are you sure about this?
Our teacher said yesterday that $p_{2}\circ p_{1}$ need not to be a covering space, but $X$ is covring space of $Z$.

it was in my exam, and I think I prove it, but maybe I made a mistake. And here there is an additional hypothesis ($Z$ has a universal covering)


Hey Guys

I have to find inverse of

f(x)=3+ln(e^3-2) where x>ln2

ne ideas????

Thnx

i'm assuming you mean $f(x)=3+\ln (e^x-2)$...

[hide="hint"]try isolating functions such as the natural log[/hide]

[hide="answer"] $f^{-1}(x)=\ln (e^{x-3}+2)$[/hide]

[hide="Simple"]Solve for x and then interchange x and y.
$\\y=3+\ln (e^x-2)\implies e^{y-3}=e^x-2\implies e^{y-3}+2=e^x\\ \implies f^{-1}(x)=\ln (e^{x-3}+2)$[/hide]
Masoud Zargar

Thnx Guys

But why is it ln(e^x-3 +2 ) and not ln(e^y-3 +2 )???

because f(x) implies finding a "y-value" for a function in terms of x


The ACT December scores are released! Who took 'em??

I was gonna take it but i got sick the day before...so i postponed it till feb.


function
algebra
IMO Shortlist
functional equation

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Please, can someone post a solution for this? Thanks!

the only thing I could proove is that $f(x)=f(\frac{1}{x})$...and
actually something like $f(x)+f(y)=f(\sqrt{\frac{x}{y}})f(\sqrt{xy})$
my hipothesis is $f(x)=x^m+\frac{1}{x^m}$ , but i still can't get it ...
it's difficult anyway ...

It can't be that hard.
The desired answer is indeed $f_m(x)=x^m+\frac{1}{x^m}$ , for any $m>0$.
It is easy to check that it satisfies our equation. I think I have an "ugly solution" for this.
Fix some $t>1$ as given and $f(t)=t^m+\frac{1}{t^m}=r>2$ , $r$-given (from this it follows some $m$). We can express $f(t^2)=f^2(t)-2$ , $f(t^3)=f^3(t)-3f(t)$ , ... and generally $f(t^n)=P_n(f(t))$, for any integer $n>0$ , where $P_n$ are polynomials ($n$-even, $P_n$-even ; while $n$-odd, $P_n$-odd and coefficients in $P_n$ are only function of $n$ - I computed them recursively and they have an "ugly" form in my oppinion). Hence  $f(t^n)$ just follows uniquely determined from the given $f(t)$ - anyway i don't need those coefficients in explicitely form, since my goal is to show that $r=P_n(x)$ has only one solution greater than $2$, $x>2$ for $r>2$ - it results by induction this important fact in the proof; here it is used the increasing condition. The equation $f(t)=P_n(f(t^{1/n}))$ has an unique solution for $f(t^{1/n})>2$ uniquely determined by that initial value $f(t)=r>2$.
So for any rational $p/q>0$ I can compute uniquely  $f(t^{p/q})$ from $f(t^p)$ which come from $f(t)$, then my function is well-determined for any rational $x=t^{p/q}>1$ as $f(x)$ starting from that $f(t)=r>2$, then it follows by extension for any real $x>1$, using again increasing condition!
Since $f(x)=f(\frac{1}{x})$ we get the only possible answer $f_m(x)=x^m+\frac{1}{x^m}$ , $f_m: \mathbb{R}^+ \longrightarrow \mathbb{R}^+$ for any $m>0$. I hope that some other "good looking" proof exists.

this solution doesn't convince me , I'll will have to look through this concise
deduction again ....

Which part of the proof you don't agree?  :cool:

[quote="heartwork"]...
The desired answer is indeed $f_m(x)=x^m+\frac{1}{x^m}$ , for any $m>0$.
Fix some $t>1$ as given and $f(t)=t^m+\frac{1}{t^m}=r>2$ , $r$-given (from this it follows some $m$). We can express $f(t^2)=f^2(t)-2$ , $f(t^3)=f^3(t)-3f(t)$ , ... and generally $f(t^n)=P_n(f(t))$, for any integer $n>0$ , where $P_n$ are polynomials ...[/quote]

I have seen both official solutions for this problem! Surprisingly, both uglier and much longer than mine!
These polynomials obey a similar recurrence as the Chebyshev polynomial of degree n:
$P_n(2x)$ similar to $p_n(x) = $ cos$(n $ arccos$(x)) $
$p_{n+1}(x) = 2xp_n(x) - p_{n-1}(x)$

[hide="Inspired by PftB"]
When I write $a,b,c$ it means that I put $x=a,y=b,z=c$
$1,1,1: f(1)=2$

$x^2,1,1: f(x^2)+2=f(x)^2$
Now we see that for every $x>0$, $f(x)>\sqrt2$. Putting that in, again we see that $f(x)>\sqrt{2+\sqrt{2}}$ Continue doing that, and soon we get that $f(x)\ge 2$ for $x>0$ (because $\sqrt{2+\sqrt{2+\sqrt{\ldots}}}=2$

That means that there exists a function $g$, such that $f(x)=g(x)+\frac{1}{g(x)}$. Now let $g(x)\ge 1\Longleftrightarrow x\ge 1$ (we can do that because we can choose between 2 values of $g$ for every $x$ and this way $g$ has some nice properties).

From the second condition we easily get that $g$ is strictly increasing on $(1,\infty)$

$x,x,\frac1x: f(x)=f\left(\frac1x\right)$
This means that $g(x)g\left(\frac1x\right)=1$, so $g$ is strictly increasing on $(0,\infty)$

$x^2,y^2,\frac{1}{x^2}: f(x^2)+f(y^2)=f(xy)f\left(\frac{x}{y}\right) (1)$
Putting $x^2=zt,y^2=\frac zt$ in $(1)$ we get $f(zt)+f\left(\frac zt\right)=f(z)f(t)   (2)$
Note that $(1)$ and $(2)$ hold for all $x,y,z,t\in\mathbb{R}^+$

Now, from $(2)$, $f\left(\frac xy\right)=f(x)f(y)-f(xy)$. Putting that into $(1)$ and $f(x^2)=f(x)^2-2$ we get $f(x)^2+f(y)^2+f(xy)^2-f(x)f(y)f(xy)=4$

It's well known that there are numbers $a,b,c$ such that $f(x)=a+\frac1a,f(y)=b+\frac1b,f(xy)=c+\frac1c$ and $abc=1$. Now it's obvious that $g(a)g(b)=g(ab)$, which is Cauchy's equation (using $ln$ function and that $g$ is strictly increasing), so $g(x)=x^k$ for some $k\ge 0$

Now it's obvious that $f(x)=x^k+\frac{1}{x^k}$ for some $k\ge 0$.

[/hide]

Lemma 1: $ f(x) \geq 2 $ for all $ x \in \mathbb{R}^{+} $.

[i]Proof:[/i]  Note that by letting $ x = y = z = 1 $ in the equation we find that $ f(1) = 2 $ and then letting $ y = z = 1 $ in the equation and replacing $ x $ by $ x^2 $ we have that $ f(x^2) = f(x)^2 - 2 $.  This implies $ f(x) > \sqrt{2} $ for all $ x \in \mathbb{R}^{+} $.  Using this result we now get $ f(x) > \sqrt{2+\sqrt{2}} $ for all $ x \in \mathbb{R}^{+} $.  Continuing in this fashion we obtain the desired result.

Lemma 2:  Let $ n $ be a nonnegative integer.  Let $ p_{n} $ be a polynomial defined by the following criteria: $ p_0 = 2, p_1 = x, p_{k + 1} = xp_k - p_{k - 1} $ for all positive integers $ k $.  Then $ f(t^n) = p_n(f(t)) $ for all $ t \in \mathbb{R}^{+} $.

[i]Proof:[/i]  It is clear that $ p_n\left(t + \frac{1}{t}\right) = t^n + \frac{1}{t^n} $ for all real $ t $ (this is easily proven by induction) so it suffices to show that $ f(t^{2n + 1}) = f(t)f(t^{2n}) - f(t^{2n - 1}) $ and $ f(t^{2n}) = f(t^n)^2 - 2 $.  We already know that $ f(x^2) = f(x)^2 - 2 $ for all $ x \in \mathbb{R}^{+} $ so the second identity follows immediately.  To prove the first identity, let $ x = t^{2n -1 } $ and $ y = z = t $ in out equation.  Then we obtain $ f(t^{2n + 1}) + f(t^{2n - 1}) + 2f(t) = f(t)f(t^{n})^2 = f(t)(f(t^{2n}) + 2) $ and so $ f(t^{2n + 1}) = f(t)f(t^{2n}) - f(t^{2n - 1}) $ as desired.

Lemma 1 implies that for any $ s \in \mathbb{R}^{+} $ we can write $ f(s) = a + \frac{1}{a} $ for some real $ a \geq 1 $.  Lemma 2 then implies that $ f(s^q) = a^q + \frac{1}{a^q} $ for all $ q \in \mathbb{Q} $.  Because the rationals are dense in the reals and because $ f $ is increasing on the interval $ [1, \infty) $ this implies that $ f(s^r) = a^r + \frac{1}{a^r} $ for all $ r \in \mathbb{R} $.  This clearly implies that $ f(x) = x^m + \frac{1}{x^m} $ for some $ m \in \mathbb{R} $ and it is easy to check that all such functions satisfy the original equation.

[hide=Solution]
We claim the only solutions are in the form $x^k+\frac{1}{x^k}$ for $k>0$. We can easily verify that these solutions work by expanding, and $x^k+\frac{1}{x^k}$ is clearly increasing for $k>0$.

Now, let $P(x, y, z)$ denote assertion. Then,
$P(1, 1, 1) \implies 4f(1) = f(1)^3 \implies f(1) = 2$(since $f(1) > 0$). (1)
$P(x^2, 1, 1) \implies 2f(x^2)+4=2f(x)^2 \implies f(x^2) = f(x)^2-2$. (2)
$P(x, x, x) \implies f(x^3)+3f(x) = f(x)^3 \implies f(x^3) = f(x)^3-3f(x)$. (3)

Lemma 1: $f(x) = x^k+\frac{1}{x^k} \iff f(x^2) = x^{2k}+\frac{1}{x^{2k}} \iff f(x^3) = x^{3k}+\frac{1}{x^{3k}}$ for $x>1$

Proof: The only if part can be done by simple expansion with either (2) or (3). The only if part is using the fact that $f(x)^2-2$ and $f(x)^3-3f(x)$ are both increasing if $f(x)>2$, as we could easily check by taking derivatives.

Now, note that since $2^k+\frac{1}{2^k}$ is continuous and increasing in $k$, and is equal to $2$ when $k=0$, we can find $k>0$ such that $f(2) = 2^k+\frac{1}{2^k}$. Now, we claim that this implies that $f(x) = x^k+\frac{1}{x^k} \forall x$.

Lemma 2: There exists arbitrarily close numbers to $1$ that are greater than $1$ among the numbers of the form $\frac{2^x}{3^y}, x, y \in \mathbb{Z}$

Proof: We consider $\log_3 \frac{2^x}{3^y}$, and prove that it can be an arbitrarily small positive number. Note that this is equal to $x\log_3(2) - y$. Now, we show that there exists nonzero integer $x$ such that  $\{x\log_3(2)\}<\epsilon \forall \epsilon > 0$,, where $\{a\}$ denotes the fractional part of $a$.  Choose $N$ such that $N > \frac{1}{\epsilon}$. Now, consider the intervals $[0, \frac{1}{N}), [\frac{1}{N}, \frac{2}{N}), …, [\frac{N-1}{N}, 1)$. Consider the numbers $\{c\}, \{2c\}, …, \{(N+1)c\}$, where $c = \log_3(2)$. Note that $2$ of the numbers must fall in the same interval by pigeon-hole, suppose they are $\{ic\}$ and $\{jc\}$, with $\{ic\} > \{jc\}$(note that they cannot be equal since $\log_3(2)$ is irrational). Then, we simply take $\{(i-j)c\}$ and this must be less than $\frac{1}{N}$, which is less than $\epsilon$. Now, we simply let $y = \lfloor x \rfloor$, proving the lemma.

Now, fix $x>1$. Now, suppose $s$ is in the form $\frac{2^i}{3^j}, i, j \in \mathbb{Z}$. Now, by repeatedly applying Lemma 1, $f(2^{s^a}) = (2^{s^{a}})^k+\frac{1}{(2^{s^a})^k}$ for $a \in \mathbb{Z}$. Furthermore, among the numbers in the form $2^{s^a}, a \in \mathbb{Z}$, we can get within an exponentiation of $s$ of $x$(otherwise we have to jump more than an exponentiation of $s$ to go from below $x$ to above $x$ or vice versa, which is a contradiction, since the numbers in this form are all an exponentiation of $s$ apart), so since $s$ can get arbitrarily close to $1$, $2^{s^a}$ can get arbitrarily close to $x$, and thus since $f$ is increasing, $f(x) = x^k+\frac{1}{x^k}$.

Furthermore, $P(x, x, \frac{1}{x}) \implies f(\frac{1}{x})+3f(x) = 4f(\frac{1}{x}) \implies f(\frac{1}{x}) = f(x)$, so $f(x) = x^k + \frac{1}{x^k} \forall x$.
[/hide]

Can sb explain the solution here?(problem 14)
http://www.imomath.com/index.php?options=341&lmm=0

Finally solved this, yay!  :trampoline: (I had tried this problem around 8 months back, but had failed to solve it then.)
[quote=orl]Let $\mathbb{R}^+$ be the set of all positive real numbers. Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ that satisfy the following conditions:

- $f(xyz)+f(x)+f(y)+f(z)=f(\sqrt{xy})f(\sqrt{yz})f(\sqrt{zx})$ for all $x,y,z\in\mathbb{R}^+$;

- $f(x)<f(y)$ for all $1\le x<y$.

[i]Proposed by Hojoo Lee, Korea[/i][/quote]
[b]Answer :[/b] $f(x) = x^k + \frac{1}{x^k}$ for some $k > 0$. One can confirm that this function does work.\\

$x=y=z=1 \implies 4 \cdot f(1) = f(1)^3 \iff f(1) = 2$. Now $x^2,1,1$ gives us that $$ 2f(x^2) + 4 = 2f(x)^2 \implies f(x^2) + 2 = f(x)^2 \dots (\star)$$

[u]Claim:[/u] $f(x) = h(x) + \frac{1}{h(x)}$ for some function $h : \mathbb{R^+} \rightarrow \mathbb{R^+}$.\\
$  $\\
[i]Proof.[/i] From $(\star)$ we can guarantee that $f \ge \sqrt{2}$. But this again gives us that $f \ge \sqrt{2+\sqrt{2}}$. Going on recursively, we have that $f \ge 2$. Hence such a function $h$ exists. Moreover define $h(x) \ge 1$ whenever $x \ge 1$. \\

So the property of $f$ asserts that $h$ is increasing on $(1 , \infty)$. Also note that such a function $h$ nicely interprets the condition $\star$.\\
$  $\\
Now consider $Q\left(x,x,\frac{1}{x}\right)$ to get that $f(x) = f\left(\frac{1}{x}\right) \implies h\left(\frac{1}{x}\right) = \frac{1}{h(x)} \ \forall x \implies h$ is increasing on $(0, \infty)$. Using $(\star)$ on $Q(x^2,y^2,1)$ we also get that $$ f(x)^2 + f(y)^2 + f(xy)^2 = f(x)\cdot f(y) \cdot f(xy) + 4$$
This is indeed a very famous setup. We get that there exist positive reals $a,b,c$ such that $abc=1$ and $f(x) = a + \frac{1}{a}$ and so on. Thus, using these substitutions we arrive at the fact that $h$ is multiplicative and considering the function $g = \log{f(a^x)}$ we get that $g$ is additive and increasing $\implies g(x) = cx \implies h(x) = x^k$ for some $k$.(This follows from $h(1)=1$). Hence we have our solution.

We claim that the solutions are $x^m + 1/x^m$ where $m$ is any positive real number. It's easy to see these work, so we'll prove these are the only solutions.
Let $P(x, y, z)$ denote the condition $f(x^2y^2z^2) + f(x^2) + f(y^2) + f(z^2) = f(xy)f(yz)f(zx)$ for all positive reals $x, y, z$ (equivalent to the first condition of the problem).

$P(1, 1, 1) \implies f(1) = 2$. $P(z, 1, 1) \implies f(z^2) = f(z)^2-2$. $P(z, z, 1/z) \implies f(z) = f(1/z)$. Subtracting $P(x, y, z)$ from $P(x, 1/y, 1/z)$ gives $f(x^2/(y^2z^2)) - f(x^2y^2z^2) = f(x/y)f(x/z)f(yz) - f(xy)f(xz)f(yz)$. Putting $y=1$ here gives $f(x^2/z^2) - f(x^2z^2) = f(x)f(z)(f(x/z) - f(xz))$. So $f(x/z) + f(xz) = f(x)f(z)$ ($f(x/z) \neq f(xz)$ if neither of $x, z$ equals $1$ (easily seen by the second condition of the problem and $f(a) = f(1/a)$). The case where one of them equals $1$ follows independently).

Let $f(2) = 2^a + 1/2^a$ where $a$ is a positive real number (note that the function $2^x + 1/2^x$ for $x > 0$ has range $(2, \infty)$ and is strictly increasing). Call a positive real number $k$ good if $f(k) = k^a + 1/k^a$. Note that if any $3$ of $x, z, xz, x/z$ are good, then the $4^{th}$ is good too $(1)$.

We claim all numbers of the form $2^{c/2^b}$ where $c, b$ are non-negative integers, are good.
We prove this by strong induction on $c$. The case $c = 0$ is trivial. For the case $c = 1$, we induct on $b$, with base case $b = 0$ trivially true, and the induction step following from $f(x) = \sqrt{f(x^2)+2}$. Returning to our induction on $c$, if the hypothesis is true till $c = l$, $l \ge 1$, the induction step follows by putting $x = 2^{l/2^b}, z = 2^{1/2^b}$ in $(1)$.

Note that the function of $x$, $x^a + 1/x^a$ with domain $[1, \infty)$ has range $[2, \infty)$ and is strictly increasing for any positive real number $a$. Since $f$ is strictly increasing over $[1, \infty)$ and the good set of numbers of the form $2^{c/2^b}$ is a dense subset of $[1, \infty)$, all numbers in $[1, \infty)$ are good, thus all positive real $x$ are good.

Nice problem!

The answer is $f(x) = x^r + \frac{1}{x^r}$ for $r \in \mathbb{R}$. These can be easily verified to work, so we show uniqueness. Let $P(x,y,z)$ denote the assertion.

We have the following preliminary observations:
[list]
[*] Plugging in $P(1,1,1)$ yields $4f(1) = f(1)^3$, or $f(1) = 2$.
[*] Plugging in $P(x^2,1,1)$ yields $f(x^2) + 2 = f(x)^2$.
[*] Plugging in $P(x^2, \tfrac1{x^2}, 1)$ yields
$$4 + f(x^2) + f(\tfrac1{x^2}) = 2f(x)f(\tfrac1x),$$
and noting the LHS is equal to $f(x)^2 + f(\tfrac1x)^2$ gives $f(x) = f(\tfrac1x)$. We thus turn out attention to defining $f$ over the interval $[1,\infty)$.
[/list]
The final push is substituting $P(x^2,y^2,1)$, which yields
$$f(x^2y^2) + f(x^2) + f(y^2) + 2 = f(xy)f(x)f(y).$$
The LHS is equal to $f(xy)^2 + f(x)^2 + f(y)^2 - 4$, so the equation is a quadratic in $f(xy)$. Solving gives
$$f(xy) = \frac{f(x)f(y) \pm \sqrt{(f(x)^2 - 4)(f(y)^2 - 4)}}{2}.$$
I claim that for $x,y \geq 1$, the $\pm$ on the RHS is always a $+$. The claim is trivial for $x = 1$ and $y=1$, so assume $x,y > 1$. Without loss of generality $x \geq y$. Because $f(xy) > f(x)$, we must have
$$f(x)f(y) - \sqrt{(f(x)^2 - 4)(f(y)^2 - 4)} > 2f(x) \implies f(x)(f(y)-2) > \sqrt{(f(x)^2-4)(f(y)^2-4)}.$$
Squaring both sides and dividing by $f(y) - 2$ gives
$$f(x)^2(f(y)-2) > (f(x)^2-4)(f(y)+2) \implies f(x)^2 < f(y) + 2,$$
or $f(y) > f(x^2)$. This is absurd, as $y \leq x < x^2$, as desired.

Finally, because $f(x) \geq 2$ for all $x \geq 1$, it makes sense to let $f(x) = g(x) + \frac{1}{g(x)}$ for a function $g : [1,\infty) \to [1,\infty)$. Substituting this into the quadratic formula yields $g(xy) = g(x)g(y)$. Because $f$ is increasing on $[1,\infty)$, so must $g$, and by Cauchy FE $g(x) = x^r$ as desired.

We claim that the answer is $f(x)=x^r+\frac{1}{x^r}$ for all $r\in\mathbb{R}_{> 0}$.

Denote the assertion as $P(x,y,z)$. First, note that $P\left(x,x,\frac{1}{x}\right)$ yields $\left(f(1)^2-3\right)f(x)=f\left(\frac{1}{x}\right)$ and $x\to\frac{1}{x}$ gives $f(1)=2,f(x)=f\left(\frac{1}{x}\right)$

Now, $P\left(z,x,\frac{1}{x}\right)$ yields $2f(z)+2f(x)=f(1)f\left(\sqrt\frac{z}{x}\right)f\left(\sqrt{zx}\right)$. If we consider $z\to rz$, the change on the RHS is independent of $x$, so we can rewrite this as $f(a)f(b)-f(ar)f(br)$ for a given $r$ is constant if $ab$ is fixed.

Setting $ab=1$, this means $f(x)^2-f(xr)f\left(\frac{x}{r}\right)=g(r)$ for some function $g$. Now, suppose $r>1$ and we fix $f(r)=k>2$. The function $h(m)=r^m+\frac{1}{r^m}$ for $m>0$ is surjective over $\mathbb{R}_{>2}$, so we can choose suitable $m$ such that $f(r)=r^m+\frac{1}{r^m}$.

Now, as $g(r)=f(1)^2-f(r)^2$, $f(r)$ must determine all values of $f(r^z)$ for $z\in\mathbb{N}$ just by applying our equation over and over again. Thus, these values of $f$ are fixed. However, setting $f$ to be $r^m+\frac{1}{r^m}$ does satisfy the equation, and therefore we have $f(r^z)=r^{zm}+\frac{1}{r^{zm}}$. Now, consider $s>1$ and $f(s)=s^n+\frac{1}{s^n}$. Then, $f(s^z)=s^{zn}+\frac{1}{s^{zn}}$. Assume $m\neq n$ and WLOG $m<n$. Then, the function $x^n+\frac{1}{x^n}$ grows faster than $x^m+\frac{1}{x^m}$, so we can find suitable $z_1,z_2$ such that $s^{z_1}<r^{z_2}$ but $f(s^{z_1})>f(r^{z_2})$, which is a contradiction to the second condition.

Thus, our assumption is wrong, and there must exist a constant $m$ such that $f(r)=r^m+\frac{1}{r^m}$ for all $r>1$, and we are done.

Solved with [b]anser[/b].

The answer is $f(x) = x^r + \frac{1}{x^r}$ for some $r > 0$. Let $P(x, y, z)$ denote the equation. First, $P(1, 1, 1)$ gives $f(1) = 2$, and $P\left(x, x, \frac{1}{x}\right)$ gives $3f(x) + f\left(\frac{1}{x}\right) = 4f(x)$, so $f(x) = f\left(\frac{1}{x}\right)$. Also $P\left(x^2, 1, 1\right)$ gives $f(x^2) + 2 = f(x)^2$. Now we focus on $x > 1$, noting that $f(x) > f(1) = 2$ for $x > 1$.

[b]Claim:[/b] If $f(a) = a^k + \frac{1}{a^k}$ for some positive integer $k$, then $f(a^c) = a^{ck} + \frac{1}{a^{ck}}$ for all positive integers $c$.

[i]Proof:[/i] We use induction on $c$ with base case $c = 1$. If $c$ is even, then
\[f(a^c) = f(a^{c/2})^2 - 2 = \left(a^{c/2} + \frac{1}{a^{c/2}}\right)^2 - 2 = a^{c} + \frac{1}{a^c}.\]
If $c$ is odd, $P(a^{c-2}, a, a)$ gives $f(a^c) + f(a^{c-2}) + 2f(a) = f(a^{(c-1)/2})^2f(a) = (f(a^{c-1})+2)f(a)$, so
\[f(a^c) = \left(a^{c-1} + \frac{1}{a^{c-1}}\right)\left(a + \frac{1}{a}\right) - \left(a^{c-2} + \frac{1}{a^{c-2}}\right) = a^{ck} + \frac{1}{a^{ck}}.\]

Now since $f(x) > 2$, for any $x$ we can write $f(x) = x^k + \frac{1}{x^k}$ for some $k > 0$. Assume that this value of $k$ is not constant over all $x > 1$. Then there exists $x < y$ with $f(x) = x^k + \frac{1}{x^k}$ and $f(y) = y^\ell + \frac{1}{y^\ell}$. If $k < \ell$, then we can replace $y$ with $y^{1/t}$ for some very large $t$ such that $y^{1/t} < x$, and swap $x$ and $y$ (since $y^{1/t}$ and $y$ have the same $\ell$-value by the previous claim). So assume $k > \ell$.

[b]Claim:[/b] If $x, y > 1$, and $k > \ell$ are positive integers, then there exists positive integers $c$ and $d$ such that $x^c \leq y^d$ and $x^{ck} > y^{d\ell}$.

[i]Proof:[/i] The second equation rearranges to \[\left(\frac{x^c}{y^d}\right)^k > y^{d(\ell - k)}.\] No matter what $d$ is, we can choose $c$ such that $\frac{x^c}{y^d} \geq \frac{1}{x}$ (as if it's less than this, we can increase $c$ by $1$), so the LHS is at least $\frac{1}{x^k}$. Meanwhile, since $y > 1$ and $\ell - k < 0$, as $d$ gets large the RHS gets arbitrarily small. So we can pick $d$ extremely large to achieve this.

But this contradicts the increasing condition -- we have $f(x^c) = x^{ck} + \frac{1}{x^{ck}}$ and $f(y^d) = y^{dk} + \frac{1}{y^{dk}}$ by the first claim. Since $x + \frac{1}{x}$ is increasing on $x > 1$, and $x^{ck} > y^{dk}$, this means $f(x^c) > f(y^d)$. But $1 \leq x^c < y^d$, so this is a contradiction.

[b]Remark:[/b] The motivation for the proof that $k$ must be constant is that if, for example, $f(3) = 3^6 + \frac{1}{3^6}$ and $f(4) = 4^5 + \frac{1}{4^5}$, then \[f(3^5) = 3^{30} + \frac{1}{3^{30}} > f(4^4) =  4^{20} + \frac{1}{4^{20}}\] even though $3^5 < 4^4$, which suggests that we can generally find a counterexample (to the second condition) of this form.

[quote=orl]Let $\mathbb{R}^+$ be the set of all positive real numbers. Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ that satisfy the following conditions:

- $f(xyz)+f(x)+f(y)+f(z)=f(\sqrt{xy})f(\sqrt{yz})f(\sqrt{zx})$ for all $x,y,z\in\mathbb{R}^+$;

- $f(x)<f(y)$ for all $1\le x<y$.

[i]Proposed by Hojoo Lee, Korea[/i][/quote]

Kinda easy.

Let \(P(x,y,z)\) denote the assertion of the given functional equation. \(P(1,1,1)\) gives us that \(f(1)=2\) and \(P(x,1,1)\) gives us that \[f(\sqrt{x})^2=f(x)+2\] Let \(f(x)=g(x)+\frac{1}{g(x)}\) (we can do this since \(f(x)>f(1)=2\) for \(x\geq1\)). Then, the above relation re-writes as \[g(\sqrt{x})+\frac{1}{g(\sqrt{x})}=\sqrt{g(x)}+\frac{1}{\sqrt{g(x)}}\] Now, \(P(x,y,1)\) combined with the fact above gives us that \[\left(\sqrt{g(x)}+\frac{1}{\sqrt{g(x)}}\right)^2+\left(\sqrt{g(y)}+\frac{1}{\sqrt{g(y)}}\right)^2+\left(\sqrt{g(xy)}+\frac{1}{\sqrt{g(xy)}}\right)^2=\left(\sqrt{g(x)}+\frac{1}{\sqrt{g(x)}}\right)\left(\sqrt{g(y)}+\frac{1}{\sqrt{g(y)}}\right)\left(\sqrt{g(xy)}+\frac{1}{\sqrt{g(xy)}}\right)+4\] and so \(g(x)g(y)=g(xy)\), \(g(xy)g(x)=g(y)\) or \(g(x)g(y)g(xy)=1\) (see PFTB chap 1). All relations except \(g(x)g(y)= g(xy)\) give us that \(g(x)=1\) for all \(x\geq 1\) which is impossible. Therefore \(g\) is multiplicative and strictly monotone after \(1\), so \(g(x)=x^k\) for all \(x\geq 1\) where \(k\) is a fixed integer, so \(f(x)=x^k+\frac{1}{x^k}\) for all \(x\geq1\). Now let \(x_0<1\) be a real. Choose \(y_0, z_0\) arbitrarily large. Then \(P(x_0,y_0,z_0)\) gives us that \(f(x_0)=x_0^k+\frac{1}{x_0^k}\) since all inputs in the first condition except \(x_0\) are greater than \(1\) and you get the functional relation as an identity.

In conclusion, the only solution is \(f(x)=x^k+\frac{1}{x^k}\) for any integer \(k\).

Let $P(x,y,z)$ be the given assertion.

$P(1,1,1)$ gives $f(1)=2.$ $P(x,1/x,x)$ gives $f(x)=f(1/x).$ $P(1,x^2,1)$ gives $f(x^2)=f(x)^2-2.$ For Euler $e$ we have $f(e)\geq f(1)$ thus for constant $k,$ $f(e)=e^k+\frac{1}{e^k}.$ By induction for any $s,t \in \mathbb{Z^+}$ $$f(e^{s\frac{1}{2^t}})=e^{ks\frac{1}{2^t}}+\frac{1}{e^{ks\frac{1}{2^t}}}.$$ Moreover $f$ is increasing provides $f(x)=x^k+\frac{1}{x^k},$ which fits.

We claim that the answer is $x^k+\frac{1}{x^k}$ which clearly follows the rules. Now to prove it is the only answer.
\\$~$\\
Let $P(x,y,z)$ denote the assertion. Now, $P(x,x,x^{-1})$ gives $3f(x)+f(x^{-1})=4f(x)$ so $f(x^{-1})=f(x).$
\\$~$\\
Thus, while $f(x)\ge 2$ on $x\ge 1$ we also must have $f(x)\ge 2$ on $x\le 1$ so let $f(x)=\frac{1}{g(x)}+g(x).$ Then, $P(1,x^2,y^2)$ gives $f(x^2y^2)+f(x^2)+f(y^2)+2=f(x)f(y)f(xy).$ $P(1,1,x^2)$ gives $f(x^2)=f(x)^2-2$ so $f(xy)^2+f(x)^2+f(y)^2-4=f(x)f(y)f(xy).$
\\$~$\\
Now, when we substitute in $g(x)$ we get that $g(xy)=g(x)g(y).$ Combined with $g$ being increasing on interval $(1,\infty)$ and decreasing on interval $(0,1)$ we have $g(x)=x^k$ for some $k$ as desired.


geometry circumcircle algebra polynomial function trigonometry ratio

Here are the problems of China TST 2003. Note that all the problems are translated by zhaoli.

[b]Test 1[/b]:
Problem 1: http://www.mathlinks.ro/Forum/viewtopic.php?t=99369
$x$, $y$ and $z$ are positive reals such that $x+y+z=xyz$. Find the minimum value of:
\[ x^7(yz-1)+y^7(zx-1)+z^7(xy-1)  \]

Problem 2: http://www.mathlinks.ro/Forum/viewtopic.php?t=99370
In triangle $ABC$, the medians and bisectors corresponding to sides $BC$, $CA$, $AB$ are $m_a$, $m_b$, $m_c$ and $w_a$, $w_b$, $w_c$ respectively. $P=w_a \cap m_b$, $Q=w_b \cap m_c$, $R=w_c \cap m_a$. Denote the areas of triangle $ABC$ and $PQR$ by $F_1$ and $F_2$ respectively. Find the least positive constant $m$ such that $\frac{F_1}{F_2}<m$ holds for any $\triangle{ABC}$.

Problem 3: http://www.mathlinks.ro/Forum/viewtopic.php?t=99371
There is a frog in every vertex of a regular 2n-gon with circumcircle($n \geq 2$). At certain time, all frogs jump to the neighborhood vertices simultaneously (There can be more than one frog in one vertex). We call it as $\textsl{a way of jump}$. It turns out that there is $\textsl{a way of jump}$ with respect to 2n-gon, such that the line connecting any two distinct vertice having frogs on it after the jump, does not pass through the circumcentre of the 2n-gon. Find all possible values of $n$.

[b]Test 2[/b]:
Problem 4: http://www.mathlinks.ro/Forum/viewtopic.php?t=52681
Convex quadrilateral with side lengths $a$, $b$, $c$ and $d$ has an incircle centred at $O$. Prove that
\[ OA \cdot OC + OB \cdot OD = \sqrt{abcd}  \]

Problem 5: http://www.mathlinks.ro/Forum/viewtopic.php?t=99374
Let $x<y$ be positive integers and $P=\frac{x^3-y}{1+xy}$. Find all integer values that $P$ can take.

Problem 6: http://www.mathlinks.ro/Forum/viewtopic.php?t=99375
The $n$ roots of a complex coefficient polynomial $f(z)=z^n+a_z^{n-1}+ \cdots + a_{n-1}z + a_n$ are $z_1, z_2, \cdots, z_n$. If $\sum_{k=1}^n |a_k|^2 \leq 1$, then prove that $\sum_{k=1}^n |z_k|^2 \leq n$.

[b]Test 3[/b]:
Problem 7: http://www.mathlinks.ro/Forum/viewtopic.php?t=99378
$m$ and $n$ are positive integers. Set $A=\{ 1, 2, \cdots, n \}$. Let set $B_{n}^{m}=\{ (a_1, a_2 \cdots, a_m) \mid a_i \in A, i= 1, 2, \cdots, m \}$ satisfying:
(1) $|a_i - a_{i+1}| \neq n-1$, $i=1,2, \cdots, m-1$; and

(2) at least three of $a_1, a_2, \cdots, a_m$ ($m \geq 3$) are pairwise distince.

Find $|B_n^m|$ and $|B_6^3|$.

Problem 8: http://www.mathlinks.ro/Forum/viewtopic.php?t=99379
Can we find positive reals $a_1, a_2, \cdots, a_{2002}$ such that for any positive integer $k$ ($1 \leq k \leq 2002$), every complex root $z$ of polynomial $a_{k+2001}x^{2001}+a_{k+2000}x^{2000}+ \cdots + a_{k+1}x+a_k$ satisfy the condition $|\text{Im} z| \leq |\text{Re} z|$, where $a_{2002+i}=a_i, i=1,2, \cdots, 2001$.

Problem 9: http://www.mathlinks.ro/Forum/viewtopic.php?t=99380
Let $x_0+\sqrt{2003}y_0$ be the minimum positive integer root of Pell function $x^2-2003y^2=1$. Find all the positive integer solutions $(x,y)$ of the equation, such that $x_0$ is divisible by any prime factor of $x$.

[b]Test 4[/b]:
Problem 10: http://www.mathlinks.ro/Forum/viewtopic.php?t=99381
In triangle $ABC$, $AB > BC > CA$, $AB=6$, $\angle{B}-\angle{C}=90^o$. The incircle touches $BC$ at $E$ and $EF$ is a diameter of the incircle. Radical $AF$ intersect $BC$ at $D$. $DE$ equals to the circumradius of $\triangle{ABC}$. Find $BC$ and $AC$.

Problem 11: http://www.mathlinks.ro/Forum/viewtopic.php?t=99382
Find all functions $f,g$:$R \to R$ such that $f(x+yg(x))=g(x)+xf(y)$ for $x,y \in R$.

Problem 12: http://www.mathlinks.ro/Forum/viewtopic.php?t=99383
Let $A= \{a_1,a_2, \cdots, a_n \}$ and $B=\{b_1,b_2 \cdots, b_n \}$ be two positive integer sets and $|A \cap B|=1$. $C= \{ \text{all the 2-element subsets of A} \} \cup \{ \text{all the 2-element subsets of B} \}$. Function $f: A \cup B \to \{ 0, 1, 2, \cdots 2 C_n^2 \}$ is injective. For any $\{x,y\} \in C$, denote $|f(x)-f(y)|$ as the $\textsl{mark}$ of $\{x,y\}$. If $n \geq 6$, prove that at least two elements in $C$ have the same $\textsl{mark}$.

[b]Test 5[/b]:
Problem 13: http://www.mathlinks.ro/Forum/viewtopic.php?t=99384
Let $S$ be the set of points inside and on the boarder of a regular haxagon with side length 1. Find the least constant $r$, such that there exists one way to colour all the points in $S$ with three colous so that the distance between any two points with same colour is less than $r$.

Problem 14: http://www.mathlinks.ro/Forum/viewtopic.php?t=40280
In isosceles right triangle $ABC$, $\angle{A}=90^o$, $AB=1$. $D$ is the midpoint of $BC$, $E$, $F$ are two additional points on $BC$. Circumcircles of $\triangle{ADE}$, $\triangle{ABF}$ intersect at $M$($\neq A$). Line $AF$ intersects the circumcircle of $\triangle{ACE}$ at $N$($\neq A$). Line $AD$ intersects the circumcircle of $\triangle{AMN}$ at $P$($\neq A$). Find $AP$.

Problem 15: http://www.mathlinks.ro/Forum/viewtopic.php?t=99387
Sequence $\{ a_n \}$ satisfies: $a_1=3$, $a_2=7$, $a_n^2+5=a_{n-1}a_{n+1}$, $n \geq 2$. If $a_n+(-1)^n$ is prime, prove that there exists a nonnegative integer $m$ such that $n=3^m$.

[b]Test 6[/b]:
Problem 16: http://www.mathlinks.ro/Forum/viewtopic.php?t=99388
Let $g(x)= \sum_{k=1}^{n} a_k \cos{kx}$, $a_1,a_2, \cdots, a_n, x \in R$. If $g(x) \geq -1$ holds for every $x \in R$, prove that $\sum_{k=1}^{n}a_k \leq n$.

Problem 17: http://www.mathlinks.ro/Forum/viewtopic.php?t=99389
Positive integer $n$ cannot be divided by $2$ and $3$, there are no nonnegative integers $a$ and $b$ such that $|2^a-3^b|=n$. Find the minimum value of $n$.

Problem 18: http://www.mathlinks.ro/Forum/viewtopic.php?t=99390
(1) $D$ is an arbitary point in $\triangle{ABC}$. Prove that:
\[ \frac{BC}{\min{AD,BD,CD}} \geq \{ \begin{array}{c} \displaystyle 2\sin{A}, \ \angle{A}< 90^o \\ \\ 2, \ \angle{A} \geq 90^o \end{array}   \]
(2)$E$ is an arbitary point in convex quadrilateral $ABCD$. Denote $k$ the ratio of the largest and least distances of any two points among $A$, $B$, $C$, $D$, $E$. Prove that $k \geq 2\sin{70^o}$. Can equality be achieved?

[b]Test 7[/b]:
Problem 19: http://www.mathlinks.ro/Forum/viewtopic.php?t=99392
There are $n$($n\geq 3$) circles in the plane, all with radius $1$. In among any three circles, at least two have common point(s), then the total area covered by these $n$ circles is less than $35$.

Problem 20: http://www.mathlinks.ro/Forum/viewtopic.php?t=99393
Given an integer $a_1$($a_1 \neq -1$), find a real number sequence $\{ a_n \}$($a_i \neq 0, i=1,2,\cdots,5$) such that $x_1,x_2,\cdots,x_5$ and $y_1,y_2,\cdots,y_5$ satisfy $b_{i1}x_1+b_{i2}x_2+\cdots +b_{i5}x_5=2y_i$, $i=1,2,3,4,5$, then $x_1y_1+x_2y_2+\cdots+x_5y_5=0$, where $b_{ij}=\prod_{1 \leq k \leq i} (1+ja_k)$.

Problem 21: http://www.mathlinks.ro/Forum/viewtopic.php?t=99394
Given $S$ be the finite lattice (with integer coordinate) set in the $xy$-plane. $A$ is the subset of $S$ with most elements such that the line connecting any two points in $A$ is not parallel to $x$-axis or $y$-axis. $B$ is the subset of integer with least elements such that for any $(x,y)\in S$, $x \in B$ or $y \in B$ holds. Prove that $|A| \geq |B|$.

[b]Test 8[/b]:
Problem 22: http://www.mathlinks.ro/Forum/viewtopic.php?t=99395
Triangle $ABC$ is inscribed in circle $O$. Tangent $PD$ is drawn from $A$, $D$ is on ray $BC$, $P$ is on ray $DA$. Line $PU$ ($U \in BD$) intersects circle $O$ at $Q$, $T$, and intersect $AB$ and $AC$ at $R$ and $S$ respectively. Prove that if $QR=ST$, then $PQ=UT$.

Problem 23: http://www.mathlinks.ro/Forum/viewtopic.php?t=99396
Let $S$ be a finite set. $f$ is a function defined on the subset-group $2^{S}$ of set $S$. $f$ is called $\textsl{monotonic decreasing}$ if when $X \subseteq Y\subseteq S$, then $f(X) \geq f(Y)$ holds. Prove that: $f(X \cup Y)+f(X \cap Y ) \leq f(X)+f(Y)$ for $X, Y \subseteq S$ if and only if $g(X)=f(X \cup \{ a \})-f(X)$ is a $\textsl{monotonic decreasing}$ funnction on the subset-group $2^{S \setminus \{a\}}$ of set $S \setminus \{a\}$ for any $a \in S$.

Problem 24: http://www.mathlinks.ro/Forum/viewtopic.php?t=99397
Let $a_{1}, a_{2}\cdots, a_{n}$ be positive real numbers ($n \geq 2$) that are not all equal, such that $\sum_{k=1}^{n}a_{k}^{-2n}=1$, prove that:
\[\sum_{k=1}^{n}a_{k}^{2n}-n^{2}\sum_{1 \leq i < j \leq n}\left( \frac{a_{i}}{a_{j}}-\frac{a_{j}}{a_{i}}\right)^{2}> n^{2}\]

For PDF file, go here: http://www.mathlinks.ro/Forum/viewtopic.php?t=26393

I have replaced two of the topics by older topics where the same problems were discussed (you will notice it as the topic numbers are much smaller than the others).

Now I'm wondering: Orl has once posted a series of problems:
http://www.mathlinks.ro/Forum/viewtopic.php?t=55868
http://www.mathlinks.ro/Forum/viewtopic.php?t=55870
http://www.mathlinks.ro/Forum/viewtopic.php?t=55873
http://www.mathlinks.ro/Forum/viewtopic.php?t=55867
http://www.mathlinks.ro/Forum/viewtopic.php?t=55881
which he claims is also from China TST 2003, but I don't find them here... Are there different kinds of TSTs? Or is there just some confusion?

  darij

There are totally 10 tests in 2003.
The last 2 tests is called "IMO Team Selection Test" officially.
The others are called quizzes.
The team members are selected according to the scores of CMO, quizzes, and TST altogether.

Hi zhaoli, long time no see!

As is known to all, the Chinese TST quizzes are extremely difficult to find. The problems after 2003 can be found in the book "Way up to IMO", but for those before are mostly still in the dark. Currently I only have the 8 sets of 2002, 8 sets of 2001 and 14.5 problems of 2001. I wonder how many problems do you have? Can you share it with me?


algebra
polynomial
analytic geometry
function
calculus
derivative
symmetry

I was wondering if there is an easy method for sketching polynomial equations without a graphing calculator. For example y= (x-2)(x+3).

$f(x)=0$ when $x=2$ or $x=-3$ so you have now two points $A=(2,0), B=(-3,0)$. We know, that $f(0)=-6$, so third point is $C=(0,-6)$. You can also calculate coordinates of vertex. Then you will have $W=(-\frac{1}{2}, -\frac{25}{4})$. I think that with four points you can draw sketch of $f(x)=(x+2)(x-3)$ :)

It's also helpful to note that since the factors are raised to an odd power (1), the graph will pass through the x-axis whereas it would bounce off of the x axis if it were to an even power. In addition, look at the end behaviors. For this one, it goes up on both sides.

the graph of the functions like $f(x) = (x-a_1)^{b_1}(x-a_2)^{b_2}...(x-a_n)^{b_n}$ are very easy. The first thing is the graph contains $n-1$ turns. secondly if the power such as $b_1$ is even then graph do not change its quadrant. but if it is odd then it changes its quadrant. the point at which the $f(x)$ is zero is the point where the graph cuts the x axis. Hope this will help you.

For polynomianls with a degress of less than 6, you can also find the maximums, minimums, and inflection points by setting the derivative equal to 0 and solving.

For example
$y=x^2+x-6$
$\frac{dy}{dx}=2x+1$
$2x+1=0$
$x=\frac{-1}{2}$
$y=(\frac{-1}{2})^2+(\frac{-1}{2})-6=\frac{-25}{4}$

The point $\frac{-1}{2},\frac{-25}{4}$ is either a maximum, minimum, or an inflection point. To find which it is, find the second derivative. In this case it is 2 which is positive so it is a minimum.

Assuming you don't know calculus you can use the formula to find the axis of symmetry -  in a degree 2, $-\frac{b}{2a}$  Factoring an equations also helps, you then know where it is zero and that between ranges it will be doing the same thing.  Also if a factor is raised to an even degree such as $(x-2)^2$ the graph will touch at that point, but it won't cross the x-axis.  Once you take calculus this hand graphing is much easier as it is easy to find where the function is increasing and decreasing as well as its concavity.

The zeros are $2$ and $-3\implies \frac{2-3}{2}=-\frac{1}{2}\implies f\left(-\frac{1}{2}\right)=-\frac{25}{4}$. So using the vertex $\left(-\frac{1}{2},-\frac{25}{4}\right)$ and the $a=1$, we obtain a quadratic equation that has zeros $2$ and $-3$ opening upward.

Masoud Zargar


geometry
vector
function
ratio
trigonometry
geometry unsolved

Let ABC be a triangle in which AB = AC. Let D be the midpoint of BC and P be a point on AD. Suppose E is the foot of perpendicular from P on AC. If $AP/PD = BP/PE = m$ and $BD/AD = n$ and $z =$ $n^{2}$($1+m$), prove that $z^{2}$ - ($m^{3}$ -$m^{2}$ $-2$)$z+1$ = 0.
Hence,show that $m\geq 2$ and m = 2 iff ABC is equilateral.

Somebody please solve this.I have written the INMO and was clueless on this one.

Hello,
    I give a rough outline of my solution to the above problem.It is a bit involved.It would be nice if somebody gave a pure geometry solution coupled with use of some trigonometry. Here,I give an outline of a vector approach to the problem.
 Let A be the origin of vectors.Denoting AB and AC as vectors b and c, it is easy to obtain the position vector of point P using section formula. Using the fact that BD/AD = n,by taking moduli and squaring, we have cos A = (1-n^2)/(1 + n^2). We can obtain the postion vector of E also by section formula and using the fact that vectors c and PE are perpendicular.
  Now, we proceed to use BP/PE = m.Taking moduli and squaring, we obtain a relation between m and n.This involves quite a bit of computation. Then, the problem simplifies as we have z as a function of m alone. What remains is to substitute for z in terms of m and prove the given expression to be zero.
  I understand that there is a small element of 'hand waving' argument here and I'll try to put up a more precise solution indicating the exact relation between m and n in a day or two.
Sathej

from similarity we have
BD/AD  = PE/AE= m

AND WE HAVE AP/PD = BP/PE = n

[b]BP^2=BD^2 + PD^2 = (nPE)^2 [/b]  ------ 1

PE^2= AP^2 - AE^2
        =AP^2 - (BP/nm)^2

WE GET [b]BP^2 = ( n^2/(m^2 + 1) ) AP^2[/b] ----- 2

MANIPULATING THE GIVEN RATIOS YIELD:

[b]BD = ( m(n+1)/n ) AP[/b] ------ 3

AND WE HAVE [b]PD = n AP[/b] ----- 4

COMBINING 1 , 2 ,3 AND 4

GIVES  AN EQUATION PURELY IN "n" AND "m"

SUBSTITUTE n^2= z / ( 1+m) AND UPON SIMPLIFYING WE GET THE DESIRED RESULT

NOW THE 2ND PART IS QUITE SIMPLE AND I'M LEAVING THAT
MY SOLUTION MIGHT APPEAR A LITTLE MESSY CONSIDERING I DO NOT KNOW HOW TO USE LATEX.... ANYHOW....

Where can I find all INMO 2007 Problems??
Please, help me.

Hello,
    Good lines of argument.But, you have interchanged m and n in writing down your solution. And moreover,the equation 4 should be PD = AP/n. and not PD = n.AP.
 Also,going by your notation,the substitution should be
         m^2 = z/(1 + n).
A good approach nevertheless.
Sathej

here is another proof the beautiful result
let
    $AP=r$  $BP=s$  $\angle B={\alpha}$
 this gives $n=cot{\alpha}$
 also,
       $AB=(r+s)cosec{\alpha}$
       $BP=PEm=rmcos{\alpha}$
       $AD=(r+s)cot{\alpha}$
NOW  apply the generalized $APPOLONIUS$ $THEOREM$ to  $ABD$ considering $BP$  as the cevian.
REST IS ALGEBRA
now as for the second part
$z+\frac{1}{z}$=$m^{3}-m^{2}-2$ $\geq 2$
REST IS  EASY

[quote="Sathej"][color=darkred]Let $ ABC$ be a $ A$- isosceles triangle. Let $ D$ be the midpoint of the side $ [BC]$ for which denote  $ \frac {BD}{AD} = n$. Define the points $ P\in (AD)$ and $ E\in (AC)$ for which $ PE\perp AC$ and $ \frac {AP}{PD} = \frac {BP}{PE} = m$.  Denote $ z = n^{2}(1 + m)$. Prove that :

$ 1\blacktriangleright z^{2} - (m^{3} - m^{2} - 2)z + 1 = 0$.

$ 2\blacktriangleright m\ge 2$ and $ m = 2\Longleftrightarrow$ the triangle $ ABC$ is equilateral.[/color][/quote]
[color=darkblue][b]Proof.[/b] I"ll suppose that $ AD = 1$. Thus, $ \{\begin{array}{c} DC = n\ ,\ AC = \sqrt {1 + n^2} \\
 \\
\frac {PA}{m} = \frac {PD}{1} = \frac {1}{m + 1}\end{array}\ .$

$ [ADC] = [PDC] + [APC]\implies$ $ n = \frac {n}{m + 1} + PE\cdot\sqrt {1 + n^2}\implies$

$ PE = \frac {mn}{(m + 1)\sqrt {1 + n^2}}\implies$ $ PB = \frac {m^2n}{(m + 1)\sqrt {1 + n^2}}\ .$ But $ PB^2 = PD^2 + DC^2\ .$

Therefore, $ n^2 + (\frac {1}{m + 1})^2 = \frac {m^4n^2}{(m + 1)^2(1 + n^2)}\implies$

$ n^2(m + 1)^2(1 + n^2) + (1 + n^2) = m^4n^2\implies$

$ z(m + 1)(1 + \frac {z}{m + 1}) + 1 = n^2(m^4 - 1)\implies$

 $ z(m + 1) + z^2 + 1 = z(m - 1)(m^2 + 1)\implies$ $ z^2 + 1 = z(m^3 - m^2 - 2)$ $ \implies$

$ m^3 - m^2 - 2 = z + \frac 1z\ge 2$ $ \implies$ $ m\ge 2\ .$

$ m = 2\Longleftrightarrow$ $ z = 1$ $ \Longleftrightarrow$ $ n = \frac {\sqrt 3}{3}$ $ \Longleftrightarrow$ $ \tan B = \frac {AD}{BD} = \sqrt 3$ $ \Longleftrightarrow$ $ B = 60^{\circ}\ .$

[b]Remark.[/b] $ z=(\frac {DB}{DA})^2\cdot \frac {DA}{DP}$ $ \implies$ $ z=\frac {DB^2}{DA\cdot DP}\ .$.

Thus, $ z=1\Longleftrightarrow DB^2=DA\cdot DP$ $ \Longleftrightarrow$ $ \widehat {DBP}\equiv\widehat {DAB}$ a.s.o.[/color]


Let $ x,y,z\in\mathbb{R}^+$ such that $ x+y+z=1$ and $ x\geq y\geq z$. Prove that
$ 1+yz\geq 2\sqrt{(1-x)(1-y)(1-z)}$

first of all note that $ x\plus{}y\plus{}z\equal{}1$ and $ x,y,z\in\mathbb{R}^\plus{}$ so $ x,y,z<1$ now by AM-GM we get that:

$ 2\sqrt{(1\minus{}x)(1\minus{}y)(1\minus{}z)}\leq (1\minus{}x)\plus{}(1\minus{}y)(1\minus{}z)\equal{}1\plus{}yz$

the condition $ x\geq y\geq z$ seems to be useless here... :maybe:


induction
inequalities
number theory unsolved
number theory

Find all $ n\in N$ such that
$ n<\tau (n^2)$

(edited because of big flaw)
$ n \equal{} 1$ is not a solution. For the rest n's, let $ n \equal{} \prod_{i \equal{} 1}^{m} p_i ^{k_i}, k_{i} \in N$ for all $ 1 \le i \le m$ - $ p_i$ are different primes.
$ \tau (n^2) \equal{} \tau (\prod_{i \equal{} 1}^{m} p_i ^{2k_i}) \equal{} \prod_{i \equal{} 1}^{m} (1 \plus{} 2k_{i})$
For every prime $ p$ and positive integer $ k \ge 3$, we have: $ p^k \ge 2^k \ge 1 \plus{} 2k$ (this can be seen with induction on k).
For k=1,2: $ p^k \ge 1\plus{}2k$ is true when $ p \neq 2$.
So:
$ \tau (n^2) \equal{} \tau (\prod_{i \equal{} 1}^{m} p_i ^{2k_i}) \equal{} \prod_{i \equal{} 1}^{m} (1 \plus{} 2k_{i}) \le \prod_{i \equal{} 1}^{m} p_i^{k_{i}} \equal{} n$ when $ 2^k \parallel{} n$, where $ k \neq 1,2$.

[u]When k=1:[/u] The inequality is equivalent to $ \prod \frac{p_i^{k_i}}{1\plus{}2k_i} < \frac {3}{2}$ (the product is only over the odd primes, so $ p_i \ge 3$). If there are no odd primes, we get $ 3>2$, which is true. Else, $ \prod \frac{p_i^{k_i}}{1\plus{}2k_i} \ge \prod \frac{3^{k_i}}{1\plus{}2k_i}$. $ 3^k \ge 1.5(1\plus{}2k)$ by induction, for natural k>1. Hence, if $ k_i>1$ for some i, $ LHS \ge RHS$.
Thus $ k_i\equal{}1$ for all i. Now $ LHS\equal{}\prod \frac{p_i}{3}$. If one of the primes is at least 5, the inequality doesn't hold ($ \frac{5}{3} > \frac{3}{2}$). Hence, there's only one odd prime, that is - 3, and then n=6.

[u]When k=2:[/u] The inequality is equivalent to $ \prod \frac{p_i^{k_i}}{1\plus{}2k_i} < \frac {5}{4}$. If there are no odd primes, we have 5>4 which is true (n=4 is a solution). Else, because the solutions of the previous case are the only possible solutions, by the same way it is left to check that case when the only odd prime is 3, with exponent of 1. We can verify that n=12 is indeed a solution.

Conclustion: the solutions are n=2,4,6,12


prove that sum of digits of 1981^n (where n>=1)  is more than or equal to 19.

well, we know that the sum of the digits is congruent to the original number mod 9, and 1981:equiv:1 mod 9, so the sum of the digits of any powers of 1981 will be 1, 10, 19, etc. It can clearly not be 1 (the only numbers whose sum of digits is 1 are 10^n). So, um, there has to be some way to show it can't be 10 either. umm..... perhaps showing there is always a 9 in the number, and at least 2 other digits meaning that the sum is :ge:11 or something...

yes, proving that 10 is not a possibility is the hard part.

To continue, 1981 :equiv: 1 (mod 99), so 1981n :equiv: 1 (mod 99).  Let A be the sum of the units digit, hundreds digit, ten-thousands digit etc. (skipping every other digit) of 1981n.  Let B be the sum of all the other digits of 1981n.  Because 100 :equiv: 1 (mod 99), it follows that 10 B + A :equiv: 1981n :equiv: 1 (mod 99).  As MysticTerminator pointed out, the sum of the digits can't be 1, so 10B + A > 1.  Thus 10B + A :ge: 100.  Also, the units digit of 1981n is 1, so A :ge: 1.  Pulling all our observations together, we get B + A = (10B + A)/10 + 9A/10 :ge: 100/10 + 9/10 = 10.9.  Thus the sum of the digits is at least 11.  As MysticTerminator showed, that means the sum of the digits is at least 19.

that is nice, but what else can you say since 1981=1(mod 99)?

We know it's 1 (mod 11) too, but I thought it was a little easier to work directly with mod 99 instead of using both mod 9 and mod 11.

There was a related problem in the Oct. 15 Math Jam.  Find digits A and B such that 88AB3 is divisible by 99.  The Math Jam treated divisibility by 99 as implying divisibility by 9 and 11.  The case work there was a little messy.  But, instead, if we work directly mod 99, we get that 3 + 10B + A + 10*8 + 8 is divisible by 99, and from there it's easy to figure out A and B.

yes, that is the one i was thinking off, nice job.

Sorry to bring up an old thread, but I don't see why 10 B + A  :equiv:  1981^n (mod 99). I see how they are congruent mod 9 though, so I guess I need to see how they are congruent mod 11. 10 B + A (mod 11) is just the sum of the digits of 1981^n, but what is 1981^n (mod 11)?

[quote]I don't see why 10 B + A :equiv: 1981^n (mod 99).[/quote]
I am using a divisibility rule for 99, analogous to the divisibility rule for 9.  For example, suppose our number is a 6-digit integer UVWXYZ.  That stands for 100,000U + 10,000V + 1000W + 100X + 10Y + Z.  Note that 100,000 :equiv: 1,000 :equiv: 10 (mod 99).  Also 10,000 :equiv: 100 :equiv: 1 (mod 99).  Thus the original number is 10(U + W +Y) + (V + X + Z) (mod 99).  Using my earlier notation, that is 10A + B (mod 99).

[quote]10 B + A (mod 11) is just the sum of the digits of 1981^n[/quote]
I think you mean the alternating (plus/minus) sum of digits (mod 11), because 10 :equiv: -1 (mod 11).

[quote]what is 1981^n (mod 11)?[/quote]
Well, 1981 :equiv: 1 (mod 11), so 1981n :equiv: 1 (mod 11).


function
floor function
inequalities
algebra
real analysis
real analysis unsolved

How do you show rigorously that if $ f$ is measurable and nonnegative and,

$ F_{n} \equal{} \{x | f(x) \geq n\}$

$ E_{ni} \equal{} \{ x| \frac {i \minus{} 1}{2^n} \leq f(x) \leq \frac {i}{2^n} \}$

then

$ s_n \equal{} \sum_{i \equal{} 1} ^ {n 2^n} \frac {i \minus{} 1}{2^n} I_{E_{ni}} \plus{} n I_{F_n}$

converges to f for every x?

EDIT: I is the indicator function

[quote="Kent Merryfield"]Suppose $ f(x) < \infty.$ Choose $ n$ large enough so that $ n > f(x).$ What estimate can you give for the size of $ |f(x) \minus{} s_n(x)|?$ You've got enough explicit information to do that.

Then you can separately look at the case of $ f(x) \equal{} \infty.$

(The sequence $ s_n$ converges pointwise to $ f$ everywhere, whether $ f$ is measurable or not. The reason for the hypothesis of $ f$ being measurable is so that each $ s_n$ will be measurable.)

This is an often-used lemma. For some uses, it may not be enough to just quote the result - we might want to get our hands dirty and go back into the details of the construction.[/quote]

Ah, I see now. That expression for s_n was just so complicated that it wasn't "clicking" for me why it made sense. When $ f(x) < \infty$ and n > f(x), we get $ s_n(x) \equal{} \frac {\lfloor f(x) 2^n \rfloor}{2^n}$.

So, then $ \left|f(x) \minus{} s_n(x) \right| \equal{} \left| f(x) \minus{} \frac{\lfloor f(x) 2^n \rfloor}{2^n}\right|$

And now there has got to be some inequality that says the last expression is $ \leq$ something that goes to 0. What inequalities involve the floor function though?

$ 0\le u\minus{}\lfloor u\rfloor<1.$

Or to be even more direct: on the set $ E_{ni},$ $ \frac{i\minus{}1}{2^n}\le f(x)<\frac{i}{2^n},$ On that set, $ s_n(x)\equal{}\frac{i\minus{}1}{2^n}.$

Hence $ s_n(x)\le f(x)<s_n(x)\plus{}\frac1{2^n}.$

[quote="Kent Merryfield"]$ 0\le u \minus{} \lfloor u\rfloor < 1.$
[/quote]

Oh yeah. So $ \left| f(x) \minus{} \frac {\lfloor f(x) 2^n \rfloor}{2^n}\right| \equal{} \frac {1}{2^n}\left|f(x) 2^n \minus{} \lfloor f(x) 2^n \rfloor\right| < \frac {1}{2^n}$

which completes the finite case.

When, $ f(x) \equal{} \infty$, $ s_n (x) \equal{} \infty$ whenever $ n > f(x)$ which completes the infinite case. Thanks for your help.


Prove that: $ \sqrt[4]{x^{4}\plus{}x\minus{}2}\plus{}\sqrt{x^{4}\plus{}x\minus{}2}\equal{}6$ has only one integer root.

[hide="Solution"]
Let $ t\equal{}\sqrt[4]{x^4\plus{}x\minus{}2}$ and $ t\geq 0$, then:
$ t^2\plus{}t\minus{}6\equal{}0$, hence:
$ t\equal{}2\vee t\equal{}\minus{}3$ but $ t\geq 0$, so:
$ \sqrt[4]{x^4\plus{}x\minus{}2}\equal{}2$
$ x^4\plus{}x\minus{}2\equal{}16$
$ x^4\plus{}x\minus{}18\equal{}0$
And now if this equation has an integer root it must be one of $ \left\{\pm 1,\pm 2,\pm 3,\pm 6,\pm 18\right\}$
And for example $ x\equal{}2$ is a root, and then $ x^4\plus{}x\minus{}18\equal{}(x\minus{}2)(x^3\plus{}2x^2\plus{}4x\plus{}9)$ and prove that equation $ x^3\plus{}2x^2\plus{}4x\plus{}9\equal{}0$ hasn't for any integer roots.[/hide]


The first four triangular numbers are pictured. The nth triangular number is formed by drawing a row of n dots below the (n-1)st triangular dot pattern. The kth triangular number is represented by 120 dots. What is the value of k?

The dot pattern is like this:
there is one dot in the first one,
second one- one dot followed by two dots below it
third one- one dot followed by two dots below followed by three dots below
fourth one - one dot followed by two dots below followed by three dots below followed by four dots below

and that goes on forever.

Question: How do you put pictures on aops? Also, how do you put a radical sign on aops, and how do you put a box around  your answer?

[hide]OK, aside from knowing the above, I can find a triangular number from the formula $ n(n \minus{} 1)/2$

If we set this equal to $ 120$, then

$ n(n \minus{} 1) \equal{} 240$
$ n^2 \minus{} n \minus{} 240 \equal{} 0$
$ (n \plus{} 15)(n \minus{} 16)$
Because we can;t use the negative answer, the positive root is
$ n \minus{} 16 \equal{} 0$
[b]$ n \equal{} 16$[/b]

I'm pretty sure this is right...

EDIT: I made a mistake, since the first row has a value of 1, then the formula becomes $ (n\plus{}1)n/2$
This changes the answer to simplify to
$ (n\plus{}16)(n\minus{}15)$, making the answer $ n\equal{}15$
Sorry about that...
Do you understand? I don't think I did a very good job explaining my process[/hide]

The answer key says the answer is 15........

The nth triangular number is given by $ \frac {n(n \plus{} 1)}{2}$, not $ \frac {n(n \minus{} 1)}{2}$, so you would get $ 15$.

edit: sorry didn't see rd's edit.