modular arithmetic
number theory
relatively prime
algebra unsolved
algebra

Suppose $ a,b,c,d$ are integers with $ 0 \le a \le b \le 99$, $ 0 \le c \le d \le 99$. For any integer $ i$, let $ n_i \equal{} 101i\plus{}1002^i$. Show that if $ n_a \plus{} n_b$ is congruent to $ n_c\plus{}n_d \pmod {10100}$ then $ a\equal{}c$ and $ b\equal{}d$.

where to start...   :maybe:

$ n_a\plus{}n_b\equiv n_c\plus{}n_d \pmod{10100} \Rightarrow$ (1) $ n_a\plus{}n_b\equiv n_c\plus{}n_d \pmod{101}$, (2) $ n_a\plus{}n_b\equiv n_c\plus{}n_d\pmod{4}$, (3) $ n_a\plus{}n_b\equiv n_c\plus{}n_d \pmod{25}$

(1) boils down to $ 1002^a\plus{}1002^b\equiv 1002^c\plus{}1002^d \pmod{101}$, i.e. $ 1002^{b\minus{}a} \equiv 1002^{d\minus{}c} \pmod{101}$. then check that the order of $ 1002 \pmod{101}$ is 100, which will imply that $ b\minus{}a \equal{} d\minus{}c$. now combine this with (2) and (3).

Hi, this is probs a noob question but how did you get mod 101 and mod 4 and mod 25? how can you split it up like that?

because $ 10100 \equal{} 101 \cdot 25 \cdot 4$ and 101, 25, 4 are [i]relatively prime[/i], divisibility by 10100 is equivalent to divisibility by 101, 25, and 4.

[quote="pluristiq"]$ n_a \plus{} n_b\equiv n_c \plus{} n_d \pmod{10100} \Rightarrow$ (1) $ n_a \plus{} n_b\equiv n_c \plus{} n_d \pmod{101}$, (2) $ n_a \plus{} n_b\equiv n_c \plus{} n_d\pmod{4}$, (3) $ n_a \plus{} n_b\equiv n_c \plus{} n_d \pmod{25}$

(1) boils down to $ 1002^a \plus{} 1002^b\equiv 1002^c \plus{} 1002^d \pmod{101}$, i.e. $ 1002^{b \minus{} a} \equiv 1002^{d \minus{} c} \pmod{101}$. then check that the order of $ 1002 \pmod{101}$ is 100, which will imply that $ b \minus{} a \equal{} d \minus{} c$. now combine this with (2) and (3).[/quote]
Ah okay, I think I get what your saying, but how did you simplify 1 and get $ b \minus{} a \equal{} d \minus{} c$? I'm not sure on the specifics when I tried myself =S

thanks

Actually the question had a typo lol, instead of $ 1002^i$ it should be $ (100)(2^i)$.

This is what I have done so far, how to complete it?

If $ n_a \plus{} n_b \equiv n_c\plus{}n_d \pmod {10100}$

Then $ n_a\plus{}n_b \equiv n_c\plus{}n_d \pmod {101}$ and $ n_a\plus{}n_b \equiv n_c\plus{}n_d \pmod {100}$

Thus $ 101a\plus{}100(2^a) \plus{}101b\plus{}100(2^b) \equiv 101c\plus{}100(2^c) \plus{}101d\plus{}100(2^d) \pmod {101}$

$ 101(a\plus{}b\minus{}c\minus{}d) \plus{}100(2^a)\plus{}100(2^b) \equiv 100(2^c) \plus{} 100(2^d) \pmod {101}$

$ 101(a\plus{}b\minus{}c\minus{}d) \equiv 0 \pmod {101}$

$ \implies 2^a \plus{} 2^b \equiv 2^c \plus{} 2^d \pmod {101}$

$ 101a\plus{}100(2^a) \plus{}101b\plus{}100(2^b) \equiv 101c\plus{}100(2^c) \plus{}101d\plus{}100(2^d) \pmod {100}$

$ 101(a\plus{}b\minus{}c\minus{}d) \plus{}100(2^a)\plus{}100(2^b) \equiv 100(2^c) \plus{} 100(2^d) \pmod {100}$

But $ 100(2^a) \equiv 100(2^b) \equiv 100(2^c) \equiv 100(2^d) \equiv 0 \pmod {100}$

$ \implies 101(a\plus{}b\minus{}c\minus{}d) \equiv 0 \pmod {100}$

$ \implies a\plus{}b \equiv c\plus{}d \pmod {100}$

$ 2^a \plus{} 2^b\equiv 2^c \plus{} 2^c \pmod{101}\Rightarrow 2^a(2^{b \minus{} a} \plus{} 1)\equiv 2^c(2^{d \minus{} c} \plus{} 1)\pmod{100}\Rightarrow 2^{b \minus{} a} \plus{} 1\equiv 2^{d \minus{} c} \plus{} 1\pmod{101}\Rightarrow 2^{b \minus{} a}\equiv 2^{d \minus{} c}\pmod{101}$. now check that the order of 2 modulo 101 is 100. this will imply that $ b \minus{} a\equiv d \minus{} c \pmod{100}$. since, however, $ b \minus{} a, d \minus{} c$ are both in the range $ [0,99]$, the congruence actually becomes an equality: $ b \minus{} a \equal{} d \minus{} c$. now, by adding this with the congruence $ b \plus{} a\equiv d \plus{} c \pmod{100}$ we get $ 2b\equiv 2d \pmod{100}$, so $ b\equiv d \pmod {50}$, which isn't enough to conclude $ b \equal{} d$. am I missing something, or are you missing something more from the problem statement?

edit: this is wrong. see my post below

Hmm that is the question.

This is how I finished it off, I'm not sure if it's right...

$ \implies 2^{a\plus{}b} \equiv 2^{c\plus{}d} \pmod {101} \cdots [2] \text{iff a\plus{}b \equal{} c\plus{}d}$ since $ q \equiv q \pmod {n}$ for any $ \text{n and some integer q}.$

$ [1] \times 2^a \implies 2^{2a} \plus{} 2^{a\plus{}b} \equiv 2^{c\plus{}a} \plus{} 2^{d\plus{}a} \pmod {101}$

$ \implies 2^{2a} \plus{} 2^{c\plus{}d} \minus{} 2^{c\plus{}a} \minus{} 2^{d\plus{}a} \equiv 0 \pmod {101}$

$ \implies 2^{2a} \plus{} 2^{c}2^{d} \minus{} 2^{c}2^{a} \minus{} 2^{d}2^{a} \equiv 0 \pmod {101}$

$ \implies 2^{2a} \minus{} 2^{a}\left(2^c\plus{}2^d\right) \plus{}2^c2^d \equiv 0 \pmod {101}$

$ \text{Let} \ x \equal{} 2^a \implies x^2 \minus{}x\left(2^c\plus{}2^d\right) \plus{}2^c2^d \equiv 0 \pmod {101}$

$ \implies \left(x\minus{}2^c\right)\left(x\minus{}2^d\right) \equiv 0 \pmod {101}$

$ \implies \left(2^a\minus{}2^c\right) \equiv 0 \pmod {101} \ \text{or} \ \left(2^a\minus{}2^d\right) \equiv 0 \pmod {101}$

$ \implies 2^a \equiv 2^c \pmod {101} \ \text{or} \ 2^a \equiv 2^d  \pmod {101} \ \text{iff a \equal{} c or a \equal{} d}$

$ \text{If a \equal{} c} \ \implies 2^b \equiv 2^d \pmod {101} \ \text{iff b \equal{} d}$

$ \text{Thus a \equal{} c and b\equal{}d}$

this actually works.

(1) $ a \plus{} b\equiv c \plus{} d \pmod{100} \Rightarrow 2^a2^b \equiv 2^c2^d \pmod{101}$

(2) $ 2^a \plus{} 2^b\equiv 2^c \plus{} 2^d \pmod{101}\Rightarrow 2^b\equiv 2^c \plus{} 2^d \minus{} 2^a \pmod{101}$

substituting (2) into (1), we get $ 2^a(2^c \plus{} 2^d \minus{} 2^a)\equiv 2^c2^d \pmod{101}$, which can be rewritten as $ (2^a \minus{} 2^c)(2^a \minus{} 2^d)\equiv 0 \pmod{101}$, which implies that $ a\equiv c$ or $ a\equiv d \pmod{100}$ (because the order of 2 modulo 101 is 100). but because of the range in which they lie, the congruence is in fact an equality: $ a \equal{} c$ or $ a \equal{} d$. now, because of $ a\plus{}b\equiv c\plus{}d \pmod{100}$, if $ a$ equals one of $ c, d$, $ b$ equals the other. because of the order restrictions, we then have $ (a, b) \equal{} (c, d)$.

[quote="pluristiq"]$ 2^a(2^{b - a} + 1)\equiv 2^c(2^{d - c} + 1)\pmod{101}\Rightarrow 2^{b - a} + 1\equiv 2^{d - c} + 1\pmod{101}$[/quote]
This isn't true.

Now a complete solution with the edited problem:

[quote]
Suppose $ a,b,c,d$ are integers with $ 0 \le a \le b \le 99, 0 \le c \le d \le 99$. For any integer $ i$, let $ n_i = 101i + 100\cdot 2^i$. Show that if $ n_a + n_b \equiv n_c + n_d \pmod {10100}$ then $ a = c$ and $ b = d$. [/quote]

we have $ 101a + 100\cdot 2^a + 101b + 100\cdot 2^b \equiv 101c + 100\cdot 2^c + 101d + 100\cdot 2^d \pmod{10100}$
so we get
$ a - c\equiv d - b \pmod{100} \Leftrightarrow a - c = b - d \Leftrightarrow 2^{a - c}\equiv 2^{d - b} \pmod{101}$ and $ 2^a + 2^b \equiv 2^c + 2^d \pmod{101} \Leftrightarrow 2^c \left( 2^{a - c} - 1 \right) \equiv 2^b \left( 2^{d - b} - 1 \right) \pmod{101}$
So we get two cases:
First case:
$ 2^{a - c}\equiv 1 \pmod{101} \Leftrightarrow a - c\equiv 0 \pmod{100}$ and $ a = c$ and also $ b = d$
Second case:
$ 2^c\equiv 2^b \pmod{101} \Leftrightarrow c\equiv b \pmod{100}$ and we get $ c = b$ and $ a = d$

[quote="pfanni"][quote="pluristiq"]$ 2^a(2^{b \minus{} a} \plus{} 1)\equiv 2^c(2^{d \minus{} c} \plus{} 1)\pmod{101}\Rightarrow 2^{b \minus{} a} \plus{} 1\equiv 2^{d \minus{} c} \plus{} 1\pmod{101}$[/quote][/quote]
This isn't true.

of course you're right. my last post, slightly edited to account for that silly error, now gives a correct solution.


This is an easy proof, even I can do it  :P , so it should belong here. I'll leave it to the moderators discretion.

Prove that the height of an equilateral triangle is $1\frac{1}{2}$ times the radius of the circle circumscribing it.

[quote="xenon_nightmare152"]an awkardly easy proof!!!!!!
Let the circle have center O and the equilateral triangle have vertices of A, B, and C. Let BD be the altitude of the triangle and r is the radius. Draw radii AO and BO. Triangle AOD is a 30-60-90 triangle. Thus, OD is r/2 by definition of 30-60-90 triangle. BO +OD=BD...r+r/2 = $1\frac{1}{2}r$  Therefore, the altitude of triangle ABC is $1\frac{1}{2}$ times the radius.[/quote]

I know, I proved it in my sleep  :rotfl:


I think the problem is
$(x^{2}+y^{2})(y^{2}+z^{2})(z^{2}+x^{2})(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^{2}\geq 8(x^{2}+y^{2}+z^{2})^{2}$

and it is equivalent to $\sum\sqrt{\frac{abc}{b+c-a}}\geq \sum a$

and this one appeared in the book old and new inequalities and it is solved by Shur $abc(a+b+c)\geq \sum a^{3}(b+c-a)$ and Holder's inequality

[quote="silouan"]I think the problem is
$(x^{2}+y^{2})(y^{2}+z^{2})(z^{2}+x^{2})(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^{2}\geq 8(x^{2}+y^{2}+z^{2})$
Maybe but the inequality I proved above is stronger   :wink:[/quote] Yours is not homogenous. So how do you derive homogenous ones from it? Please show  :lol:
[hide]But of course I'll believe you if you write $8(x^{2}+y^{2}+z^{2})^{2}$ on the RHS  :)  :) [/hide]

besides the typos [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14569]here[/url] is the old discussion of the problem

For  above soory I made a typo .It is $(a^{2}+b^{2}+c^{2})^{2}$ as you found  :) .


Gauss
Euler
number theory
advanced fields
advanced fields theorems

http://arxiv.org/PS_cache/math/pdf/0510/0510081.pdf

Please, I'd like comments about it.

[quote="José"]anyone? It's a short proof[/quote]
It's a false proof.
I'm not an expert in number theory but this looks too short
and too simple to be missed by Gauss,Euler,Ramanujan and others during 300 years of the problem history.
Real question is where are the errors in the proof? :)


In convex hexagon $ABCDEF$, triangles $ABC, CDE, EFA$ are similar.
$\angle BAC = \angle DCE = \angle FEA$
$\angle BCA = \angle DEC = \angle FAE$.
For what condition of these three triangle,
that triangle $ACE$ is equilateral if and only if triangle $BDF$ is equilateral?

$\angle BAC \not=30$ or $\angle BCA \not= 30$
I have a solution using sin, con calculation  :(
Are there other solution?

Well, is the answer really only that? The problem looks cyclic in $A,C,E$ and $B,D,F$, so if you have $\angle BAC \not= 30$, then you should have $\angle DCE, \angle FEA$, etc. Or am I wrong?

Anyway, I think it's quite a funny problem, asking to find a condition that an iff condition is true.

Also, please post your solution.

Yes, you are right.

If $ACE$ is equilateral, then $BDF$ is equilateral, too.
it is enough to consider only case that $BDF$ is equilateral.
Let $CA: BC: BA=1: p: q$, $CE=a, AC=b, AE=c$. then we can calculate BF by 2nd cosine law.
BF=FD=DB -> $(3p^{2}-1)^{3}(a^{2}-b^{2})(b^{2}-c^{2})(c^{2}-a^{2})=(3q^{2}-1)^{3}(a^{2}-b^{2})(b^{2}-c^{2})(c^{2}-a^{2})$

if $p=q=\frac{1}{\sqrt{3}}$, then $ACE$ don't have to be a equilateral.
else, it is easy to show that $a=b=c$

[quote="Little Gauss"]then we can calculate BF by 2nd cosine law.
BF=FD=DB -> $(3p^{2}-1)^{3}(a^{2}-b^{2})(b^{2}-c^{2})(c^{2}-a^{2})=(3q^{2}-1)^{3}(a^{2}-b^{2})(b^{2}-c^{2})(c^{2}-a^{2})$
[/quote]

I have no idea how you can calculate that, and how this equation comes...

I solved it by using the new method of triangle super imposing of triangles ABC and CDE, then they must be equaled. I'm advised to not put a link here, and I don't know how to add graphics to this site. So I can not post the details.


HMMT
geometry
perimeter
inradius
trigonometry
trapezoid
3D geometry

$1)$ Octagon $ABCDEFGH$ is equiangular. Given that $AB = 1$, $BC = 2$, $CD = 3$, $DE = 4$, and $EF = FG = 2$, compute the perimeter of the octagon.

$2)$ Suppose $ABC$ is a scalene right triangle, and $P$ is the point on hypotenuse $AC$ such that $\angle ABP = 45^\circ$. Given that $AP = 1$ and $CP = 2$, compute the area of $ABC$.

$3)$ Let $A$, $B$, $C$, and $D$ be points on a circle such that $AB = 11$ and $CD = 19$. Point $P$ is on segment $AB$ with $AP = 6$, and $Q$ is on segment $CD$ with $CQ = 7$. The line through $P$ and $Q$ intersects the circle at $X$ and $Y$. If $PQ = 27$, find $XY$.

$4)$ Let $ABC$ be a triangle such that $AB = 2$, $CA = 3$, and $BC = 4$. A semicircle with its diameter on $BC$ is tangent to $AB$ and $AC$. Compute the area of the semicircle.

$5)$ Triangle $ABC$ has side lengths $AB = 2\sqrt{5}$, $BC = 1$, and $CA = 5$. Point $D$ is on side $AC$ such that $CD = 1$, and $F$ is a point such that $BF = 2$ and $CF = 3$. Let $E$ be the intersection of lines $AB$ and $DF$. Find the area of $CDEB$.

$6)$ A circle of radius $t$ is tangent to the hypotenuse, the incircle, and one leg of an isosceles right triangle with inradius $r = 1 + \sin \frac \pi 8$. Find $rt$.

$7)$ Suppose $ABCD$ is an isosceles trapezoid in which $AB || CD$. Two mutually externally tangent circles $\omega_1$ and $\omega_2$ are inscribed in $ABCD$ such that $\omega_1$ is tangent to $AB$, $BC$, and $CD$ while $\omega_2$ is tangent to $AB$, $DA$, and $CD$. Given that $AB = 1$, $CD = 6$, compute the radius of either circle.

$8)$ Triangle $ABC$ has a right angle at $B$. Point $D$ lies on side $BC$ such that $3\angle BAD =\angle BAC$. Given $AC = 2$ and $CD = 1$, compute $BD$.

$9)$ Four spheres, each of radius $r$, lie inside a regular tetrahedron with side length $1$ such that each sphere is tangent to three faces of the tetrahedron and to the other three spheres. Find $r$.

$10)$ Triangle $ABC$ has side lengths $AB = 65$, $BC = 33$, and $AC = 56$. Find the radius of the circle tangent to sides $AC$ and $BC$ and to the circumcircle of triangle $ABC$.

[hide="2"]
Let $AB=p$.
Since $BP$ bisects $\angle ABC$, $AB: BC=AP: PC=1: 2$
Hence $\triangle ABC = \frac{1}{2}\cdot p\cdot 2p =p^2$

By Pythagorean Theorem, we have $p^2+(2p)^2=5p^2=3^2\Longleftrightarrow p^2=\boxed{\frac{9}{5}}$
[/hide]

[hide="5"]
By Stewart's Theorem,
$(AF)^2(BC)+(AC)^2(BF) = (CF)(AB)^2 + (CF)(BF)(BC)$
$\Longleftrightarrow (AF)^2+25\cdot 2= 3\cdot 20 +3\cdot 2\cdot 1$
$\Longleftrightarrow (AF)^2=16$
$\therefore AF=4$

$\triangle AFC$ is a 3-4-5 triangle, so the area of $\triangle AFC$ is $6$.

Using mass points, we obtain $DE: EF=2: 5$.

$\triangle BEF=6\cdot \frac{DC}{AC}\cdot \frac{EF}{DF} \cdot \frac{BF}{CF} = 6\cdot \frac{1}{5}\cdot \frac{5}{7}\cdot \frac{2}{3} =\frac{4}{7}$

Therefore, the area of quadrilateral $CDEB$ is
$\triangle CDF-\triangle BEF =\frac{1}{5}\cdot 6 -\frac{4}{7} =\boxed{\frac{22}{35}}$

[/hide]

HMMT is a one day competition.  They cannot get all the tests graded and results out on the same day if they put proofs on the test.

[quote="Altheman"]these are good problems! one thing though that #9 i have seen many times before...it is pretty difficult to, i think it was imo shortlist

just curious, how much time do you get for all of those problems? i feel like i can do all of them, but i don't know about how fast[/quote]

You would get 50 minutes for this set of quetions, so... not a lot of time  :P

[hide="Number 4"]We know that the radius at the point of tangency makes right angles, and so we know that if we draw the radius tangent to both sides and drop a segment from $A$, we have two different triangles. If the radius is $x$, we know that the area of $\triangle{ABC}=\frac{3\sqrt{15}}{4}$ (after Heron's) is also equal to $\frac{2x+3x}{2}$. So then $x=\frac{3\sqrt{15}}{10}$, and the area is $\boxed{\frac{27}{40}\pi}$
[/hide]

[hide="Number 7"]If we call the radius of each of the two circles $x$, we see that if we call the tangent points of $\omega_{1}$ $E$, $F$, and $G$ to $AB$, $BC$, and $DC$ respectively, we see that $EB=BF=\frac{1}{2}-x$ and $GC=CF=3-x$, also, if we call the center of the circle $O$, we have that $\triangle{OBC}$ is right at $O$, and so we know that since $OF$ is the altitude of $\triangle{BOC}$, and  $x^2=(\frac{1}{2}-x)(3-x) \Longrightarrow x=\boxed{\frac{3}{7}}$. [/hide]

[hide="Number 3"]I don't feel like explaining it, it's simple if you use power of point and get that the little segments are $1$ and $3$ and the answer is $\boxed{31}$[/hide]

[hide="8"]Applying Pithagoras on $\Delta BAD$ and $\Delta BAC$ we get $AD^2 + 2BD = 3$. Put $\angle BAD = \alpha$. Also $\frac{2}{\cos \alpha} = \frac{1}{\sin 2\alpha}$ $\Longrightarrow$ $\sin \alpha = \frac{1}{4}$. Hence $AD = 4BD$. Subsitute this and get $16BD^2 + 2BD - 3 = 0$ from which $BD = \frac{3}{8}$.[/hide]

[hide="Problem 1"]Since the octagon is equiangular, all interior angles have measure $135^{\circ}$.  Let $A=(0,0)$ and $B=(1,0)$.  Since $BC=2$, we have $C=(1+\sqrt{2},\pm\sqrt{2})$.  Without loss of generality, let $C=(1+\sqrt{2},\sqrt{2})$.  Then $D=(1+\sqrt{2},3+\sqrt{2})$, $E=(1+\sqrt{2}-2\sqrt{2},3+\sqrt{2}+2\sqrt{2})=(1-\sqrt{2},3+3\sqrt{2})$, $F=(1-\sqrt{2}-2,3+3\sqrt{2})=(-1-\sqrt{2},3+3\sqrt{2})$, and $G=(-1-\sqrt{2}-\sqrt{2},3+3\sqrt{2}-\sqrt{2})=(-1-2\sqrt{2},3+2\sqrt{2})$.  We know that $H$ is directly below $G$, and the $x$-coordinate of $H$ is the opposite of the $y$-coordinate.  Therefore, $H=(-1-2\sqrt{2},1+2\sqrt{2})$.  Therefore, $GH=2$ and $HA=4+\sqrt{2}$.  The perimeter is $\boxed{20+\sqrt{2}}$.[/hide]

[hide="9"]
Consider a regular tetrahedron with side length $s$. Drawing a perpendicular along one of the faces, an altitude of a face is $\frac{s\sqrt{3}}{2}\Rightarrow$ the distance from a vertex to the center of a face containing that vertex is $\frac{s\sqrt{3}}{3}$.

Now drop an altitude of the tetrahedron, which will intersect a face at its center. The altitude is then $\frac{s\sqrt{6}}{3}$. Finally, since the coordinates of any tetrahedron in space is given by $\frac{{\bf a}+{\bf b}+{\bf c}+{\bf d}}{4}$, where those letters represent the coordinates of the vertices, the center of the regular tetrahedron lies $\frac{s\sqrt{6}}{12}=\frac{s}{2\sqrt{6}}$ above any face.

This can be applied to the problem because the given tetrahedron and the one formed by connecting the centers of the spheres (this has side length $2r$) have the same center and have a distance of $r$ between a pair of parallel faces. So we get

\[ r+\frac{r}{\sqrt{6}}=\frac{1}{2\sqrt{6}}\Rightarrow 2r\left( 1+\sqrt{6}\right) =1\Rightarrow r=\frac{1}{2\left( 1+\sqrt{6}\right)} =\boxed{\frac{\sqrt{6}-1}{10}} \]
[/hide]


Hello,

find the smallest positive integer $ n$ such that the number of squares of divizors of $ n$ (including $ 1$ and $ n$) is equal to $ (n\plus{}3)^2$.

[quote="Tomekk"]the number of squares of divizors of $ n$[/quote]  What does this mean?  (Unrelatedly, the English word is "divisors.")

the number of squares of divisors means that you sum the squares of divisors of n.

For example, let n=6, then we have 1,2,3,6 dividing 6.

The number of squares of divisors is $ 1^2 \plus{} 2^2 \plus{} 3^2 \plus{} 6^2$.

I apologise for my english, but I simply didnt know how to express myself. I hope this is clear now.

[quote="Tomekk"]Hello,

find the smallest positive integer $ n$ such that the number of squares of divizors of $ n$ (including $ 1$ and $ n$) is equal to $ (n \plus{} 3)^2$.[/quote]

I'm getting 287... doesn't seem like a very friendly question.

Can you show me your way how to solve it, please? :)


prove that if $ a,b,c,d$ are 4 consecutive terms of the fibonacci sequence ,
then, $ ad \minus{}bc  \equal{} 1$ or $ \minus{}1$

Let
$ a \equal{} F_k$
$ b \equal{} F_{k \plus{} 1}$
$ c \equal{} F_{k \plus{} 2}$
$ d \equal{} F_{k \plus{} 3}$

$ ad \minus{} bc$
$ \equal{} F_{k}F_{k \plus{} 3} \minus{} F_{k \plus{} 1}F_{k \plus{} 2}$
$ \equal{} F_{k}F_{k \plus{} 1} \plus{} F_{k}F_{k \plus{} 2} \minus{} F_{k \plus{} 1}F_{k \plus{} 1} \minus{} F_{k \plus{} 1}F_{k}$
$ \equal{} F^2_k \plus{} F_{k}F_{k \plus{} 1} \minus{} F^2_{k \plus{} 1}$

Now $ F_{n \minus{} 1}F_{n \plus{} 1} \minus{} F^2_{n} \equal{} ( \minus{} 1)^{n}$ which can be proven by using the formula for the n'th Fibonacci number

Using that by letting $ F^2_{n} \equal{} F^2_{k \plus{} 1}$ now we get
$ \equal{} F^2_k \plus{} F_{k}F_{k \plus{} 1} \minus{} F^2_{k \plus{} 1}$
$ \equal{} F^2_k \plus{} F_{k}F_{k \plus{} 1} \plus{} ( \minus{} 1)^{k\plus{}1} \minus{} F_{k}F_{k \plus{} 2}$
$ \equal{} F^2_k \plus{} F_{k}F_{k \plus{} 1} \plus{} ( \minus{} 1)^{k\plus{}1} \minus{} F^2_{k} \minus{} F_{k}F_{k \plus{} 1}$
$ \equal{} ( \minus{} 1)^{k\plus{}1}$

[hide="Matrix Approach"][b]Setup:[/b] Define a sequence of four-by-four matrices as:
\[ M_n = \left( \begin{array}{cc} F_{n + 3} & F_{n + 1} & F_{n + 2} & F_n \end{array} \right)\]
One can show that:
\[ M_{n + 1} = \left( \begin{array}{cc} 1 & 1 & 1 & 0 \end{array} \right) M_n\]
So, we can show by induction that:
\[ M_n = \left( \begin{array}{cc} 1 & 1 & 1 & 0 \end{array} \right)^n \left( \begin{array}{cc} 2 & 1 & 1 & 0 \end{array} \right)\]
[b]Problem statement:[/b] Show that $ \text{det} \; M_n = \pm 1$ for all $ n$.

[b]Solution:[/b] For square k by k matrices $ A_1 \ldots A_m$, it is true that:
\[ \text{det} \; \left( \prod_{i = 1}^m A_i \right) = \prod_{i = 1}^m \left( \text{det} \; A_i \right)\]
Thus:
\[ \text{det} \; M_n = \left(\text{det} \; \left( \begin{array}{cc} 1 & 1 & 1 & 0 \end{array} \right) \right)^n \cdot \text{det} \; \left( \begin{array}{cc} 2 & 1 & 1 & 0 \end{array} \right) = ( - 1)^n \cdot ( - 1) = ( - 1)^{n + 1}\]
[/hide]


[quote="Arne"]Who was in the problem selection committee?[/quote]

Ngo Viet Trung, Le Tuan Hoa, Tran Nam Dung, Géza Kós and Ilya Bogdanov

You missed one person: Ha Huy Khoai (chairman).

[quote="Arne"]Who was in the problem selection committee?[/quote]

Why are you asking?

It's listed on the cover sheet of the ISL ;)


Find the real number $ a$ such that the complex roots of the equation $ 2x^2 \minus{} (a \plus{} 2)x \plus{} 2 \equal{}$ have $ |x_1| \equal{} |x_2| \equal{} 1$.

Direct calculation yields the solution space $ \{a(t)\equal{}4\cos t\minus{}2\,|\,t\in[0,2\pi [\,\}$


calculus
integration
trigonometry
logarithms
real analysis
real analysis unsolved

$ \int_{0}^{\frac{\pi}{2}}\frac{1}{5+4 \tan x}\hspace{1 mm}dx = \int_{0}^{\frac{\pi}{2}}\frac{\cos x}{5 \cos x+4 \sin x}\hspace{1 mm}dx$

Now we could write that this is equal to:

$ \log |5 \cos x+4 \sin x|+\int_{0}^{\frac{\pi}{2}}\frac{5 \sin x-3 \cos x}{5 \cos x+4 \sin x}\hspace{1 mm}dx$

but that doesn't quite work.  But we can play around with constants and get:

$ I= \frac{4}{41}\log |5 \cos x+4\sin x|+\frac{1}{41}\int_{0}^{\frac{\pi}{2}}\frac{20 \sin x+25 \cos x}{5 \cos x+4 \sin x}\hspace{1 mm}dx = \frac{4}{41}\log |5 \cos x+4 \sin x|+\frac{1}{41}\int_{0}^{\frac{\pi}{2}}5 \hspace{1 mm}dx = \frac{4}{41}\log |5 \cos x+4 \sin x|+\frac{5}{41}x$

evaluated at the endpoints $ 0$ and $ \frac{\pi}{2}$.

This should simplify to:  $ \frac{4}{41}\log{\frac{4}{5}}+\frac{5 \pi}{82}$

[quote="misan"]solve :D

$ \int_{0}^{\frac{\pi}{2}}\;\frac{1}{5+4\tan x}\;\textbf dx$[/quote]
No complex analysis.  :P

Use the formula $ A\cos x+B\sin x = \sqrt{A^{2}+B^{2}}\sin (x+\theta)$ where $ \tan \theta = \frac{A}{B}$ given that $ A,B\not = 0$.

So,
$ \int_{0}^{\pi/2}\frac{1}{5+4\tan x}dx = \int_{0}^{\pi/2}\frac{\cos x}{5\cos x+4\sin x}dx$
Thus,
$ \int_{0}^{\pi/2}\frac{\cos x}{\sqrt{41}\sin (x+\theta)}dx$ where $ \tan \theta = \frac{5}{4}$.
Let $ t=x+\theta$ so,
$ \int_{\theta}^{\pi/2+\theta}\frac{\cos (t-\theta)}{\sqrt{41}\sin t }dt = \frac{1}{\sqrt{41}}\int_{\theta}^{\pi/2+\theta}\frac{\cos t \cos \theta+\sin t \sin \theta}{\sin t}dt$
So,
$ \frac{1}{\sqrt{41}}\int_{\theta}^{\pi / 2+\theta}\cot t \cos \theta+\sin \theta dt = \frac{1}{\sqrt{41}}\left( \int_{\theta}^{\pi/2+\theta}\cot t \cdot \frac{4}{\sqrt{41}}+\frac{5}{\sqrt{41}}dt\right)$
So,
$ 4\int_{\theta}^{\pi/2+\theta}\cot t dt+\frac{5\pi}{82}$
Almost,
$ 4 \cdot \ln |\csc t-\cot t| \big|_{\theta}^{\pi/2+\theta}+\frac{5\pi}{82}$

Can you finish the last step yourself? It kinda boring.

$ \int \frac{m \sin x+n \cos x}{a \sin x+b \cos x}\, dx = Ax+\ln |a \sin x+b \cos x|+C$

differentiating and solving linear system

$ A = \frac{am+bn}{a^{2}+b^{2}}$

$ B=\frac{an-bm}{a^{2}+b^{2}}$


summer program
Mathcamp
IMC
college contests

since there is a topic who's coming to mexico, here's one for the students..

I'm wondering who of those people who were in greece 2004 is coming to blagoevgrad?

See ya..

Unfortunately I won't be coming this year :( I'm on duty at MathCamp during that time :) But I wish you all good luck and to have fun in the competition! :)

[quote="Matija"]I'm wondering who of those people who were in greece 2004 are coming to blagoevgrad? [/quote]
well, i'm going to bulgaria though i've not been in greece (but i've been in japan in 2003 ;))

I'm coming, but I didn't go to greece either.

I think it would be nice if we (= people from mathlinks) could find eachother there :)

[at the moment I don't know anyone who's going...]

Any germans leaving from munchen airport btw? :)

(it looks like this forum is dead... is the IMC still too far off, or is no one here participating?)

[quote="Peter VDD"]
(it looks like this forum is dead... is the IMC still too far off, or is no one here participating?)[/quote]

I'm coming to IMC and I hope to see all of you in Bulgary

I'm coming to IMC.I haven't been in Greece,but as makc I have been in Japan.By the way it is the first year that a team from Armenia participates.Good luck to all of you and us.   :)

If you see a guy on the airport, seekign desperately for intelligent people to guide him to blagoevgrad, looking like the pic below, you have a good chance it's me. :P

Good luck at this years IMC guys! I am sorry I cannot be with you :( We will surely meet next year though!! :P

Do not forget posting the IMC 2005 problems.
Thank's


trigonometry
geometry
circumcircle
geometry solved

Let A', B', C' be points symmetric to the vertices A, B, C of triangle ABC with respect to its sides BC, CA, AB. What relationship between the angles A, B, C of triangle ABC is necessary and sufficient for the straight lines AB'. BC', CA' to pass through one point?

<i>Source: M.S.Modenov - Problems in Geometry</i>

By trigonometric form of Ceva's theorem we can derive that:

cos a * cos b * cos c = -1/8 where a, b and c are the three angles of triangle.

Is this the relation that you are asking for?

They are concurent iff (by trig Cheva)

(sin 2a sin 2b sin 2c)/(sin a sin b sin c)=-1   <==>

cosA cos B cos C=-1/8.

You are right. What about the condition for ABC such that A', B', C' are colinear?

Namdung

The problem Namdung is asking about was proposed for some Romanian TST, I think 1999, but I have to remember the condition, since I don't have the problems here with me. Maybe someone who took part in that test can tell us what the condition was.

the Romanian TSTs from 1998 to 2003 are available in the downloads section and you can vew them at any time :D

the condition is OH=2R (proposed by France, ISL 1998).

Yes, I remember now. All I need to do is remember my solution :D... Here's something that might help: the condition OH=2R is equivalent to cosA*cosB*cosC=-3/8 (for example, the triangle with angles 120,30,30 is good).

A nicer way to write the condition for the previous problem:  \pi cosA=-1/8 iff OH=\sqrt2*R. I used an inversion of pole H to prove it. I'll come back later with more details.

I think I've got it now. Here's an outline:

Let N be the nine-point circle (I used to call it E from Euler's point, but I see this is the convention here). Let d(X,l) be the distance between line l and point X. Let H be the orthocenter. Try to prove that HX'=4d(N,x), where X can be any of the points A, B, C and x is the side a, b, or c, which corresponds to X (when X is A x is a etc.) We can also show that the angles between HA', HB', HC' are the same as the angles between the distances from N to the sides of the triangle, so we will eventually reach the conclusion: the feet of the perpendiculars from N to the sides of the triangle ABC are collinear, so, from the reciprocal os Simson's thm we get that N is on the circumcircle of ABC, so ON=R, so OH=2ON=2R, and we're done.

Hope it's clear enough.


find all the soluations of the system

$y^{2}=x^{3}-3x^{2}+2x$

$x^{2}=y^{3}-3y^{2}+2y$

[quote="Rust"]
We need solution $x^{6}-5x^{5}+11x^{4}-14x^{4}+16x^{2}-8x+4=0.$ Last equation had not rational roots.[/quote]

This equation has no real roots since
$x^{6}-5x^{5}+11x^{4}-14x^{4}+16x^{2}-8x+4=x^{4}(x^{2}-5x+\frac{25}{4})+x^{2}(\frac{19}{4}x^{2}-14x+\frac{196}{19})+\frac{108}{19}x^{2}-8x+4>0$

because all trinomials are nonnegative and the final is postive if I am not wrong


[quote="CircleSquared"][quote="serialk11r"][quote="massnet"]Two players move their two "cavaliers" in turn starting from the opposite corners of an $ 8$x$ 8$ chessboard. The piece "cavalier" can be moved like a chess king or a chess knight. The player who "checks" the cavalier of the opponent wins. Does anybody have a winning strategy?[/quote]
Um, technically you can't check with either piece because it would be an illegal move...[/quote]

just assume you check them first
[hide]
1st player wins i believe
start off with any move
whenever the opponent's cavalier moves, make a move onto a square of the opposite color
eventually you'll check them
[/hide][/quote]
Hmmm. I don't think there can be a winner. When they get within a certain distance of each other, what happens is you can't move any closer or else the other "cavalier" will "check" you, and therefore both players must move back and forth if they don't want to lose. The first player may be able to get pushy, but I'm pretty sure there's always a square the other one can go to such that he can escape.

then technically wouldn't that be stalemate?  :P

[quote="worthawholebean"]If you can get to the following position, where B is the black cavalier, W is the white cavalier, and X's are the spaces attacked by white, with black against the wall:

OXOXO
XXXXX
BXWXO
XXXXX
OXOXO

then White wins, as Black cannot move without losing. The key fact is that there's that hole two spaces horizontally and vertically in each direction.[/quote]

There are two things wrong with your reasoning.

1: Have you proved that B can be cornered into an edge?

2: Black would win because it would "check" the white cavalier the next turn by the statement of the problem.

Each move doesn't necessarily change the color the cavalier stands on... it can move diagonally from a white space to another white space (or black to black).

[quote="bertram"]Each move changes the color your cavalier is standing on... in the beginning, they are both on different colors; when one checks the other, they are obviously on the same color. Thus, White can't lose... it remains to prove that the game cannot continue indefinitely :)[/quote]
I really think that isn't true...
I think this proves that its not possible for anyone to win (unless they're stupid).
Assume there were a way for one of the pieces to check the other without getting checked first. It would have to move into a square where in 1 move it can get into attacking range. However, both pieces have the range of movement, so if the first piece were to move to that square where it can supposedly win, it would get "checked" and lose, so neither player can win.

[quote="MellowMelon"]If you draw out the area that can be reached in two moves (the forbidden area for the other player), you get something like this.

[code]..XXXXX..
.XXXXXXX.
XXXXXXXXX
XXXXXXXXX
XXXXOXXXX
XXXXXXXXX
XXXXXXXXX
.XXXXXXX.
..XXXXX..[/code]

This isn't rigorous, but you can show using this shape that a win occurs only if a player is cornered. And it's impossible to make a player's only choice to move into a corner.[/quote]

I think this works, but your diagram is a little off.  Eight of the X's on the sides need to be removed.  However I think this can be made into a rigorous argument, but it would be annoying to try to write up.


Solve this following equation:
                                               $ sinx\plus{}sin2x\plus{}sin3x\equal{}1$

[quote="Nbach"]Solve this following equation:
                                               $ sinx \plus{} sin2x \plus{} sin3x \equal{} 1$[/quote]
It is an old problem from PHAN HUY KHAI 's book,I think it is not hard because we can solve:$ f(x)\equal{}1$ wiht $ degf\equal{}3$
and many other way

[quote="Nbach"]Solve this following equation:
                                               $ sinx \plus{} sin2x \plus{} sin3x \equal{} 1$[/quote]

[color=darkblue]You can get a other solution by use:
$ (sinx \plus{} sin2x \plus{} sin3x).2sin\frac {x}{2}$

$ \equal{} 2sinx.sin\frac {x}{2} \plus{} 2sin2x.sin\frac {x}{2} \plus{} 2sin3x.sin\frac {x}{2}$

$ \equal{} cos\frac {x}{2} \minus{} cos\frac {3x}{2} \plus{} cos\frac {3x}{2} \minus{} cos\frac {5x}{2} \plus{} cos\frac {5x}{2} \minus{} cos\frac {7x}{2}$

$ \equal{} cos\frac {x}{2} \minus{} cos\frac {7x}{2}$[/color]

[i]Oh , old problem
   This is a good application of Comlex number :

     Frist , I will give a hint , If nobody can continue , I will post full solution :

            [hide]Denote $ t \equal{} cos x \plus{} isinx$[/hide] [/i]

i would like to see your intelligent solution for this nice solution nguyenvuthanhha thank you!


geometry
analytic geometry
graphing lines
slope

The vertices of a triangle $ABC$ are the points $A(0, 6)$, $B(12, 0)$, and $C(0, 0)$.  A line through the point $(3, 0)$ bisects the area of the triangle.  Compute the slope of the line.

[hide]area of triangle: $\frac{12*6}{2}=36$
area of new triangle: $18$
base = $9$, height = $4$
slope = $4$ (see diagram)[/hide]

Short and sweet solution.  When I drew my diagram, the bisecting line looked perpendicular to AB, and I had to keep fighting the urge to draw a right-angle symbol there.


For positive integers $ n$, let $ f(n),\ g(n)$ be non negative integers such that :

(i) $ g(99)\equal{}1,\ g(100)\equal{}0$

(ii) $ f(100)\equal{}1$

(iii) $ f(n)\plus{}f(n\plus{}g(n))\equal{}f(n\plus{}1)$

Find the values of $ f(99),\ g(101),\ f(2008)$.

-) $ f(n\plus{}1)\geq f(n)$ then $ f(m)\geq f(n)\forall m\geq n$, it mean $ f(n\plus{}1)\geq f(n\plus{}g(n))\Rightarrow n\plus{}1\geq n\plus{}g(n)\Rightarrow g(n)\equal{}0$ or $ 1$.

-) $ g(m)\equal{}0\Rightarrow f(m\plus{}1)\equal{}2f(m)$ ; $ g(m)\equal{}1\Rightarrow f(m)\equal{}0$

-) $ g(99)\equal{}1\Rightarrow f(99)\equal{}0$

-) $ g(100)\equal{}0\Rightarrow f(101)\equal{}2f(100)\equal{}2\neq 0$
$ \Rightarrow g(101)\equal{}0$ (Because if $ g(101)\equal{}1$ then $ f(101)\equal{}0$) $ \Rightarrow f(102)\equal{}2f(101)\equal{}4\neq 0$
$ \Rightarrow g(102)\equal{}0\Rightarrow f(103)\equal{}2f(102)\equal{}8\neq 0$...

Continue by this way we have $ f(2008)\equal{}2^{1908}$

Your answer is correct, tdl. The problem would be difficult for the applicant.
How do you think about this respect?


Positive REALS $a,b,c$ satisfied $a+b+c=1$.

Prove: $\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}\leq 2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)$

thanks, but i think the solution should be nicer.

http://www.mathlinks.ro/Forum/viewtopic.php?t=25780

  darij


calculus
integration
geometry
analytic geometry
trigonometry
function
calculus computations

Hey yall, I have been using this forum for a while now, and it has been to alot of help. ;)  But today i really need some help for this specific problem, and therefor I hope someone here can help.

$\int \int_R \sqrt{x^2 + y^2} dA$
$R = {\{(x,y)\in R^2 | x^2-4x+y^2\leq0}\}$

I have been studying double integrals for a couple of weeks now, but I have still not gotten the hand of it.. I would really appriciate if anyone could help me with this one for example, and perhaps tell me some guidelines when finding the sollution to this double integral. The thing I struggle the most with is finding the limits and actually picturing how the area looks like in a coordinate system (cartesian coordinates).

I really appriciate the help, and thanks in advance. :)

The region $R$ is actually a circular disk, centered at $(2,0),$ of radius $2.$

Try writing it in polar coordinates:

$r^2\cos^2\theta-4r\cos\theta+r^2\sin^2\theta\le0$ becomes

$0\le r\le4\cos\theta,\,-\frac{\pi}2\le\theta\le\frac{\pi}2.$

Now try changing the integral to polar coordinates.

ok.. tnx..

But after I change to polar coordinates, is it then possible to change the double integral to:

$\int\int_T x+y |detRG(t,\theta)| drd\theta$

where $T$ is the region in the $r\theta$-plane that is mapped by $G$ onto the disk $R$.

..since it says sqrt(x^2+y^2) ? Or cant that be done when dealing with an integral?

And by the way, am I integrating a half disk, since $-\frac{\pi}{2} \leq\theta\leq \frac{\pi}{2}$ ?

tnx in advance..

No. Two things:
The square root of a sum is [b]not[/b] the sum of the square roots.
If you're integrating with respect to $r$ and $\theta$, the function you integrate should be written in terms of $r$ and $\theta$.

Now, what is $\sqrt{x^2+y^2}$ in polar coordinates?
Also, that Jacobian is pretty simple and definitely worth knowing; polar coordinates have lots of applications.

It's not a half-disk, but it's all in the right half-plane; that's why we limit the angles.

lol... i am sorry.. my bad.. :blush:  its 4am here in Norway so i am kindof done for the day actually  :lol: ..

what i ment was the right half-plane..

and about the integral and the squareroot sum, i get it ..  :P  it was really obvious.. hehe

and $\sqrt{x^2+y^2}$ is in polar coordinates $\sqrt{r^2cos^2\theta+r^2sin^2\theta}$

ok.. now im off to bed. ill continue in the morning..

Edit:

cant sleep.. might aswell get something clear..

since $r^2 = x^2+y^2$, then we can write $\sqrt{x^2+y^2}$ as $\sqrt{r^2}$ wich can be written as just $r$ ?

then the integral is $\int\int_T r |detRT(r,\theta)|drd\theta$ ?

and for the Jacobian:

$x^2=r^2cos^2\theta = \sqrt{r^2cos^2\theta} = rcos\theta$
and
$y^2=r^2sin^2\theta = \sqrt{r^2sin^2\theta} = rsin\theta$

and by the looks of it now, the Jacobian seems to be $r$, so the final integral to evaluate is:
$\int\int_Tr^2drd\theta$ with the limits

am I right or am I just not ?  :P

You're right, $\sqrt{x^2+y^2}$ is $\sqrt{r^2\cos^2 \theta + r^2\sin^2\theta}$ in polar coordinates, but can't you simplify this to maybe... $r$?

So, your new integral is:
\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^{4\cos\theta} r^2 \ drd\theta = \frac{256}{9}  \]

The reason it's $r^2$ and not $r$ is because the Jacobian for a polar transformation is $r$.

Edit: Sorry, didn't see your edit.  Looks like I just repeated you. :)

hehe.. no problem  :)  Im just glad i got it right.. now im of to bed again..

tnx for the help!

Hmm.. Strange..  When I was working with this one yesterday, I felt that I finally understood the method of finding the limits and converting to polar coordinates. Although today I cant really make any sense of it.  :?  Or should I say, my method did not work to well on other similar tasks.

Can someone explain how to find the limits to the double integral, perhaps step by step?

I have looked in my textbook, but it does not cover much of it..


algebra
polynomial
function
induction
number theory unsolved
number theory

Let $p$ be a prime number and $n$ be any natural number $\leq p-1$. Show that $p$ divides $\sum a_{1}a_{2}...a_{n}$, where the sum is taken for all $1\leq a_{1}<a_{2}<...<a_{n}\leq p-1$

I was thinking about this problem recently, and I found two
different solutions, none of them is pretty but, man's gotta do
what a man's gotto do :)

First one: Look at polynomial $P(x)=(x-1)(x-2)...(x-(p-1))$. This
polynomial in $Z_{p}[x]$ is the same as $x^{p-1}-1$, because they
have the same roots $1,2,...,p-1$. So every coefficient in
polynomial $P$ except the first and the last one is divisible by
$p$, and those coefficients are in fact those sums.

Second one(as ugly as the first one): If you don't know, look
somewhere for Newton formulas for symmetric polynomials. If
$\sigma_{k}=\sum a_{1}a_{2}...a_{k}$ then Newton formula calculate
$\sigma_{n}$ in the function of
$\sigma_{n-1},\sigma_{n-2},...,\sigma_{1}$, and
$s_{i}=1^{i}+2^{i}+...+(p-1)^{i}$. Now use induction and prove that $s_{i}$
is divisible with $p$ when $p-1\not |i$.

Third one(but I didn't complete it): If $g$ is primitive root
modulo $p$ than $\sum a_{1}a_{2}...a_{n}\equiv_{p}\sum g^{a_{1}+a_{2}+...+a_{n}}$ (this is nice way to transform products to
sums). Then all you have to do is to go through all this sums and
prove that there are the same number of that that are $i \mod p$
for every $i$, so there you go.

Now choose wisely :)
Cheers!


From the picture, find the largest force $ F$ such that the box $ 2$ kg stays on the floor.

Also, if $ F \equal{} 100$ newtons, what is the rope tension.

Assume the pulley and the rope to be weightless, and the pulley doesn't rotate.

[hide="solution"] as the pulley is massless and there is no friction(Otherwise the pulley would have rotated) between the rope and teh pulley tension $ T\equal{}\frac{F}{2}$
also as u increase the magnitude of force  $ F$ the normal tries to decrese it's value to keep the block fixed to the ground but if $ T\equal{}20 N\rightarrow F \equal{} 40 N$ the normal has to be $ 0$ and the block lifts so at approx. $ F\equal{}40 N$ the the block just stays on the floor
alsoa s shown above at $ 100 N$the block lifts and tension $ T\equal{}50 N$ [/hide]

[quote="pardesi"] as the pulley is massless and there is no friction(Otherwise the pulley would have rotated) between the rope and teh pulley tension $ T \equal{}\frac{F}{2}$
[/quote]

Well, actually the force is on the pulley, and I don't think the pulley needs to be in balance...

[quote="Radio"]Well, actually the force is on the pulley, and I don't think the pulley needs to be in balance...[/quote]

The pulley must be in balance, because we are considering it to be weightless. Suppose, according to your picture, that the pulley is not in equilibrium. Then we must have, according to Newton's second law,

$ Ma \equal{}\sum F_{i}$

where M is the pulley's mass, a its acceleration, and Fi the forces applied to it. But we are considering that M = 0, and so

$ \sum F_{i}\equal{} 0$,

which means that the pulley is in equilibrium.

Umm.. OK

Can't I think like this:

The force $ F$ on the pulley acts like moving the whole system up, so considering this force only, $ F\equal{}3a$.
We have $ 20\minus{}T\equal{}2a, T\minus{}10\equal{}a$, so considering the forces other than $ F$, we get $ a\equal{}\frac{10}{3}$. So $ F \equal{} 10$

Why F = 3a? What is the meaning of your "a"?

[quote="radio"]Umm.. OK

Can't I think like this:

The force $ F$ on the pulley acts like moving the whole system up, so considering this force only, $ F \equal{} 3a$.
[/quote]
why do u think the entire system ha sthe same accelaration . :?:

Well i think it is because the two blocks are attached to string and must remain the same distance apart. Therefore if one block accelerates downwards then the other has to accelerate upwards to stay the same distance apart.

yes that clearly implie s that they don't have the same accelaration(direction and magnitude)

The only body that can have acceleration is box 1, since box 2 is supposed to stay on the floor, and I have already demonstrated that the pulley must be in equilibrium.


geometry
3D geometry
pyramid
sphere
tetrahedron
trapezoid

Ok so, 14 spheres with radii of 1 are stacked so that there are nine spheres in the base, four spheres on top of those spheres, and one sphere on top of those spheres. All of the spheres are externally tangent to its neighbors. A square pyramid is circumscribed around the 14 spheres. What is the volume of this pyramid? Express your answer in simplest radical form.

Are you sure that this should not be square pyramid, not tetrahedron?

[quote="Ihatepie"]well,  whats the radius of each sphere?[/quote]
um, you need to answer this question as well

[quote="ZhangPeijin"]Umm I edited it, it says with a radii of 1.[/quote]

So you made this question without finding the answer to it. Since I am not able to find the answer, can you do it? Strange if you create a problem that you don't know how to do yourself.

Maybe you should ask Mr. Frost to approve it if he hasn't done yet.

hmm... The only things I can say are...

- The base of the pyramid is not a 6 x 6 square.
- The vertices of the pyramid do not touch and spheres.

If your answer meets the above, then...
I will  get stumped on this question on your test, and I'm not afraid to admit it.

[quote="ZhangPeijin"]The edges of the pyramid are tangent to the squares,[/quote]
That's very confusing. Do you mean spheres?

[quote="alanchou"]you probably should post the solution, because I wonder if you really meant that the spheres were tangent to the edges[/quote]
Do you know the answer? If not then I wonder that frost would say.

[hide="Solution"]Consider one triangular face of the pyramid.  Obviously all 4 of the faces are congruent, and it's easy to see that all the faces are equilateral.  We thus have a regular square pyramid.  The area of a pyramid is given by $ \frac {1}{3}Bh$, where $ B$ denotes the area of the base.
Draw one of the isosceles trapezoids by connecting the radii of three circles and then connecting the endpoints of that segment to the vertices of the triangle (not clear, I know).  Draw the altitudes of the trapezoid; we then have two congruent $ 30 \minus{} 60 \minus{} 90$ right triangles.  Since the angle opposite the $ 30^\circ$ is a radius and thus of length 1.  We can find the length of the other leg by multiplying by $ \sqrt {3}$.

We now have that the length of one side of the square base is $ 6 \plus{} 2\sqrt {3}$.
To find the height, we draw the altitude of one of the faces and connect the foot of it to the center of the square base.  Then we draw the height from the vertex of the pyramid to this center.

Then, $ h^2 \plus{} (\frac {s}{2})^2 \equal{} a^2$, where $ s$ denotes the length of one side of the square base, and $ a$ denotes the length of the altitude of one of the triangular faces.
$ a$ is simply $ (3 \plus{} \sqrt {3})\sqrt {3} \equal{} 3 \plus{} 3\sqrt {3}$ (this is found using the well-known $ 30 \minus{} 60 \minus{} 90$ triangle.

We also know that $ s \equal{} 6 \plus{} 2\sqrt {3}$.

So $ h^2 \plus{} 12 \plus{} 6\sqrt {3} \equal{} 36 \plus{} 18\sqrt {3}$
$ h^2 \equal{} 24 \plus{} 12\sqrt {3}$
$ h \equal{} 2\sqrt {6 \plus{} 3\sqrt {3}} \equal{} 2\sqrt {\frac {3}{2}(4 \plus{} 2\sqrt {3})} \equal{} 2(1 \plus{} \sqrt {3})\frac {\sqrt {6}}{2}$ (lol)

So now... our answer is $ \frac {1}{3}*4*(12 \plus{} 6\sqrt {3})*2*(1 \plus{} \sqrt {3})*\frac {\sqrt {6}}{2} \equal{} \boxed{40\sqrt {6} \plus{} 72\sqrt {2}}$[/hide]
Probably wrong considering I'm doing this without scratch paper.

And... I'm pretty sure he means that the spheres are tangent to the faces of the pyramid.

I think you did your last step wrong.
Because I get $ 16\sqrt{6}\plus{}72\sqrt{2}$
Or maybe I can't distribute right.

Eh I guess I can't distribute right then.
Though I did do that all in my head too....

[quote="ZhangPeijin"]Actually Jason, it's simple if you draw 2-d drawings.
So yeah, I do have a solution.[/quote]

Um, drawing pictures and doing everything in your head?  That doesn't really make sense

OMG, I couldn't get the answer cause I was using a square pyramid with 4 faces  :huh:

I see...
Well I can draw pictures in my head too.
Look
I just drew a circle.
Yay....


If there exist $p_0,q_0 \in (0,\infty)$, such that $L^{p_0}(E) \subset L^{q_0}(E)$. Prove that it holds $L^p(E)\subset L^q(E)$ for all $0<p<q<\infty$.

Of course, $p_0<q_0$ is part of the condition.

I suspect this condition is equivalent to $E$ being an atomic measure space - that is, it consists only of points of positive measure.


calculus
derivative
function
real analysis
real analysis unsolved

Hi, am I right if I say that it takes more than 2 pages to solve this:

$ f$ is a function from $ \mathbb{R}^2$ to $ \mathbb{R}$ and $ f(x,y)\equal{}\frac{xy(x^2\minus{}y^2)}{x^2\plus{}y^2}$ and $ f(0,0)\equal{}0$.
Show that $ \frac{\partial^2f}{\partial x \partial y}(0,0) \neq \frac{\partial^2f}{\partial y \partial x}(0,0)$.

I remember doing this problem, or one really similar.  It's pretty annoying, but it's nothing more than algebra.  It took me about one half of a piece of paper to write it out, but I write small.


calculus
integration
function
calculus computations

I'm not quite sure how everything is related so can you guide me through some of these problems

9. Find the position function s(t) given acceleration a(t) = 3t if v(2) = 0
and s(2) = 1.
10. An object in rectilinear motion has acceleration a(t) = 12t - 6. If the
velocity at time t = 2 is -3, and the position at time t = 1 is 4, nd
the position function, s(t), for the object.
11. A particle starts moving along the x-axis from the point (100, 0),
(distance in meters) and with an initial velocity of 25 m/min. If the
acceleration is given by the equation a(t) = 13 sqrt(t), what is the equation
of motion of the particle?
The next two questions involve the following situation: An
astronaut stands on a platform 3 meters above the moon's surface and
throws a rock directly upward with an initial velocity of 32 m/s.
12. Given that the acceleration due to gravity on the moon's surface is 1.6
m/sec^2, derive an equation which gives the height of the rock at time
t seconds (valid while the rock is in the air).
13. How high will the rock travel?
14. A bocce ball is accidentally dropped from a building 98m high. How
long does it take for the bocce ball to hit the gound, given that the
acceleration due to gravity is 9.8 meters per second per second?
The next two questions involve the following situation: A
particle moving along the number line has acceleration given by a(t) =
2t - 1. We also know that s0 = 2 and v0 = -2.
15. What is the net distance traversed from time t = 0 to t = 3?
16. What is the total distance traversed from time t = 0 to t = 3?

Don't post homework problems. This is easy kinematics, all you really need to know is;

$ x(t) \equal{} \int v(t) \;dt \equal{} \int \int a(t) \;dt$

Assuming a(t) is a constant (which is the case for every problem you posted except 11 and 14), we get;

$ v(t) \equal{} v_{0} \plus{} at$
$ x(t) \equal{} x_{0} \plus{} v_{0}t \plus{} \frac {1}{2} at^{2}$

Apply those equations to get your answers.

Actually that formula was all I really needed to know to do the problems.
Thanks,


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $K$ be a field with 81 elements. Show that K does not have a subfield with 27 elements. Show also that $L = \{a\in K | a^{9}= a \}$ is the unique subfield of $K$ with 9 elements.

so...i think you can easily do this by taking the associated (K,*)\{0} group...it has 80 elements, therefore it doesn't have a 26 element subgroup...for the second part we use Sylow's theorem.. :maybe:

This is a problem from an exam in a very introductory course in abstract algebra. No sylow!

It is very straight forward to check the field-axioms for L. We have to prove, that L has 9 elements and is the only subfield with this property.

First $|L|=9$:
$L^{*}$ is cyclic and we have $\forall a\in L^{*}: a^{9}=a \Rightarrow  a^{8}=1$. Hence the order of a generating element is  a divisor of 8 and $|L|\leq 9$.
The group $K^{*}$ is cyclic too. So there is an element y with order $|K^{*}|=80$, so $y^{10}$ has order 8 and $\langle y^{10}\rangle$ is therefore a subgroup of $K^{*}$ with 8 elements. Now we have $a\in\langle y^{10}\rangle \Rightarrow  a^{8}=1 \Rightarrow  a^{9}=a \Rightarrow  a\in\ L$, so $L^{*}$ has at least 8 elements and $|L|\geq 9$.

Now uniqueness:
Let F a subfield with 9 elements, then $F^{*}$ is a group with 8 elements, so we have $\forall x\in F^{*}: x^{8}=1 \Rightarrow  \forall x\in F: x^{9}=x \Rightarrow  F\subseteq L\Rightarrow F=L$.


Let $ a,b,c\geq 1$ and $ a^2\plus{}2b^2\plus{}3c^2\equal{}24$. Find Max,Min of value
$ P\equal{}a\plus{}b\plus{}c$ :lol:

Cauchy-Schwartz kills it!

$ (a^2 \plus{} 2b^2 \plus{} 3c^2 ) (1 \plus{} \frac {1}{2} \plus{} \frac {1}{3}) \geq (a \plus{} b \plus{} c)^2$.

Therefore, $ (a \plus{} b \plus{} c)^2 \leq (24)(\frac {11}{6}) \equal{} 44$,

$ a \plus{} b \plus{} c \leq \sqrt{44}$

[quote="mufc"]Let $ a,b,c\geq 1$ and $ a^2 \plus{} 2b^2 \plus{} 3c^2 \equal{} 24$. Find Max,Min of value
$ P \equal{} a \plus{} b \plus{} c$ :lol:[/quote]
The maximum value is actual easy. Let I post my solution for the minimum value
Setting $ a^2\equal{}x\plus{}1,b^2\equal{}y\plus{}1,c^2\equal{}z\plus{}1$ then $ x,y,z \ge 0$ and $ x\plus{}2y\plus{}3z\equal{}18$. From this, it follows
\[ 3(x\plus{}y\plus{}z) \ge x\plus{}2y\plus{}3z \equal{}18\Rightarrow x\plus{}y\plus{}z \ge 6\]
Now, Notice that
\[ \sqrt{m\plus{}1}\plus{}\sqrt{n\plus{}1}\ge 1\plus{}\sqrt{m\plus{}n\plus{}1}\qquad \forall m,n \ge 0\]
We get
\[ P\equal{}\sqrt{x\plus{}1}\plus{}\sqrt{y\plus{}1}\plus{}\sqrt{z\plus{}1} \ge 1\plus{}\sqrt{x\plus{}y\plus{}1}\plus{}\sqrt{z\plus{}1} \ge 2\plus{}\sqrt{x\plus{}y\plus{}z\plus{}1} \ge 2\plus{}\sqrt{7}\]
Equality holds for $ x\equal{}y\equal{}0,z\equal{}6$ or $ a\equal{}b\equal{}1,c\equal{}\sqrt{7}$.
We get the conclusion
\[ \min P\equal{}2\plus{}\sqrt{7}.\]


How would you prove or disprove that for x,y integers:

if x^3 divides y^3 then x divides y
if x^3 divides y^2 then x divides y
if x^2 divides y^3 then x divides y

A little bit of reasoning about the prime factorizations of $x, y$ provides a quick solution.

[quote="Alope"]
if x^2 divides y^3 then x divides y[/quote]
I think it is not true x=8,y=4

I think you should post these problems in pre-olympaid section,Alope. :)

Sounds good, thank you.  I wasn't sure where to post because I couldn't find "number theory" anywhere else.

Umm, do not repost, but ask a moderator to move it.


trigonometry
calculus
derivative
algebra unsolved
algebra

Let $ y(x) \equal{} |\cos x \plus{} \alpha\cos 2x \plus{} \beta\cos 3x|$. Find
\[ \min_{\alpha,\beta\in \mathbb R}\left(\max_{x\in \mathbb R} y(x)\right)
\]

Could this work?:

convert the cos2x and cos3x to cosx and then take the derivative to see when it is maximum. Then cosx = some expression involving a and b. then put it back into y(x). Then find the minimum of that expression by partial derivatives.

So... post solution and we will see...  :)


inequalities
algebra
polynomial
geometry
geometric transformation
function
inequalities unsolved

please translate http://zgbdsyjxz.nease.net/fcd/lbq12-8.pdf into english.

In the document, it gave the following proof:

Assume $z\leq x \leq y$, and let $x = z+a$ and $y=z+a+b$. Plug them into the inequality, and expand. Then we get a huge polynomial in $z, a, b$ whose coefficients are all positive. This proves the inequality.

Of course, we MathLinkers would like to see more insightful solutions than that.

Also in the document, there were several comments/questions regarding the method (such as, why is the degree of $z$ in the expansion lowered). I don't see the significance behind these comments, so I'm not going to bother translating them.

Here is a proof of the inequality which I, personally, find nice, but I must admit this is quite a subjective judgement. I guess (and hope) it can be simplified in some of its steps, however I am rather sure it cannot be made much shorter because the inequality itself is about 3 times as long and complicated as the inequalities we usually consider on MathLinks. Anyway, have fun with my opus :D .

[color=red][b]1. Proving inequalities by "parallel translation"[/b][/color]

First we show a trivial but useful fact, which allows us to reduce inequalities in three nonnegative variables to the case when one of these variables is 0.

[color=blue][b]Lemma 1.[/b] Let $f: \mathbb{R}^3\to\mathbb{R}$ be a function of three real variables such that:

[i]Assertion 1.[/i] We have $f\left(x;\;y;\;z\right)\geq 0$ for any three nonnegative reals x, y, z such that (at least) one of these reals x, y, z equals 0.

[i]Assertion 2.[/i] We have $f\left(x+d;\;y+d;\;z+d\right)\geq f\left(x;\;y;\;z\right)$ for any four nonnegative reals x, y, z, d.

Then, $f\left(x;\;y;\;z\right)\geq 0$ for any three nonnegative reals x, y, z.[/color]

[hide="Proof of Lemma 1"][i]Proof of Lemma 1.[/i] Consider three nonnegative reals x, y, z. We can WLOG assume that x is the smallest of these reals, so that $x\leq y$ and $x\leq z$, and thus the reals y - x and z - x are nonnegative. Hence, Assertion 1, applied to the nonnegative reals 0, y - x, z - x, one of which equals 0, yields $f\left(0;\;y-x;\;z-x\right)\geq 0$, and Assertion 2, applied to the nonnegative reals 0, y - x, z - x, x, yields $f\left(0+x;\;\left(y-x\right)+x;\;\left(z-x\right)+x\right)\geq f\left(0;\;y-x;\;z-x\right)$. Thus

$f\left(x;\;y;\;z\right)=f\left(0+x;\;\left(y-x\right)+x;\;\left(z-x\right)+x\right)\geq f\left(0;\;y-x;\;z-x\right)\geq 0$,

and Lemma 1 is proven.[/hide]

[color=red][b]2. Algebraic manipulation of the inequality in question[/b][/color]

In the following, we will use the sign $\sum$ for cyclic sums.

We have to prove the inequality

$\sum x\left(x-y\right)\left(x-z\right)$
$\geq xyz\left(1-\frac{8xyz}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\right)+\frac{4\left(y-z\right)^2\left(z-x\right)^2\left(x-y\right)^2}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}$

for any three nonnegative reals x, y, z. Multiplication by (y + z) (z + x) (x + y) transforms this inequality into

[color=green][b](1)[/b][/color] $\left(y+z\right)\left(z+x\right)\left(x+y\right)\sum x\left(x-y\right)\left(x-z\right)$
$\geq xyz\left(\left(y+z\right)\left(z+x\right)\left(x+y\right)-8xyz\right)+4\left(y-z\right)^2\left(z-x\right)^2\left(x-y\right)^2$.

Now, simple calculations show that

[color=green][b](2)[/b][/color] $\left(y+z\right)\left(z+x\right)\left(x+y\right)-8xyz=\sum x\left(y-z\right)^2$.

What is a bit more difficult (not with a computer, of course, but that's not the way to go) is to show that

[color=green][b](3)[/b][/color] $\left(y+z\right)\left(z+x\right)\left(x+y\right)\sum x\left(x-y\right)\left(x-z\right)=\sum yz\left(y+z\right)^2\left(y-z\right)^2$.

This nice identity (which, by the way, yields another proof of the Schur inequality $\sum x\left(x-y\right)\left(x-z\right)\geq 0$), can be proven, e. g., as follows:

First note that $2\left(x-y\right)\left(x-z\right)=\left(z-x\right)^2+\left(x-y\right)^2-\left(y-z\right)^2$, so that

$2\sum x\left(x-y\right)\left(x-z\right)=\sum x\cdot 2\left(x-y\right)\left(x-z\right)=\sum x\cdot\left(\left(z-x\right)^2+\left(x-y\right)^2-\left(y-z\right)^2\right)$
$=\sum x\left(z-x\right)^2+\sum x\left(x-y\right)^2-\sum x\left(y-z\right)^2$
$=\sum z\left(y-z\right)^2+\sum y\left(y-z\right)^2-\sum x\left(y-z\right)^2=\sum\left(y+z-x\right)\left(y-z\right)^2$.

So we have obtained the identity

[color=green][b](4)[/b][/color] $2\sum x\left(x-y\right)\left(x-z\right)=\sum\left(y+z-x\right)\left(y-z\right)^2$.

Thus,

$2\left(\left(y+z\right)\left(z+x\right)\left(x+y\right)\sum x\left(x-y\right)\left(x-z\right)-\sum yz\left(y+z\right)^2\left(y-z\right)^2\right)$
$=\left(y+z\right)\left(z+x\right)\left(x+y\right)\cdot 2\sum x\left(x-y\right)\left(x-z\right)-2\sum yz\left(y+z\right)^2\left(y-z\right)^2$
$=\left(y+z\right)\left(z+x\right)\left(x+y\right)\cdot\sum\left(y+z-x\right)\left(y-z\right)^2-2\sum yz\left(y+z\right)^2\left(y-z\right)^2$
$=\sum\left(y+z\right)\left(\left(z+x\right)\left(x+y\right)\left(y+z-x\right)-2yz\left(y+z\right)\right)\left(y-z\right)^2$
$=\sum\left(y+z\right)\left(\left(\left(x+y+z\right)x+yz\right)\left(y+z-x\right)-2yz\left(y+z\right)\right)\left(y-z\right)^2$
$=\sum\left(y+z\right)\left(\left(x+y+z\right)x\left(y+z-x\right)+yz\left(y+z-x\right)-2yz\left(y+z\right)\right)\left(y-z\right)^2$
$=\sum\left(y+z\right)\left(\left(x+y+z\right)x\left(y+z-x\right)-yz\left(x+y+z\right)\right)\left(y-z\right)^2$
$=\sum\left(y+z\right)\left(x+y+z\right)\left(x\left(y+z-x\right)-yz\right)\left(y-z\right)^2$
$=\sum\left(y+z\right)\left(x+y+z\right)\left(x-y\right)\left(z-x\right)\left(y-z\right)^2$
$=\left(x+y+z\right)\left(y-z\right)\left(z-x\right)\left(x-y\right)\underbrace{\sum\left(y+z\right)\left(y-z\right)}_{=\sum\left(y^2-z^2\right)=0}=0$,

so that the identity [color=green][b](3)[/b][/color] is proven.

Using the identities [color=green][b](2)[/b][/color] and [color=green][b](3)[/b][/color], we can rewrite the inequality [color=green][b](1)[/b][/color] as

$\sum yz\left(y+z\right)^2\left(y-z\right)^2\geq xyz\sum x\left(y-z\right)^2+4\left(y-z\right)^2\left(z-x\right)^2\left(x-y\right)^2$,

or, equivalently, as $f\left(x;\;y;\;z\right)\geq 0$, where

$f\left(x;\;y;\;z\right)=\sum yz\left(y+z\right)^2\left(y-z\right)^2-xyz\sum x\left(y-z\right)^2-4\left(y-z\right)^2\left(z-x\right)^2\left(x-y\right)^2$.

Now,

$\sum yz\left(y+z\right)^2\left(y-z\right)^2-xyz\sum x\left(y-z\right)^2=\sum\left(yz\left(y+z\right)^2-xyzx\right)\left(y-z\right)^2$
$=\sum\left(x+y+z\right)\left(y+z-x\right)yz\left(y-z\right)^2=\left(x+y+z\right)\sum\left(y+z-x\right)yz\left(y-z\right)^2$,

so that

$f\left(x;\;y;\;z\right)=\left(x+y+z\right)\sum\left(y+z-x\right)yz\left(y-z\right)^2-4\left(y-z\right)^2\left(z-x\right)^2\left(x-y\right)^2$.

[color=red][b]3. Application of Lemma 1[/b][/color]

Now, we want to prove that $f\left(x;\;y;\;z\right)\geq 0$ for any three nonnegative reals x, y, z. By Lemma 1, this will be done once Assertions 1 and 2 for this function f(x; y; z) will be proved.

[i]Proof of Assertion 1.[/i] We have to prove that $f\left(x;\;y;\;z\right)\geq 0$ for any three nonnegative reals x, y, z such that (at least) one of these reals x, y, z equals 0. We can WLOG assume that x = 0. Then, in the formula for f(x; y; z), all terms containing x as a multiplicative factor vanish:

$f\left(x;\;y;\;z\right)=\sum yz\left(y+z\right)^2\left(y-z\right)^2-xyz\sum x\left(y-z\right)^2-4\left(y-z\right)^2\left(z-x\right)^2\left(x-y\right)^2$
$=yz\left(y+z\right)^2\left(y-z\right)^2-4\left(y-z\right)^2\left(z-0\right)^2\left(0-y\right)^2=yz\left(y+z\right)^2\left(y-z\right)^2-4\left(y-z\right)^2z^2y^2$
$=yz\left(y-z\right)^2\left(\left(y+z\right)^2-4yz\right)=yz\left(y-z\right)^2\left(y-z\right)^2\geq 0$,

and Assertion 1 is proven.

[color=red][b]4. Proof of Assertion 2[/b][/color]

[i]Proof of Assertion 2.[/i] We have to show that $f\left(x+d;\;y+d;\;z+d\right)\geq f\left(x;\;y;\;z\right)$ for any four nonnegative reals x, y, z, d.

But as we have

$f\left(x;\;y;\;z\right)=\left(x+y+z\right)\sum\left(y+z-x\right)yz\left(y-z\right)^2-4\left(y-z\right)^2\left(z-x\right)^2\left(x-y\right)^2$,

we have

$f\left(x+d;\;y+d;\;z+d\right)$
$=\left(\left(x+d\right)+\left(y+d\right)+\left(z+d\right)\right)\sum\left(\left(y+d\right)+\left(z+d\right)-\left(x+d\right)\right)\left(y+d\right)\left(z+d\right)\left(\left(y+d\right)-\left(z+d\right)\right)^2$
$-4\left(\left(y+d\right)-\left(z+d\right)\right)^2\left(\left(z+d\right)-\left(x+d\right)\right)^2\left(\left(x+d\right)-\left(y+d\right)\right)^2$
$=\left(x+y+z+3d\right)\sum\left(y+z-x+d\right)\left(y+d\right)\left(z+d\right)\left(y-z\right)^2-4\left(y-z\right)^2\left(z-x\right)^2\left(x-y\right)^2$;

hence, in order to prove that $f\left(x+d;\;y+d;\;z+d\right)\geq f\left(x;\;y;\;z\right)$, it is enough to show that

[color=green][b](5)[/b][/color] $\left(x+y+z+3d\right)\sum\left(y+z-x+d\right)\left(y+d\right)\left(z+d\right)\left(y-z\right)^2\geq\left(x+y+z\right)\sum\left(y+z-x\right)yz\left(y-z\right)^2$.

Now, we will prove the following inequalities:

[color=green][b](6)[/b][/color] $\sum\left(y+z-x\right)yz\left(y-z\right)^2\geq 0$;
[color=green][b](7)[/b][/color] $\sum\left(y+z-x\right)\left(y+d\right)\left(z+d\right)\left(y-z\right)^2\geq\sum\left(y+z-x\right)yz\left(y-z\right)^2$.

These two inequalities will yield

$\left(x+y+z+3d\right)\sum\left(y+z-x+d\right)\left(y+d\right)\left(z+d\right)\left(y-z\right)^2$
$\geq\left(x+y+z+3d\right)\sum\left(y+z-x\right)\left(y+d\right)\left(z+d\right)\left(y-z\right)^2$    (since $d\geq 0$)
$\geq\left(x+y+z+3d\right)\sum\left(y+z-x\right)yz\left(y-z\right)^2$    (by [color=green][b](7)[/b][/color])
$\geq\left(x+y+z\right)\sum\left(y+z-x\right)yz\left(y-z\right)^2$     (by [color=green][b](6)[/b][/color]),

and thus the inequality [color=green][b](5)[/b][/color] will be proven. Thus, it remains to verify the inequalities [color=green][b](6)[/b][/color] and [color=green][b](7)[/b][/color].

We can WLOG assume that $x\geq y\geq z$. Then, $z+x\geq z+y\geq y$, so that $z+x-y\geq 0$, and $x+y\geq x+z\geq z$, so that $x+y-z\geq 0$. If we also have $y+z-x\geq 0$, then the inequalities [color=green][b](6)[/b][/color] and [color=green][b](7)[/b][/color] are trivial ([color=green][b](6)[/b][/color] is obvious since all three terms on the left-hand side are $\geq 0$, and [color=green][b](7)[/b][/color] is trivial because $y+d\geq y$ and $z+d\geq z$ yield that every term on the left-hand side is $\geq$ to the corresponding term on the right-hand side). So troubles are only posible in the case y + z - x < 0. Hence, we will consider only this case.

We have $y\leq x$, hence $yz\leq zx$. Upon multiplication with the negative number y + z - x, this inequality changes its sign: $\left(y+z-x\right)yz\geq\left(y+z-x\right)zx$. On the other hand, $y\geq z$ yields $xy\geq zx$. Upon multiplication with the nonnegative number x + y - z, this inequality doesn't change its sign: $\left(x+y-z\right)xy\geq\left(x+y-z\right)zx$. Hence,

$\sum\left(y+z-x\right)yz\left(y-z\right)^2$
$=\left(y+z-x\right)yz\left(y-z\right)^2+\left(z+x-y\right)zx\left(z-x\right)^2+\left(x+y-z\right)xy\left(x-y\right)^2$
$\geq\left(y+z-x\right)zx\left(y-z\right)^2+\left(z+x-y\right)zx\left(z-x\right)^2+\left(x+y-z\right)zx\left(x-y\right)^2$
$=zx\left(\left(y+z-x\right)\left(y-z\right)^2+\left(z+x-y\right)\left(z-x\right)^2+\left(x+y-z\right)\left(x-y\right)^2\right)$
$=zx\sum\left(y+z-x\right)\left(y-z\right)^2=zx\cdot 2\sum x\left(x-y\right)\left(x-z\right)$    (by [color=green][b](4)[/b][/color])
$\geq 0$

by Schur, and the inequality [color=green][b](6)[/b][/color] is proven. As for the inequality [color=green][b](7)[/b][/color], it rewrites as

$\sum\left(y+z-x\right)\left(y+d\right)\left(z+d\right)\left(y-z\right)^2-\sum\left(y+z-x\right)yz\left(y-z\right)^2\geq 0$;

but

$\sum\left(y+z-x\right)\left(y+d\right)\left(z+d\right)\left(y-z\right)^2-\sum\left(y+z-x\right)yz\left(y-z\right)^2$
$=\sum\left(y+z-x\right)\left(\left(y+d\right)\left(z+d\right)-yz\right)\left(y-z\right)^2$
$=\sum\left(y+z-x\right)d\left(y+z+d\right)\left(y-z\right)^2=d\sum\left(y+z-x\right)\left(y+z+d\right)\left(y-z\right)^2$,

so it remains to prove that

$\sum\left(y+z-x\right)\left(y+z+d\right)\left(y-z\right)^2\geq 0$.

We have $y\leq x$, hence $y+z+d\leq z+x+d$. Upon multiplication with the negative number y + z - x, this inequality changes its sign: $\left(y+z-x\right)\left(y+z+d\right)\geq\left(y+z-x\right)\left(z+x+d\right)$. On the other hand, $y\geq z$ yields $x+y+d\geq z+x+d$. Upon multiplication with the nonnegative number x + y - z, this inequality doesn't change its sign: $\left(x+y-z\right)\left(x+y+d\right)\geq\left(x+y-z\right)\left(z+x+d\right)$. Hence,

$\sum\left(y+z-x\right)\left(y+z+d\right)\left(y-z\right)^2$
$=\left(y+z-x\right)\left(y+z+d\right)\left(y-z\right)^2+\left(z+x-y\right)\left(z+x+d\right)\left(z-x\right)^2+\left(x+y-z\right)\left(x+y+d\right)\left(x-y\right)^2$
$\geq\left(y+z-x\right)\left(z+x+d\right)\left(y-z\right)^2+\left(z+x-y\right)\left(z+x+d\right)\left(z-x\right)^2+\left(x+y-z\right)\left(z+x+d\right)\left(x-y\right)^2$
$=\left(z+x+d\right)\left(\left(y+z-x\right)\left(y-z\right)^2+\left(z+x-y\right)\left(z-x\right)^2+\left(x+y-z\right)\left(x-y\right)^2\right)$
$=\left(z+x+d\right)\sum\left(y+z-x\right)\left(y-z\right)^2=\left(z+x+d\right)\cdot 2\sum x\left(x-y\right)\left(x-z\right)$    (by [color=green][b](4)[/b][/color])
$\geq 0$

by Schur, and the inequality [color=green][b](7)[/b][/color] is proven. This completes the proof of our inequality. $\blacksquare$

  Darij


While Bob was selecting a beverage at the pizza shop, he noticed that the ice tea company was giving out free merchandise when selected bottles of there drink were bought.  Bob looked at one bottle and read the following: "One in 12 people will win".  Bob reasoned as follows:  If I buy 12 bottles of ice tea I am garunteed to win a prize.  So he did.  But he ended up winning no prizes at all.  What is the flaw in Bob's reasoning and what is the correct probability of winning at least one prize among 12 bottles of ice tea?  Addtionally, give a brief explanation, with reasoning, as to why your chances of winning a prize increases as the number of bottles of ice tea you buy increases.

I'm not going to answer this, but I will say that Bob is extremely unlucky to not win after buying 12 bottles. :P

I'm not going to do all that reasoning, but the correct probability (I think) is

[hide]$(\frac{11}{12})^{11}\cdot\frac{1}{12}\cdot12\approx0.38$

nvm, I think it's wrong...[/hide]

[hide="answer"]
Don't have a calculator on hand, but the answer is:

1 - (11/12)^12.
[/hide]

[hide="Hint"]The key words are [b]at least one prize[/b]. Basically, it's $1-P$, where $P$ is the probability of not winning any prizes.[/hide]

[b]EDIT:[/b] Correction: Treething has the right answer.

Treething's looks more logical, but why is mine wrong?  :?

Well... as I said, unless I'm making a mistake, I know that the answer is higher than that... I don't really understand what you did, for one thing. It looks like you might be thinking the probability of winning [b]exactly[/b] one prize.

As for Treething's... my mistake. I was actually doing the wrong thing there. Sorry about that. :P

[quote="JesusFreak197"]Well... as I said, unless I'm making a mistake, I know that the answer is higher than that... I don't really understand what you did, for one thing. It looks like you might be thinking the probability of winning [b]exactly[/b] one prize.
[/quote]

Yeah i did  :roll: that means Treething was correct  :blush:


If $ x,y,z \geq 0$ then:
$ \sum x^8 \geq x^{2}y^{2}z^{2}( \sum xy ) $ .

(xxx).

cheers! :D :D

Well, in the form

2 sum(x^8) >= 2 sum(x^3 y^3 z^2),

it follows from Muirhead...

  Darij


A lamppost in a park can illuminate an area of radius 40 feet. What is the minimum number of lampposts that must be installed in the park below so that the entire region is illuminated?
[asy]draw((0,0)--(0,60)--(100,60)--(100,0)--cycle);
label("200$\sqrt{2}$\ ft", (50,60), N);
label("120$\sqrt{2}$\ ft", (100,30), E);[/asy]

Partition the rectangle into a 3x5 square grid of 40sqrt(2) ft sides.
Then put a lamppost on the center of each square. 15.

If anyone has a more efficient method, please speak up.

i beleive this is a glitch on AOPS. you see, a radius of 40ft means that the area that each lamppost can illuminate is about 5026.5 sqft. the area of the lot is 48,000sqft so you need at leat ten?

 :|

no its not a glitch, if you think about it, the lampposts illuminate a circle shaped area so you would need more than just 10. Four of the circles would have to be tangent to the corners of the rectangle and some of the illumination would be outside of the park

I have finally solved this problem. Let me explain, it might be rough... I recommend you to read through the whole thing and draw a diagram with a circle with an inscribed square inside it.

First off, it asks for the minimum number of lampposts so you would need to tactically place them so they illuminate the most area without wasting any.(What i mean by waste area is if the illumination goes outside of the park or if it overlaps with another lampposts illuminated area.)

So... the way you would need to think about it is the maximum area each lamppost can cover without having too much overlap with other lampposts.

In this problem, they gave you the radius that each lamppost can cover which an area of radius 40 feet. (Now is when you make the radius of your circle that you drew, 40! ) This is where it gets slightly confusing....
Now, you have to find the largest square that can fit inside this circle of 40 feet radius because this will guarantee the minimum number of lamppost needed. (I hope that wasn't too confusing :-D) If you think about it, the largest 4 sided shape that can fit in a circle is a square and we need to find that area of the square. Because the radius of the circle is 40, the length from the center of the square (or the circle) is 40, because that is the radius of the circle.

Simply to find the side length of the square is using trigonometry and realizing that the diagonal of a square splits the square into 2  45-45-90 triangles. Therefore 80sqrt2 (the length of the diagonal of the square or the diameter of the circle) equals x sqrt2 where x is the length of the side. The length of the side can be solved; 40sqrt2.

This is the last part i need to explain and might be the hardest.
You can fit 5 of these squares you drew at the top because 200sqrt2/40sqrt2=5.
You can also fit 3 of these squares of the side because 120sqrt2/40sqrt2=3

AND FINALLY, (after all of this blibber-blabber) the number of lampposts needed to light up the park would be 5*3=[size=150]15
[/size]

If you still don't understand, PLEASE REPLY TO ME, i want to help you  :D

the thing is is that he won't understand just because the last time he visited AoPS was 2013

[quote=fishy15]the thing is is that he won't understand just because the last time he visited AoPS was [size=200]2013[/size][/quote]

seriously???????????????????????

[quote=WUMASTER]Therefore 80sqrt2 (the length of the diagonal of the square or the diameter of the circle) equals x sqrt2 where x is the length of the side. The length of the side can be solved; 40sqrt2.
[/quote]

There's a mistake there. The hypotamoose of the 45-45-90 triangle is x sqrt 2, so it would make more sense to just take 40 and multiply it by sqrt2.

I still got the same answer, though. [hide=By the way...]Remember when you were in 8th grade and you did mathcounts? There was one dumb kid who tried to offer you 5000 dollars so you would let him go to state. He has the same last name as you. To let you know, I am that kid :P[/hide]

Fishy15, are you talking about WUMASTER?

no, i am talking about [url = https://artofproblemsolving.com/community/user/68604]rvd101[/url]


Please mail names of greatest mathematical geniuses whose performance are sensational at IMOs.

All IMO contestants who have won Fields Medals are the best contestants:
Gregory Margulis (RUS) (Fields Medal 1978, IMO 1959 silver)
Vladimir Drinfeld (RUS) (Fields Medal 1990, IMO 1969 gold)
Jean Christophe Yoccoz (FRA) (Field Medals 1994, IMO 1974 gold)
Richard Borcherds (UNK) (Field Medals 1998, IMO 1977 silver,1978 gold)
Timothy Gowers (UNK) (Field Medals 1998, IMO 1981 gold)
Laurant Lafforgue (FRA) (Fields Medal 2002, IMO 1984,1985 silver)

Ciprian Manolescu (ROM) is an outstanding contestant because he has perfect score 42/42 in all three IMO that he participated, IMO 95,96,97. Later he won Morgan Prize for his outstanding undergraduate research. Other Morgan prize winners who participated in IMO include: Kiran Kedlaya (USA) (IMO 90-92), Lenhard Ng (USA) (IMO 91-93), Jacob Lurie (USA) (IMO 94-95), Sounderrajan Kannan (IND) (IMO 91)

Reid Barton (USA) won the most gold medals, he won 4 IMO gold medals in IMO 98,99,2000,2001, he is also the highest scorer in IOI 2001, he is Putnam Fellow 2001,2002 and is a member of the second placed team at ACM ICPC 20001-2002

Gabriel Carroll (USA) is Putnam Fellow 2000 (top 5) when he is still a high school student! he also have 2 IMO gold (1998,2001) and 1 IMO silver(1999). while he was in high school, he also submit a mathematics research project to Intel Science Talent Search, he won the third place in STS 2000-2001

Eugenia Malinnikova (RUS) is the best female contestant in IMO history, she has 3 gold medals in IMO 1989 (41/42), IMO 1990 (42/42) and IMO 1991 (42/42), losing only 1 point!

What is that a "Field Medal"? ..Putnam Fellow also?

What is that a "Fields Medal"?
..and a Putnam Fellow?
Ciprian Manolsecu gets for 3 times from 3 participations 42/42,
anyone gets for many times or the same.. maximum points?

Hello. Ciprian Manolescu is the only 3 time perfect scorer at IMO. There are many who managed the feat of getting full scores 2 times.
       Field Medal is the highest honour for any mathematician , it is equivalent of noble prize in mathematics.
                                                Happy Problem Solving  !!!

imam why don't you join us in this "happy problem solving activity" ? :)

we could use some new indian flavored problems, what do you say?

Well, I'd just like to include an Australian into that list if I may:

Terry Tao (IMO 86,87,88)
He won bronze, silver, gold respectively in those years.

Not very impressive...but you have to take into account of the fact that he was 12 years-old when he won the gold medal in 88. To this day, he is still the youngest ever gold medalist at the IMO.  :D

wow! very young and very smart! :D :D

what happened to him? did he stop participating in 1988?

[quote="Valentin Vornicu"]if you win a gold medal at the IMO at the age of 12, I think you start looking for some more challenging stuff :D - he's a professor at Berkeley University (so I've heard) - he's only 25 (or so) now.[/quote]

Berkeley, are you sure? In my memory, he's not teaching berkeley

A perfect score is a great achievement, but keep in mind that the test can vary in difficulty from year to year. So some years a perfect score is more rare and difficult than in other years.

Anyhow, here's an interesting link that I found, though its somewhat outdated: http://www.iisc.ernet.in/mocell/PEOPLE/halloffame.html

I think Terry Tao received field prize last year.


can someone kindly explain the uniqueness proof of that particular theorem please?

Is the following of any use?

http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic

i even have a book what i want to know is why is such a proof needed and an explanation of the part where division algorithm comes in play in short see this

For uniqueness, let us assume that there is at least one positive integer n that
has two distinct prime factorizations; that is,
n = p1 p2 · · · pk = q1q2 · · · qh
where p1, p2, . . . , pk , q1, q2, . . . , qh are primes with p1 ≤ p2 ≤ · · · pk and q1 ≤
q2 · · · qh such that the k-tuple (p1, p2, . . . , pk ) is not the same as the h-tuple
(q1, q2, . . . , qh). It is clear that k ≥ 2 and h ≥ 2. Let n be the minimal such
integer. We will derive a contradiction by finding a smaller positive integer that
also has two distinct prime factorizations.
We claim that pi
= q j for any i = 1, 2, . . . , k, j = 1, 2, . . . , h. If, for
example, pk = qh = p, then n = n/p = p1 · · · pk−1 = q1 · · · qh−1 and 1 <
n
< n, contradicting the minimality of n. Assume without loss of generality that
p1 ≤ q1; that is, p1 is the least prime factor of n in the above representations. By
applying the division algorithm it follows that
q1 = p1c1 + r1,
q2 = p1c2 + r2,
... qh =
p1ch +
rh,
where 1 ≤ ri < p1, i = 1, . . . , h.
We have
n = q1q2 · · · qh = (p1c1 + r1)(p1c2 + r2) · · · (p1ch + rh).
Expanding the last product we obtain n = mp1 +r1r2 · · · rh for some positive
integer m. Setting n = r1r2 · · · rh we have n = p1 p2 · · · pk = mp1 + n. It
8 104 Number Theory Problems
follows that p1 | n and n = p1s. As we have shown, s can be written as a
product of primes. We write s = s1s2 · · · si , where s1, s2, . . . , si are primes.
On the other hand, using the factorization of r1, r2, . . . , rh into primes, all
their factors are less than ri < p1. From n = r1r2 · · · rh, it follows that n
has a factorization into primes of the form n = t1t2 · · · t j , where ts < p1,
s = 1, 2, . . . , j . This factorization is different from n = p1s1s2 · · · si . But
n
< n, contradicting the minimality of n.


geometry
circumcircle
power of a point
radical axis
geometry proposed

Let $ ABC$ be an isosceles triangle with $ AB\equal{}AC$, and $ D$ be midpoint of $ BC$, and $ E$ be foot of altitude from $ C$. Let $ H$ be orthocenter of $ ABC$ and $ N$ be midpoint of $ CE$. $ AN$ intersects with circumcircle of triangle $ ABC$ at $ K$. The tangent from $ C$ to circumcircle of $ ABC$ intersects with $ AD$ at $ F$.
Suppose that radical axis of circumcircles of $ CHA$ and $ CKF$ is $ BC$. Find $ \angle BAC$.

1. suppose$ \angle A \leq 90$
line $ l$ parallel to $ CE$ from$ A$ intersect circumcircle at $ S$ we have $ A(ECNS)\equal{}\minus{}1$so $ (BCKS)\equal{}\minus{}1$
now assume that $ FK$ intersect circumcircle at $ T$.we have $ (BCKT)\equal{}\minus{}1$ so $ S \equiv T$
we have$ DB.DC\equal{}DH.DA\equal{}DC^2$ means $ DC$ is tangent to circumcircle of $ AHC$ so it should be tangent to circumcircle $ FKC$
let$ \angle KFC\equal{}\angle BAK\equal{}\alpha$ but $ \angle ASK\equal{}\alpha \plus{}\angle B$ .and $ \angle AFK\equal{}90\minus{}\alpha\minus{}\angle A$
but$ \angle AFK\plus{}\angle SAK\plus{}\angle FSA\equal{}180$
so $ 90\minus{}\alpha\minus{}\angle A\plus{}\angle B\plus{}\alpha \plus{}\angle B\equal{}180$ means $ \angle A\equal{}45$

2.if $ \angle A> 90$ we have  $ \angle A\equal{}135$


Some time ago I remember seeing a problem in From Erdos to Kiev concerning the sequence of integers of the form $\left \lfloor n\sqrt{2} \right \rfloor$.  As I recall the solution they gave involved writing $\sqrt{2}$ in base 2.  Could someone post the problem?

Its to prove there is an infinate number of 'powers of two' in that sequence

Thanks seamusoboyle!  Now I'll see if I can recreate the given solution

is there a simple way to determine f(n;m;a) which stands for the nth integer not occuring in the sequence of  floor (a*b*root(m))  :?:
(sequence runs over all integers b )
i guess so but im to lazy to think about it myself now :P

i guess you could always try to take x , divide by a*root(m), take the floor and ceiling of that and then recompute .

eg 3 = floor (n*root(2))? -> 3/root2 is between 2 and 3; floor(2*root(2)) and   floor (3*root(2)) are 2 and 4, so the answer is no.

but perhaps theres an easier way or even better an expression for f(n;m;a)

x/(a root(m)) should approximate an integer though so an expression in mods seems reasonable.


advanced fields
advanced fields unsolved

Let $\mathcal{A} = \{ E \subseteq \mathbb{R} : E \, \text{or} \, E^{c} \, \text{has a finite number of elements} \}$ and let $\mu$ be the counting measure restricted to $\mathcal{A}$. Note that $\mathcal{A}$ is an algebra of subsets of $\mathbb{R}$ and $\mu$ is a measure on $\mathcal{A}$.

(i) Let $\mu^{*}$ be the associated outer measure. Describe $\Sigma$, the collection of $\mu^{*}$-measurable sets.

(ii) Show that there is a set E in $\Sigma$ so that there does not exist F in the $\sigma$-algebra generated by $\mathcal{A}$ with $E \subseteq F$ and $\mu(F \cap E^{c})=0$.

I know that usually we derive the measure from the outer measure by selecting the measurable sets $E$ which satisfy $\mu^{*}(A)=\mu^{*}(A \cap E)+\mu^{*}(A-E)$ for every set $A$ on which $\mu^{*}$ is defined. But I am not aware of the way to define a unique outer measure given its induced measure: which associated outer measure are we talking about ?

It is defined as follows:
Let $\mathcal{A}$ be an algebra of subsets of a set $\Omega$ and let $\mu : \mathcal{A} \rightarrow [0,\infty]$ be a measure. Define $\mu^{*} : \mathcal{P}(\Omega) \rightarrow [0,\infty]$ by $\mu^{*}(E) = \inf \sum_{k=1}^{\infty} \mu(E_{k})$ where the infimum is taken over all sequences $(E_{k})_{k=1}^{\infty}$ of sets in $\mathcal{A}$ such that $E \subseteq \cup_{k=1}^{\infty} E_{k}$. We call $\mu^{*}$ the outer measure associated with $\mu$.

For (i), I have found that $\Sigma = \mathcal{P}(\mathbb{R})$.
For (ii), using the result in (i), I see that it suffices to find a set $E \subseteq \mathbb{R}$ that is not in $\sigma(\mathcal{A})$.

Note that you can characterize $\sigma (A)$ as the set of subsets $E$ of $R$ such that $E$ or $E^{c}$ is countable (meaning idempotent to $N$, finite or empty). Now you can tell $[0,1]$ is not in $\sigma (A)$.


[quote="juancarlos"]Muchas gracias a mi amigo hucht por invitarme a participar en este foro.
Este problema en:
http://www.mathlinks.ro/Forum/topic-81081.html
La primera vez que lo propuse en 1974, lo extraje de la configuración del original problema de Langley. Tengo 4 soluciones adicionales que las enviaré más adelante, todas sin usar trigonometría. Creo que se pueden encontrar mas soluciones.[/quote]

Tengo dos soluciones pero las dos estan en el post en ingles  :?  seguire buscando otra solucion  :?

[quote="hucht"][quote="José"]Una preguntita: este problema tiene alguna importancia especial????[/quote]

No se  :?  pero me gustaria conocer el problema mas general, aunque has de saber que cualquier problema que tenga un nivel de dificultad aceptable es bien recibido ya que en este foro estamos caido de problemas, hemos de animarnos para sacarlo adelante.[/quote]

Perfecto, gracias!!!!

Solución1: Tomamos el punto E sobre AD de tal forma que AE=EB, luego: <EAB=<EBA=40º, también tomamos el punto F sobre ED de tal modo que BE=BF, así: <FEB = <EAB + <EBA = 40º + 40º = <EFB = 80º. Además tomamos el punto G sobre BC de tal manera que el triángulo BFG sea equilatero, luego: BF=FG=BG=BE=AE, también: <BFG=<FBG=<FGB=60º. Entonces: <GFC=40º y como el triángulo ECB es isosceles: EC=BC donde AD=AE+ED=EC=ED+CD = BC entonces AE=BE=BF=FG=BG=CD. Finalmente en el triángulo FGC:  FG= CD y <GFC=2<FCG=40º, por lo tanto: GD=CD=FG y <FDG =<DFG = 40º además <DGC = 20º y GD=BG entonces: <DBG+<GDB=<DGC=20º finalmente: <DBC=10º.
Solucion2:Prolongamos AB hasta el punto E de tal forma que el triángulo BEC sea equilátero, luego: BE=BC=CE=AD,donde : <CBE=BCE=<BEC=60º.Tenemos ahora <ECA=2<EAC=80º y AD=CE, por lo tanto: <DEA=<DAE=40º y <DEC=20º, luego E es circuncentro del triángulo BDC entonces: <DBC=<DEC/2=10º.
Splucion3:Sobre la prolongacion de CA tomamos el punto E de tal manera que AE=AB y <AEB=<ABE=20º además construimos el triángulo equilátero EFB donde F está del mismo lado que A con respecto a BE, luego:<AEF=<ABF=40º y AF es bisectriz de <EFB, es decir: <AFE=<AFB=30º.También: BE=EF=BF=AD=BC, por lo tanto,los triángulos ABF y ABD son congruentes, entonces: <ADB=AFB=30º y como <DBC=<ADB-<DCB=30º-20º=10º.
Solucion4: En la prolongación de CA tomamos el punto E de tal manera que AE=AB, así <AEB=<ABE=20º, además BE=BC=AD. Construimos el triángulo equilátero ADF con F del mismo lado que B con respectoa AD.Luego: <BAD=40º, <BAF=20º donde los triángulos ABF y ABE son congruentes,luego: <AFB=20º y AE=AB=BF, por lo tanto BD es bisectriz de <ADF, entonces: <ADB=30º= <DBC+<DCB=<DBC+20º es decir : <DBC=10º.

juancarlos cual es el problema original?

Este problema lo "extraje" derivandolo del problema de Langley, que se puede ver en:
http://agutie.homestead.com/files/index.html
Ver: Langley's problem
También en:
http://www.personal.us.es/rbarroso/trianguloscabri/curso0203.htm
Ver Problema 90.
Como verás en la solución1 aparece el triángulo isósceles 20º-80º después de los primeros trazos auxiliares. Hay más problemas de angulos adventicios derivados del mismo triángulo.


Let ABC be weighting right-angled triangle  . M is inbuilt triangle such that $ MA: MB: MC\equal{}1: \sqrt{2}: \sqrt{5}$
Find $ \widehat{AMB}$
 :lol:

to algadin

could you be more specific? what do you mean by "weighting"?

many thanks
sunken rock


$ \boxed{1}$ (1) Factorize $ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc.$

(2) Factorize $ bc(b \minus{} c) \plus{} ca(c \minus{} a) \plus{} ab(a \minus{} b).$

(3) Suppose that $ a \plus{} b \plus{} c \equal{} 0$. Simplify the following expression.
\[ \left(\frac {b \minus{} c}{a} \plus{} \frac {c \minus{} a}{b} \plus{} \frac {a \minus{} b}{c}\right)\left(\frac {a}{b \minus{} c} \plus{} \frac {b}{c \minus{} a} \plus{} \frac {c}{a \minus{} b}\right).
\]
$ \boxed{2}$ (1) Let $ m,\ n$ be real constants. The function of $ l(x) \equal{} mx \plus{} n$ satisfies $ l(\alpha) \equal{} l(\beta) \equal{} 0$ for distinct two values $ \alpha ,\ \beta$. Find the expression of $ l(x)$.

(2) The graph of the quadratic function $ y \equal{} g(x)$ passes  through three points $ A\left( \minus{} 1,\ \frac {1}{2}\right),\ B\left(0,\ \minus{} \frac {1}{2}\right),\ C\left(1,\ \frac {1}{2}\right)$. Find the expression of $ g(x)$.

(3) Let $ a,\ b$ be real constants. The graph of the quadratic function $ y \equal{} f(x) \equal{} x^2 \plus{} ax \plus{} b$ always satisfies $ |f(x)| \leq \frac {1}{2}$ for $ \minus{} 1\leq x\leq 1$. Use $ g(x)$ found in (2) to find the expression of $ f(x)$ by considering $ h(x) \equal{} f(x) \minus{} g(x)$.

hello kunny, i think it is well-known:
(1)$ a^3\plus{}b^3\plus{}c^3\minus{}3abc\equal{}(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}ac\minus{}bc)$.
Sonnhard.

hello kunny, to (2) we have
$ (ab(b \minus{} c) \plus{} ca(c \minus{} a) \plus{} bc(b \minus{} c) \equal{} a^2b \minus{} ab^2 \minus{} a^2c \plus{} b^2c \plus{} ac^2 \minus{} bc^2 \equal{}$
$ ab(a \minus{} b) \minus{} c(a^2 \minus{} b^2) \plus{} c^2(a \minus{} b) \equal{} (a \minus{} b)(ab \minus{} ac \minus{} bc \plus{} c^2) \equal{}$
$ (a \minus{} b)(b(a \minus{} c) \minus{} c(a \minus{} c)) \equal{} (a \minus{} b)(a \minus{} c)(b \minus{} c)$.
Sonnhard.

hello kunny, to (3), inserting $ c\equal{}\minus{}a\minus{}b$ in your term we get
$ \left(\frac{2b\plus{}a}{a}\plus{}\frac{\minus{}2a\minus{}b}{b}\plus{}\frac{a\minus{}b}{\minus{}a\minus{}b}\right)\left(\frac{a}{2b\plus{}a}\plus{}\frac{b}{\minus{}2a\plus{}b}\plus{}\frac{\minus{}a\minus{}b}{a\minus{}b}\right)$, after an ugly algebraic computation i have got $ 9$.
Sonnhard.

hello kunny, what is the function $ f(x)$ in the condition $ l(\alpha)\equal{}f(\beta)$ in number 2. 1?
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello kunny, what is the function $ f(x)$ in the condition $ l(\alpha) \equal{} f(\beta)$ in number 2. 1?
Sonnhard.[/quote]

Sorry, $ l(\alpha)\equal{}l(\beta)\equal{}0$ was right.

kunny

hello, to 2.1), from the given condtion we get $ m\alpha\plus{}n\equal{}m\beta\plus{}n \Leftrightarrow m(\alpha\minus{}\beta)\equal{}0$ and from here $ m\equal{}0$ since $ \alpha\ne\beta$, so your solution is $ l(x)\equal{}n$ with $ n \in \mathbb{R}$.
Sonnhard.

hello, to 2.2) by the ansatz $ g(x)\equal{}ax^2\plus{}bx\plus{}c$ and the given conditions we get the system
$ a\minus{}b\plus{}c\equal{}\frac{1}{2}$
$ c\equal{}\minus{}\frac{1}{2}$
$ a\plus{}b\plus{}c\equal{}\frac{1}{2}$
solving this we get $ a\equal{}1,b\equal{}0,c\equal{}\minus{}\frac{1}{2}$ and $ g(x)\equal{}x^2\minus{}\frac{1}{2}$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, to 2.1), from the given condtion we get $ m\alpha \plus{} n \equal{} m\beta \plus{} n \Leftrightarrow m(\alpha \minus{} \beta) \equal{} 0$ and from here $ m \equal{} 0$ since $ \alpha\ne\beta$, so your solution is $ l(x) \equal{} n$ with $ n \in \mathbb{R}$.
Sonnhard.[/quote]

Kon-ni-chi-wa, Dr Sonnhard Graubner, That's incorrect.

[quote="dgreenb801"]Three is in a Lithuania Olympiad, you can use the factorization in #1 to solve it!
[url]http://www.mathlinks.ro/viewtopic.php?t=83378[/url][/quote]

Oh, really? That was a well-known problem, wasn't that?

[hide="Number 1 part 3"]

First factor:
$ \frac{a(c\minus{}a)(a\minus{}b)\plus{}b(b\minus{}c)(a\minus{}b)\plus{}c(b\minus{}c)(c\minus{}a)}{(a\minus{}b)(b\minus{}c)(c\minus{}a)}\equal{}\frac{a^2c\plus{}ac^2\plus{}a^2b\plus{}ab^2\plus{}b^2c\plus{}bc^2\minus{}a^3\minus{}b^3\minus{}c^3\minus{}3abc}{(a\minus{}b)(b\minus{}c)(c\minus{}a)}$, since $ a\plus{}b\plus{}c\equal{}0$, so $ a^3\plus{}b^3\plus{}c^3\equal{}3abc$

$ \equal{}\frac{(ab\plus{}bc\plus{}ac)(a\plus{}b\plus{}c)\minus{}9abc}{(a\minus{}b)(b\minus{}c)(c\minus{}a)}\equal{}\frac{\minus{}9abc}{(a\minus{}b)(b\minus{}c)(c\minus{}a)}$

Second factor:
$ \frac{ab(a\minus{}b)\plus{}bc(b\minus{}c)\plus{}ac(c\minus{}a)}{abc}\equal{}\frac{\minus{}(b\minus{}c)(c\minus{}a)(a\minus{}b)}{abc}$ (from number 1 part 2)
so

$ (\frac{a\minus{}b}{c}\plus{}\frac{b\minus{}c}{a}\plus{}\frac{c\minus{}a}{b})(\frac{a}{b\minus{}c}\plus{}\frac{b}{c\minus{}a}\plus{}\frac{c}{a\minus{}b})\equal{}\frac{\minus{}(b\minus{}c)(c\minus{}a)(a\minus{}b)}{abc}.\frac{\minus{}9abc}{(a\minus{}b)(b\minus{}c)(c\minus{}a)}\equal{}9$

[/hide]

[quote="Dr Sonnhard Graubner"]hello, to 2.1), from the given condtion we get $ m\alpha \plus{} n \equal{} m\beta \plus{} n \Leftrightarrow m(\alpha \minus{} \beta) \equal{} 0$ and from here $ m \equal{} 0$ since $ \alpha\ne\beta$, so your solution is $ l(x) \equal{} n$ with $ n \in \mathbb{R}$.
Sonnhard.[/quote]

But, since it's a horizontal line as you determined, $ n\equal{}0$ or else it would never cross the x-axis, and would therefore have no zeroes.  So assuming the rest of your work is correct, it should be $ l(x)\equal{}0$.

All problems except $ \boxed{2}$ (3) has  been solved. :lol:

[quote="kunny"]$ \boxed{2}$ (1) Let $ m,\ n$ be real constants. The function of $ l(x) \equal{} mx \plus{} n$ satisfies $ l(\alpha) \equal{} l(\beta) \equal{} 0$ for distinct two values $ \alpha ,\ \beta$. Find the expression of $ l(x)$.

(2) The graph of the quadratic function $ y \equal{} g(x)$ passes  through three points $ A\left( \minus{} 1,\ \frac {1}{2}\right),\ B\left(0,\ \minus{} \frac {1}{2}\right),\ C\left(1,\ \frac {1}{2}\right)$. Find the expression of $ g(x)$.

(3) Let $ a,\ b$ be real constants. The graph of the quadratic function $ y \equal{} g(x)$ always satisfies $ |f(x)| \leq \frac {1}{2}$ for $ \minus{} 1\leq x\leq 1$. Use $ g(x)$ found in (2) to find the expression of $ f(x)$ by considering $ h(x) \equal{} f(x) \minus{} g(x)$.[/quote]

I'm very sorry that I made a typo in (3). I correct it as follows.

Let $ a,\ b$ be real constants. The graph of the quadratic function $ y \equal{} f(x) \equal{} x^2 \plus{} ax \plus{} b$ always satisfies $ |f(x)| \leq \frac {1}{2}$ for $ \minus{} 1\leq x\leq 1$. Use $ g(x)$ found in (2) to find the expression of $ f(x)$ by considering $ h(x) \equal{} f(x) \minus{} g(x)$.

[color=blue] 1st place: Mnjm   :winner_first:  :first:
  2nd place: Fji  :winner_second:
3rd place: Mikity :winner_third: [/color]

$ \displaystyle \displaystyle \displaystyle \begin{tabular}{|c|c|c|c|c|c\parallel{}c|}\hline Name & Problem 1 & Problem 2 & Total Score \\
\hline Mnjm & 50 & 48 & 98 \\
\hline Fji & 50 & 44 & 94 \\
\hline Mikity & 36 & 43 & 79 \\
\hline Average & 22.8 & 14.6 & 37.4 \\
\hline\end{tabular}$

Entry 50


A security firm is designing an attache case which will hold $ \$1,000,000$ in $ \$100$ bills. A packet of $ 25$ $ \$100$ bills from the bank measures $ 6 \frac{1}{8}$" x $ 2 \frac{5}{8}$" x $ \frac{3}{16}$". To the nearest cubic inch, find the minimum volume of the interior of an attache case which can hold these bills.

[quote="GameBot"]A security firm is designing an attache case which will hold $ \$1,000,000$ in $ \$100$ bills. A packet of $ 25$ $ \$100$ bills from the bank measures $ 6 \frac {1}{8}$" x $ 2 \frac {5}{8}$" x $ \frac {3}{16}$". To the nearest cubic inch, find the minimum volume of the interior of an attache case which can hold these bills.[/quote]

We need $ 4000$ packets.

Thus, $ 4000 \cdot \frac{4381}{1024} \approx 17114$.

We actually need $400$ packets since $\frac{1,000,000}{100\times25}=\frac{(10,0\cancel{00})\cancelto{4}{(100)}}{\cancel{100}\times\cancel{25}}=400$.  The volume of $25$ $\$100$ bills is not $\frac{4381}{1024}$, but $\left(6\frac18\right)\left(2\frac58\right)\left(\frac3{16}\right)=\left(\frac{49}8\right)\left(\frac{21}8\right)\left(\frac3{16}\right)=\frac{3087}{1024}$.  Please see Wolfram Alpha for confirmation :P.
[url]http://www.wolframalpha.com/input/?i=(49/8)(21/8)(3/16)[/url]

We multiply that by $400$ to get the desired volume:
\[\left(\frac{3087}{1024}\right)\left(400\right)\approx\boxed{1206\text{ cub. in.}}\]

JSGandora your answer boggled me when I initially saw it, but then I just solved it myself and it was the exact same way. Except I used a calculator. :P

Actually, you don't have to calc, I solved this in 30 secs no calc
From JSGandora's Sol:
Don't count the $\frac{49}{8}*\frac{21}{8}*\frac{3}{16}$ and just straight to multiply it with $400$, this will help you cancel quickly, eventually you get $\frac{77175}{64}$ Then just count this as quick as you can


logarithms
inequalities
calculus
probability
real analysis
real analysis solved

Of course, Peter meant to say that $f(x)=-x\ln x$ is concave. (Better, say that $f(x)=-x\log_{2004}x$ is concave.)

I'm using "concave" to mean the negative of a convex function. In first-year calculus terms, that's "concave downward."

This is Jensen's inequality. I'm moving it to "solved."

[quote="kunny"]Suppose that $x_1>0,x_2>0,\cdots\cdots,\ x_{2005}>0$ and $x_1+x_2+\cdots+x_{2005}=1$.
Prove the following inequality.

\[-x_1\log _{2004}{x_1}-x_2\log _{2004}{x_2}-\cdots\cdots -x_{2004}\log _{2004}{x_{2005}}\leqq\log _{2004}{2005}\][/quote]
greeting from uec.ac.jp :)

FYI, this is the basic theorem in information theory delevoped by Shannon.

For a probability [tex]P=(p_1,p_2, ....,p_n)[/tex]. Define entropy of H: [tex]H(P) = \sum -p_i \log p_i[/tex] where we will define [tex] 0\log 0 [/tex] as 0. Then we have:

[tex] H(P) \leq \log n[/tex]

Yes.This is the background of this problem.