geometry
IMO Shortlist
power of a point
perpendicular bisector
circles

A circle $S$ bisects a circle $S'$ if it cuts $S'$ at opposite ends of a diameter. $S_A$, $S_B$,$S_C$  are circles with distinct centers $A, B, C$ (respectively).
 Show that $A, B, C$ are collinear iff there is no unique circle $S$ which bisects each of $S_A$, $S_B$,$S_C$  . Show that if there is more than one circle $S$ which bisects each of $S_A$, $S_B$,$S_C$  , then all such circles pass through two fixed points. Find these points.

[b]Original Statement:[/b]

A circle $S$ is said to cut a circle $\Sigma$ [b]diametrically[/b] if and only if their common chord is a diameter of $\Sigma.$
Let $S_A, S_B, S_C$ be three circles with distinct centres $A,B,C$ respectively. Prove that $A,B,C$ are collinear if and only if there is no unique circle $S$ which cuts each of $S_A, S_B, S_C$ diametrically. Prove further that if there exists more than one circle $S$ which cuts each $S_A, S_B, S_C$ diametrically, then all such circles $S$ pass through two fixed points. Locate these points in relation to the circles $S_A, S_B, S_C.$

Let the radical axis of the (non-concentric) circles $S_A \equiv (A),\ S_B \equiv (B)$ with radii $r_A \ge r_B$ meet their center line AB at a point R and let M be the midpoint of the segment AB. Since the point R has the same power to both the circles (A), (B),

$(RA + r_A)(RA - r_A) = (RB + r_B)(RB - r_B),\ \ RA^2 - r_A^2 = RB^2 - r_B^2$

Using oriented line segments, when factoring the difference of the 2 squares,

$r_A^2 - r_B^2 = RA^2 - RB^2 = (AR + RB)(AR - RB) =$

$= AB\ [(AM + MR) - (MB - MR)] = 2\ AB \cdot MR$

Let P be the center of a circle (P) with radius r intersecting both the circles (A), (B) at diametrally opposite points $D_A, E_A$ and $D_B, E_B$, respectively. The center A of the circle (A) is on the radical axis of the circles (A), (P), and similarly, the center B of the circle (B) is on the radical axis of the circles (B), (P). Hence,

$r^2 - PA^2 = (r - PA)(r + PA) = AD_A \cdot AE_A = r_A^2$

$r^2 - PB^2 = (r - PB)(r + PB) = BD_B \cdot BE_B = r_B^2$

$r^2 = PA^2 + r_A^2 = PB^2 + r_B^2,\ \ PB^2 - PA^2 = r_A^2 - r_B^2$

Let Q be the foot of a normal from the point P to the center line AB. From the right angle triangles $\triangle APQ, \triangle BPQ$,

$PQ^2 = PA^2 - QA^2 = PB^2 - QB^2,\ \ PB^2 - PA^2 = QB^2 - QA^2$

Using oriented line segments when factoring the difference of the 2 squares, it follows that

$r_A^2 - r_B^2 = QB^2 - QA^2 = (QB + AQ)(QB - AQ) =$

$= AB\ [(QM + MB) - (AM - QM)] = 2\ AB \cdot QM = -2\ AB \cdot MQ$

As a result, we see that the position of the foot Q of the normal from P to the center line AB does not depend on the point P. Hence, the locus of the centers of circles (P) intersecting the circles (A), (B) at diametrally opposite points is the normal to the center line AB at a fixed point Q. Moreover, it is clear that MQ = -MR, i.e., this normal is a reflection of the radical axis of the circles (A), (B) in the perpendicular bisector of the segment AB. If P is an arbitrary point on the locus normal $QP \perp AB$, the radius r of the circle (P) is obtained by erecting normals to the segments PA or PB at the points A, B intersecting the circles (A), (B) at diametrally opposite points $D_A, E_A$ and $D_B, E_B$ and $r = PD_A = PE_A = PD_B = PE_B$. Let the circle (P) intersect the center line AB at points U, V, so that the points U, A, B, V follow on the line AB in this order.

$QU^2 = PU^2 - PQ^2 = r^2 - (PA^2 - QA^2) = r_A^2 + QA^2 = r_A^2 + (MA - QM)^2 =$

$= r_A^2 + \left(\frac{AB}{2} - \frac{r_A^2 - r_B^2}{2\ AB}\right)^2 = r_A^2 + \frac{AB^2}{4} - \frac{r_A^2 - r_B^2}{2} + \frac{r_A^4 - 2r_A^2r_B^2 + r_B^4}{4\ AB^2} =$

$= \frac{AB^2}{4} + \frac{r_A^2 + r_B^2}{2} + \frac{r_A^4 + 2r_A^2r_B^2 + r_B^4}{4\ AB^2} - \frac{r_A^2r_B^2}{AB^2} = \frac{(AB^2 + r_A^2 + r_B^2)^2 - 4 r_A^2r_B^2}{4 AB^2}$

which is an expression symmetrical with respect to $r_A, r_B$ (as expected) and completely independent on the position of the point P on the locus normal QP. Since $QP \perp AB \equiv UV$, QV = -QU. Therefore, all circles intersecting the circles (A), (B) in diametrally opposite points (i.e., centered on the locus normal) intersect the center line AB at the same 2 points U, V, i.e., they belong to the same pencil.

If the (pairwise non-concentric) circle $S_A \equiv (A),\ S_B \equiv (B),\ S_C \equiv (C)$ have their pairwise radical axes intersecting at the radical center, their centers A, B, C are not collinear. The locus normals of the circle pairs (A), (B) and (B), (C) then intersect at a single point P, the center of a unique circle (P) intersecting the circles (A), (B), (C) at diametrally opposite points. By the transitivity of equivalence, the locus normal of the circle pairs (C), (A) also passes through the intersection P.  If the circles (A), (B), (C) have their radical axes parallel, their centers A, B, C are collinear. Their pairwise locus normals are then also parallel, i.e., they intersect at infinity and the circle intersecting all 3 circles (A), (B), (C) at diametrally opposite points has infinite radius, i.e., it is a line, obviously their common center line $AB \equiv BC \equiv CA$, but this is not a proper circle. A proper circle can exist only if the locus normals of the circle pairs (A), (B) and (B), (C) coincide and by the transitivity of equivalence, the locus normal of the circle pair (C), (A) then coincides with the first two. In this case, we have a pencil of circles with the base points $U, V \in AB$ found earlier, intersecting all 3 circles (A), (B), (C) at diametrally opposite points.

$r_i$ is the radius of circle $i$.
1. Collinear ==> No unique circle
Note that if $A$, $B$, $C$ are collinear, then given one circle $\omega$ that cuts $S_A$, $S_B$, $S_C$ diametrically, we can always find another by reflecting $\omega$ over line $ABC$.
2. No unique circle ==> Collinear
Say there are two circles $\omega_1$ and $\omega_2$.  Consider the power of $A$, $B$, $C$ with respect to $\omega_1$ and $omega_2$.  Because $\omega_1$ and $omega_2$ diametrically cur $S_A$, $S_B$, $S_C$, we have that each of $A$, $B$, $C$ lies on the radical axis of $\omega_1$ and $omega_2$, implying that $A,B,C$ are collinear.
3. >1 circle==>  two fixed points
Note that, by the above, $A,B,C$ must be collinear.  Now, if a circle $\omega_1$ diametrically cuts $S_A,S_B,S_C$, then it must intersect the line $ABC$ in two points (Otherwise, it could not pass through two diametrically opposite points on any of $S_A,S_B,S_C$ as $ABC$ is their line of centers).  Further note that all the circles that diametrically cut $S_A,S_B,S_C$ form a pencil.  Thus, the pencil passes through two fixed points $X,Y$ on $AB$.
We have by PoP on each of $A,B,C$ with respect to $\omega_1$ that $XA*AY=r_A^2$ and similarly for $B,C$. Note that, if $B$ is to the right of $A$, we get that
\begin{align*}
&(XB)(BY)=r_B^2\\
&\Rightarrow (XA+AB)(AY-AB)=r_B^2\\
&\Rightarrow (XA)(AY)+AB(AY-AX)-AB^2=r_B^2\\
&\Rightarrow AY-AX=\frac{r_B^2+AB^2-r_A^2}{AB}
\end{align*}
implying that $AY-AX$ and $(AY)(AX)$ are known values. We can thus solve explicitly for them and therefore find $X,Y$'s relation to $A,B,C$.


Let $ f_{k}(x)=\sqrt{x}+^{4}\sqrt{k-x^{2}}$ a real value function, whit $ k>0$ a  parameter.

A) find the maximun value $ P$, $ f_{k}(x)$ takes.

B) find $ x$ such that $ f_{k}(x)=P$

 :)

By the way, how do I write the fourth root symbol?...

$ y=\sqrt[4]{x^{2}}+\sqrt[4]{k-x^{2}}\ (0\leq x\leq \sqrt{k})$, use Jensen's inequality and A.M.-G.M. ,
$ y^{2}=\sqrt{x^{2}}+\sqrt{k-x^{2}}+2\sqrt[4]{x^{2}(k-x^{2})}$

$ \leq \sqrt{2}\sqrt{x^{2}+(k-x^{2})}+2\sqrt{\frac{x^{2}+(k-x^{2})}{2}}$

$ =2\sqrt{2k}$.

Thus the maximum value is $ \sqrt[4]{8k^{2}}$.

The equality holds for $ x=\frac{k}{\sqrt{2}}$, which satisfies $ 0\leq x\leq \sqrt{k}$.

[quote="kunny"]
The equality holds for $ x=\frac{k}{\sqrt{2}}$, which satisfies $ 0\leq x\leq \sqrt{k}$.[/quote]

then $ f_{k}(x)$ takes it's maximun when $ x=\frac{k}{\sqrt{2}}$ :?:

are you  saying that $ x=\frac{k}{\sqrt{2}}\leq \sqrt{k}$ for all possible values of $ k$ :?:

also I think the maximun isn't right

Rewrite as $ f_{k}(x)=\sqrt[4]{x^{2}}+\sqrt[4]{k-x^{2}}$.  Use power-mean inequality to find $ f_{k}(x) \leq 2 \sqrt[4]{\frac{k}{2}}$.  By the equality condition of the power-mean we also find that $ \sqrt[4]{x^{2}}= \sqrt[4]{k-x^{2}}$, so equality occurs at $ x=\sqrt{\frac{k}{2}}$.


algebra
polynomial
Rational Root Theorem
algebra unsolved

let f[x] = px^5 + qx^4 + rx^3 + sx^2 + tx + 1, such that f[1] = 4 and f [2] = 11 and all the numbers p, q, r, s and t are integers. Show that f(x)=0 has no integer roots.

By the rational root theorem and the fact that the constant term is one, the only possible integer roots are $x=\pm 1$.  $x=1$ is not a root, since $f(1)\ne0$.  $x=-1$ is not a root, since if it were we could write $f(x)=(x+1)g(x)$, where $g(x)$ is polynomial of degree $4$ with integer coefficients.  This cannot be the case, since $(2+1)g(2)=3g(2)=11$, and hence $g(2)$ is not an integer.

So $f(x)=0$ has no integer roots.


geometry
ratio
parallelogram
trigonometry
analytic geometry

Find $k$ if $P$,$Q$,$R$ and $S$ are points on the sides of quadrilateral $ABCD$ such that:
$\& \& \frac{AP}{PB}= \frac{BQ}{QC}= \frac{CR}{RD}= \frac{DS}{SA}=k \&$
and the area of quadrilateral $PQRS$ is exactly 52% of the area of quadrilateral $ABCD$

[quote="cosinator"][hide="is it ..."]$\frac{2}{3}$    (or $\frac{3}{2}$ depending how you label it)[/hide][/quote]
Yes, you are right :D
Can you show your proof please?

[quote="cosinator"][hide]I just said, WLOG, let the quadrilateral be a square and then it's pretty easy to go from there. I wouldn't call it the best proof there could be for this, but it gives the right answer.
[/hide][/quote]
Can you explain how can you assume that WLOG, let the quadrilateral be a square?

[quote="cosinator"]From my interpretation of the question, the problem implies that all quadrilaterals have this property (as it did not specify some specific set of quadrilaterals $ABCD$).  If you are looking for a proof that this value for $k$ will be the same for all quadrilaterals, then you are correct, assuming WLOG is not rigorous.[/quote]
i dont think that , "find k" means "find constant k"  :D

Well, I guess you could coordinatize; (0,0), (1,0), (a,b), (c,d), use shoelace... but I don't know what that'll do for concave quadrilaterals, since a set of four points can define up to 3 concave quadrilaterals.

Do any one have other solutions to the probelm?

[quote="Virgil Nicula"][color=darkred]Let $P\in (AB)$,$Q\in (BC)$,$R\in (CD)$ and $S\in (DA)$ be four points on the sides of quadrilateral $ABCD$

such that  $\frac{PA}{PB}=\frac{RC}{RD}=\alpha$ and $\frac{QB}{QC}=\frac{SD}{SA}=\beta$. Prove that $[PQRS]=\frac{1+\alpha\beta}{(1+\alpha)(1+\beta)}\cdot [ABCD]$.

Particularly, $\alpha =\beta =k\Longrightarrow [PQRS]=\frac{1+k^{2}}{(1+k)^{2}}\cdot [ABCD]$ and $2\cdot [PQRS]\ge [ABCD]$.[/color][/quote]
[color=darkblue][b]Here are three similar problems.[/b]

$1\blacktriangleright$ Let $ABCD$ be a convex quadrilateral and the points $\{M\ ,\ N\}\in (AB)$, $\{P\ ,\ Q\}\in (CD)$

such that $N\in (MB)$, $Q\in (PD)$. Prove that $\frac{AM}{CP}=\frac{MN}{PQ}=\frac{NB}{QD}$ $\Longrightarrow$ $[MNPQ]=\frac{MN}{AB}\cdot [ABCD]$.

$2\blacktriangleright$ Let $ABCD$ be a convex quadrilateral and the points $M\in (AB)$, $N\in (CD)$

so that $\frac{MB}{AB}+\frac{NC}{CD}=1$. Prove that $[MCD]+[NAB]=[ABCD]$.

$3\blacktriangleright$ Let $ABCD$ be a parallelogram and the points $M\in (AB)$, $N\in (BC)$, $P\in (CD)$, $R\in (DA)$

so that $MA=m\cdot MB$, $NB=n\cdot NC$, $PC=p\cdot PD$, $RD=r\cdot RA$.

Prove that $[MNPR]=\frac{1}{2}\left[1+\frac{(1-mp)(1-nr)}{(1+m)(1+n)(1+p)(1+r)}\right]\cdot [ABCD]$.[/color]

Hey guys,

MASSIVE ANALYTIC GEOMETRY IS ALWAYS FUN

$[PQRS]=\frac{1}{2}\left|\left(\frac{k\overrightarrow{BD}+\overrightarrow{AC}}{k+1}\right)\times\left(\frac{-k\overrightarrow{AC}+\overrightarrow{BD}}{k+1}\right)\right|$.

DO SOME COOL STUFF

$[PQRS]=\frac{k^{2}+1}{(k+1)^{2}}\cdot [ABCD]$.

THEN USE THE 0.52 THING

$6k^{2}-13k+6=0$.

[quote="vietkhoa"]OK, let $\frac{m}{n}=k$. You can easily have $\frac{\Delta BPQ}{\Delta BAC}=\frac{mn}{(m+n)^{2}}$. Similarly, you can make three similar ratios with $\Delta CQR; \Delta DRS; \Delta ASP$. Adding the four ratios gives $\frac{mn}{(m+n)^{2}}=\frac{12}{25}$. Then $(m;n)=(2;3);(3;2)$. Another solution is always welcome!!![/quote]

Can someone go through this more explicitly? I'm trying to follow it and not really succeeding, basically from "you can easily have..." onward.

[hide="hint"]
If you have a quadrilateral $ABCD$ and points $P$ on $\overline{AB}$, $Q$ on $\overline{BC}$, $R$ on $\overline{CD}$, and $S$ on $\overline{DA}$, then we can find the area of $PQRS$ in terms of the area of $ABCD$ by subtracting off the four outer triangles.

Now try expressing the area of $\triangle SAP$ in terms of $\triangle DAB$, etc. using the $\frac{1}{2}ab\sin C$ area formula.
[/hide]

This problem is old IMTS problem.

http://www.mathlinks.ro/Forum/viewtopic.php?p=339634#339634

Locked.


geometry
3D geometry
tetrahedron
circumcircle
sphere
trapezoid
Pythagorean Theorem

2 questions:

1) Given a tetrahedron with side lengths a, b, c, d, d, d (a-b-c in the same plane). Find its volume.

2) If you know the sides of any tetrahedron and their respective orientations, can you find its volume?

The first isn't too bad. As for the second one, I've no idea how to go about solving. Anyone know it? Is there a Heron's Theorem equivalent for tetrahedra?

Thanks.

Okay, so here we go:

[hide]For number 1, the key lies in the fact that the point above which the top vertex lies is going to be the circumcenter of the triangle.  Since we know the edges, we know by Heron's the area and we know by K = abc/(4R) the circumradius, so we can use the Pythagorean theorem to find the height and this (with the area) gives us the volume.  It works out to something like :sqrt: ((Kd) :^2: - (abc) :^2:)/3 where K is the area of the triangle with sides a, b, c.[/hide]

Now, for the second one, a tetrahedron is definitely defined uniquely by its edges since the base triangle is uniquely defined and then we get the three top edges to have only two possible locations by looking at spheres centered at the vertices of the triangle.  So the volume is uniquely defined.  But can we find it?  Hmm ...

Well, I can tell you for sure that it's nasty.  I don't see any simple way to go about the question at all.  While I mull it over, try this special case:  define an iscoceles tetrahedron as one in which opposide edges are equal.  Find the volume of an iscoceles trapezoid with edges a, b, c, a, b, c.

According to MathWorld, there is a formula, called the Cayley-Menger determinant, for expressing the volume in terms of a 5x5 determinant involving the edge lengths.  Here is a link: [url]http://mathworld.wolfram.com/Cayley-MengerDeterminant.html[/url]

Wow.  Go figure.  At least I am vindicated by knowing that if one were to write that out, it would be very ugly :).

[quote="JBL"]the key lies in the fact that the point above which the top vertex lies is going to be the circumcenter of the triangle.[/quote]

Why is this?

If you draw the altitude to a-b-c, you'll notice that you have nice congruent triangles between the apex and the three base vertices. That's why it's the circumcenter.

Hmm, that Cayley-Menger determinant looks intimidating. But thanks for the link, I'll try to understand it.

Unfortunately, mathworld doesn't provide proofs for its theorems.  Perhaps you could find a good proof of it somewhere online.

Interesting note:  this website http://sep.princeton.edu/papers/index.shtml claims to use the Cayley-Menger determinant to prove an n-dimensional variation of the Pythagorean Theorem, of which the 3-D case is a trirectangular tetrahedron.  There was talk of this on another thread somewhere which I couldn't find.

He also appears to include a proof of Cayley-Menger, which I haven't yet checked for readability, but I'll see:
http://sep.princeton.edu/papers/gp-r.pdf
(there's also a postscript version on the first site I liked to, if you prefer.)

[quote="wyra"]If you draw the altitude to a-b-c, you'll notice that you have nice congruent triangles between the apex and the three base vertices. That's why it's the circumcenter.
quote]

So if we have tetrahedron ABCD with ABC being the triangle with sides a, b, c and the altitude from D meeting ABC at point O, are you refering to triangles DOA, DOB and DOC?

Yes.  They have SSA, which wouldn't be enough for congruence, but the angle happens to be a right angle, which makes congruence happen.

<b>Wow</b> I need to brush up on linear algebra.  :)  But I can follow most of the pdf. Thanks, Joel!

Also, could you give a small hint on the a-b-c-a-b-c volume?

erm, well ...

I didn't have any way of doing it in advance.  I just thought it might be nicer.


A fair coin is tossed $n$ times. What is the expected length of the longest chain of flipping the same side face up? What is the expected length of the longest chain of heads?

Please move this if it is in the wrong forum. :D

Um.. i'm stupid, but is it supposed to be 1 and 1?

I believe it's $\lfloor0.5+\log_{2}{n}\rfloor$ for heads and $\lfloor0.5+\log_{2}{2n}\rfloor$ for either side.

Could you please post a solution? Thanks. :D

Also, why are there only integral answers? For $n=1$, we get $0$ for heads and $1$ for longest chain, both obviously incorrect.

eh, i thought that you'd want an integral approximation

if you don't want one, just take away the floor and the 0.5


In our school test,I get this problem
Let $ a,b,c,x,y,z\in R$ $ (x\plus{}y)c\minus{}(a\plus{}b)z\equal{}\sqrt 6$
Find the min of $ a^2\plus{}b^2\plus{}c^2\plus{}x^2\plus{}y^2\plus{}z^2\plus{}ax\plus{}by\plus{}cz$
It is hard,
My work that day was bad,and I think I will fail :(  :(
Good luck

I think I will fail :( !I'm very sad now!
This problem is very hard at skill!
very strong and intelligent to solve it!But I think it is not well for us!

[quote="Allnames"]In our school test,I get this problem
Let $ a,b,c,x,y,z\in R$ $ (x \plus{} y)c \minus{} (a \plus{} b)z \equal{} \sqrt 6$
Find the min of $ a^2 \plus{} b^2 \plus{} c^2 \plus{} x^2 \plus{} y^2 \plus{} z^2 \plus{} ax \plus{} by \plus{} cz$
It is hard,
My work that day was bad,and I think I will fail :(  :(
Good luck[/quote]
It is not hard as you think, boy. We can solve it as follow:
Put $ a\plus{}b\equal{}2t,x\plus{}y\equal{}2s$, then by Cauchy Schwarz Inequality, we have
\[ a^2\plus{}b^2\plus{}x^2\plus{}y^2 \plus{}ax\plus{}by \equal{}\left( a\plus{}\frac{x}{2}\right)^2\plus{}\left( b\plus{}\frac{y}{2}\right)^2 \plus{}\frac{3}{4} (x^2\plus{}y^2) \ge \frac{1}{2}\left( a\plus{}b\plus{}\frac{x\plus{}y}{2}\right)^2\plus{}\frac{3}{8}(x\plus{}y)^2 \equal{}2(t^2\plus{}ts\plus{}s^2)\]
Now, we will show that
\[ 2(t^2\plus{}ts\plus{}s^2) \plus{}c^2\plus{}cz\plus{}z^2 \ge \sqrt{6} (cs\minus{}tz)\]
which is equivalent to
\[ c^2 \plus{}\left( z\minus{}\sqrt{6} s\right) c\plus{}2t^2 \plus{}\left( 2s \plus{}\sqrt{6}z\right) t \plus{}2s^2\plus{}z^2 \ge 0\]
\[ \Leftrightarrow \left( c\plus{}\frac{z\minus{}\sqrt{6}s}{2}\right)^2 \plus{}2\left( t\plus{}\frac{2s\plus{}\sqrt{6}z}{4}\right)^2 \ge 0\]
which is true.
From now, we can easily obtain that
\[ a^2\plus{}b^2\plus{}c^2\plus{}x^2\plus{}y^2\plus{}z^2\plus{}ax\plus{}by\plus{}cz \ge 3\]
;) :)

Yes,It is not hard,But I got it in a contest,7 hard problems in a day with only 180 minutes to kill all


geometry
circumcircle
trapezoid
power of a point
radical axis
perpendicular bisector
cyclic quadrilateral

A circle, $\Gamma$, (not the circumcircle) centre $P$ passes through the vertices of $A$ and $B$ of $\triangle ABC$ and intersects $AC$ and $BC$ again at points $K$ and $L$ respectively. The circumcircles of $\triangle ABC$ and $\triangle CKL$ intersect at two distinct points $C$ and $M$. Prove $\angle OMC = \frac{\pi}{2}$.

[hide="Hints"]
[b]Remember, don't look at the hints until you've had a real go at the question.[/b]

[hide="Hint 1"]Radical Axis Theorem: $AB, KL, CM$ are concurrent at a point $N$.[/hide]
[hide="Hint 2"]Prove: $BLMN$ is cyclic.[/hide]
[hide="Hint 3"]Power of a Point: $C$ and $N$ with respect to $\Gamma$.[/hide]
[/hide]

That's a Geometry problem from [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=%2AIMO+1985%2A&t=60787]IMO 1985[/url]

It should be $\angle PMC = \frac \pi 2$. Let O, Q be the circumcenters of the triangles $\triangle ABC, \triangle LKC.$ Since these 2 triangles are oppositely similar, CO, CQ are isogonals of CB, CA. Let CO, CQ meet the circle (O) again at B', A' and let CO, CQ meet the circle (Q) again at L', K'. The quadrilaterals $AA'B'B,\ KK'L'L$ are both cyclic isosceles trapezoids, hence the AB, A'B' have a common perpendicular bisectors and KL, K'L' also have a common perpendicular bisector intersecting at the center P of the circumcirle (P) of the cyclic quadrilateral ABLK. Since CB', CK' are diameters of the circles (O), (Q), the angles $\angle B'MC = \angle K'MC = \frac \pi 2$ are right, hence the points B', K', M are collinear. For the same reason, the angles $\angle K'A'B' \equiv CA'B' = \frac \pi 2$ and $\angle K'L'B' = \pi - \angle K'L'C = \frac \pi 2$ are right, hence the quadrilateral $A'B'L'K'$ is cyclic with the circumcenter P at the midpoint of the segment B'K', which makes all 4 points B', P, K', M collinear. As a result, $\angle PMC \equiv \angle K'MC = \frac \pi 2$ is right.

Here is my solution (yet another solution...) - quite a nice and simple one, IMHO.

It is easy to see that $BPKM$ is cyclic, since \begin{eqnarray*} \angle BMK + \angle BPK & = & \angle BMC - \angle KMC + \angle BPK \\ & = & \angle BMC - \angle KLC + \angle BPK \\ & = & (180^{\circ} - \angle BAC) - \angle BAC + 2 \angle BAC \\ & = & 180^{\circ}. \end{eqnarray*} Hence \[ \begin{eqnarray*} \angle CMP & = & \angle CMK + \angle KMP \\ & = & \angle CLK + \angle KBP \\ & = & \angle BAC + (90^{\circ} - \angle BAC) \\ & = & 90^{\circ}.\ \blacksquare  \]

It's very easy to rewrite this using oriented angles (in order to cover all possible configurations) by the way :)


See: http://www.amazon.com/gp/product/product-description/0199205604/ref=dp_proddesc_0/002-2788598-3301647?ie=UTF8&n=283155&s=books

Does anyone have access to this book, and could they comment on it?

Hm, I looked briefly through it once, and it seemed kind of basic, maybe I have no clue what I'm talking about though.

[quote="tokenadult"]It's a good book for seeing his thought process. It's not as much of a how-to book as one might guess. I own it and like it.[/quote]I agree, that's why the book is also titled 'A Personal Perspective"

This book was written years ago, and was well known among Australian math olympiad participants.   It has not produced any new IMO masters, not even within Tao's own family where one of his younger brothers was on AUS IMO team.   So to put it mildly, there is something else to Tao's accomplishments on the IMO beyond any special teachable techniques that one might or might not find in his book.  (edit:  this is not to say that the book is better or worse than others on the same subject; none of those are likely to be a shortcut to excellence, either.)

I would be interested to know whether the writing of this book was at Tao's initiative, or part of the prodigy-worship that surrounded him during his time in Australia.  It is clear that Tao likes to write expositions of many mathematical subjects including some for a very broad audience, so maybe it was his idea, but the opposite situation is also easy to imagine.


n is a natural number such that 2 :phi:(n) =n-1. Show that:
1. 3 doesn’t divide n
2. if p divides n and q divides n, with p and q primes, then p  :!=:  1 (mod 2q)
3. n has at least 11 prime disctint divisors.

If there is a prime p such that p^2 divides n then left side is divisible by p but right one isn't. So this number is square-free. We can write it as a product of distinct primes: $n=p_1p_2\ldots p_m$ and we are solving: $2(p_1-1)(p_2-1)(p_3-1)\ldots(p_m-1)=p_1p_2p_3\ldots p_m-1$.

1) Suppose that $p_1=3$. We know that n is odd (if not, there's an obvious contradiction). So other primes are bigger than 3 so they are all in the form 6k+1 or 6k-1.

$p_1p_2p_3\ldots p_m-1=3p_2p_3\ldots p_m-1$ it's in form 3k-1.

$2(p_1-1)(p_2-1)(p_3-1)\ldots(p_m-1)=\newline 4(p_2-1)(p_3-1)\ldots(p_m-1)\equiv\newline (p_2-1)(p_3-1)\ldots(p_m-1) (mod 3)$
Each member of this product is in form 3k+1 or 3k so also whole product is in form 3k or 3k+1.

So these two sides of equation cannot be equal.

And what about 2) and 3) ? Can you solve them?

:arrow: [b]2) [/b]Suppose for absurd that p  :equiv:  1 (mod2q) then p-1 = 2q*t with t natural number, and then from n-1 = 2  :phi: (n) we have that   p….q….pk –1 = 2(p-1)….(q-1) = 4q*t….(q-1)…(pk –1), we have then that q divides 1 (because divides p…q…pk and 4q*k….(q-1)…(pk –1)) that is absurd.
[b]3) [/b]For have n-1 = 2  :phi: (n) we must have that   2  :phi: (n)/n = 1-1/n less than 1, that is  2(1-1/p1)(1-1/p2)…(1-1/pk) (*)  less than 1, and now we must prove that (*) is true if and only if n has at least 11 primes that is k is more than 11

Note: k in “p….q….pk-1?,for example, is not a factor but it is an exponent and also p1, p2 in
“2(1-1/p1)(1-1/p2)?.

I hope you will understand what I wrote because in this form it’s too difficult to understand!!

 :)

We can't have 5 and 11 simultaneously! See 2) ;)

You're right. I've forgotten about it  :blush:


1. Prove that exist $\large{x}$  for any Z numbers $\large{a,b,c}$ :

         $\large{a^x+x\,\equiv\,b\,(\textrm{mod}\,c)}$ .


floor function
modular arithmetic
inequalities
number theory unsolved
number theory

Find all the ordered triples $ (a,m,n)$ of positive integers such that $ a \ge 2$ , $ m \ge 2$, and $ a^m\plus{}1$ divides $ a^n\plus{}203$

Let $ k\equal{}n\bmod m$ and $ t\equal{}\lfloor n/m\rfloor\equal{}(n\minus{}k)/m$.
Since $ a^m \plus{} 1$ divides $ a^n \plus{} 203 \equal{} (a^{n\minus{}k} \minus{} (\minus{}1)^t)a^k \plus{} (\minus{}1)^t a^k \plus{} 203$ and $ a^m \plus{} 1$ divides $ a^{n\minus{}k} \minus{} (\minus{}1)^t\equal{}a^{tm}\minus{}(\minus{}1)^t$, we have that $ a^m \plus{} 1$ divides $ (\minus{}1)^t a^k \plus{} 203$. Hence, $ a^m \plus{} 1\leq a^k \plus{} 203\leq a^{m\minus{}1} \plus{} 203,$ implying that $ a^{m\minus{}1}(a\minus{}1)\leq 202$ and $ a^{m\minus{}1}\leq 202$, leaving only a finite number of variants for pairs $ (a,m)$.

The rest is pure computational. Here is the complete list of solutions:
[code][2, 2, Mod(1, 4)]
[2, 3, Mod(2, 6)]
[2, 4, Mod(0, 8)]
[2, 6, Mod(9, 12)]
[3, 2, Mod(3, 4)]
[4, 2, Mod(0, 4)]
[5, 2, Mod(1, 4)]
[8, 2, Mod(3, 4)]
[10, 2, Mod(2, 4)][/code]

For example, the solution [b][2, 6, Mod(9, 12)][/b] is treated as $ a\equal{}2$, $ m\equal{}6$, and $ n\equiv 9\pmod{12}$. It is easy to see that $ a^m\plus{}1\equal{}65$ divides $ a^9\plus{}203\equal{}715$, $ a^{9\plus{}12}\plus{}203\equal{}2097355$, etc.

You missed $ (203,m,km\plus{}1)$ with $ k$ odd.

Yes, I've missed one case here:

[quote="maxal"]we have that $ a^m \plus{} 1$ divides $ ( \minus{} 1)^t a^k \plus{} 203$. Hence, $ a^m \plus{} 1\leq a^k \plus{} 203\leq a^{m \minus{} 1} \plus{} 203,$[/quote]

If $ ( \minus{} 1)^t a^k \plus{} 203 \equal{} 0$ (i.e., $ k\equal{}1$, $ a\equal{}203$ and $ t$ is odd), we don't have the aforementioned inequality.
This missed case corresponds to the solutions $ (203,m,tm\plus{}1)$ with odd $ t$.

i don't blame u...this problem is really annoying...when i solved it, i misseed it too, but when i was writing the solution, i realized that inequality didn't need to hold in every case...i had used ur list to check my solutions the first time, so i knew u'd missed it too...if u allow $ m$ to be 1, u get some more solutions, but that makes it even more annoying


If $a+b+c=3$ a,b,c positive reals find the maximum price of

$\sum\dfrac{ab}{(a^2+b)(a+b^2)}$

I proved that the maximum is $\frac{3}{2}.$
 for $( x\rightarrow{0}, y\rightarrow{0}, z\rightarrow{3})$

I prove that $\sum\frac{ab}{(a^{2}+b)(a+b^{2})}<\frac{3}{2}.$
Firstly, we have $\sum\frac{ab}{(a^{2}+b)(a+b^{2})}\leq \sum\frac{1}{(a+1)(b+1)}.$
From condition $a+b+c=3$ we obtain $a=\frac{3x}{x+y+z}, b=\frac{3y}{x+y+z}, c=\frac{3z}{x+y+z}$ with $x,y,z>0.$
It follows $\sum\frac{1}{(a+1)(b+1)}=\sum\frac{(x+y+z)^2}{(4x+y+z)(x+4y+z)}<\frac{3}{2}\Leftrightarrow$
$\Leftrightarrow(1+\frac{3x}{x+y+z})(1+\frac{3y}{x+y+z})(1+\frac{3z}{x+y+z})>4,$ obvious, because $(1+m)(1+n)(1+p)>1+m+n+p,\forall m,n,p>0.$

Yes. The ineq at the end is Wairstrass ineq. Your solution is nice.

[hide="my solution"]We will prove $\sum_{cyc}\frac{ab}{(a^2+b)(a+b^2)}<\frac{3}{2}$.
Since $(a^2+b)(a+b^2)\geq (a^2+a)(b^2+b)\Leftrightarrow a^3+b^3\geq a^2b+ab^2$,
$\sum_{cyc}\frac{ab}{(a^2+b)(a+b^2)}\leq\sum_{cyc}\frac{ab}{(a^2+a)(b+b^2)}=\sum_{cyc}\frac{1}{(a+1)(b+1)}$
We only need to prove $\sum_{cyc}\frac{1}{(a+1)(b+1)}\leq\frac{3}{2}$
$\Leftrightarrow (c+1)+(a+1)+(b+1)\leq\frac{3}{2}(a+1)(b+1)(c+1)$
$\Leftrightarrow (a+1)(b+1)(c+1)\geq 4$
$\Leftrightarrow (3a+3)(3b+3)(3c+3)\geq 4\cdot 3^3$
$\Leftrightarrow (4a+b+c)(a+4b+c)(a+b+4c)\geq 4\cdot 3^3$
By Generalized Holder Inequality, we have $(4a+b+c)(a+4b+c)(a+b+4c)\geq (\sqrt[3]{4a^3}+\sqrt[3]{4b^3}+\sqrt[3]{4c^3})^3=4(a+b+c)^3=4\cdot 3^3$
Therefore we have proved that $\sum_{cyc}\frac{ab}{(a^2+b)(a+b^2)}\leq\frac{3}{2}$. In order to have the equality hold, the equality in the Holder inequality must also hold, which means $4a: a: a=b: 4b: b=c: c: 4c$$\Rightarrow$two of them become zero. So equality cannot hold.[/hide]


[quote="frt"]Find all possible values of $a$ such that the equations $x^{2}+ax+1=0$ and $ax^{2}+x+1=0$ have at least one common root.[/quote]
[hide]
From the second equation, we derive $a$:
$a=-\frac{x+1}{x^{2}}$.
Now sub in this value into the first equation, we get:
$x^{2}-(\frac{x+1}{x})+1=0$
or
$x^{2}-\frac{1}{x}=0$.
This is equal to $x^{3}=1$.
So the roots for the above equation are $1$ and $\frac{-1\pm i\sqrt{3}}{2}$.
Thus, $a=-2,-1$.
[/hide]

[quote="pkerichang"][hide="hmm"]
let $r$ be the common root for both equations.  then $r^{2}+ar+1 = 0 = ar^{2}+r+1$, rearrange to get $(a-1)r^{2}= (a-1)r$, so we have $a = 1$, $r = 0$, or $r = 1$.  Check $r = 0$; it never works.  If $r = 1$, then $a =-2$.
[/hide]
kimby...I think you have a typo on the final answer part.[/quote]
Yes, u got me this time  :D
I shouldve checked my solution...


1. In how many ways can a photograoher at a weeding arrange 6 people in a row, including the bride and the groom, if the bride is positioned somewhere to the left of the grooM?

2. In how many different 6-letter arrangements can be made from the word AARDVARK?

[hide="hint for a solution to 1"] First, consider only placing the bride and the groom in a way that the bride is to the left of the groom. Then place the other people.[/hide]

[hide="hint for another solution to 1"] By symmetry, this is half of the number of ways to place everyone with no restrictions.[/hide]

You can probably solve 2 by casework on the number of As.

[hide="Question 1"]Q1. There are 6!=720 ways to place 6 people in a row with no restrictions. In exactly half of these, the bride is to the left of the groom, so there are [b]360 [/b]ways the photographer can line them up.[/hide]

[hide="Question 2"]Q2. There are 12 distinct ways to choose 6 letters from AARDVARK.

For (ARRDVK) there are 6!/2! ways of arranging them = 360
For (AAADVK) there are 6!/3! ways of arranging them = 120
For (AARDVK) there are 6!/2! ways of arranging them = 360
For (AARRVK) there are 6!/2!2! ways of arranging them = 180
For (AARRDK) there are 6!/2!2! ways of arranging them = 180
For (AARRDV) there are 6!/2!2! ways of arranging them = 180
For (AAARVK) there are 6!/3! ways of arranging them = 120
For (AAARDK) there are 6!/3! ways of arranging them = 120
For (AAARDV) there are 6!/3! ways of arranging them = 120
For (AAARRK) there are 6!/3!2! ways of arranging them = 60
For (AAARRV) there are 6!/3!2! ways of arranging them = 60
For (AAARRD) there are 6!/3!2! ways of arranging them = 60

The total number of distinct 6 letter arrangements is therefore 360+120+360+180+180+180+120+120+120+60+60+60= [b]1920[/b]

[b]There are 1920 distinct 6-letter arrangements that can be made from AARDVARK.[/b]

I hope this is correct because the method is very crude...[/hide]

It's better for a topic being too easy for the forum it's in than for it to be too difficult.

#1 is almost exactly the same as a Mathcounts National problem, and #2 is annoying. I suppose it'd be fine in High School Basics, but it doesn't really matter.

[quote="orangefronted"]2. In how many different 6-letter arrangements can be made from the word AARDVARK?[/quote]
There are $ 3$ A's, $ 2$ R's, $ 1$ D, V, K, and we require $ [x^6] (1\plus{}x\plus{}x^2\plus{}x^3)(1\plus{}x\plus{}x^2)(1\plus{}x)^3$ WITHOUT accounting for the permutations (for example, we have only counted AARDVA once but there should be $ 6!/3!$ in total)... so what should the actual generating function be for this problem?

You want the exponential generating function  $ \left[\frac {x^6}{6!}\right]\left(1 \plus{} x \plus{} \frac {x^2}{2} \plus{} \frac {x^3}{6}\right)\left(1 \plus{} x \plus{} \frac {x^2}{2}\right)(1 \plus{} x)^3$.  It's not incredibly hard (if you're familiar with generating functions) to see how this builds in the case-based approach.  This is probably not worth the trouble, though.


[quote="Scrambled"]is it ok if a parent checks your math to see if you have careless computatino errors or a bungled approach?[/quote]

No.  On the entry form you signed a statement saying that the work is entirely your own.  That even means that parents can't pitch in.

Do the work, set it aside for a few days, then check it yourself.


Putnam
combinatorics unsolved
combinatorics

Let $ S\equal{}\{1,2,...,n\}$. Subset $ A$ of $ S$ is called [b]nice[/b] if and only if arithmetic mean of the elements of $ A$ is an integer. Let $ S(n)$ be the number of nice subsets of $ S$. Prove that $ S(n)\minus{}n$ is an even number.


We put 0 or 1 to every cell of a $ m \times n$ chessboard. If number of $ 0$ and number of $ 1$ is equal in a putting, then we call putting [b][i]equal putting[/i][/b]. Let $ a$ be a real number; If we can find $ m$ and $ n$ at a [i][b]equal putting[/b][/i] such that for each of $ m$ row and $ n$ column, percentage of $ 1$'s in this column or row isn't less than $ a$ or more than $ 100 \minus{} a$ then $ a$ is a [b][i]good number[/i][/b]. Find the greatest [b][i]good number[/i][/b].

Sorry for my bad english.

[quote="pluristiq"]obviously $ a\leq 50$, and 50 is achieved by even $ n$ and $ m$ and a standard checkerboard coloring, so I really don't get what there is to prove here...?[/quote]

You don't understand becouse  my bad english. sorry.. percentage of 1's isn't less than $ a$, or percentage of 1's isn't more than $ 100\minus{}a$ for this column or row (for this $ n\plus{}m\minus{}1$ cells)


Let ABC be an acute-angled triangle, and let D;E be the feet of the perpendiculars from A;B to BC;CA respectively. Let P be the point where the line AD meets the semicircle constructed outwardly on BC, and Q be the point where the line BE meets the semicircle constructed outwardly on AC. Prove that CP = CQ.

$CP^2 = CD \cdot CB$ and $CQ^2 = CE \cdot CA$.  but $CD \cdot CB = CE \cdot CA$ because $BDEA$ is cyclic.  then $CP^2 = CQ^2$ and the problem's done.

Denote the triangles sides $a = BC, b = CA, c = AB$ and and angles $\alpha = \angle A, \beta = \angle B, \gamma = \angle C$.

$CP^2 = CD^2 + PD^2 = CD^2 + CD \cdot BD = CD(CD + BD) = CD \cdot BC = ab \cos{\gamma}$

$CQ^2 = CE^2 + QE^2 = CE^2 + CE \cdot AE = CE(CE + AE) = CE \cdot CA = ab \cos{\gamma}$

$CP = CQ$


Hi Y'all,

I dont really know if this is the best place to express this comment. If it is not, I beg forgiveness from the moderators and administrators.

I was playing around in the Members list, and I realized that form the 6581 user registrated, 4498 of them have never placed a post, that is more than 68% of them. Now, let me say something before I get a crowd screaming for my execution!, I totally agree and encourage people to be silent participants of a forum if they prefer not to have an active roll on it, but I suppose that from that 68% of people that have never placed a post, more than half of those have never logged in the site again since they registered.

Therefore, I suggest to have a rule of expiration on the membership. I mean, for example, if someone has logged in the site just once or twice, and has not logged in a year or so, chances are that this person never log in again. Therefore, the data in "Who is online" is not really acurate in some sences. I suppose that a year is something everyone can handle, and I am talking about time without logging in the site (data that you somehow manage because you have number of users online, registered and hidden).

Just a comment.

Best regards,

Thanks for the suggestion, but this is probably something we won't implement in any form primarily because our to-do list is already packed with several years worth of more pressing tasks.  Those members who have never posted and haven't visited the site in over a year merely occupy one row in a database with zillions of rows, so they're not really hindering the site.

[quote="djimenez"]Therefore, I suggest to have a rule of expiration on the membership. I mean, for example, if someone has logged in the site just once or twice, and has not logged in a year or so, chances are that this person never log in again. Therefore, the data in "Who is online" is not really acurate in some sences. I suppose that a year is something everyone can handle, and I am talking about time without logging in the site (data that you somehow manage because you have number of users online, registered and hidden).

Just a comment.

Best regards,[/quote]This will not be implemented. Users can keep their unread posts / favorite topics as long as they wish, even if they log in just once a year. We have no problem having extra users with 0 posts, they don't take almost any space.


geometry
geometric transformation
reflection
ratio
trigonometry
circumcircle
angle bisector

Let P and P' be two isogonal conjugate points with respect to triangle ABC. Let the lines AP, BP, CP meet the lines BC, CA, AB at the points A', B', C', respectively.

Prove that the reflections of the lines AP', BP', CP' in the lines B'C', C'A', A'B' concur.

Synthetic proof? :maybe:

  Darij

Did you manage to prove this very nice property?

I have absolutely no idea how to do it...  :(

Thanks for the picture. Actually, I have been searching for a synthetic proof for a long time and I have really no reasonable approach. (The word "synthetic" may be irritating: I don't have even an analytic proof doable by hand. The correctness of the result has been verified by computer algebra by Paul Yiu.) I'll be grateful for [b]any ideas[/b] on the problem.

  Darij

Here is a possible idea but I'm unable to find a synthetic proof of the following Lemma (it can be proved by computation)

Lemma : for any point M, let T(M) = A(M)B(M)C(M) the precevian triangle of M; M* and I(M) the isogonal conjugates of M wrt ABC and T(M) resoectively.
Then the line A(M*)I(M*) is the reflection of the line A(M)I(M) wrt BC,... cyclic

Now, if we come back to your problem; let U,V,W be the isogonal conjugates of A, B, C wrt the triangle A'B'C' and Q the isogonal conjugate wrt UVW of the isogonal conjugate of P wrt A'B'C'. If we apply the Lemma with P instead of M and A'B'C' instead of ABC, we get that your three lines concur at Q.

[quote="jpe"]Lemma : for any point M, let T(M) = A(M)B(M)C(M) the precevian triangle of M; M* and I(M) the isogonal conjugates of M wrt ABC and T(M) resoectively.
Then the line A(M*)I(M*) is the reflection of the line A(M)I(M) wrt BC,... cyclic[/quote]

This is just great! I have to reply very briefly now, since I am running out of time.

Since it is easily shown that A(M*) = (A(M))*, your above lemma follows from the following theorem, applied to P = A(M):

[color=blue][b]Theorem.[/b] Let P be a point in the plane of a triangle ABC, and P* its isogonal conjugate with respect to triangle ABC. Then, the reflection of the line AP in the angle bisector of the angle BPC and the reflection of the line AP* in the angle bisector of the angle BP*C are symmetric to each other with respect to the line BC.[/color]

This theorem looks easy, but I don't see any simple proof at the first sight. Here is, anyway, a sketch of a synthetic proof due to Peter Scholze and me: First, show that the two reflections meet on the line BC (this can be done using ratios of segments; use the identity $AP\cdot\sin\measuredangle BPC=AP^{\ast}\cdot\sin\measuredangle BP^{\ast}C$, which itself looks trivial but I don't see a proof simpler than replacing AP and AP* by the A-sides of the pedal triangles of P and P* and using the fact that these two pedal triangles have a common circumcircle - I guess I am just unable to see obvious things). The rest is angle chasing, and it must work. Wait, I think there is something more conceptual: Let X be the reflection of the point P in the line BC, and let the line PX meet the line BC at D. (Of course, D is simply the point where the inconic of triangle ABC with focii P and P* touches the side BC, but I'm trying to keep everything elementary.) Since you know the angle between the lines P*X and BC (this is a heritage of the synthetic - non-Ceva - proof that the isogonal conjugate P* of P exists: construct the reflections X, Y, Z of the point P in the sides BC, CA, AB of triangle ABC, then show that the circumcenter of triangle XYZ is exactly the isogonal conjugate of P), you basicly know all angles necessary to show that this new point D lies on the reflection of the line AP in the angle bisector of the angle BPC and on the reflection of the line AP* in the angle bisector of the angle BP*C, and that these two reflections are symmetric to each other with respect to BC. So the Theorem is proved again.

Great work, Jean-Pierre, I am glad that the problem finally turns out to be so "simple". (I already feared that the first step of the solution would be to draw the isogonal cubic through A, B, C, M, M*, A(M), A(M*), B(M), B(M*), C(M), C(M*)...)

  darij

Dear Darij, many thanks for your nice answer.
In fact, your theorem is very easy to prove :
Let  U be the reflection of P* wrt BC (P* = isogonal of P wrt ABC)
We have <UBC = <CBP* = <PBA and
<UCP = <UCB + <BCP = <BCP* + <BCP = <BCA
The relations <UBC = <PBA and <UCP = <BCA prove that U is the isogonal conjugate of A wrt BPC.
Same way the reflection of P wrt BC is the isogonal conjugate of A wrt BP*C and your theorem.
FRiendly. Jean-Pierre

[quote="jpe"]Now, if we come back to your problem; let U,V,W be the isogonal conjugates of A, B, C wrt the triangle A'B'C' and Q the isogonal conjugate wrt UVW of the isogonal conjugate of P wrt A'B'C'. If we apply the Lemma with P instead of M and A'B'C' instead of ABC, we get that your three lines concur at Q.[/quote]

Very nice solution. I don't understand one thing though. (it is not a vital part of this problem, the solution works without mentioning it) Why is $UVW$ the precevian triangle of $P*$  :? how do you show smth like: "the map $P \mapsto P*$ also maps the precevian triangle's vertices to their isogonal conjugates" ?

sorry if it's trivial, but i can't prove it  :blush:

[quote="perfect_radio"]Why is $UVW$ the precevian triangle of $P*$  :? how do you show smth like: "the map $P \mapsto P*$ also maps the precevian triangle's vertices to their isogonal conjugates" ?[/quote]

Indeed, this is not vital to the solution of the problem (which, by the way, I am quite glad you understood - I actually wrote for Hyacinthians and mainly for JPE, so I didn't expect anybody else to read :) ). Anyway, let A'B'C' be the precevian (or anticevian, as I am used to calling it) triangle of a point P with respect to a triangle ABC. Then, the line AB' is the harmonic conjugate of the line AP with respect to the angle CAB. Similarly, if P* is the isogonal conjugate of the point P with respect to the triangle ABC, and A*'B*'C*' is the precevian triangle of the point P* with respect to the triangle ABC, then the line AB*' is the harmonic conjugate of the line AP* with respect to the angle CAB. Now, since the points P and P* are isogonal conjugates, the lines AP and AP* are symmetric to each other with respect to the angle bisector of the angle CAB; thus, the lines AB' and AB*', being their respective harmonic conjugates with respect to the angle CAB, must also be symmetric to each other with respect to the angle bisector of the angle CAB. Similarly, the lines CB' and CB*' are symmetric to each other with respect to the angle bisector of the angle BCA. Thus, the points B' and B*' are isogonal conjugates with respect to triangle ABC. Similarly for the points A' and A*', and for the points C' and C*'.

  Darij

[quote="darij grinberg"]Then, the line AB' is the harmonic conjugate of the line AP with respect to the angle CAB.[/quote]

How do you prove this?

I saw this result being used in a solution of a ml contest problem (in the 5th edition - smth with radical axis) and there was said to use menelaus, but how?

[quote="perfect_radio"][quote="darij grinberg"]Then, the line AB' is the harmonic conjugate of the line AP with respect to the angle CAB.[/quote]

How do you prove this?[/quote]

Okay, since you probably define the precevian triangle implicitely (as the triangle such that the cevian triangle of the point P with respect to this triangle is the triangle ABC), let me start with the precevian triangle as the base triangle. Then, we have to prove that:

[color=blue][b]Theorem.[/b] Let P be a point in the plane of a triangle ABC, and let $A^{\prime}=AP\cap BC$, $B^{\prime}=BP\cap CA$ and $C^{\prime}=CP\cap AB$. Then, the line BC is the harmonic conjugate of the line AP with respect to the angle C'A'B'.[/color]

[i]Proof.[/i] Let the line B'C' meet the lines BC and AP at the points T and U, respectively. Then, since the lines AU, B'B and C'C concur (namely, they concur at the point P), the Ceva theorem for the triangle AB'C' yields $\frac{AC}{CB^{\prime}}\cdot\frac{B^{\prime}U}{UC^{\prime}}\cdot\frac{C^{\prime}B}{BA}=1$. On the other hand, since the points C, B and T are collinear, the Menelaos theorem for the triangle AB'C' yields $\frac{AC}{CB^{\prime}}\cdot\frac{B^{\prime}T}{TC^{\prime}}\cdot\frac{C^{\prime}B}{BA}=-1$. Comparing these two equations, we get $\frac{B^{\prime}T}{TC^{\prime}}=-\frac{B^{\prime}U}{UC^{\prime}}$; thus, the point T is the harmonic conjugate of the point U with respect to the segment B'C'. Thus, the line A'T is the harmonic conjugate of the line A'U with respect to the angle B'A'C'. In other words, the line BC is the harmonic conjugate of the line AP with respect to the angle C'A'B'. And the theorem is proven.

  darij

one last question: can one construct the anticevian triangle of a point wrt a triangle? (with ruller & compass)

ps; nice proof

[quote="perfect_radio"]one last question: can one construct the anticevian triangle of a point wrt a triangle? (with ruller & compass)[/quote]

Of course, as you can construct the harmonic conjugate of a line with respect to an angle...

  darij

Darij, I understood your theorem:

Theorem. Let P be a point in the plane of a triangle ABC, and P* its isogonal conjugate with respect to triangle ABC. Then, the reflection of the line AP in the angle bisector of the angle BPC and the reflection of the line AP* in the angle bisector of the angle BP*C are symmetric to each other with respect to the line BC.

JPE solution:

Let U be the reflection of P* wrt BC (P* = isogonal of P wrt ABC)
We have <UBC = <CBP* = <PBA and
<UCP = <UCB + <BCP = <BCP* + <BCP = <BCA
The relations <UBC = <PBA and <UCP = <BCA prove that U is the isogonal conjugate of A wrt BPC.
Same way the reflection of P wrt BC is the isogonal conjugate of A wrt BP*C and your theorem.

But I can't understand final step, could you explain jpe words or make some steps:

Lemma : for any point M, let T(M) = A(M)B(M)C(M) the precevian triangle of M; M* and I(M) the isogonal conjugates of M wrt ABC and T(M) resoectively.
Then the line A(M*)I(M*) is the reflection of the line A(M)I(M) wrt BC,... cyclic

Now, if we come back to your problem; let U,V,W be the isogonal conjugates of A, B, C wrt the triangle A'B'C' and Q the isogonal conjugate wrt UVW of the isogonal conjugate of P wrt A'B'C'. If we apply the Lemma with P instead of M and A'B'C' instead of ABC, we get that your three lines concur at Q.

How do you use $I(M^{*})$ - isogonal conjugate wrt $A_{1}B_{1}C_{1}$ of the $M^{*}$ -isogonal conjugate of $M$ ?

Sorry for the late reply. As this thread got confusing, I completely rewrote the proof; it now can be seen in the paper "Isogonal conjugation with respect to a triangle", which can be found as an attachment at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=569871#569871]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=18472 post #6[/url] and on [url=http://www.cip.ifi.lmu.de/~grinberg/geometry2.html]my website[/url] (Theorem 9).

  Darij


geometry
trigonometry
cyclic quadrilateral

I have a question on the naming of a set of related formulae:

Brahmagupta's formula is generally used to name the formula
"The area of a cyclic quadrilateral with side lengths $a,b,c,d$, and semiperimeter $s$ is $\sqrt{(s-a)(s-b)(s-c)(s-d)}$." (1)

However, what is the generalized form of the formula, which gives the area of a general quadrilateral, called?  The formula I speak of is
"The area of a quadrilateral with side lengths $a,b,c,d$, opposite angles $B,D$, and semiperimeter $s$ is $\sqrt{(s-a)(s-b)(s-c)(s-d) - abcd \cos^2 \left( \frac{B+D}{2} \right) }$." (2)

It seems to have been proven first by Bretschneider, but Bretscheneider's formula I believe refers to still another formula:
"The area of a quadrilateral with side lengths $a,b,c,d$, diagonal lengths $p,q$, and semiperimeter $s$ is $\sqrt{(s-a)(s-b)(s-c)(s-d) - \frac14 (ac+bd+pq)(ac+bd-pq) }$." (3)

So what is (2) called?

Actually, I have never seen a name for (2) other than [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15973]"Generalized Brahmagupta formula"[/url]. And the formula (3) is new to me. I remember a different formula which is often referred to as Bretschneider's formula:

$p^2 q^2 = a^2 c^2 + b^2 d^2 - 2abcd \cos\left(B+D\right)$.

This formula is a kind of cosine law for quadrilaterals. And actually, once this formula is known, the formula (3) easily follows from (2).

  Darij


[color=darkblue]I'm confused.  Where does everyone get these MathCounts competition questions that appeared perhaps more than ten years ago?  I can only find competitions at most as early as 1998-1999.  Could anyone tell me?  Thanks! :D  [/color]

[quote="xpmath"]You buy it from the Mathcounts website :P[/quote]

That's only as early as 1998-99, you should read better.

See the sticky at the top of the forum.

What sticky?  Sorry, the sticky has yet to be found. :oops:


Find the remainder when $2^{1000}$ is divided by $3$
What we want here are indigenous solutions :lol:
Method 1:-MOD
Method 2-Binomial theorem.
any others?

Calculate the number and divide it by 3  :D

[quote="sen"]Umm process of elimination?

$2^{1000}$ either leaves a remainder of ${0, 1, 2}$ when divided by $3$

Now it can't leave a remainder of $0$ because $3k \neq 2^{1000}$

It can't leave a remainder of $2$ which would

$\implies 2^{1000}$ is odd but clearly it isn't, leaving $1$ as our only answer.[/quote]

Yes, $1$ is the correct one.

Careful!

A remainder of 2 would not imply that $2^{1000}$ is odd.  $2^3$ has a remainder of 2 when divided by 3.


[b]I have to admit that I am no genius in math or any subject…just incredibly curious.  Is there an equation for my question:  Mathematical repercussions of “Resist not evil” To either obey or not… Jesus’ command to all people on earth…I would love to see it and explained.  Regardless of your religious beliefs.  Thank you. :? [/b]

Sir, $\frac{a+b^n}{n}=x$, hence God exists; reply!


A quadrilateral has vertices at $ (0,1)$, $ (2,5)$, $ (5,2)$ and $ (7,0)$. Its perimeter can be expressed in the form $ a\sqrt2\plus{}b\sqrt{5}$ with $ a$ and $ b$ integers. What is the sum of $ a$ and $ b$?

hello, let $ A(0;1), B(2;5), C(5;2),D(7;0)$ then we get
$ \overline{AB}\equal{}2\sqrt{5}$
$ \overline{BC}\equal{}3\sqrt{2}$
$ \overline{CD}\equal{}2\sqrt{2}$
$ \overline{AD}\equal{}5\sqrt{2}$
and the searched perimeter is $ p\equal{}2\sqrt{5}\plus{}10\sqrt{2}$ and from here we get $ a\plus{}b\equal{}12$.
Sonnhard.


Let me start off with an ultra simple problem

Identify X, Y, Z

[hide]well, actually i wanted to test the ACD chemsketch software....its good  :lol: [/hide]

i think this a condensation rexn (1st step),isn't it?? :D


What is the arithmetic mean of all of the positive two-digit integers
with the property that the integer is equal to the sum of its first digit
plus its second digit plus the product of its two digits?

The integer $ \underline{a}\, \underline{b}$, where $ a$ and $ b$ are digits, must satisfy $ 10a\plus{}b\equal{}a\plus{}b\plus{}ab \Rightarrow 9a\equal{}ab \Rightarrow b\equal{}9$. Therefore, the integers must have a units digit of 9, and the average of these is 59.


inequalities
calculus
calculus computations

Let $ A$  be a non-empty set of positive integers and $ A$  is infinte, $ b$  and $ c$  are positive integers.
Is this inequality true?

$ lim_{n \to \infty} \frac {|A \cap (1,2,...,[\frac {n \minus{} c}{b}]|}{|A \cap (1,2,...,n)|} \ge \frac {1}{b}$
with $ [x]$ is the integer part of $ x$

I 've edited this problem. Please help me!


geometry
geometric transformation
reflection
parallelogram
geometry proposed

[quote="Omid Hatami"]Let $ ABC$, $ l$ and $ P$ be arbitrary triangle, line and point. $ A',B',C'$ are reflections of $ A,B,C$ in point $ P$. $ A''$ is a point on $ B'C'$ such that $ AA''\parallel l$. $ B'',C''$ are defined similarly. Prove that $ A'',B'',C''$ are collinear.[/quote]

http://www.cut-the-knot.org/Curriculum/Geometry/Zaslavsky.shtml (interchange triangles ABC and A'B'C').

  darij

My solution: :blush:
Let $ A_3,B_3,C_3$ be points on $ BC,CA,AB$ respectively,such that $ A_2A_3 \| BC',B_2B_3 \| CA',C_2C_3 \| AB'$
Easy to see that $ AB' \equal{} A'B \equal{} C_2C_3,BC' \equal{} B'C \equal{} A_2A_3,CA' \equal{} C'A \equal{} B_2B_3$.Then apply Thales's theorem and Menelaus's theorem we have:$ A_2,B_2,C_2$ are collinear if and only if $ A_3,B_3,C_3$ are collinear.
Let $ d_1,d_2,d_3$ be lines pass through $ A,B,C$ and parallel to $ A_2A_3,B_2B_3,C_2C_3$
,respectively.Then $ d_1,d_2,d_3$ are concurrent at Q (the proof is simple,just take $ Q' \equal{} d_1 \cap d_2$ then $ QB \equal{} CA'$ and $ QB \| CA'$ so $ QBA'C$ is parallelogram and from it we have $ CQ \| BA' \| C_2C_3$.
$ QAA_2A_3,QBB_2B_3,QCC_2C_3$ are parallelograms so $ QA_3 \| QB_3 \| QC_3 \| l$,then $ Q,A_3,B_3,C_3$ are collinear and that complete the proof  :wink:
[img]http://img9.imageshack.us/img9/6431/parax.jpg[/img]

Trivial by moving point.Move the point $A''$ on $B'C'$ then define $B''$and $C''$ such that $AA''//BB''//CC''$
$deg(A'')=deg(B'')=deg(C'')=1$ we want to show that $A''-B''-C''$ collinear so 4 case is enough.Take $A''$
$C''$,$B''$ ,$AA''//BB'$ and $\infty_{BC}$ .$\blacksquare$


cool. :D
[hide]
$\frac{1}{2} \times 4 = 2$

$\text{The Answer Is 2}$[/hide]

[quote="nonie"][quote="236factorial"]What is the value of $\frac{1}{2}+0.5+(2-\frac{3}{2})-(-\frac{105}{210})$?[/quote]
1+1=2 That was SOOO hard!  ;)  :lol:[/quote]

You want a hard one?  ;)


Prove that in the sequence (10^n+3) (n=1,2,3,...) are infinitely many primes.
 :maybe:

Problems of this type are, as far as I know, wide open.


trigonometry
geometry
incenter
circumcircle
geometry proposed

Leta ABCD be a square, with side a. Let E and F be on the sides BC and AB, such that <EDF=45. Let r be the radius of the inscribed circle in EBF. Prove that r+EF=a

let I the incenter of FEB and p the semiperimeter of FEB. Then clearly $ IF \perp FD$ and $ IE \perp ED$ so D is the B-excenter of EFB and so $ AB\equal{}p$. But $ r \plus{} EF \equal{} p \minus{} EF \plus{} EF \equal{} p$

[quote="primoz2"]I dont think that the 2 perpendiculars are obvious....[/quote]
I'm not sure if the perpindicular fact is quite necessary; we can prove directly that $ D$ is the $ B$-excenter of $ \triangle BEF$ as follows:
[hide="Proof"]
Notice that the incenter, $ I$, of triangle $ BEF$ lies on $ BD$ since it must lie on the bisector of $ \angle ABC$ and $ BD$ is the bisector of $ \angle ABC$ since $ ABCD$ is a square.  Now,\[ \angle EIF \equal{} 90 \plus{} \frac {\angle FBE}{2} \equal{} 135 \equal{} 180 \minus{} 45 \equal{} 180 \minus{} \angle EDF\]so $ EIFD$ is cyclic, which means that $ D$ is the second intersection between the circumcircle of $ \triangle EIF$ and $ BD$.  Yet, it is well known that the $ B$-excenter of $ \triangle FBE$ is also the second intersection between $ BD$ and the circumcircle of $ \triangle EIF$, so $ D$ is the $ B$-excenter of $ \triangle EFB$. [/hide]


Well, chess players amongst you know that the Reunification Match for the World Title in chess is being played in Elista, Russia. You also probably know everything about the scandal concerning Kramnik's visits to the toilet... For those who aren't informed, here's a link or two [url]http://www.worldchess2006.com/main.asp[/url]

[url]http://www.chessgames.com/perl/chess.pl?tid=52037[/url]

Will the match continue?
What about Game 5?

Well not that I am that into the game meaning I don't follow t but I've tried to play it once or twice so......Topalov if he is really concerned about Kramnik will lose his mind and therefore his game.

[quote="fredbel6"][url]http://en.wikipedia.org/wiki/Vladimir_Kramnik#Reunification_Match[/url]
I know next to nothing about this chess, but this definitely is a remarkable story.  That Wikipedia is maybe a bit more accessible to newbies to the chess world. :blush:  :maybe:

So basically the point is to bring two lineages of Chess World Champions back together? :!:[/quote]

Basically, that is it. The match started fine, with two really exciting games and two dull games :) Then, it was revealed that Kramnik went to his toilet 50 times in 6 hours, during one of the matches. Because of that, Topalov's second, Danailov, made a request to the Organizing board to opn a common toilet for both players (to remove all suspicion concerning illegal actions?). Danailov revealed that the videotape of the camera in Topalov's break room is missing... And that two suspiciously looking men were in Topa's room during the match...
Board made it's decision and opened a common toilet...
But then, the toilet became a matter of principles for Kramnik, and he refused to play the Game 5...

[quote]The match started fine, with two really exciting games and two dull games[/quote]
Is this a tournament with tons of players or just an event concerning two "champions"?  I'm not at all familiar with the competition, why do they play "two exciting games" and "two dull games"?  Don't they just have to play one game? :maybe:  :|

Apart from that, it seems that Kramnik is right... if they find that suspicious they should have thought of that first, and apply new rules next time.

This isn't a tournament: it's a match, consisting of 12 games (for more info: [url]http://en.wikipedia.org/wiki/FIDE_World_Chess_Championship_2006[/url]).

[quote="ZetaX"]Then you play an additionall number of games (mostly also even).
At some point, it will end (except both know a winning strategy).[/quote]
In Belgian politics, when there is a complete draw, the oldest is sometimes favoured, maybe this is their solution after 20 extra games? :P

Okay seriously : when there is no proof the candidate has done anything against the rules as they were declared before the competition, I see no reason to change them?! :!:

[quote="fredbel6"][quote="JBL"][quote="fredbel6"]Don't they just have to play one game?[/quote]

Because there is a small advantage in playing white, chess matches typically consist of an even number of games with the players alternating white and black.  Most major chess championships involve some moderately large (6, 8, or more) games.  For instance, Kasparov played 6 games against Deep Blue in both of their matches (in 1996 and the rematch in 1997).[/quote]
For me, the world of chess still seems to hold many secrets. :maybe:   White has an advantage?
But... what if both parties win half of the games?  Isn't that a quitely likely problem with an even number of games? :huh:[/quote]

Umm. I've only tried chess since we have like 3 boards here so....white has an advantage since it starts first, right?

I voted Topalov since I do not like Kramnik much.

I think now Topalov is first, Anand is second and Kramnik third in FIDE ratings.

White has an advantage?  :huh: I know many players who are more comfortable with the black pieces.

[quote="Koda"][quote="SplashD"][quote="Koda"]
White has an advantage?  :huh: I know many players who are more comfortable with the black pieces.
[/quote]

I think that's because if you're black, you can counter your opponent's moves instead of attacking.[/quote]

Oh really? So you think in something like, say, the Sicillian, white is the only one who attacks and black sits back and defends? Take a walk!
[/quote]

black could frustrate white with their defenses and then maybe white will make a mistake and black could capitalize on it, it's just a thought, sheesh

It seems Kramnik has won, pity... Preferred Topalov's bolder style

Daniel

There you go... Topa was better... But he lost.

[quote="dhernandez"][quote="Koda"]
White has an advantage?  :huh: I know many players who are more comfortable with the black pieces.[/quote]

Well, not likely professional players. White has the advantage in chess. This is simply statistics. Players playing  white, win more often than players with black. Actually in the first compasses of any game of chess, the white player has what is called, the initiative. In practice this means that white can afford some careless moves without actually falling in a lost position. Black must in turn play very accurately if he wants not to end in a very constrained position or fall victim of a ruthless attack. Nevertheless, the theory of openings (the first moves in any game of chess for newbies) is more developed each day and the main variants are studied beyond the 20th move for both players. So usually, a professional player [i]knows already by heart[/i] which ones are the best moves in the first fifteen or so in order to achieve [i]equality[/i] at a certain point in the game. From that point on, players are on their own and the game becomes more interesting.

As a curiosity, the question of the draws have been a recurrent one in chess championships and the rules have varied almost every time. There have even been matches along chess history where it has been suggested thar the challenger should win by more than one game, that is, a 2-3 result for challenger was no good, the champion retained the title. Kasparov managed to retain the title against Karpov in Sevilla by winning the very last game with black (he was losing by one game).

Daniel[/quote]

If white has the initiative, it does not mean that he can afford careless moves. That is a very basic piece of chess lesson.

And as for the statistics, they only say that more players are comfortable with white. There are players who have won more with black than white.

And coming back to Sicillian, it is very difficult to assert with certainty who has the initiative, especially if black goes for a (good) novelty early on.

I am horrible at chess. I am in 8th grade and have a rating of 1255 USCF. I haven't been in a chess tournament for like 1 year, and I have no idea how to improve since I can't get a coach. Besides I like math a lot more.


On page 126 of Introduction to Geometry, there is a section that says "Thing To Watch Out For!"
It says that you cannot assume $ \bigtriangleup ABC$~ $ \bigtriangleup ADE$ or that $ \bigtriangleup JKL$~ $ \bigtriangleup BLM$. I don't understand why you can't assume either.

The only things I thought of were
1. line AC and AE are not the same line (notice I said line and not line segment)
2. line AB and line AD are not the same line
3. line JL and JM are not the same line
4. KL and KN are not the same line...

Otherwise, I dont' see why the triangles aren't similar

Attached is my recreation of the triangles I'm referring to.

Also, what program do you recommend that I can get for free to draw prettier geometry pictures with?

The book is probably asking you to prove those similarities, though I'm not sure as you didn't post the problem and I dont have my copy of the book with me...

And my favorite geometry program is CaRMetal. It's awesome [i]and[/i] open-source (=free  :D  :D )

You misread what the book wrote.  For the first diagram, the book says that the following argument is incorrect: $ \triangle ABC\sim\triangle ADE$, so $ AB / BD \equal{} BC / DE$.  The book is objecting to the proportion conclusion, not to the similarity condition.

I think I get it. BD is not a corresponding side like ED is to BC so AB / BD definitely does not equal BC / DE.


This might have already been posted but yea

If $ p$ is a prime $ \ge 5$, prove that $ p^{2}$ is divisible by $ 24$.

My proof for it was pretty long and convoluted so let's see what you guys come up with... xP

Hehe ... how about $ p^{2}-1$, instead?

[hide="Hint"]Consider the possible values of $ p$, and thus $ p^{2}$, in some specific mods.  [hide="Like, for example"]mod 6 and 8.[/hide][/hide]

This has probably appeared on the forums plenty times before, incidentally.

see my post here: http://www.mathlinks.ro/Forum/viewtopic.php?t=156110


Does anyone know any good, free tutorials on learning C++?

This golden link will save your 40$\$$  for a book.

Enjoy! :)

[hide]http://newdata.box.sk/bx/c/index.htm[/hide]

[hide][quote="math92"]thanks alot. :)
Where can i find some free software to run C++ programs?[/quote]

I recommend you use devc++ or MINGW

make a google search ,both are free[/hide]

you could get Visual C++ 2005 Express Edition for free from Microsoft (it's a download on their site).

you could also use gcc, but that takes longer to learn, and is probably more complicated for new users.


topology
complex analysis
real analysis
real analysis unsolved

Let $ U$ and $ V$ be two open homeomorphic subsets of $ \mathbb{C}$. Under what condition (on $ U$ and $ V$) does there exist a homeomorphism $ f$ between $ U$ and $ V$ which extends to an homeomorphism between $ \overline{U}$ and $ \overline{V}$?

*Edit: I forgot [i]open[/i]*

Can you apply Riemann's mapping theorem?

Doesn't this theorem apply to simply connected open sets?

Right.  :) They are biholomorphically isomorphic to the unit disk D. I just thought - without really thinking more - that then the boundary of U and V looks like the boundary of D. ... but this is just some wild guess  :P Maybe it helps.

Well, actually, this question is motived by the following fact:
If $ U$ is bi-holomorphic (via $ f$) to the open unit disque $ D(0,1)$ and if $ U$ has a simple boundary, then $ f$ can be extended to a homeomorphism between $ \overline{U}$ and $ \overline{D(0,1)}$.

Edit: typo corrected

Ok, then my guess wasn't too bad. ;) What does mean "simple boundary" in this context?

A point $ x$ of the boundary of $ U$ is called [i]simple[/i] if for every sequence $ (x_n)$ (in $ U$) tending to $ x$ there exists a continuous curve in $ U$ passing by all the points $ x_n$ and ending in $ x$. $ U$ has simple boundary if all boundary points are simple. For instance, the open unit disc without a radius has not a simple boundary (but is homeomorphic to $ D(0,1)$ and any homeomorphism between these two cannot be extended).


Prove the inequality which a,b,c are positive numbers:
$\frac{a}{a+b})^{2}+(\frac{b}{b+c})^{2}+(\frac{c}{c+a})^{2}\geq \frac{3}{4}$

I've post it here long time ago! If you want, i'' post my solution
P/s :It is ChinaTST 2005 with some substitutions and $xyz1=1$ where $xyz=1$ :D

inq equalvelent to :Prove that $\sum_{cyc}\frac{1}{1+x}\geq \frac{3}{4}$ where $x= \frac{b}{a}, y=\frac{c}{b}, z=\frac{a}{c}$ or $xyz=1$ .Suppose that  $xy\geq 1$ then $\frac{1}{1+x}+\frac{1}{1+y}\geq \frac{1}{1+xy}$
  so we have  $\sum_{cyc}\frac{1}{1+x}\geq  \frac{1}{1+xy}+\frac{1}{z}=\frac{z}{1+z}+\frac{1}{(1+z)^{2}}\geq  \frac{3}{4}$ as desired


I have an equation array that is quite big. I starts halfway in the middle of a page, but is long enough that it needs to have a few of the last lines on the next page. When I compile, however, LaTex is keeping the whole eqnarray together and starting a new page with it. So I have a blank half page, which does not look very good.

How do I make it so that LaTex just continues the array in the next page, putting some of the lines on the first page, and some on the next.

thanks,
brian

Try putting the \nopagebreak command just before the eqnarray code starts.

i tried it, but im afraid it did not work.
any other suggestions?

Have a look at the packages discussed in The TeX FAQ at http://www.tex.ac.uk/cgi-bin/texfaq2html?label=longtab

You can also make the table into a float. That is, if there's room on the page it will put the table there, if not, the text will continue (so not leaving a blank half page) and the table is started on the next page - exactly as happens with figures.

To do this you use
\begin{table}
Put header text here
Put table here
Put footer text here
\end{table]


geometry
3D geometry
sphere
rotation
calculus
integration
analytic geometry

Hello everybody, my name's Michael, and this is my first post here!

I'm trying to find the (hyper?)volume of a hypersphere, a sphere rotated along a fourth spatial axis.  I am going by the same method I used to derive the volume of a sphere, by simplifying the double integral by replacing it with the volume of a sphere.

radius is 3 (cuz i like 3 and it's small enough to not create large numbers but large enough to avoid trivialities or misunderstandings in formulas that you can get with 1 and 2).

I did it on a calculator and got (81 :pi: ^2)/2, or the general equation r^4/2
Erm... Why don't I see an integral sign in the emoticons? Should there be one?

So here's the integral I came up with, and how I tried to solve it.

(integral) 4 :pi: /3 *  :sqrt: (r^2-x^2)^3 dx,  from -3 to 3

Since I've got no idea how to tackle that in that form, I hit the tables and after a few hours of critical thinking, I had a Eureka moment and expressed it as this:

4 :pi: /3    (integral) (r^2-x^2)^2/ :sqrt: (r^2-x^2) dx, from -3 to 3

since the value squared over the value sqrt'd is the same

then I foiled it and then the first part is just arcsin x/r * r^4
the next part w/ the -2r^2*x^2 I found in my calc book for if you have u^2 over :sqrt: a^2-u^2, so I used that, which gave me:

r^2(x :sqrt: (r^2-x^2) - r^2*arcsin x/r)

Then the third part I tried u-substitution w/ u equaling x^2, which makes

(integral) x^4/ :sqrt: r^2-x^2 dx

go to

(integral) u^(3/2) / 2 :sqrt: r^2-u du, which has another table property for u^n / :sqrt: a + bu

it gives: 1/2(2/-4(u^(3/2)* :sqrt: (r^2-u) - 3r^2 (integral) :sqrt: u  / :sqrt: r^2-u  du   (didn't bother redoing limits, cuz i saw a good opportunity w/ this, or so I thought)

So with the integral part, if you plug back in x = :sqrt: u and du = 2xdx, it looks kinda nice, but I can't figure that one out.  Any suggestions from here, or alternate routes from the start?

I have several suggestions, here, any of which could be extended to the more general problem of the $n$-volume of the $n$-ball. One is to integrate a Guassian - that is, $e^{-(x^2+y^2+z^2+w^2)}$ - in the 4-dimensional analogue of spherical coordinates.

Here's another. If $B=\{(x_1,x_2,x_3,x_4) : x_1^2+x_2^2+x_3^2+x_4^2<R^2\}$, we are trying to compute

$V=\int\int\int\int_B1\,dx_1dx_2dx_3dx_4$

Fix $x_1$ and $x_2$ and convert them to polar coordinates, so that $x_1^2+x_2^2=r^2.$ Then $x_3^2+x_4^2<R^2-r^2.$ If we integrate over $x_3$ and $x_4$ we get the area of a circle of radius $\sqrt{R^2-r^2}$, namely $\pi(R^2-r^2).$

Now integrate over $x_1$ and $x_2$, converted to polar coordinates:

$V=\int_0^R\int_0^{2\pi}\pi(R^2-r^2)r\,d\theta\,dr = 2\pi^2\int_0^R(R^2- r^2)r\,dr=\frac{\pi^2R^4}2$

This method (slicing off two dimensions at a time) can be used to give an inductive computation of the $n$-volume of the $n$-ball.

If $V_n$ is the $n$ dimensional measure of the unit ball in $\mathbb{R}^n,$ then by this method:

$V_n=\frac{2\pi}n\cdot V_{n-2}.$

Since we know that $V_1=2$ (or if you prefer, $V_3=\frac{4\pi}3$) and that $V_2=\pi,$ we can use this recursion to determine $V_n$ for any $n.$

I'll get right on it once I learn about this stuff in Calc BC.

Perhaps I'm getting too far ahead of myself...

I apologize if you were being sarcastic but this stuff is way beyond the Calculus BC level. BC is only equivalent to a full year of calculus in a single variable. Thus, you only do differentiation and integration in 2 dimensions (xy plane). The problem in this thread deals with methods of multivariable calculus far beyond my realm of understanding. Hopefully i will work my way up to that level next year, but i doubt it.


there's no choice for $\displaystyle\text{\LaTeX}$

oops, I forgot, but that's a too common language in this forum :D

lol,
well, if you search programming languages in Wikipedia, you would find Latex
[url]http://en.wikipedia.org/wiki/Alphabetical_list_of_programming_languages[/url]

[quote="furious"]i was just kidding. it's not a real programming language.[/quote]Of course it is a programming language. Most of the people don't use the programming in $\LaTeX$, but you can do a lot of stuff there.

By the way, this topic is quite old in the forum, you can look at it (with a lot more options and variety) in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15636]this post[/url].

[quote="furious"]i was just kidding. it's not a real programming language.[/quote]

Well, TeX is turing-complete, so I don't see why it shouldn't be considered one. (Although I personally can't remember other programs the docstrip being implemented in TeX. I have often toyed with the idea of creating an X-SAMPA parser in TeX.)

click on the link that says LaTeX on the left side of this page. there is a lot of good information that brings you a long way in knowing the language. it's very helpful

I have completely the art of javascript.  But I don't think web languages count.

I can program on TI calculator(Basic). I have made some games.
I'm also going to take programming next year.

I know Basic, VB, C++, PHP, and Flash Actionscript.

I know C, C++, and Java, to different extents.  However, C++ is my favorite and is the language that I use the most for computing contests and general programming.

um...there's no choice for qbasic...it's a newbie language.  I've just barely started programming... :blush:

I started out with QB when i was really young, but moved to visual basic. Then I got my calculator, and started making a bunch of programs that together could solve any problem in our Algebra I class. Thats why we couldnt use our own calcs for our final.  :P

I know python, scheme, haskell, and I'm learning java.

I know Python, but I picked JAVA in the poll, since it appears to be the closest option to Python in the poll.

[quote=IPERFERPI]This poll only apply to people who know programming. I know you may know more than one, but please chose the one you know the best
just want to see how many people actually knows programming :D[/quote]

Python

Moderately advanced python. Been a while, though.

Visual C++, C# (not included on the list), some JavaScript and Python. That's pretty much it (oh yes, and some LaTeX as well).

c++, python, js, java

most comfortable with using js

actually html and css but i absolutely suck at css

Python up to ML. Java up to APSCA 5 level. The website languages up to creating a basic functional form-based website.


For positive integers n, define $ S_n$ to be the minimum value of the sum
\[ \displaystyle\sum_{k = 1}^n \sqrt {(2k - 1)^2 + a^2_k}
\]
as the $ a_1, a_2,...,a_n$ range through all positive values such that
\[ a_1 + a_2 + ... + a_n = 17.
\]
Find $ S{}_1{}_0$.

This is either an AMC or an AIME problem and is definitely in the contest archives; check there first.

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=246218[/url]


function
inequalities
trigonometry
real analysis
real analysis unsolved

Can you find a real valued function $ f$ satisfying a uniform Lipschitz condition of order $ 0<\alpha<1$ on the interval $ [0,1]$ such that $ f$ is NOT of bounded variation on $ [0,1]$. A real valued function $ f$ is said to satisfy a uniform Lipschitz condition of order $ \alpha>0$ on $ [a,b]$, if there exists $ M>0$, such that for any $ x\not\equal{}y\in[a,b]$, the inequality $ |f(x)\minus{}f(y)|<M|x\minus{}y|^\alpha$ holds. This is one of the Apostol's [i]Mathematical Analysis[/i]. Thanks.

For $ \alpha\le \frac12$ the example $ f(x) \equal{} x^{2\alpha}\sin\left(\frac1x\right)$ (with $ f(0)\equal{}0$) should work.

We'll have to work a little harder than that for $ \frac12 < \alpha < 1.$

OK, same basic idea, but a little more general.

For $ k\in\mathbb{N},$ consider the points $ \left(k^{1 \minus{} \frac1{\alpha}},(\minus{}1)^kk^{ \minus{} 1}\right).$

Let $ f(x)$ be the piecewise linear continuous function with these nodes, and extend it to be continuous at $ 0.$ (Hence $ f(0) \equal{} 0.$)

Look at the linear segment from the $ k$th point to the $ (k \minus{} 1)$st point. We have $ \Delta x\approx c\cdot k^{ \minus{} 1/\alpha}$ and $ \Delta y\approx 2k^{ \minus{} 1}.$ That gives us $ \Delta y\approx c\cdot(\Delta x)^{\alpha},$ which means that this satisfies the Lipschitz condition.

We should also check that with zero at one endpoint. Then the required condition is that $ k^{ \minus{} 1}\le c\cdot k^{\alpha \minus{} 1},$ which is true since $ \alpha > 0.$

But the total variation is the sum $ \sum_k\Delta y$ for the approximations given above, and that diverges by comparison to the harmonic series.


algorithm
advanced fields
advanced fields unsolved

Find the the first seven terms of the sequence {a_n}={3+(-1)^n}. ;)

What is this doing here? (The sequence is 4, 2, 4, 2, 4, 2, 4, ...)

No actually the answer is 2,4,2,4... not 4,2,4,2..., but good job though. I calculated the answer of yours, but it just didn't  fit in, so i did it again and this is how i did it for the first four terms:
3+(-1)^1,  3+(-1)^2,  3+(-1)^3, 3+(-1)^4,...
2,4,2,4

Depends, I let $n$ start at 0 and then my answer is correct. Doesn't matter!!

But this question should go to one of the Middle School forums or something like that.

More generally, I don't understand your posts.  Do you know the answers to all of the questions you're asking, or are you just randomly asking them?  If it's the latter, I might want to consider changing my posting habits, because it seems that most people are not interested in your problems.  Finally, if you want to elicit more answers to your problems, make an effort to explain some of the technical terms contained within - for example, the 196 algorithm doesn't mean anything to me.

Now, I could easily look it up on mathworld, but you should probably do that in your posts to save us all some trouble.

Your recent posts 'long answer problem' and 'fundamental problem' are examples of posts which make absolutely no sense unless you explan some of the terminology.  For example, find the values for which Z=22 is contained in the unit square?  That doesn't make any sense.


Hello friends,
I am studying in 10th class. Actually I have a question and I’m unable to solve this question. My question is: How can we find the square root of a number by hand? How about cube roots? If anybody can solve my question I will grateful. Thanks in advance!

Look up square root algorithm. You'll find more than enough info.

I like this way, it's a tad complex, but do-able by hand

[url]http://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Digit_by_digit_calculation[/url]


trigonometry
calculus
derivative
function
inequalities unsolved
inequalities

Let $0 < \alpha, \beta, \gamma < \frac{\pi}{2}$ and $\sin^{3} \alpha + \sin^{3} \beta + \sin^3 \gamma = 1$. Prove that
\[ \tan^{2} \alpha + \tan^{2} \beta + \tan^{2} \gamma \geq \frac{3 \sqrt{3}}{2} . \]

has anyone have an idea what to do here? Leo Telosa, my team mate from the Philippines and i had the same idea/ draft solution. But this idea/ draft solution of ours was something we invented and had a few assumptions i'm sure we're not allowed to make. I just want to have a starting point for this thing. Btw, the only training in Trig i had was the day before the contest, make that the night before the contest... whatta crammer. Anyway, what we did was since the equation we had to prove had 3 addends, we decided to prove that each addend was supposed to be greater than or equal to sin60 so that when we add the three, we get the result. Yeah, it's a lousy solution, definitely wrong. So, basically i need help. nyehe...

I think we have discussed this type of problems several times on the forum. The idea is to write that $\frac{x}{1-x^2}\geq Ax^2$ where, of course, the constant $A$ has the marvelous property that we have equality exactly when $x^2+y^2+z^2=1$.

[quote="indybar"][quote="_el_doopa"]
$(x(1-x^2))'=1-3x^2=0$
just chek:
$x={\sqrt{3} \over 3}$

and we are done...[/quote]

What do you mean?[/quote]
have you learnt calculus?
actually he means that the extremum is found when equating first derivative to 0. Second derivative test shows that the function attains its maximum when $x=\frac{\sqrt{3}}{3}$. So we only need to check x for that value to prove that inequality.

[quote="indybar"]Second deriv. is $-6x$, right?[/quote]
yes.

[quote="zhaoli"]Let $0 < \alpha, \beta, \gamma < \frac{\pi}{2}$ and $\sin^{3} \alpha + \sin^{3} \beta + \sin^3 \gamma = 1$.
Prove that $\tan^{2} \alpha + \tan^{2} \beta + \tan^{2} \gamma \geq \frac{3 \sqrt{3}}{2}$.[/quote]
let $x=\sin \alpha$,and $y,z$ similar.
then $x^3+y^3+z^3=1$
we should to prove:$\frac{x^2}{1-x^2}+\frac{y^2}{1-y^2}+\frac{z^2}{1-z^2} \geq \frac{3 \sqrt{3}}{2}$
which is equal to $\frac{x^3}{x(1-x^2)}+\frac{y^3}{y(1-y^2)}+\frac{z^3}{z(1-z^2)} \geq \frac{3 \sqrt{3}}{2}$
it is easy by $2x^2(1-x^2)^2 \le (\frac{2x^2+2-2x^2}3)^3$

but I think the equality can't be hold...
so what is the minimum??

I think that the minimum is about 2.598 by using the Jensen's inequality.


A penny, a nickel, a dime and a quarter are tossed simultaneously. What is the probability that at least two heads show and one of the heads showing is the dime? Express your answer as a common fraction.

This could be worked out by just looking through all the cases. There are only $ 2^4\equal{}16$ cases.

But:
[hide]Let the asked probability be A.

B = "dime shows head"
C = "at least two heads show"

A= B and C

Two things must happen at the same time. First, the dime must show heads and the others must also sum up at  least one head. These events are not independent, so we can just multiply the probabilities together.

P(B)=1/2
P(C|B)=? (that means, what is the probability of C if B has already happened.)

If the dime is already head, then the question becomes: what is the probability that the remaining three show at least one head all together. P(C|B)=$ \frac{{3 \choose {3}} \plus{} {3 \choose 2} \plus{}{3 \choose 1}}{2^3}\equal{}\frac{7}{8}$

So, $ P(A)\equal{}P(B)*P(C|B)\equal{}\frac{1}{2}*\frac{7}{8}\equal{}\frac{7}{16}$
[/hide]


Let $f(x) = 3x + 2.$ Prove that there exists $m \in \mathbb{N}$ such that $f^{100}(m)$ is divisible by $1988$.

It's easy to show that $f^n(x)=9^nx+2\sum_{i=0}^{n-1}3^i=9^nx+3^n=3^n(3^nx+1)$ (for instance by induction). Because $gcd(3^{100},1988)=1$ thus taking $x=-(3^{100})^{-1}$ $(\mod 1988)$ we get infinitely many such $m$.

I think you missed a $-1$ in the computation, so that in fact $f^n(x) = 3^n(x+1)-1.$
Anyway, the idea is the same by choosing $x = -1 + 3^{-100} \mod [1988].$

Pierre.

I used the obvious lemma:
$a, b, c$ are positive integers, if $(a, b) = 1$, then $am + c$ (mod $b$) are all distinct for $m = 1, 2, \ldots, b$.

Thus, let $m = 1, 2, \ldots, 1988$, which are all distinct (mod $1988$), since 3 is prime, $(3, 1988) = 1$, then using the lemma. We have $f(1), f(2), \ldots, f(1988)$ are all distinct (mod $1988$), that is a permutation of $1, 2, 3, ..., 1988$.

Repeat above disscusion 100 times, then the conclusion follows easily:
Among $f^{100}(1), f^{100}(2), \ldots, f^{100}(1988)$, just one value is divisible by 1988.

By the way, from the above solution, we can say the minimum value of $m$ satisfying the condition is no larger than 1988.

It isn't a particularly hard problem but that's a very beautiful way to do it.  It looks to me like the other answers get a solution < 1989 as well.


Prove that there is a sequence of integer numbers a(n) for which:
1) a(n) -> +oo and
2) the sequence cos(a(n)+t) is uniformly convergent for each t in [-1,1].
Also, prove that limcos(a(n)+cos(a(n)+cos(a(n)))) exists in [-1,1]

Please guys, I'd be grateful to anyone helping...

Hint: The sequence $ 1, 2, 3, ...$, taken $ \mod 2\pi$ is dense in $ [0, 2\pi)$.

i think $ a_n\equal{}2\pi n\plus{}\frac{1}{n}$ is satisfies the conditions..

[quote="anik_andrew"]i think $ a_n \equal{} 2\pi n \plus{} \frac {1}{n}$ is satisfies the conditions..[/quote]

Except $ a_n$ was to be an integer sequence.

Thanks a lot WWW, I think I have something to start now...  :)

[quote="Ricko Rickito"]Unfortunately I didn't manage to do much with it... could you give some other hint?[/quote]

First, note that your 2) doesn't make sense as stated. It should read $ \cos(a(n)\plus{}t)$ converges uniformly on $ R$. Second, it follows from what I said that we can find positive integers $ a_n \equal{} 2\pi m_n \plus{} \epsilon_n$, where $ m_n$ is an increasing sequence of positive integers and $ \epsilon_n \rightarrow 0$.


Suppose that $ a,b,c$ are sidelengths of an acute triangle. Prove the inequality
\[ ab(a+b)^{2}+bc(b+c)^{2}+ca(c+a)^{2}\geq a^{4}+b^{4}+c^{4}+3abc(a+b+c). \]

turned out very nice
[hide]
one liner factorization
assume $ c\geq b\geq a$
the inequality is equivalent to
$ (c^{2}+2c(a+b)-(a+b)^{2})(a-b)^{2}+(a^{2}+b^{2}-c^{2})(c-a)(c-b)\geq 0$ which is true
equalit iff $ a=b=c$ or $ c=a\sqrt{2}=b\sqrt{2}$ and cyclic cases.
[/hide]


[quote="epitomy01"]find all functions $f$ from the reals to the reals  s.t. $f(x) f(y) = f(xy)$.[/quote]
I think this problem can't be solve without any other coditions,may you have forgotton to write that this function is increasing or continiue?

Well it can be solved...
By some stuff we get that $f(x)=0 , x \neq 0$ implies $f \equiv 0$.
Similar, when $f(0) \neq 0$, then $f \equiv 1$.
Thus lets neglect that, so lets assume $f(x) =0 \iff x=0$.
Now you could do the same as for $g(x+y)=g(x)+g(y)$:
Take a base $B$ of $(\mathbb{R}^+ , \cdot )$ and define it arbitrary $\neq 0$ on the elements of $B$. Then also define $f(-1) \in \{\pm 1\}$ arbirary, and all those give solutions by continuing to all nonzero reals.

I need help, please reply, i need to no how many kilolitres there are in 43.541 Cubic litres squared

the question is

It is known that 1000 litres of water occupie 1 cubic meter.  How many kilolitres of water are there in the silo

there are 43.541 Cubic Litres squared  of water in the silo.... please reply....


geometry
geometric transformation
articles
conics

Hey guys,

After the AWESOME feedbacks I got from valedictorian speech topic, I came here again to ask for advices. You see, I'll be probably taking my placement test at my college (Washington and Jefferson) and well, I'm really nervous about them.  I'm not worried about math because obviously, everyone here is a phenomenal mathlete (I just like it  :roll: ) so I'm not worrying about it. Now, I have to take one in Spanish... Well, my high school Spanish (4 years) was not intense at all and I'm afraid that I won't do so good on it. The thing is this:

If I do bad, I end up in Spanish 105 and take Spanish 106 on second semester.

If I do semi-bad (or good), I end up in Spanish 106 and take an Intermediate level class Spanish 207 on second semester.

If I do well, I'll just take one semester of Spanish 207.

I'm trying to place on Spanish 207 but I heard it's very hard. But I got 100s on my Spanish classes (I mean, they have to mean something) and well, I wonder how I can study/review for them. Any good suggestion?

I'm in something vaguely like your situation-- I have to review a language I've mostly forgotten so that I can get around this summer, and then pass a translation exam in the fall.

What I'm planning to do is this (though not necessarily in this order):

Grammar:  Read through the textbooks I used at the time, along with a laminated quick grammar fact sheet, make an outline of the grammar that I need to know, figure out what I don't know, and review what I don't remember.

Vocabulary:  Reading through the texts and vocabulary lists in the textbook helps, but since I'm re-learning this language to travel in a foreign country, I try to think through the situations I'll encounter when I'm there (e.g., taking the train) and then think through the vocabulary I'll need to know to do that (e.g., what's the word for train, ticket, platform, departure, arrival, board...)

Speaking:  If you can find a native speaker to practice with, that is the best thing to do.  Since I can't, I try to "think" in the language as much as I can, and then fill in gaps in my knowledge as I encounter them.  I guess you could try holding conversations with yourself, but I can't bring myself to do that because I feel too stupid!

Listening:  I don't know.  I don't really have any resources at my disposal other than movies where everyone talks to fast and uses too many words I don't know!  Since I'm going to be talking with people instead of taking a test, I can always ask them to please say that again, but slower this time!   :oops:

Reading:  Find something you're interested in reading in that language and read it.  I have a stash of science magazines in various foreign languages for this!

These are more strategies for reaching a functional level with the language than taking a particular test.  For that, you should probably ask the testing center if they have a syllabus of the material covered on the test (or in the classes you're trying to place out of), and follow that for the specific topics you need to know.

Thanks.

I don't think I can ask the college though because um, they probably will not reveal how. I do need to ask if I need to know about listening part though.

Hmm...  I like the idea of reading article in Spanish. That would be actually pretty good thing (since I'll be used to new words that I might need to know and yet have not learned.. also, doing that lets me to face the real-world Spanish conjugation)! Any specific idea to which magazine?

And more advices will be great...

Make sure you can identify which tenses are which with no effort at all. I'm only in Spanish III, and we tripled our tense knowledge this year alone (from two or three at the beginning to about seven or eight). This will definately help you understand reading portions and even writing portions of the placement test.

Also, make sure to study the irregular forms of the tenses if you haven't had them memorized yet.

My old spanish teacher recommended that you watch a spanish TV channel for a while. (Place the closed captions on so you can read too). By listening (hopefully they won't go too fast) and reading you can improve your ability to identify certain words/phrases.

Hope this helped  :).

[quote]I don't think I can ask the college though because um, they probably will not reveal how.[/quote]

Some information about the content may be available with the placement test (for example, at my university, they give you a list of topics covered on the test).  The courses themselves may have course websites, which may give syllabi or homework assignments or some other indication as to the level of the course.  Alternately, you can try to find out (from the college bookstore?) which books are used in the courses you're interested in, and that should give you an idea of the level of difficulty and what's expected of you.  Figuring these things out often takes some creativity    :roll:

[quote]Any specific idea to which magazine? [/quote]

Not really.  I travel internationally quite a bit, so I just buy these things at bookstores and airport newsstands.  I know that there is a German edition of Scientific American, so I looked around to see if there's a Spanish version, and there is, called "Investigacion y Ciencia":  [url]http://www.investigacionyciencia.es/[/url]  Maybe you could start there.

I'd be a little cautious about practicing reading from any science type magazine. The vocab is probably skewed from what you will be tested on. I had to do this when I entered college and did well. My prep for trying to think and quickly conjugating was to look at either comics from the paper or just online comics and make up stories about the pictures (ignore the actual words). Just tell the story in Spanish, or write it down so that you can critique it later (or maybe even post it here).

As for reading...hmmm...I'm trying to think if there were easy things from my Spanish literature course. I guess if you want a magazine, try People or something that is more general. [url=http://www.ciudadseva.com/textos/cuentos/esp/ggm/dia.htm]This[/url] is a famous short story by Garcia Marquez. It might be tough going, but lots of important words (and a bit of culture, too!).

For listening. Almost any DVD will have an option for Spanish. Usually you can do both in Spanish and with subtitles. Of course, being able to catch everything someone says in a movie is probably way more than you'll need to do. I still can't do it, but I aced the listening section.

Most importantly would be to not panic during the test if you feel like you are doing poorly. They have to test for all levels. If it is anything like ours, then you can miss tons (they expect non-native speakers to) and still get placed really high. The mental game of feeling like you are doing horribly messes a lot of people who aren't expecting it up.

Edit: Also, if you have a 501 verbs book, the newer ones have quizzes for grammar (with answers) in the back. These are amazing. There are tons of practice questions to see if you know common phrases and grammar.

For a placement test, don't be dishonest with yourself. If you study hard and do better than you "should", all it means is that you'll be placed in a class you'll probably do badly in since you're not prepared. Recall what you learned, and let that place you where it will.


Dear All My Friends,

How are conditions of positive real $ a$, $ b$, $ c$ such that system:
$ a^{2}\plus{}x^{2}\equal{}b^{2}\plus{}y^{2}\equal{}(x\minus{}y)^{2}\plus{}c^{2}$
has positive real solutions $ x$, $ y$?

Best regards,
Bui Quang Tuan

[quote="Quang Tuan Bui"]Dear All My Friends,

How are conditions of positive real $ a$, $ b$, $ c$ such that system:
$ a^{2} \plus{} x^{2} \equal{} b^{2} \plus{} y^{2} \equal{} (x \minus{} y)^{2} \plus{} c^{2}$
has positive real solutions $ x$, $ y$?

Best regards,
Bui Quang Tuan[/quote]
$ c\equal{}a\plus{}b.$
Try a geometric way. :wink:

Is this an inequality?
I wouldn't say so.
It would be better placed in Algebra.

[quote="micf"]arqady, can you show me the geometric way?[/quote]
Thank you, micf! I solved another problem:
Let $ a,$ $ b,$ $ c,$ $ x$ and $ y$ are positive numbers such that $ ay\plus{}bx\plus{}\sqrt3(ab\minus{}xy)\equal{}0$ and
$ a^{2}\plus{}x^{2}\equal{}b^{2}\plus{}y^{2}\equal{}(x\minus{}y)^{2}\plus{}c^{2}.$ Prove that $ c\equal{}a\plus{}b.$

I think the condition to be

$ a<c\ \vee \ b<c.$

[quote="Quang Tuan Bui"]
How are conditions of positive real $ a$, $ b$, $ c$ such that system:
$ a^{2} \plus{} x^{2} \equal{} b^{2} \plus{} y^{2} \equal{} (x \minus{} y)^{2} \plus{} c^{2}$
has positive real solutions $ x$, $ y$?
[/quote]

Suppose $ z\equal{}x\plus{}y$ then z is also positive real. The system is now:
$ a^{2}\plus{}(y\minus{}z)^{2}\equal{}b^{2}\plus{}(z\minus{}x)^{2}\equal{}c^{2}\plus{}(x\minus{}y)^{2}$
How are the solution conditions now?


group theory
abstract algebra
superior algebra
superior algebra unsolved

Prove that a group of order $ 255\equal{}3\times 5\times 17$ is commutative.

Hint:

$ G \cong G_{85} \times G_3 \cong \mathbb{Z}_{85} \oplus \mathbb{Z}_3 \cong \mathbb{Z}_{255}$, and so it is cyclic. Sylow subgroups will come in useful.

Yes, with Sylow theorems I've proved that the 17-sylow subgroup is normal, but I cant do the same with the 3-group and the 5-group... And if i can prove that, i can do the rest...

Hum... I was trying to solve this with an elementary approach, as it is in the book, through the Sylow theorems.

But that theorem is a very nice result.

There is a normal Sylow 17-group, call it $ P$.  Let $ Q$ be a Sylow 5-group, then $ PQ$ is a subgroup of index 3 in your group.  Since 3 is the smallest prime that divides 255, $ PQ$ is normal in your group. $ PQ$ is cyclic of order 85, and 3 does not divide $ |Aut(PQ)|$.

Steve

:?: Why does $ 3$ have to divide $ |Aut(PQ)|$?

Let $ R$ be a Sylow 3-group.  Up until that point, we've shown our group is a semidirect product $ (PQ) \rtimes R$.  Since $ Aut(PQ)$ has no element of order 3, $ R$ must centralize $ PQ$, so $ R$ is normal, and we have a direct product.

Steve

[quote="coquitao"]- In fact it is a well-known result from the theory that a group or order $ n$ is cyclic if and only if $ (\varphi(n),n) \equal{} 1$:

[url]http://yiminge.wordpress.com/2009/01/22/all-groups-of-order-n-are-cyclic-iff/[/url][/quote]
It should be "every group of order", not a group of order. For example $ (\varphi (4),4) \equal{} 2$ but there do exists a cyclic group of order $ 4$. I was wondering why the link says "finite group of order n" as every group of finite order is finite.


Prove that exist infinity solutions in the positive integers to the equation:
$2x^{2}+3y^{2}-5z^{2}=1997$

There are infinetely many solutions with x=31. It give Pells equation $y^{2}-15t^{2}=25=5^{2}$ (z=3t). Solution of $y^{2}-15t^{2}=1$ are $y+t\sqrt{15}=(4+\sqrt{15})^{k}$, therefore solutions are
$x=31, y+\frac{z}{3}\sqrt{15}=5(4+\sqrt{15})^{k}.$


How can I prove that $ \infty$ is unstable for the following ODE's:

1: $ \frac {dz}{dt} \equal{} \minus{} z \plus{} x^2$

2:$ \frac {dx}{dt} \equal{} y$

Is this even the right forum?  :oops:

Thanks.

[quote="JRav"]Are $ z$ and $ y$ functions of $ x$, or just $ t$?  A nice method is to plot the equation and observe what it does around its critical points.  If the graph is positive, then draw arrows to the right.  If it is negative, draw arrows to the left.  It is stable where arrows converge and unstable where they diverge.[/quote]

Can you show some kind of example so I will have a starting point of how to do these kind of problems?  :)