geometry
circumcircle
inradius
incenter
trigonometry
ratio
geometric transformation

Hi,

Can someone tell me or direct me to any proofs of the three relationships at the bottom of the page: http://mathworld.wolfram.com/Orthocenter.html

Many thanks!

PS. I would prefer easy-to-follow proofs, as I just learned about isogonal conjugates and such yesterday :)

It's always better to post the problem itself rather than just a link to the problem. The three relationships in the form as provided in MathWorld are

$a^2+b^2+c^2+AH^2+BH^2+CH^2 = 12R^2$;
$AH+BH+CH=2\left(r+R\right)$;
$AH^2+BH^2+CH^2=4R^2-4Rr$,

where ABC is an arbitrary triangle, H is its orthocenter, R is its circumradius and r is its inradius. But actually, the second of these three relationships is not absolutely correct - in the form given, it holds only for acute-angled triangles ABC -, and the third relationship is wrong - it actually should be

$AH^2+BH^2+CH^2=4R^2-4Rt$,

where t is the inradius of the orthic triangle of triangle ABC, and it is also true only for acute-angled triangles ABC.

I've got no time to write detailed proofs of these formulas now, but at least the first one immediately follows from the equations $a^2+AH^2=4R^2$, $b^2+BH^2=4R^2$, $c^2+CH^2=4R^2$, which are quite easy to prove. The second one (for acute triangles ABC) is a consequence of Carnot's theorem. I'm leaving the complete proofs to other mathlinkers.

  Darij

[quote="darij grinberg"]
I'm leaving the complete proofs to other mathlinkers.
[/quote]

[hide="Notations"]
I shall make use of the following notations:

$\bullet \alpha, \beta, \gamma$ : angles of the triangle $\triangle ABC$ with vertices A,B,C respectively

$\bullet a, b, c$ : lengths of the sides $BC, CA, AB$

$\bullet O,R$ : circumcenter and circumradius

$\bullet I,r$ : incenter and inradius

$\bullet H$ : orthocenter, the point of intersection of the three altitudes

$\bullet I_a, I_b, I_c$ : excenters, i.e., centers of excircles

$\bullet r_a, r_b, r_c$ : exradii, i.e., the radii of excircles.

$\bullet p$ : semiperimeter, $p = \frac{a+b+c}{2}$

$\bullet \triangle$ = area of  $\triangle ABC$

$\bullet \left[XYZ\right]=$ area of the triangle $\triangle XYZ$
[/hide]

[b]Lemma 1.[/b] If is ABC is an arbitrary triangle we have

$(1) \ \ AH = 2R\cos \alpha \ \ \text{if} \ \ 0 < \alpha \le \frac{\pi }{2}$

$(2) \ \ AH = -2R\cos \alpha \ \ \text{if} \ \ \alpha > \frac{\pi }{2}$

and similar relations are worth for BH and CH.

[hide="Proof."]
Let D,E,F be the orthogonal projections of A,B,C onto the sidelines BC,CA,AB.

If $\angle BAC$ is acute then:   $AH = \frac{{AF}}{{\sin \beta }} = \frac{{b\cos \alpha }}{{\sin \beta }} = 2R\cos \alpha$

If $\angle BAC$ is obtuse then:   $AH = \frac{{AE}}{{\sin \beta }} = \frac{{b\cos \left( {\pi - \alpha } \right)}}{{\sin \beta }} = - 2R\cos \alpha \ \ \ \blacksquare$[/hide]

[b]Lemma 2.[/b] Let $\triangle ABC$ be an acute-angled triangle. If $D,E,F$ are the feed of the altitudes onto the sideline $BC,CA,AB$ then we have

$(1) \ \ \ \left[ {DEF} \right] = \frac{\Delta }{{4R^2 }}\left( {a^2 + b^2 + c^2 - 8R^2 } \right)$

$(2) \ \ \ DE + EF + FD = \frac{{2\Delta }}{R}$

$(3) \ \ \ t = \frac{{a^2 + b^2 + c^2 - 8R^2 }}{{4R}}$

where $\Delta,R,t$ are the area of $\triangle ABC$, the circumradius and  the radius of its orthic triangle.

[hide="Proof."]
[b](1)[/b] We have

$\left[ {AFE} \right] = \frac{1}{2}AF \cdot AE = \frac{1}{2}\left( {b\cos \alpha } \right)\left( {c\cos \alpha } \right)\sin \alpha = \frac{{abc}}{{4R}}\cos ^2 \alpha = \Delta \cos ^2 \alpha$

and similar relations are worth for $\left[ {BDF} \right],\left[ {DCE} \right]$ . Thus

$\left[ {DEF} \right] = \left[ {ABC} \right] - \left[ {AFE} \right] - \left[ {BDF} \right] - \left[ {DCE} \right] =$

$= \Delta - \Delta \left( {\cos ^2 \alpha + \cos ^2 \beta + \cos ^2 \gamma } \right) =$

$= \Delta \left( {\sin ^2 \alpha + \sin ^2 \beta + \sin ^2 \gamma - 2} \right) =$

$= \Delta \left( {\frac{{a^2 + b^2 + c^2 }}{{4R^2 }} - 2} \right) = \frac{\Delta }{{4R^2 }}\left( {a^2 + b^2 + c^2 - 8R^2 }\right)$

[b](2)[/b] Let $O,H$ be the circumcenter and the orthocenter of $\triangle ABC$. The quadrilateral $AFHE$ is cyclic, so that, by chordal theorem and lemma 1

$EF = AH\sin \alpha = 2R\sin \alpha \cos \alpha= a \cos \alpha$

Similarly $DF = b \cos \beta$ , $DE = c \cos \gamma$ . Therefore

$DE + EF + FD = a\cos \alpha + b\cos \beta + c\cos \gamma = \frac{{aR\cos \alpha + bR\cos \beta + cR\cos \gamma }}{R} = \\ \\ = \frac{{2\left[ {OBC} \right] + 2\left[ {OAC} \right] + 2\left[ {OAB} \right]}}{R} = \frac{{2\Delta }}{R}$

[b](3)[/b] Using the relations (1) and (2) we have:

$t = \frac{{2\left[ {DEF} \right]}}{{DE + EF + FD}} = \frac{\Delta }{{2R^2 }}\left( {a^2 + b^2 + c^2 - 8R^2 } \right)\frac{R}{{2\Delta }} = \frac{{a^2 + b^2 + c^2 - 8R^2 }}{{4R}} \ \ \blacksquare$

[/hide]

[b]Remark 1.[/b] The relation (1) of previous lemma is true also for acute-angled triangles. The proof can be made using the formula on the area of the triangle pedal of a generic point $P$

$\Delta _1 = \frac{{R^2 - OP^2 }}{{4R^2 }}\Delta$  (see Crux Mathematicorum 1999, pag 499)

and the formula on the distance between the circumcenter and the orthocenter $OH^2 = 9R^2 - \left( {a^2 + b^2 + c^2 } \right)$ .

[b]Theorem 1. [/b]If $\triangle ABC$ is an arbitrary triangle then   $AH^2 + BH^2 + CH^2 = 12R^2 - \left( {a^2 + b^2 + c^2 } \right)$

[b]Proof.[/b] From the lemma 1 we have:

$\begin{array}{c} AH^2 + BH^2 + CH^2 = 4R^2 \left( {\cos ^2 \alpha + \cos ^2 \beta + \cos ^2 \gamma } \right) = 4R^2 \left( {3 - \sin ^2 \alpha - \sin ^2 \beta - \sin ^2 \gamma } \right) = \\ = 4R^2 \left( {3 - \frac{{a^2 }}{{R^2 }} - \frac{{b^2 }}{{R^2 }} - \frac{{c^2 }}{{R^2 }}} \right) = 12R^2 - \left( {a^2 + b^2 + c^2 } \right) \ \ \ \blacksquare \end{array}$

[b]Theorem 2.[/b] If $\triangle ABC$ is an acute-angled triangle then

$(1) \ \ \ AH + BH + CH = 2 \left( {R+r } \right)$

$(2) \ \ \ AH^2 + BH^2 + CH^2 = 4R^2 - 4Rt$

where t is the inradius of the orthic triangle of $\triangle ABC$.

[b]Proof.[/b] The first relation follow from lemma 1 and the well know identity $\sum {\cos \alpha } = \frac{{R + r}}{R}$

$AH + BH + CH = 2R\left( {\cos \alpha + \cos \beta + \cos \gamma } \right) = 2R\frac{{R + r}}{R} = 2\left( {R + r} \right)$

The second relation follows from Theorem 1 and lemma 2:

$AH^2 + BH^2 + CH^2 = 12R^2 - \left( {a^2 + b^2 + c^2 } \right) = 4R^2 - 4Rt \ \ \blacksquare$

[quote="darij grinberg"]

$a^2+b^2+c^2+AH^2+BH^2+CH^2 = 12R^2$;
$AH+BH+CH=2\left(r+R\right)$;
$AH^2+BH^2+CH^2=4R^2-4Rr$,

But actually, the second of these three relationships is not absolutely correct - in the form given, it holds only for acute-angled triangles ABC -, and the third relationship is wrong - it actually should be

$AH^2+BH^2+CH^2=4R^2-4Rt$,

where t is the inradius of the orthic triangle of triangle ABC, and it is also true only for acute-angled triangles ABC.

[/quote]

[b]Remark 2.[/b]  If $\triangle ABC$ is an obtuse-angled assuming WLOG $\alpha > \frac{\pi }{2}$ it is easy to prove (using lemma 1) that:

$(3) \ \ - AH + BH + CH = 2\left( {R + r} \right)$

$(4) \ \ \ AH + BH + CH = 2\left( {r_a - R} \right)$

Thanks, Leon, for your detailed writeups - they will be surely a helpful source for future referencing!

Just to note that there is a simpler approach to the formula $AH^2  + BH^2  + CH^2  = 12R^2  - \left(a^2  + b^2  + c^2 \right)$ (your Theorem 1):

[b]Theorem 3.[/b] If A', B', C' are the midpoints of the sides BC, CA, AB of triangle ABC, and O is the circumcenter of triangle ABC, then $AH=2\cdot OA^{\prime}$, $BH=2\cdot OB^{\prime}$ and $CH=2\cdot OC^{\prime}$.

[hide="Proof of Theorem 3"]
[i]Proof of Theorem 3.[/i] Since the points B' and C' are the midpoints of the sides CA and AB of triangle ABC, we have B'C' || BC and $B^{\prime}C^{\prime}=-\frac12\cdot BC$, where segments are directed (and the parallel lines B'C' and BC are directed in the same sense). From B'C' || BC and from the similarly verifiable relations C'A' || CA and A'B' || AB, it follows that the triangles A'B'C' and ABC are homothetic, and since $B^{\prime}C^{\prime}=-\frac12\cdot BC$, the ratio of homothety is $-\frac12$. Thus, corresponding segments of these two triangles are parallel to each other, and the ratio of their directed lengths is $-\frac12$.

On the other hand, since the circumcenter O of triangle ABC lies on the perpendicular bisector of its side BC, we have $OA^{\prime}\perp BC$. Combining this with B'C' || BC, we get $OA^{\prime}\perp B^{\prime}C^{\prime}$. This means that the point O lies on the perpendicular from the point A' to the line B'C'; in other words, the point O lies on the altitude of triangle A'B'C' issuing from its vertex A'. Similarly, the point O lies on the other two altitudes of triangle A'B'C', and thus, O is the point of intersection of the altitudes, i. e. the orthocenter of triangle A'B'C'. Hence, the segment OA' is the segment joining the orthocenter O of triangle A'B'C' with its vertex A'. On the other hand, the point H is the orthocenter of triangle ABC, and thus, the segment HA is the segment joining the orthocenter H of triangle ABC with its vertex A (note that the vertices A' and A are corresponding vertices of the homothetic triangles A'B'C' and ABC). Thus, the segments OA' and HA are corresponding segments of the homothetic triangles A'B'C' and ABC. Since we know that corresponding segments of triangles A'B'C' and ABC are parallel to each other, and the ratio of their directed lengths is $-\frac12$, it follows that $\frac{OA^{\prime}}{HA}=-\frac12$, thus $OA^{\prime} = -\frac12\cdot HA = \frac12\cdot AH$, and $AH=2\cdot OA^{\prime}$. Similarly, we can prove $BH=2\cdot OB^{\prime}$ and $CH=2\cdot OC^{\prime}$, and Theorem 3 is established.[/hide]

[b]Theorem 4.[/b] For the circumradius R of triangle ABC, we have $a^2+AH^2=4R^2$, $b^2+BH^2=4R^2$ and $c^2+CH^2=4R^2$.

[hide="Proofs of Theorem 4"]
[i]First proof of Theorem 4.[/i] If A' is the midpoint of the side BC of triangle ABC, then clearly $A^{\prime}C=\frac{BC}{2}=\frac{a}{2}$. On the other hand, Theorem 3 yields $AH=2\cdot OA^{\prime}$, where O is the circumcenter of triangle ABC. In other words, $OA^{\prime}=\frac{AH}{2}$. Since the point O, as the circumcenter of triangle ABC, lies on the perpendicular bisector of its side BC, we have $OA^{\prime}\perp BC$, and thus, the triangle OA'C is right-angled. Thus, after the Pythagoras theorem,

$CO^2 = A^{\prime}C^2 + OA^{\prime}\ ^2 = \left(\frac{a}{2}\right)^2+\left(\frac{AH}{2}\right)^2=\frac{a^2+AH^2}{4}$,

so that $a^2+AH^2=4\cdot CO^2$. But since O is the circumcenter of triangle ABC, while R is the circumradius, we have CO = R, so that $a^2+AH^2=4R^2$. Similarly, we can prove $b^2+BH^2=4R^2$ and $c^2+CH^2=4R^2$, and Theorem 4 is proven.

[i]Second proof of Theorem 4.[/i] After Lemma 1, we have either $AH=2R\cos A$ or $AH=-2R\cos A$. In both cases,

$AH^2=\left(2R\cos A\right)^2 = 4R^2\cos^2 A$.

On the other hand, the Extended Sine Law yields $a=2R\sin A$, so that

$a^2+AH^2=\left(2R\sin A\right)^2+4R^2\cos^2 A =4R^2\sin^2 A +4R^2\cos^2 A$
$=4R^2\left(\sin^2 A+\cos^2 A\right) = 4R^2$,

and similarly, $b^2+BH^2=4R^2$ and $c^2+CH^2=4R^2$. Thus, again, Theorem 4 is proved.[/hide]

Of course, if we take the equations $a^2+AH^2=4R^2$, $b^2+BH^2=4R^2$ and $c^2+CH^2=4R^2$ of Theorem 4 and sum them up, we get

$\left(a^2+AH^2\right)+\left(b^2+BH^2\right)+\left(c^2+CH^2\right)=12R^2$;

in other words,

$\left(a^2+b^2+c^2\right)+\left(AH^2+BH^2+CH^2\right)=12R^2$,

so that

$AH^2+BH^2+CH^2=12R^2-\left(a^2+b^2+c^2\right)$,

and Theorem 1 is proven.

Note that your Lemma 2 (1) yields a simple proof of the assertion in your thread http://www.mathlinks.ro/Forum/viewtopic.php?t=25080 post #3:

[b]Theorem 5.[/b] [b](a)[/b] The directed area of the orthic triangle of a triangle ABC is never equal to the directed area of triangle ABC.
[b](b)[/b] The directed area of the orthic triangle of triangle ABC is oppositely equal to the directed area of triangle ABC if and only if triangle ABC is harmonic.

[i]Proof of Theorem 5.[/i] After Lemma 2 (1), we have

$\left[DEF\right]=\frac{\Delta}{4R^2}\left(a^2+b^2+c^2-8R^2\right)$,

where $\Delta$ is the directed area of triangle ABC, and [DEF] is the directed area of the orthic triangle of triangle ABC. Thus,

$\frac{\left[DEF\right]}{\Delta}=\frac{a^2+b^2+c^2-8R^2}{4R^2}$.

Hence, we have $\left[DEF\right]=\Delta$ if and only if $\frac{a^2+b^2+c^2-8R^2}{4R^2}=1$, what is equivalent to $a^2+b^2+c^2-8R^2=4R^2$, i. e. to $a^2+b^2+c^2=12R^2$, what is impossible since Theorem 1 yields

$12R^2-\left(a^2+b^2+c^2\right)=AH^2+BH^2+CH^2> 0$.

Hence, we can never have $\left[DEF\right]=\Delta$, i. e. the directed area of the orthic triangle of triangle ABC can never be equal to the directed area of triangle ABC. This proves Theorem 5 [b](a)[/b].

On the other hand, we have $\left[DEF\right]=-\Delta$ if and only if $\frac{a^2+b^2+c^2-8R^2}{4R^2}=-1$, what is equivalent to $a^2+b^2+c^2-8R^2=-4R^2$, i. e. to $a^2+b^2+c^2=4R^2$, what is simply the condition of triangle ABC to be harmonic. Consequently, the directed area of the orthic triangle of triangle ABC is oppositely equal to the directed area of triangle ABC if and only if the triangle ABC is harmonic. This proves Theorem 5 [b](b)[/b].

Thus, the proof of Theorem 5 is complete.

As a consqeuence of Theorem 5 [b](b)[/b], we see that all harmonic triangles are obtuse - in fact, an acute triangle has the same orientation as its orthic triangle, so that the directed area of the orthic triangle can never be oppositely equal to the directed area of the initial triangle, and thus, an acute triangle can never be harmonic.

  Darij

Thanks, Darij, for your acute observations and for Theorems 4,5 that increase my acquaintances on the theory of the harmonic triangles !

[quote="darij grinberg"]
As a consequence of Theorem 5 [b](b)[/b], we see that all harmonic triangles are obtuse - in fact, an acute triangle has the same orientation as its orthic triangle, so that the directed area of the orthic triangle can never be oppositely equal to the directed area of the initial triangle, and thus, an acute triangle can never be harmonic.
Darij[/quote]

To such purpose it can be demonstrated that, if $\alpha$ is the obtuse-angle of a harmonic triangle $\triangle ABC$ it must turn out $128^\circ 10'38'' < \alpha  < 135^\circ $ . I do not know the proof, but it is brought back in following articles:

1) R. Muller - Ueber die Dreiecke deren Umkreis der 9 Punkte orthogonal schneidet ,Zeitschrift fur Matematischen und Naturwissen schaftilichen unterrecht, vol.36(1905)

2) J. Neuberg - Sur le triangles dont le cercle circonscrit et le cercle des neuf points se coupent orthogonalement,  Mathesis (1906)

Leon


logarithms
probability
probability and stats

I have no idea how to solve the following problem.

Let $ X_n$ be the length of the longest connected subsequence of heads in n throws of a fair coin (eg. for a sequence HHTTHTHHHTH, $ X_{11}$ = 3). Prove
a) $ P(X_n  \geqslant a\log_2 n)  \rightarrow  0$  for $ a > 1$,
b) $ P(X_n  \geqslant a\log_2 n)  \rightarrow  1$  for $ 0< a < 1$,

You have $ n$ coin flips and you need to choose a block of $ a \log n$ flips to be all heads.  You can do this in roughly $ n \minus{} a\log n$ ways.  For the other $ n \minus{} a\log n$ coin flips, we have $ 2^{n \minus{} a\log n} \equal{} 2^n\cdot n^{ \minus{} a}$ possibilities.  This means the probability we're looking at behaves like $ \frac {1}{2^n} \cdot (n \minus{} a\log n) \cdot (2^n n^{ \minus{} a}) \equal{} (n \minus{} a\log n)n^{ \minus{} a}$, and it's clear why $ 1$ is the dividing point.  This is essentially enough to conclude part (a) (although it's not very formal), but for part (b) the serious overcounting we've done is a problem (because we're trying to find a lower bound that goes to 1 instead of an upper bound that goes to 0) and so some additional idea is needed.

For part (b), break the $ n$ throws into $ \frac{n}{a \log_2 n}$ blocks of $ a \log_2 n$ throws each.  These blocks are independent, so we can exactly calculate the probability that one of them is all heads; the probability approaches 1.

Thanks a lot for replies and ideas how to deal with the problem. Now everything seems simply.

Ravi and JBL have a way of doing that to problems!! :) I always admire their approach to problem solving.


trigonometry
function
quadratics
algebra
quadratic formula

[quote="draconis"]What is the minimum value of:

$\frac{1}{2}(cot\frac{A}{2}+3tan\frac{A}{2})$

for $0<A<180$?[/quote]

wait is it set equal to something? or do you want the minimal answer for which the least value of A is 1?

The question's asking the minimum value that expression can have with an A value that's between 0 and 180 degrees.

[quote="draconis"]What is the minimum value of:

$\frac{1}{2}(cot\frac{A}{2}+3tan\frac{A}{2})$

for $0<A<180$?[/quote]

AM-GM

Please elaborate trig is not my forte. Thanks.

[hide]
By AM-GM

$\frac{\cot\frac{A}{2}+3\tan\frac{A}{2}}{2}\ge \sqrt{3\cot\frac{A}{2}\tan\frac{A}{2}}$

Since the tangent and cotangent functions are reciprocals:

$\frac{\cot\frac{A}{2}+3\tan\frac{A}{2}}{2}\ge \sqrt{3}$

Now we know it is atleast $\sqrt{3}$, but we need to prove it is possible to be $\sqrt{3}$. Let $\tan\frac{A}{2}=x$

$\frac{\frac{1}{x}+3x}{2}=\sqrt{3}$
$\frac{1}{x}+3x=2\sqrt{3}$
$1+3x^{2}=2\sqrt{3}x$
$3x^{2}-2\sqrt{3}x+1=0$
By the quadratic formula:
$x=\frac{2\sqrt{3}\pm\sqrt{(2\sqrt{3})^{2}-4(3)(1)}}{2}=\frac{2\sqrt{3}\pm\sqrt{0}}{2}=\sqrt{3}$

$\tan60^\circ=\sqrt{3}$ so $A=120^\circ$
[/hide]


limit
logarithms
integration
real analysis
real analysis unsolved

The limit is $\frac{(\ln 2)^{2}}{2}$.
Let $a_{n}$ be the sequence in question. Then
$\frac{1}{n+i}<\ln\frac{n+i}{n+i-1}<\frac{1}{n+i-1}$
It follows that
$\sum_{i=1}^{n}\frac{1}{n+i}\ln \frac{n+i}{n}<a_{n}<\sum_{i=1}^{n}\ln \frac{n+i}{n}\frac{1}{n+i-1}<\frac{n+1}{n}\sum_{i=1}^{n}\ln \frac{n+i}{n}\frac{1}{n}\frac{1}{1+i/n}$.
Thus
$a_{n}\rightarrow \int_{0}^{1}\frac{\ln(1+x)}{1+x}dx=\frac{(\ln 2)^{2}}{2}$.


WHat is the largest amount of posts you have posted in a day.  Today i had 48.

When I look at who's posted last in each topic in G&F, I find that very easy to believe.

Spamming the F & G forum is frowned upon :noo:

omg....who actually counts the number of times they post in a day?? Its not something to be "proud" of anyways.

Well if 20 people answer a question in the exact same way, it's not just the last person that's spamming...

Aside from that, math92 please read the guidelines on the announcement thread of this board. It seems that some of your posts may be viewed as excessive and unnecessary.

As for posts in a day, I'm sure that there were times when Tare made more than 150. 200 may be a stretch...but definitely 150. Those were the days when you could hit refresh for the Etc. board (which is now F&G) every 10 seconds or so and there'd be 5 new posts. With only 50 or so consistent Etc. spammers, you'd see that they were considerably active. Of course, most of those posts are no longer part of our post counts so a remarkable page in the history of AoPS (plus more than just a few hundred of my own spam posts) were erased. Quite unfortunate if I do say so myself...

Guys don't push it. I'm very close into installing a personal post count manager. Some people, like say math_77 (just a random example :D) could find his post count stuck at say ... umm .. 10 posts? :)

Do not spam the boards. It will only make me do that!

Something tells me you're having fun with all your power, Valentin.  (jk--or am I?)  Well, I guess it's necessary to enforce the rules, though.

[quote="bubala"]Something tells me you're having fun with all your power, Valentin.[/quote]Maybe I am having fun, maybe not. But does it really matter? The point was made. Or would you rather have me say the following instead:

[quote]Stop spamming the boards  :mad:  :furious:  :cursing: or else :censored:  :bomb:[/quote]
:roll: I think while we are all having fun and enjoying without breaking the rules of common sense everything is going to be great! :)

I agree :)  I never said that spamming the boards was good thing; to the contrary, spamming the boards undermines what AoPS-MathLinks stands for, and you are doing a great job making sure it doesn't happen.  It's just that your post had sort of a "muahahahahahaha" kind of tone :D

Yah I gotta get my posts up I want to be a yang-mills theory it just looks cool, the mos9t I got is 13, kinda wimpy but I don't post much.

Could you please try and write grammatically-correct sentences? I had to break it to you, but writing coherently helps other people read what you are saying.

And for the record, I think I post too much in the F&G Forum.

You mean "I HATE to break it to you..." :D

If you post too much in F+g, then I post waaaaaay too much in f+g.  However, most of my f+g posts are in mafia, which is fun!

I don't think Mafia counts for post count. Does it?

[quote="MithsApprentice"][quote="bubala"]Something tells me you're having fun with all your power, Valentin.[/quote]

I honestly don't know of very many online forum admins/mods who actually believe it's fun to sit there and delete spam posts or work on codes to prevent spam.

bubala, watch some of your own posts. Sometimes we're lenient, but that doesn't mean we encourage spamming.

As for the Mafia question, I do delete the threads after Mafia games are over. And I believe this is quite reasonable since a majority of the posts on those threads are one word "aye" or "nay" that would normally be cleared anyways. Valentin can probably answer the question of whether your post count actually goes down when mods delete threads.[/quote]

I didn't mean for my comment to be taken so seriously.  The only point was that Valentin's tone was kind of funny.

Second, have I been spamming?  If you'd give me some examples (via PM, please), then I'd certainly stop.  I apologize if I have been, but as of now, I'm unaware of it.

No, the post count actually doesn't go down when you delete threads; Valentin answered this question a few days ago.  According to him, he can re-sync the post counts whenever he wants to, but he is too busy working on more important projects like WOOT :w00t: He says if people don't spam it'll never become an issue (good, because I'd decrease by about 100 from my mafia posts).  By the way, most mafia posts, if you read the games that are going on, consist of a lot more than "AYE" or "NAY" and in fact, they don't contribute any more nonsense to this forum than Your Favorite Number, Your Favorite Word, Name an Element, Stupid [star wars] Game, etc.   Of course, I am not the moderator; you are, so of course it's your judgement call.

I usuallly get 12 posts on an active day.  I also restrict to two new topics.  Some people who joined for a week have already twice my post #.

[quote="bubala"]Second, have I been spamming?  If you'd give me some examples (via PM, please), then I'd certainly stop.  I apologize if I have been, but as of now, I'm unaware of it.  [/quote]

Quite often, spamming happens without awareness. I'm sure that everyone's done it at some time or other.  :)

:o Spamming! Maybe I would call it obsession of a topic? lol

A better question would be:

What's the largest number of time you posted a solution to a problem or gave some valuable insight to a hard problem (in one day)?  :P

[quote="G-UNIT"]I agree with GoBraves, some people are so concerned about how many posts they have, they forget about the purpose of aops to help encourage and promote mathematical culture.  But instead these people spam the servers and say things like " I agree"  or post answers to problems after 20 people have already answered it.[/quote]

You're absolutely right, I know a person on this forum who waits for everyone to answer and then just says, "yes, I got this too". It drives me crazy  :mad: because it just ruins the thread....don't post unless you have a solution to the problem along with it.

it's probably me :P, if it is i'll write my solutions.

[quote="math92"]it's probably me[/quote]

agreed

[quote="Valentin Vornicu"][quote="Iversonfan2005"]You're absolutely right, I know a person on this forum who waits for everyone to answer and then just says, "yes, I got this too"[/quote]Who is this person?[/quote]

I don't wanna say the name of this person, and no math92, it's not you. I don't wanna snitch on the guy you know, but Valentin, if you really wanna know..I guess I'll say it....

[hide]4everwise, just look at all his posts (find all posts by this person on his little profile thing). None of them have solutions or anything...and he always answers after like 5 people answer the question.[/hide]

[quote="math92"]yay, its not me!!! :lol:[/quote]

Yes, but that's not an excuse to make comments without an sort of constructive content. Si?

please do not take this the wrong way, but i can also testify that both "math92" and "4everwise" simply post numerical anwers to problems after several people have already solved them using full solutions.  First of all, you should always be posting full solutions to problems (unless it is bluntly obvious) moreover, once ( this is just a random figure) 15 people lets say, have answered the problem correctly it is not generally appropriate to post the answer once again, this will most often cause situations like this to arise.  To keep things simple, follow the guidelines and dont abuse the forums, there here to help you.

i don't count the post but it think i've got 15 post in a day at the most

I think this thread has turned into a "DONT SPAM :bomb: " thread.

ok lets just not be [b]pointing[/b] fingers minding into someone's business, let the mods and admins check and determine it, mind your own business please. i may sound rough and authoritative but thats the better way because people start to dislike other users and thats not good for overall of the community to be hating each other. hmm lets say it this way, control your posting.

these are my guidelines that i follow when i post
1) say useful helpful constructive things, if not, then dont post
2) if it is a free-style topic, just one post humourous or teasing post, unless a reply is required
3) posting a new problem
4) do not repeat what others said beforehand, its a waste of time and space
5) do not make zigurats (my new rule :P)
6) post a solution with different approach or maybe same approach but can explain a little better, but no posting if more than say 5 already given solutions
7) no flaming
8) no double posting (i always check)
9) no double topics in different forums (very irritating but some new users still do)
10) no pointing fingers
11) no personal comments about others, including implicit comments like 'rubbish' (yea i learnt keep it to myself, sorry djshowdown2)
maybe some more but overall is... common sense people, we are not animals that can only think of basics, we are way beyond that, be more responsible to yourself and to others. hah i sound like my father...  :police:

Yeah I agree with Palytoxin -  but if you feel that the person is going under the radar you can always PM an admin to make sure they know of the situation, but it is always best to keep the finger pointing off of posts.

Yeah, don't point fingers. It makes the person feel bad  :(

[quote="paladin8"][quote="Palytoxin"]
6) post a solution with different approach or maybe same approach but can explain a little better, but no posting if more than say 5 already given solutions
[/quote]

I agree with everything except maybe this line. Even if there are already 5 solutions, if you have a valuable approach you want to share (key word: VALUABLE) then go ahead and post it. Other than that, I think everyone appreciates your work in compiling this pretty comprehensive list. Maybe there should be a sticky with an ongoing list...[/quote]
yea my line already contains your meaning, basically different approaches to a problem are valuable but if we see the same approach, either paraphrasing or in some other twist, is boring and dull. i gave 5 as an example, i even get bored if i see 2 or 3 solutions done the same way, which i see that a lot recently. i think if you want to talk about on the same approach, at least try to improvise the previous poster's solution or perhaps do a simple 1 or 2 sentence comment/generalization/insight.

the most iha ve ever had is 5 i am not very active :play_ball:


geometry
circumcircle
perimeter
geometric transformation
reflection
perpendicular bisector

let $\omega$be  the circumcircle of triangle $ABC$ and two tangents of circumcircle at  $B$&$C$ meets each other at piont $N$ and $M$is on the perimeter of circumcircle and $AM$ is parallel with $BC$.let k be the point where $MN$ meets circumcircle $\omega$prove that $AK$ divide $BC$into two equal parts.

Call $P$ the midpoint of arc $BC$, and $R$ the intersection of $AN$ and $\omega$.
Call $l$ the perpendicular bisector of $BC$.

Clearly, $A$ is the reflection of $M$ in $l$. Therefore, also $R$ and $K$ are each other's reflection with respect to $l$.

Hence, because $P$ is on $l$, we know that $\angle KAP = \angle PAR$. We know that $AN$ is the symmedian of angle $A$, so $AK$ must be the median, $\mathbb{QED}$.


limit
trigonometry
calculus
calculus computations

Compute
$ \boxed{\lim_{x \rightarrow 1} \frac {Arccos(\frac {2x}{1 \plus{} x^2})}{x \minus{} 1}}$

 :)

[hide="Answer"]Changing variables according to $ x \equal{} \tan \frac{t}{2}$ we have that $ \sin t \equal{} \frac{2x}{1\plus{}x^2}$, and when x goes to 1, t goes to pi/2. Therefore:

$ L \equal{} \lim_{x \to 1} \frac{\arccos\left(\frac{2x}{1\plus{}x^2}\right)}{x\minus{}1} \equal{} \lim_{t \to \pi/2} \frac{\arccos(\sin t)}{\tan \frac{t}{2} \minus{} 1} \equal{} \lim_{t \to \pi/2} \frac{\frac{\pi}{2} \minus{} t}{\tan \frac{t}{2} \minus{} 1}$

$ \equal{}> L \equal{} \minus{} \lim_{t \to \pi/2} \frac{1}{\frac{\tan \frac{t}{2} \minus{} 1}{t \minus{} \frac{\pi}{2}}} \equal{} \minus{} \frac{1}{(\tan \frac{t}{2})'_{t \equal{} \pi/2}} \equal{} \minus{} \frac{1}{\left(\frac{1}{2} \sec^2 \frac{t}{2}\right)_{t \equal{} \pi/2}}$

$ \equal{}> L \equal{} \minus{}\left(2 \cos^2 \frac{t}{2}\right)_{t \equal{} \pi/2} \equal{} \minus{}2 \cdot \cos^2 \frac{\pi}{4}$

$ \equal{}> L \equal{} \minus{}1$.[/hide]

Nice, Carcul,
but you made a small mistake here
$ \lim_{t \to \pi/2} \frac {\arccos(\sin t)}{\tan \frac {t}{2} \minus{} 1} \equal{} \lim_{t \to \pi/2} \frac {\frac {\pi}{2} \minus{} t}{\tan \frac {t}{2} \minus{} 1}$

In fact, $ \arccos(\sin t)$ is equal to $ \frac {\pi}{2} \minus{} t$ only if $ t\le \frac {\pi}{2}$ , so we should compute the limit in $ \frac {\pi}{2}^ \plus{}$ and $ \frac {\pi}{2}^ \minus{}$.

We can solve this limit by $ x \equal{} tg(t)$ or we consider $ Arcos(\frac {2x}{1 \plus{} x^2})$ and we simplify it to $ 2arctg(x) \minus{} \frac {\pi}{2}$, $ \forall x \ge 1$ by computing it's derivate.

 :)


trigonometry
geometry
circumcircle
trig identities
Law of Cosines
combinatorial geometry
geometry proposed

Given $n$ points in the plane such that the distance between each two of them is less than $1$, find the minimal value of $r$ such that there exists a circle with radius $r$ with all the points in its interior.

[hide="Solution"]This problem simplifies to finding the radius of the smallest circle that circumscribes a regular n-gon of side length 1.  Which can be solved using the law of cosines $r^2+1-2rcos(\frac{360}{n})=r^2$ and $r=\frac{1}{2cos(\frac{360}{n})}
$ :D

Also, if that distance were p, $r=\frac{p}{2cos(\frac{360}{n})}$[/hide]

Your aolution is wrong, diophantus!

Actually, the problem refers to the Young circle. The answer is $\frac{\sqrt 3}{3}$. Thy to prove this!

Myth is right, now it remains to prove that there is such circle and there is not a smaller one, which is not as easy as it looks but not too hard anyway.

P. S. Myth, I didn't know that this thing has a name, could you please say something more about Young circle?

Oh, there is a great theory devoted to this theme. The whole problem is called Goldbach (? I am really not sure, I will check it later) optimal covering. They try to find the "minimal" figure s.t. that it is possible to cover an arbitrary set of diametr 1 with the figure. The problem is still unsolved.

It suffices to restrict our attention to the convex hull of the system of points.

We have a convex polygon in which the lengths of all sides and diagonals are $<1$. Start with a large circle containing the polygon, and then make it progressively smaller, until you reach the minimal radius. There are two possible situations: either two of the vertices of the polygon form a diameter of the circle, in which case the circle has radius $<\frac 12$, which is even better than what we want, or there are three vertices of the polygon forming an acute-angled triangle on the circle. After we show that every acute-angled triangle with sides less than $1$ has circumradius $<\frac {\sqrt 3}3$ (this can be done by taking $2R=\frac a{\sin A}$, where $A$ is the largest angle of the triangle, which must then be $\ge\frac\pi 3$), we are done.

To show that the value can't be improved, take equilateral triangles with sides as close to $1$ (but $<1$) as we want.

Ok, I may be wrong, but I think it is solved by Helly's Theorem.
Probably the bound I am going to find isn't the best, but the idea is the following:
Plave a circle of radius $R$ with center in each of those $n$ points. Now, if $R$ is big enough, for any $3$ circles, they will meet, so all circles will meet (by Helly). Now, take the point at their intersection and notice that a disk with radius $R$ will cover all of them.
Now, I think Myth, it should have been $\frac{\sqrt{3}}{2}$ since an equilateral triangle is an counterexample...
I hope that $R = \frac{\sqrt{3}}{2}$ works :)
I hope there are no mistake in my calculations (that I didn't do :D)
[Edit : Post above made by grobber didn't show when I wrote my solution, also sorry Myth for saying the bound is bad, as usually, my calculations are bad :( ]

Dust, the circumradius of an equilateral triangle of side $1$ is $\frac{\sqrt 3}3$. It's altitude is $\frac{\sqrt 3}2$.

Indeed, the proof for Young circle is very simple and classical.

Now the second question: what if we want to cover our set with regular hexagon of minimal size? ;)


function
calculus
derivative
geometry
parallelogram
complex analysis

Hi, I am going over my notes from my complex course last year, I was hoping someone could explain something to me, b/c my notes are unclear.

I am considering the set X in C^2, parametrized by: y^2 = 4x^3 - 20a_2x + 28a_4, ie. elliptic curve parametrized by the general p-function. My notes state that this is clearly a 1 complex dimensional submanifold via. the implicit function theorem.

Now I get that when y =/= 0, then obviously the partial derivative wrt. y does not vanish hence the curve can be para. via y = f(x). But I am confused how to show that the curve can be parametrized x = g(y) when y = 0. I know that when y = 0, we have x = p(e_1 / 2) or p(e_2 / 2) or p((e_1 + e_2)/2) which is a value of multiplicity 2 for p and p only assumes a value twice in a period parallelogram but I'm not sure how to conclude what I want from that. Thanks for any help!

[quote="Spacecow"]But I am confused how to show that the curve can be parametrized x = g(y) when y = 0. I know that when y = 0, we have x = p(e_1 / 2) or p(e_2 / 2) or p((e_1 + e_2)/2) which is a value of multiplicity 2 for p and p only assumes a value twice in a period parallelogram but I'm not sure how to conclude what I want from that. Thanks for any help![/quote]Can you show that p(e_1 / 2), p(e_2 / 2) and p((e_1 + e_2)/2) are pairwise different?


An error!!!  I would have beaten this guy if the error wasn't there!! (No fault of you, bluecarneal). 5^(n-5) = 1/125
Answer: 9 ???
How could the answer be nine???
Post on the poll for this forum at the top of the page!

What do YOU thinkthe answer is?

Uhhh, you should of asked on the problem discussion place. Anyways,  the answer is 2. $ 5^{(2 \minus{} 5)} \equal{} 5^{( \minus{} 3)} \equal{} 0.008 \equal{} 1/125.$

$ 5^{(9 \minus{} 5)} \equal{} 5^{4} \equal{} 625 \equal{} 625$

Yea, just checked my work; edited.

If you know the problem number, please PM it to me so that I can fix the problem.

mrsaturns, you wrote 9-5=-4.

The answer would be 2, so the exponent becomes -3, resulting in 1/125.


Often, when I try to post, it says "Invalid Session. Please resubmit the form." I keep trying to resubmit it, but the same message appears. I've even tried clearing my cookies, logging out and back in, etc. What causes it, and how do I stop it?

Invalid session cookie. Your browser settings are most likely the culprit here.


I just got this letter to go to the ARML contest in the mail this afternoon.

I was wondering if any of you guys know how the ARML team (15 people) for each state is chosen.

Does anyone know?

I think that in Texas, the top eight finishers at state mathcounts get invited to ARML. Also, anyone who went to ARML the previous year gets invited again.

Last year, I finished in the top 4 in Mathcounts State, but I didn't get invited.

As I know, ARML is a high school contest. Does anyone know how popular it is, though?

Ahhhh, I believe I know who you are now! :P
[i]Tong![/i]

09husbri, did you get an invitation to participate in ARML (although I feel obliged to call you by your real name)?

Go ahead, call me Brian.

I didn't get anything that I know of. But it may be different in Texas and in Ohio.


Suppose that $0<a,b,c,d<\frac{\pi}{2}$ satisfy $cos2a+cos2b+cos2c+cos2d=4(sina\cdot sinb\cdot sinc\cdot sind-cosa\cdot cosb\cdot cosc\cdot cosd)$.
Find all possible values of $a+b+c+d$.

I think you might be able to make progress by using the identity $\cos^2 a + \cos^2 b + \cos^2 (a+b) - 2\cos a \cos b \cos (a+b) = 1$ for a and b, then c and d...

Divide both sides by \prod \cos \alpha. To avoid the sum on the LHS and obtain a product, use the prostepheresis formulas

Hey i didn`t manage to solve it. Help. Solution? :?

Expanding......I'm not sure but It maybe availlable...

Do you mean $cos2x=cos^2x-sin^2x$
Hey please post solutions not hints.  :)

i think it isn't hard :roll:

just use these identities
(1) $2cosacosb = cos(a+b)+cos(a-b)$
(2) $2sinasinb =-cos(a+b)+cos(a-b)$
(3) $cosa+cosb = 2cos(\frac{a+b}{2})cos(\frac{a-b}{2})$

Let $a+b=x, a-b=y, c+d=z, c-d=w$

By (1),(2)
$RHS =-2(cosxcosw+cosycosz)$

By (3)
$LHS = 2(cosxcosy+coszcosw)$

And we get

$(cosxcosy+coszcosw)+(cosxcosw+cosycosz) = 0$
$(cosx+cosz)(cosy+cosw) = 0$

If $cosy+cosw = 0$ then $y=\pi-w$ or $y=\pi+w$
it means WLOG, $a+b>\pi$ and contradicts with the condition.
so $cosy+cosw \neq 0$ and $cosx+cosz$ must be 0.
then $x=\pi-z$ or $x=\pi+z$, so $a+b+c+d = \pi$ or $a+b-c-d = \pi$
second case contradicts so we finally get

$a+b+c+d = \pi$

It's easy to check it. :D


number theory
relatively prime
number theory unsolved

I was wondering if anyone knows how to show that there are an infinite number of primes such that

\[2^{p-1}\not\equiv 1\bmod p^{2}.\]

$2^{p-1}\not\equiv 1\ mod\ p^{2}\ \Longleftrightarrow \ p|ord_{p^{2}}(2)$

[quote="amcavoy"]Thank you for the responses.  What does $\textrm{ord}_{p}\left(2\right)$ mean?  I haven't seen this notation before.[/quote]

It means the order of an element.

Given $2$ and prime $p$ we define $\mbox{ord}_{p}(2) =k$ where $2^{k}\equiv 1 (\bmod p)$ and $k$ is minimal exponent $k$. This notation is only defined for integers relatively prime to $p$.

[quote="Rust"][quote="aviateurpilot"]$2^{p-1}\not\equiv 1\ mod\ p^{2}\ \Longleftrightarrow \ p|ord_{p}(2)$[/quote]
What is $ord_{p}(a)?$
I mean $ord_{p}(a)=max\{n: \ p^{n}|a\}.$[/quote]
I mean $ord_{p^{2}}(2)=min\{k|\ 2^{k}\equiv \ 1\ mod\ p^{2}\}$

[quote="sylow_theory"]
It means the order of an element.

Given $2$ and prime $p$ we define $\mbox{ord}_{p}(2) =k$ where $2^{k}\equiv 1 (\bmod p)$ and $k$ is minimal exponent $k$. This notation is only defined for integers relatively prime to $p$.[/quote]
minimal nonnegative exponent is 0.
If minimal positive exponent , then never $p|ord_{p}(a)$.

[quote="Rust"]Let $P_{n}=\{p\in P|ord_{p}(2^{p-1}-1)=n\}$.
If  $P_{1}$ is finite for infinetely many primes, then $2^{p-1}=1+x^{2}$ - give contradition. It is obviosly if $P_{n}$ is empty for $n\ge 3$.[/quote]
$P_{1}=\{p\in P|\ p^{2}\not| 2^{p-1}-1\}$
if $|P_{1}|<+\infty$
where is th contradiction in $2^{p-1}=1+x^{2}$ and what si $x$?

edit, i say $p|ord_{p^{2}}(a)$

Let $P_{1}=\{p_{1},p_{2},...,p_{k}\}$. let $a>1$ and $(a,m)=1, \ m=p_{1}(p_{1}-1)p_{2}(p_{2}-1),...,p_{k}(p_{k}-1)$.
There are infinetely many primes p: $P_{a,m}=\{p=a(mod m)\}$. For any prime $p\in P_{a,m}$ we get $rad(2^{p-1}-1)<2^{(p-1)/2}$. It give contradition with abc conjecture ($(2^{p-1}-1)+1=2^{p-1}$ and $rad(1*(2^{p-1}-1)2^{p-1})<2^{(p+1)/2}.$
If $P_{n}, n\ge 3$ are empty for any $p\in P_{a,m}$ exist even divisor $d|p-1$, suth that $2^{d}-1$ is perfect square, because $2^{d}$ is pervect square too, it give contradition.

If $a=\prod_{i}p_{i}^{k_{i}}$ then $rad(a)=\prod_{i}p_{i}$. For example $rad(12)=6$.
$Rad(aZ)=rad(a)Z$.


Spread the virus, go find fanatic and usaha, and some other random people... and then, threaten to kill their families if they don't join

u should pm them, cuz it seems like u know them personally. i would do it, but i guess it would seem kinda awkward...

oh. the first three letters are usa, so i figured his intials must be
HA. i guess my intense genious (is that how u spell it?) failed me! lol, jk. i'm kinda retarded.

BTW, indianamath, weren't u 8th in states? can u pm me ur score? i'll pm u mine.

that is g-e-n-i-u-s, genius.(i was in the spelling bee :D )

Lol, i was in the spelling bowl ^^, and i only know three people in Indiana on this forum so -_-...

i was 13th in states.  :wink:

if you still can't remember me, then think about countdown... did anything unusual happen? :wink: it sucked.

i already knew who you were... richard told me... lol, you could've easily gotten into the semis for countdown, but you raised your hand lol ^^.... i made the stupidest mistake in the world EVER... and i could've completely owned -_- but i didnt lloll

you keep saying "lol" like it was funny or something. it wasn't. i didn't even know what ppl were laughing at! :furious:  i am soooooooo pathetic!!

[quote="mwpl11"][quote="undefined117"]hahh it's pretty cool how no one knows me  :roll:

and mwpl11, i didn't realize people were laughing..? forget about them.. you owned them on the AMC 8![/quote]

thanks. i'm over it now, though. and btw, i don't really believe the thing about biology (milad said)... but who cares? everyone does it  :lol: . jk.[/quote]

O_O riiiiiiight..

he's trying to sabotage me.. i swear

i dunno why i kept saying lol either, its just automatically integrated within my talking i guess lol... see what i mean?!?! lol -_- XD, so are these though XD -_- lol <(^^)><(^^<)(>^^)> -_- aaaggghhhh the stupid stupid world of god ***** urban slang noooo

are you in 6th grade or 7th grade mathcrazed?

[quote="undefined117"][quote="mwpl11"][quote="Fanatic"]Lyndon is actually really strong =/. MUCH stronger than I am TT[/quote]

ya, i know what it's like. i'm pretty strong even though i'm tiny also (but i also play sports aka swimming, tennis) no offense, but i can't stand track. i'm only good at sprints (lol, running  :lol: )...[/quote]

haha but i bet i can outrun you :]

and i don't think you can have AIME scores in your sig yet, even if it's a guess..[/quote]

Soo-Hyun Mrs. Hufford said you got a 6 on the AIME.

until lyndon joined we were perfectly normal.

dont be hatin.

till lyndon joined we were totally boring!

[quote="cognos599"]i never said that normal was good. lyndon has been an "interesting' addition to the forum.[/quote]

interesting, indeed... but it has helped improve this forum a lot. i only wish we had gotten this forum going before states so i couldve met all u guys. that would've been interesting. oh, well. expect me to be at the ICTM awards ceremony. i trust every one of u will be there too.

[quote]i trust every one of u will be there too.
[/quote]

Probably not me because Im stupid (indianamath can tell you), and does Sycamore even do ICTM?

no one on this forum is stupid, with the exception of me and Ubemaya (lol)

This isn't fair. All of you get perfect scores on the algebra test, and i get a less-than-perfect score on the comprehensive test, and possibly don't get to go? :(

[quote="mwpl11"][quote="undefined117"][quote="mwpl11"]that's ur fault for taking pre-calculus. haha

BTW, i still hate u. at least i beat karen. michael will never be able to live it down...*evil grin*.

and i still hate u *narrowing eyes*. i got a four when i studied for literally 14 days. wait till next year, soo-hyun. i won't even have mathcounts to weigh me down. i'll go all out and get a perfect...[/quote]

O_O you scare me..

and see what I mean? You don't really have much.. so basically I'm joining the robotics club and own everyone at the competition, the ISEF science fair and make it to nat'ls, and I think I might start swimming again.. so that leaves about two hours a day to finish all my homework as well as extra math. I LOVE MY LIFE.. not.[/quote]

robotics? ooookay. i wanna try to get on the varsity tennis team next year. r u gonna start swimming for school?? and u are not going to have 2 hrs. each day if u join swimming. u'll be in senior and they practice twice a day for like 3 hrs each practice. i don't even know how alisa manages, u have to give her some credit. i'm amazed at how good she is at everything... she was a couple seconds off making the Olympics i heard...[/quote]

ok then i WON'T start swimming

then maybe i can try out for tennis (I SUCK) but I won't cuz i know you'll laugh at me

-_-

only choice left for personal workout is bike rides around lafayette, which isn't too exciting..

[quote="undefined117"]
ok then i WON'T start swimming
quote]

ya, good decision. at first i thought i would keep swimming through high school, but then i decided against it for these reasons:

1) i am totally not good enough
2) no time
3) i would get killed in meets (a.k.a. i suck compared to those high school kids)

ya, i've seen those senior kids practice. it's scary

im probly playing tennis in high school cuz i rock mihika's sox at it.

OMFG, no way! u suck! plus ur gonna have to compete against Joey. good luck!


This might have been asked before, but if find and submit two different solutions to a given problem, how exactly will it be graded?

Submitting two different solutions in two different submissions causes paperwork confusion. Don't do that.

I have graded submissions where the submitter wrote up two separate solutions for one problem back to back. My habit is to glance over both solutions and grade the one I like best. Other graders have their own ways of handling double solutions. I have seen a few beginning graders mistake a double solution for one long solution and try to grade it that way.

There is no advantage in showing more than one way to solve a problem. A single correct and complete solution earns the full five points. An elegant correct and complete solution can earn a commendation too, but adding a second method of solution does not add elegance.

Erin Schram
USAMTS grader


algebra
polynomial
linear algebra
matrix
superior algebra
superior algebra solved

Let A, B be two matrices of size n.
Prove that E+AB and E+BA have the same characteristic polynomial, where E is the identity matrix of size n.

[url]http://www.mathlinks.ro/viewtopic.php?p=18693#18693[/url]

$\det(XE-AB-E)=\det(XE-BA-E) \Leftrightarrow \det(XE-AB)=\det(XE-BA)\Leftrightarrow P_{AB}=P_{BA}$

You will find a nice proof in the link above.


ARML
logarithms
geometry
rectangle
trigonometry
analytic geometry
inequalities

1. Find the greatest integer $n$ such that $n\log_{10} 4$ does not exceed $\log_{10} 1998$.

2. Rectangle $ABCD$ has sides $AB=CD=12/5$, $BC=DA=5$. Point $P$ is on $AD$ with $\angle BPC = 90^\circ$. Compute $BP+PC$.

3. Compute the number of sequences of four decimal digits $(a,b,c,d)$ (each between 0 and 9 inclusive) containing no adjacent repeated digits. (That is, each digit is distinct from the digits directly before and directly after it.)

4. Solve for $t$, $-\frac{\pi}{4}\leqslant t\leqslant \frac{\pi}{4}$, if \[ \sin^3 t + \sin^2 t\cos t+\sin t\cos^2 t+\cos^3 t=\frac{\sqrt{6}}{2}. \]

5. Find all integers $n$ such that $n-3$ divides $n^2+2$.

6. Find the maximum number of bishops that can occupy an $8\times 8$ chessboard so that no two of the bishops attack each other. (Bishops can attack an arbitrary number of squares in any diagonal direction.)

7. Points $A,B,C,D$ are on a Cartesian coordinate system with $A=(0,1)$, $B=(1,1)$, $C=(1,-1)$, and $D=(-1,0)$. Compute the minimum possible value of $PA+PB+PC+PD$ over all points $P$.

8. Find the number of distinct real values of $x$ which satisfy \[ (x-1)(x-2)(x-3)\cdots (x-9)(x-10)+\frac{1^2\cdot 3^2\cdot 5^2\cdot 7^2\cdot 9^2}{2^{10}}=0. \]

[quote="chess64"]1. Find the greatest integer $n$ such that $n\log_{10} 4$ does not exceed $\log_{10} 1998$.[/quote]

OK then I guess i will start it off...

[hide]
$4^n<1998$
$n\le \boxed {5}$
[/hide]

[hide="3"]

There are 10 possibilities for the first digit.  There are then 9 options for the second (can't be the same as the first).  Similarly, there are 9 options for the third and fourth.  So 10*9*9*9=7290 total.

  [/hide]

try factoring it and then figure out a way of simplifying a sine and a cosine to a lesser trig function.

Was my work correct in the #25 Maryland Maht Competition 2006.

[hide="Number 6"]
Place 16 bishops on the board, 8 on Row 1, 8 on Row 8.  Then remove the 2 corner bishops on Row 1, and see that you have $\boxed{14}$ bishops on the board. (Furman stole this problem in 2006 for their individual contest).
[/hide]

[hide="Number 4"]

Simplify the given expression to
$\frac{\pi}{4}\leqslant t\leqslant \frac{\pi}{4}= (\sin^{2}t+\cos^{2}t)(\sin t+\cos t)$

Invoke the identity $\sin^{2}x+\cos^{2}x = 1$ (IDENTITY A)

$(\sin^{2}t+\cos^{2}t)(\sin t+\cos t) = (\sin t+\cos t)$

We (transitively) know from the problem that thus $(\sin t+\cos t) = \frac{\sqrt{6}}{2}$.  Now we use more identities:

$(\sin t+\cos t)^{2}= \frac{6}{4}$ (squaring both sides)
$2\sin t \cos t = \frac{1}{2}$ (expanding above, and subtracting identity A)
$\sin 2t = \frac{1}{2}$ (another trig identity)
Thus $2t = 30$, and $t = \boxed{15}$

[/hide]

That was probably the poorest written solution ever, but it's 12:30 in the morning on a school night, so forgive me.

That looks good...

And for number 7 find a way of using Triangle Inequality in what you do,

[hide="7"]
We want to minimize $AP+PC$.  By Triangle Inequality, this is minimized when P is on AC.  Also, $BP+DP$ is minimized when P is on BD.  So the sum is minimized when P is at the intersection of AC and BD.  So the sum we are looking for is the sum of the diagonals, which is (by simple geometry) $\boxed{2\sqrt{5}}$.

NOTE: Triangle Inequality in this sense is basically just putting that third point on the line created by the first two points, creating a degenerate triangle, in which all three points lie on the same line.
[/hide]

[quote="mathgeniuse^ln(x)"]Ah you have been tricked...notice that the coordinates aren't what you think they are...[/quote]

Oh sorry, part typo, part careless mistake.  Fixed.  Also, is the answer to #8?:
[hide]9 (by seeing double root and repeat application of middle value theorem)[/hide]
If so, I'll post a (very badly thought out) solution.

EDIT: Also, can someone post a solution (or hint) to #2.  I know I'm going to feel stupid when I see it, but I suck at geometry...

#2:

[hide="Hint"]

Look for complementary angles.  You'll find a lot of them.

[/hide]

[hide="Solution"]

Set $m\angle PCB = x$.  Then clearly $m\angle PCD = m\angle PBC = 90-x$.  From there we quickly get $m\angle PBA = m\angle DPC = x$ and $m\angle APB = 90-x$.

These angle relations give us $\tri BPC \sim \tri PAB \sim \tri CDP$.  From side relations, if $AP = r$, $\frac{r}{\frac{12}{5}}= \frac{\frac{12}{5}}{5-r}$.

We get a quadratic and find that $r=\frac{16}{5}$.  $BP=\sqrt{\frac{12}{5}^{2}+\frac{16}{5}^{2}}=4$, and thus $PC=\sqrt{5^{2}-4^{2}}=3$.  So $BP+PC = 3+4=7$.

[/hide]

Yep, I feel really stupid, as predicted.  OK, I'll post a new problem:

Given triangle ABC, with AB = 15 and CA = 12. The median from B, the altitude from C, and the angle bisector from A are all concurrent inside the triangle. Compute the length of BC.

(SC ARML Practice 2004)

[hide]Hint:
Ceva's Theorem.

Answer:
13[/hide]

I will post a solution if needed...


We know that if $f(x)=(g(x))^2+(h(x))^2$, where $g(x)$ and $h(x)$ are real polynomial (polynomial with real coefficients), then $f(x)\geq 0$ for all real number $x$. Prove that the converse of the statement above is also true, that is, if $f(x)$ is a real polynomial with $f(x)\geq 0$ for all real number $x$, we can always find real polynomials $g(x)$ and $h(x)$ such that $f(x)=(g(x))^2+(h(x))^2$

take a look at:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33975


ratio
number theory
prime numbers
calculus
calculus computations

prove :D

$ \frac{\zeta(s)^{2}}{\zeta(2s)}\equal{}\prod_{p}\Big(\frac{1\plus{}p^{\minus{}s}}{1\minus{}p^{\minus{}s}}\Big)$

( where $ p$ are the prime numbers )

$ \zeta(s)\equal{}\prod_{p}(1\minus{}p^{\minus{}s})^{\minus{}1}$ is Euler's representation of the zeta function.

Therefore $ \frac{1}{\zeta(2s)}\equal{}\prod_{p}(1\minus{}p^{\minus{}2s})\equal{}\prod_{p}(1\minus{}p^{\minus{}s})\cdot\prod_{p}(1\plus{}p^{\minus{}s})$ .

Hence $ \frac{\zeta^{2}(s)}{\zeta(2s)}\equal{}\prod_{p}\frac{1\plus{}p^{\minus{}s}}{1\minus{}p^{\minus{}s}}$


AMC
AIME
geometry
perimeter
ratio
quadratics
Pythagorean Theorem

14.  Triangle APM has ∠A = 90 degrees and perimeter 152. A circle center O (on AP) has radius 19 and touches AM at A and PM at T. Find OP.

I solved this by managing to simplify this equation

$x+19+\frac{38\sqrt{x^2-19^2}}{x-19}+\sqrt{x^2-19^2}=152$, where x=OP

However I suspect there's a more clever approach to this one.

I recall solving this problem by an uninspiring trigonometric method. Kalva's solution isn't very nice either. I'm curious to see if the official solution is more ingenious and elegant.

Notice how triangle AMP is similar to TOP.  Thus, if OP = x, then AM/19 = (19 + x)/TP = PM/x
Using Pythagorean Theorem plus the information about the perimeter combined with some nasty substitution will give you your answer...

Hmm... how about this?

Clearly, triangles $APM$ and $TPO$ are similar.  Let $MA=MT=a$, and note that $OA=OT \Rightarrow OP+OT = AP$, so $OP+OT+TP = 152-2a$.  Because the ratio of the sides of two similar triangles is equivalent to the ratio of their perimeters, we have

\[\frac{a}{19} = \frac{152}{152-2a} \iff a^2-76a+19\cdot 76 = 0 \iff (a-38)^2 =0 \iff a=38.\]

Now, with $OP=x$, we get $2\sqrt{x^2-19^2} = x+19 \iff 3x^2-38x-5\cdot 19^2 = 0$.  With all the multiples of 19, we can factor this by inspection as $(3x-5\cdot 19)(x+19)=0$, and hence $x=95/3$.

Not too much better, but ehh...

Um, well I'll do this once. I have the book solution, though I kind of did it the worst way possible... sigh.

[hide]T and B are points where teh circle meets PM and AP. Triangles POT and PAM are right, and similar. Let $p_1$ and $p_2$ be their respective perimeters. Then, $OT/AM = p_1/p_2$. AM = TM, so $p_1 = p_2 - (AM + TM) = 152 - 2AM$. Thus, 19/AM = (152 - 2AM)/152, and so AM = 38, and $p_1 = 76$.

$OP/PM = p_1 / p_2$, so 1/2 = OP/PM = OP/(152 - (38 + 19 + OP). OP = 95/3.

Wow... nice solution.[/hide]

Hmm. That really killed the problem. Though it was pretty easy for a #14 in the first place... just similar triangles + ugly algebraic manipulation, nothing especially intuitive is really required (but it is preferred :P and the official solution usually requires some).

Just wondering, where does the T come in?  Is that just telling you how to label it or what?

Let AM = y, AP = x.

By similar triangles,

$\frac{PT}{OT} = \frac{x}{y} $
$\implies \frac{152 - x - 2y}{19} = \frac{x}{y}$

Also,

$\frac{OP}{OT} = \frac{152 - x - y}{y} $
$\implies \frac{x - 19}{19} = \frac{152 - x - y}{y} $
$\implies \frac{x}{19} = \frac{152 - x - y}{y} + \frac{y}{y} $
$\implies \frac{x}{19} = \frac{152 - x}{y}$

Adding these two equations:

$\frac{152 - 2y}{19} = \frac{152}{y}$

Now it's down to a quadratic (maybe an essentially equivalent one to ThAzN1's).


[quote="cooly291"]Ok, I have an exam and I am doing the review work and I have the answers (so I am not trying to get the answers ^.^) but I cannot get these three problems to work out.

[img]http://img.photobucket.com/albums/v486/cooly291/math852.jpg[/img]

Ok, I know how to solve the radical equations (isolate the radical, etc.) however, it's is the negative exponents that I do not know how to do at all. As I look through my lessons and text book, those type of problems were not in there, but they are on my review >_<. So, please, I don't need to answer, I need more of an explanation!

Thanks much![/quote]

[hide]Negative exponents just means you take the recipricol of the fraction in the base. For example $(\frac{2}{3})^{-1}=\frac{3}{2}$  and $4^{-2}=\frac{1}{16}$ Hope this helps.  :) [/hide]

For the other 2... [hide] It's just $\frac{n-v}{\sqrt n-\sqrt v}$, so multiply by the conjugate.

Get the $y$'s on 1 side, factor out the $y$ and divide. Then, rationalize the denominator. [/hide]


geometry
3D geometry
sphere
analytic geometry

If a solid sphere contains $n$ lattice (integer) points in the 3-D coordinate system, what are the maximum and minimum volumes of that sphere?

Anyone? :) I'm really wondering how to figure this out...

You might have better luck with this if you post it in the pre-olympiad section or the olympiad forum.

Ok.

Can a mod move this thread to the pre-olympiad section? Or should I just re-post it there?

If the sphere is centered at the origin, the minimum volume occurs when it is circumscribed around the cube with side $\sqrt[3]{n}$ (I think). I wonder if there's an analogy to Pick's for three dimensions?

Your best bet is probably to just re-post it.


How did everyone do on this?

I got

[code]           1  2  3  4  5  6  7  8  9 10 11
lineup      *  *  *  *  *  *  *  *  *  *  *
moat        *  x  x  x  x  x  x  x  x  x
pie1        *  *  *  *  *  *  *  *  *  *[/code]

The ones I got seemed pretty easy to me, but I got the last one wrong cause I didn't know convex hull.  And then I didn't really do the first three.

I hope this will be enough to get into gold.

Here's mine:
                        -- case number --
           1  2  3  4  5  6  7  8  9 10 11 12 13 14 15
acng        *  *  *  *  *  *  *  *  *  *  *  *  *  *  *
guess       *  *  x  *  x  x  x  x  x  x
pie1        *  *  *  *  t  t  t  t  t  t

For some reason my guess didn't quite get it right, and pie I sort of bruteforced, thus the overtime.

[quote="ThAzN1"]Hmm...

[code]           1  2  3  4  5  6  7  8  9 10 11 12 13 14 15
acng        *  *  *  *  *  *  *  *  *  *  *  *  *  *  *
guess       *  *  *  *  *  *  *  *  *  *
lineup      *  *  *  *  *  *  *  *  *  *  *
moat        *  *  *  *  *  *  *  *  *  *
pie1        *  *  *  *  *  *  *  *  *  *
skate       *  *  *  *  *  x  *  *  s  s[/code]

So wait, we didn't have to do all the problems to make gold? (This is my first year doing this.)

Typo in the title btw...[/quote]

What algorithm did you use to do convex hull? I tried the one I learned from TC, but it timed out after the first four cases.

What books/materials would you recommend for learning algorithms? Because unfortunately, as of now, I don't know very many.  :(

http://train.usaco.org
http://activities.tjhsst.edu/sct/

These are the two sources where I have learned all of my Compsci.

Well, that and

http://www.topcoder.com/tc but that is more problem-based.


For all reals $x,y,z$ we have $(\sum{x^{2}})^{2}\geq{xyz\sum{x}}$, thus from the condition, $(\sum{x^{2}})^{2}\geq{(xyz)^{2}}$ and this implies $(\sum{x^{2}})\geq{(xyz)}$.

we have by mac laurin $(\frac{a^{2}+b^{2}+c^{2}}{3})^{2}\geq{(\frac{ab+ac+bc}{3})^{2}}\geq{\frac{a+b+c}{3}abc}\geq{\frac{(abc)^{2}}{3}}$.
so $a^{2}+b^{2}+c^{2}\geq{\sqrt{3}abc}$ and we find a stronger one :wink:

ThnQ.I did the same way fermat3 did. I typed  the Q a bit wrong ,sorry to have done that way.I edited it


limit
group theory
superior algebra
superior algebra unsolved

Does a pro-$ p$-group have nontrivial center?

I'm not familiar with the theory, but I think I have a counterexample.
Define the [url=http://en.wikipedia.org/wiki/Dihedral_group]dihedral group[/url] of order $ 2^n$ for $ n\in\mathbb{N}$ as
$ D_{2^n}: \equal{} \langle x,y|x^2 \equal{} y^2 \equal{} (xy)^{2^{n \minus{} 1}} \equal{} 1\rangle$
and $ f_{m,n}: D_{2^m}\to D_{2^n},x\mapsto x,y\mapsto y$ for all $ m\ge n$.
Then the inverse limit $ D: \equal{} \lim_{\longleftarrow}(D_{2^n},f_{m,n})_{m\ge n}$ is pro-$ 2$ finite group (I think  :maybe: ) with $ Z(D) \equal{} 1$.
(the projections $ D\to D_{2^n}$ are all surjective, so $ Z(D)$ is mapped into $ Z(D_{2^n})$,
which has order $ 2$ for $ n > 2$, and is mapped to $ 1$ under $ f_{n,n \minus{} 1}$)

I was wondering.
Can somebody find some fields $ F_n$ with $ \mathcal{G} (F_n/\mathbb{Q})\cong D_{2^n}$?


trigonometry LaTeX inequalities function

let x,y,z: positive and $\tan^2{x}+\tan^2{y}+\tan^2{z}=1$. Proof: $\tan{2x}+\tan{2y}+\tan{2z} \geq \frac{3\sqrt{3}}{2}$
Greater than what?

EDIT: Equation completed

:lol: yes.that's right. I don't know how typing formula. Can everybody help me?

Click LaTeX on the side bar of this website to see a guide on how to write math stuff in the forums.

it is $\frac{3\sqrt{3}}{2}$ and not just $3\sqrt{3}$?

sorry!
Equation happen when x=y=z=pi/6.
Proof: tan2x+tan2y+tan2z>=3sqrt3

nobody answer me???? :rotfl:
this problem made from easy inequality

cauchy gives $(\tan{x}^2+\tan{y}^2+\tan{z}^2)(1+1+1) \geq (\tan{x}+\tan{y}+\tan{z})^2$, so $\sqrt{3}\geq \tan{x}+\tan{y}+\tan{z}$ (i).

$\tan{2x}=\frac{2\tan{x}}{1-\tan{x}^2}$

So $\tan{2x}+\tan{2y}+\tan{2z}=\frac{2\tan{x}}{1-\tan{x}^2}+\frac{2\tan{y}}{1-\tan{y}^2}+\frac{2\tan{z}}{1-\tan{z}^2}$.  Cauchy gives:

$\frac{2\tan{x}}{1-\tan{x}^2}+\frac{2\tan{y}}{1-\tan{y}^2}+\frac{2\tan{z}}{1-\tan{z}^2} \geq \frac{(\sqrt{2\tan{x}}+\sqrt{2\tan{x}}+\sqrt{2\tan{x}})^2}{3-(\tan{x}^2+\tan{y}^2+\tan{z}^2)}$.

expand the numerator, then to figure out the max of the rhs use the fact that $\tan{x}+\tan{y}+\tan{z}$ is maximal when it is equal to $\sqrt{3}$ [from eqn. (i)] and then use am-gm on $\tan{x}+\tan{y}$, etc.

cats made mistake!!! rhs max not correct. We need rhs min for your solving!

sorry cats...! I'm see you. Because tanx,tany,tanz<=1.

im confused now... why does the fact they're <=1 matter?

[quote="mathcat"]this problem can be solved by function[/quote]
Can you please show it?

That problem becomes:
let $x,y,z$is positive reals and $x^2+y^2+z^2=1$
Prove that $\sum \frac{x}{1-x^2} \geq \frac{3\sqrt{3}}{2}$
Solution:
we have$\frac{x}{1-x^2}=\frac{x^2}{x-x^3}$
we find: with $x\in(0,1)$ $x-x^3\leq \frac{2}{3\sqrt{3}}$
Then,problem is too easy!
 :D

[quote="mathcat"]let x,y,z: positive and (tgx)^2+(tgy)^2+(tgz)^2=1. Proof: tg2x+tg2y+tg2z>=(3sqrt3)/2 \sqrt{3}[/quote]

Proof by mathcat is nice, but the problem is a little sloppy.  Instead  of  "x,y,x" positive one must say " they are between  $0$ and $\pi /4 $  .. or something like that.  (That is  $\tan x $ positive  and not just $x$ )

As the orginal problem stands, one can choose,  $x=y=z= (11 \pi /6)   $ and  the expression you get is something negative, which is obviously less than
$ \frac {3 \sqrt 3} 2$
JMT

:lol: Oh!!! I want to say sorry again. I made mistake. I'm newcomer


function
algebra
domain
algebra proposed

Shouldn't that be $\forall x, y \in I$, not $\forall x, y \in \mathbb{R}$, since the domain of $f$ is $I$ and not $\mathbb{R}$.

[quote="perfect_radio"]Let $I \subseteq \mathbb R$ be an interval and $f : I \to \mathbb R$ be a function s.t. $\left| f(x)-f(y) \right| \leq \left| x-y \right|, \, \forall x,y \in I$.

Prove that $f$ is monotonic on $I$ if and only if $\forall x,y \in I$ we have either $f(x) \leq f \left( \frac{x+y}2 \right) \leq f(y)$ or $f(y) \leq f \left( \frac{x+y}2 \right) \leq f(x)$.[/quote]
I think it is easy ???
because:
[hide]f'(x)=1 $\to$ f(x)=x+c......[/hide]

I don't understand what you're trying to say.


I filled 49 packages with big and small pens. No package was left empty, and all the packages were filled differently. For example, a package could have been filled with 3 pens in only 4 ways: 3 big, 2 big and 1 small, 1 big and 2 small, and 3 small. What is the least total number of pens I could have put in these 49 packages?

there are 2 ways to fill a box with a total of 1 pen (1 big\0 small or 0 big\1 small)
there are 3 ways to fill a box with a total of 2 pens (2 big\0 small or 1 big\1 small or 0 big\2 small)
there are 4 ways to fill a box with a total of 3 pens
.
.
.

so in general, there are n+1 ways to fill a bow with a total of n pens
you want to find the smallest total for 49 boxes
use the 1's first
$ 1\plus{}1\plus{} \ldots$
thats 2 boxes down
then the ways to use 2 pens
$ 1\plus{}1\plus{}2\plus{}2\plus{}2 \ldots$
blah
so eventually, you figure out that you want to know when the sum $ 2\plus{}3\plus{}4\plus{} \ldots$ is less than 49 for the last time
this happens at $ 2\plus{}3\plus{}4\plus{} \ldots \plus{}7\plus{}8\plus{}9\equal{}44$
you ned 5 more boxes, so fill them with 9 pens each (because there are normally 1-0 different ways to fill a box with _pens, so _=10-1=9)
the sum is
$ 2(1)\plus{}3(2)\plus{}4(3)\plus{}5(4)\plus{}6(5)\plus{}7(6)\plus{}8(7)\plus{}9(8)\plus{}5(9)$
either evaluate this or find a clever way
evaluating this isnt too hard
the sum is
$ \boxed{285}$


The triangle ABC has incenter I. A circle lies inside the circumcircle and touches it. It also touches the sides CA, BC at D, E respectively. Show that I is the midpoint of DE.

Va baaz ham enekaase makoos...  :yinyang:

[quote="mahsa mokh"]shoma tala darin? :o chejoori gereftin?[/quote]
ehtemalan be sakhti. :rotfl:

ye soale asoon:
dar dayere $P$chahar zeli $ABCD$ ra mohat kardeim
sabet konid markaze dayerehaye mohati mosalashaye $ABC,ADC,ABD,BDC$tashkile yek mostatil midahand :?:


linear algebra
matrix
function
algebra
polynomial
integration
calculus

Let $ P$ be a linear transformation from $ \mathbb{R}^n$ tp $ \mathbb{R}^n$.
Is it true that $ \parallel{}P\parallel{} * \parallel{}P^{\minus{}1}\parallel{} \equal{}1$?
$ \parallel{}P\parallel{}$ is defined as $ sup \{\frac{\parallel{}Pv\parallel{}}{\parallel{}v\parallel{}}, v \in \mathbb{R}^n\}$

[quote="romanovodka"]Let $ P$ be a linear transformation from $ \mathbb{R}^n$ tp $ \mathbb{R}^n$.
Is it true that $ \parallel{}P\parallel{} * \parallel{}P^{ \minus{} 1}\parallel{} \equal{} 1$?
$ \parallel{}P\parallel{}$ is defined as $ sup \{\frac {\parallel{}Pv\parallel{}}{\parallel{}v\parallel{}}, v \in \mathbb{R}^n\}$[/quote]

Nope. It is only true if $ P$ is an orthogonal matrix (i. e., it preserves distances).
In all other cases, $ \left|\left|P\right|\right|\cdot\left|\left|P^{\minus{}1}\right|\right|>1$.

  darij

The quantity $ \kappa\equal{}\|P\|\cdot\|P^{\minus{}1}\|$ is called the [i]condition number[/i] of a matrix $ P.$ This is a numerical analysis concept, and a matrix for which the condition number is large is called [i]ill-conditioned[/i].

The numerical analysis idea: suppose we're trying to solve $ Px\equal{}y$ for $ x.$ We have an approximate solution $ \tilde{x},$ and we compute that $ P\tilde{x}\equal{}\tilde{y}$ where the [i]relative error[/i] of $ y,$ namely $ \frac{\|\tilde{y}\minus{}y\|}{\|y\|}$ is small. Does that mean that $ \tilde{x}$ is a good approximation of $ x?$ Not necessarily. We can say that the relative error of $ x$ is bounded by $ \frac{\|\tilde{x}\minus{}x\|}{\|x\|}\le\kappa\cdot\frac{\|\tilde{y}\minus{}y\|}{\|y\|},$ but if $ \kappa$ is large - namely, if the matrix $ P$ is ill-conditioned - then the relative error of $ x$ can be that much larger than the relative error of $ y.$

The $ n\times n$ Hilbert matrix, which is to say the matrix $ P$ whose $ j,k$ entry is $ \frac1{j\plus{}k\minus{}1},$ is notoriously ill-conditioned.

[color=green][Unnecessary quote removed by moderator.][/color]

interesting.. does this number "consistent" for every other matrix norm? i mean if we use other norm, does it also tell us about how "ill" the approximation?

also is this thing can be found in almost every number theory books?
thanks before.  :)

In detail, the condition number varies from norm to norm. But if we look past exact values, it is generally true that a well-conditioned matrix in one norm is well-conditioned in another norm, and an ill-conditioned matrix will show as so in any norm.

By the way, the subject is "numerical analysis", not "number theory". You won't find this concept in a number theory book! But do look in either numerical analysis textbooks or linear algebra books with an applied or numerical tilt to them.

I'll be back here later today to discuss the following problem:

Suppose $ f$ is a continuous function on $ [0,1].$ Find an $ n$th degree polynomial $ p(x)\equal{}\sum_{k\equal{}0}^nb_kx^k$ that minimizes the quantity
\[ \|f\minus{}p\|_2^2\equal{}\int_0^1|f(x)\minus{}p(x)|^2\,dx\]
I have  a discussion of that problem that ties into the concern with condition numbers of matrices.

Returning to finish my thought from this morning.

The problem stated above is a least-squares approximation problem. We'll have a solution when we can write $ f(x) = p(x) + r(x)$ where $ p(x)$ is the polynomial we seek and $ r(x)$ is orthogonal to every polynomial of degree $ n$ or less, which means that $ \int_0^1x^kr(x)\,dx = 0$ for $ 0\le k\le n.$

Write $ p(x) = \sum_{k = 0}^nb_kx^x,$ where the $ b_k$ are yet to be determined.

$ \int_0^1x^mf(x)\,dx = \sum_{k = 0}^mb_k\int_0^1x^mx^k\,dx + \int_0^1x^mr(x)\,dx.$

That last integral will be zero, and the other integral on the right hand side can be readily computed. The result is $ n + 1$ linear equations for the $ n + 1$ unknowns $ b_0,b_1,\dots,b_n:$
\begin{align*}b_0 + \frac12b_1 + \frac13b_2 + \cdots + \frac1{n + 1}b_n & = \int_0^1f(x)\,dx \\
\frac12b_0 + \frac13b_1 + \frac14b_2 + \cdots + \frac1{n + 2}b_n & = \int_0^1xf(x)\,dx \\
\frac13b_0 + \frac14b_1 + \frac15b_2 + \cdots + \frac1{n + 3}b_n & = \int_0^1x^2f(x)\,dx \\
& \vdots \\
\frac1{n + 1}b_0 + \frac1{n + 2}b_1 + \frac1{n + 3}b_2 + \cdots + \frac1{2n + 1}b_n & = \int_0^1x^nf(x)\,dx\end{align*}
That equation has a unique solution. So we've solved the problem, right? Well, not so fast. Theoretically we do have a solution, but if we try to compute this in practice, numerically, with all of the roundoff errors of floating point arithmetic, then take a look at the matrix of this system. That's exactly what I just called a Hilbert matrix, and I just said that it was notoriously ill-conditioned. Any small error in computation is likely to result in a large error in the values of the coefficients $ b_k.$

Is there a way around this problem? Yes, there is. What we need to do is find an orthogonal set of polynomials with the same span as $ 1,x,x^2,\dots x^n.$ We can do this by Gram-Schmidt, or we can take advantage of the fact that this is well-known. (It's a version of the Legendre polynomials.) $ p_k(x) = c_k\left(\frac {d}{dx}\right)^k[x^k(1 - x)^k]$ should do it, where the $ c_k$ are constants chosen to normalize this in some way. We do exactly what we did above: write $ f(x) = \sum_{k = 0}^nb_kp_k(x) + r(x),$ multiply by $ p_m(x),$ and integrate. As before, we get $ n + 1$ linear equations in $ n + 1$ unknowns - but now the matrix of the system isn't ill-conditioned. It's diagonal. In fact, if we choose $ c_k$ so that the $ p_k$ are orthonormal, the matrix of this system is the identity. No more numerical difficulties.

=======

Here's a strictly matrix version of the same idea. Suppose $ A$ is an $ m\times n$ matrix with $ m > n$ and we wish to solve the equation $ Ax = b.$ Well, with more equations than unknowns, that's an overdetermined system and most likely inconsistent. What we really want is the best compromise solution. The least squares problem is to find $ x$ that minimizes $ \|Ax - b\|_2^2.$

The most straightforward way to express the solution to this problem is to state the [i]normal equations.[/i] To start with, let's assume that $ A$ has full rank (rank $ n$), which in practical terms is a pretty mild requirement. Then $ A^TA$ is invertible (along with being positive definite). We solve the equation $ A^TAx = A^Tb.$ In other words, we let $ x = (A^TA)^{ - 1}A^Tb.$

That is our solution. There's only one problem, and it's exactly the problem from above: $ A^TA$ is invertible (assuming $ A$ is full rank) but it may be ill-conditioned. That potentially calls into doubt the numerical accuracy of our answer.

The cure is basically the same as the cure for the polynomial approximation problem, and it involves Gram-Schmidt. Find an orthogonal $ m\times m$ matrix $ Q$ and an upper triangular $ m\times n$ matrix $ R$ such that $ A = QR.$ (Saying $ Q$ is orthogonal is saying that $ Q^TQ = I$ or $ Q^T = Q^{ - 1}$.)

Now, minimizing $ \|Ax - b\|_2^2$ is minimizing $ \|QRx - b\|_2^2$ is minimizing $ \|Rx - Q^Tb\|_2^2.$ Note that $ Q$ is a well-conditioned as possible: its condition number is $ 1.$ We then simply ignore the bottom $ m - n$ rows and solve the upper $ n\by n$ upper triangular system.

(I could have talked about the SVD for $ A$ but decided not to go there.)

Oh gosh, i typed "number theory".. What i actually meant is "numerical analysis"..sorry, lack of concentration.. Thanks.


The boiling point of diethyl ether is 34.6˚C. Which is true
for the vaporization of diethyl ether at 25.0˚C?
(A) ∆G˚vap > 0
(B) ∆H˚vap < 0
(C) Kvap = 1
(D) ∆S˚ vap  < 0

The correct answer is A. How can you immediately tell that you gain more free energy in the reaction?

I'm also not really familiar with having Kvap=1/∆G˚vap=0. When would that ever happen and what would happen with the reaction?

A is the correct answer because at 34.6 degrees Celcius, the reaction is at equilibrium.  Lower than this temperature, the reaction will not proceed normally so the reaction is nonspontaneous so ∆G˚vap > 0.

B is incorrect because you always have to add heat to vaporize something.

C is incorrect because the reaction is nonspontaneous which means the prescence of reactants is favored so K<1.

D is incorrect because when something vaporizes the particles become more random and disoriented, so entropy actually increases.

Hope this helped.


logarithms
floor function
integration
ARML
limit
calculus
function

So, who's going?

Time to place bets on rankings!  I bet my already broken UGA puzzle that Edward will be in the top 2 or 3...again!

Ugh, I have a feeling that someone is going to start flaming me for that statement by saying that:

1). Math shouldn't be cause to gamble.
2). It's too early to be sure of anything.
3). You should care more about doing well yourself than how well others will do.
4). All of the above and this are true and that it's useless anticipating the arguments of others.

But oh well!  Good luck to everyone who needs it!  (Someone's probably going to start flaming me by saying that performance at math tourneys is independent of luck, unless if you're Heng :D )

I'll say my team will be in the top 5 in GISA. :)

Sitan, actually I will bother you about word choice.

Why did you say top 2 [b]or[/b] top 3?

If we let $ T_n$ be the set of the top $ n$, then we see that $ T_2 \subset T_3$ and thus $ T_2 \bigcup T_3\equal{}T_3$, meaning your statement is equivalent to "top 3". tsk tsk

Long live completely unneccesary posts.

And you misspelled unnecessary...

I predict Norcross actually has a choice in which four scores it will use this time!

I'm hoping that I will get top 30.

There is usually a 60% chance I spell it right.

You should come!  I don't think the tests are too hard, but there are a few pretty problems nonetheless.  You may even learn something. :D

Anyway, there are many "random" factors, such as what you have eaten recently, how much you've been sleeping, what your mood is, etc., but you can actually influence many of them if you try.  Especially sleep.  You might be surprised by how much difference this can make.

Yes, Boy Soprano is right. It [i]is[/i] very important not to sleep the night before a big tournament. Similarly, be sure [i]not to eat[/i] good food before and during the tournament, but start eating food that is bad for you and that you're not used to eating. If you have a girl/boyfriend, be sure to have a big fight on your Friday-night date immediately previous, and don't make up before the meet; the added stress will help you focus more and go through the problems more quickly, and see shortcuts you wouldn't see otherwise. These three steps will ensure you do well~

:rotfl: Billy's such a 51\/14127455. :P

But seriously, I find that the better I sleep on the day before a math tourney, the better I do.

Walton will be there with mostly a full team. No Henry, No Allen's brother. :-)

Is there some other meet tomorrow?  It seems that a lot of science/robotics people aren't attending.

The December SAT, and an FLL Qualifier hosted by Wheeler are both on tomorrow.

Congrats to all members who participated.

Miles, I'm disappointed.  I thought you would make the test super hard, but it was actually doable.  You even gave us a definition of e to help us on the last free response.

Congrats to Eric for writing a perfect paper and Chamblee for winning in varsity!

[quote="Xantos C. Guin"]Anyone want to put up results for those who weren't there?[/quote]
I second this request.

Varsity Team:
1. Chamblee - 904
2. Walton - 844
3. Northview - 836
4. Alpharetta - 715
5. Milton - 688

Varsity Individual:
1. Eric - 240
2. Oliver - 233
3. Andrei - 228
4. Sitan - 225
5. Robert - 225
6. Billy - 218
7. Boru - 218
8. Allen - 216
9. Vladimir - 211
10. Ben - 205

Congrats to Chamblee and Eric. They did a really nice job today.

[quote="Criticalline1859"]lol
Sitan lost his broken UGA puzzle. Edward wasn't there.[/quote]

And of course, we shall give our thanks to his fabulous brother, Allen, for being there.

Edward was at some honor band audition thing scheduled for 1:00 that actually took only 10 minutes.

[quote="Math Geek"]Question about a problem from the 2006 Lassiter test:

Find the greatest integer less than the base 2 logarithm of the base 10 logarithm of 2006! without certain ugliness.[/quote]
Ah, that year, Phillip was running the tournament.

We could also have something like this :
\begin{align*}
\log (2006!) &= \sum_{k=1}^{2006} \log k = \sum_{k=1}^{2006} \log k \\
&=  \sum_{k=1}^{2006} \int\limits_{1}^{k} \frac{dx}{x}\\
&= \sum_{k=1}^{2006} \sum_{j=1}^{k-1} \int\limits_{j}^{j+1} \frac{dx}{x} \\
&= \sum_{j=1}^{2006} (2006-j) \int\limits_{j}^{j+1} \frac{dx}{x} = \sum_{j=1}^{2006} \int\limits_{j}^{j+1} \frac{(2006-j)dx}{x} \\
&= \sum_{j=1}^{2006} \int\limits_j^{j+1} \frac{2006dx}{x} - \sum_{j=1}^{2006} \int\limits_j^{j+1} \frac{jdx}{x} = 2006 \log 2006 - \sum_{j=1}^{2006} \int\limits_j^{j+1} \frac{jdx}{x} \\
&\approx 2006 \log 2006 - \sum_{j=1}^{2006} 1 \\
&= 2006 \log 2006 - 2006 .
\end{align*}
Simpson's rule with $ n=1$ gives us $ \log 10 \approx 2$, so
\[ \log_{10}(2006!) \approx \frac{2006 \log 2006 - 2006}{\log 10} \approx 2006 \cdot 4 - 1000 \in (2^{12}, 2^{13}) . \]
Those estimates are kind of crude, but who cares?  It's nowhere close to outside that interval.  I actually find it easier than the arithmetic in Santhosh's first solution.  And believe it or not, it's actually very close to his second solution.  If you exponentiate the beginning and end of my big approximation (replacing 2006 with $ n$), you get
\[ n! \approx e^{n \log n - n} = n^n/e^n = (n/e)^n . \]
The $ \sqrt{2\pi n}$ comes from the fudging I did with
\[ \int\limits_{j}^{j+1} \frac{j dx}{x} \approx 1 . \]
Since we only want an order of magnitude, it's irrelevant.

I'm glad the test was doable.  (Actually, I wish I had made the hardest questions harder.)  It's bad for business to slaughter people.  Before you complain that it was too easy, though, note that there were no perfect power questions. :moose:

Yes; if the definition of $ e$ is taught in the Georgia math curriculum, it's pretty obscure.  But this way, I could have put it on the JV test from a pure knowledge point of view. :wink:

The turnout was excellent.  I think we could easily have picked an A-division ARML team from the contestants.  Congratulations also to Milton, who went from 25th to 5th place since last year.  Also, the top two JV were only 2 points away and 45 points ahead of everybody else---Grant Carlson and Hyan Kim.  I think I don't usually see this sort of attendance except at UGA and the state tournament.  Thanks for coming!

[quote] Yes; if the definition of $ e$ is taught in the Georgia math curriculum, it's pretty obscure.  But this way, I could have put it on the JV test from a pure knowledge point of view. :wink: [/quote]

Actually, the definition of $ e$ taught in my Georgia text book was $ \lim_{x\to\infty} (1\plus{}\frac{1}{x})^x$.

Grr...I can't believe I got twelfth by only nine points.  I missed $ 24$ and $ 25$, which were both A.  Ironically, my math teacher neglected to tell me that when in doubt, I should guess A for this test. :mad:

[quote="azn6021023"]
Actually, the definition of $ e$ taught in my Georgia text book was $ \lim_{x\to\infty} (1 \plus{} \frac {1}{x})^x$.

Grr...I can't believe I got twelfth by only nine points.  I missed $ 24$ and $ 25$, which were both A.  Ironically, my math teacher neglected to tell me that when in doubt, I should guess A for this test. :mad:[/quote]
Yes, I know this definition is taught, but to me it isn't quite as useful and natural as $ \sum 1/k!$.  And it holds a rather obscure place in the curriculum---to be honest, I wasn't sure whether it was taught universally, or just something Ms. Poss was adding for us.

I try to randomize my multiple-choice answers (uniformly, of course).  This happened for most of the questions on the Lassiter test this year, and it was the case for almost all of last year's test.  (Ever wonder why many of the answers are in strange orders?)  On such a test, you will have no better strategy than asking your calculator for a pseudorandom integer between 1 and 5, other than reading the problem and actually working it.  (There are a few exceptions.  On the problem that asked for a remainder mod 5, I saw no point in putting the five possibilities in anything but increasing order.  But actually, the answer was randomized for that problem as well, because I asked Python for two random integers between 6 and 15, I think.  It is possible that I might have rejected two multiples of 5, so perhaps you shouldn't have guessed 0 on that one.)

[quote="Boy Soprano II"][quote="azn6021023"]
Actually, the definition of $ e$ taught in my Georgia text book was $ \lim_{x\to\infty} (1 \plus{} \frac {1}{x})^x$.
[/quote]
Yes, I know this definition is taught, but to me it isn't quite as useful and natural as $ \sum 1/k!$.  And it holds a rather obscure place in the curriculum---to be honest, I wasn't sure whether it was taught universally, or just something Ms. Poss was adding for us.[/quote]

When I've taught Analysis in the past, I usually mention the limit definition of $ e$. It definitely gets covered in Calculus. I usually save the Taylor series definition for the Taylor Series section in Calculus. Both of these seem somewhat like magic to me until we have the Calculus techniques to prove them. (Although I can't think of how to show that the series converges to $ e$ right now.)

We could also have something like this :
\begin{align*} \log (2006!) & = \sum_{k = 1}^{2006} \log k = \sum_{k = 1}^{2006} \log k \\
& < \sum_{k = 1}^{2006} \int\limits_{k}^{k + 1} \ln(x){dx} \\
& = \int\limits_{1}^{2007} \ln(x){dx} \\
& = 2007\ln(2007) - 2006
\end{align*}
which is a bit sharper; within 2 of the actual value. Then we divide by ln(10) for fun and profit.

I actually memorized ln(2009) and ln(10) in case you did that problem for this year; 7.605 and 2.3025 :).

[quote="befuddlers"]Can anyone tell me how to do problem number 16. Since the answer is 9 i guess there are a pair of congruent triangles somewhere. :)  I seem to be missing a crucial  piece of information. :blush:[/quote]

You can assume the lines are parallel....... or you can extend the lines and use power of a point to prove they're the same.

Billy, I do not copy questions, especially not from past editions of the tournament I'm making!  Any untoward resemblance to other questions is due to unconcious recollection, other contributing authors, or sheer coincidence.  Also, shouldn't that be an \approx at the beginning of the second line?

Also, it more difficult to approximate $ \log 2007$ by hand than it is to approximate $ \log 10$, or even $ \log_{10} 2006$.  Your error is actually 3.88, which is not that much better than my 4.72.  Our estimates have roughly the same long-run error.

Re the definitions of $ e$ :  I think that the $ \sum_k 1/k!$ series is more natural because the definition $ \exp(t) \equal{} \sum_k t^k/k!$ seems a natural definition for the exponential function, whereas $ \lim_{n \to \infty} \left( 1 \plus{} \frac{t}{x} \right) ^{x}$ does not.

It is not hard to show that $ \sum t^k/k!$ converges.  Let $ N$ bet the least positive integer such that $ N \ge \lvert t \rvert$.  Then
\[ \biggl\lvert \sum_{k \ge N} \biggr\rvert \le \sum_{k \ge N} \frac{ \lvert t \rvert ^N}{N!} ( \lvert t \rvert / N)^{k\minus{}N} , \]
which is a geometric series of ratio between 0 and 1.  The other part is a finite sum.

I am not sure how one would show that $ \lim_{n \to \infty} \left( 1 \plus{} \frac{t}{n} \right)^n$ exists without calculus.  It looks like the easiest way is probably to show that
\[ n \bigl[ \log(n\plus{}t) \minus{} \log n \bigr] \equal{} n \int\limits_{n}^{n\plus{}t} \frac{dx}{x} \in \left[ \frac{nt}{n\plus{}t}, t \right] , \]
which obviously uses calculus.

Anyway, what exactly are we proving?  How do we articulate the idea of $ e$ without referring to calculus?

There are a few ways to show that
\[ \sum_{k\ge 0} t^k / k! \equal{} \lim_{n \to \infty} \left( 1 \plus{} \frac{t}{n} \right)^n . \]
One is to note that
\[ \frac{d}{dt} \lim_{n\to \infty} \left(1 \plus{} \frac{t}{n} \right)^n \equal{} \lim_{n \to \infty} \frac{1}{n} \cdot n \left( 1 \plus{} \frac{t}{n} \right)^{n\minus{}1} \equal{} \lim_{n\to \infty}\left( 1 \plus{} \frac{t}{n} \right)^n , \]
so this function is its own derivative, and go from there.

Or you can note that the coefficient $ t^k$ in $ \left( 1 \plus{} \frac{t}{n} \right)^n$ is
\[ \binom{n}{k} \cdot \frac{1}{n^k} \equal{} \frac{1}{k!} \left( \frac{n!}{n^k (n\minus{}k)!} \right), \]
and go from there.

Using the series definition of $ \exp(t)$, it's not incredibly hard to show that $ \exp(a) \exp(b) \equal{} \exp(a\plus{}b)$ combinatorially, no calculus touched.  With this, you can dig pretty deeply into it without really using calculus.

Apologies, your problems are indeed unique and awesome. And you're right, it's a <.

[quote="iniquitus"]
You can assume the lines are parallel....... or you can extend the lines and use power of a point to prove they're the same.[/quote]

Yes since it is a multiple choice question one can probably assume that the lines are parallel. Can you possibly give a solution using power of point of intersection of the $ 2$ line segments because i seem to be nowhere close to the answer. :blush:

I can try.  But I don't have the points memorized.

Extend the segments so that they meet at a new point.  Call the length of the secants to the first circle $ a\plus{}6$ and $ b\plus{}6$.  By power of a point, we have that $ a^2\plus{}6a\equal{}b^2\plus{}6b$.  If you recall the ciphering question today, you know that $ a\equal{}b$.  Sure, you could solve to get $ 6,0$, but it's not important.  Then, since you've proven that those lengths are equal, you can do power of a point again for the second circle.  Or intuitively, it seems that it will be $ 9$.

[quote="Boy Soprano II"]

It is not hard to show that $ \sum t^k/k!$ converges.[/quote]
I agree that it's easy to show the convergence, but just what does it converge to? That's a much harder question.

[quote]Anyway, what exactly are we proving?  How do we articulate the idea of $ e$ without referring to calculus?[/quote]
I like to talk about the function whose slope is the same as the value of the function. I know that it's still a reference to Calculus concepts, but easily understandable by non-calculus students.

While you guys are discussing the test, I must point out:

I'm still laughing thinking about Billy's reaction to placing.

I'm still embarrassed about how Arvind, Billy, and I were calling for a speech from Eric, and how, of course, Eric was too 1337 to reply, so our cries of "speech" were reduced to my feebly shouting the word while everyone in the building looked at me like I was an idiot, [sarcasm]which I'm clearly not[\sarcasm].

Anyways, concerning this talk about $ e$, I must point out:

I thought Miles just named the thing $ e$ for kicks and giggles, not to actually refer to the $ e$ that we have come to love.  So I spent half my time trying to find a value to which the sum would converge, and then I was like, oh crap, it's THE $ e$!

EDIT: I thought the thread ended on the first page, so I responded to Allen's post, not to the above discussion about e and my question about a problem from 2006. :lol:

[quote="Math Geek"]
I'm still embarrassed about how Arvind, Billy, and I were calling for a speech from Eric...[/quote]

Um...why is my name there?

All I remember doing is sitting between you and Billy, amused by your...eh...enthusiasm.

Fine.  Be mature. X{

Anyways, Billy's personality is veeeeeeeeeerrrrrrrrryyyyyyyy infectious.  I was definitely not like this before Duke...


linear algebra
matrix
linear algebra unsolved

nice decomposition :) due to Bruhat:
Let $A$ be a nonsingular matrix of order $n$ over a ﬁeld $F$. Then their exist nonsingular lower triangular matrices $L_{1}$ and $L_{2}$ of order$n$ and a unique permutation matrix $P$ of order $n$ such that:$A=L_{1}.P.L_{2}$

You know that beautiful (and very hard) problem posted by Alekk: for any two flags there exists an adapted basis for both of them? This decomposition is immediate from there. Is there an easier proof?

yes i know exactly the proof that you pinpoint .actualy i want to know whether there is purely algebraic solution.

I don't think this problem is difficult at all. When sam-n first posted it I just used straightforward matrix manipulations, but the proof turned out to be too hard to explain in writing for me to post it.

Ok, thanks, I got what you mean. It's just adaptation of operations on lines and columns.


Find values for k and d so that there is no real solutions (x,y) for the eqautions.
x^3+y^3 = 2
y = k*x+d

I posted a similar problema few days ago , but I noticed that I miswrote something! (not x^2 + y^2, but x^3+y^3.)

there's at least one solution if k \neq -1  !!!

Actually, I found out that I can calculate the values of k and d which satisfies both equations.

But I cannot find the values for which there is no real solutions.
Or Can I simply suggest that all the values (k,d) except for the pair that satisfies the equations are the values of k and d that has no real solution for the equations?

if $ k \neq \minus{}1$ then $ deg(x^3\plus{}(kx\plus{}d)^3\minus{}2)\equal{}3$ so there's at least one real root.
if $ k\equal{}\minus{}1$ then (except d=0) $ x^2\minus{}dx\plus{}\frac{d^3\minus{}2}{3d}\equal{}0$ hasnt real root iff $ \frac{d^3\minus{}8}{3d}<0$.theonly solutions are (k,d)=(-1,d) with $ d \le 0 \text{ or } d>2$

bye


Let f(x) = 4x/(4x + 2) with x \in R. Evaluate \sum 1994k=1  f(k/1995).

created by Sefket Arslanagic

Indeed this is the classical trick

f(x) + f(1-x) = 1

f(1995) = 997


Points $ K,L,M,N$ are repectively the midpoints of sides $ AB,BC,CD,DA$ in a convex quadrliateral $ ABCD$.Line $ KM$ meets dioganals $ AC$ and $ BD$ at points $ P$ and $ Q$,respectively.Line $ LN$ meets dioganals $ AC$ and $ BD$ at points $ R$ and $ S$,respectively.
Prove that if $ AP\cdot PC\equal{}BQ\cdot QD$,then $ AR\cdot RC\equal{}BS\cdot SD$.

Call $ PQ\cap AD\equal{}T$
Use Menelaus theorem for $ 2$ triangle $ ABD,ACD$ to prove $ \frac{AP}{PC}\equal{}\frac{BQ}{QD}$.
Then $ AP.PC\equal{}BQ.QD$ mean $ AP\equal{}BQ,PC\equal{}QD,AC\equal{}BD$
Then OK!

Misollar unchalik qiyin emaskanu, lekin Uzbekistan faqatgina bitta bronze medal olgani chatoq bo'lipdida!
I think my solutin isn't nice.
Let $ AC\cap BD\equal{}O$.Sice points$ K,P,Q$ are collinear using  Menelaus theorem we have $ \frac{BQ}{AP}\equal{}\frac{QO}{OP}$, because $ AK\equal{}KB$.Similarly $ \frac{PC}{QD}\equal{}\frac{OP}{OQ}$.Then condition of the problem states that $ QO\equal{}OP$.Then
$ \angle LKM\equal{}\angle OPQ\equal{}\angle OQP\equal{}\angle LMK$.Hence $ KLMN$ is a rhombus, i.e $ LN\perp KM$.Then $ \angle ORS\equal{}90^\circ\minus{}\angle OPQ\equal{}\angle OSR$.Hence $ OR\equal{}OS$.(1)
Since points $ R,S,N$ are collinear using  Menelaus theorem we have  $ \frac{AR}{SD}\equal{}\frac{OR}{OS}$ (2)
, because $ AN\equal{}ND$.Similarly$ \frac{BS}{RC}\equal{}\frac{SO}{OR}$ (3).From (1),(2) and (3) we have $ 1\equal{}\frac{AR}{SD}\equal{}\frac{BS}{RC}$ or $ AR\cdot RC\equal{}BS\cdot SD$ and we are done.

[quote="Ibrogimov"]Misollar unchalik qiyin emaskanu, lekin Uzbekistan faqatgina bitta bronze medal olgani chatoq bo'lipdida![/quote]
Please,write in english.
My solution to this problem is very short.Hope some of you will find it.

[hide="Another way to prove that AP/PC=BQ/QD from the official solution"]
Denote $ d(X,YZ)$ by the distance between point $ X$ and a line $ YZ$. Note that $ \frac{AP}{PC}\equal{}\frac{d(A,KM)}{d(C,KM)}\equal{}\frac{d(B,KM)}{d(D,KM)}\equal{}\frac{BQ}{QD}$. Thus $ \frac{AP}{PC}\equal{}\frac{BQ}{QD}$.[/hide]

In our team one boy solved this by trigonometric methods.[/quote]

let $U,V$ be the intersections of AB and CD;AD and BC.
by Menelaus,$AP*PC=\frac{AP*PC}{AC^2}*AC^2=\frac{CM*UK*MU*KA}{CM*UK+MU*KA)^2}*AC^2=\frac{DM*MU*UK*KB}{(DM*UK+MU*KB)^2}*BD^2=BQ*BD$
hence $AC=BD$
analagously,$AR*RC=BS*SD\leftrightarrow AC=BD$.

Here's a bashy way to get to the magical $AC = BD$.

Let $X$ and $Y$ be the midpoints of segments $AC$ and $BD$, respectively, and $O = \overline{AC} \cap \overline{BD}$. Let $G$ be the centroid of $ABCD$, or otherwise $G = \overline{XY} \cap \overline{KM} \cap \overline{NL}$. WLOG, we can assume that $G$ lies inside $\triangle BOC$. Let $Z = \overline{GX} \cap \overline{MN}$ and $T = \overline{GY} \cap \overline{NK}$.

The idea of this setup is to interpret $AP\cdot PC$, $BQ \cdot QD$, $AR \cdot RC$, and $BS\cdot SD$ through the midpoints $X$ and $Y$ as the powers of the points $P, Q, R, S$ with respect to the circles with diameters $AC$ and $BD$. That is, $AP \cdot PC = AX^2 - XP^2$, $BQ\cdot QD = BY^2 - YQ^2$, $AR \cdot RC = AX^2 - XR^2$, and $BS \cdot SD = BY^2 - YS^2$. Given that $AP \cdot PC = BQ \cdot QD$, proving $AR\cdot RC = BS \cdot SD$ is equivalent to showing that $XP^2 - XR^2 = YQ^2 - YS^2$. For that reason, we introduced points $Z$ and $T$, and using Thales's theorem repeatedly, we get that:
\begin{align*}
&XP^2 - XR^2 = YQ^2 - YS^2 \\
&\iff \left(\frac{GR}{GN}\right)^2 \cdot (ZM^2 - ZN^2) = \left(\frac{GY}{GT}\right)^2 \cdot (TK^2 - TN^2) \\
&\iff \left(\frac{\text{dist}(G, \overline{AC})}{\text{dist}(G, \overline{MN})}\right)^2 \cdot (ZM^2 - ZN^2) = \left(\frac{\text{dist}(G, \overline{BD})}{\text{dist}(G, \overline{NK})}\right)^2 \cdot (TK^2 - TN^2)
\end{align*}
However, we have that
\[\text{dist}(G, \overline{AC}) = \frac{1}{2}\left(\text{dist}(B, \overline{AC}) - \text{dist}(D, \overline{AC})\right)\quad \text{and} \quad \text{dist}(G, \overline{MN}) = \text{dist}(G, \overline{AC}) + \frac{1}{2}\text{dist}(D, \overline{AC}) = \frac{1}{2}\text{dist}(B, \overline{AC})\]
Similar expressions can be obtained for $\text{dist}(G, \overline{BD})$ and $\text{dist}(G, \overline{NK})$, whence
\begin{align*}
&XP^2 - XR^2 = YQ^2 - YS^2 \\
&\iff \left(\frac{\text{dist}(B, \overline{AC}) - \text{dist}(D, \overline{AC})}{\text{dist}(B, \overline{AC})}\right)^2 \cdot (ZM^2 - ZN^2) = \left(\frac{\text{dist}(C, \overline{BD}) - \text{dist}(A, \overline{BD})}{\text{dist}(C, \overline{BD})}\right)^2 \cdot (TK^2 - TN^2)\\
&\iff \left(1 - \frac{DO}{BO}\right)^2 \cdot (ZM^2 - ZN^2) = \left(1 - \frac{AO}{CO}\right)^2 \cdot (TK^2 - TN^2)
\end{align*}
Now, by Menelaus's theorem for $\triangle MNL$ and the line $Z, G, T$, we have that $\frac{KT}{TN} \cdot \frac{NZ}{ZM} = 1$ as $G$ is the midpoint of $MK$. Substituting in $r = \frac{TK}{TN}$ for brevity implies $\frac{ZM}{ZN} = r$ and so we get:
\begin{align*}
&XP^2 - XR^2 = YQ^2 - YS^2 \\
&\iff  \left(1 - \frac{DO}{BO}\right)^2 \cdot (ZM^2 - ZN^2) = \left(1 - \frac{AO}{CO}\right)^2 \cdot (TK^2 - TN^2)\\
&\iff  \left(1 - \frac{DO}{BO}\right)^2 \cdot \frac{r^2-1}{(r+1)^2}\cdot  MN^2= \left(1 - \frac{AO}{CO}\right)^2 \cdot \frac{r^2-1}{(1-r)^2} \cdot NK^2 \\
&\iff  \frac{\left(1 - \frac{DO}{BO}\right)^2}{\left(1 - \frac{AO}{CO}\right)^2} \cdot \frac{AC^2}{BD^2} = \left(\frac{1+r}{1-r}\right)^2
\end{align*}
The initial condition from the problem statement should imply the last line, but
\begin{align*}
&AP\cdot PC = BQ \cdot QD\\
&\iff AC^2 - 4XP^2 = BD^2 - 4YQ^2 \\
&\iff AC^2 - BD^2 = 4\left(\frac{GR}{GN}\right)^2\cdot ZM^2 - 4\left(\frac{GY}{GT}\right)^2 \cdot TK^2 \\
&\iff AC^2 - BD^2 = 4\left(1 - \frac{DO}{BO}\right)^2 \cdot \frac{r^2}{(1+r)^2} \cdot MN^2 - 4 \left(1 - \frac{AO}{CO}\right)^2 \cdot \frac{r^2}{(1-r)^2} \cdot NK^2 \\
&\iff AC^2 - BD^2 = \left(1 - \frac{DO}{BO}\right)^2 \cdot \frac{r^2}{(1+r)^2} \cdot AC^2 -  \left(1 - \frac{AO}{CO}\right)^2 \cdot \frac{r^2}{(1-r)^2} \cdot BD^2 \\
&\iff AC^2 \cdot \left(1 - \frac{r^2}{(1+r)^2}\cdot  \left(1 - \frac{DO}{BO}\right)^2\right) = BD^2 \cdot \left(1 - \frac{r^2}{(1-r)^2}\cdot \left(1 - \frac{AO}{CO}\right)^2\right)
\end{align*}
Comparing this equation with the previous $ \frac{\left(1 - \frac{DO}{BO}\right)^2}{\left(1 - \frac{AO}{CO}\right)^2} \cdot \frac{AC^2}{BD^2} = \left(\frac{1+r}{1-r}\right)^2$ shows that the two are equivalent if and only if $\frac{AC}{BD} = 1$. Having conjectured this claim, it isn't that hard to prove. If $\overline{MK}\cap\overline{AD} = E$, then by Menelaus's theorem applied on $\triangle ACD$ and the line $E, M, P$ and $\triangle ABD$ and the line $E, Q, K$ implies that $\frac{AP}{PC} = \frac{BQ}{QD}$, which readily implies $AC = BD$ because $AP \cdot PC = BQ \cdot QD$.