IMO Shortlist
trigonometry
geometry
circumcircle
hexagon

An acute triangle $ ABC$ is given. Points $ A_1$ and $ A_2$ are taken on the side $ BC$ (with $ A_2$ between $ A_1$ and $ C$), $ B_1$ and $ B_2$ on the side $ AC$ (with $ B_2$ between $ B_1$ and $ A$), and $ C_1$ and $ C_2$ on the side $ AB$ (with $ C_2$ between $ C_1$ and $ B$) so that

\[ \angle AA_1A_2 \equal{} \angle AA_2A_1 \equal{} \angle BB_1B_2 \equal{} \angle BB_2B_1 \equal{} \angle CC_1C_2 \equal{} \angle CC_2C_1.\]

The lines $ AA_1,BB_1,$ and $ CC_1$ bound a triangle, and the lines $ AA_2,BB_2,$ and $ CC_2$ bound a second triangle. Prove that all six vertices of these two triangles lie on a single circle.

[quote="orl"]An acute triangle $ ABC$ is given. Points $ A_1$ and $ A_2$ are taken on the side $ BC$ (with $ A_2$ between $ A_1$ and $ C$), $ B_1$ and $ B_2$ on the side $ AC$ (with $ B_2$ between $ B_1$ and $ A$), and $ C_1$ and $ C_2$ on the side $ AB$ (with $ C_2$ between $ C_1$ and $ B$) so that
\[ \angle AA_1A_2 \equal{} \angle AA_2A_1 \equal{} \angle BB_1B_2 \equal{} \angle BB_2B_1 \equal{} \angle CC_1C_2 \equal{} \angle CC_2C_1.
\]
The lines $ AA_1,BB_1,$ and $ CC_1$ bound a triangle, and the lines $ AA_2,BB_2,$ and $ CC_2$ bound a second triangle. Prove that all six vertices of these two triangles lie on a single circle.[/quote]
We assume that,
\[ \angle AA_1A_2 \equal{} \angle AA_2A_1 \equal{} \angle BB_1B_2 \equal{} \angle BB_2B_1 \equal{} \angle CC_1C_2 \equal{} \angle CC_2C_1 \equal{} \theta
\]
At first we notice that for any given point $ A_1$ the constructions of the other points are unique. To construct $ A_2$ we have to take point symmetric to the foot of perpendicular from $ A$. And analogously we have to construct other points by rotating the altitudes by angle $ \frac \pi2 \minus{} \theta$.

Now we chase the angles. We get easily that, $ \angle P_1Q_1R_1 \equal{} \pi \minus{} (\pi \minus{} \theta) \minus{} (\frac \pi2 \minus{} B \minus{} (\frac \pi2 \minus{} \theta)) \equal{} B$. Analogous angle chasing assures that $ \triangle ABC \sim \triangle P_1Q_1R_1 \sim \triangle P_2Q_2R_2$

Now We shall prove that $ P_1R_2 \parallel AC$. We have $ BB_1 \equal{} BB_2$. So it is enough to prove that $ BP_1 \equal{} BR_2$. We have $ \angle R_2AB \equal{} \angle CP_1B \equal{} \pi \minus{} B \minus{} \theta \equal{} \phi$.
So, Sine law in $ \triangle R_2AB,\triangle CP_1B$ gives,
\[ \frac {BR_2}{\sin \phi} \equal{} \frac {AB}{\sin C} \text{ and } \; \frac {BP_1}{\sin \phi} \equal{} \frac {BC}{\sin A}
\]
Dividing the first equation by the second one we get, $ BZ_2 \equal{} BP_1$

Now we shall prove $ R_1P_2 \parallel CA$  i.e.  $ BR_1 \equal{} BP_2$.

Sine law in $ \triangle BP_2C_2, \triangle BR_1A_1$ gives,
\[ \frac {P_2B}{\sin \theta} \equal{} \frac {BC_2}{\sin A} \; \text{ and } \frac {BR_1}{\sin \theta} \equal{} \frac {A_1B}{\sin C}
\]
Let, $ \angle C_2CB \equal{} \angle AA_1B \equal{} \theta \minus{} B \equal{} \alpha$

Again by sine law,
\[ \frac {P_2B}{\sin \alpha} \equal{} \frac {BC}{\sin \theta}\; \text{ and } \frac {A_1B}{\sin \alpha} \equal{} \frac {AB}{\sin \theta}
\]
From these four equations we get, $ BR_1 \equal{} BP_2 \iff R_1P_2 \parallel AC \parallel P_1R_2$.

And also $ R_2Q_1 \parallel BC$. Now straight forward angle chasing gives, $ \angle P_1R_2Q_1 \equal{} \pi \minus{} (\theta \minus{} A) \minus{} (\pi \minus{} \theta \minus{} B) \equal{} A \plus{} B \equal{} \pi \minus{} C$.
Which proves that $ R_2$ lies on the circumcircle of $ \triangle P_1Q_1R_1$. Analogously, $ Q_2,R_2$ lies on the circle proving the claim.

EDIT: Fixed typo.
I hope that I haven't made any more typo :)

This is my solution  :)
We have $ \angle BC_1J \equal{} \angle AA_1A_2 \Rightarrow BC_1JA_1$ is cyclic $ \Rightarrow AJ.AA_1 \equal{} AC_1.AB$
Similarly, $ AK.AA_2 \equal{} AB_2.AC$
But $ \angle BC_1C \equal{} \angle BB_2C \Rightarrow BC_1B_2C$ is cyclic.
$ \Rightarrow AC_1.AB \equal{} AB_2.AC$
$ \Rightarrow AJ.AA_1 \equal{} AK.AA_2$
Similarly, $ AN.AA_1 \equal{} AM.AA_2$
We get $ AJ.AN \equal{} AK.AM \Rightarrow NJKM$ is cyclic.
But $ JA_1A_2K$ and $ LC_1C_2M$ are cyclic then $ \angle JKM \equal{} \angle AA_1A_2 \equal{} \angle CC_2C_1 \equal{} \angle JLM$
$ \Rightarrow JKLM$ is cyclic. Similarly, $ JPNM$ is cyclic.
$ \rightarrow$ QED

@[b]livetolove212[/b]: Nice solution, but hard for me to think of. :)

BTW there is another solution, faster than ours. For this we need to show that $ H$ is the center of the circle. That can be done by calculating the radius using trigonometry. :)

[quote="Moonmathpi496"]
BTW there is another solution, faster than ours. For this we need to show that $ H$ is the center of the circle. That can be done by calculating the radius using trigonometry. :)[/quote]

Can you post this solution? Thanks a lot  :)

Actually this solution is from IMO compendium. This is surprisingly short for an ISL 04.

Triangles $ AHB, BHC, CHA, ABC$ have the same circumradius $ R$. $ BCP_1H$ is concyclic and therefore $ HP_1 \equal{}2R \cos \theta$. The same holds for the other radii and the conclusion follows.

I’m going to use the second post’s picture and write my solution using it.
 First it’s easy to see that $AR_2HR_1B$ , $CQ_2HQ_1A$ and $CP_1HP_2B$ lie on one circle.Then we know that $AH$ and $BH$are the angle bisector of $\angle A_1AA_2$ and $\angle B_1BB_2$ so $R_2H=HR_1$ ( using the same way $P_1H=HP_2$ & $Q_1H=HQ_2$ )  $(1)$
By angle chasing we find that $\angle AB_1C_2=\angle AQ_2C_2=\angle B$ and also $\angle B_1R_1A= \angle B_1C_2A= \angle C$ thus $B_1,Q_2,R_1,C_2,A$ lie on one circle. Then we have $B_1Q_2R_1C_2$ and also $B_1C_2BC$ are concyclic using this it’s easy to see that: $Q_2R_1 \parallel CB$
Thus $AH$ is the perpendicular bisector of $Q_2R_1 \Rightarrow HQ_2=HR_1$ (with the same way: $HQ_1=HP_2$ & $HP_1=HR_2$) combining with $(1)$ $\Rightarrow$ the points $P_1,P_2,Q_1,Q_2,R_1,R_2$ lie on a circle which $H$ is it’s center.
[img]https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYS9lLzIwMzNlNjNlYzNlZThiY2RhZTI1MmJhM2I5MmI1NmNiNTE3YWM4LnBuZw==&rn=RmlndXJlLnBuZw==[/img]


Prove that if $ a, b, c and d$ are non negative reals :

$ \sum \frac{1}{a^2\plus{}b^2\plus{}c^2} \ge \frac{12}{(a\plus{}b\plus{}c\plus{}d)^2}$

I don't know the topic. Can you say clearly?

[quote="Aravind Srinivas L"]Prove that if $ a, b, c and d$ are non negative reals :

$ \sigma cyclic \frac tion 1/(a^2 + b^2 + c^2) > = \frac tion 12/(a + b + c + d)^2$

I am sorry for the latex.[/quote]

You say $ a,b,c,d$are non negative reals. prove that
$ \sum \frac{1}{a^2+b^2+c^2} \ge \frac{12}{(a+b+c+d)^2}$?

the q has been rectified.
The topic meant that this is a good example of an inequality which can be solved using the Cauchy lemma[or Titu Lemma].

Well basically it boils down to
$ \sum a^2  \ge \sum ab$


Do the awards listed under intramural awards on the AMC website get sent to the schools?  If so, any idea when?

what are inturmeral awards?

       sorry for not answering your question :(

I'm not sure, because last year at Longfellow my friend and I each received a pin at our school, but my teacher said he just had a bunch of them in his desk that he forgot to give to the students from previous years. :/

This question can be answered by a careful reading of the AMC 10/AMC 12 Teachers' Manual.
In fact, every policy and adminstration question for the AMC 8, AMC 10, AMC 12, AIME and USAMO can be answered by reading the corresponding Teachers' Manual.  Each Manual is posted on our website before it is printed and remains there indefinitely.

Your school Contest Manager has a paper copy of the 2006 AMC 10/AMC 12 Teachers' Manual.  An exact electronic copy of precisely the same source is on our website at
http://www.unl.edu/amc/d-publication/d1-pubarchive/2005-6pub/2006AMC1012/html/06tm-04-text.html

All intramural awards for the A-date contests were sent to schools about Feb14, 2006.  All intramural awards for the B-date contests were sent to school about Feb 25, 2006.  All intramural awards for the AIME were sent (at latest) earlier this week.  National and regional awards will be sent this month and next as we get them made.     Your Contest Manager should have ALL the school awards in his/her possession, unless they have already been distributed.

On page 8 of the Manual you will find:
XI. Intramural and National Awards

Intramural Awards
The AMC Office will send you Intramural awards, along with your school results. Your registration fee entitles you to pins, medals, certificates, and a copy of the National Summary of Results and Awards. An order form for additional intramural awards will be included with your results.

Winner Pin — given to the top scoring student in your school for both the AMC 10 and AMC 12. Medals will be given for consecutive wins in both contests mentioned above. Details of how medals are awarded will be included with your school’s results.

Certificate of Distinction — awarded to all students who qualify for the AIME.

Honor Roll of Distinction Pin — given to the top 1% of the AMC 12 and to the top 1% of AMC 10 participants.

AMC 12 Certificate of Achievement — given to students in grade 10 and below who score 90 or above on the AMC 12 Contests.

AMC 10 Certificate of Achievement — given to students in grade 8 and below who score 90 or above on the AMC 10 Contests.

School Certificate of Honor — awarded to schools with a team score (AMC 12) of 400 or greater.

School Certificate of Merit — awarded to schools with a team score (AMC 12) between 300 and 399, inclusive.

State/Provincial Awards
Plaques –– Top Scoring Student Plaque awarded to a student in each state or province.

Some State/Provincial Directors offer various awards or scholarships in addition to the plaques.

National Awards
In each of the eleven regions the AMC divides the United States and Canada into, the five schools with the highest team scores (sum of the highest three scores by participants) are recognized by Donor or Committee Awards.

The nationally highest scoring team will receive the Charles T. Salkind Memorial Silver Cup, a silver Trophy Cup awarded to the school with the highest team score over all, donated by the L.G. Balfour Company, Attleboro, MA.

The second nationally highest scoring team will receive the William H. Fagerstrom Memorial Silver Cup, a silver Trophy Cup awarded to the school with the second highest team score over all, also donated by the L.G. Balfour Company, Attleboro, MA.

In each of the regions in which a silver cup is not awarded the Committee on the American Mathematics Competitions provides a Committee Bronze Cups to the school with the highest team score.

Awards for Schools not receiving a Trophy Cup:
The four remaining schools in each region will receive one of these awards. Distribution of awards is arranged so that schools do not receive duplicates of awards made to them in prior years.

CAMC Mathematics Books — in each of the eleven regions, five books are donated by the Committee on the American Mathematics Competitions to one school having a high team score.

W. H. Freeman Books — in each of the eleven regions, a set of books, donated by W. H. Freeman Company, San Francisco, CA is awarded to one school having a high team score.

Mathematics Magazine — in each of the eleven regions, a one-year subscription is donated by the Committee on the American Mathematics Competitions to two or more schools having a high team score.

Most Improved Team Score Award
Pedagoguery Award — awarded to one school in each of the 50 states, US Territories, Military Schools abroad, and Canada with the “Most Improved” Team Score on the AMC 10 and AMC 12.

Steve Dunbar
AMC Director

Question: Are all schools who competed on AMC receiving this?

The reason is that last year, I scored highest on AMC-10. I didn't make AIME but I was still highest on my school. But I got NOTHING.

This year, I made AIME and scored highest on AMC-12. Should I get something? Or my teacher didn't receive one? Or is that my teacher received one and not giving to me?


Let m and n be positive integers. Find a closed formula for the following sum :)

$ \displaystyle\sum_{i=0}^{n} \sum_{j=1}^{m} \sum_{k=0}^{n} \dbinom {n-k-(m-j)(i-1)-i-1}{m-j-1} \dbinom{k-i(j-1)-1}{j-2}$

Where,
$ \dbinom nr = \left\{\begin{array}{cc}& \dbinom {-1}{-1}=1 \
& \text {Binomial Coefficient  } \forall {  n,r } \ge 0\
& 0,\text{ otherwise.}\end{array}$

Just to remove ambiguities...

Binomial Coefficient = $ \binom nr = \left\{\begin{array}{cc}& \frac {n!}{r!(n - r)!}\text{ if }n\ge r, \
& 0,\text{ otherwise.}\end{array}$


Find every integer solution(s) $ (x;y;z)$ of the equation: $ x!\equal{}y!z!$

Oh!What an interesting problem!
Anyone solve it?It's quite hard for me. :maybe:

This is listed in Guy's "Unsolved Problems in Number Theory", so unless surprising progress was made in the last few years, I suspect it may be an open problem.


inequalities
rearrangement inequality
inequalities proposed

Let a,b,c>0.Prove that:

$ \frac{a^3}{(b\plus{}c)(a\plus{}c)}\plus{}\frac{b^3}{(b\plus{}a)(a\plus{}c)}\plus{}\frac{c^3}{(b\plus{}c)(a\plus{}b)}>\equal{}\frac{a\plus{}b\plus{}c}{4}$

this might be ok...

$ \frac {a^3}{(b \plus{} c)(a \plus{} c)} \plus{} \frac {b^3}{(b \plus{} a)(a \plus{} c)} \plus{} \frac {c^3}{(b \plus{} c)(a \plus{} b)} \geq \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{\sum a(b\plus{}c)(c\plus{}a)}$, now using $ a^2\plus{}b^2\plus{}c^2 \geq \frac{1}{3}(a\plus{}b\plus{}c)^2$ we get:

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

So it's enough to prove $ \frac {(\frac{1}{3}(a \plus{} b \plus{} c)^2)^2}{\sum a(b\plus{}c)(c\plus{}a)} \geq \frac{a\plus{}b\plus{}c}{4}$, which simplifies to

$ 4\sum a^3 \plus{} 3\sum a^2b \geq 3abc\plus{} 6 \sum a^2b$.

Now it's enough to have $ \sum a^3 \plus{} \sum a^2b \geq 2\sum ab^2$, which is equivalent to $ \sum a^3 \plus{} (a\minus{}b)(a\minus{}c)(b\minus{}c) \geq \sum ab^2$.

We may assume that $ a$ is the maximal element of $ a,b,c$. Inequality is obvious if $ b \geq c$.

So we only need to prove it for $ a \geq c \geq b$.
Now $ \frac {1}{(a\plus{}c)(a\plus{}b)} \leq \frac{1}{(a\plus{}c)(c\plus{}b)} \leq \frac {1}{(a\plus{}b)(c\plus{}b)}$.
By rearrangement inequality
$ \frac {a^3}{(b\plus{}c)(a\plus{}c)}\plus{}\frac {b^3}{(b\plus{}a)(a\plus{}c)}\plus{}\frac {c^3}{(b\plus{}c)(a\plus{}b)} \geq \frac {a^3}{(a\plus{}b)(a\plus{}c)}\plus{}\frac {b^3}{(b\plus{}a)(b\plus{}c)}\plus{}\frac {c^3}{(a\plus{}c)(c\plus{}b)}$.
But $ \frac {a^3}{(a \plus{} b)(a \plus{} c)} \plus{} \frac {b^3}{(b \plus{} a)(b \plus{} c)} \plus{} \frac {c^3}{(a \plus{} c)(c \plus{} b)} \geq \frac{a\plus{}b\plus{}c}{4}$ is well known and many times posted on this forum.

[quote="cuopbientcd"]Let a,b,c>0.Prove that:

$ \frac {a^3}{(b \plus{} c)(a \plus{} c)} \plus{} \frac {b^3}{(b \plus{} a)(a \plus{} c)} \plus{} \frac {c^3}{(b \plus{} c)(a \plus{} b)} > \equal{} \frac {a \plus{} b \plus{} c}{4}$[/quote]

Not hard. Just AM-GM

$ \frac {a^3}{(b \plus{} c)(a \plus{} c)} \plus{} \frac {b \plus{} c}{8} \plus{} \frac{a \plus{} c}{8} \geq \frac {3}{4}a$

This inequality is an immediate consequence of the well known inequality


geometry
analytic geometry
vector
absolute value
complex numbers
AMC

Let z=-1+16i. let w be a complex number with absolute value 1. wz can be written as a+bi, where a,b are reals. Find the maximum value of b.

[hide]Let w = c + di. After multiplying that expressions with z = -1 + 16i, we find wz = -(c + 16d) + (16c - d)i. Because the absolute value of w is 1, we have Sqrt(c^2+d^2)=1<=>c^2+d^2=1. To maximize 16c - d, we let d = 0, so c^2+0^2=c^2=1<=>c=+/- 1. We obviously only want the positive value, and so the maximum value of 16c - d is 16*1-0=16.[/hide]

I believe you assumed d :ge: 0 in that proof, but that is not at all necessary.

Here's a good thought for this one:  [hide]think about the geometry of complex numbers, not the algebra.  What happens when you multiply by a number with absolute value 1?  This is a pretty question, I think.[/hide]

Quote:I believe you assumed d  0 in that proof, but that is not at all necessary.

Yes, my mistake.

[hide]And as for your thought, it is sorta hard to picture the geometry of complex numbers here when you consider that the absolute value of a+bi is Sqrt(a^2+b^2). How would you picture that on the complex plane?[/hide]

:sqrt:(a:^2: + b:^2:) doesn't look familiar to you?

Do you know about the polar form of complex numbers?  (Or, for that matter, the exponential form?)  Demoivre's formula?  If not, then this won't be helpful.  But you need to know two things:  what is the set of all points with absolute value 1?  What happens when you multiply one complex number by another?

Oh, I thought you meant imagine them in the complex plane (x-axis real, y-axis imaginary). Its easier with polar coordinates, yeah.

Both planes are the same.  They both give a geometric representation of the complex numbers and they both identify the same complex number with the same point.  The only difference is the coordinates used to describe that point.  Multiplication by a complex number looks the same if you're using x- and y-axes as if you're using polar coordinates -- it's just that it looks messier in x's and y's.

Wait, you have me thoroughly confused.

How can |z| where z is a complex number be less than 0? Because my answer was what Fyty got.

[quote]How can |z| where z is a complex number be less than 0? [/quote]

It can't be, although both c and d can be when z=c+di

But.. the answer is 16!!

when you multiply two complex numbers, you add the number of degrees, and multiply the  absoloute values. So the absolute value of a+bi will be the same is -1+16i. Now draw a circle with that equation. The maximum of i value is at -1+16i, so it's 16!!!

Everything you said was true, Southpaw, except your last sentence.

i don't think anyone answered this one yet, so here goes
[spoiler]multiplying two complex numbers means adding their angles and multiplying their lengths. clearly, the length of the product will be the length of 16i-1, since the length of the other is 1. the angle of the second vector can be anything we want, so we choose it to make the product point straight up, maximizing b. (as the second vector moves around the unit circle, the product vector moves around the circle centered at the origin with radius :sqrt:257), so the maximum possible value of b is :sqrt: 257.

mystic is correct
for an easy solution to this question:
check out Cauchy-Schwarz

Joel, im surprised u didn't see it....

[quote]check out Cauchy-Schwarz [/quote]

Huh? How can we do this?

How is that easier than saying it's a circle with radius sqrt(257) centered at 0, so the maximum of the coefficient of i must be sqrt(257)?

If you happen to own AOPS, it's in book 2, chapter 14.  If not, sniff around on this site -- it's been discussed before, fairly often actually.  Look at the Advanced Forum -- it should be at least mentioned under the thread "Useful Inequalities."

[quote="teratomato"][quote="darkquantum"]but its harder to figure out that that cirlce has radius root 257[/quote]
If you seriously think using the distance formula is harder than using the Cauchy-Schwarz inequality... then you are a BIG DUMMY HEAD HAHAHAHAHAHA.[/quote]
...although I agree with you that's not really appropriate nor is it nice...

Probably the easiest way to remember Cauchy-Schwarz (honestly, the way I remember it) it to remember that the dot product of two vectors is <= the product of the magnitudes.


geometry
trigonometry
angle bisector
algebra
system of equations

You have triangle $ \triangle ABC$, where $ m\angle ABC = m\angle BAC = 40^{\circ}$. The angle bisector of $ \angle ABC$ intersects side $ \overline {AC}$ at $ D$, and is extended to $ E$ such that $ DE = DA$. Find the $ m\angle DEC$.

And before you tattle on me by saying this is from http://www.artofproblemsolving.com/Forum/viewtopic.php?p=975012 , it's not. It's altered slightly, making the method and problem entirely different.

(Don't read that topic; it has my solution to this problem in it.)

Your solution in the thread you provided is incorrect.

(1) Your system of equations is not independent; any of the four equations can be deduced from the other three. For example, we can have y=50,x=10,x_1=70,y_1=20 work just the same.
(2) If <DEC=20, then EC=CB=AC, meaning ECA is isosceles, but this is not the case as <DAE=60 and <CEA=80.
(3) You didn't even use the definition of E, should worry you slightly.

[quote="Temperal"]$ \left\|\begin{array}{c} \triangle\ ABC\ : \ B = C = 40^{\circ} \\
= = = = = = = = = = = = = = = \\
\ D\in (AC)\ ,\ \widehat {ABD}\equiv\widehat {CBD} \\
\ D\in (BE)\ ,\ DE = DA\end{array}\right\|$ $ \implies$ $ m(\widehat {DEC}) = 80^{\circ}\ .$ .[/quote]
[color=darkblue][b]Proof.[/b] Denote $ m(\widehat {DEC}) = x\ .$ In the quadrilateral $ ABCE$ apply the well-known relation

$ \boxed {\sin \widehat {CAB}\cdot\sin \widehat {EBC}\cdot\sin\widehat {ACE}\cdot\sin\widehat {BEA} = \sin \widehat {ABE}\cdot\sin \widehat {BCA}\cdot\sin\widehat {CEB}\cdot\sin\widehat {EAC}}$

Ascertain easily the angles which appear in the upper relation and it becomes

$ \sin 100\sin 20\sin (120 - x)\sin30 = \sin 20\sin 40\sin x\sin 30\ \ (*)\ .$

Therefore, $ \cos 10\cos(30 - x) = \sin 40\sin x$ $ \Longleftrightarrow$

$ \cos (40 - x) + \cos (x - 20) = \cos(40 - x) - \cos (40 + x)$ $ \Longleftrightarrow$

$ \cos(x - 20) = \cos (140 - x)$ $ \Longleftrightarrow$ $ x - 20 = 140 - x$ $ \Longleftrightarrow$ $ x = 80^{\circ}\ .$

[b]Remark.[/b] This nice problem is easily because in the relation $ (*)$

two angles ($ 20$ and $ 30$) appear twice left and right ![/color]


[quote="Beat"]Show that $(\sum_{i=1}^{n}x_iy_iz_i)^2 \le (\sum_{i=1}^{n}x_i^3) (\sum_{i=1}^{n}y_i^3) (\sum_{i=1}^{n}z_i^3)$ for any positive reals $x_i, y_i, z_i$.[/quote]
Are you sure you mean $(\sum_{i=1}^{n}x_iy_iz_i)^2$ but not $(\sum_{i=1}^{n}x_iy_iz_i)^3$? :huh:

Well, I was asking myself the same question but see this:
http://www.kalva.demon.co.uk/aus-pol/ap89.html

This problem is wrong,if you want I can sand you solution of this ineq.
$(\sum_{i=1}^{n}x_iy_iz_i)^3 \le (\sum_{i=1}^{n}x_i^3) (\sum_{i=1}^{n}y_i^3) (\sum_{i=1}^{n}z_i^3)$ :lol:

[quote="leepakhin"][quote="Tiks"]This problem is wrong.[/quote]
How do you know for sure that the problem must be wrong?[/quote]
If $a_i=x_i=y_i=z_i$ we have that $1\leq {a_1}^3+{a_2}^3...+{a_n}^3$for any positiv$a_i$,isn't it,but this ineq isn't tru :(  :D

The corrected version is just Hölder:
$\left(\sum_{i=1}^{n}x_i^3\right)^\frac13\left(\sum_{i=1}^{n}y_i^3\right) ^\frac13\left(\sum_{i=1}^{n}z_i^3\right)^\frac13 \ge \sum_{i=1}^{n}x_iy_iz_i$ :)

Use Cauchy Schwartz twice to get  ∑p^4 . ∑q^4 . ∑r^4 . ∑t^4  ≥  ( ∑pqrt) ^4
Then set p=x^ 3/4     q=y^ 3/4   r=z^ 3/4 and t= (xyz)^ 1/4

It is similar to the trick used for proving the AM-GM theorem.

The original question is wrong.  It's easy to demonstrate that by setting each y and each z as 1, and each x as 1/n^2


group theory
abstract algebra
number theory
greatest common divisor
superior algebra
superior algebra unsolved

Suppose that $n=p_1,p_2\dots p_n$ which $p_i$ are distinct prime numbers. Prove that there exist $i$ that $G$ has a normal $p_i$-sylow subgroup.

A stronger version:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=normal&t=55461[/url]

The proof given there is incorrect :).

Here is a great solution from Adel Javanmard one of my greatest friends.

[b]Lemma 1[/b]
Suppose $G$ is a group which $|G|$ is free-square then $G$ is not simple.
[b]Proof[/b]: Suppose $|G|=p_1p_2\dots p_s$ and $H_i$ is a $p_i$-sylow sybgroup of $G$. If $H_i\ntrianglelefteq G$ then suppose $[G: H_i]=m_i (m_i=\frac{p_1p_2\dots p_s}{p_i})$. Easily you could see that there is a homorphism $\varphi$ from $G$ to $S_{m_i}$ which $\mbox{ker}(G)$ is the biggest subgroup of $H_i$ which is normal in $G$. Because $|H_i|=p_i$ then its subgroups are $\{1\}$ and $H_i$. Because $H_i\ntrianglelefteq G$ so $\mbox{ker}(\varphi)=\{1\}$. So $\varphi$ is a monomorphism from $G$ to $S_{m_i}$. And $\varphi(G)\leq S_{m_i}$ and is simple, but the only simple subgroup $S_{m_i}$ is $A_{m_i}$. So $G\cong A_{m_i}$. So $|G|=\frac{m_i!}2$ for each $i$. This is contradiction beacause $m_1\neq m_2\neq\dots$.

[b]Lemma 2[/b]
If $H$ is a normal subgroup of $G$ and $p$divises $|G|$ and $gcd([G: H],p)=1$, then every $p$-sylow subgroup of $G$ is a subset of $H$.
[b]Proof:[/b] Suppose $|G|=p^mq$ which $gcd(p,q)=1$. So there exist $|H|=p^mq_1$. Now because of the first sylow theorem $H$ has a $p$-sylow subgroup like $K$ that is $p$-sylow subgroup of $G$. Suppose $K'$ is another $p$-sylow subgroup of $G$. Beacuse of second Sylow theorem ther exist $x\in G$ which $K'=xKx^{-1}$. So $K'=xKx^{-1}\subset q xHx^{-1}\subset q H$.
 So every $p$-sylow subgroup of $G$ is a subgroup of $H$.

Suppose $|G|=p_1p_2\dots p_s$ because of Lemma 1 $G$ has a non-obvious normal subgroup $H_1$. WLOG suppose $|H_1|=p_1p_2\dots p_{s_1}$ and for each $1\leq i\leq s_1$ all of $p_i$-sylow subgroups of $G$ are subgroups of $H_1$. There exist $H_2\unlhd H_1$ that $|H_2|=p_1p_2\dots p_{s_2}$ and for every $1\leq i\leq s_2$ every $p_i$-sylow subgroup of $G$ is a subgroup of $H_1$, so it's a subgroup of $H_2$. Continuing this algortim we reach to a group $H_r$ which $|H_r|=p_1p_2$ which every $p_1$-sylow subgroup of $G$ and every $p_2$-sylow subgroup of $G$ is a subgroup of $H_r$. Now Suppose $p_1<p_2$. If we prove that the number of $p_2$-sylow subgroups of $H_r$ is 1 then the unique $p_2$-sylow subgroup of $G$ is normal subgroup of $G$ and proof is complete. Suppose $n_{p_2}$ is the number of $p_2$-sylow subgroups of $G$. From the third sylow theorem $n_{p_2}|p_1$ and $p_2|n_{p_2}-1$. IF $n_{p_2}\neq1$ then $p_1<p_2+1\leq n_{p_2}\leq p_1$. Contradiction proves that $n_{p_2}=1$ and everything is proven. Proof is complete. :D  :D

[quote="Omid Hatami"]but the only simple  subgroup $S_{m_i}$ is $A_{m_i}$ [/quote] i  think this is not true ,any way this problem is an easy corrollary of burnside complement theorem

Yes its true. You could find it in the book Algbera from Bautacharia. And I'd like to see your solution.

There is a proof using Hall's theorem in post #7 of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=72003]this topic[/url].

omid ,i dont know if i say somthing stupid but you can just say the only normal subgroup of $S_n$ is $A_n$ ,the only simple is not correct because with your claim every simple group of order $n$ should be of cardinal $n!/2$:? . :(

I was thinking about a similar problem, and I realized that the following generalization is true:

If $|G|=p_1^{\alpha_1}\cdot\ldots\cdot p_k^{\alpha_k}$ where $p_i$ are primes with $p_1<\ldots<p_k$ and the Sylow subgroups of $G$ are cyclic, then there is only one $p_k$-Sylow subgroup of $G$ (in other words, $G$ has a normal subgroup of size $p_k^{\alpha_k}$).


function
logarithms
algorithm
algebra open
algebra

I know there is a solution to the equation x^x = 2 but, how is the solution done?

You have to use a special function called Lambert W function. It has the property that if

$ x e^x \equal{} y$

then

$ x\equal{}W(y)$

So, back to your problem. Use the nathural logarithm on both sides. This gives you:

$ x \ln x \equal{} \ln2$

Now, substitute

$ \ln x \equal{} y$

This gives you

$ e^y y \equal{} \ln2$

So

$ y \equal{} W(\ln2)$

$ \ln x \equal{} W(\ln2)$

$ x \equal{} e^{W(\ln2)}$

But, since we know that

$ W(x) e^{W(x)} \equal{} x \Rightarrow e^{W(x)} \equal{} \frac{x}{W(x)}$

we can write

$ x \equal{} \frac{\ln2}{W(\ln2)} \approx 1.55961$

A few years ago, one of my friends told me their dad had a real life problem that they couldn't solve. It was similar to: 1000 dollars are in a bank gaining 5% compounded interest. Once they take it out, they must pay 10 dollars for each year it was in the bank. How long is it before they have can have 1500 dollars after taking it out?

So the equation is $ 1000(1.05)^x\minus{}10x\equal{}1500$. Is there some special function to solve this?

To tell you the truth, the problem was really that he had to pay 10 dollars each year, but I think that's too hard. I forgot the exact amounts of money I just made the ones I gave up.

No, you use the same function, but some manipulation is necessary. Here's how you do it! First, denote a general situation:

$ A b^x \minus{} B x \equal{} C$

$ A \ln b  e^{x\ln b} \equal{} (B x \plus{} C)\ln b$

Substitute: $ y \equal{}(B x \plus{} C)\ln b \Rightarrow x \equal{} \frac{y}{B \ln b}\minus{}\frac{C}{B}$

$ A \ln b  e^{\left(\frac{y}{B \ln b}\minus{}\frac{C}{B}\right)\ln b} \equal{} y$

$ \frac{A \ln b}{e^\frac{C\ln b}{B}}  e^{\frac{y}{B}} \equal{} y$

$ \frac{A \ln b}{B e^\frac{C\ln b}{B}}  e^{\frac{y}{B}} \equal{} \frac{y}{B}$

$ \minus{}\frac{A \ln b}{B e^\frac{C\ln b}{B}} \equal{} \minus{}\frac{y}{B} e^{\minus{}\frac{y}{B}}$

$ W\left(\minus{}\frac{A \ln b}{B e^\frac{C\ln b}{B}}\right)\equal{} \minus{}\frac{y}{B}$

$ \minus{}B W\left(\minus{}\frac{A \ln b}{B e^\frac{C\ln b}{B}}\right)\equal{} y$

$ \minus{}B W\left(\minus{}\frac{A \ln b}{B e^\frac{C\ln b}{B}}\right)\equal{} (B x \plus{} C)\ln b$

$ \minus{}\frac{1}{\ln b} W\left(\minus{}\frac{A \ln b}{B e^\frac{C\ln b}{B}}\right)\minus{}\frac{C}{B}\equal{}x$

$ x\equal{}\minus{}\frac{1}{\ln b} W\left(\minus{}\frac{A \ln b}{B b^\frac{C}{B}}\right)\minus{}\frac{C}{B}$

In you case, the result would be somewhat about $ x \approx 9.58$, but there is a problem getting it because the Lambert W function is defined for positive numbers only, and you would get a negative one. So, my advice would be to use an approximation algorithm to solve this.


Although direct computation is possible, knowing the sum-and-difference pattern of factoring can some a significant amount of thinking and calculating time.

$ (17\plus{}10)^2\minus{}(17\minus{}10)^2$
$ \equal{}27^2\minus{}7^2$
$ \equal{}(27\plus{}7)(27\minus{}7)$
$ \equal{}34(20)$
$ \equal{}\fbox{680}$


trigonometry
algebra
polynomial
Irrational numbers

Show that $\cos \frac{\pi}{7}$ is irrational.

See http://www.problem-solving.be/pen/viewtopic.php?t=213
Alternatively, it's not hard to find the minimal polynomial, it's then standard to check that it is irreducible (it has degree $3$, so it suffices to show that there are no roots).

Just as addendum the elementary way: the minimal polynomial is $8x^{3}-4x^{2}-4x+1$, we see that it is one by setting $2 \cos \frac{\pi}{7} =  -\zeta_7-\zeta_7^{-1}$ (here $\zeta_7=e^{\frac{8 \pi i}7}$ which results in $-\zeta_7^3-\zeta_7^2-\zeta_7-1-\zeta_7^{-1}-\zeta_7^{-2}-\zeta_7^{-3}$, being zero by $0=\zeta_7^7-1=(\zeta_7-1)(\zeta_7^6+\zeta_7^5+\zeta_7^4+\zeta_7^3+\zeta_7^2+\zeta_7+1)$.

The polynomial is irreducible because if we set $x=y-1$ we get $8y^{3}-28y^{2}+28y-7$, being irreducible by Eisenstein.


The PA State Mathcounts competition was today.  Does anyone know who qualified for nationals from PA?

i assume matt tanzer (he qualified last year) and some unknowns.

I know who qualified last year.  ;)    Does anyone know about this year?


[quote="Arne"]For any positive integer n, define n5 as the multiple of 5 which is closest to n.
For example, 55 = 5, (-7)5 = -5, 85 = 10.
Solve the following equation in positive integers:
(x5)2 + (y4)5 = 2xy2 + 51.

Have fun :)[/quote]
We have that 2xy2 + 51 = 0 mod 5
xy2= 2 mod 5
Since 5 is prime:
x mod 5 * y^2 mod 5 = 2 mod 5
y2 mod 5 = 0, 1 or 4 mod 5, and from the above, y2 is clearly not 0 mod 5, so we have (x=2mod5, y=1mod5) or (x=3mod5, y=4mod5).
Case: (x=2mod5, y=1mod5)
(x5)2 = (x-2)2
(y4)5 = y4 - 1
The original equation then reduces to
x2 - 4x + y4 = 2xy2 + 48
Substitute z = y 2 to get
x2 -4x+z2 -2xz = 48, z > 0
(x-z)2 - 4x = 48
Examine that mod 4 to get rid of things I don't like:
(x-z)2 = 0 mod 4
So (x-z) = 0 or 2 mod 4, and we also have that z = 0, 1, or 2 mod 4 (because z=y^2). Unfortunately, that places no restrictions on x mod 4.
x^2 mod 4 + z^2 mod 4 - 2xz mod 4 = 0 mod 4
z^2 = 0 or 1 mod 4

Time to get ready for the whole "Valentine's Day" thing. I'll (hopefully) finish this up later - I feel like I have a decent start.

it done in Romania :( ... oh well there is still next year :D:D

Ok, after a great Valentine's Day, and a few other days, I'll have another go at this, rather than doing homework for AP US History.

Solve the following equation in positive integers:
(x5)2 + (y4)5 = 2xy2 + 51.

Case:  (x=3mod5, y=4mod5)
(x+2)2=(x5)2
y4+1= (y4)5
So the original eqn =
x2+4x+y4=2xy2+46
Again, substitute z=y2 to get
x2+4x+z2-2xz=46, z > 0
(x-z)^2+4x=46
(x-z)^2=2 mod 4, which clearly isn't possible since x and z are integers and all perfect squares are 0 or 1 mod 4. So back to what I had before...
--------------------
We have that z^2 = y^4 = 0 or 1 mod 4, and that y=1 mod 5, implying that y^4=1 mod 5. If y^4=1 mod 5, y^4=1 mod 4 because 1mod5 is always odd. So we have from that y^4=1 mod 20. So y^2=1,9,11 or 19 mod 20, from which we have that y^2=1 mod 4. Going around again, we have y^2=1 mod 20, so y=1,9,11, or 19 mod 20, or y=1 mod 4, from which we have y=1mod20, assuming there wasn't an error in my logic.
So x=2mod5=2 or 7 mod 10=2,7,12,or 17 mod 20, and y=1 mod 20
(x-y^2)^2 = 0 mod 4 = 0, 4, 8, 12, or 16 mod 20 from the above, which implies x-y^2 is even, and that x=7 or 17 mod 20. I guess I get to postpone this until later... again, since I don't know where to go.

[quote="me!"]So x=2mod5=2 or 7 mod 10=2,7,12,or 17 mod 20, and y=1 mod 20
(x-y^2)^2 = 0 mod 4 = 0, 4, 8, 12, or 16 mod 20 from the above, which implies x-y^2 is even, and that x=7 or 17 mod 20. I guess I get to postpone this until later... again, since I don't know where to go.[/quote]
Ok, so now I have x=7 or 17 mod 20, and I think that there must be something that I've been missing, since I haven't solved the problem yet (generally a good sign that something is missing)
Case: x=7 mod 20
x5=5 mod 20, so (x5)2=5 mod 20 too.
(x5)2 + (y4)5 = 2xy2+51
5 mod 20 + 0 mod 20 = 2*5*1 mod 20 + 11 mod 20 = 1 mod 20, a contradiction.

case x=17 mod 20
x5=15 mod 20, so (x5)2=5 mod 20
(x5)2 + (y4)5 = 2xy2+51
5 mod 20 + 0 mod 20 = 2*5*1 mod 20 + 11 mod 20, the same contradiction as before, and I'm out of cases, so I'd say I've probably messed up... likely numerous times.


analytic geometry
graphing lines
slope
calculus
trigonometry

Real numbers $ x,\ y$ satifies $ x^2 \plus{} y^2 \equal{} 1$. Find the maximum and minimum value of $ \frac {y \minus{} 4}{x \minus{} 7}$.

[hide]The problem is dual to optimizing $ a/b$ subject to $ (a \plus{} 7)^2 \plus{} (b \plus{} 4)^2 \equal{} 1$.
So the slope is obviously optimized when it is the slope of one of the two tangents passing the origin.
Let its coordinates be $ (x,y)$ and the equation of the tangent $ y \equal{} mx$. Then we have
$ x^2 \plus{} 14x \plus{} 49 \plus{} m^2x^2 \plus{} 8mx \plus{} 16 \minus{} 1 \equal{} 0$.
The discriminant is $ (14 \plus{} 8m)^2 \minus{} 4(1 \plus{} m^2)*64$.
Setting that to $ 0$ we get
$ 64m^2 \plus{} 2*8*14m \plus{} 14^2 \minus{} 4*46 \minus{} 4*46m^2 \equal{} 0 \equal{} \minus{} 3*64m^2 \plus{} 28*8m \plus{} (14^2 \minus{} 4*46)$.
$ \minus{} 3*16m^2 \plus{} 2*28m \plus{} 7^2 \minus{} 2*46 \equal{} 0$

Does this test allow calculators, because I don't feel like doing the tedious arithmetic without a calculator!  :P [/hide]

I believe you made some minor errors. :( But you got the general idea. :)

[hide="Solution"]
Let $ n \equal{} y \minus{} 4$ and $ m \equal{} x \minus{} 7$ so we want to find the extrema of $ k \equal{} \frac nm\implies n \equal{} km$. For such extreme $ k$, the line passing through the origin with slope $ k$ (in the $ m \minus{} n$ plane) must be tangent to the circle satisfying $ (m \plus{} 7)^2 \plus{} (n \plus{} 4)^2 \equal{} (m \plus{} 7)^2 \plus{} (km \plus{} 4)^2 \equal{} (k^2 \plus{} 1)m^2 \plus{} (8k \plus{} 14)m \plus{} 65$
$ \equal{} 1\implies(k^2 \plus{} 1)m^2 \plus{} (8k \plus{} 14)m \plus{} 64 \equal{} 0$, and thus there must be only one possible value for $ m$ for these $ k$ (if the line is not a tangent, it will intersect in two places if it touches the circle). This corresponds to a discriminant of zero, i.e. $ (8k \plus{} 14)^2 \minus{} 4(k^2 \plus{} 1)(64) \equal{} 192k^2 \minus{} 224k \plus{} 60 \equal{} 0$
$ \implies48k^2 \minus{} 56k \plus{} 15 \equal{} (4k \minus{} 3)(12k \minus{} 5) \equal{} 0\implies\boxed{(\max k,\min k) \equal{} \left(\frac34,\frac5{12}\right)}$.
[/hide]

Yeah you're right; I have no idea how I changed 64 to 46... LOL
(That's one of the reasons why I hate the no-partial-credit system so much.)

We wish to find the maximum and minimum of $ \frac {sin A \minus{} 4}{cos A \minus{} 7}$. By calculus, this occurs when $ 4sinA \plus{} 7cosA \equal{} 1$. Substituting, the given is equivalent to $ \frac {15sinA \plus{} 28cosA}{48cosA \plus{} 28sinA} \equal{} \frac {15tanA \plus{} 28}{28tanA \plus{} 48}$. Also, we have $ 4tanA \plus{} 7 \equal{} secA$. Squaring, $ 16tan^2a \plus{} 56tanA \plus{} 49 \equal{} tan^2A \plus{} 1$. Factoring, $ (5tanA \plus{} 12)(3tanA \plus{} 4) \equal{} 0$. Substituing, we get $ \frac {3}{4}$ and $ \frac {5}{12}$.
It seems strange that you sub in the negative recipriocals of the answers for tanA. Is there a reason for this?

Indeed,
[asy]size(160); pen sm = fontsize(7);
pair A = (7,4), B=expi(atan(-4/3)); D(A--(0,0)--B); D((0,0)--expi(atan(-12/5) + pi)); D(rightanglemark((0,0),B,A));D(anglemark(B,(0,0),(1,0)));
D(unitcircle);D(A);D(A--A-9*(4,3)/abs((4,3)) );D(A--A-9*(12,5)/abs((12,5)) ); D((0,-2)--(0,5),Arrows(4)); D((-2,0)--(8,0),Arrows(4)); MP("O",(0,0),SW,sm); MP("B",B,SE,sm); MP("A\ (4,7)",A,NE,sm); MP("8",(A+B)/2,SE,sm);MP("1",B/2,SW,sm);[/asy]

The answer is the slope of the tangents, or the negative reciprocal of the slope of the radii, whose respective slopes are
\[ \tan \left(\tan^{ \minus{} 1} \left( \frac 47\right) \pm \angle AOB\right) \equal{} \frac {4/7 \pm 8}{1 \mp 8 \times 4/7} \equal{} \frac { \minus{} 12}{5}, \frac { \minus{} 4}{3}
\]


summer program
MathZoom
geometry
number theory
greatest common divisor
quadratics
AwesomeMath

[quote="cheeesec4k3"]Hi everyone,
I'm new to this whole math camp thing, and I'm wondering if I should go to AMSP or Mathzoom. I heard the curriculum is pretty much the same, and one of the coaches from AMSP now teaches at Mathzoom. I'm leaning towards AMSP, but could you guys share your opinions? Thanks.[/quote]
ummmmm yeah go to amp definitely
that guy from amp who made mathzoom...ehhh lets just say me and my buddies didnt like him very much

[quote="cheeesec4k3"]I heard the curriculum is pretty much the same...[/quote]

That is probably not true. AMP is much more intensive academically, and it basically gets as hard as you want it to get. MathZoom is geared toward younger students, so you can't really say that about it. If you got in to AMP, go to AMP.

This is hopefully a more constructive and helpful comment than Taylor Han's. Who by the way has a very feminine name.

[quote="CatalystOfNostalgia"]
This is hopefully a more constructive and helpful comment than Taylor Han's. Who by the way has a very feminine name.[/quote]
shuddup karel haha

Perhaps I can formalize what Taylor & Peter are saying.

Here is the situation: during AMSP last year, me, Taylor (junggi), and Peter (mustafa) all roomed together. Additionally, we all were in track 2 (which was geometry in the morning and number theory in the afternoon). The first week, we had a really good number theory lecturer. Mr. Josh (Nichols Barrer) taught us a lot of interesting proofs such as the development of GCD in integers and other number systems. We also learned a bit about classical proofs relating to integers (like the fact that 1 is the smallest positive integer).

Possibly the next week, we got Dr. Kevin as our number theory lecturer. I know that me, Taylor, and Peter were all disappointed by his lectures because all he did was give us a set of 10 to 15 easyish problems. These lectures were way less interesting than those before because we felt that part of the lecture component should be for new techniques.  Instead, we just got a collection of problems. I would say that I still learned a fair amount. I remember a few particular classes of problems that I learned to solve as a result of those 'lectures.' In particular, we went over the proofs for showing that the $ n$th harmonic number is not an integer and assorted generalizations of that. However, I was generally disappointed because I had wanted a lecture in possibly quadratic reciprocity (which other students got in another track from Oleg, I believe). Furthermore, there are numerous problems that are interesting in Diophantine equations. There was no shortage of things to teach, I just felt that Dr. Kevin did not present these to us.

To be fair, I think that this method of teaching is good for new students. But I also feel that you should try and expand your knowledge by listening to lectures that are challenging for you rather than just doing problems. Along with what Taylor said, I agree that the curricula are not the same. Of course, I'm partial towards AwesomeMath, but even without that bias, I would recommend AMSP over MathZoom based on my limited knowledge of the latter camp because you have more opportunity to learn more difficult problem solving if you want to.

Oh man that's a wall o' text...I wish this could be my English essay that I'm procrastinating :D.

Well, take all of these opinions with a grain of salt because they come from AMPpers. If you didn't go to MathZoom, then you probably are not in a position to say which one is better.

Well we did have Dr. Kevin (the head of MathZoom) as a teacher for a week so we have some basis for discussing this topic.

[quote="Phelpedo"]Does he even teach at MathZoom? He's not listed under faculty.[/quote]
dr kevin=dr wang

Well, I've decided to go to AMP... just mainly because of the curriculum, and it seems a bit more organized than MathZoom. Thanks to everyone who replied! =]

It is generally not a good idea to revive an old thread, but since (1) I only saw this now; (2) this thread is referenced by another thread; and (3) my comments will be highly contextual to this thread, I will post a comment here.  I am not going to promote Math Zoom Summer Program here (I do that in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=453]MathZoom[/url] forum  :D ), but as a teacher, I would give a response to any feedback I receive, even though the feedback is about a year old.

Positive feedback can be heard everyday, but negative ones are rare, and that makes negative feedback all the more valuable.  Especially, Altheman's detailed descriptions of how he and his friends felt and expected are greatly appreciated.  I feel sorry that you felt my classes less interesting.  I could have paid closer attention to you guys and remedied that.

I'll also add some of my own perspective :)  That year (2007), I taught at AwesomeMath for two tracks: Track 2 Number Theory, and Track 5 Modular Arithmetic.  I discussed with my TAs, Oleg and Keone, about what to teach, in order to coordinate with their corresponding sessions.  For Track 2, the topics I covered were residue classes, diophantine equations, and powers of integers.  For track 5, I covered floor function, diophantine equations, and quadratic reciprocity.  My style of teaching was mostly covering the concepts through discussion of problems, and more lecture is given when the depth is needed.  The quadratic reciprocity lecture in Track 5 took the whole session, but other sessions contained a lot of problems, because my focus was teaching the techniques of solving those types of problems.

About half of the problems I presented were on the Math Olympiad level.  Some students said the power of integer problems were diabolical. :) If you felt they were easyish, that was because you were good :). In any case, it is noted that I would always prepare some more challenging materials for the best students in my classes.
On the other hand, I also heard from students from AwesomeMath 2007 that my diophantine equation class was the only class they understood during the whole camp :)

I usually don't post in this forum, and seldom do forum-surfing---just too busy: a whole camp of students to take care of during summer, and a lot of year-round classes during other seasons.  I welcome any comments, PM or email, or posting in the MathZoom forum.  Feedbacks (+ or -) are most appreciated.


superior algebra
superior algebra solved

I see that you like problems with matrices, I shall try to post some, for a starter here is one given at a contest in 1998:
Let X and Y be two matrices with complex elements. Prove that we have the following:
a)(trivial) det(X+Y)+det(X-Y)=2(det(X)+det(Y));
b)(trivial) if XY=I_2 then YX=I_2 (here X and Y have real elements)
c)generalize point b)
d)prove that (XY)^2=XY^2X iff (YX)^2=YX^2Y.

have fun with it! :)

I think you mean that X and Y are in M_2(C).
a) is done really simple :
consider
Y=
a b
c d

and
X=
x y
z t

and compute!

I think that the hard part is d)

cheers! :D :D

You are right Lagrangia, please help with d), I have managed to do the other subpartes :)
Let X and Y be two matrices with complex elements, X and Y in M_2(C)  prove that (XY)^2=XY^2X iff (YX)^2=YX^2Y.

thank you!

(XY) 2 =XY 2 X <=>
XY(XY-YX)=0 (1).

We have by Hamilton-Cayley
(XY-YX) 2 =Tr(XY-YX)*(XY-YX)-det(XY-YX)*I2.
We know Tr(XY-YX)=0, so (XY-YX) 2 =-det(XY-YX) *I2. (2) If we multiply this to the left by XY, we get by (1)
0=0*(XY-YX)=XY(XY-YX)(XY-YX)=-XY*det(XY-YX).

If det(XY-YX)<>0 then XY=0 and (YX) 2 =Y(XY)X=0=(YX)(XY)=YX 2 Y.
If det(XY-YX)=0, we get from (2) (XY-YX) 2 =0, so
(XY) 2-XYYX-YXXY+(YX) 2 =0. Since (1), we get (YX) 2 =YX 2 Y.

For the converse replace X by Y and Y by X in the proof for the direct.


parameterization
probability
expected value
advanced fields
advanced fields unsolved

Got a little problem with this ones. The first part is easy but then I could not get farther

Ex: The number of oil tankers, arriving at a certain refinery on one day, follow a poisson distribution with the parameter
 µ=2. The present harbour facilities can serve 3 oil tankers a day. If more tha 3 oil tankers arrive on a given day, the additional tankers ( in excess of the 3) are sent to another refinery.

1: What is the probability that oil tankers are sent to another refinery

2: By how much should the capacity of the refinery harbour be extended, if it is desired that on a given day the probability of sending away arriving oil tankers does not exceed 0.05? and 0.01?

3: What is the expected number of oil tankers arriving on a given day?

4: Whar is the most probable number of oil tanker arriving on a given day?

5: What is the expected number of oil tankers served on a given day?

6: What is the expected number of oil tankers sent on to other refineries on a given day?

Im having trouble with 5 and 6.

Thanks

The first thing I'd do is make a numerical table. The Poisson distribution lends itself well to recursive definition, and hence is very easy to construct on a spreadsheet. (Spreadsheets are wonderful, flexible mathematical tools.)
[code] x   P(X=x)     P(X≤x)     P(X>x)
 0   .13533528  .13533528  .86466472
 1   .27067057  .40600585  .59399415
 2   .27067057  .37667642  .32332358
 3   .18044704  .85712346  .14287654
 4   .09022352  .94734698  .05265302
 5   .03608941  .98343639  .01656361
 6   .01202980  .99546619  .00453381
 7   .00343709  .99890328  .00109672
 8   .00085927  .99976255  .00023745  [/code]
So the answer to the first question, the probability that tankers are diverted, is $P(X>3)$ which is approximately $.14287654.$

(Note that the answers to #2 and #4 are easy to find in that table.)

Now for question 5: the expected number of tankers serviced. That would be

$E(\min(X,3))=1\cdot P(X=1)+2\cdot P(X=2)+3\cdot P(X=3)+3\cdot P(X>3)$

$\approx 1\cdot0.27067057+2\cdot0.27067057+3\cdot0.18044704+ 3\cdot0.14287654$

$\approx1.78198245.$

And the answer to #6 is the difference between $2$ and that number, or about $.21801755.$

Hey thanks for helping but why is it 2 - the number you got in ex 5. I have trouble to understand this.

The expected number of tankers that appear on a given day is simply the mean of the Poisson distribution, which is 2. All tankers are either serviced locally or diverted, so the sum of the mean number serviced locally and the mean number diverted must be equal to 2, the overall mean.


Person A tosses n+1 fair coins, with a heads face and a tails face.

Person B tosses n fair coins.

What is the probability that Person A tosses more heads than person B.

isn't it just 1/2? Since we can just discount all the coins tossed before the $n+1$ coin.

I think I have seen this problem before.
Try to use it to prove:
$\frac{1}{2} = \displaystyle\sum_{i=0}^{n}\sum_{j=i+1}^{n+1}\frac{{n \choose i} \cdot {n+1 \choose j}}{2^{i+j}}$

Try to prove this summation equals 1/2 some other way too.  :lol:

I believe this approach must be credited to DTS? but I can't find it anywhere.


let ABC be a triangle with circumcenter O. a line passes through O and meet AB ,AC and BC in the points X ,Y and Z such that OX=OY. the line AO meet the circumcircle in the point H. prove that ZH is tangent on the circumcircle

It is same problem with http://www.mathlinks.ro/viewtopic.php?t=178873


MATHCOUNTS logarithms trigonometry AMC AIME USA(J)MO USAMO

Find Im(cos(12)+i*sin(12)+cos(48)+i*sin(48)), where Im(z) is the imaginary part of z.

Why isn't the answer just sin(12)+sin(48)?

WHY THE HECK IS THIS IN GETTING STARTED??????

Case in point when I say Getting Started is no longer for beginers and the level is way higher than MATHCOUNTS stuff.

Err...well what is the answer then?

BTW-I agree that this should probably be in Intermediate...

Well, first of all I don't know how hard it is, though it initially looked quite easy though my answer is wrong. Secondly, it might too easy for Intermediate perhaps so I threw it in Getting Started. I don't know what MC stuff looks like, but I'm guessing national stuff is a lot harder than regional stuff.

Basically if it has logarithm or trigonometry, then it's automatically in Intermediate (even if it is only identities; course not everyone follows this :wink:)

Well, someone should talk about the problem..

The answer is sin 12 + sin 48. If this isn't the answer given, then it will be a simplification of this.

[quote="WarpedKlown1335"]Umm...do you mean to say Im[(cos 12+ i sin 12 + cos 48 +i sin 48)^6]?  Because I recall a problem about that.[/quote]

Hah, yes, that is in fact correct. I was working on this problem last night but for some reason I didn't see the exponent. Seriously, it wasn't there last night. :)

Sorry about that folks.

[quote="tokenadult"][quote="Fierytycoon"]Well, first of all I don't know how hard it is, though it initially looked quite easy though my answer is wrong.[/quote]

Is there a moderator here who can move this into Intermediate?[/quote]
Nope, both of them have left like a month ago, in fact we really ought to reassign mods and boards but Mr. Rusczyk is on vacation.

One shouldn't keep sin12+sin48 unsimplified. sin12 has a closed form in radicals which I could derive if I needed to for some reason.

[quote="Fierytycoon"]Well the original problem (without the exponent) is easy so it belongs here.[/quote]

Are complex numbers to be found in Mathcounts problems at the chapter level? I know they are NOT to be found in AMC 8.

It seems to me (not that I am the final authority on the matter) that "getting started" level begins with VERY easy problems (the kind I can do  :)  )and goes up to problems about at the level of AMC 8 or the chapter problems in Mathcounts--because that is the level most people START on in math contests. Some people have their first taste of math competition in elementary school, solving problems MUCH easier than most of what is now posted on this forum.

Then intermediate level includes state and national Mathcounts level contests (the kind that people only get to if they score high enough to go beyond chapter competition), AMC 10 (which, remember, is STRICTLY math before precalculus in the usual high school sequence), and possibly a few of the statewide high school level math competitions.

Then advanced level, in my humble opinion, includes AIME, USAMO, ARML, IMO, Putnam, and any contest that includes calculus or post-calculus mathematics, difficulty level unbounded above. I would include any contest in which the questions are "Prove . . . " in the advanced level.

Does my general impression of where the difficulty levels lie square with everyone else's impression? I think that there has been a BAD problem of too difficult problems being posted here on Getting Started, problems that really belong in Intermediate, and even some that properly belong in Advanced. Rather than merely complain, I will try to dig out some problem books for elementary math contests and adapt problems from those for posting here, to exemplify what I think is the proper level for the Getting Started forum.

So do we need new moderators for this forum? I thought the listed moderators were still active here, but maybe just not logging in recently. I would volunteer. My own lack of practice in solving math competition problems gives me a pretty good sense, I dare say, how hard is too hard for a beginner. I can stay on the board daily far into the future, and I can get some practice out of looking at the problems here that I can apply to parenting my three younger children. My oldest son should be beyond this level by next year--he is right about at this level right now, expecting to go to his first Mathcounts chapter meet on 21 February 2004 and having taking the AMC 8 for the last two years.

Yes this forum is too difucult for me to work in right now.

tokenadult - I respectfully disagree. If we limit problems according to your analysis, the Intermediate forum, which I moderate, would be left with a small, small group of problems. After all, there are not that many problems which are harder than mc - regionals/state but dont involve concepts on the AMC 12. In my opinion, the Getting Started forum is for people who indeed are just getting started, but also for people who are at a level so that they are competitive to make the second round of contests. I mean the Getting Started forum is for those people who are between just getting started and are pretty confident they can make AIME. If begginers here run into concepts they are not familiar with, those who have more experience can explain. This applies to this problem involving complex numbers. If begginers just dont look at any problems that involve concepts they dont know, how will they ever learn? The intermediate forum is, in my opinion, for students who are at least fairly capable of reaching AIME and other 2nd rounds in concepts. It is here we should put problems at the AIME level, as well as questions from free response contests that are often the second parts of state olympiads. Easier national olympiad questions that ask for rigorous proofs also belong here. After all, students need to learn proofs somehow. Where better than the Intermediate forum? Of course, problems involving less known concepts which are however not too difficult belong here also, to introduce students to the concepts. In Advanced, problems with difficult concepts and difficult solutions should be posted. This includes difficult National Olympiad problems, IMO, Putnam problems, and the like. tokenadult, it sounds like you are reducing the getting started and intermediate forums to the point that only Advanced will have non-trivial problems. If the goal of these forums is to increase the love of math in students, how better than challenging them just a little farther than they can currently do?

BHorseMath - why is this forum too difficult for you to work in right now? Im just looking at the last few questions. Fiery's complex number question is simple. If you dont understand the complex numbers, just ask and someone will give you a very detailed and informative explanation.  I also see "really ez problem", "simple patten", "100 numbers" , "pythagorean theorem', ... etc. It looks to me like these are all good problem for the begginer. However, if you want easier problems, why dont you just ask? I personally will post some if you wish. But saying you want to take the problems that are too difficult for you out does not make sense. How will you improve if you dont take a look at those problems which are too hard for you right now?


function
induction
algebra proposed
algebra

The function $ f : Z^ \plus{} \rightarrow R$ satisfies $ f(1) \equal{} 1, f(2) \equal{} 2$ and $ f(n \plus{} 2) \equal{} f(n \plus{} 2 \minus{} f(n \plus{} 1) ) \plus{} f(n \plus{} 1 \minus{} f(n) )$. Show that $ 0\leq f(n \plus{} 1) \minus{} f(n)\leq 1$. Find all $ n$ for which $ f(n) \equal{} 1025$

-) We prove by induction that $ 0\leq f(n\plus{}1)\minus{}f(n)\leq 1$
 +) $ 0\leq f(2)\minus{}f(1)\leq 1$
 +) Assume that we have $ 0\leq f(k\plus{}1)\minus{}f(k)\leq 1$ for all $ k\leq n$, we prove that $ 0\leq f(n\plus{}2)\minus{}f(n\plus{}1)\leq 1$
$ f(n\plus{}2)\minus{}f(n\plus{}1)\equal{}f(n\plus{}2\minus{}f(n\plus{}1))\minus{}f(n\minus{}f(n\minus{}1))$
Because $ n\plus{}2\minus{}f(n\plus{}1)\minus{}(n\minus{}f(n\minus{}1))\equal{}2\plus{}f(n\minus{}1)\minus{}f(n\plus{}1)$ then $ 0\leq n\plus{}2\minus{}f(n\plus{}1)\minus{}(n\minus{}f(n\minus{}1))\leq 2$
And $ n\minus{}f(n\minus{}1)\leq n\plus{}2\minus{}f(n\plus{}1)\leq n\plus{}1$ then $ 0\leq f(n\plus{}2)\minus{}f(n\plus{}1)\equal{}f(n\plus{}2\minus{}f(n\plus{}1))\minus{}f(n\minus{}f(n\minus{}1))\leq 2$
If $ f(n\plus{}2\minus{}f(n\plus{}1))\minus{}f(n\minus{}f(n\minus{}1))\equal{}2$ then $ n\plus{}2\minus{}f(n\plus{}1)\equal{}n\minus{}f(n\minus{}1)\plus{}2\Leftrightarrow f(n\plus{}1)\equal{}f(n)\equal{}f(n\minus{}1)$ and $ f(n\plus{}2\minus{}f(n\plus{}1))\equal{}f(n\plus{}1\minus{}f(n))\plus{}1\equal{}f(n\minus{}f(n\minus{}1))\plus{}2$
It mean $ f(n\plus{}1)\equal{}f(n\plus{}1\minus{}f(n))\plus{}f(n\minus{}f(n\minus{}1))$ is odd.
But if exists $ f(n\plus{}1)\equal{}f(n)\equal{}a\Leftrightarrow f(n\plus{}1\minus{}f(n))\equal{}f(n\minus{}f(n\minus{}1))\equal{}f(n\minus{}1\minus{}f(n\minus{}2))$ then $ a$ is even! Contradiction!
It mean $ 0\leq f(n\plus{}2)\minus{}f(n\plus{}1)\leq 1$
-) Now we prove that $ f(n)\equal{}2^k\plus{}1\Leftrightarrow n\equal{}2^{k\plus{}1}$ by induction!
 +) $ f(4)\equal{}3$
 +) If $ f(n)\equal{}2^k\plus{}1\Leftrightarrow n\equal{}2^{k\plus{}1}$ we have:
$ 2^{k\plus{}1}\plus{}1\equal{}f(n)\equal{}f(n\minus{}f(n\minus{}1))\plus{}f(n\minus{}1\minus{}f(n\minus{}2))$
$ \Leftrightarrow f(n\minus{}f(n\minus{}1))\equal{}2^k\plus{}1, f(n\minus{}1\minus{}f(n\minus{}2))\equal{}2^k$ (Because $ 0\leq (n\minus{}f(n\minus{}1))\minus{}(n\minus{}1\minus{}f(n\minus{}2))\leq 1$)
$ \Leftrightarrow n\minus{}f(n\minus{}1)\equal{}2^{k\plus{}1}, n\minus{}1\minus{}f(n\minus{}2)\equal{}2^{k\plus{}1}\minus{}1$ then $ f(n\minus{}1)\equal{}f(n\minus{}2)\equal{}2^{k\plus{}1}, n\equal{}2^{k\plus{}2}$ (Because $ f(n)$ is odd then $ f(n\minus{}1)\equal{}f(n)\minus{}1$)
It mean $ f(n)\equal{}2^{k\plus{}1}\plus{}1\Leftrightarrow n\equal{}2^{k\plus{}2}$


Our teacher performed this experiment:

He put a few tablets of solid sodium hydroxide in a beaker of liquid ammonium hydroxide. The contents were pulled up through a dessicant and a gas was captured. The gas turned litmus paper blue (so we assumed it was ammonia). Then, he held the flask with the gas upside down above a beaker of water. The flask had a rubber stopper in the opening with two holes, one holding a pipet containing some water and the other holding a tube that was put into the beaker of water. By squeezing the pipet and shooting some water into the flask, suddenly water began being drawn up through the other tube to form a fountain of magenta liquid.

How is this explained? (I might have missed some details.)

What were you covering in class?  Magenta suggests either phenolphthalein or permaganate ion, but you didn't mention any of those.  Not sure.  Think about what you were doing in class.

Well one thing that needs to be explained is how the water is drawn up (he just squeezed a pipet with some water in it, and the other tube started to suck in water). I assumed this was due to a sudden change in pressure, but what caused it?

He didn't mention whether there was any indicator used, but it does make sense since phenolphthalein is magenta in alkaline conditions (i.e. ammonia).


Λοιπόν  βλέπω  το  φόρουμ  απέκτησε  πολύ  κίνηση .......... :lol:  ...... Καλό  είναι  που  μπαίνουν  αρκετές  ασκήσεις  αλλά  ,   αν  κάποιος  δεν  μπαίνει  καθημερινά  στο  φόρουμ  χάνει  τις  εξελήξεις .....

 :D .... Ρε  παιδιά  ας  γίνει  κάποιος  moderator του  ελληνικού  φόρουμ ( κάποιος  που  ίσως  ξέρει  καλύτερα  τον  Vornicu ) και  ας  δημιουργήσει  υποφόρουμ  όπως  σε  αυτό  της    Ρουμανίας   . Λέω  τώρα   για  να  διωγκώσω  το φαίνόμενο  να  προτείνω  και  εγώ  ένα  πρόβλημα  .  :D

      α) Θεωρούμε  ένα  τρίγωνο  ΑΒΓ  και  τον περιγεγραμμένο  κύκλο  αυτού .  Να  δείξετε  οτι  η  ευθεία  που  ενώνει  το  ορθόκεντρο  με   τυχαίο  σημείο  Μ  της  περιφέρειας   διχοτομείται  από  την ευθεία  simson  του  σημείου   Μ .
      β) Να δείξετε  οτι  τα  τέσσερα  ορθόκεντρα  πλήρους  τετραπλέυρου    βρίσκονται  επί  της  ίδιας  ευθείας ... (προφανώς  το  β είναι  απόρια  του  α) )

Λοιπόν η λύση στο 1ο ώστε το δεύτερο να γίνει προφανές .
Έστω Μ' το συμμετρικό του Μ ως προς την ΒΓ , με Α'' και Η' τα κοινά σημεία της περιγεγραμμένης περιφέρειας, αντίστοιχα με τις ΜΜ'  ΑΗΔ ευθείας όπου Η το ορθόκεντρο .Έστω τώρα Α', Β' , Γ' οι προβολές του Μ στις πλευρές του τριγώνου
Επειδή ΑΑ''  ||Α'Β'Γ'  ευθεία έχουμε ότι  $\widehat{MA'\Gamma '}=\widehat{MA''A}$ (1)
Τώρα εύκολα βρίσκουμε ότι το ΗΗ'Μ'Μ είναι ισοσκελές τραπέζιο και ότι τα τόξα ΑΜ και ΗΑ'' είναι ίσα.Από όλες αυτές θα έχουμε οτι ΡΑ' || ΗΜ' όπου Π το σημείο τομής των ΗΜ και Α'Γ' .
Άρα ΗΡ=ΡΜ επειδή ΜΑ'=Α'Μ'       :D


inequalities
trigonometry
geometry
parallelogram
triangle inequality
trig identities
Law of Cosines

Given a triangle ABC with M the midpoint of BC, show that

[tex]AB +AC \geq 2(AM)[/tex]

From the Appollonius theorem (proved easily by Law of Cosines), we have:

ab^2 + ac^2 = 2am^2 + bc^2/2

bc >= (ac - ab)
bc >= (ab - ac)

(the last two come from the triangle inequality)

So:

ab^2 + ac^2 >= 2am^2 + ab^2/2 + ac^2/2  - abac

1/2ab^2 + 1/2ac^2 + abac >= 2am^2

1/2(ab+ac)^2 >= 2am^2

(ab+ac)^2 >=4am^2

ab+ac >= 2am

QED

[quote="zabelman"]Complete parallelogram BACA'. Then triangle inequality on triangle ABA' gives us:

AB+BA' > AA'

But AB+BA'=AB+AC and AA' = 2AM, so we're done.[/quote]

Nicely done, zabelman.

In any event, it should be obvious in looking at this problem that equality (probably) only occurs when ABC becomes degenerate and BC vanishes.

So, what classical inequality becomes an equality iff one member of the equality goes to zero?  The triangle inequality, of course.  This observation provides a very convienient way to start on the problem (whether you use a more analytical approach as I did, or a nicer 'classical' approach as did zabelman).


I don't know how to solve exercise 2 page 2 with inversion:
http://www.nevada.edu/~baragar/inversion.pdf

[color=red]EDIT by moderator:[/color]

3 congruent circles intersect at a point $ H$ in pairs at $ A, B$ and $ C$ as shown in figure 2. Prove that $ H$ is orthocenter of the $ \triangle ABC.$

For three congruent circles w1(O1),w2(O2),w3(O3)  let (A1,K)=W2∩W3,(A2,K)=W3∩W1,(A3,K)=W1∩W2.
And let M1=A1K∩O2O3,M2=A2K∩O3O1,M3=A3K∩O1O2

A(i)K bisects O(i+1)O(i+2) perpendicularly (i=1,2,3).
2M(i+1)M(i+2)=O(i+1)O(i+2)=A(i+1)A(i+2) (i=1,2,3)
M(i)M(i+1)∥A(i)A(i+1),O(i)O(i+1)∥M(i)M(i+1) (i=1,2,3).

A(i)K⊥O(i+1)O(i+2)
A(i)K⊥A(i+1)A(i+2) (i=1,2,3)
Thus K is orthocenter of the triangle A1A2A3.

I have made my own first reply  :)


Solve the equation: $ 3^{x\minus{}1}\plus{}3^{\frac{1}{\sqrt[3]{x}}}\equal{}4,\ x\in\mathbb{R}$ .

Should there be more solutions except trivial one ($ x\equal{}1$)?

hello, $ x\equal{}1$ is the only real solution of the given equation.
Sonnhard.


We will be hosting a [url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]Math Jam[/url] to discuss the MathZoom program and to answer questions.  The Math Jam will be moderated by Dr. Kevin Wang, the director of the program.

The Math Jam is on [b]Wednesday, April 25[/b] at the special time of [b]10 PM Eastern / 7 PM Pacific[/b].


This is fairly simple, but it reminded me of an AIME problem (not a specific one, just in general):

What is the smallest possible value of $n$ such that $\displaystyle\prod_{k=1}^n\dfrac{n}{10} >1$?

$\displaystyle\prod_{k=1}^n\dfrac{n}{10} >1$

$\dfrac{n!}{10^n}>1$

From n to n+1, a factor of 10 is always added to bottom, but one of n+1 is added to top. Clearly 1/10 <1, so we want to get to the point at which the numerator is increasing more than the denomiator. $n\ge10$ is clearly it, because the expression is equal for n=10 or 9. After that it's simple checking to see that n=11 is $\dfrac{399168}{10^9}$ and is approximately $\dfrac{4}{10^4}$. Very quickly from there you can find that $n=25$ gives a value over 1.


If $ xy \equal{} b$ and $ \frac{1}{x^2} \plus{} \frac{1}{y^2} \equal{} a$, then $ (x \plus{} y)^2$ equals:

$ \textbf{(A)}\ (a \plus{} 2b)^2\qquad
\textbf{(B)}\ a^2 \plus{} b^2\qquad
\textbf{(C)}\ b(ab \plus{} 2)\qquad
\textbf{(D)}\ ab(b \plus{} 2)\qquad
\textbf{(E)}\ \frac{1}{a} \plus{} 2b$

hello, we have $ x^2\plus{}y^2\equal{}a^2(xy)^2$ or $ (x\plus{}y)^2\equal{}a^2b\plus{}2b\equal{}b(ab\plus{}2)$, so the answer is
$ \boxed{C}$.
Sonnhard.


A rectangular prism with whole number dimensions has faces with areas 54, 72, and 108 square inches. how many cubic inches are there in its volume?

Let the dimensions of the prism be $ x$, $ y$, and $ z$.

We get the following:

$ xy\equal{}54$
$ xz\equal{}72$
$ yz\equal{}108$

Multiplying all three equations gives

$ x^2y^2z^2\equal{}54*72*108\equal{}9*9*6*8*9*12\equal{}3^8*2^6$
$ xyz\equal{}3^4*2^3\equal{}81*8\equal{}\boxed{648}$ cubic inches, which is the volume.


inequalities
algebra
polynomial
algebra solved

Prove that if a,b,c are reals and a^2+b^2+c^2<=1, then |a+b+c-abc|<=(8*sqrt(3))/9.

Thank Harazi for putting my name in title. It's only the variation of Polish and Vietnamese one. You can find it in section Solved problems. I just rewrite the solution.

(a+b+c-abc)^2 = ((a+b).1 + (1-ab).c)2 <= ((a+b)^2+(1-ab)^2)(1+c^2)
= (1 + a^2 + b^2 + a^2b^2)(1+c^2) <= (1 + a^2 + b^2 + (a^2+b^2)^2/4)(2-a^2-b^2) = RHS
Let t = a^2 + b^2 then RHS = (1+t+t^2/4)(2-t) = 2+ t - t^2/2 - t^3/4.
We want to show that
            2+ t - t^2/2 - t^3/4 <= 64/27
But by an easy computation this equivalent to
           (t-2/3)^2(t+10/3) >= 0
which is true b'se t>=0.

We can see, this method works also for this variant. Now let think about the generalization.

Namdung

I posted it since I had a nice solution, which can lead to very difficult inequalities:
  Consider the polynomial p=X^3-(a+b+c)X^2+X(ab+bc+ca)-abc. Then
  |p(ix)|^2=|-ix^3+x^2(a+b+c)+ix(ab+bc+ca)-abc|^2>=(x^2(a+b+c)-abc)^2. Take x=1 and you will find that |a+b+c-abc|<=|p(i)|=|(i-a)(i-b)(i-c)|=sqrt((a^2+1)(b^2+1)(c^2+1)). Now apply AM-GM.
  It's strange that this method cannot be applyed to the polish problem.

Seeing your solution, I found that I could shorten mine:
(a+b+c-abc)^2 = ((a+b).1 + (1-ab).c)2 <= ((a+b)^2+(1-ab)^2)(1+c^2)
= (1 + a^2 + b^2 + a^2b^2)(1+c^2) = (1+a^2)(1+b^2)(1+c^2) and then apply AM-GM as you.

Namdung


linear algebra
matrix
function
induction
superior algebra
superior algebra unsolved

It's a bit unfair to post them here since I don't know solutions for all of them, but I have the official solutions :D

(1)
Let p>=2 be a natural number and A=(a_(i,j)) be a square matrix of order n with integer elements. Prove that for any permutation f in S_n there is a function g:{1,2,..,n}->{0,1} s.t. by replacing the elements a_(f(i),i) from the matrix A with a_(f(i),i)+g(i) for all i from 1 to n, the determinant is not divisible by p.

(2)
Let A be a square matrix of odd order (>=3), with integer odd elements. prove that if A is invertible then it's not possible for all the minors of the elements of a row to have the same modulus.

(3)
Let M be the set of all derivable functions f:R->R, s.t. f(x)>=f(x+sinx), for all real x.

a) Prove that M contains at least one nonconstant function.
b) Prove that if f is in M then the set of the solutions of the equation f'(x)=0 is infinite.

(4)
Let f,g:R->R be 2 bijective and continuous functions s.t.
f(g -1 (x))+g(f -1 (x))=2x for all real x. Prove that if there is an x_0 s.t. f(x_0)=g(x_0) then f=g.

Hope you enjoy them!

Here is what I have for 2, hope it's okay! :D :D

Well let's say that:
A=(a_(i,j)) and let x_(i,j) be the cofactor of a_(i,j). So if for some i every |x_(i,j)|=d then d<>0.
Now let's take for some k<>i and get that sum(j=1,2n+1) x_(i,j)*a_(k,j)=0 and after this we divide it by d(d<>0) and we get that  \pm a_(k,1) \pm ..  \pm a_(k,2n+1)=0, we get imposobility for any randomly chosen case (I mean all cases :D ) because (a_(k,j) are odd). Problem solved, I think!

What's the official sol?

cheers! :D :D

errrr.. The 1st one is a simple induction!!!

[quote="grobber"]
(2)
Let A be a square matrix of odd order (>=3), with integer odd elements. prove that if A is invertible then it's not possible for all the minors of the elements of a row to have the same modulus.

Hope you enjoy them![/quote]

I used your ideas Lagrangia thx for writing another solution
 A=(a(i,j)) \in M_(2n+1)(Z) with a(i,j) odd intergers
A is invertible in M_(2n+1)(Z) => detA= \pm 1

Suppose there exist a row h such that all minors have the same
modulus

Expand the determinant for h-row
detA=(-1)^(h+1)a(h,1)D(h,1)+(-1)^(h+2)a(h,2)D(h,2)+...+(-1)^(h+2n+1)a(h,2n+1)D(h,2n+1)

but |D(h,1)|=|D(h,2)|=...=|D(h,2n+1)|=D
(*) detA=( \pm a(h,1) \pm a(h,2) \pm....\pm a(h,2n+1))D

All coefficients in the determinant  D(h,k) are odd=>
D(h,k) is congru to 0 (mod 2) => D is congru to 0 (mod 2)

Taking (*) modulo 2 we get \pm 1=0 (mod 2) contradiction.

I really enjoy that one, thx grobber.

[quote="grobber"]You would've gotten 7 points for problem 2! Even the notations are the same in the official solution! :D[/quote]

That is dangerous!! :D :D

I shall be thinking more on the 4th one tomorrow! :D :D

[quote="grobber"]

(3)
Let M be the set of all derivable functions f:R->R, s.t. f(x) \geq f(x+sinx), for all real x.

a) Prove that M contains at least one nonconstant function.
b) Prove that if f is in M then the set of the solutions of the equation f'(x)=0 is infinite.

Hope you enjoy them![/quote]

a)

Constant functions are obviously solution, let f a non constant function f in
M then is even :
Suppose f is odd f(-x) \geq f(-x+sin(-x)) => f(x) \leq f(x+sinx)
f is not in M.

Some properties on g(x)=x+sinx is derivable, increasing, g(0)=0
on A(k)= [2k \pi,(2k+1) \pi ] g(x) \geq x
on B(k)= [(2k-1) \pi , 2k \pi ] g(x) \leq x

What can we say about the monotony of f  \in M :

on A(k) f must be decreasing indeed
x \leq x+sinx => f(x) \geq f(x+sinx)

on B(k) f must be increasing indeed
x \geq x+sinx => f(x) \geq f(x+sinx)

f(x)= cosx works

I enjoyed a) I 'll hope b) also

[quote="Lagrangia"]

I shall be thinking more on the 4th one tomorrow! :D :D[/quote]

Lagrangia where is your proof for 4th ?

[quote="grobber"]

(3)
Let M be the set of all derivable functions f:R->R, s.t. f(x) \geq f(x+sinx), for all real x.

a) Prove that M contains at least one nonconstant function.
b) Prove that if f is in M then the set of the solutions of the equation f'(x)=0 is infinite.

[/quote]

Hello oubi,
b)
If f'(x)=0 has a finite number of solutions a_1<...<a_p for x> a_p
f'(x) =/=0
If f' is continuous for x>a_p, f'(x) will have a constant sign => f will be monotone on ]a_p;+oo[ wich contradict that the variation of f change on A_k, B_k

----------------------------------------------------
Some observation:

g(x)=x+sinx has some others properties
g(x) is 2-lipschitzienne |g(x)-g(y)|=|x-y+sinx-siny| \leq |x-y|+|x-y|

g^2(x)=x+sinx+sin(x+sinx)=g(x)+sing(x)
g^3(x)=g(g(x))+sin(g^2(x))=g(x)+sin(g(x))+sin(g^2(x))
by induction
g^n(x)=g(x)+sing(x)+...+sing^n(x) where g^n=gogo...og

f(x) \geq f(g(x)) \geq f(g^2(x)) \geq ... \geq f(g^n(x))

g is bijective f(x) \geq f(g(x)) => f(g^(-1)(x)) \geq f(x) by induction
for any n \in N f(g^(-n)(x)) \geq f(x) \geq f(g^(n)(x))


trigonometry
inequalities
algebra
polynomial
calculus
calculus computations

Determine whether the series $ \sum_{n\equal{}1}^{\infty}\frac{1}{n^{1.7\plus{}\sin n}}$ is convergent or divergent.

It diverges.

First, note that all of the terms of the series are positive, so we may deal with direct comparisons.

Define $ n_k$ to be the unique integer closest to $ 2\pi k \minus{} \frac {\pi}2.$ Note that $ \left|2\pi k \minus{} \frac {\pi}2 \minus{} n_k\right| < \frac12.$

$ \sin n_k\le\sin\left(\frac { \minus{} \pi}2 \minus{} \frac12\right) \equal{} \minus{} \cos\frac12\le \minus{} .875.$

That last inequality comes from a Taylor polynomial: $ \cos x\ge 1 \minus{} \frac {x^2}2$ for $ x$ in the first quadrant (at least).

Hence $ n_k^{1.7 \plus{} \sin n_k}\le n_k^{1.7 \minus{} .875} \equal{} n_k^{.825}.$

Note also that $ n_k < 2k\pi.$

Since all the terms are positive, if we use only certain terms, that makes the sum smaller.

$ \sum_{n \equal{} 1}^{\infty}\frac1{n^{1.7 \plus{} \sin n}} > \sum_{k \equal{} 1}^{\infty} \frac1{n_k^{1.7 \plus{} \sin n_k}}$

$ \ge \sum_{k \equal{} 1}^{\infty}\frac1{n_k^{.825}}$ $ \ge\frac1{(2\pi)^{.825}}\sum_{k \equal{} 1}^{\infty}\frac1{k^{.825}} \equal{} \infty.$

[quote="§outh§tar"]Try comparing with n^-0.7[/quote]
Useless.

$ \frac1{n^{1.7 \plus{} \sin n}}\le \frac1{n^{0.7}}$ and $ \sum\frac1{n^{0.7}} \equal{} \infty.$

Being shown to be smaller than a divergent series contains no information.

Sorry. You are right Kent. I should start writing these down..

The OP uses a plus sin n though.

[quote="§outh§tar"]The OP uses a plus sin n though.[/quote]
My argument above was using the correct sign, but a typo was creeping in towards the end. I've edited and fixed.


The arithmetic mean of two numbers is 12. Their geometric mean is 8. Find the sum of the square roots of the two numbers.

[hide="Solution"]
$ \frac{a\plus{}b}{2}\equal{} 12$
$ \sqrt{ab}\equal{} 8$
$ a\plus{}b \equal{} 24, ab \equal{} 64$

We want $ \sqrt{a}\plus{}\sqrt{b}$
$ a \equal{} 24\minus{}b\rightarrow b(24\minus{}b) \equal{} 64$
$ \minus{}b^{2}\plus{}24b\minus{}64 \equal{} 0\rightarrow b \equal{}\frac{\minus{}24\pm\sqrt{576\minus{}256}}{\minus{}2}\rightarrow \equal{} 12\pm 4\sqrt{5}$

So
$ b \equal{} 12\pm 4\sqrt{5}$, $ a \equal{} 12\mp 4\sqrt{5}$

We have  $ \sqrt{a}\plus{}\sqrt{b}$ = $ \sqrt{12\plus{}4\sqrt{5}}\plus{}\sqrt{12\minus{}4\sqrt{5}}$
That doesn't seem right..
[/hide]

That's because it's not simplified.

[hide="Note that..."] $ (\sqrt{a}\plus{}\sqrt{b})^{2}\equal{} a\plus{}b\plus{}2\sqrt{ab}$ [/hide]

this problem was already posted in this forum once:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=164375


Is my answer wrong on this. It comes out diff. then the answer key.

1. There are 12 diagonals that can be drawn on the faces of a cube.  Two different diagonals are selected at random.  What is the probability that the diagonals selected share a common point.
My answer: 4/11
Real answer: 5/11

Actually, it should be 5/11, because there are also six different points [b]on[/b] the faces at which two lines could meet.

Well what Monkey said basically is that besides the end points of the diagonal you choose, there is another diagonal on the same face that will intersect diagonal you choose, so those two would also share a common point as well as the diagonals sharing the endpoints.

But why is it 5/11
please show me the work
it can't be that much

Haha, I missed the cross thingies.  This is what happens when you don't do mathcounts for a year :D.

so did I! It took me hours of intense thinking to finally get that problem.
2002-2003 handbook stretch I believe.

NO
It has just the answer not the solution


Let $ ABCD$ be a quadrilateral,$ AC$ cuts $ BD$ at $ O$
Circumcirlce triangle $ AOB$ cut circumcircle $ COD$ at $ M$
Prove that $ P_{M\/(AOC)} \equal{} P_{M\/(BOD)}$
$ (AOC)$;($ BOD$) are circumcircles of triangles $ AOC$ and $ BOD$

Please check your problem. Note that A,O,C are collinear.


powers of 10
powers
arithmetic
Why the bump

[quote="AIME15"]$ \frac {10^7}{5 \times 10^4} \equal{}$
\[ \textbf{(A)}\ .002 \qquad \textbf{(B)}\ .2 \qquad \textbf{(C)}\ 20 \qquad \textbf{(D)}\ 200 \qquad \textbf{(E)}\ 2000
\]
[/quote]

[hide]Cancel 10s first to get $ \frac{10^3}{5}$. Then cancel the five out with a ten: $ \frac{10^2 \cdot 2}{1}$. Then just expand the rest out: $ \boxed{200}$[/hide]

[hide="Solution"]
This is equivalent to $\frac{10^3}{5}=2 \cdot 10^2=\boxed{\text{(D)} 200}$
[/hide]

trick question... did anyone else end up with 2000 the first time doing this

[hide=Post #10 by yackspit][url=aops.com/community/user/775443][b]yackspit[/b][/url] · Yesterday at 8:44 AM [url=aops.com/community/p20983523](view)[/url][color=transparent]helo[/color]
Yep I got this one too. It's just subtracting the exponents and then dividing by 5. So it's 200.

-----------
[color=#5b7083][aops]x[/aops] 10[color=transparent]hellloolo[/color] [aops]Y[/aops] 0 [color=transparent]hellloolo[/color] [/hide]

[tip=@yackspit][img]https://avatar.artofproblemsolving.com/avatar_775443.png[/img]
AoPS User[/tip] what a bump

Wow this is 12 years old.
[hide=solution] Cancel out the 10^7 with 10^4 to leave 10^3 in the numerator. 10^3=1000. 1000/5=200 or D[/hide]

I really hope I didn't mess this up. Is the answer 200?

[quote=Malu_2020]I can't believe this is 12 years old!  :o[/quote]

Some threads are 18 years old!

[hide=Post #1 by AIME15][url=aops.com/community/user/41047][b]AIME15[/b][/url] · Jan 11, 2009, 9:54 PM [url=aops.com/community/p1373346](view)[/url][color=transparent]helo[/color]
$ \frac{10^7}{5 \times 10^4}\equal{}$

\[ \textbf{(A)}\ .002 \qquad
\textbf{(B)}\ .2 \qquad
\textbf{(C)}\ 20 \qquad
\textbf{(D)}\ 200 \qquad
\textbf{(E)}\ 2000
\]

-----------
[color=#5b7083][aops]x[/aops] 1[color=transparent]hellloolo[/color] [aops]Y[/aops] 3 [color=transparent]hellloolo[/color] [/hide]

[tip=@AIME15][img]https://avatar.artofproblemsolving.com/avatar_41047.png[/img]
AoPS User[/tip]

[hide=Solution]$10^{7-4}=10^3 \implies \dfrac{10^3}{5}=\boxed{200}$[/hide]

[quote=mewto]Who keeps bumping all these old threads?[/quote]

time stamps exist

[quote=PlaneGod]This is equal to
$\frac{1}{5} \times \frac{10^7}{10^4} = 200$.[/quote]

Please hide your solution.

[quote=PlaneGod]How do you do that?[/quote]

Use [code][hide][/hide][/code] tags


Um...I know this is kind of a dumb question, but...
How do you make a blog?
Yeah I know, I'm retarded.

EDIT: Never mind...


Consider the vector space $C[-1, 1]$ with inner product defined by \[ \ < f,g > = \int^1_{-1} f(x) g(x) dx  \]
Show that $v_1(x) = \frac 1{\sqrt{2}}$ and $v_2(x) = \frac{\sqrt{6}}2x$ form an orthonormal set of vectors.

You should know how to integrate polynomials...
Just calculate the pairwise inner products.

We went over this so quick at school, but if I understand right $u_1$ & $u_2$ form orthonormal set when:

<$u_1$, $u_2$> = 1 , <$u_1$, $u_1$> = 0 and <$u_2$, $u_2$> = 0

is that all I do to prove, because I found that all are true.

Can someone explain who I use results of this to find best least squares approximation to

$h(x) = x^{\frac 13} + x^{\frac 23}$ by a linear aproximation


Hi,

I need to find out whether the reduced model is a better fit than the complete model.

In a complete model I have 4 variables. In the reduced model there are 2.

The regression analysis shows that the p-values of the variables in the complete model are 0.020, 0.001, 0.022, 0.011, i.e. all variables seem to be significant as p-values under the threshold of 0.05.

F-test of the complete model is slightly higher than the F-test of the reduced model: 9.88 vs. 9.44.

R^2 and R^2 (adj) values of the complete model are much higher than those of the reduced model.

So, I have all indications that the complete model is a much better fit than the reduced model, so I should be using the fitted regression equation of the model with 4 variables.

However, the BREG command outputs indicate, that 2 variables (with p-values: 0.001 and 0.011) are a much better fit than those with p-values 0.020 and 0.022.

I am not sure if I should drop 2 variables (with p-values 0.020 and 0.022) because of the BREG output or if I should stick to the complete model?

Thanx.

I don't know what all of these tests are, so I'm not really sure about this, but probably you should stick with the complete model. 4 variables is still pretty few for a linear regression model, so it's not like you're doing anything ridiculous. There's nothing wrong with 2 of the variables having higher p-values -- it just means that they're not quite as significant as the other two variables. Nobody said that all of the variables have to be equally significant.


geometry
circumcircle
inradius
incenter
inequalities
trigonometry
perpendicular bisector

... when you see the trick:

Let R be the circumradius and r the inradius of an acute-angled triangle ABC. Assume that the angle CAB is the greatest of the three angles of triangle ABC. Let M be the midpoint of the side BC, and let X be the point of intersection of the tangents to the circumcircle of triangle ABC at the points B and C. Prove that

$\frac{r}{R}\geq\frac{AM}{AX}$.

[color=red][EDIT: I am just seeing that this problem was posted before: http://www.mathlinks.ro/Forum/viewtopic.php?p=19365 .][/color]

  darij

Isn't this very similar to the problem from Korea 2004  :) ?

[quote="grobber"]How nice! :) Just show that $\frac{AM}{AX}=\frac{OM}R$, where $O$ is the circumcenter. It's easy from here.[/quote]

Let $P, Q$ be the intersections of the lines $AM, AX$ with the circumcircle $(O)$ other than the vertex $A$.

$\frac{OM}{OC} = \frac {OC}{OX} = \frac {OQ}{OX}$

because the right angle triangles $\triangle OMC \sim \triangle OCX$ have the common angle $\measuredangle OMC$ and $OQ = OC = R$ is the circumradius. The lines $AM, AX$ are the median and symmedian, respectively, from the vertex $A$ (Honsberger's Episodes, Chapter 7). Hence the bisector of the angle $\measuredangle A$ is also the bisector of the angle $\measuredangle DAE$. This bisector cuts the arcs $BC, PQ$ opposite the vertex $A$ at their common midpoint $N$. Thus the angle $\measuredangle QOX \equiv \measuredangle QON = \measuredangle POQ / 2 = \measuredangle PAQ \equiv \measuredangle MAX$. Since the triangles $\triangle OQX, \triangle AOX$ have another angle $\measuredangle AXO \equiv \measuredangle QXO$ common, they are similar and

$\frac {AM}{AX} = \frac{OQ}{OX} = \frac{OM}{OC}$

Let $I$ be the incenter and $T$ the tangency point of the incircle $(I)$ with the side $BC$. From the condition that the angle $\measuredangle A$ is the largest one in the triabgle $\triangle ABC$, $\measuredangle A + \measuredangle B / 2 > 90^o$, $\measuredangle A + \measuredangle C / 2 > 90^o$ and

$\measuredangle IBT = \frac{\measuredangle B}{2} > 90^o - \measuredangle A = \measuredangle OBM$

$\measuredangle ICT = \frac{\measuredangle C}{2} > 90^o - \measuredangle A = \measuredangle OCM$

The locus of incenters $I$ of all triangles $\triangle ABC$ with the fixed base $BC$ and fixed value of the angle $\measuredangle A$ is the shorter arc $BC$ of a circle $(N)$ centered at the midpoint $N$ of the shorter arc $BC$ of the circumcircle $(O)$ and with radius $NB = NC$. Indeed, from the circumcircle $(O)$, the angle $\measuredangle BNC = 180^o - \measuredangle A = \measuredangle B + \measuredangle  C$. On the other hand, the angle $\measuredangle BIC = 180^o - \frac{\measuredangle B + \measuredangle C}{2} = 90^o + \frac{\measuredangle A}{2}$. The central angle of the shorter arc $BC$ of the desired locus is then twice this value, or $180 + \measuredangle A = 360^o - (\measuredangle B +  \measuredangle C)$, i.e., the locus circle is centerd at the point $N$. Under the condition that the angle $\measuredangle A$ is the largest one, the locus of incenters $I$ is just a part of the shorter arc $BC$ of the circle $(N)$, namely, the arc $EF$, where $E = OB \cap (N)$, $F = OC \cap (N)$ are the intersections of the lines $OB, OC$ with the shorter arc $BC$ of the circle $(N)$. From this, it is clear that

$\frac r R = \frac{IT}{R} \ge \frac{OM}{R} = \frac {AM}{AX}$

The equality can occur only if the acute-angle triangle $\triangle ABC$ is equilateral.

:rotfl:

Yetti, your solution is nice but it can be simplified four times  :) . For instance, once the inequalities $\measuredangle IBT \geq \measuredangle OBM$ and $\measuredangle ICT \geq \measuredangle OCM$ are shown, it follows that the point O lies inside or on the boundary of the triangle BIC (well, of course, to draw this conclusion, we have used the fact that the point O lies inside the triangle ABC - what is clear, since the triangle ABC is acute-angled). But when a point lies inside or on the boundary of a triangle, the distance from this point to a side of the triangle is $\leq$ to the distance from the opposite vertex of the triangle to this side. Hence, the distance from the point O to the side BC of the triangle BIC is $\leq$ to the distance from the vertex I to this side BC. In other words, $OM\leq IT$. This rewrites as $IT\geq OM$, and thus

$\frac{r}{R} = \frac{IT}{R} \geq \frac{OM}{R} = \frac{AM}{AX}$.

So far about your solution. My solution of the problem is less elegant, it goes as follows:

Let O be the center of the circumcircle of triangle ABC. Since the radius of this circumcircle is R, we have OA = OB = OC = R.

In http://www.mathlinks.ro/Forum/viewtopic.php?t=19720 post #3, I showed that the point M is the inverse of the point X in the circumcircle of triangle ABC. Thus, the point M lies on the line OX and satisfies $OX\cdot OM=R^2$ (since O is the center and R is the radius of the circumcircle of triangle ABC). Using OA = R, the equation $OX\cdot OM=R^2$ becomes $OX\cdot OM=OA^2$. In other words, $\frac{OM}{OA}=\frac{OA}{OX}$. Since we also have < MOA = < AOX, it follows that the triangles MOA and AOX are similar. Thus, $\frac{MA}{AX}=\frac{OM}{OA}$. Since OA = OB, this becomes $\frac{MA}{AX}=\frac{OM}{OB}$.

Since the point O is the center of the circumcircle of triangle ABC, we have $\measuredangle BOC=2\measuredangle BAC=2A$ by the central angle theorem. Now, consider the isosceles triangle BOC (isosceles because of OB = OC, of course). Its base angle < OBC must thus equal

$\measuredangle OBC=\frac{180^{\circ}-\measuredangle BOC}{2}=\frac{180^{\circ}-2A}{2}=90^{\circ}-A$.

Since the points M and O both lie on the perpendicular bisector of the segment BC (the point M as the midpoint of this segment BC, and the point O as the circumcenter of triangle ABC), we have $OM\perp BC$; thus, in the right-angled triangle OMB, we have

$\frac{OM}{OB}=\sin\measuredangle OBM=\sin\measuredangle OBC=\sin\left(90^{\circ}-A\right)=\cos A$.

Thus, the equation $\frac{MA}{AX}=\frac{OM}{OB}$ rewrites as $\frac{MA}{AX}=\cos A$. In other words, $\frac{AM}{AX}=\cos A$. This is the trick which I promised in my first post and which, at least, made [i]me[/i] laugh out loudly :D .

Now as we have seen that $\frac{AM}{AX}=\cos A$, the inequality which we must prove, $\frac{r}{R}\geq\frac{AM}{AX}$, rewrites as $\frac{r}{R}\geq\cos A$. But it is well-known that $\frac{r}{R}=\cos A+\cos B+\cos C-1$. Thus, we have to prove that $\cos A+\cos B+\cos C-1\geq \cos A$. This immediately simplifies to $\cos B+\cos C-1\geq 0$; in other words, we have to show that $\cos B+\cos C\geq 1$. Since we know from the problem hypothesis that the angle < CAB = A is the greatest angle of the triangle ABC, the angles B and C are the two smallest angles of triangle ABC. And thus, in order to solve our problem, it remains to show the following fact: The sum of the cosines of the two smallest angles of a triangle is always $\geq 1$. But this was proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=4379 .

And thus, the problem is solved.

  darij

[quote="darij grinberg"]Yetti, your solution is nice but it can be simplified four times  :) .[/quote]

I appreciate the simplification, it can always be used in a more complex proof. Thanks for a different solution of the problem. What made me :rotfl: was the fact that the line $AX$, being a symmedian from the vertex $A$, immediately leads to the equality of the angles $\measuredangle QOX = \measuredangle MAX$. I did not remember this fact and had to look it up.

yetti


calculus
integration
trigonometry
calculus computations

It seems to that i went wrong way.
[b]1. The problem statement, all variables and given/known data[/b]

$ \int \frac {arcsin(e^x)dx}{e^x}$

[b]3. The attempt at a solution[/b]
$ \equal{} \minus{} \frac {arcsin(e^x)}{e^x} \plus{} \int \frac {dx}{\sqrt {1 \minus{} e^{2x}}}$

$ \int \frac {dx}{\sqrt {1 \minus{} e^{2x}}} \equal{} \frac {x}{\sqrt {1 \minus{} e^{2x}}} \plus{} \int \frac {xe^{2x}dx}{(1 \minus{} e^{2x})^{\frac {3}{2}}}$

Let $ u \equal{} e^{ \minus{} x}$; it remains to compute

$ \minus{} \int \arcsin \left( \frac {1}{u} \right) \, du \equal{} \minus{} \int \text{arccsc } u \, du$

Integration by parts should give an answer from here.

Like this?
$ -\int arcsin(\frac{1}{u})du=-\int arccsc(u)du=-uarccsc(u)-\int\frac{du}{\sqrt{u^2-1}}$

$ u=sec\theta\rightarrow du=-\frac{sin\theta d\theta}{cos^2\theta}$

$ \int\frac{du}{\sqrt{u^2-1}}=-\int \frac{cos^3\theta d\theta}{sin^2\theta}$

$ =-\int \frac{(1-sin^2\theta)cos\theta d\theta}{sin^2\theta}=-\int \frac{cos\theta d\theta}{sin^2\theta}+\int cos\theta d\theta{}$
$ =\frac{1}{sin\theta}+sin\theta$

My mistake

$ \int\frac{du}{\sqrt{u^2\minus{}1}}\equal{}\minus{}\int \frac{d\theta}{cos\theta}$
 But I don't know,actually,how to integrate last one.

Three comments on that last post:

(1) In the integral $ \int\frac{du}{\sqrt{u^2\minus{}1}},$ it would not be appropriate to make the substitution $ u\equal{}\sin\theta,$ since the integral only makes sense (in real terms) for $ |u|>1.$ Use $ u\equal{}\sec\theta$ or $ u\equal{}\cosh t$ instead.

(2) Even if you did ignore square roots of negative numbers and apply that substitution anyway, you still wouldn't get $ \int\frac{d\theta}{\cos\theta}.$ There's something wrong with your substitution.

(3) $ \int\frac{d\theta}{\cos\theta}\equal{}\int\sec\theta\,d\theta\equal{}\ln|\sec\theta\plus{}\tan\theta|\plus{}C$

(Something you really need to have memorized. It is easy to prove that this is the correct formula, provided you already know the answer.)

[quote="Carcul"][quote="Kent Merryfield"]It is easy to prove that this is the correct formula, provided you already know the answer[/quote]

What if we did not knew the answer? That is, how to show that the primitive of sec(x) is ln|sec(x) + tan(x)| without assuming the answer?[/quote]

Here is two approaches.

1. $ \int \frac{1}{\cos \theta}\equal{}\int \frac{\cos \theta}{1\minus{}\sin ^ 2 \theta}\ d\theta$,

2. Let $ u\equal{}\tan \frac{\theta }{2}$.

hello, i have
-$ \frac{\arcsin(e^x)}{e^x}\minus{}artanh\left(\frac{1}{\sqrt{1\minus{}(e^x)^2}}\right)$.
Sonnhard.


[size=150]1. $ f(x) \equal{} \frac {3 \plus{} 5sin(x)}{\sqrt {5 \plus{} 4cos(x) \plus{} 3sin(x)}}$
find the domain of $ f(x)$
2.  How many even numbers are there in the set $ S \equal{} {C^{n}_{2008} (n \equal{} 0,1,...2008)}$
3.Simplification $ \sum^{n}_{k \equal{} m}C^{k}_{n}C^{m}_{k}x^{k \minus{} m}(1 \minus{} x)^{n \minus{} k}$

4. $ \frac {1}{2008} \equal{} 0.000 498...625 498...625.....$
find the length of $ 498...625$

[/size]

:P
thanks

[hide="1)"]$ f(x) \equal{} \frac {3 \plus{} 5sin(x)}{\sqrt {5 \plus{} 4cos(x) \plus{} 3sin(x)}}$

Minimum value of $ \sqrt {5 \plus{} 4cos(x) \plus{} 3sin(x)} \equal{} 5 \minus{} \sqrt {16 \plus{} 9} \equal{} 0$

Therefore,we need to find $ x$ such that $ \sqrt {5 \plus{} 4cos(x) \plus{} 3sin(x)} \ne 0$

$ \Longrightarrow 3\sin x \plus{} 4\cos x \ne \minus{} 5 \\
\frac {3}{5}\sin x \plus{} \frac {4}{5}\cos x \ne \minus{} 1 \\
sin(x \plus{} k) \ne \minus{} 1 \\
x \ne n\pi \plus{} ( \minus{} 1)^n\left( \minus{} \frac {\pi}{2}\right) \minus{} k$

($ k \approx 37$)

Therefore,domain of $ f(x)$ is $ R\minus{} \{n\pi \plus{} ( \minus{} 1)^n\left( \minus{} \frac {\pi}{2}\right) \minus{} 37\}$[/hide]

1
I use the calculator and the answer goes to  $ ( \minus{} \frac {4}{5}\sqrt {10},\sqrt {10}]$
I have a outline
$ sin(x) \equal{} \frac {2t}{1 \plus{} t^{2}}$
$ cos(x) \equal{} \frac {1 \minus{} t^{2}}{1 \plus{} t^{2}}$
$ t \equal{} tan(\frac {x}{2})$
and then $ f(x) \equal{} \frac {3 t^{2} \plus{} 10t \plus{} 3}{(t \plus{} 3)\sqrt {1 \plus{} t^{2}}} \equal{} \frac {3t \plus{} 1}{\sqrt {1 \plus{} t^{2}}}$
$ f'(x) \equal{} 0 ,x \equal{} 3 ,f(3) \equal{} \sqrt {10}$
but the min is not solved

[quote="admire9898"]$ f'(x) \equal{} 0 ,x \equal{} 3 ,f(3) \equal{} \sqrt {10}$[/quote]
I guess,you have mixed up $ t$ and $ x$ because neither $ f'(x) \equal{} 0$ nor $ f(3) \equal{} \sqrt{10}$

[quote="Bictor717"][quote="picture74"][hide="1)"]$ f(x) \equal{} \frac {3 \plus{} 5sin(x)}{\sqrt {5 \plus{} 4cos(x) \plus{} 3sin(x)}}$

Minimum value of $ \sqrt {5 \plus{} 4cos(x) \plus{} 3sin(x)} \equal{} 5 \minus{} \sqrt {16 \plus{} 9} \equal{} 0$

Therefore,we need to find $ x$ such that $ \sqrt {5 \plus{} 4cos(x) \plus{} 3sin(x)} \ne 0$

$ \Longrightarrow 3\sin x \plus{} 4\cos x \ne \minus{} 5 \\
\frac {3}{5}\sin x \plus{} \frac {4}{5}\cos x \ne \minus{} 1 \\
sin(x \plus{} k) \ne \minus{} 1 \\
x \ne n\pi \plus{} ( \minus{} 1)^n\left( \minus{} \frac {\pi}{2}\right) \minus{} k$

($ k \approx 37$)

Therefore,domain of $ f(x)$ is $ R \minus{} \{n\pi \plus{} ( \minus{} 1)^n\left( \minus{} \frac {\pi}{2}\right) \minus{} 37\}$[/hide][/quote]
You shouldn't mix degrees and radians.[/quote]Agreed, it would've been better if he just wrote $ k\equal{}\arcsin\frac45$.

And what's the deal with the 'Urgent!' from OP?

these questions are exercises that my teacher gave to us but I counldn't solve.

[quote="admire9898"]what about the forth?[/quote]Prime factorize 2008 and the period is equal to the reciprocal of one of the primes.


trigonometry
inequalities
trig identities
Law of Cosines

Let $M$ be an arbitrary point inside the equilateral triangle $ABC$. Denote by $A_{1},B_{1},C_{1}$ the intersection points of the traight lines $AM$ and $BC$ , $BM$ and $CA$, $CM$ and $AB$ respectively. Prove that $A_{1}B_{1}.B_{1}C_{1}.C_{1}A_{1}\geq\frac{1}{2}(A_{1}B.B_{1}C.C_{1}A+AB_{1}.BC_{1}.CA_{1})$

[hide]Let $A_{1}B=x, A_{1}C=y, CB_{1}=z, B_{1}A=a, AC_{1}=b,$ and $C_{1}B=c$

Then we seek to prove that

$A_{1}B_{1}\cdot B_{1}C_{1}\cdot C_{1}A_{1}\ge \frac{1}{2}(xzb+acy)$

Ceva is kind enough to tell us that xzb=acy

We thus rewrite the desired as the more symmetric

$(A_{1}B_{1}\cdot B_{1}C_{1}\cdot C_{1}A_{1})^{2}\ge abcxyz$

Now the law of cosines is enough to demonstrate that

$A_{1}B_{1}^{2}=z^{2}+y^{2}-zy$

Similarly rewriting $B_{1}C_{1}$ and $C_{1}A_{1}$ we transform the inequality yet again into

$(z^{2}+y^{2}-zy)(a^{2}+b^{2}-ab)(c^{2}+x^{2}-cx) \ge abcxyz$

Now $z^{2}+y^{2}\ge 2zy$, $a^{2}+b^{2}\ge 2ab$, and $c^{2}+x^{2}\ge 2cx$

The desired inequality immediately follows[/hide]


probability
ratio
inequalities
geometry
triangle inequality

Source: Illinois(ICTM) Algebra II Regional AA Test 2002

A line segment of length 6 is divided at random into three parts with the restrictions that the longest part has a length greater than 2 and less than 4 and that one of the remaining parts is twice the other remaining part. Find the probability that the three lengths of the three parts can serve as the lengths of the 3 sides of a triangle. Express your answer as a common fraction reduced to lowest terms.

Usually the ICTM problems are easy, but I can't seem to put a handle on this problem. Help would be nice, it's bothering the heck out of me.

$\frac{3}{8}$?
the longest peice cant be shorter than, $\frac{12}{5}$ because other wise you wont have the 2:1 ratio of the remaining pieces.
so the longest peice can be $\frac{12}{5}<L<4$ so the interval is $\frac{8}{5}$ wide. but  the longest peice cant be longer, than 3, otherwise you wont havea triangle. $\frac{12}{5}<L<3$
so the probability is:$\frac{\frac{3}{5}}{\frac{8}{5}}=\frac{3}{8}$

[hide]Form a bunch of inequalities

$2<x<4$

$\frac 23 < y < \frac 43$

From the triangle inequality

$y<x$

$y> \frac 13 x$

Graphing it we find the intersection points of all the lines and find the area of the intersection of the inequalities over the area given by $2 < x < 4$ and $\frac 23 < y < \frac 43$

Using the shoestring method and the intersection points $(\frac 23, \frac 23), (\frac 43, \frac 43), (4, \frac 43), (2, \frac 23)$ we find the area of the smaller section is $\frac 43$ out of a total area of 4 gives a final answer of $\frac 13$[/hide]

Is it well defined?  Is the probability distribution of the longest part uniform over $(2,4)$?

I took towersfreak's approach, and got 3/10, but the given answer is 3/8, and amirtlusa's approach looks good to me.

I used towersfreak's and amirhtlusa's approach and got different results as well. By instinct, amirhtlusa's solution seems more correct to me though.

[quote="towersfreak2006"]The lengths we're looking at are $l, 2l, 6-3l$ where $2<6-3l<4$ and $6-3l>2l$, so $\frac{6}{5}>l>\frac{2}{3}$

Triangle inequality:
$l+2l>6-3l\Rightarrow l>1$
$l+6-3l>2l\Rightarrow l<\frac{3}{2}$
$2l+6-3l>l\Rightarrow l<3$
$1<l<\frac{3}{2}$ will create three sides of a triangle.

Since $l<\frac{6}{5}$, of the possible lengths of $l$, $1<l<\frac{6}{5}$ satisfy the conditions.
Thus, we have $\frac{\frac{6}{5}-1}{\frac{4}{3}-\frac{2}{3}}=\frac{\frac{1}{5}}{\frac{2}{3}}=\frac{3}{10}$.[/quote]
i know why towersfreak2006 got different answer: if you check in last part
[quote="towersfreak2006"]$\frac{\frac{6}{5}-1}{\frac{4}{3}-\frac{2}{3}}$[/quote]
instead of $\frac43$ it should be $\frac65$, then towersfreak2006's answer is same as amirhtlusa's answer. the reason is simple: if $\frac43>l>\frac23$ then we only have $2<6-3l<4$ but it does not satisfy condition that it is largest piece. for example, imagine a piece with $\frac43>l=\frac{19}{15}>\frac65$ then $2l=\frac{38}{15}$ but $6-3l=\frac{33}{15}$ which shows $2l>6-3l$ :!:
did this help!!!!??? cuz it took me a while to find what was wrong in towersfreak2006. im going to have  :icecream:  now

Oops. Hehe, I actually had $\frac{6}{5}>l>\frac{2}{3}$ at the top. I guess I just forgot to change it at the bottom.  :blush:

Nice catch though. I've been scratching my head for a while.  :)


Buenos dias,
somo sabran....
la OLIMPIADA DE MAYO de este año
se ha rehalizado el dia sabado 13 de mayo al menos en PERU....

Ya cuento con el enunciado de las preguntas de los
dos niveles ...pero por disposiciones de la Organizacion del evento...
se evita su publicacion hasta un lapso de 5 dias apartir
de la celebracion del examen...

Por tanto, ruego a los jovenes miembros de esta comunidad,
evitar la divulagacion de todo tipo de los examenes dentro de este lapso.

Gracias.

alguien podría publicar ahora el examen de la olimpiada de mayo 2006???