algebra
polynomial
Vieta
arithmetic sequence
sum of roots

Determine $ m$ so that the equation \[ x^4  \minus{} (3m \plus{} 2)x^2  \plus{} m^2  \equal{} 0
\]
has four real roots in arithmetic progression

We only have $ x^4$ and $ x^2$, so our polynomial is $ (x^2 \minus{} a)(x^2 \minus{} 9a) \equal{} 0$.

This means our restriction is $ 10a \equal{} 3a \plus{} 2$, and so $ m \equal{} \frac {2}{7}$.

Clearly if m is a real number, then the only solution is $ m\equal{}6$

1) It is poor expository form to use the word "clearly" (though this is a common problem among math people, myself not excluded).
2) This is especially true when something is not clear, as in the present case.
3) This is especially especially true when the claimed statement is not only not clear, but actually false, also as in the present case.

Incidentally, the modifier "real" is irrelevant to the problem -- if we allow complex roots, there are still only the same two solutions, one of which was found by the preceding poster.  The first half of Differ's solution is fine, but something went awry in solving the equations $ 9a^2 \equal{} m^2$ and $ 10a \equal{} 3m \plus{} 2$.

@differ...
well, you assumed that a and m are the same?
you mean that the polynomial is in the form $ (x^2 \minus{} a)(x^2 \minus{} 3a) \equal{} 0$ but then you skipped a step from 10a=3a+2 to m=2/7.
I think you meant that -10a=-(3m+2) and $ 9a^2 \equal{} m^2$

EDIT:  beaten to it.

[hide="the other answer should be"]
6/19
[/hide]
just solve the two equations, i guess.

@zapi2007:

$ m\equal{}\frac{6}{19}$ is not correct.

@JBL:

I don't want to argue with people who don't know maths at all, because it's useless.  For the record,
you can check for yourself that if $ m\equal{}6$, then the equation becomes $ x^4\minus{}20x^2\plus{}36\equal{}0$

This equation has four real roots, namely $ \minus{}3\sqrt{2},\minus{}\sqrt{2},\sqrt{2},$ and $ 3\sqrt{2}$.  And I think that even a middle school student can clearly see that these numbers form an arithmetic progression.

If you can prove me wrong, just reply to my message and prove me wrong.  As for the solution, maybe you can figure it out yourself since it's very trivial.

zapi2007 is almost right and you're still wrong.  $ m \equal{} 6$ is [i]a[/i] solution, but it's not the [i]only[/i] solution, as you claimed.  Have you tried solving when $ m \equal{} \minus{}\frac {6}{19}$?   :roll:

Edit: A PM titled "you suck"?  Very classy.   :roll:

[hide]

Let the 4 roots be $ a\minus{}3d,a\minus{}d,a\plus{}d,a\plus{}3d$ where $ d$ is the common difference of the A.P.
Using vieta's we get,sum of roots=0 which gives $ a\equal{}0$
Now $ (a\minus{}3d)(a\minus{}d)(a\plus{}d)(a\plus{}3d)\equal{}m^{2}$
Form here we get $ 9d^{4}\equal{}m^{2}$

Taking sum of roots two at a time,again using vieta's,
$ 10d^{2}\equal{}3m\plus{}2$ solving we get the answers.
[/hide]

@ JBL:

sorry dude.  my bad!   :o
you're right, there are indeed two solutions, $ m\equal{}6$ and $ m\equal{}\minus{}\frac{6}{19}$

oh well, i guess i have to study more algebra.   :oops:


No calculators, as you shouldn't need them. They're probably not even mathcounts level and they have no creativity since I came up with most of them probably =S

1. There exists a circle with area $A$ and radius $r$. What is the radius (in terms of r) if the area of a circle is $2A$?

2. Let m and n be the greatest positive integers such that the following equations are true: $(1+2+3+...+m)^2=(1+8+27+...+1331)$ and $(1+2+3+...+n)^2<(1+8+27+...+2197)$

Find $\frac{m} {n}$

3. What is the number of positive factors of $6^6$?

4. How many perfect squares are products of two or more other perfect squares?

5. Let a, b, c, and d be consecutive increasing terms in an arithmetic sequence such that the common difference is 2006. Find the value of $\frac{d^2-a^2} {c^2-b^2}$

[hide="Answers"]1. $r\sqrt{2}$
2. $\frac{11} {12}$
3. 49
4. Infinitely many
5. 3[/hide]

1.  r^2 * pi = A
2 * r^2 * pi = 2A
so [b]sq. rt. 2 * r[/b]

2.That formula is the sum of cubes.  m = 11 and n = 12
so [b]11/12[/b]

3.6^6 = 3^6 * 2^6 and the formula for the factors is add one to the powers and multiplly
so 7 * 7 = [b]49[/b]

4.There is no limit becuase there a pattern but there is nothing to stop at.

5.a^2 + b^2 is (a+b)(a-b)
so ((2a+6018)(6018))/((2a+6018)(2006))
2a+6018 cancels out
so 6018/2006 = [b]3[/b]

[quote="arkmastermind"]note that in number 5, you can factor it as $\frac{(d+a)(d-a)}{(c+b)(c-b)}$. $d+a$ and $c+b$are both equal because we have an arithmetic sequence, and they can be canceled out. we have to compute $\frac{d-a}{c-b}$. the denominator is the common difference, whereas the numerator is the three times the common difference. The answer is thus $\boxed{3}$.

note that in this problem the common difference could have been anything, all that we needed to know was that $a,b,c,d$ formed an arithmetic sequence.[/quote]

The way I did it is just to call:

a - 2006
b - 2(2006)
c - 3(2006)
d - 4(2006)

Square those to get $\frac{16(2006)-2006} {9(2006)-4(2006)}$

Which all cancels out to get $\frac{15} {5}$ which equals $3$.

I'm not a moderator, but I'd just like to remind people to put their answer in hide tags so others may have a chance to solve the problems without seeing the solutions.


1) Let $ a,b,c$ sides of a triangle. Prove that
$ \frac{\sum_{cyc}a\left( b\plus{}c\minus{}a\right) }{\prod_{cyc}\sqrt{b^{2}\plus{}c^{2}\minus{}a^{2}}}\geq \sum \frac{1}{\sqrt{b^{2}\plus{}c^{2}\minus{}a^{2}}}$

2) Let $ a,b,c\in\mathbb{R}^\plus{}$ such that $ ab\plus{}bc\plus{}ca\equal{}1$. Find the biggest $ C$ such that $ C\leq \frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{a^{2}b^{2}\plus{}b^{2}c^{2}\plus{}c^{2}a^{2}}$

3) Let $ a,b,c\in\mathbb{R}$ such that $ a\plus{}b\plus{}c>\sqrt{a^2\plus{}b^2\plus{}c^2}$ and $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}>0$
Prove that $ a,b,c>0$

[quote="bambang"]3) Let $ a,b,c\in\mathbb{R}$ such that $ a \plus{} b \plus{} c > \sqrt {a^2 \plus{} b^2 \plus{} c^2}$ and $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} > 0$
Prove that $ a,b,c > 0$[/quote]
From $ a \plus{} b \plus{} c > \sqrt {a^2 \plus{} b^2 \plus{} c^2}$%=>$ a\plus{}b\plus{}c>0$$ ;ab\plus{}bc\plus{}ca>0$;$ abc>0$
Suppose that $ a<0$=>$ bc>0$=>$ a(b\plus{}c) >\minus{}bc>0$=>$ b\plus{}c<0$
=>$ a\plus{}b\plus{}c<0$
So $ a>0$
Similar we have $ b;c>0$
Done!

Please don't post reply here.
Some of the problems are from Indonesia journal and can't be discussed now.
Bambang : don't do this again!!!!


Can anyone post here the results of the romanian team at the IMC 2006? Thank you very much  :)

Claudiu Raicu(Bucharest) got a great first prize, with 180 points. Ioan Marcut(Cluj) got a first prize, and the rest third or below. I got a third with 100 points.

[url=http://www.imc-math.org.uk/imc2006/results2006.htm]Results of IMC 2006[/url]


Find the range of $ f(x) \equal{} \dfrac{x}{\sqrt {x \minus{} 9}}$

[hide]Clearly $ x > 9$, and the function is never negative.  There is a vertical asymptote at $ x \equal{} 9$ and as $ x\to \plus{} \infty$, $ f(x)\to \sqrt {x}$, which also tends to infinity.
This means there is some minimum value of $ f(x)$, to find it we write $ f(x)$ as:

$ f(x) \equal{} \frac {x \minus{} 9 \plus{} 9}{\sqrt {x \minus{} 9}} \equal{} \sqrt {x \minus{} 9} \plus{} \frac {9}{\sqrt {x \minus{} 9}}$

By AM-GM:
$ \sqrt {x \minus{} 9} \plus{} \frac {9}{\sqrt {x \minus{} 9}} \geq 6$

So the range of $ f(x)$ is $ [6, \plus{} \infty)$
[/hide]

[hide="..."]... And equality can be achieved by letting $ x\equal{}18$.[/hide]


geometry
videos
3D geometry
AMC
USA(J)MO
USAMO
AIME

Like what type of games are we supposed to bring? I live in CA, so I can't bring much with me.

I have apples to apples, monopoly and scrabble, but I probably can't take scrabble because the board is a huge rotating one.   I'll bring whatever I'll have room for, either apples to apples or monopoly.

[quote="korammath"]
i live in dallas, actually within walking distance of UTD, so if we need anything else i know where to go, but whatever.[/quote]

Walking distance? Aww. I'm about a 20-30 minute drive.

I think someone should bring playing cards.

I could bring playing cards, and maybe uno...I also have monopoly and scrabble, but they're probably too big.

ill handle scrabble.
ill also bring monopoly.
bring apples to apples, anyone who has it.

ok......... heres  log of whos bringing what

korammath:
scrabble, playing cards, monopoly, ping pong paddle(s)? and balls.

infinity4ever
apples to apples

carbon57
monopoly?, playing cards,Uno

distracted523
playing cards?

Set, Chess, Parcheesi, Gordian's Knot, sudoku book.

I love Apples to Apples.

I have a box of a dozen unopened decks of cards on my shelf, so I'll bring those. I'll also take a scrabble set and probably a go board.

And if anyone else wants to, we can play Nim. :D

[quote="korammath"]ok.......... so who will bring what?

i live in dallas, actually within walking distance of UTD, so if we need anything else i know where to go, but whatever.[/quote]

I don't think that people were allowed to wander off campus randomly like that. I suppose that nobody can stop you but generally I think that we were supposed to stay around and if you wanted something from like walmart or whatever, then you could ask a counselor to buy it for you (like soap, some snacks, etc).

huh?

The message is too small. Please make the message longer before submitting.

huh whoa...  I have a question.

How are we going to get all of these games together?  Unlike you guys, I'm a girl, so I probably won't be able to come over to your rooms with apples to apples.  Is there some specific room where we're going to drop off all of our games at?

The dorms that we had last year were like three bedrooms for students, two bathrooms, another bedroom for the mentor and then like a living room area. I know that I hung out in one of the girls' living room areas almost every night so I don't think thats a problem :D. There is a rule that somebody cannot be in the bedroom of somebody of the opposite sex, but there are other places to hang out. The UTD student union has a place with ping-pong, ddr, etc, so some people hung out there. Others I think hung out in this building sorta in the middle with pool and a tv maybe. There are also places that people practiced violin or something.

[quote="junggi"]also anyone play tennis here
i suck at it but i figure i need some exercise[/quote]

I play tennis although I suck too. We'll have a tourney to see who sucks the most. Wow I haven't posted since forever.

[quote="blackbelt14253"]what is starcraft :?:
I'm asian. (yes, really. and I have no idea what starcraft is.)[/quote]

Yeah, same here.

Basically a tactical game with future as background and 3 clans fighting.

It's pretty fun and brain-using.

Dude. you play on bnet? You should play the UMS games cuz those pwn melee/tvb/1v1/w/e else you play xD. Like spawn d woot =D...

lol i just play stupid online maps on starcraft like massing, golems, yugi-oh, halo, turret D, mixed D, sunken D, etc with my friends occasionally...I don't know any build orders for SC but I guess that I know RTS games pretty well.

um
no
ums games take like
near 0 actual skill
why would you ever play them
it just like defeats the whole purpose of starcraft

hmm
get iccup
http://sc.iccup.com/info/scstart/

err uh. ums games. are fun. that's the point. xD. and um. they take 0 actual skill. but. it takes. lots of noob skill. =D

starcraft is a game about REAL skill
not that i have any but
still
ums is dumb

Are we supposed to bring a calculator? (two more days! two more days!)

I'm sure in the olympiad classes, you won't have any use for a calculator. Perhaps it could be used in some of the earlier classes but really you don't need one. Maybe like a pocket calculator but there is no reason to bring a TI 89 or whatever.

And taylor, one can play starcraft for fun and do crazy maps...

blargh
ums like turrent defense and stuff are like
meh
are they any different from random flash games

you can play them with friends...like i've played fair number of 6 way TD, etc and its fun when you are around people and stuff...


Prove that Nagel point of a triangle lies on its incircle iff one of its side is equal to its semiperimeter

Hmm... if one of its sides is equal to its semiperimeter, then $a = \frac{a+b+c}2$, then $2a = a+b+c$ or $a=b+c$, impossible in a triangle... did I misundarstand anything?

If one of the sides is the semiperimeter, then the triangle is degenerate, so the incircle has radius zero. I believe though, that this isnt even true. If it isosceles and degenerate then this is true, but I think its false otherwise. As far as the only if part goes, I have no idea.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=22461]Correct problem with solutions[/url].

  darij


geometry
circumcircle
cyclic quadrilateral

I think this was solved on http://www.mathlinks.ro/Forum/viewtopic.php?t=19099 .

  Darij

I think $EF$ should be the diameter of the circumcircle instead of the incircle. I drew a dynamic sketch from which I could see that it's not true if $EF$ is the diameter of the incircle (as I understand from your post). [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=19099]Here[/url]'s a similar (if not the same :)) problem.

[i]wrong post. check the following...[/i]

Sorry for the misunderstanding, but EF is the diameter of the outer circle, not the incircle.

Also, isn't an incircle supposed to touch each side at a tangent. From the link that you (Grobber) has posted,

[quote]Let O,I be the centers of the circumscribed and inscribed quadrilaterals of PQRS respectively.[/quote]

Is this implying that PQRS is the only quadrilateral and it has both an outer circle and an incircle?

[quote="eagle"][quote]Let O,I be the centers of the circumscribed and inscribed quadrilaterals of PQRS respectively.[/quote]

Is this implying that PQRS is the only quadrilateral and it has both an outer circle and an incircle?[/quote]

Of course not. This was a typo. Read "circumscribed and inscribed circles" instead of "circumscribed and inscribed quadrilaterals".

  Darij

Oh, sorry for that :D, but certainly one can imagine that it has to be circles instead of quadrilaterals..

A simple but related problem: suppose $ABCD$ is as described above (a cyclic quadrilateral with an inscribed circle). Prove that the area of the quadrilateral is given by
\[area=\sqrt{AB\cdot BC\cdot CD\cdot DA}.\]


1.  For which property is the value greater for a solution of a nonvolatile solute than for the pure solent.
A. Boiling point
B. Freezing point
C. Triple point
D. Vapor pressure

2. Which statement is true about a substance that is subjected to a lower external pressure at a constant temperature
A. A liquid will boil at a lower temperature
B. A liquid will exhibit a lower vapor pressure
C. A gas in an insulated container will change into a liquid
D. A gas in a nonrigit container will exhibit a smaller volume

3. A white solid dissolves in water to form a solution that does not conduct electricity.  What type of bonding is most likely present in the solid.
A. Ionic
B. Metallic
C. Network covalent
D. Polar covalent

4. The chemicals in a lightstick give off light as they react.  When the lightstick is placed in warm water the glow increases.  This is because the
A. activation energy for the process is lowered
B. average kinetic energy of the reactants increases
C. high temperature catalyzes the reaction
D. high temperature changes the wavelength of light emitted

5. Which pair of solutions forms a buffer solution when equal volumes of each are mixed
A. 0.2 M HCl and 0.20 M NaOH
B. 0.4 M HC2H3O2 and 0.20 M NaOH
C. 0.20 M HCl and 0.20 M NH3
D. 0.40 M HCl and 0.20 M NH3

6. The Ka of hydrocyanic acid, HCN, is 5.0x10^-10.  WHat is the pH of 0.50 M HCN (aq)
A. Between 3.5 and 4.5
B. Between 5.0 and 5.5
C. Between 9.0 and 9.5
D. Between 10.5 and 11.0

Thanks, and explanations would be nice

1. [b]A[/b] For a solution with a nonvolatile solute, vapor pressure decreases so boiling point becomes greater (by Raoult's law.)

2. [b]A[/b] A lower external pressure decreases atmospheric pressure and so more vapor forms, resulting in a lower boiling point.

3. [b]C[/b] The most common nonconductive type of bonding in solution there is polar covalent. The only other type of nonconductive bonding would be network covalent, but that's not as common. --> check on this. I'm not TOO sure.

4. [b]B[/b] Not much to say here. Increased temperature increases speed of reactants.

5. dont know on this one. I thought buffers only formed with weak acid and base pairs?

6.
HCN->H+ + CN-
[H+]=[CN-]=x, [HCN]=0.5-x
x^2/0.5=5.0*10^-10, x=1.58*10^-5, pH=4.8 ??? Not a solution :\

The correct answer for [b]3[/b] is D (for example, a sample of sugar). Network covalent solids, like diamond and graphite, are not water soluble.

[b]5[/b] The correct answer is B: for example, mixing one liter of each solution, 0.20 moles of NaHO will convert completely (since this is a strong base) an equal number of moles of acetic acid into acetate ion, and we end up with 0.20 moles of acetic acid and an equal number of moles of acetate ion.

[b]6[/b] This is a very weak acid (which rules out C and D), so B is very likely the correct pH range. We can confirm this by computing $ pH \equal{} \minus{} \log(\sqrt{0.5 \times 5.0 \times 10^{\minus{}10}}) \equal{} 4.8$. There is, therefore, some typo on the given data.


[quote="madness"]D?
i knw R is true
to contradict A take area = 4 and consider a rectangle of sides 4 and 1, P = 10
for circle $ P \equal{} 4 \sqrt {\pi}$ < 10  :D[/quote]

read the question properly

OMG
Assertion Reasoning (Framed by yours truly with some inputs online)

A: For a given fixed area, the smallest perimeter is that of a circle.
R: For a given fixed perimeter, the largest area is that of a circle

A) Both are True and the reason is correct explanation
B) Both are True and reason is not correct exp
C) A is true and R is wrong
D) A is wrong and R is true

Then the ans is B da  :D

I will post the answer as soon as some others try and give a proper reasoning.

dont worry ill post the reasoning today sooner or later  :D

r is $ \frac {a}{\sqrt{\pi}}$ da n not as you said  :huh:
dont worry, i think its B only  :)

Dei i copied the same ques again da  :D  Lol u thought that it was my own?
that was shre's n even i said B  :)

Sorry
prob 129 a parkave illa :rotfl:  :rotfl:  :rotfl:

dei fraud shre, i found this link http://www.cut-the-knot.org/do_you_know/isoperimetric.shtml
here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=194065  :P   :rotfl:   :D

I think you should think of the solutions by yourself than to search AOPS for the answer to all the questions and if indeed you already have seen the soln, then keep shut and let others try. Bad karma.

You can assume prove the assertion using the reason.

Assume that R is true and A is false.

Then for the same area, there exists another figure with lesser perimeter than a circle. So now, draw another circle with the same perimeter as the other figure. This 2nd circle has a lesser area than the figure which contradicts our R because for the same perimeter, circle has largest area.

Note that in this question, A and R are interchangable.

err i was reading thro sanjith's posts chumma to see his solns when i came across this as the first thing  :D so boy what's the ans accord to you  :maybe:

Isn't that obvious. One knows (by simple knowledge that both are true) and also using one, one can prove the other.


Harvard
college
MIT
Princeton
algorithm
Yale
pigeonhole principle

Hey all when this is me in 3 years:

National Honor Society, Mu Alpha Theta, Cum Laude Society
National Merit Scholar
2350 SAT, 35 ACT
4.0 GPA
Captain of Scholars Bowl Team, Science Bowl team, Science Olympiad, Math Team
#1 Singles player on Varsity Boys Tennis Team
Piano for 6 years
Few hundred hours of volunteering
10 or 11 APs, all 4+ on exam
Valedictorian

Will I be able to get into Harvard or MIT or any of those? If not, why?

Probably, but not necessarily. Why? That's a good start. But you might not get it, and you wouldn't get it if you fail to demonstrate on your essays that your numerical credentials are matched by your ability to think critically and be an interesting person.

Also, some of those might be completely worthless, depending on your school. One time when I was in high school I tutored a college student in physics. He was top of his class in high school, but he needed help in the intro physics class at a mediocre state school. Why? His high school just wasn't that great. So don't assume that admissions officers are dumb enough to view "valedictorian" or "captain of XXX" really highly without bothering to check how good the other people at your school probably were.

Personally, I think if you're at a school where all of those actually mean something, there's not a snowball's chance in h3ll that you'll actually manage to obtain every last one of those credentials.

[hide="More details"]Here's my personal opinion of each of those:
National Honor Society: worthless. At most schools you can be in this by doing almost nothing.

Mu Alpha Theta: almost worthless. Admissions officers will be able to figure out that you're good at math from more important criteria, probably. (Hopefully you will have some decent competition results to mention, and good AP and SAT math scores.)

Cum Laude Society: I'm not sure what this is, but I'm assuming it's similar to NHS, so the same comment applies.

National Merit: It's not that hard. Way more people get this than get into Ivies+MIT/Caltech/Stanford. It certainly helps, but it's not a buy-in.

2350 SAT / 35 ACT: Come on, you're so optimistic about everything else, why not assume you're going to get a perfect score? Anyway, all you really need to do is not bomb the standardized tests. If they have two students that are pretty much equivalent, I'll wager the [i]last[/i] thing they'd use to choose between the two of them would be a difference of 20 or 30 points in SAT score. By the way, lots of students with perfect standardized test scores get rejected. Why? Because they don't have anything else to show, since they spent too much time studying for standardized tests and not having any kind of life.

4.0 GPA: This really depends a lot on the grading system at your school. Based on my knowledge of typical American high school grading standards, this is probably almost worthless. My school gives a +1.0 weight to honors and AP classes, and sometimes the number of people who graduate summa cum laude (>4.0) is more than the top 10% of the class. And I went to a decent high school.

Captain of Scholars Bowl Team: Not sure what this is. If it's similar to quiz bowl or science bowl, then being the captain of the team isn't much different from being on the team, really. It's a position of slightly superior responsibility. It won't mean much unless your team also has good results to show for your efforts.

(Captain of?) Science Bowl team: see what I said above.

(Captain of?) Science Olympiad: again, this is a contest that a lot of people do, so it's not really going to mean much at a place like MIT unless you and your team actually do pretty well at it.

(Captain of?) Math Team: as above.

#1 singles player on varsity tennis team: Is your tennis team actually any good? It is very helpful that you play a varsity sport. But again, this isn't a shoe-in even if your team is pretty good, and if your team is a no-name in the tennis world then it's probably almost no different than if you say "I play tennis a lot."

Piano for 6 years: Can't hurt. Not a buy-in. I'd say about 80% of the people I know played an instrument when they were in high school. It's not an unusual thing. (Also, just so you know, piano is the most common instrument. If you want to get admitted to a school because of the instrument you play, play something weird like a bassoon or a bagpipe. And be good at it.)

Few hundred hours of volunteering: Probably good. Volunteering at what? If it's BS volunteering, like tutoring for extra credit after class or playing piano for your church/other religious institution/some social institution, then it's probably not really worth that much. If it's like Habitat for Humanity and stuff, serious volunteering, then that's pretty good. But again, no single thing can buy your way in.

10 or 11 APs, all 4+: Ho hum. Do 20 APs and that'll be pretty spectacular. Or get 5s on all 11 and that might draw a bit of attention. But it's not really that hard to get a 4 (or even a 5, in most cases) on an AP test, so unless you do it for a LOT of them, it's not going to make a big difference. If you took ones that you had to study on your own for (and did well), this might help somewhat because it shows that you're motivated to learn things on your own (and can actually succeed too).

Valedictorian: As I said before, depends entirely on your school. At my school that would be pretty good -- it's a class of 500+ and we tend to have 10+ National Merit finalists, so there's decent competition. At the high school across town, though, probably any of the top 20 people at my school could have been valedictorian if they transfered over there. Maybe more than that. And that high school wasn't even that bad.

Sorry if I sound overly critical. I just have the impression that at top schools, it's personality and motivation and passion that are the strongest traits, not numbers. If you put in too much effort to craft a perfect resume, admissions officers might realize this and doubt whether you have actual useful skills to back up what you've done. They don't want someone who appears to be a perfect student on paper, but is actually only good at making themselves look good on paper. They want people who actually are good at things, even if they aren't so great at making themselves look better than they really are.

For the record, I'm at Princeton, and here's how little I had of your list:

National Merit
Good SAT/SATII scores
High GPA
Captain of Science Bowl team; vice president or president or something of Math/Science Team. (Note: these teams had good results in state and national contests.)
Swimming (not varsity. no good times or competition results or anything.)
17 AP tests. One 3, the rest 4 or 5.

That's not even half of what you "plan" to have. I think I got in because I had good essays and recommendations, and because they thought that I was motivated and would take advantage of the opportunities available at Princeton. Not because my list of accomplishments got a high score in their "applicant rating equation". [/hide]

Do what you like, not what you think will get you into a good college. The first is more likely to get you into college than the second, and you'll enjoy it more too.

Xev left out the most important thing that I think people neglect to take into account: college admissions are a probabilistic process.  There is no grand deterministic algorithm applied by Harvard (or MIT, or Yale, or (nowadays) the University of Michigan).  For any statistic you like, each of these schools gets more applicants above the mean of its student body than it has spots available to freshmen.  (My personal favorite is version of this is that Harvard gets enough applicants that it could fill its freshman class entirely with 1600 SAT scores* (probably not 2400 in the new system, but I don't know) or entirely with high school valedictorians, if it so chose.)  As a result, "anything you do" is likely to enhance you chances of getting in and nothing you is likely to guarantee you a place at any given school.  The trick, of course, is that you don't have to apply to just a single school -- people who raise their likelyhood of getting into Harvard are probably also raising their likelyhood of getting into MIT, and Yale, and the University of Michigan.

* Apparently this is not (maybe "is no longer"?) true -- see Tokenadult's post that follows.  Having actual data is a wonderful thing.

Mathematically, based on the size of its entering class, Harvard could definitely NOT fill its class with students who have 2400s on the three-section SAT.

http://www.collegeboard.com/prod_downloads/highered/ra/sat/composite_CR_M_W_percentile_ranks.pdf

It couldn't even quite fill its class with students who score 1600 based on two-section consideration of math and critical reading scores.

http://www.collegeboard.com/prod_downloads/highered/ra/sat/SAT_composite_CR_M_percentile_ranks.pdf

But even though the pigeonhole principle "forces" Harvard to admit some students with less-than-perfect SAT I scores, what I find interesting is that Harvard (and every other college worthy of note in this regard) passes over some perfect scorers to admit some students with other desirable characteristics. The linear extension considered by admission offices of the partially ordered set of college applicants does not just lie along the dimension of test scores.

JBL has it right: you can do things to increase your probability of admission, but you can't do much of anything to guarantee admission. Apply widely to all of the colleges that are of interest to you. While you are still in high school, enjoy your studies and your activities and do your personal best in them. When you apply, be sure to mention anything unique about you that makes you a good match for each college. Then submit the applications, and enjoy high school while you wait for the response of each admission office.

Lol wow thanks for the truth Xevarion. Heres some extra info to clear up:

I say I'd have a close to perfect if not perfect standardized test scores because i scored a 30 on the ACT in 7th grade, 2026 on SAT in 8th grade, 31/32 PLAN in 9th, and 221 on PSAT in 9th. Look im not bragging or boasting or anything, im just telling you.

By 4.0 GPA I mean unweighted, if it was weighted, it would only be a 4.6 because thats the highest my school gives with APs (kinda stupid)

I can only take 10 or 11 APs because you can only start taking APs in 11th grade at my school (another stupid thing)

Google Cum Laude Society

I am in Habitat for Humanity, but im not doing so much right now because im only 14

I agree national honor societ and MAT are pretty worthless

Currently, our Science Olympiad team almost always goes to national, Scholars Bowl team finishes as finalists at state tourney, Science Bowl team almost always goes to nationals, Math Team the best in the state for our school size.

Our tennis team was runner up at state last year, i was #5 ( but ya ure probably right our team wont be that great when im #1)

We have a class size of 72, and we have approximately have 10 naitonal merit semifinalists every year(thats amazing) And my grade has always had the highest test scores and is expected to have almost 20% of the class be national merit semifinalists (with me being the smartest in our grade, and one of the only AZNs XD)

And of course, i perfectly know that statistics and numbers arent the most important, but they're up there.
I just want a heads-up

So what else did u do that made Princeton accept you?

Btw I live in Alabama and attend Randolph School

Sounds like you do go to a pretty good school. Did your school go to NSB in 2004 or 2005? I don't remember playing your school, but there were a lot of matches and it's been a while...

As JBL and tokenadult and I already said, chance is a part of it. I had good essays and good recommendations. It probably helped that my dad went to Princeton. I had some leadership kind of stuff (like I created a chess club). That's about it, I guess...

[quote="Xevarion"]at a mediocre state school.[/quote]
That's a harsh assessment, Mr. Princeton. :cool:

[quote="mlok"]That's a harsh assessment, Mr. Princeton. :cool:[/quote]
Sorry. :P You should take it only as a criticism of the intro physics class, which (in my opinion) was a bit horrific.

[quote="lifesapain"]2026 on SAT in 8th grade[/quote]

This is the most impressive statistic on your list. :wink:

Anyway, everything you have is good, but do you see something that if someone (an admissions officer, specifically) read they would be like "Wow, I really want this guy at my school." Once you do, I think you'll have a much better chance at getting into Harvard/MIT. But even if you don't, it's not that big of a deal. The point is that you do something unique that you enjoy and that's what sets you apart.

I'm going to echo what a couple of other people have said and tell you to do what interests you, not what you think will get you into college.  I'll add that you don't look like the kind of person who needs to worry about tests and other statistics - just don't slack off too much in your classes and you'll be fine in that regard.  :)  There are more important things to worry about!

As it's been said before in this thread (and tons more times in this forum) the top schools really like to admit students that have something on their application that stands out in some way.  From the people I know at MIT, here are a few things I've noted that probably attributed a lot to why they're here (the rest of their app is also solid, of course):

Research - If you can do some research in high school, that'll go a long way.  If you can get into RSI, then you're doing [i]really[/i] well (clearly, since they only accept the top of the top).  But doing research at a state school near you for a summer or so does a lot as well.  Some places have programs that pair up students with mentors.  If there's nothing like that around, try emailing a prof at a local university to see if you they'd be willing to mentor you.  I admit that it'd be hard to find a prof that'll accept a high school student, but if you were successful, just think about how impressed elite schools would be.

Starting a Business - Starting a business and getting it of the ground requires a lot of hard work and determination.  Colleges definitely want this type of person.

So that's just a few ideas.  There are other ways to stand out.  Be creative and good luck.

joml88 is right about research. There are also lots of research contests you can participate in; some of those are team contests, so you could do them with your friends. I think Toshiba has a contest, there are the famous Siemens and Intel competitions, and probably other corporate-sponsored ones too that I don't know about. You can look for more these kinds of opportunities yourself, of course. Find something that looks fun, and then do a really good job of it. That's the way to go! ;-)

[quote]If there's nothing like that around, try emailing a prof at a local university to see if you they'd be willing to mentor you. I admit that it'd be hard to find a prof that'll accept a high school student, but if you were successful, just think about how impressed elite schools would be.[/quote]

Several professors at my local college (not a slouchy one, either) mentored me while I was still a high school student.  It was one of the best things that ever happened to me.  The intellectual stimulation was wonderful, the experience I gained helped me immensely in college, and I'm still friends with some of the people I met (in fact, one of the professors still writes letters of recommendation for me).

I'd say, if you want to try to find a professor to work with, you need to show him or her that you already have an exceptionally good knowledge base and are willing to work hard, that you know what the professor does and are really interested in it (reading some of their papers really helps), and that you can't get any of this from your high school.  So, basically, you should show why you'd be rewarding to mentor and why you wouldn't be wasting their time.   :D

Oh yea definitely, dont' get me wrong i enjoy scholar's bowl and math team and such. I like academic competitions and competing with other students across the nation. those are my interests. And i keep my grades up because i always want to be #1 anyway, not just so it will look better for college. " i strive to be the best" -lol i know sounds corny

Anyway, im only taking the non-science and no-math APs purely for college ( i h8 english and social studies)

What can I do during the summer to help me out? The researching part?

Q: What's sillier than asking someone to handicapping your chances of getting into Harvard?

A: Asking someone to handicap the chances of your hypothetical future self.

Sorry, I don't mean to be so obnoxious.  I really do sympathize with today's students.  It's a ridiculous academic arms race out there, and it's only getting harder and harder to get in to these schools.

Here's my honest answer:  Sounds like a strong application, but it's no shoe-in.  Almost nothing is.  These top schools do not admit by the numbers.  Your raw info is more impressive than that of lots of people who get accepted and less impressive than that of lots of people who get rejected.  I've been doing interviews for Harvard for 5 years now, and I scarcely have a clue what Harvard admissions is doing.  You would be terrified to know about some of the students I've seen Harvard reject.  The fact is that Harvard admits a *class*.  It is not exactly a meritocracy, and it's not even really intended to be, as far as I can tell.

I don't believe in giving advice on how to get into college, because I don't believe I have any useful advice to offer.  I think that's what they talk about over at College Confidential.

okay from now on everyone can refer to this person as "Bob" and not me. This is random person bob's application and he needs help.

not too shoot down your hopes or anythingk, but you better do something much more impresive than just be cdaptain of your math team. i know randolph school, and its pretty mediocre at best when ti comes to math. kurt was leaps and bounds ahead of everyone else on that math team and he wasn't even that good just in teh spectrum of AL math students.

ummm seva? i didnt say i was "captain" of the math team. There is no captain and it wouldnt mean anything either. ya and unlike vestavia hills, which is only good at math (cuz of ure 24/7 practicing), randolph has other strengths too...

Let's not make dumb college threads, mmk?

I'm not usually one to defend Vestavia, but shoot. They aren't even that good in the spectrum of national mathematics. They are good as a product of practice. Just like you are good at tennis as a product of practice. I think most college admissions officers can spot a filler extra-curricular activity. NHS/MAO are pretty useless. Cum Laude Society? really. Can you quantify a few hundred hours of volunteering? (my suggestion for that is to have a set cause and work towards it.)

Also, you're banking a lot on things remaining the same. I understand that Randolph is decent at Scholars Bowl, to some extent, behind the beast that is Brindlee Mountain, but is your current class going to be that good in a few years? You could be an amazing quiz bowl player that a place like MIT would want. Eh.

I think focus is kind of important. There are thousands upon thousands of people who are exactly like you or better based on what you've said.

What makes you special in the eyes of the college admissions officer?

Thats harsh, but you know, there is some truth to it. You can bring out alot of that in your essays but it needs to show on your application as well.

Interesting information, i have similar credentials to "Bob" and i found this very helpful. "Be yourself" is the new catch phrase of college admisssions advisers everywhere yet no one seems to understand this simple approach.
 I personally have been doing things that interest me and have found that i do well in school because i like school, every part from the romantic language of shakespeare to the beautiful understanding that came with physics.
However, because this is such a long post i stayed up reading this rather my new john grisham book, The Appeal, therefore hindering me from my pursuit to do what i like, as many have suggested =(.
By the way I enjoyed the criticism put forth by the princeton guy, xev. It seems that too many high schools lack any real challenges and therefore instill an intellectual bravado in its myriad of 4.0 students.
I look forward to applying to Harvard next year and wish "bob" luck as well.


geometry
circumcircle
inradius
inequalities
inequalities proposed

[quote="Ligouras"]Let $ R$ and $ r$ be the circumradius and inradius, rispectively, of an triangle $ ABC$. Let $ h_a,h_b,h_c$ the heights, $ l_a,l_b,l_c$ the bisectors and $ m_a,m_b,m_c$ the medians of $ ABC$. Prove that

$ \frac {l_ah_a \minus{} 1}{l_a^2} \plus{} \frac {l_bh_b \minus{} 1}{l_b^2} \plus{} \frac {l_ch_c \minus{} 1}{l_c^2}\geq\frac {12r^3 \plus{} 3r^2R \minus{} 1}{3r^2R}$[/quote]

The inequality doesn't hold

$ a\equal{}3.464101616, b\equal{}3.334345103, c\equal{}4.748558666$

??????????????????

[quote="hedeng123"][quote="Ligouras"]Let $ R$ and $ r$ be the circumradius and inradius, rispectively, of an triangle $ ABC$. Let $ h_a,h_b,h_c$ the heights, $ l_a,l_b,l_c$ the bisectors and $ m_a,m_b,m_c$ the medians of $ ABC$. Prove that

$ \frac {l_ah_a \minus{} 1}{l_a^2} \plus{} \frac {l_bh_b \minus{} 1}{l_b^2} \plus{} \frac {l_ch_c \minus{} 1}{l_c^2}\geq\frac {12r^3 \plus{} 3r^2R \minus{} 1}{3r^2R}$[/quote]

The inequality doesn't hold

$ a \equal{} 3.464101616, b \equal{} 3.334345103, c \equal{} 4.748558666$

??????????????????[/quote]

Sorry my friends

$ \frac {l_ah_a \minus{} 1}{l_a^2} \plus{} \frac {l_bh_b \minus{} 1}{l_b^2} \plus{} \frac {l_ch_c \minus{} 1}{l_c^2}\geq\frac {12r^3 \plus{} 3r^2R \minus{} R}{3r^2R}$

[quote="Ligouras"]Let $ R$ and $ r$ be the circumradius and inradius, rispectively, of an triangle $ ABC$. Let $ h_a,h_b,h_c$ the heights, and $ l_a,l_b,l_c$ the bisectors of $ ABC$. Prove that

$ \frac {l_ah_a \minus{} 1}{l_a^2} \plus{} \frac {l_bh_b \minus{} 1}{l_b^2} \plus{} \frac {l_ch_c \minus{} 1}{l_c^2}\geq\frac {12r^3 \plus{} 3r^2R \minus{} R}{3r^2R}$[/quote]

The inequality holds
 :)


I'm not sure how to represent this in the format that AoPS uses, so I'll just go ahead.

Problem 10: Find the exact value of:

$ ((2 \sin (140)\sec (280)/\sec (220))\plus{}(\sec (340)/\csc (20)))/((\cot (200)\minus{}\tan (280))/\cot (200))$

Is there a way to make a big fraction sign? And why is there so much space around the + and - signs?

[quote="Jebus McAzn"]Is there a way to make a big fraction sign? And why is there so much space around the + and - signs?[/quote]  Try \frac{top}{bottom} to get $ \frac {top}{bottom}$.  The space is because LaTeX is struggling with the line break.  (You also won't need so many parentheses when you write it like this, which will be a lot easier to read.)

By the way, on AoPS there's a Test Forum where you can try out new LaTeX stuff, a LaTeX forum where you can ask questions (or search for answers -- many common questions have already been answered there), and you can click on other people's LaTeX images to see the code they've used.


geometry
geometric transformation
homothety
ratio
conics
parabola
geometry proposed

Given two points O,A. A point B moves on the line d passes O and perp to OA. Prove that the circle that touches d at B and passes through the midpoint of AB, always tangent to a fixed circle.

an horrible and simple solution:

let $ \overline{OB}\equal{}x$ and $ \overline{AO}\equal{}d$ and call D the center of the circle, R his radius and M the midpoint of AB. That center of the circle is the point where the perpendicolar axes of MB meet the perpendicolar line from B to d. so we have:

$ R\equal{}\frac{\sqrt{x^2\plus{}d^2}}{d} \cdot \frac{\sqrt{x^2\plus{}d^2}}{4}\equal{}\frac{x^2\plus{}d^2}{4d}$

With an homothety with center A and ratio $ \minus{}\frac{1}{4}$ O go on a point N, the we have:

$ \overline{ND}\equal{}\sqrt{\left ( \frac{x^2\plus{}d^2}{4d}\minus{}\frac{5d}{4} \right )^2 \plus{} x^2} \equal{} \sqrt{\frac{x^4\plus{}16d^4\minus{}8d^2x^2\plus{}16d^2x^2}{16d^2}}\equal{}\frac{x^2\plus{}4d^2}{4d}$

Then $ \overline{ND}\minus{}R\equal{}\frac{3d^2}{4d}\equal{}\frac{3}{4}d$ that is fixed, so is thangent to a circle with center N and radius $ \frac{3}{4}d$.

Line d is the x-axis. $ A \equal{} (0, a), B \equal{} (b, 0).$ Equation of the circle with radius $ r$ tangent to $ d$ at $ B$ is $ (x \minus{} b)^2 \plus{} y^2 \minus{} 2yr \equal{} 0.$ The circle passes through $ M \equal{} (b/2, a/2)$ $ \Longrightarrow$ $ r \equal{} \frac {b^2}{4a} \plus{} \frac {a}{4}.$ Center $ P \equal{} (b, r)$ is on parabola with vertex tangent $ y \equal{} \frac {a}{4}$ and focal distance $ a.$ The focus is $ F \equal{} (0, 5a/4)$ and directrix $ y \equal{} \minus{} \frac {3a}{4}.$ Circle concentric with $ (P)$ and tangent to the directrix passes through $ F.$ Its radius is $ R \equal{} r \plus{} \frac {3a}{4}$ $ \Longrightarrow$ $ (P)$ is tangent to circle $ (F)$ with radius $ \frac {3a}{4}.$


Hola amigos, mi 1er. post.

Les dejo el 3er. problema de Sangaku que en verdad no es difícil.

[img]http://img160.imageshack.us/img160/1531/sangaku3sq9.png[/img]

Se tienen 3 circunferencias de radios a, b y c como se muestra en la figura, todas tangentes entre sí y tangentes a una línea horizontal.

Pruebe que $\frac{1}{{\sqrt a }}+\frac{1}{{\sqrt b }}= \frac{1}{{\sqrt c }}$

Saludos.

Denote $L(R,\ r)$ by the length of the common external tangent line for two circles with radii $R,\ r$, by Pythagorian theorem we have that $(R+r)^{2}-(R-r)^{2}=\{L(R,\ r)\}^{2}\Longleftrightarrow L(R,\ r)=2\sqrt{Rr}.$
Since $L(c,\ a)+L(b,\ c)=L(a,\ b)\Longleftrightarrow 2\sqrt{ca}+2\sqrt{bc}=2\sqrt{ab}$
$\Longleftrightarrow \sqrt{c}(\sqrt{a}+\sqrt{b})=\sqrt{ab}$, yielding $\frac{1}{\sqrt{c}}=\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}.$ Q.E.D.

Expondré una demostración en español para que así sea todo más expedito.

Sean $A$, $B$ y $C$ los puntos de contacto (de izquierda a derecha) de las circunferencias con la respectiva línea tangente.

De acuerdo con la propiedad de la tangente común a dos circunferencias tangentes exteriormente, se tiene que:

\[\begin{equation*}\begin{aligned}\overline{AB}&= 2\sqrt{a \cdot c}\hfill \\ \overline{BC}&= 2\sqrt{b \cdot c}\hfill \\ \overline{AC}&= 2\sqrt{a \cdot b}\hfill \\ \end{aligned}\end{equation*}\]
Pero $\overline{AC}= \overline{AB}+\overline{BC}$ (suma de segmentos), y por tanto:

\[\begin{equation*}\begin{aligned}2\sqrt{a \cdot b}&= 2\sqrt{a \cdot c}+2\sqrt{b \cdot c}\hfill \\ \sqrt{ab}&= \sqrt{ac}+\sqrt{bc}\hfill \\ \frac{{\sqrt{ab}}}{{\sqrt{abc}}}&= \frac{{\sqrt{ac}}}{{\sqrt{abc}}}+\frac{{\sqrt{bc}}}{{\sqrt{abc}}}\hfill \\ \frac{1}{{\sqrt c }}&= \frac{1}{{\sqrt b }}+\frac{1}{{\sqrt a }}\hfill \\ \end{aligned}\end{equation*}\]

Este es el conocido Primer teorema de Mikami y Kobayashi. La formula extendida teniendo en cuenta la tangencia externa del círculo con radio $ c$  es la siguiente:
\[\left | \frac {1}{\sqrt {r_b}} \pm \frac {1}{\sqrt {r_a}} \right| \equal{} \frac {1}{\sqrt {r_c}}\]


linear algebra
matrix
modular arithmetic
Gauss
quadratics
algebra
polynomial

Let $ p$ be an odd prime. Let $ \zeta$ be a primitive $ p$th root of unity in $ \mathbb{C}.$ Let $ M$ be the $ p\times p$ complex matrix whose $ j,k$ element is $ \zeta^{jk}.$ (I personally would prefer to let the indices $ j,k$ run from $ 0$ to $ p \minus{} 1.$ That's not vital - any consistent notation will work.)

1. Show that the only possible eigenvalues of $ M$ are $ \pm\sqrt {p},\pm i\sqrt {p}.$

Given 1, let $ a,b,c,d$ be the algebraic multiplicities of $ \sqrt {p}, \minus{} \sqrt {p},i\sqrt {p}, \minus{} i\sqrt {p}$ respectively. Clearly $ a,b,c,d$ are nonnegative integers with $ a \plus{} b \plus{} c \plus{} d \equal{} p.$

2.a) If $ p\equiv 1\pmod{4},$ show that $ \text{trace}\,M$ is real. Conclude that $ c \equal{} d.$

b) If $ p\equiv 3\pmod{4},$ show that $ \text{trace}\,M$ is pure imaginary. Conclude that $ a \equal{} b.$

3. Does a subfield of $ \mathbb{C}$ that contains $ \zeta$ necessarily also contain $ \sqrt {p}\,?$

hint: Kent is a specialist in DFT.

10000 posts for him ; does t0rajir0u look at his rear-view mirror ?

I should add that computing the trace of $ M^2$ has its uses in this.

Here is a related (how?) topic : http://www.mathlinks.ro/Forum/viewtopic.php?t=32348

[hide="Index"]Gauss sums -- Discrete Fourier Transform[/hide]

I think the point of this topic is to come up with an alternate proof of some classic results.  Very interesting, Kent.

[hide="1"] With $ j, k$ running from $ 0$ to $ p \minus{} 1$, the orthogonality relations imply that $ M^2$ is a Toeplitz matrix with entry $ a_{jk} \equal{} p$ if $ j \plus{} k \equiv 0 \bmod p$ and $ 0$ otherwise.  The underlying permutation is a disjoint union of transpositions, so $ M^4 \equal{} p^2 I$.  In other words, the (usual normalization of) the DFT has period $ 4$. [/hide]
[hide="2"] If $ \minus{} 1$ is a quadratic residue $ \bmod p$, then $ \sum_{j \equal{} 0}^{p \minus{} 1} \zeta^{j^2}$ is equal to its conjugate, hence purely real.  Otherwise, $ \sum_{j \equal{} 0}^{p \minus{} 1} \zeta^{j^2} \plus{} \zeta^{ \minus{} j^2} \equal{} 2 \sum_{j \equal{} 0}^{p \minus{} 1} \zeta^j \equal{} 0$, hence it is purely real. [/hide]
[hide="3"] Certainly when $ p \equiv 1 \bmod 4$; by the above, $ \text{tr } M$ is an integer multiple of $ \sqrt {p}$.  On the other hand, when $ p \equiv 3 \bmod 4$, again by the above $ \text{tr } M$ is an integer multiple of $ i \sqrt {p}$.  If $ \sqrt {p}$ were also a polynomial in $ \zeta$, then $ i$ would have to be in the subfield.  I'm not sure how to get a contradiction from here, but it probably has something to do with the fact that the order of the Galois group isn't necessarily divisible by $ 4$. [/hide]

Have you been able to compute the magnitude or the sign of the Gauss sum using this method?  Computing $ \text{tr } M^2$ gives bounds, but it looks like you should be able to push this method the rest of the way.  What do the eigenspaces look like?

Edit:  [hide="Remarks"] Since $ \text{tr } M^2 \equal{} p$ (not hard to verify), we obtain $ a \plus{} b \minus{} c \minus{} d \equal{} 1$.  This gives $ a \plus{} b \equal{} \frac {p \plus{} 1}{2}, c \plus{} d \equal{} \frac {p \minus{} 1}{2}$.  We can also verify this by computing the determinant of $ M^2$.  The eigenspace corresponding to the eigenvalue $ p$ is spanned by $ e_0$ and $ e_{i} \plus{} e_{p \minus{} i}, 1 \le i \le \frac{p\minus{}1}{2}$, whereas the eigenspace corresponding to the eigenvalue $ \minus{} p$ is spanned by $ e_{i} \minus{} e_{p \minus{} i}$.  It's unclear how these eigenspaces split up when we take the square root.

Since $ M$ is a Vandermonde matrix, it has determinant

$ \sum_{0 \le i < j < p} (\zeta^i \minus{} \zeta^j) \equal{} \sqrt {p}^p ( \minus{} 1)^b i^{c \minus{} d}$

but I haven't thought of a particularly slick way of determining its sign.

Anyway, [url=http://en.wikipedia.org/wiki/Discrete_Fourier_transform#Eigenvalues_and_eigenvectors]the Wiki article[/url] has some interesting remarks. [/hide]

[quote="t0rajir0u"]I think the point of this topic is to come up with an alternate proof of some classic results.[/quote]
Sure.
Well, as a matter of fact, this is pretty much the content of the "purely algebraic solution" from that ENS exam mentioned by [b]harazi[/b] in the topic I posted in my last message.

The setup :
let $ n > 1$ be an odd integer and let $ \zeta \equal{} \exp \left( \frac {2 i \pi}n \right).$
Let $ V$ be the $ n\minus{}$dimensional $ \mathbb{C}\minus{}$ vector space of the functions from $ \mathbb{Z}/n\mathbb{Z}$ to $ \mathbb{C}.$
Let $ T$ be the endomorphism on $ V$ which maps the function $ f$ to the function $ T(f)$ defined by :
$ \begin{array}{cccc} T(f) : & \mathbb{Z}/n\mathbb{Z} & \to & \mathbb{C} \\
 & x & \mapsto & \sum_{y \in \mathbb{Z}/n\mathbb{Z}} f(y) \zeta^{xy} \end{array}$

Note that the sum from [b]alekk[/b]'s post is the trace of $ T.$
Of course, this is just a rephrasing of Kent's matrix vision.

As it was proved by [b]t0rajir0u[/b], $ a\plus{}b \equal{} \frac {n\plus{}1}2$ and $ c\plus{}d \equal{} \frac {n\minus{}1}2.$

Besides, the result from that other topic is : $ | \text{Tr}(T) | \equal{} \sqrt{n},$ so that we also get $ (a\minus{}b)^2 \plus{} (c\minus{}d)^2 \equal{} 1.$
Therefore either $ a\equal{}b$ and $ c \equal{} d \pm 1$ or $ c\equal{}d$ and $ a \equal{} b \pm 1.$

The next step is indeed, as suggested by [b]t0rajir0u[/b], to compute $ \det T$ in order to get $ a, b, c, d$ in terms of $ n.$
This is a somewhat tedious piece of computation, whose final result is $ \det T \equal{} \sqrt{n}^n i^{\frac {n(n\minus{}1)}2}.$
But since the determinant is the product of the eigenvalues (counted with multiplicity), we also have $ \det T \equal{} \sqrt{n}^n (\minus{}1)^{b\plus{}d} i^{c\plus{}d},$ so that $ (\minus{}1)^{b\plus{}d} i^{c\plus{}d} \equal{} i^{\frac {n(n\minus{}1)}2}.$
Then a bit of work yields that when $ n \equiv 1 \pmod 4,$ $ a \equal{} \frac {n\plus{}3}4$ and $ b \equal{} c \equal{} d \equal{} \frac {n\minus{}1}4,$ while when $ n \equiv 3 \pmod 4,$ $ a \equal{} b \equal{} c \equal{} \frac {n\plus{}1}4$ and $ d \equal{} \frac {n\minus{}3}4.$

Finally, since $ \text{Tr}(T) \equal{} \sqrt{n} ((a \minus{} b) \plus{} i (c \minus{} d)),$

$ \sum_{x \in \mathbb{Z}/n\mathbb{Z}} \zeta^{x^2} \equal{}
\begin{cases}
 \sqrt{n} & \text{if } n \equiv 1 \pmod 4 \\
 i\sqrt{n} & \text{if } n \equiv 3 \pmod 4
\end{cases}$

Amendment to (3): no, we don't have $ i$ in the subfield generated by $ \zeta.$ Hence instead of asking for $ \sqrt {p}$ to be in that subfield, we ask for exactly one of $ \sqrt {p}$ or $ \sqrt { \minus{} p}$ to be in that subfield.

I think now I'm interested in making the proof that $ |\text{Tr}(M)| \equal{} \sqrt {p}$ as slick as possible.


How many roots does the equation
\[ |\ldots\parallel{}|x|\minus{}1|\minus{}2|\ldots \minus{} 1995|\equal{}1
\]
have?

$ | | \cdots | | |x| \minus{}1 | \minus{}2 |\cdots| \minus{}n | \equal{} 1$
$ \Rightarrow x\equal{} \pm \left(\frac{n(n\plus{}1)}{2}\plus{}1\right), \pm \left(\frac{n(n\plus{}1)}{2}\minus{}1\right), \pm \left(\frac{n(n\plus{}1)}{2}\plus{}1\minus{}2n\right)$

The inverse of absolute values is $ \pm$, so for each absolute value we undo we end up with two times as many solutions.  Let $ f(n) \equal{} |\ldots\parallel{}|x| \minus{} 1| \minus{} 2|\ldots \minus{} n|$, so then we have $ f(1995) \equal{} 1 \implies f(1994) \equal{} 1995 \pm 1 \implies f(1993) \equal{} 1994 \pm (1995 \pm 1) \cdots$ so $ f(0) \equal{} |x| \equal{} 1\pm(2\pm(3\pm \cdots 1995 \pm 1))\cdots)$ has 1995 plus or minuses on the RHS and thus $ 2^{1995}$ possibilities so we have a total of $ 2^{1996}$ roots.

Only problem... are these roots all distinct?  I suspect they might be...

[quote="facis"]are these roots all distinct?  I suspect they might be...[/quote]
Verify that $ \parallel{}|x| \minus{} 1| \minus{} 2| \equal{} 1$ has five roots $ x \equal{} 4, 2, 0, \minus{} 2, \minus{} 4$.

You're solving for $ |x|$, not $ x$.  In solving for $ x$ you need to discard the negative values as well as the repeated ones.

oops... I think I ignored the fact that an absolute value cannot give a negative number.  So $ f(n) \ge 0$.  Trying again...

$ f(1995) \equal{} 1 \implies f(1994) \equal{} 1995 \pm 1 \equal{} {1993,1995} \implies f(1993) \equal{} {1994 \pm 1993, 1994 \pm 1995} \equal{} {1, 3987, \minus{}1, 3989}$ but $ f(1993)\equal{}\minus{}1$ is impossible, so we have $ f(1993) \equal{}{1, 3987, 3989} \implies f(1992) \equal{} {1993 \pm 1, 1993 \pm 3987, 1993 \pm 3989} \equal{} {1992, 1994, 5980, 5982}$ so we have four valid solutions here.  The pattern is to add 1 valid solution for every decrease of n by 1, because the smallest value $ f(n)$ can equal is always less than or equal to $ n$, giving two solutions, and the others are all larger than n, only giving a valid solution with the plus.

Wait, now we have $ f(1991) \equal{} {0, 3986, 7972, 7974} \implies f(1990) \equal{} {1991, 5977, 9963, 9965}$ so the pattern does not hold.  From here on, all values $ f(n)$ can take on are greater than $ n$, so the $ \minus{}$ case in $ f(n) \equal{} (n\plus{}1) \pm [f(n\plus{}1)]$ (where [f(n+1)] represents the values which f(n+1) can take on) cannot give positive solutions.  Thus, $ f(0)\equal{}|x|$ has 4 possible values for a total of $ 8$ solutions for $ x$.

But if you draw a graph of the equation, I believe we have only 6 roots  :maybe:

oops, I had a few numbers wrong (1995+1=1995? 1995-1=1993?) Anyway, corrected it should be:

$ f(1995) \equal{} 1 \implies f(1994) \equal{} 1995 \pm 1 \equal{} {1994,1996} \implies f(1993) \equal{} {1994 \pm 1994, 1994 \pm 1996} \equal{} {0, 3988, \minus{}2, 3990}$ But $ f(1993)\equal{}\minus{}2$ is extraneous so we now continue $ f(1992) \equal{} 1993 \pm 0, 1993 \pm 3988, 1993 \pm 3990 \equal{} 1993, 5981, 5983$.  $ 1993$ is repeated here, and now all values which $ f(n)$ can equal are greater than $ n$, so in $ f(n\minus{}1)\equal{}n\pm[f(n)]$ (where $ [f(n)]$ represents the possible values for $ f(n)$) only the plus sign gives positive values.  Thus, with each successive reduction of n, the possible values for f(n) only get greater and thus we can never use the minus case.  Hence, there are always 3 valid solutions for $ n \le 1992$.  So $ f(0)\equal{}|x|$ has 3 solutions, and the equation has plus or minus these solutions as roots, for a total of $ 6$ roots.


two nonoverlapping squares of sides a and b are placed inside a square of side 1.
prove that a+b<=1

[hide]if we place a squre with side a along the vertex then 'a' will occupied by the square on two sides
the remaining are area would be a polygon and to a maximum a square of side 1-a or b can placed because thats the length and width available for square, making it a+b=1
  if the first square isplaced without touching the sides, then the second squares length would be less than 1-a
so a+b=<1[/hide]


Let $ a,b,c\in[0;1]$. Prove that:
$ \frac{a}{b^3\plus{}c^3\plus{}7}\plus{}\frac{b}{c^3\plus{}a^3\plus{}7}\plus{}\frac{c}{a^3\plus{}b^3\plus{}7}\leq\frac13$

[quote="tdl"]Let $ a,b,c\in[0;1]$. Prove that:
$ \frac {a}{b^3 \plus{} c^3 \plus{} 7} \plus{} \frac {b}{c^3 \plus{} a^3 \plus{} 7} \plus{} \frac {c}{a^3 \plus{} b^3 \plus{} 7}\leq\frac13$[/quote]
[hide="Hidden by mod"]$ \sum_{cyc}\frac {a}{b^3 \plus{} c^3 \plus{} 7} \equal{} \sum_{cyc}\frac {a}{b^3 \plus{} 2 \plus{} c^3 \plus{} 2 \plus{} 3}\leq\sum_{cyc}\frac {a}{3b \plus{} 3c \plus{} 3a} \equal{} \frac {1}{3}.$[/hide]

[quote="tdl"]Nice solution! It's my solution, too! Now we consider a harder problem:
Find $ r$ max so that:
$ \frac {a}{b^r \plus{} c^r \plus{} 7} \plus{} \frac {b}{c^r \plus{} a^r \plus{} 7} \plus{} \frac {c}{a^r \plus{} b^r \plus{} 7}\leq\frac13 \ \forall a,b,c\in [0;1]$[/quote]

$ r\equal{}\frac{9}{2}$, but my method involves calculus.  Is there an elementary proof?

[quote="arqady"]
$ \sum_{cyc}\frac {a}{b^3 \plus{} c^3 \plus{} 7} \equal{} \sum_{cyc}\frac {a}{b^3 \plus{} 2 \plus{} c^3 \plus{} 2 \plus{} 3}\leq\sum_{cyc}\frac {a}{3b \plus{} 3c \plus{} 3a} \equal{} \frac {1}{3}.$[/quote]

I can't understand..
Tell me, please..

Oh, I forgot the conditions that $ a,b,c \in [0,1]$  :blush:  :blush:

[quote="JoeBlow"][quote="tdl"]Nice solution! It's my solution, too! Now we consider a harder problem:
Find $ r$ max so that:
$ \frac {a}{b^r \plus{} c^r \plus{} 7} \plus{} \frac {b}{c^r \plus{} a^r \plus{} 7} \plus{} \frac {c}{a^r \plus{} b^r \plus{} 7}\leq\frac13 \ \forall a,b,c\in [0;1]$[/quote]

$ r \equal{} \frac {9}{2}$, but my method involves calculus.  Is there an elementary proof?[/quote]I think we can use contradiction by assuming $ r>\frac92$. I haven't solved it yet, but I have found that $ \frac {a}{b^r \plus{} c^r \plus{} 7} \plus{} \frac {b}{c^r \plus{} a^r \plus{} 7} \plus{} \frac {c}{a^r \plus{} b^r \plus{} 7}<\frac32$, which is useless. :oops:

If $ r>\frac92$ then we consider case $ a\equal{}b\equal{}c$:
$ \frac{3a}{2a^r\plus{}7}\leq\frac13\Leftrightarrow 2a^r\minus{}9a\plus{}7\geq 0$
$ f(a)\equal{}2a^r\minus{}9a\plus{}7\Rightarrow f'(a)\equal{}2ra^{r\minus{}1}\minus{}9$
$ f'(a)\equal{}0\Leftrightarrow a\equal{}a_0\equal{}\Big(\frac{9}{2r}\Big)^{\frac{1}{r\minus{}1}}\in(0;1)$ with $ r>\frac92$ and then $ f(a_0)<f(1)\equal{}0$. Contradiction!


function
graph theory
combinatorics solved
combinatorics

Let n, m be integers such that m>n>= 1. Consider the set of n+m elements. Let X be the collection of all its subsets, which have exactly n elements and Y be collection of all its subsets, which have exactly n+1 elements. Show that there exists an injective (1 to 1) function f:X->Y such that every x is a subset of f(x).

This could be looked at from a more general point of view (I think):

Construct a bipartie graph with two sets of vertices $V_1,V_2$. The vertices in $V_1$ are the sets with $n$ elements, while those in $V_2$ are the sets with $n+1$ elements. Two vertices $x_i\in V_1,y_j\in V_2$ are connected iff $x_i\subset y_j$. All vertices in $V_1$ have the same degree, $m$, and all vertices in $V_2$ have the same degree, $n+1$. I think a more general result holds:

Given a bipartite graph with sets of vertices $V_1,V_2$ s.t. $|V_1|\le|V_2|$ and s.t. all vertices in $V_i$ have the same degree $d_i$ (we could have $d_1\ne d_2$, of course), then there is an injection $f:V_1\to V_2$ s.t. $x$ is connected with $f(x)$.

Well, do you have any idea how to show this?

Not really.. I would have posted the complete solution if I had an idea :).

this is just an application of Hall's theorem.
consider any collection of k sets,A_1,A_2...A_k, each of size n.we wish to prove that the set of subsets of the original set X(|X|=m+n) which contain atleast one of the A_i's has atleast k many elements in it.
for each of teh sets A_i, consider the list of all distinct sets of the form A_i U {x}.
there are m such choices for each i.On the other hand,any set in this complete listing can occur atmost n+1 times,since there are only n+1 choices for the element x,in such a representation.hence the number of distinct such sets in this entire collection is atleast km/(n+1)>=k(since m>n).
now use Hall's theorem and conclude that there is a system of distinct representatives for the n-sets of X.

I never heard about the Hall's theorem, but now I see that it is pretty common in the internet.

Thanks very much, seshadri!

And the grobber generalization as well follows nearly immediately from the theorem. We just must control the theorem condition, whereas the smaller part $V_1$ of the graph is a family of sets and another part $V_2$ is a collection of their elements.(as in the theorem). For every $a \in V_1$, $x \in V_2$ put that $x \in a$ iff $a$ is connected to $x$.

If all the vertices in $V_1$ have degree $d_1$, all the vertices in $V_2$ have degree $d_2$ and $|V_1|\leq|V_2|$ then $d_1\geq d_2$.

Consider union $A$ of $n$ sets in $V_1$.  Assume that $|A|=m$. Then $nd_1 \leq md_2$ because the vertices in $A$ have all the edges from $n$ vertices mentioned above and maybe some more. But then $n\leq m$ because $d_1\geq d_2$.


Eight singers participate in an art festival where m songs are performed.
Each song is performed by 4 singers, and each pair of singers
performs together in the same number of songs. Find the smallest
m for which this is possible.


[quote="othman"]solve in positive integers :

   $x^3+31=y^2$[/quote]
This equation reminds me this which i solved very easily:
$x^5+31=y^2$ which appeared on 1st international zhatykov mathematical olympiad
held in Kazakhstan.
Solution:$x^5+32=y^2+1$
$(x+2)T=y^2+1$
we know that for every natural number $n$ all prime divisors of $n^2+1$ are
of the form $4k+1$.proof comes immediately from Fermats theorem.
Back to equation,Looking mod 4 we get contradiction!
Maybe you can solve it like this methods.I will also think about this problem.

i dont think that your method can solve the problem but it a  good method anyway


Is it true that $\frac{1}{1-ab}+\frac{1}{1-bc}+\frac{1}{1-ca}\ge 3$ if $a,b,c$ are reals, such that $a^{2}+b^{2}+c^{2}=1$

@aviateurpilot : I think u've made a typo.
That must be $a^{2}+b^{2}+c^{2}$.
And if $a = b = c = \frac{1}{\sqrt3}$, $\sum_{cyc}\frac{1}{1-ab}= \frac{9}{2}> 3$....

@Beat : Counter Example : $a = 0, b = \frac{3}{5}, c =-\frac{4}{5}$

The real problem I know is
$\sum_{cyc}\frac{1}{1-ab}\geq \frac{9}{2}$ for $a,b,c \in \mathbb{R^{+}}$

I'm sorry Beat if I've made a mistake in my last post
But if $a=0, b=\frac{3}{5}, c=-\frac{4}{5}$, we have $\sum_{cyc}\frac{1}{1-ab}= \frac{99}{37}< 3$, and your ineq. is wrong


Let $a,b,c\in\mathbb{R}$\{1,2,3}. Solve the system of equation:
$\{\begin{array}{c}x\frac{x}{a-1}+\frac{y}{a-2}+\frac{z}{a-3}=1\ \frac{x}{b-1}+\frac{y}{b-2}+\frac{z}{b-3}=1\ \frac{x}{c-1}+\frac{y}{c-2}+\frac{z}{c-3}=1\end{array}$

Must have $(a-b)(b-c)(c-a)\not = 0$, then $a,b,c$ are all roots of an equation has form $t^{3}+A(x,y,z)t^{2}+B(x,y,z)t+C(x,y,z)=0$. Now use Viette's theorem!

I'm sory for my stupid, but really i don't what do you mean? Please explain it more clearly for me? Thanks

Now, i understand, thanks for help me to solve it. The first i think it very complicated, but you really have killed it :huh:

What is your solution? Sorry , because I see you post it in Algebra Proposed & Own Problems, I think you know at least one solution :)

Okay, this is my solution

Let $f(X)=\frac{x}{X-1}+\frac{y}{X-2}+\frac{z}{X-3}-1$ then $f(X)$ is rewriten as:
$f(X)=\frac{g(X)}{(X-1)(X-2)(X-3)}$ where $g(X)$ is a third degree polynomial with variable $X$ has a first coeeficient equal $-1$.
From assumtion of problem, we have $f(a)=f(b)=f(c)=0\Rightarrow g(a)=g(b)=g(c)=0$ then $g(X)$ is writen as: $g(X)=-(X-a)(X-b)(X-c)$.
Therefore $f(X)=\frac{(X-a)(X-b)(X-c)}{(X-1)(X-2)(X-3)}$ (1)
Multiplying two hands of (1) with $X-1$, we get:
$x+(X-1)\left(\frac{y}{X-2}+\frac{z}{X-3}-1\right)=-\frac{(X-a)(X-b)(X-c)}{(X-2)(X-3)}$ (2)
Put $X=1$ into (2), we has $x=-\frac{(1-a)(1-b)(1-c)}{2}$
Analogously, we get $y=(2-a)(2-b)(2-c)$ and $z=-\frac{(3-a)(3-b)(3-c)}{2}$

Done :)

[quote="hien"]
Put $X=1$ into (2), we has $x=-\frac{(1-a)(1-b)(1-c)}{2}$
[/quote]
Nice!  :wink: But I think this must is $X\to 1$.


Considering $(mod4)$ or otherwise , prove that if $a,b,c,d$ are integers satisfying
                   $(3a+5b)(7b+11c)(13c+17d)(19d+23a)=2001^2001$
then $a$ must be even

Use the fact that the sum of the all  parentheses  is divisible by 4.


I am entering my first year as a MATHCOUNTS coach and wanted some pointers to help build a successful program.  How should I keep track of student progress?  Should I soley pic students based on School test or how many questions they get correct in practice.  Should I teach or allow the students to work and help each other.  Thanks for any help that you can give me.

It is usually a good idea to do more than just give tests because then the students may get bored and what you really need are motivated students.
I would suggest teaching a formula and seeing how much they study it and whether they can apply it by giving them a test the class after you teach it to them.

The most important thing to keep in mind, is that mathcounts should be enjoyable. The students should mostly want to be there. Bring snacks. Get to know the students individually! As for teaching - the process is more important than the final result. Regarding level, it will help to have an idea of the school's mathcounts history - how they did at chapters, states, etc. Then you can base the difficulty of the current problems based on past performance. Do a variety of problems, AMC 8's and 10's are useful also. What my team emphasized, and what helped us the most: after the problems, we'd go over the ones we got wrong - and make sure we understood the idea behind the solution. Have the students be teachers to each other - they can share insight and practice communicating their solutions. As you develop your coaching method, just remember to be dedicated and show the students you care. Good luck!

What keeps a program thriving isn't just making a community and trying to maintain it, it's to make a math [i]culture[/i]. Students usually drop MathCounts when they enter the 9th grade. What you should do is have them help coach. You may have to play teacher for the first year, but then some people will have moved on to high school, and they can help teach with the knowledge you taught then a year or so ago.

That's what happened with my team.

P.S. Pick your group solely on what they get in the competition. How they do in practice doesn't affect on how they would do under pressure. Last year, I flunked all of the practices, but then get 2nd chapter and state. Just because someone gets perfect on practices doesn't mean that they get perfect in the competition. Some schools take the AMC 8 into matter as well, but my school only takes the School Competition into matter.

We did 5 tests this year to pick the team...

And I thought kids have to drop mathcounts in 9th grade, in terms of competing

I'm in same situation as you!

I'm currently a senior in high school and I have been involved in mathematics since 7th grade. Using my experience and qualification to state (first one in my school), I volunteered to be an assistant coach and I go there about once a week to coach the kids who are in gifted (we're a small school).

I basically make lessons and focus on topics that I feel are necessary. I try to make them think that Mathcounts is not another dreadful school work, and also try to add some humor to my words (like I said today, try to do these 10 problems instead of watching America's Funniest Videos  :rotfl:  :rotfl: ). I use actual Mathcounts problems than those workouts they make because to me, they're bit unrealistic to actual competitions.

As with checking scores, I don't do that. I have a website which I maintain, and post up the solutions (not problems since they're copyrighted). I initially planned to choose a captain based on the score but since not every kid is in the meeting, I decide to throw that away and choose captain later.

The most important thing is to tell them that don't get pressured and just have fun. I keep reminding how important that this is for them because this is like, you do Mathcounts, and rest of your school math becomes piece of cake. That comes from experience, and I did get 100 through all my math classes in high school except one midterm, which I got 99. So, yeah, you can be pressuring coach but if this is the first year, the most important point is to get kids to like you and then go from here.

Best Wishes,

Silverfalcon

P.S. Feel free to contact me and talk about coaching.  :)

[quote="jboesrelpehant"]I would suggest teaching a formula and seeing how much they study it and whether they can apply it by giving them a test the class after you teach it to them.[/quote]

I [u]strongly[/u] disagree with that, as may many others. First of all, you seem to be suggesting that one should be able to memorize a formula , then use it in a routine situation. That is NOT the point of MathCounts, and problem solving in general.

Emphasis should be placed on understanding (complete rigor isn't totally necessary) [i]why[/i] things are true, not 'learning' as many things as possible. It seems counter-intuitive that this is helpful at all, but in the long run, it allows students to become better thinkers, which will ultimately boost them.

As a mathcounts participant, i've seen lots of 6th graders join and then quit rite away cuz it was "boring."  U can't force any1 to join, but if they're good at math and/or like it, it's good to keep them entertained.  Try teaching some info every week, combinatorics, triangles, fast multiplication, divisibility rules, etc.  My school does a countdown every friday which adds some fun and competition.  Towards chapter time we start giving tests, and choosing the team.


For what value of k is the number:

$ {\sqrt(2+\sqrt3)}*{\sqrt(2+(\sqrt(2+(\sqrt3))))}*{\sqrt(2+(\sqrt(2+(\sqrt(2+(\sqrt3))))))}*{\sqrt(2-(\sqrt(2+(\sqrt2+(\sqrt3)))))}$

the solution for the equation: $ 3x^{4}+5x+2k = 0$?

im sorry, i couldn't understand ur solutions given..how many solutions are there? does that period after one solution mean the next set of numbers is another solutions?

btw, the LaTeX code for sqrt is \sqrt.
so \sqrt{2}+\sqrt{3}$ \Longrightarrow\sqrt{2}\plus{}\sqrt{3}$ that might help!

I've tried to bring in the $ /sqrt\&$ symbol in the question, though I'm not sure if that helps. What I'm trying to convey is that there is a square root within another. and the entire expression equals 'k'

LaTeX help:
\sqrt{blah} gives $ \sqrt{blah}$.  (In general, use curly brackets to delimit arguments of expressions in LaTeX, e.g. \frac{numerator}{denominator} or \binom{n}{r}.  Also, \cdot gives $ \cdot$, which looks much nicer for multiplication than $ *$.


George has an unfair six-sided die. The probability that it rolls a 6 is $ \frac{1}{2}$, and the probability that it rolls any other number is $ \frac{1}{10}$. What is the expected value of the number shown when this die is rolled? Express your answer as a decimal.

maybe the question maker should add that the rest of the numbers are 1, 2, 3, 4, 5

It is assumed the other numbers are 1, 2, 3, 4, and 5 since that are the numbers on a die.
@redsky: What answer did you get? Then we can help you on what you are confused on.

The EV is always the sum of the probability of getting the numbers

So to get a 1 it is: $ \frac{1}{10} \times 1 \equal{} \frac{1}{10}$

So to get a 2 it is: $ \frac{1}{10} \times 2\equal{} \frac{1}{5}$

So to get a 3 it is: $ \frac{1}{10} \times 3\equal{} \frac{3}{10}$

So to get a 4 it is: $ \frac{1}{10} \times 4\equal{} \frac{2}{5}$

So to get a 5 it is: $ \frac{1}{10} \times 5\equal{} \frac{1}{2}$

So to get a 6 it is: $ \frac{1}{2} \times 6\equal{} 3$

So the sum is $ \frac{1}{10} \plus{} \frac{1}{5} \plus{} \frac{3}{10} \plus{} \frac{2}{5} \plus{} \frac{1}{2} \plus{} 3 \equal{} \boxed{4.5}$

I thought you had to add $ \frac{1}{10}\cdot5$ and $ \frac{1}{2}$ together...

Wow this was all the way back in 2009! Anyway: [hide]I just think it's simpler to use the distributive property when adding the fractions, yielding our answer to be 1/10(1 + 2 + 3 + 4 + 5) + 3, instead of adding the fractions separately, so 15/10 + 3, and we have our answer 4.5.


pigeonhole principle
number theory
relatively prime

1)Five points are situated inside an equilateral triangle whose side has length one unit. Show
that two of them may be chosen which are less than one half unit apart. What if the equilateral
triangle is replaced by a square whose side has length of one unit?

2) Given any n+2 integers, show that there exist two of them whose sum, or else whose difference,
is divisible by 2n?

3) Given any n + 1 distinct integers between 1 and 2n, show that two of them are relatively
prime. Is this result best possible, i.e., is the conclusion still true for n integers between 1
and 2n?

4) Prove that there is some integer power of 2 that begins 2006 · · ·

5) Given any 2n integers, show that there are n of them whose sum is divisible by n. (Though
superficially similar to some other pigeonhole problems, this problem is much more difficult
and does not really involve the pigeonhole principle.)

3) Pair them like (1, 2), (3, 4), ..., (2n - 1, 2n). There are $ n$ pairs, so in choosing $ n \plus{} 1$ elements, by the Pigeonhole Principle, there must be one pair with both elements chosen, and since they are 1 apart, they are relatively prime.

If we choose $ n$ integers, we could choose 2, 4, 6, ... $ 2n$, and no two of them would be relatively prime, as all are divisible by 2.

1) Connect the medians of this triangle to get 4 smaller congruent equilateral triangles. Let these be your "holes". Routine from here.

2) If any two of the integers are congruent $ \bmod 2n$ then take their difference. We are done. Otherwise, if all of the integers are different $ \bmod 2n$, pair the residues $ \bmod 2n$ like so: $ (0,0), (1,2n\minus{}1), (2,2n\minus{}2), \ldots , (n, n)$. There are $ n\plus{}1$ pairs and $ n\plus{}2$ different residues $ \bmod 2n$, thus one of the pairs has integers corresponding to both of its residues. Each pair's sum is $ 0 \bmod 2n$.


Hi, im new to USAMTS.

Can you write your own program to help solve a problem?

I read in the USAMTS FAQ that: "Computer solutions should be accompanied by an outline of the program as well as a printout of the outcome."  Does the "outline of the program" mean the program's source code or just pseudocode?

Thanks.

Yes you definitely can, although it usually makes your solution inelegant and sometimes destroy the whole point of the problem.

Usually I write a program to check my answer tho, that always helps :D

The rules [url=http://www.usamts.org/About/U_AbRules.php]here[/url] also state that you must prove that your program correctly solves the problem.


When a positive three-digit divident with a units digit of 2 is divided by a positive one-digit divisor, the result is a whole-number quotient with a remainder of 1. How many distinct values are possible for this three-digit divident?

because it ends in two and has a remainder of one, one less than it (ending in 1) is divisible by the divisor. This means the divisor must be either 3 or 7, because no others would fit the criteria. Then you go through the 3 digit mulitiples of 3 and 7 finding all the dividends.

[quote="Hamster1800"]Then you go through the 3 digit mulitiples of 3 and 7 finding all the dividends.[/quote]

What you left is exactly what I like to see. Any body has a way to solve the problem in 2 minutes? This is a Mathcounts Sprint problem.

[quote="tarquin"]It's a bit unclear ...
Does it mean divided by any postiive 1-digit integer? Or do we go case by case for each 1-digit integer individually? The former doesn't yield a three-digit answer.[/quote]

This is a question of 2003 National Sprint Round.

2 words: divisibility tricks. For the dividend to meet the criteria, the number xy1 where x and y are digits must be divisible by either 3 or 7. Divisibility trick for 3 is easy, add up the number so:

x+y can be 2, 5, 8,11,14, or 17

this gives 111, 201, 141, 231, 321,411, 501, 171,261,351,441,531,621,711,801 291, 381,471,561,651,741,831,921,591,681,771,861,951,891, 981.

The divisibility trick for 7 is take away double the last digit from the remaining digits, so 10x+y-2 is divisible by 7.

2 digit multiples of 7:
14
21
28
35
42
49
56
63
70
77
84
91
98

This gives 161, 231, 301, 371, 441, 511, 581, 651, 721, 791, 861, 931

Thus the set of all 3 digit dividends is the union of these two sets which is (i forgot to add one to each number so i will here)

[b]112, 142, 162, 172, 202, 232, 262, 292, 302, 322, 352
372,382,412,442,472,502, 512, 532, 562, 582,592,
622,652,682,712,722,742, 772, 792,802,
832,862,892,922,932,952,982[/b]

the list is complete

by the way, that's 38 values

[quote="Hamster1800"][hide]2 words: divisibility tricks. For the dividend to meet the criteria, the number xy1 where x and y are digits must be divisible by either 3 or 7. Divisibility trick for 3 is easy, add up the number so:

x+y can be 2, 5, 8,11,14, or 17

this gives 111, 201, 141, 231, 321,411, 501, 171,261,351,441,531,621,711,801 291, 381,471,561,651,741,831,921,591,681,771,861,951,891, 981.

The divisibility trick for 7 is take away double the last digit from the remaining digits, so 10x+y is divisible by 7.

2 digit multiples of 7:
14
21
28
35
42
49
56
63
70
77
84
91
98

This gives 161, 231, 301, 371, 441, 511, 581, 651, 721, 791, 861, 931

Thus the set of all 3 digit dividends is the union of these two sets which is (i forgot to add one to each number so i will here)

[b]112, 142, 172, 202, 212, 232, 262, 282, 292, 322,
352,382,412,422,442,472,492,502,532,562,592,
622,632,652,682,702,712,742,772,802,
832,842,862,892,912,922,952,982[/b]

the list is complete

by the way, that's 38 values[/hide][/quote]

Which is what I did, except you did the 7s, too  :D

Hamster1800 and chess64 got the correct answer. However,  their way will not work out with the time limit.

Does anyboy have a faster way doing this?

well yes. of course :D

the first value on my list for 3s is
112
add three to the first two digits to get the remaining ones.
142
172
202
232
262
292
322
352
382
412
442
472
502
532
562
592
622
652
682
712
742
772
802
832
862
892
922
952
982

The first one for 7s is 162 adding 7 to the first 2 digits givs us:
162
232
302
372
442
512
582
652
722
792
862
932

There are 12 values for 7 and 30 for 3 minus the 4 that are in both giving a total of $\boxed{38 \text{ values}}$
 :jump:  :jump:  :jump:  :jump:  :jump:

faster way:

100 - 10 = 90 (two digit numbers)

90/3 = 30
90/7 = 12 (take off fraction)
90/21 = 4 (take off fraction)

so the answer = 30 + 12 - 4 = $\boxed{38}$
:jump::jump:

I agree with both his answer and his solution.


integration
calculus
derivative
calculus computations

In my calc book, it uses the differential equation $y'=x-y$, passing through the point $(0,1)$ as an example in the section on Euler's Method, and it says you can verify the exact particular solution to be $y=x-1+2e^{-x}$.  I wanted to see how they got that solution, but I can't figure it out.  Rewriting it, you get $dy+(y-x)dx=0$, but $f(x,y)=1$ is homogeneous of degree 0, while $f(x,y)=y-x$ is homogeneous of degree 1.  This means that you can't use the substitution $y=vx$ to rewrite it in a form that allows separation of variables, and I can't come up with any other way to do it...

This is a first order linear differential equation, and there's a systematic method, the method of multiplying factors.

Write the equation in the standard form

$y'+p(x)y=q(x).$

The let $\mu(x)=e^{\int p(x)\,dx}.$ Multiply both sides of the equation by $\mu(x),$ identify the left side as the derivative of a product, and proceed from there. So let's do that:

$y'+y=x$

$\mu(x)=e^{\int 1\,dx}=e^{x}.$

$e^{x}y'+e^{x}y=xe^{x}$

$\left(e^{x}y\right)'=xe^{x}$

$e^{x}y=\int xe^{x}\,dx=x{}^{e}{}^{x}-e^{x}+C.$

Let's put in the initial condition $x=0,y=1$ right here:

$1\cdot 1=0\cdot 1-1+C$ so $C=2.$

$e^{x}y=xe^{x}-e^{x}+2$

$y=x-1+2e^{-x}.$

[Redundant derivation deleted]

Methods like Kent's post are taught in a differential equations course and appear in differential equations textbooks. I assume you saw this in a calculus textbook, which only mentions separation.
For the problem, you don't really need to know how the solution was obtained; it can be verified by simple differentiation.

The multiplying factor method for first order linear equations is often in calculus books. For instance, I know there's a section on it in Stewart. But a "short version" of the textbook used in a high school AP course may have omitted that chapter.

It's in chapter 9; separation is early in the chapter and integrating factors are at the end. The class I've been teaching stops in the middle.

The book we're using is Larson/Hostetler/Edwards (and yes, this is for a high-school AP course, although I'm working ahead of the class...)  Anyway, thanks for the explanation, Kent.

Btw, I just found out my book has that explanation in the next section, lol.

How about this problem?

Solve the differential equation $xy'-y=x^{2}e^{x}.$

One note: the "standard form" I mentioned has $y'$ with a coefficient of $1.$ So (in teaching that calculus class) the first thing I would do with kunny's problem is divide both sides of the equation by $x.$

(The point $x=0$ is a "singular point" - we'll later on call it a "regular singular point" - for this equation. We shouldn't be too surprised if some of our solutions behave badly as $x\to 0.$)

Dr. Kent Merryfield, you are right.

We can rewrite the differential equation as $\left(\frac{y}{x}\right)'=e^{x}$ for $x\neq 0.$


function
combinatorics proposed
combinatorics

Let $F$ be the family of all $k$- element subsets of the set ${\{1, 2, ..., 2k+1}$. Prove that there exists a bijective function $f: F\to F$ such that for every $A\in F$, the sets $A$ and $f(A)$ are disjoint

Same idea http://www.mathlinks.ro/Forum/viewtopic.php?p=839454#839454  :wink:


[quote="olimpic cocuk"]other than the limits of Cauchy equations have a way Display?[/quote]

Sorry, but could you explain a bit your question, please ?

[quote="olimpic cocuk"]other than the limits of Cauchy equations have a way Display?[/quote]

what does that mean exactly??

i think about Cauchy equation  :(  ? $ f(x) \equal{}ax$


Five years ago I spent a lot of time( many weeks) trying to proof this beautiful theorem

[quote]On an arbitrary continuous closed convex curve you can always fix four points as the vertex of a square.[/quote]

using the magic 1858 Holditch's Theorem (sorry if I have exagerate defining it "magic" but it is one of my favourite theorem).

Do you think is it possible?

But doesn't Holditch's theorem require the extra crucal condition that the curve be convex?

Are you sure that it is a theorem and not an open question?
As far as I know, the result has been proved by Stromquist that :
- Each smooth simple closed curve in $R^d$ admits an incribed quadrilateral with equal sides and diagonals.
- Each locally monotone Jordan curve in $R^2$ admits an inscribed square.

Pierre.

I have read the proof, but it is very hard so  I think a proof for convex closed curves using Holditch could be a good winning post.

But I don't know if it is possible?

My open question is : "Is it possible to prove the closed convex curve theorem using Holditch ? "

P.S. You are perfecly right, Grobber. I missed to say the curve must be convex. Sorry, very much. :blush: I will correct my post.

Is there any web source for the Russian mathematician's proof?  I only read English and hope to find its English version. Could anyone help please? Thanks a lot.

I do not want to give a new proof but wish to use the proof.


I have three questions....Help on any of them is highly appreciated

1.
Consider each of the following sets $S$. Describe what $S$ is. Is $S$ open? Find $int(S)$. Is $S$ cloased? Briefly justify each answer.

(a) $S = \cap_{n=1}^{\infty} (-1, \frac{1}{n})$ in $\mathbb{R}$

(b) $S = [{(x,y) \in \mathbb{R}^2 : x + y \geq 0]}$

2.
Let $A$ and $B$ be open sibsets of $\mathbb{R}$. Prove $ A \times B =  [{(x,y) \in \mathbb{R}^2 : x \in A}$ and ${y \in B}]$ is an open subset of $\mathbb{R}^2$.

3.
Let $S = [ (x,y) \in \mathbb{R}^2 \mid xy > 1 ]$ and $B = [ d((x,y),(0,0) \mid (x,y) \in S ]$. Find $inf(B)$.

Thanks again. I think this is in the right section...if not i'm sorry please move it....

With such questions, you have a better chance for getting an answer in Calculus Computations and Tutorials, so I'm moving it there. Let me answer #1(a) to give it a start.

$S=(-1,0]$ (clearly, this interval is a subset of every interval in the intersection and, for any $x>0$, there exists $n$ such that $1/n<x$, so $x\notin S$).
$S$ is not open because no neighborhood of $0\in S$ is contained in $S$. It is not closed because its accumulation point $-1$ does not belong to $S$. The interior of $S$ is the open interval $(-1,0)$.


Wanda, Darren, Beatrice, and Chi are tutors in the school math lab. Their schedule is as follows: Darren works every third school day, Wanda works every fourth school day, Beatrice works every sixth school day, and Chi works every seventh school day. Today they are all working in the math lab. In how many school days from today will they next be together tutoring in the lab?

$ \textbf{(A) }42\qquad\textbf{(B) }84\qquad\textbf{(C) }126\qquad\textbf{(D) }178\qquad\textbf{(E) }252$

Sorry, I was posting these fast for the resource section.  I didn't read it, just glanced at it, and figured it was PIE.

[hide]Here we are looking for the lowest common mulitple of 3, 4, 6, and 7, which is $\boxed{84}$[/hide]

[hide]$LCM(3,4,6,7)=84$[/hide]

Shouldn't this be in classroom math?

[quote="Mike Tang"]too easy..[/quote]
dude, wtf, why did you awaked this thread?

[quote="13375P34K43V312"]Um four solutions to the same problem, which are exactly the same; please don't post solutions that have already been posted unless yours has a step that is significantly different or more elegant.[/quote]
The whole point of posting problems is to let others have fun solving them; if only one person does it, what's the point in that?

I'm not saying don't do it, I'm saying don't clog up the thread by typing up your solution.

By all means do the problem

but if your solution has already been posted it doesn't benefit anyone

if you feel the compulsive need to type lcm(blah)=blahblah, do it in microsoft word.

I agree..  unless you have an alternate solution don't clog up the thread.

All right, I won't in the future. I just see 97% of the forum doing it and so I conclude it's what to do. Back on topic, please.

[quote="Temperal"][quote="Art of Owna"][quote="rd5493"]Wait, if it says school days, would that make a difference or am I just thinking too hard?[/quote]
but the question also asks how many [i]school[/i] days from not, so you dont need to account for weekends.[/quote]

It doesn't specify what day of the week it is. Therefore, the "school" part must have no effect, or the problem would not have enough info to solve it.  :idea:[/quote]

I dont follow, today must be a school day because they are all working..so it doesnt really matter what day it is.

[quote="Art of Owna"][/quote][quote="Temperal"][quote="Art of Owna"][quote="rd5493"]Wait, if it says school days, would that make a difference or am I just thinking too hard?[/quote]
but the question also asks how many [i]school[/i] days from not, so you dont need to account for weekends.[/quote]

It doesn't specify what day of the week it is. Therefore, the "school" part must have no effect, or the problem would not have enough info to solve it.  :idea:[/quote][quote="Art of Owna"]

I dont follow, today must be a school day because they are all working..so it doesnt really matter what day it is.[/quote]
But if weekends are not counted, as they must if 'school days' is a factor, then we don't know when/how many weekends will occur in the time period.

Just erase the word "school" in the problem (since they are all the same) and there is no confusion hopefully;;

Wanda, Darren, Beatrice, and Chi are tutors in the school math lab. Their schedule is as follows: Darren works every third [color=yellow]school[/color] day, Wanda works every fourth [color=yellow]school[/color] day, Beatrice works every sixth [color=yellow]school[/color] day, and Chi works every seventh [color=yellow]school [/color]day. Today they are all working in the math lab. In how many [color=yellow]school[/color] days from today will they next be together tutoring in the lab?

$\text{(A) }42\qquad\text{(B) }84\qquad\text{(C) }126\qquad\text{(D) }178\qquad\text{(E) }252$[/quote]

[quote="Altheman"][hide]it just might be lcm(3,4,6,7)=84[/hide][/quote]

[hide="response"]It just is lcm(3,4,6,7)=84[/hide]

Rather hilariously (in a sad way), despite the fact that there have been so many repeatedly posts answering the problem, and so many add-on complaints, no one was given the letter to answer the question, which indeed should be part of the solution shown. So here it is. $ \boxed{\text{B}}$.


probability
number theory
relatively prime

Many states use a sequence of three letters followed by a sequence of three digits as their standard license-plate pattern.  Given that each three-letter three-digit arrangement is equally likely, the probability that such a license plate will contain at least one palindrome (a three-letter arrangement or a three-digit arrangement that reads the same left-to-right as it does right-to-left) is $m/n$, where $m$ and $n$ are relatively prime positive integers.  Find $m+n$.

[hide="Solution"]
P(at least one palindrome)=1-P(no palindromes)=$1-\left(\frac{25}{26}\right)\left(\frac{9}{10}\right)=\frac{7}{52}.$  The answer is $7+52 = \boxed {59}$.
[/hide]

Sorry for bumping an old topic, but I don't get how chess got (25/26)(9/10)

the only way you will have a palindrom is that the last letter (or digit), is the same as the first one.

[quote=JesusFreak197][hide="Answer"]For a letter palindrome, we have $26$ possibilities for the first letter and $26$ for the second, so $676$. For the number palindrome, we have $10$ possibiities for the first number and $10$ for the second, so $100$. Now, we have $676\cdot 10^3$ possibilities with a letter palindrome and $100\cdot 26^3$ possibilities with a number palindrome, but each of these include number and letter palindromes, respectively, when we counted all of those separately. So, we subtract the number of number and letter palindromes from one of the two, since the overlap is equal, and we have $26^2\cdot 10^3$ and $10^2\cdot 26^2\cdot 25$. Therefore, we have $\frac{26^2(10^2)(25+10)}{26^3(10^3)}=\frac{35}{260}=\frac{7}{52}$, and $7+52=\boxed{59}$.[/hide]

[b]EDIT:[/b] Oops, I subtracted the overlap twice. That was dumb. :roll:[/quote]

I don't understand what the overlap is. Could you give an example please?


linear algebra
matrix
linear algebra unsolved

Give an example of a real inner-product space V and T an operator of V and real numbers a,b with $ a^2<4b$ such that $ T^2\plus{}aT\plus{}bI$ is not invertible.

$ \begin{bmatrix}0&1\\\minus{}b&\minus{}a\end{bmatrix}$

You're trying to make this way too complicated.

$ V \equal{} \mathbb{R}^2,a \equal{} 0,b \equal{} 1,$ and $ T \equal{} \begin{bmatrix}0 & \minus{} 1 \\
1 & 0\end{bmatrix}.$

One thing to note: any such $ T$ cannot be self-adjoint.

Added in edit: OK, my example is essentially a special case of jmerry's example.

\[ \begin{pmatrix}
0& b/a\\
\minus{}a & \minus{}a
\end{pmatrix}
\]


geometry
circumcircle
radical axis
perpendicular bisector
power of a point
geometry proposed

Given triangle $ ABC$ with circumcircle $ (O). P$ lies inside triangle $ ABC, AP, BP, CP$ intersect $ BC, CA, AB$ at $ A', B', C'$ respectively such that $ A'B'C'$ is pedal triangle. $ B'C'\cap (O)\equal{}\{A_1, A_2\}$, similar for $ B_1, B_2, C_1, C_2. P'$ is isogonal conjugate of $ P$ wrt $ ABC$. Prove that the radical center of three circles $ (A'A_1A_2), (B'B_1B_2), (C'C_1C_2)$ lies on $ OP'$

What does  "isogonal conjugate " mean? :(

I think that the problem is true since it states that $ \bigtriangleup A'B'C,$ as the [b][size=100]Cevian triangle[/size][/b] of $ P$, is also the  [b][size=100]Pedal triangle[/size][/b] of another point so be it $ Q,$ which ( I see by the drawing ) lies also on the line segment $ OP'.$

An interested problem here is how to construct in general, points with the property as the problem states ( such that their [b][size=100]Cevian triangle[/size][/b], to be also the [b][size=100]Pedal triangle[/size][/b] of another point ).

The only known to me points with this property, in the configuration of a given triangle, are the [b][size=100]Orthocenter[/size][/b] and the [b][size=100]Centroid[/size][/b] and the [b][size=100]Gegrgone point[/size][/b].

Also, for every point which has the property as the proposed problem states, its [b][size=100]Isotomic point[/size][/b] with respect to the given triangle, has also the same property ( easy to prove by [b][size=100]Ceva theorem[/size][/b] and [b][size=100]Carnot theorem[/size][/b] ).

Kostas Vittas.

OK!, or if P=G(centroid), then $ \triangle A^\prime B^\prime C^\prime$ -is the pedal triangle of O.

[quote="vittasko"]An interested problem here is how to construct in general, points with the property as the problem states ( such that their [b][size=100]Cevian triangle[/size][/b], to be also the [b][size=100]Pedal triangle[/size][/b] of another point.[/quote]
Dear Kostas, the locus of the points that satisfy that their pedal triangle is also a cevian triangle is known as the Darboux cubic of the triangle ABC.

[quote="mihai miculita"]$ \mbox{If R radical center, then: }R\notin{OP'. } \mbox{Something 's wrong, view figure...}$[/quote]
Dear mihai miculita,
I think in your figure $ A'B'C'$ is not pedal triangle!
Two special cases of this problem are at here:
P is [b]Gergone[/b] point: http://www.mathlinks.ro/viewtopic.php?t=253661
P is [b]centroid[/b]: http://www.mathlinks.ro/viewtopic.php?t=297152

Thank you dear [b][size=100]Luis[/size][/b] for the reference about the [b][size=100]Darboux cubic[/size][/b].  :ewpu:

So, there are some well known points outwardly of a given triangle and their [b][size=100]Isotomic points[/size][/b], with the property as we want. But, is there an elementary way to construct points of this cubic?

Sometimes, although a locus is an unusual curve, however we can define points belong to it, by elementary way.

Best regards, Kostas Vittas.

Let $ T \equiv B'C'\cap BC$ and let $M_a$ be the midpoint of $BC.$ From  $(B,C,A',T)=-1,$ it follows that $ TA_1 \cdot TA_2 \equal{} TB \cdot TC \equal{} TM_a \cdot TA'$ $\Longrightarrow$ $ M_a$ lies on $ \odot(A'A_1A_a).$ Similarly $ M_b,M_c$ lie on the circles $ \odot(B'B_1B_2)$ and $ \odot(C'C_1C_2).$ Center $ O_a$ of  $ \odot(A'A_1A_2)$ is  the intersection of the perpendicular bisector of $ A'M_a$ with the perpendicular to $ A_1A_2$ from $ O.$ Thus, $ O_a$ is the midpoint of $ OA'',$ where $ A''$ is the intersection of the perpendicular to $ A_1A_2$ from $ O$ with the perpendicular to $ BC$ at $ A'.$ Cyclically, $ O_b,O_c$ are midpoints of $ OB'',OC''$ $\Longrightarrow$ $ \triangle O_aO_bO_c$ and $ \triangle A''B''C''$ are centrally similar with factor 2.

Radical axis $ \mathcal{L}_c$ of $ (O_a),(O_b)$ goes through $ C'$ and is orthogonal to $O_aO_b \parallel A''B''.$ Similarly, $ \mathcal{L}_a, \mathcal{L}_b$ go through $ A',B'$ orthogonally to $ B''C''$ and $ A''C''.$ If $ Q$ is the common point of the perpendiculars to $ BC,CA,AB$ at $ A',B',C',$ then $ \triangle A'B'C'$ and $ \triangle A''B''C''$ are perspective with perspector $Q$ and orthologic, one orthology center is $ O$ and the other is the radical center of $ (O_a),(O_b),(O_c)$ $\Longrightarrow$ $ R \equiv \mathcal{L}_a \cap \mathcal{L}_b \cap \mathcal{L}_c.$ By Sondat's theorem $ O,Q,R$ are collinear. $\blacksquare$

P.S. It remains to prove the following lemma:

[b]Lemma:[/b] $ P$ is a point on the plane of $ \triangle ABC.$  $ \triangle A'B'C'$ is the cevian triangle of $ P$ such that $ \triangle A'B'C'$ is pedal triangle of $ Q.$ Then  the isogonal conjugate $ P'$ of $ P$ lies on $ OQ,$ where $ O$ is the circumcenter of $ \triangle  ABC.$

[quote][b]Lemma:[/b] $ P$ is a point on the plane of $ \triangle ABC.$  $ \triangle A'B'C'$ is the cevian triangle of $ P$ such that $ \triangle A'B'C'$ is pedal triangle of $ Q.$  Then the isogonal conjugate $ P'$ of $ P$ lies on $ OQ,$ where $ O$ is the circumcenter of $ \triangle  ABC.$[/quote]
See the proof of this supporting lemma as a remark in the topic [url=http://www.artofproblemsolving.com/viewtopic.php?p=1717214#1717214]Concurrent 2 (PC point again)[/url]. Hence the proof of the proposed problem above is finally completed.


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