geometry
trigonometry
ratio
geometric transformation
reflection
homothety
angle bisector

[b](a)[/b] Let the incircle of a triangle ABC have center I and touch its sides BC, CA, AB at the points D, E, F. Let the interior angle bisector of the angle BIC meet the line BC at a point M, and let the interior angle bisector of the angle FDE meet the line EF at a point P. Prove that the points A, M and P are collinear.

[b](b)[/b] Let H be the orthocenter and X, Y, Z the feet of the altitudes of a triangle ABC. Let the H-symmedian of triangle BHC meet the line BC at a point Q, and let the X-symmedian of triangle XYZ meet the line YZ at a point R. Prove that the points A, Q and R are collinear.

Part [b](a)[/b] is a rephrase of problem 3 of the 1st Mediterranean Mathematical Olympiad 1998 (see Crux Mathematicorum 8/29 pages 503-504); part [b](b)[/b] is a variation of mine. Note that the problems are by no means difficult - I just find them pretty neat.

[b]EDIT:[/b] Thanks to Stergiu for the following reference for problem [b](a)[/b]: GM 1991 - Nr. 3, page 111.

  Darij

[quote="darij grinberg"][b](a)[/b] Let the incircle of a triangle ABC have center I and touch its sides BC, CA, AB at the points D, E, F. Let the interior angle bisector of the angle BIC meet the line BC at a point M, and let the interior angle bisector of the angle FDE meet the line EF at a point P. Prove that the points A, M and P are collinear.

  Darij[/quote]

I didn't find at all this problem easy. For this reason here follows a cumbersome proof.

Let's call i the incircle of ABC. Let J the point where DP cuts i. The tangent t' through J to i is parallel to FE. Let G and H be the points where FE and t cut the line t"=AB and finally let's call p the line AH and call q the line HP. It is a well-known fact that the quadruple {t';t";p;q} is an harmonic quadruple. Then, if L=p^FE, it holds that PG=GL. Since t'//FE, the triangle PGD is similar to JHD then PG=DG. Therefore <PDL = 90°.
As J is the midpoint of arch FE the antipode N of J wrt i lies on line LD. By easy angle chasing one ca see that IM//DP and then NM is tangent to i.
From duality, since we have that FE (wich is the polar of A wrt i), ND (wich is the polar of M wrt i)  and p (wich is the polar of P wrt i) concurr at L, then A, M and P are collinear points.

PS
I guess that there should be a simpler proof.

[quote]I guess that there should be a simpler proof.[/quote]
Yes, there is.

Let $\alpha$, $\beta$, $\gamma$ be the angles of the triangle ABC ($\alpha=\angle{BAC}$, $\beta=\angle{ABC}$ and $\gamma=\angle{ACB}$).

Since $AC \cdot CB \cdot \sin{\gamma} = 2S_{\triangle{ABC}} = AB \cdot BC \cdot \sin{\beta}$ we have $\frac{\sin{\alpha}}{\sin{\beta}}=\frac{AB}{AC}$.

Clearly, $ED=2EC\sin{\gamma/2}=2CI\sin{\gamma/2}\cos{\gamma/2}=CI\sin{\gamma}$. Similarly, $FD=BI\sin{\beta}$.

By the two last observations we have $\frac{ED}{FD} \cdot \frac{AC}{AB}=\frac{CI}{BI}$.

Since DP is a bisector of $\angle{EDF}$, we have $\frac{EP}{FP}=\frac{ED}{FD}$. Similarly, $\frac{CM}{BM}=\frac{CI}{BI}$.

And the previous equality now looks like $\frac{EP}{FP} \cdot \frac{AC}{AB}=\frac{CM}{BM}$.

Observe that $\frac{\sin{\angle{PAE}}}{\sin{\angle{PAF}}}=\frac{EP}{FP} \cdot \frac{AF}{AE}=\frac{EP}{FP}$ and $\frac{\sin{\angle{MAC}}}{\sin{\angle{MAB}}}=\frac{CM}{BM} \cdot \frac{AB}{AC}$. By the previous we have that the right sides of the last two equalities are the same. Therefore, the left sides are the same. So, $\frac{\sin{\angle{PAE}}}{\sin{\angle{PAF}}}=\frac{\sin{\angle{MAC}}}{\sin{\angle{MAB}}}$. And now it is easy to see that $\angle{PAE}=\angle{MAC}$ and the points A, M and P are collinear.

[quote="Remike"][b]darij grinberg[/b], wouldn't you be so kind to post the idea of solution of part (b)?[/quote]

I wonder why you write this - it's exactly the same idea as you use for part [b](a)[/b]! Just note that the symmedian from a vertex of a triangle divides the opposite side in the ratio of the squares of the adjacent sides. (This is Theorem 5 in http://www.mathlinks.ro/Forum/viewtopic.php?t=58872 .)

It is possible (although not sure) that when I will find time, I will post a note with my proofs of both parts on my website (or, at least, here on ML). I am also still thinking about a simplification of Sprmnt21's proof.

  Darij

[quote="darij grinberg"][quote="Remike"][b]darij grinberg[/b], wouldn't you be so kind to post the idea of solution of part (b)?[/quote]

I wonder why you write this - it's exactly the same idea as you use for part [b](a)[/b]! Just note that the symmedian from a vertex of a triangle divides the opposite side in the ratio of the squares of the adjacent sides.[/quote]
I didn't know this fact. Thank you.

Darij could you post your proof  ?Or could someone find a solution without trigo .My solution is as  Remike's one ...

We have $\angle IMC=180-\angle MIC-\angle ICB=180-(90+\angle A/2)/2-\angle C/2$$=(90-\angle A/2)/2+(90-\angle C/2)$
$=\angle FIE/4+\angle EDC=\angle EDP+\angle EDC=\angle PDC$
so $IM//DP$ $(*)$
$DP$ meets circle $(I)$ at $K$, $IA$ meets $EF$ at $N$.
Easily $A,K,N,I$ are collinear
We have $\angle MID=\angle IDK=\angle IKD$ and $\angle KNP=\angle IDM=90$ so triangle $KNP$ and $IDM$ are similar
So $KP/IM=KN/ID=KN/IK$ $(1)$
we aslo have $IN*IA=IF^{2}=IK^{2}$ so $1-IN/IK=1-IK/IA$ or $KN/IN=AK/IA$ $(2)$
From (1) and (2) we have $KP/IM=AK/IA$ and with (*) so by Thales theorem we have $A,P,M$ are collinear $q.e.d$

Huyền Vũ  very nice solution , congratulations .It is quite elementary and simple .Thank you for sharing this with us. :lol:

I'll give a solution to b):

Since a reflection on the angle bisector of $\angle BAC$ + homotety brings $AB_{1}C_{1}$ in $ABC$, and by Steiner's theorem, we have just to prove that:
$\frac{BQ}{QC}\frac{B_{1}R}{RC_{1}}= \frac{c^{2}}{b^{2}}$
But from the properties of the symmedian, this equality becomes:
$\frac{BH^{2}}{CH^{2}}\frac{B_{1}A_{1}^{2}}{C_{1}A_{1}^{2}}= \frac{c^{2}}{b^{2}}$
$\frac{BH}{CH}\frac{B_{1}A_{1}}{c}= \frac{C_{1}A_{1}}{b}$
But, by similitude of $ABC, A_{1}B_{1}C, A_{1}BC_{1}$ we get:
$\frac{B_{1}A_{1}}{c}= \frac{CH}{2R}$
$\frac{C_{1}A_{1}}{b}= \frac{BH}{2R}$

Substituting, we get a big $\mbox{QED}$.  :)

Using Huyền Vũ's diagram i present a slightly different solution
We easily get $A,K,N,I$ collinear and also $KD\parallel IM$
So now we have $\angle{ DIM}=\angle {IDK}=\angle {DKI}$
Let $KI$ cut the incircle at a point $Q$
then $\angle {DIQ}=2\angle{DKI}\Leftrightarrow \angle{DIM}=\angle{MIQ}$
Now since $MD$ is tangent to the incircle we will also have $MQ$ tangent to the incircle $\Rightarrow MQ\perp IK \Rightarrow MQ\parallel FE $
Hence we conclude that the triangles $\bigtriangleup NPK$ and $\bigtriangleup IMQ$ have all their sides respectively parallel
Therefore $MP,IK,QN$ intersect in the honothetic center of the two triangles , so $A,M,P$ collinear

My solution for [b]part (a)[/b] :

Let $ T=AI \cap BC, S=AI \cap EF $ .

Since
$ A(B,C;T,M)=(B,C;T,M)=\frac{BT}{CT} \cdot \frac{CM}{BM}=\frac{AB}{AC} \cdot \frac{IC}{IB} $

$ =\frac{IC \cdot \sin \angle BCA }{IB \cdot \sin \angle ABC}=\frac{DE}{DF}=\frac{EP}{FP}=(F,E;S,P)=A(F,E;S,P) $ ,

so we get $ A, P, M $ are collinear .

Q.E.D

Dear Mathlinkers,

1. P, Q the points of intersection of AI with (I) (P between A and I)
2. R the point of intersection of AI and EF
3. the tangent to (I) at Q goes through M
4. the triangles MQI and NRP are homothetic
5. considering (P, Q, R, A) harmonic, we conclude that A is the homothetic center
and we are done…

Sincerely
Jean-Louis


logarithms
limit
calculus
derivative
LaTeX
calculus computations

[quote="kunny"]Let $f(x)>0,f'(x)>0$

\[ \dfrac{f''(x)}{f'(x)}=\ln f(x) +1,f'(0)=f(0)=e  \]

Find

\[ \lim_{n\to\infty} n\ln \left [\ln \left\{\ln f\left(1+\dfrac{f(1)}{n}\right)\right\}\right]  \][/quote]

The road is long and rather treacherous, so please do not be surprised if I trip; and in the event that I do, feel free to intervene and assume control of the reins.

\[ \dfrac{f''(x)}{f'(x)} = \ln f(x) + 1, f'(0) = f(0) = e  \]

\[ p = \dfrac{d(f'(x))}{dx}  \]

\[ \dfrac{d^2f}{dx^2} = \dfrac{dp}{dx} = \dfrac{dp}{df}\dfrac{df}{dx} = p \dfrac{dp}{df}  \]

\[ p\dfrac{dp}{df} = p\ln f + p  \]
\[ \dfrac{dp}{df} = \ln f + 1  \], \[ p(e) = e  \]

\[ p = \dfrac{df}{dx} = f\ln f  \]
\[ f(x) = e^{e^x}  \]

After applying L'Hopital's rule and treading through some tedious differentiation using the chain rule, we obtain:

\[ \lim_{n\to\infty} n \ln \ln \ln (e^{e^{1+\dfrac{e^e}{n}}}) = e^{e}  \]

P.S. Thank you, kunny, for creating the fine example of Latex that I used to learn the format of Latex used by this message board.

Thank you,mathisfun1.Your answer is corret.

kunny


function
real analysis
real analysis unsolved

[quote="F-o-u-r-i-e-r"]Find: $ \displaystyle\max_{\substack{f\in C(0, 1) \\
\|f\|_{1} \equal{} 1}}\|f\|_2$.[/quote]

Hint: Suppose first the continuity assumption is dropped. What would the answer be?

[quote="F-o-u-r-i-e-r"]Thanks for your reply,
If $ f$ is not continuous then $ \|f\|_2$ can be infinity.  :huh:[/quote]

Right, now take such a gadget, like $ 1/(2\sqrt x)$, and consider modifications of it.

Thanks! You are right! :)

What about the following one:

Find: $ \displaystyle\min_{f\in C(0, 1)}\frac {e^{\|f\|_1}}{\|f\|_2^2}$

Just use the same arguments as in your previous question to construct a function whith "small" $ L_1$ norm and "big" $ L_2$ norm :wink:
And observe that in this question it is better to write $ inf$ instead of $ min$.

Ah :blush: ,
thanks!

It is really easy and I have a solution. Yes, I have to write of course $ inf$. I was thinking about another problem.............. :blush:

I am really confused.
What does C(0,1) mean?

Assuming that it means a function continuous on the interval [0,1], the answer is infinite because a triangle of height h, base 2/h and 0 everywhere else has a l1 norm of 1 and a l2 norm that grows without bound as h is increased.
Since that makes the problem trivial, I assume that C(0,1) means something else.

That's completely correct, 0714446459923. That is what $ C(0,1)$ (or $ C([0,1]),$ or $ C[0,1]$ ) means. As long as you know how to produce and manipulate examples, which you seem to know how to do, there's nothing too difficult about this.


List all formulas for pi that you know.  I know 2:
$\sqrt{\sum_{x=1}^{\infty}{\frac{6}{x^2}}}$

and

$\lim_{n\to{\infty}}{\frac{n\sin{\frac{360}{n}}}{2}}$

Any more?

please don't post topics that are already up.
this one has been posted like three times, during the past couple of weeks
take a look here:(that one got locked...)
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39730

I think someone has proved in this forum my first equation, but I don't know about the others

My bad, I didn't see the other topics


superior algebra
superior algebra unsolved

Suppose that the action of $G$ on $X$ is transitive. Prove that at least $\frac{|G|}{|X|}$ elements of $G$ do not fix any element of $X$.

Let $S_{x}$ be the stabilisator of $x\in X$, then $|X|=|G: S_{x}|$ because the action is transitive.
No $S_{x}$ can't be the whole group, because x can be moved by some g in G (by the way: $|X|$ has to be greater than 1. otherwise this is obviously false).

$\bigcup_{x\in X}S_{x}=\bigcup_{g\in G}gS_{y}g^{-1}$ for some fixed y. There are $|G: N_{G}(S_{y})|\leq|G: S_{y}|$ conjugates of $S_{y}$ and they have at least one element in common, namely the neutral element.
$\Rightarrow  \left|\bigcup_{x\in X}S_{x}\right|\leq |G: S_{y}|\cdot(|S_{y}|-1)=|G|-|G: S_{y}|=|G|-|X|$

Hence there are $\left|G\backslash\bigcup_{x\in X}S_{x}\right|\geq|X|$ elements that don't lie in any stabilisator and therefore move all elements.

[quote="Gockel"]
Hence there are $\left|G\backslash\bigcup_{x\in X}S_{x}\right|\geq|X|$ elements that don't lie in any stabilisator and therefore move all elements.[/quote]

And why is that $\geq \frac{|G|}{|X|}$ ?


Given an acute triangle XYZ, prove that the perimeter of the inscribed triangle X'Y'Z' connecting the intersections of the 3 altitudes of XYZ with opposite sides is less than twice any altitude.

Thnx a mil.


calculus
integration
trigonometry
calculus computations

Find $ \int_0^\frac {\pi}{2} \frac {\sin^n x}{\sin^n x \plus{} \cos^n x} dx$

I can't figure out how to do this.  My book suggested $ u \equal{} \pi/2 \minus{} 1$ for the case where $ n \equal{} 1$, which worked out nicely.  $ n \equal{} 2$ is obvious since the denominator disappears, but I haven't be able to figure out the general case.  I think the substitution that really does the job is $ u \equal{} \tan x/2$: I ended up with a rational function.  But I was unable to find a general partial fraction decomposition.  Please help!

Integral is equal to $ \frac{\pi}{4}$ for all $ n$. Look at [url=http://www.mathlinks.ro/viewtopic.php?t=260035]this[/url].

nice.  Thanks.  I didn't expect that to be so simple.


+Construct a line l which passes through $ T_a$ and perpendicular to $ AE_a$. This line contains B and C.
+Construct a ray Ax such that $ \angle xAT_a \equal{} \angle E_aAT_a$.
+Construct a line which passes through $ E_a$ and parallel to Ax, it cuts l at M.
+Construct a line which passes through M and perpendicular to l, it cuts Ax at O.
+Construct $ (O, OA)$. It cuts l at B and C.
We are done.


LaTeX
inequalities
Olimpiada de matematicas

i guess something may be wrong, because equality is achieved also for triangle with angles 90, 90, 0... and you should be more careful when posting the solutions, this is the spanish community...


Show that in a tournament of 799 teams(every team plays with every other for a win or a loss), there exists 14 teams such that the first seven have each defeated the remaining 7 teams.


Find $ n$, such that the sum of digits of $ n^2$ is equal to 2008.

An answer** in exponential form is

n = $ 10^{223} \minus{} 9.$

See this pattern:

91^2 = 8281 = (100 - 9)^2 = 10,000 - 1,800 + 81

991^2 = 982081 = (1,000 - 9)^2 = 1,000,000 - 18,000 + 81

9991^2 = 99820081 = (10,000 - 9)^2 = 100,000,000 - 180,000 + 81

This pattern continues; the number of nines and the number
of zeroes increases by one apiece in the square.

The digits to the right of the nines add up as follows:
8 + 2 + 0 + ... + 0 + 8 + 1 = 19

Subtract:  2008 - 19 = 1989

Divide:  1989/9 = 221

There must be 221 9s in the square.

Note:

 (10^2 - 9)^2 has zero 9s.
 (10^3 - 9)^2 has one 9.
 (10^4 - 9)^2 has two 9s.

And so forth.

The number of nines in the square of n
is two less than the exponent of 10 as
I typed it here.  Or, turned around,
because I need to have the exponent
be two greater than the number of
nines in the square of n, I must ADD
2 to 221 to get 223.

n = $ 10^{223} \minus{} 9.$

** (I do not know if there are other answers.)


[quote="worthawholebean"]([b]4[/b]) Determine all real numbers $ a$ such that the inequality $ |x^2 \plus{} 2ax \plus{} 3a| \leq 2$ has exactly one solution in $ x$.[/quote]

[hide]If there is one solution then it must exist at $ 2$ otherwise there would be a point where the expression crosses 2, giving us more solution.

Therefore $ x^2 \plus{} 2ax \plus{} 3a \ge 2$ or $ \minus{} x^2 \minus{} 2ax \minus{} 3a \leq \minus{}2$

Considering the first case
$ (x \plus{} a)^2 \minus{} a^2 \plus{} 3a \geq 2$ The minimum occurs when $ x \equal{} a$. We need to chose $ a$ therefore such that  $ \minus{} a^2 \plus{} 3a \equal{} 2 \Rightarrow a \equal{} 2,a \equal{} 1$

Consider the second case

$ \minus{}[(x\plus{}a)^2\minus{}a^2\plus{}3a]\leq \minus{}2 \Rightarrow (x\plus{}a)^2\minus{}a^2\plus{}3a \geq 2$ Which is the first case? Too tired to work out whats going on here[/hide]

This is identical to HMMT 2008 - Algebra #3, and has already been thoroughly discussed -- [b][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1053168]click here to see the thread[/url][/b].


Find the area of a shape bounded by the lines x=0, x=6, y=0 and the shape bounded by the y= absolute value of (x-2)

I think the shape is a right triangle. The legs of the triangle measure 4 and 4. So the area is 4*4/2=8.

Isn't there two shapes bounded by those lines?

Either x=6 or x=0 isn't necessary or the problem is wrong.

[quote="vishalarul"]Yes, so then the area would be $9$.[/quote]

Wouldn't it be $10$?


geometry
trigonometry
perimeter
conics
ellipse
area of a triangle

Find the greatest possible area for an isosceles triangle whose equal sides are each 4 units long.

[hide="trig solution"]The area of a triangle can be expressed as $\frac12AB\sin C$, where $\angle C$ is the angle opposite $\overline{AB}$.

Thus, the area is maximized when $\sin C$ is maximized, or when $\sin C=\frac{\pi}2$.

$\frac124\cdot4\sin\frac{\pi}2=8$[/hide]

[b]Follow-up Question:[/b]

Find ([i]with proof[/i]) the least possible perimeter of a triangle with area $8$.

fix one side...using a recently posted result [trivial considering an ellipse, etc], the perimeter is reduced when the other sides are equal...this is true when all sides are equal...equilateral...the rest is trivial


category theory
topology
Galois Theory
superior algebra
superior algebra unsolved

Forgive the terminology; I may be abusing it.  There are many categories in which one can define the notion of an "automorphism group."  In some categories, such as $ \text{Graph}$, every group arises as the automorphism group of some object in the category.  (Similarly, the same is true of the fundamental group construction on topological spaces.)  In other words (and forgive me again if I am misusing the terminology), the "automorphism group" functor is onto.

Can we say anything general about categories of this type?  For example, it appears to be an open question whether the category of field extensions of $ \mathbb{Q}$ has this property, and it doesn't appear to be trivial to prove that the category of graphs has this property.

The notion of 'automorphism group' generally refers to a group of morphisms in $ Hom(A,A)$ for some object $ A$ in the category. The group of automorphisms of a graph and the Galois group are an automorphism groups in this sense of the word, and similarly is the functor that gives the group of automorphisms of a group or the group of bijections on a set. However, the fundamental group can't be realized as some set of homeomorphisms from a topological space to itself (as far as I know); it's just a certain functor from [b]Top [/b]to [b]Grp[/b].

For a general category, you could take a map which, for each object $ A$ of the category, gives the group of invertible elements of $ Hom(A,A)$. However, I'm not sure that this necessarily gives a functor from that category. (If $ a$ is an invertible element of Hom(A,A), and $ f \in Hom(A,B)$, can you somehow define f(a) in Hom(B,B)?)

If a category has enough objects, one could create a functor that arbitrarily assigns a group to each object and assigns the zero homomorphism to each morphism. (Of course, this is kind of boring.)

I wonder if the Yoneda Lemma is of any use here.


At 25 degree celcius the specific conductance of a saturated solution of $ \ce{AgCl}$ after subtracting the specific conductance of water is $ 1.82 \times 10^{ \minus{} 4}Sm^{ \minus{} 1}$.Given
$ \Lambda_{\ce{AgNO3}}^{0} \equal{} 0.1333 Sm^{ \minus{} 1}$
$ \Lambda_{\ce{HNO_{3}}^{0}} \equal{} 0.0421 Sm^{ \minus{} 1}$
$ \Lambda_{\ce{HCl}}^{0} \equal{} 0.0426Sm^{\minus{}1}$
Calculate the $ E^{0}$ for the half cell:$ \ce{Pt}$ |50 ml 0.5 M $ \ce{KCl}$in which a few drops of $ \ce{AgNO_{3}}$ is added to precipitate some $ \ce{AgCl}$
$ E^{0}_{\ce{Ag}^{ \plus{} }|\ce{Ag}} \equal{} 0.8 V$

Will all the Cl- be precipitated?? Or are you given the Ksp of AgCl or something?

sorry this is all the question say s  :(


AMC
AIME
summer program
Mathcamp
USA(J)MO
USAMO
topology

Is a person who scored 132 on AMC 10 and 7 on AIME (me) this year capable of going to this camp?

Certainly. When I first went to Mathcamp (in 2005), I had lower scores than that: 127.5 on the AMC 10 and 6 on the AIME. Mathcamp has all sorts of classes at different difficulty levels. So everyone should be able to find [i]interesting[/i] classes that are accessible to them. Mathcampers have a very wide range of skill levels. There are USAMO winners and people who have trouble making the AIME.

And there's not really that big of a difference between them. Mathcamp is not focused on competitions, and so everyone's scores on various competitions don't really mater much when you're at camp. The thing that really matters is how interested you are in math. If you're genuinely interested in math and you want to learn, then you'll have a great time at Mathcamp.

What you said could be interpreted in several ways, so I'll go ahead and answer all of them.

If you mean to ask whether you could understand the classes at camp, learn tons of math, and have a great time, then the answer is certainly yes.  Camp has a wide range of classes available at varying difficulty levels.  Some of these classes, such as introductory Combinatorics and Number Theory, will already be familiar to you, while others, such as Algebraic Topology and Category Theory, focus on bits of mathematical arcana that you're unlikely to find elsewhere until late in your college years.  Since classes are never mandatory, you can go to classes that interest and challenge you.  For more info, take a look at some of the older class schedules at http://www.mathcamp.org/academic.php.

If you're asking about whether you can get in with your AMC scores, the answer is both yes and no.  First, admissions are mostly based upon your qualifying quiz, personal essays, and teacher recommendations.  High contest scores are helpful, but not enough to get you in on their own.  That having been said, based upon your AMC scores, you know enough math to tear up the qualifying quiz, and thus have a very good chance of getting in if you put an honest effort into the quiz and essays (and be careful -- as with the USAMO and USAMTS, rigorously proving that your answer is correct is a big part of the game).  For your reference, I think I had around 120 on the AMC 12 and a 6 on the AIME when I applied and got in, so you've definitely got a good shot.

I hope that helped!

Edit:  JAM's post above (*waves*) hits the nail right on the head.  One's excitement for math and their motivation to learn are far more important at camp than contest scores and math experience.

Thanks for the insights and suggestions. By the way, eyefragment, in what year did you get 120/6?

2006, I believe.

Incidentally, I applied in '06 and got in with the exact same scores that you have, so don't worry too much about it.  If you really love math, you'll do fine at camp.  And you'll probably get in, too, because you'll put plenty of time into the Qualifying Quiz.

Heh, I only barely qualified for the AIME in my 9th and 10th grade years (when it was 2.5 points for a blank) and scored like 1 or 2, and have scored less than 100 in 11th grade and this year on the AMC 12. I did fine at Mathcamp and found my prefered difficulty level of class to be about ** (they range from *-****).

[quote="eyefragment"]e^pi*i is correct, the scores I listed were from 2006 (so 2.5 points were still given for blank answers), but you shouldn't put much stock in them -- qualifying quiz performance matters most.[/quote]The qualifying quiz is kinda hard...cuz I can't count...

Well, don't get discouraged. You still have over a month to work on them. The problems are intended to take a while. It took me ages to solve everything I did back in 2005.

It's also true that ability on timed exams like the AMC series doesn't correlate perfectly to the quiz/what ultimately happens at Mathcamp.  We've rejected plenty of USAMO qualifiers; we're looking for something different, about your creativity and ability to think about hard problems over extended periods.  Speed isn't as important, and knowing lots of problem-solving techniques isn't as important as what you can come up with and how deeply you can understand abstract mathematics.


[quote="Silverfalcon"]The degree of $(x^2+1)^4 (x^3+1)^3$ as a polynomial in $x$ is

A. 5
B. 7
C. 12
D. 17
E. 72[/quote]

[hide]
The degree of the first polynomial is 2, so it would be 8 raised to the 4th power.  The degree of the second polynomial is 3 so it would be 9 raised to the 3rd power.  The product of a 8th degree polynomial and a 9th degree polynomial is a 17th degree polynomial.

D
[/hide]

[hide]Take the highest degrees you can find, combine them to get 4*2+3*3=17.[/hide]

[quote="Silverfalcon"]The degree of $(x^2+1)^4 (x^3+1)^3$ as a polynomial in $x$ is

A. 5
B. 7
C. 12
D. 17
E. 72[/quote]

[hide]
The highest degree of the first part is 8, because (x^2)^4=x^8.
The highest degree of the second part is 9, because (x^3)^3=9.
8+9=17, D.
[/hide]

[hide]
You get the highest polynomial form each one, 4*2=8 & 3*3=9, so the answer is 8+9=17, or $D$[/hide]


Let $ y_n \equal{} 1 \minus{} \frac {2}{\pi}arcsin x_n$, then $ 0 < y_n < 1, y_{n \plus{} 1} \equal{} \frac {4}{\pi}arcsin\frac {y_n}{\sqrt 2 }.$
Because $ \frac {siny}{y}$ decrease, when $ 0 < y < \frac {\pi}{2}$, $ \frac {y_{n \plus{} 1}}{y_n} < \frac {4}{\pi}arcsin{\frac {y_n}{2}} \equal{} 1$.
It mean $ y_{n \plus{} 1} < y_n$, therefore $ y_n\to 0$ as $ (c \plus{} \epsilon)^n$, were $ c \equal{} \frac {2\sqrt 2}{\pi} < 1$.
Therefore $ x_n\to 1$.

I can't see all your post,but it is a strange proof
my solution is quite long and my result is similar to yours
At all,thanks


function
algebra
partial fractions
combinatorics unsolved
combinatorics

How many triples do they satisfy the condition:
 $ 0<x< y< z$;$ x,y,z\in \mathbb N$ and
a)$ x\plus{}y\plus{}z\equal{}n$
b)$ x\plus{}2y\plus{}3z\equal{}n$

a) Consider the genrating function for the triples $ (x,y,z)$
$ g(x) \equal{} (x \plus{} x^2 \plus{} x^3 \plus{} x^4 \plus{} ...)(x \plus{} x^2 \plus{} x^3 \plus{} x^4 \plus{} ...)(x \plus{} x^2 \plus{} x^3 \plus{} x^4...)$

b) Denote $ (a,b,c) \equal{} (x,2y,3z)$. The condition is equivalent to: $ a \plus{} b \plus{} c \equal{} 3n$, where $ 2|b, 3|c$.
Therefore the generating function for the triples $ (x,y,z)$ is:
$ g(x) \equal{} (x \plus{} x^2 \plus{} x^3 \plus{} x^4 \plus{} ...)(x^2 \plus{} x^4 \plus{} x^6 \plus{} ...)(x^3 \plus{} x^6 \plus{} x^9 \plus{} x^{12} \plus{} ...)$

Continue!, mathVnpro, I am not good at generating function  :blush: .
I only have some ideas for a):
$ g: (x_1;y_1;z_1) \to (x_1;y_1 - 1;z_1 - 2) =$. So we can find the number of solution of the equation $ x + y + z = n - 3$ and $ x\le y\le z$.
Denote $ S_m$ be the number of triples $ (x;y;z)$ satisfying our condition when $ n = m$.
The number of triples satisfying $ x = y = z$ is only $ f(m)$ with $ f(m)$ is defined by $ f(3k) = 1$;$ f(3k + 1) = f(3k + 2) = 0$
Again, denote $ a_m$;$ b_m$; $ c_m$ be the number of triples $ (x;y;z)$ such that $ x = y$; and $ y = z$ and $ x\neq y$;$ y\neq z$, respectively (of course, now $ x,y,z$ are different ). So that $ S_m = a_m + b_m + c_m + f(m)$. After some arguments, I got $ S_{m + 1} = 2(a_m + b_m) + 3{}_c{}_m + f(m)$. I found $ a_m = [\frac {m - 1}{3}]$; $ b_m = [\frac {m - 1}{2}] - [\frac {m + 1}{3}]$.
Thus $ S_{m + 1} = 3S_m + m - 3[\frac {m}{3}] - 1 - [\frac {m - 1}{2}]$.
The last step, I can't. So I post it on unsolved forum.
For the second; I don't have any ideas.
 Please give  me full solutions for both a) and b). I also expect solutions using  generating function as mathVN pro.
@)  :( . Actually, I  solve neither a) nor b).
Thank you JBL

As far as I can see, mathVNpro's comment completely fails to take account of the condition $ x < y < z$.  The first question is equivalent to your earlier question http://www.artofproblemsolving.com/Forum/viewtopic.php?t=282465 -- a triple $ (x, y, z)$ satisfies $ x < y < z$ and $ x \plus{} y \plus{} z \equal{} n$ if and only if $ (a, b, c) \equal{} (x, y \minus{} x, z \minus{} y)$ satisfies $ 3a \plus{} 2b \plus{} c \equal{} n$.

When you have questions like this, it's always worth a look in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].  The second sequence is [url=http://www.research.att.com/~njas/sequences/A025835]A025835[/url] -- a solution $ (x, y, z)$ gives us a solution $ (a, b, c) \equal{} (x, y \minus{} x, z \minus{} y)$ to $ 6a \plus{} 5b \plus{} 3c \equal{} n$.  These have generating function $ \sum x^{6a} \cdot \sum x^{5b} \cdot \sum x^{3c}$, which is where the description of the sequence I've linked comes from.  You could in principle use partial fractions to actually write down a formula here, but it won't be pretty.

Your recursion for part (a) appears to be incorrect (it gives $ S_6 \equal{} 5$, but the only actual solutions are 114, 123 and 222), but I haven't tried to figure out what you've done wrong.


Let $ d_1,d_2,\ldots,d_n$ be positive divisors of 1995. Prove that there exists two of them, $ d_i$ and $ d_j$, for which the numerator of the reduced fraction$ \frac{d_i}{d_j}$ is at least $ n$.

are you sure the question is really right?
isn´t it obvious?just pick $ d_i\equal{}1995$ and $ d_j\equal{}1$.

[quote="Johan Gunardi"]Why must there be 1995 and 1? What if, for example, I only take $ 5,19,35$? :D  And, yes, I'm sure the problem is correct.[/quote]

sorry, i misread the problem. :blush:

[hide="Hint 1"]$ 1995\equal{}3\cdot5\cdot7\cdot19$[/hide]
[hide="Hint 2"]If there exist $ d_i,d_j$ where $ 19\not|d_i,19|d_j$, the numerator of $ \frac{d_i}{d_j}$ is at least 19.[/hide]


How many integers between 1 and 200 are multiples of both 3 and 5 but not of either 4 or 7?

The solution on alcumus involves listing out the multiples of 15 and then crossing off those that are also multiples of 4 and 7.

Better/easier solution:

(These are the floor values
200/15 = 13
200/(15)(4) = 200/60 = 3
200/(15)(7) = 200/105 = 1

So there are 13 - 3 - 1 = 9 multiples.

I think the solution can be improved taking advantage of the fact that 3, 4, 5, and 7 are relatively prime.  Then all we need to do is count the numbers from 1 to 13 that are not divisible by 4 or 7.

The solution on Alcumus is way too slow! If you are doing a Mathleague SPRINT test, you won't have the time to list everything out. Better solution: find how many multiples of fifteen there are that are less than 200. Then subtract the number of multiples of 15 time 4 and 15 times 7. That would be 200/15, dropping the remainder, minus 200/60 and then subtracting 200 over 105, which gives you 13-3-1=9.

I get the Alcumus solution, but I don't really get any of the solutions above- could someone explain?

Wow... how do you make this work "faster"?


quadratics
number theory unsolved
number theory

Prove or disprove:For any odd prime p=4k+3, if each prime divisor of a positive integer n is a quadratic residue modulo p, then n can be written in the form of $n=x^{2}+py^{2}$, where x and y are integers.


If $ a,b,c,d > 0$ such that $ abcd \equal{} 1$ and $ a\le 1\le b\le c\le d$ prove that $ \left(a \plus{} 1\right)\left(d \plus{} 1\right) > 3$.

Is it $ a\le1\le b\le c\le d$ or $ 1\le a\le b\le c\le d$?

It`s $ a\le 1 \le b\le c \le d$. I`m sorry.

I improve this inequality. Let me state the problem:

[i] If $ a,b,c,d$ are positive real numbers such that $ a \ \leq \  b \ \leq  \ c \ \leq \ d$ with $ abcd\equal{}1$ then the following inequality holds:[/i]

$ (a\plus{}1)(d\plus{}1)\geq 3\plus{}\frac{3}{4d^3}.$

When does equality occurs?

For the original problem, for a fixed a, d is lowest when $ b\equal{}c\equal{}d$. So we can assume $ ad^3\equal{}1$. Then,
$ (a\plus{}\frac{1}{5}\plus{}\frac{1}{5}\plus{}\frac{1}{5}\plus{}\frac{1}{5}\plus{}\frac{1}{5})(d\plus{}1) \ge 6 \sqrt[6] {\frac{a}{5^5}} \cdot 2 \sqrt {d} \equal{}12 \sqrt[6] {\frac{ad^3}{5^5}}\equal{}\frac{12}{\sqrt [6] {5^5}} > 3$

[quote="Maxim Bogdan"]I improve this inequality. Let me state the problem:

[i] If $ a,b,c,d$ are positive real numbers such that $ a \ \leq \ b \ \leq \ c \ \leq \ d$ with $ abcd \equal{} 1$ then the following inequality holds:[/i]

$ (a \plus{} 1)(d \plus{} 1)\geq 3 \plus{} \frac {3}{4d^3}.$

When does equality occurs?[/quote]

[b]Solution.[/b] It's an easy inequality ( for $ 7^{th}$ grade i think). From $ a \ \leq \ b \ \leq \ c \ \leq \ d$ we get that: $ a\geq\frac{1}{d^3}.$

We'll prove that : $ \boxed{(a \plus{} 1)(d \plus{} 1)\geq \left (\frac{1}{d^3}\plus{}1\right )(d\plus{}1)\geq 3\plus{}\frac{3}{4d^3}}$

We have: $ (a\plus{}1)(d\plus{}1)\geq\left (1\plus{}\frac{1}{d^3}\right )(1\plus{}d)\geq 3\plus{}\frac{3}{4d^3}\Longleftrightarrow$

$ \Longleftrightarrow(d^3\plus{}1)(d\plus{}1)\geq 3d^3\plus{}\frac{3}{4}\Longleftrightarrow$

$ \Longleftrightarrow \left [d(d\minus{}1)\right ]^2\ \minus{} \ [d(d\minus{}1)]\ \plus{} \ 1\ \geq \ \frac{3}{4}\Longleftrightarrow$

$ \Longleftrightarrow \left [d(d\minus{}1)\minus{}\frac{1}{2}\right ]^2\geq 0.$

Equality occurs iff $ a\equal{}\frac{1}{d^3}$ and $ d(d\minus{}1)\minus{}\frac{1}{2}\equal{}0$ which yields : $ \boxed{d\equal{}\frac{1\plus{}\sqrt{3}}{2}}$

@: Maxim Bogdan
 Every problem is easy to solve if you know the solution ,but if you don't ...it may be really hard .


Low tide occurred at 10:45 a.m. and high tide occurred exactly 6 hours later. High tide was 194 cm above low tide. If the low water mark is taken to be zero and the height of the water follows a cosine curve, find a function that describes the height of the water at time t given in hours past midnight.

I get...

y = (195/2)(cos(pi/6(x+4.75))) + 97.5

Is this correct??

THanks!!

[hide]I got $f(t)=97cos\frac{\pi(t-4.75)}{6}+97$[/hide]
It seems that you accidentally changed the high tide to $195$ cm rather than $194$ and mixed up a sign.


Okay-- I was doing some AMC problems and some of the answers did not quite make sense. Could someone clarify them for me?

(paraphrased from the problem)

1.  Does "If A, then B" imply "a necessary condition for A to be true is B?"

Question: Shouldn't "necessary" be replaced with "sufficient?"

2. A car travels a certain distance at 50 mph and then returns to its starting point at 45 mph. What is the average speed of the car?

Question: The answer says: let the average speed = A.

1/A = 1/2(1/50 + 1/40)

A = 47 7/19

How do they get this solution? I have no idea what they did.

Thanks.

2) Well, you could do it this way:
Assign a number for the number of miles in the trip. 450mi will give you nice numbers.
450/50=9 hours
450/45=10 hours
total = 19 hours, 900 miles
900/19=47.36842 mph

on 2: They're doing it a way that is probably familiar to you, but they did the reciprocals of the fractions.
All I did was use the average speed formula, which I'm pretty sure you know:
2xy / (x + y)

2(45)(50)/ (45+50)
90(50)/(95)
= 900/19 = 47 7/19

For the first one, necessary is correct. Because if B isn't true then A can't be - it is necessary for B to be true when A is, from the first statement.

necessary and sufficient is covered in the "Prove It" chapter of aops v1 if you need to refresh

I thought "necessary" means "only if" and sufficient means "if."

Well, thats one way of thinking about it, but that would get kinda confusing.. I prefer just thinking the words like they are..
but anyway, it works in this case, since replacing necessary by only if asks you

Does "If A, then B" imply "A is true only if B is true" which is true. (because if B wasn't true, and A was true, the first statement is false).

I'm just going to try to restate the answer to number 1, in case anyone is still confused.

If B is true, then:  A could be true (since T -> T) or A could be false (since F -> T).

If B is false, then: A could be false (since F -> F) but A could not be true (since "if T, then F" is false).

So: If B is true, A could be true or false.   Thus, it is not sufficient.  If B is false, A cannot be true.  Thus, it is necessary.  (In order for A to be true, B has to be true.  But B being true does not mean than A is true.)


group theory
abstract algebra
superior algebra
superior algebra solved

Prove that the simetric group or order 5 does not have a subgroup with 30 elements. Very nice problem.

I it ha a group with 30=2*3*5 elements, then there woul exist elements of rder 2 an 5, namely the cycle of order 5 and a transposition. Let's assume the transposition is (1 2). Then (2 3 4 5 1)*(1 2)=(1 3 4 5 2)= a cycle of order 4, but this is impossible since 4 doesn't divide 30.

Moreover, I think it can be proved that or a prime p, the group Sp is generated by a cycle of order p and a transposition (I don't know iff it works for non-primes).

I knew that product of disjoint transpositions also has order 2.

What's the connection with this problem? You don't have 2 transpositions, but a transposition and a cycle of order 5. And besides, I don't think what you said is true for any 2 transpositions, just for those which are disjoint. Generally, the order of the composition of 2 disjoint cycles is the smallest common multiple of the orders of the cycles.

Read carefully and you will find that I said they are disjoint. The conection with this problem? You say the subgroup must have an element of order 2 and you take it to be a transposition. But why should it contain a transposition as the element of order 2 and not the product of 2 disjoint transpositions?

Sorry!! :( I guess I should go back to the drawing board then, huh? :D

Yeah. And yes, the group S_n is generated by the transposition (1,2) and the cycle (1,2,...,n) for any n.

Here's something else, but I won't post solutions for the following 2 lemmas, because they're long and I'm sure there's a better approach :):

(1) A5 is a simple non-abelian group (in fact, it's true for n>=5).

(2) For any simple non-abelian group G, if H=/=G and H is a subgroup G, then |G:H|>=5.

Let's assume there is such a subgroup G of S5. Then it's obvious that |G\capA5| is either 30 or 15, but in both cases |A5:G\capA5|<5 (it's 2 or 4), so we have a contradiction.

For Moubi:

By that multiplication I mean the permutation (2 3 4 5 1) composed with (2 1 3 4 5) (the second one applied to the first one), so in (2 3 4 5 1) we reverse 1 and 2, thus getting (1 3 4 5 2). Isn't this correct?

Here's another approach, but I still used Sylow's 3'rd thm:

There is a group of 3 elements. Let n be the number of such groups. Then n|30/3=10 and n=1 (mod 3) (Sylow's 3'rd thm), so n=1. This means that the unique group of 3 elements is normal. Let a be an element of order 3 and b an element of order 5. Then we have 1 of 3 cases:

bab-1=1, which is obviously impossible;

bab-1=a. This is impossible, because it would mean that ab=ba, so the order of ab is 15 (=3*5), but every permutation can be written as the composition of disjoint cycles, so the maximum order an element in Sn could have is the maximum product of elements whose sum is n, so in our case that's 6 (6=2*3 and 5=2+3);

bab-1=a2 => ba2b-1=a4=a => b2ab-2=ba2b-1=a=> b2 and a commute, but that's also impossible because b2 is also of order 5 and the demonstration from the previous paragraph works.

Thus we have a contradiction.


integration
calculus
calculus computations

If $ I_{m,n}$ denotes $ \int_{0}^{\infty} \frac{x^{m}}{(x^{2}\plus{}a^{2})^{n}}\ dx$, find a relation between $ I_{m,n}$ and $ I_{m\minus{}2, n\minus{}1}$, and state the values of $ m$ and $ n$ for which this is valid. Hence evaluate (a) $ \int_{0}^{\infty} (\frac{x}{x^{2}\plus{}a^{2}})^{7}\ dx$, (b) $ \int_{0}^{\infty} \frac{x^{4}}{(x^{2}\plus{}a^{2})^{\frac{7}{2}}}\ dx$.


LaTeX blogs logarithms trigonometry limit

1)Solve for x

Here numbers in [] are base of log

log[3/4]log[8](x^2 +7)+log[1/2]log[1/4]((x^2 +7)^-1) =-2

2)
log[b]a.log[c]a+log[a]b.log[c]b+log[a]c.log[b]c=3

here a,b,c are different posituve numbers and not equal to 1

Find abc

Sorry i dunno latex

what is the answer for ur second problem..... :)

[quote="RAGHAV GROVER"]For the second problem:
apply A.M>G.M
log[b]a.log[c]a+log[a]b.log[c]b+log[a]c.log[b]c>=3{log[b]a.log[c]alog[a]b.log[c]blog[a]c.log[b]c}^1/3..

log[b]a.log[c]a+log[a]b.log[c]b+log[a]c.log[b]c>=3
equality occurs when log[b]a.log[c]a=log[a]b.log[c]b=log[a]c.log[b]c

from here we get a=b=c ..
dont know how to proceed further ..[/quote]

You can't apply AM-GM the way you did, since not all the terms are guaranteed to be positive: either of $ a,b,c$ can be less than $ 1$

[quote="Arrange your tan"][quote="papu24"]1)Solve for x

Here numbers in [] are base of log

log[3/4]log[8](x^2 +7)+log[1/2]log[1/4]((x^2 +7)^-1) =-2

[/quote]
I entered:
log(3/4)log(8)(x^2 +7) + log(1/2)log(1/4)((x^2 + 7)^-1) = -2

at the http://www.quickmath.com site in the Equations section.

It came back with 4 exact complex solutions ("monsters") that
no "reasonable" user would come up with, so possibly there
are unwanted additions, typos, or misunderstandings on my
part about the nature of the particulars of this equation for
it to have a "reasonable" solution or solutions.  ??[/quote]

The solution to the equation is 3 or -3...u can check it also. I guess u didn't see that the number in square brakets[] are bases of log.......u can help by writing the question properly 'cause i dunno latex

and what about the second  problem ???
what is the answer to it..  :?:

I have got solution to the first problem also...
solution is quiet lengthy so i am just posting what i actually did::

log[2] (x^2+7)=y
you can substitute this easily using the properties of the log..

next we get log[3/4](y/3) + log[1/2](y/2)=-2
now,log[3/4](1/3) + log[3/4](y)+log[1/2](y)=-3

using the change of base property in log[3/4](y) we get
log[2](y)*log[3/4](2)  again substituting log[2](y)=z

we get log[3/4](1/3) + z{log[3/4](2) - log[3/4](3/4)}=-3

simplifying the above we get z=2 so y is 4

log[2] (x^2+7)=y so x=+3,-3.
done

[quote="RAGHAV GROVER"]No farenhajt its given that the a,b,c are different postive numbers and neither of them is equal to 1..

so according to me we can definitely apply am>g.m[/quote]  Read again what Farenhajt wrote: if one or more of $ a, b, c$ is [i]less that[/i] 1, some of the log terms will be negative and so AM-GM cannot be applied.

[[LaTeX]] is easy -- read about it at that link and at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690

$ \text{\LaTeX}$'ed .. I hope I wrote it correctly.

[quote="papu24"]1)Solve for x

$ \log_{\dfrac{3}{4}}(\log_{8}(x^2 + 7)) + \log_{\dfrac{1}{2}}(\log_{\dfrac{1}{4}}(x^2 + 7)^{ - 1}) = - 2$

2)
$ \log_{b} a \cdot \log_{c} a + \log_{a} b \cdot \log_{c} b + \log_{a} c \cdot \log_{b} c = 3$

$ a,b,c > 0 \text{ } | \text{ } a,b,c \not = 1$

Find $ abc$

[/quote]

Is this what he meant?

Use \log for the logarithm (similarly \sin, \cos, \lim, etc.)  Also, you appear to have more parentheses in the first one than the original problem intended.

Second problem...

Sub, $ x=\ln a, y=\ln b, z = \ln c$

$ \frac{x^2}{yz} + \frac{y^2}{zx} + \frac{z^2}{xy} = 3$

$ \therefore \frac{1}{2}(x+y+z)((x-y)^2 + (y-z)^2 + (z-x)^2)) = 0$

$ \therefore x=y=z$ or $ x+y+z=0$

Since $ x\not = y\not = z$ from the condition that $ a,b,c$ are ditinct we have
$ x+y+z = 0 \Longrightarrow \ln(abc) = 0 \Longrightarrow \boxed{abc = 1}$

[quote="ocha"]Second problem...

Sub, $ x = \ln a, y = \ln b, z = \ln c$

$ \frac {x^2}{yz} + \frac {y^2}{zx} + \frac {z^2}{xy} = 3$

$ \therefore \frac {1}{2}(x + y + z)((x - y)^2 + (y - z)^2 + (z - x)^2)) = 0$

$ \therefore x = y = z$ or $ x + y + z = 0$

Since $ x\not = y\not = z$ from the condition that $ a,b,c$ are ditinct we have
$ x + y + z = 0 \Longrightarrow \ln(abc) = 0 \Longrightarrow \boxed{abc = 1}$[/quote]

Thanks....ur answer id absolutely right


function
trigonometry
real analysis
real analysis unsolved

let f be differentiable function in R s.t. f`(x) is bounded for all real nos. with modulus < 1,x.show f has atleast  one fixed point and its lifschitz function. :)
(sorry for putting wrong pro before)

No fixed point for you: $f(x)=x+1$.

To prove that $f$ is Lipschitz, use the Mean Value theorem.

[quote="mlok"]No fixed point for you: $f(x)=x+1$.[/quote]

Or $f(x)=x+\frac1{10}\sin x,$ which has infinitely many fixed points.


Jim the beetle is trapped in a circle with radius 1. When he is squished, his squished body makes a circle with a radius of 0.25. If when he is squished, the center of his squished remains is in the circle of radius 1, find the probability that none of his remains fall outside of the circle.

[quote="lotrgreengrapes7926"]What if the bug was trapped in a square with side length 1? :D[/quote]

[hide]$\frac14$[/hide]

[hide]The beetle's center must be within the circle of radius 3/4. The area of the circle is pi. $\frac{\frac{9\pi}{16}}{\pi}=\frac{9}{16}$.[/hide]


Point Q lies Midway between points O and P. Line Segment OQ is X. An arc is drawn from O to P with center Q. Another arc is drawn from Q to P with center R, which lies midway between Q and P. This "seashell" shape is Figure A. A line, Y , is drawn from point Q that devides Figure A into two sections, with the same perimeter. What is the degree of angle created by line segment X and line Y?

sorry if that was confusing. Draw it out to see what I'm saying.

I only have time to give you an answer for now.

[hide]124.65 degrees[/hide]

I will post the solution later.


Would it be helpful for a math student to take the AoPS courses even after they have bought the AoPS books?

This is because I'm planning to take a course but not sure if I should although it seems to be very benficial to me if I do take a course.

Yeh.. its still very good. The courses go way beyond what is offered in the book.

Yes I agree with jelyman.  Just make sure that you take a course that is at your level.  They present different problems and more techniques than what the books have.

I have another question.

Does enrolling in evaulated give you a better advantage than audited?

My question is more like, is it really helpful to receive comments from the instructor or is it just fine by only taking the lesson and working on problems without any comments.

I believe it is helpful in a sense that they will guide you to understanding everything fully and being able to explain it well. It is up to you to decide if for that particular course yuo need that.

We intend the courses to both cover additional topics to those covered in the AoPS books and to take those topics to a deeper level.  We also included 200-400 problems in each course that serve to solidify students' abilities to use the algorithms, theorems, and problem solving techniques covered.


Keiko tosses one penny and Ephraim tosses two pennies. What is the probability that Ephraim gets the same number of heads that Keiko gets?  Express your answer as a common fraction.

CASE 1: keiko gets $ 0$ heads
the probability that keiko gets $ 0$ heads is $ \frac {1} {2}$
the probability that Ephraim gets $ 0$ heads, is $ \frac {1} {4}$
so the probability that this is the case is $ \frac {1} {8}$

CASE 2: keiko gets $ 1$ head
the probability that keiko gets $ 1$ head is $ \frac {1} {2}$
the probability that Ephraim also gets $ 1$ head is $ \frac {1} {2}$
so the probability that this is the case is $ \frac {1} {4}$

to get our answer, we must add the probabilities of our two cases together
$ \frac {1} {8} \plus{} \frac {1} {4} \equal{} \frac {3} {8}$

our answer is $ \boxed {\frac {3} {8}}$


suppose (ABCD)=-1

O is a point in wich:angle AOB=90.

prove that:

               OC and OD are the angle bisectores of AOB and COD

ohhh...im sorry.because i shouldent post this question to this case.because its hard for this case.

i mean harmonic,in line 1.suppose A,C,B,D are on aline respectivly.and we have:AC.BD=BC.AD

then we call these four points,harmonic points. :blush:

I dun get it still, OC or OD is the bisector of COD? is that possible?


If H is a proper subgroup of a finite group G, then the set union
$\bigcup_{x\in{G}}x^{-1}Hx$ is not the whole of G.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22729]Here you go[/url] :).


MIT
college
AMC
AMC 12
USA(J)MO
USAMO
summer program

How many juniors typically make MOP? The reason I am asking is that since MOP and RSI and MIT run concurrently for some time, a junior that got into RSI that also makes MOP can only choose one, and so I am trying to determine how many alternates will get to go (since some will turn down RSI for MOP).

Last year, the RSIers just left MOP like 2 or 3 days early to make RSI at Caltech.

Elessar, what about freshmen? There are total 38 of them, and the manual says that they will take approximately thirty. Do you think that all 38 will make to the freshmen MOP? If not, how are they going to make the cut? Twenty two out of 38 were qualified by making the floor. You cannot simply index them, cuz some of them took AMC12 and others AMC10. What is you opinion?

I believe that the 9th graders will not all be invited (unless they all happen to do extremely well on the USAMO).  It is not yet clear how the top 9th graders will be determined, but you can safely assume that the USAMO score will be the most important input.

[the top 9th graders .[/quote]

You mean top 30 out off 38? I don't think that all top 30 9th grade qualifiers can break the ground (I mean getting a non zero score)

[quote="bubala"]This is a little random, but WHY does MOP run in early June to July?  My school runs through the last week of June.  I'm a USAMO qualifier this year, say Akamai gives the grant and I make MOP next year, what would I do?[/quote]

That just makes it even better! If I make MOP (a very big IF), I get to miss 2 weeks of school :P. How awesome is that, missing school to do math!

But the actual reason is that the IMO is usually mid-July, so they need to finish MOP before the IMO.

[quote="qvc122"][the top 9th graders .

You mean top 30 out off 38? I don't think that all top 30 9th grade qualifiers can break the ground (I mean getting a non zero score)[/quote]

I don't think the number of extra 9th graders is set in stone.  It will depend on how they do as a group, how many are among those who are invited to MOP outside the 9th grade expansion, etc.

Well, in response to your first post, qvc122, Mr. Rusczyk has many times my experience in math competitions so you can expect that his answer will be a better guesstimate than mine.

When the AMC says "approximately 30" I don't believe they intended to say "approximately 40". By some means, the freshmen will most likely be reduced so that there is adequate funding to cover everyone invited. However, it has been mentioned that none of these numbers are set in stone and that is indeed the case. I don't think anyone can say that exactly 30 "normal" MOPpers, and to some extent the amount of freshmen MOP funding depends on what goes towards the normal MOP.

Remember that the point of MOP, unlike most other math camps, is to select and train an IMO team for the United States; to find potential candidates for future IMO teams; and to train potential candidates for future IMO teams. (The movies, card games, ultimate frisbee, etc. are extra.) Keeping that in mind, you'll see that the exact dates and length of MOP actually vary from year to year. From my limited knowledge, it's assumed that the dates are established so that MOP concludes a week or so before the Team has to leave for IMO. Timing for MOP also conveniently dodges most other math camps that MOPpers go to (PROMYS, Canada/USA MathCamp, etc.).

[quote="probability1.01"]Just a very slight note, I think it's actually the other camps like MathCamp, PROMYS, etc. that dodge MOP. :D

As for the MOP qualification for freshmen, I would guess it would be top USAMO scores (since that's how they did it for the extra 5 sophomores last year as far as I know).[/quote]

They had extra sophomores come??? That, I dont think is written on the manual??

Well, it wasn't planned, but since there were only 25 or so 9th grade qualifiers to begin with, they decided to take 5 extra sophomores.  Unless all my sources are wrong :P.

But very likely a few of the freshmen won't be able to go, and possibly 1 or 2 will qualify the regular way, that should get to something around 30(and also one of the freshman is from Canada). But still they might cut a few more freshmen to get in a few sophomores or something like that.

I don't know, Feiqi...none of last year's frosh rejected the MOP invitation so it seems that the chances of rejections may be small (but definitely not impossible) this year.

I definitely agree that other math programs try to dodge MOP. I know for a fact that the USACO organizers have during the last few years (although not previously).  RSI, however, is generally a conflict. It only worked out last year that 4 students were able to do both because the Caltech RSI started a week later than MIT RSI which lessended the amount of overlap with MOP. I've never heard of any other year when any students did both. No one should figure that will work for them.

[quote="darktreb"]I wonder how they count Freshmen that would have made it anyway. I remember my freshman year there was 3 other 9th graders and an 8th grader at MOP (which was the last 30 person MOP). So I wonder how that contributes to their "approximately 30" or whatever.[/quote]

I believe they are not counted as among the 'extras', but don't hold me too that!

[quote="rrusczyk"]I believe they are not counted as among the 'extras', but don't hold me too that![/quote]

Do you mean that if 3 freshmen made to top 30 fro regular MOP, there would be additional 30 extra freshmen, which would be total 33 freshmen? I hope that is the case.

Another thing is when they rank the freshmen, they might use something called USAMO index = AMC10/12+10*AIME+9*USAMO. That formula was used before. However, this USAMO index has a flaw, since it basically punishes those freshmen who took AMC12. Generally, if a freshman can get a score of 120 on AMC12, he/she may very well be able to get 140 or near perfect score on AMC10. If that is the scenario, it is just not fair.

I am not going to the MOP (I am too old for that!). But I am keeping my fingers crossed for a freshman girl. I really hope that they would take all 38 freshmen.

[quote="qvc122"]
Another thing is when they rank the freshmen, they might use something called USAMO index = AMC10/12+10*AIME+9*USAMO. That formula was used before. However, this USAMO index has a flaw, since it basically punishes those freshmen who took AMC12. Generally, if a freshman can get a score of 120 on AMC12, he/she may very well be able to get 140 or near perfect score on AMC10. If that is the scenario, it is just not fair.
[/quote]

Hmmm really?? AMC had always been encouraging people to take AMC12 though. If that's the case, why would they even use this "USAMO Index"?

[quote]In advising students who desire to be selected for the USAMO whether to take the AMC 12 level contests or the AMC 10 level contests, it will be to their advantage to take the AMC 12 level contests. [/quote]
Presumably that's USAMO, and not MOP.

Why don't they do that for Sophomores...?? i mean not all sophomores who qualify, but for some that are at borderline... I mean they are the futures too... i dont see tghat many Sophomores being IMO participant, so they shoolud be considered future as well...

I think it's pretty tough to find a good system for the index. If they only take the AMC10/12 score, it'll obviously be an advantage for those who took the easier AMC10, which isn't right. But I don't think giving those who took the AMC12 an automatic 150 is right either. Of course, I'm a little biased, since I took the AMC10, but I still think it should be fair. I'm certainly pretty curious to see how they adjust it.

[quote="qvc122"][quote="probability1.01"]I think they will use the index and then somehow adjust for the AMC 10/12 thing.  Maybe AMC 12 people get automatic 150 as some have suggested.[/quote]

I am almost sure when there are more than 30 freshmen qualifiers, they will use so called USAMO index. How are they going to make adjustment for amc12 vs. amc10, or no adjustment at all, I have no idea. I would suggest people with potential interest or with this issue in their mind PM the AMC director and ask for the adjustment, since there are roughly 20-25 points advantage (amc10)/ disadvantage (amc12). Another suggestion would be discard amc score and only put 10*aime+9*usamo. One more suggestion is whether someone can solicit Akamia to support all freshmen qualifiers, since only 8 more people, which may only cost the company $20,000 more. That is a small number for a company like Akamai.[/quote]

There is a good chance that the selection process will be only based on USAMO score.

Maybe they'll compare USAMO scores, then if there is a tie, compare AIME scores, then AMC scores.  If it's still tied they'll probably take both :P.

As far as I've heard, the mosp index is only for juniors.  So they can only take AMC 12 so there is no problem.  And even then, USAMO is the heaviest weighted factor by far.  Considering that the USAMO cutoff was so high, the highest advantage anybody could have over anyone else is about 1 problem on the USAMO.  And back to the thread topic, in general, about how many juniors make mosp?

[quote="pdiao"] And back to the thread topic, in general, about how many juniors make mosp?[/quote]

In general, I'd like to say between 10 and 20...closer to the 10 than the 20.

Last year there were 11 that I remember (someone correct me if I miscounted).


Prove that for a,b,c>0, the following inequality holds:

1/[a(b+1)]+1/[b(c+1)]+1/[c(a+1)] \geq 3/(1+abc).

Just an observation:

This would solve the inequality manlio posted:

( \sum 1/a)( \sum 1/(a+1)) >= 9/(abc+1) because we use the inequality you posted and rearrangement (1/a,1/b,1/c and 1/(a+1) etc. are ordered the same way and the ineq you posted also works for a cyclic permutation of a,b,c). I don't have a soln for this one.

This is a famous and well known inequality which appeared in many mathematical journals.The official solution of the proposer was based on calculus.  The nicest proof was found by a sixteen student who was praised for it by the proposer. This proof is a "mathematical gem" for its simplicity.

THE PROOF:

Rewrite
                     (1+abc)/a(1+b)=(1+a)/a(1+b) +b(1+c)/(1+b) -1

                     (1+abc)/b(1+c)=(1+b)/b(1+c) +c(1+a)/(1+c) -1

                     (1+abc)/c(1+a)=(1+c)/c(1+a) +a(1+b)/(1+a) -1

Remembering that    t + 1/t  \geq 2 we obtain

   (1+abc)/a(1+b)+(1+abc)/b(1+c) + (1+abc)/c(1+a)  \geq 2+2+2-3 =3

Grobber, you are right. This inequality solve the other inequality I posted.
I also used this method.
But there is an elegant and neat proof.
Use the AM-GM inequality twice and then the HM-GM inequality.


inequalities
geometry
circumcircle
trigonometry
trig identities
Law of Sines
geometry proposed

Let $ABC$ be an acute triangle. The tangents in $A$ and $B$ to the circumcircle of $ABC$ meet at $S$. Let $D$ and $E$ be the Midpoints of $BC$ and $CA$. Prove that $\angle ACB + \angle DSE < 90^\circ$

First $\measuredangle ACB = \measuredangle AOS$. Also $\measuredangle ACB + \measuredangle BAO = 90^{\circ}$. Hence we want to prove $\measuredangle DSE < \measuredangle OSA$. By the law of sines $\frac{DE}{\sin \measuredangle DSE} = \frac{ES}{\sin \measuredangle SDE}$ and $DE = AS\sin \measuredangle OSA$.

Then $\frac{ES\sin \measuredangle DSE}{\sin \measuredangle SDE} = AS\sin \measuredangle OSA$, from which $\frac{\sin \measuredangle OSA}{\sin \measuredangle DSE} = \frac{ES}{AS} \cdot \frac{1}{\sin \measuredangle SDE}$.

Clearly $ES > AS$ and $0 < \sin \measuredangle SDE < 1$. Hence $\frac{\sin \measuredangle OSA}{\sin \measuredangle DSE} > 1$ and $\measuredangle OSA > \measuredangle DSE$.

[img]C:\Documents%20and%20Settings\All%20Users\Desktop\inequality%20of%20angles.png[/img]

Nice proof!

Btw, a 'little' angle-chasing shows, that this problem is equivalent to IMO1998/5


geometry
perimeter
calculus
integration
trigonometry
congruent triangles

Find the perimeters of all non-congruent triangles in which the lengths of all three sides are integral, the cosine of one angle is $ -\frac{1}{4}$, and the length of one of the two sides not opposite this angle is $ 16$.


If $\frac{3}{x^{2}-8x+12}=\frac{A}{x-6}+\frac{2B}{x-2}$, find the value of $2A+5B$. Express your answer as a common fraction reduced to lowest term.
[hide]
Letting $x=2$ and then $x=6$, I got $B=\frac{-8}{3}$ and $A=\frac{3}{4}$. I got [b]-11 5/6[/b]. Yet my solution book says [b]-3/8[/b]???[/hide]


function
geometry
perimeter
calculus
trigonometry
calculus computations

Now i am working with $sin$, $cos$ and $tan$ in differentiation. I have these two functions:

$f(x)=3\cdot sin(x)$

$f'(x)=3\cdot cos(x)$

and:

$g(x)=x\cdot cos(x)$

$g'(x)=cos(x)-sin(x)\cdot x$

For $f(x)$:
Putting in $3$ instead of $x$ gives $2.9959$. But if i write the function into my calculator and then ask what $\frac{dy}{dx}$ in $x=3$ is, then the answer is $0.0523$.

For $g(x)$
Putting in $3$ instead of $x$ gives $0.8416$. But if i write the function into my calculator and then ask what $\frac{dy}{dx}$ in $x=3$ is, then the answer is $0.9959$.

Do you see what is wrong here? I am working in degrees...

Plz someone :(

When i put a function with $sin$ into my calculator like:

$3\cdot sin(x)$

and i then view the graph on my calculator it shows a straight line equal to $y=0.5233x+0$, and i am pretty sure $3\cdot sin(x)=0.5233x+0$ is not true. As far as i know we will see more like a wave rather than a straight line.

Can someone explain what is wrong here?

[quote="Sporally"]
Do you see what is wrong here? I am working in degrees...[/quote]

That's your problem, change to radians.

Ah, ok, that helped. Never been working with radian though. What is the relationship between degree and radian? It is nice to know what you are working with... :)

Say you have an angle that has a meansure of $x^\circ$, then in radians it has a measure of $\frac{\pi\cdot x}{180}$.

OK. Will try and keep that i mind. Here in Europe we say a circle is $360\circ$, do you have anything eqillient to that? Or what does radian say?

Take the unit circle, which has 360 degrees.  The perimeter of that circle is $2\pi$.  Therefore a radian is just the length of the arc of the cirlce that is subtended by an angle.  Take a 90 degree angle, that subtends a quarter of the circle which has a length of $\frac{\pi}{2}$, so in radians 90 degrees is $\frac{\pi}{2}$.  I know I worded this badly, but I hope you can understand what I'm saying.

It was so badly written that i understood it perfectly the first time :P Ok, i have no problem in understanding what you mean, and i have to say it doesn't sound like the worst thing to work with at first glance. Give me some time and i will know :)  Thx a lot for your help... couldn't be better.

The calculus of trig functions is [b]always[/b] in radians, and calculus is the reason we use radians. You can use degrees for everything else, but don't try to differentiate or integrate with them.

I think i will soon get used to this new thing :)


algebra
polynomial
Vieta
complex numbers

For certain real values of $a, b, c,$ and $d,$ the equation $x^4+ax^3+bx^2+cx+d=0$ has four non-real roots.  The product of two of these roots is $13+i$ and the sum of the other two roots is $3+4i,$ where $i=\sqrt{-1}.$  Find $b.$

[hide]well since the sum of the those two roots is nonreal, those two are not clearly conjugates, so the product roots must be conjugates of the summed roots.
so :
$m\cdot n=13+i,m'+n'=3+4i\Rightarrow m'\cdot n'=13-i,m+n=3-4i$
by viet's we know: $b=mm'+nn'+mn'+nm'+mn+m'n'=(m+n)(m'+n')+mn+m'n'=\boxed{51}$[/hide]

Nice solution amir.  :)

[hide="Solution"]Let the roots of the polynomial be $r,$ $s,$ $t,$ and $u.$ Without loss of generality, assume $rs=13+i$ and $t+u=3+4i.$ By Vieta's, \begin{eqnarray*}a &=&-(r+s+t+u) \\ b &=& rs+rt+ru+st+su+tu=rs+(r+s)(t+u)+tu \\ d &=& rstu, \end{eqnarray*} where $a,b,d\in \Re.$

Note that $d=(13+i)tu.$ Hence, $tu=0$ or $tu=(13-i).$ However, both roots are imaginary, so $tu=(13-i).$

Similarly, $a=-(3+4i+r+s)\implies r+s=-(a+3)-4i.$ However, we know that $(r+s)(t+u)=(3+4i)(-(a+3)-4i) \in Re,$ (from $b$) so $r+s$ must equal $3-4i.$

Hence, our answer is $51,$ as amir found. Kinda redundant, but it worked. :D
[/hide]

[quote="cdymdcool"]when two of the roots' sum is 3+4i, how could you conclude that the other two roots' sum is 3-4i?[/quote]

If two of the roots are $ r_1\equal{}a\plus{}bi$ and $ r_2\equal{}c\plus{}di$, then $ r_1\plus{}r_2\equal{}(a\plus{}b)\plus{}(c\plus{}d)i$.

By the complex conjugate root theorem, $ \overline{r_1}\plus{}\overline{r_2}\equal{}(a\plus{}b)\minus{}(c\plus{}d)i$ is also a root.

Let $ a\plus{}b\equal{}3$ and $ c\plus{}d\equal{}4$, of course, and you have your answer.

The problem states that the four roots are not real.  Unfortunately, $55$ and $-\frac{1}{4}$ are both real.

In fact, that's why the problem has to state that all four roots are not real:  you can conclude immediately that the four roots must come in two complex conjugate pairs.  From there, you must note that when the sum of two of the roots is complex and not real, they must be from DIFFERENT conjugate pairs.  Same for the product.

[quote="droid347"]How do you know the products of the conjugates is $13-i$ if the product of the roots is $13+i$?[/quote]

Set the roots as $a + bi$ and $c + di$. Their product is $(ac - bd) + (bc + ad)i$, and if you multiply the conjugates $a - bi$ and $c - di$ you get $(ac - bd) - (bc + ad)i$ which is the conjugate of the first product.

[hide=Solution]Let the quartic equation have roots of $x,\overline{x},y,\overline{y}$, where $xy=13-i$ and $\overline{x}+\overline{y}=3+4i$. In addition, $\overline{x}\overline{y}=\overline{xy}=13+i$ and $x+y=\overline{\overline{x}+\overline{y}}=3-4i$. Then, we use Vieta's to express $b$ in terms of these roots:
\begin{align*}b&=x\overline{x}+xy+x\overline{y}+\overline{x}y+\overline{x}\overline{y}+y\overline{y}\\
&=(13+i)+(13-i)+(x+y)(\overline{x}+\overline{y})\\
&=26+3^2+4^2=26+25=\boxed{51}\end{align*}[/hide]

Let $w, z, \overline{w}, \overline{z}$ be the roots. WLOG, we can say that $wz=13-i$ and $\overline{w}+\overline{z}=3+4i$ (products like $w\overline{w}$ will never give a non-real result). We can also say that $\overline{wz}=13+i$ and that $w+z=3-4i$. Using Vieta's Formulas, we have that
\begin{align*}
b&=wz+\overline{wz}+w\overline{w}+w\overline{z}+z\overline{w}+z\overline{z}\\
&=(13-i)+(13+i)+(w+z)(\overline{w}+\overline{z})\\
&=26+|3+4i|^2=26+25=\boxed{51}\\
\end{align*}

The roots must come in conjugate pairs. Therefore, the sum of the other two must be the conjugate of the sum of these two roots. We can see that the sum of these two roots are not conjugates because the imaginary terms did not cancel out. Thus, we have $$x+y=3-4i$$
$$xy=13+i$$

The product of the other two roots must be $13-i$. Thus, we can simply calculate $b$ by evaluating $$\frac{1}{2}\left(x+y+\overline{x}+\overline{y}\right)^2-\frac{1}{2}\left(x^2+y^2+\overline{x}^2+\overline{y}^2\right)=18-(-33)=\boxed{51}$$


[hide]
They buy 1000 at 5 for $\$2$ each.
So that is 1000/5=200 sets of 5.
So $\$400$ is the amount they spend to buy them
Selling them they sell 500 sets of 2, each for a dollar.
That is $\$500$.
They spent $\$400$ and got $\$500$.
That is a total of $\$100$ profit.
$A$[/hide]

[hide]They buy 1000 candy bars for a price of 5 for 2. The cost is 1000/5 * 2,
which is 400. Then they sell the candy bars for 2 for 1. They earn 1000/2,
which is 500. Subtract 400 from 500, so their total profit is 100.
The answer is [b]A[/b][/hide]

[hide]The scout troop paid $\$$400 for the candy bars. They sold all the candy bars for $\$$500. Their total profit is $\$$100, answer A.[/hide]

[quote="catcurio"]A scout troop buys $1000$ candy bars at a price of five for $\$2$. They sell all the candy bars at a price of two for $\$1$. What was their profit, in dollars?

$\text{(A)}\ 100 \qquad\text{(B)}\ 200 \qquad\text{(C)}\ 300 \qquad\text{(D)}\ 400 \qquad\text{(E)}\ 500$[/quote]

[hide]THey pay 400 for the bars. They get 500 for the bars. THus they earn 100 A[/hide]

[quote="catcurio"]A scout troop buys $1000$ candy bars at a price of five for $\$2$. They sell all the candy bars at a price of two for $\$1$. What was their profit, in dollars?

$\text{(A)}\ 100 \qquad\text{(B)}\ 200 \qquad\text{(C)}\ 300 \qquad\text{(D)}\ 400 \qquad\text{(E)}\ 500$[/quote]

[hide]The troop makes: $1000*2/5$. That is $\$400$. They sell for $50$ cents each $1 dollar/2 bars$ That makes 500 dollars ($(.5*1000)$).) Subtract 400 from 500 and that is the profit, which is 100. or $\text{(A)}$[/hide]


function
geometry
incenter
national olympiad

[size=150]First Day[/size]
1st April 2005
Time allowed : 4h

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=202785#202785][b][u]Problem 1 :[/u][/b][/url]

Find all the functions $f:R\rightarrow R$ stisfying :
$f(x^2)-f(y^2)=(x+y)(f(x)-f(y))$ for all real numbers $x,y$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=202787#202787][b][u]Problem 2 :[/u][/b][/url]

$ABC$ is a triangle such that $\angle{A}<\angle{B},\angle{C}$. Let $C$ be the circumencircle of $ABC$ with radius $R$. and $D$ a point of the arc $BC$ which does not contain $A$. The bicectors of the segments $[AB]$ and $[AC]$ meets line $(AD)$ in $M$ and $N$ respectively. Lines $(BM)$ and $(CN)$ intersects at $T$.
Prove that : $BT+CT \leq 2R$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=202788#202788][b][u]Problem 3 :[/u][/b][/url]

consider $n$ a positive integer $n>1$ and $p$ a prime number such that $p-1$ is divisible by $n$ and $n^3-1$ is divisible by $p$.
Prove that $4p-3$ is a perfect square.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30952][b][u]Problem 4 :[/u][/b][/url]

Let $a,b,c$ be positive real numbers such that $abc=8$. Prove that :
$\frac{a^2}{\sqrt{(1+a^3)(1+b^3)}} +\frac{b^2}{\sqrt{(1+b^3)(1+c^3)}} +\frac{c^2}{\sqrt{(1+c^3)(1+a^3)}} \geq \frac{4}{3}$

[size=150]Second Day[/size]
2nd April 2005
Time allowed : 4h

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=202791#202791][b][u]Problem 1 :[/u][/b][/url]

$a,b,c$ are positive real numbers different from $0$ and satisfy : $ab+bc+ca+2abc=1$
Prove that : $\frac{1}{4a+1} +\frac{1}{4b+1} +\frac{1}{4c+1} \ge 1$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=202793#202793][b][u]Problem 2 :[/u][/b][/url]

Find all the positive integers $a,b,c$ such that :
$a \leq b \leq c$
$gcd(a,b,c)=1$
$a^3+b^3+c^3$ is divisible by $a^2b$ , $b^2c$ and $c^2a$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=202796#202796][b][u]Problem 3 :[/u][/b][/url]

Find all the injective functions $f:N\rightarrow N$ satisfying :
$f(f(n)) \leq \frac{n+f(n)}{2}$ For all positive integer $n$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=202797#202797][b][u]Problem 4 :[/u][/b][/url]

let $ABCD$ be a square. Consider $M$ a point on segment $[BC]$ and $K$ on segment $[CD]$ such that $CM+MK+KC=2AB$. Find $\angle{MAK}$

[size=150]Third Day[/size]
3rd April 2005
Time allowed : 4h

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=202799#202799][b][u]Problem 1 :[/u][/b][/url]

Find all the functions $f:R\rightarrow R$ stisfying :
$f(xf(x)+f(y))=(f(x))^2+y$ For all real numbers $x,y$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=202802#202802][b][u]Problem 2 :[/u][/b][/url]

$a,b,c,d$ are positive integers different  from $0$ and stisfying :
$a<b \leq c <d$    and    $ad=bc$   and    $\sqrt{d}- \sqrt{a} <1$
Prove that $a$ is a perfect square.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=202800#202800][b][u]Problem 3 :[/u][/b][/url]

Given a triangle $ABC$ with $\angle{B}=30$ and $AC$ is the shortest side. Points $D$ and $E$ are on $BA$ and $BC$ respectively such that $DA=EC=AC$. Prove that $OI$ is perpendicular to $DE$, where $O$ and $I$ are the circumcentre and incentre respectively.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=202804#202804][b][u]Problem 4 :[/u][/b][/url]

Given $n$ a positive integer diffferent from $0$, consider the set $E=\{1,2, \ldots,3n\}$. We partionate $E$ into three sets $A$ , $B$ and $C$. each one contains $n$ elements.
Prove that there exist $(x,y,z)$ from $A*B*C$ such that one of the numbers $x,y,z$ is the sum of the two others.

i wonder wt if contestants have seen the questions before.

So all or almost all problems are already on forum... Yes?

[quote="erdos"]If I knew the forum a year before, I think I would have got the perfect score. But even though, I am qualified for the final selection tests for the IMO.  :lol:[/quote]

Maybe your trainers were on ML before you...  :D


geometry
3D geometry
sphere
geometry unsolved
3-Dimensional Geometry

On the surface of the unit sphere, $n$ points are provided which determine a convex polyhedra. We want to find all such positions of the points so that the polyhedra's volume has a maximum. For $n = 4$ it is clearly a tedrahedron. Determine the respective figures for $n = 5,6,7$.

if n=6 it should be an octahedron.
I'm not sure if 5 gives you a quadrangular pyramid, but it feels like it.


Prove(without using Dirichlet's Theorem) that if $a$ and $b$ are coprime,for each positive integer $m$ exist infinately many positive integers $k$ such that $ak+b$ and $m$ are relatively prime.

Let $m=tm'$ where $m'$ is largest positive integer such that $m'|m$ and $(m', a) = 1$. Then for any $k$ we have $(ak+b, t)=1$ since if $p$ is prime and $p|t$ then also $p|a$ (otherwise we could change $m'$ with $pm'$ and we would have $m' < pm'$). But of course $p \not | b$ because $(a, b)=1$. Since $(m', a)=1$ we can take $k \equiv \frac{1-b}{a}\mod{m'}$ and then: $ak+b \equiv 1-b+b \equiv 1 \mod{m'}$. So we are done.


calculus
integration
calculus computations

Write down a double iterated integral that represents the volume of the
solid under the surface $ \sqrt{x^3 \plus{} y^3}\minus{}z \plus{} 5 \equal{} 0$ and above the triangle with vertices (0, 0), (2, 0), and (0, 3).

Drawing a picture really helps.  After drawing a picture, we see that the triangle is bounded by $ x\equal{}0$ and $ x\equal{}2$ on the left and right, and $ y\equal{}0$ and $ y\equal{}\minus{}\frac{3}{2}x\plus{}3$ on the bottom and top.  We can also rearrange the surface to be $ f(x,y)\equal{}z\equal{}\sqrt{x^{3}\plus{}y^{3}}\plus{}5$.  So, as an iterated integral,

$ \int_{0}^{2}\int_{0}^{\minus{}\frac{3x}{2}\plus{}3}\sqrt{x^{3}\plus{}y^{3}}\plus{}5\mbox{ }dy\mbox{ }dx$

Evaluate that and its game over.


Radius of external circles' $ a, b, c$.
What is the radius of inner circle interms of $ a, b, c$?

If the given circles are exterior tangent to the circle $ C(r)$ and are interior tangent to the circle $ C(R)$ , then

$ r\equal{}\frac {p}{2\sqrt {mp}\plus{}n}$ and $ R\equal{}\frac {p}{2\sqrt {mp}\minus{}n}$ , where $ a$ , $ b$ , $ c$ are the roots of the equations $ x^3\minus{}mx^2\plus{}nx\minus{}p\equal{}0$ .

In this case $ \frac {1}{\sqrt a}$ , $ \frac {1}{\sqrt b}$ , $ \frac {1}{\sqrt c}$ are  the lengths of the sides of a triangle and $ \left(\frac 1R\plus{}\frac 1r\right)^2\equal{}\frac {16(a\plus{}b\plus{}c)}{abc}$ .


algebra
polynomial
analytic geometry
linear algebra
matrix

Find a quartic equation in standard form the runs through the points $ (1,6)$, $ (7,4)$, $ (8,4)$, $ (10,2)$ and $ (12,0)$.

$ y\equal{}ax^4\plus{}bx^3\plus{}cx^2\plus{}dx\plus{}e$

$ \begin{cases}a\plus{}b\plus{}c\plus{}d\plus{}e\equal{}6\\
2401a\plus{}343b\plus{}49c\plus{}7d\plus{}e\equal{}4\\
4096a\plus{}512b\plus{}64c\plus{}8d\plus{}e\equal{}4\\
10000a\plus{}1000b\plus{}100c\plus{}10d\plus{}e\equal{}2\\
20736a\plus{}1728b\plus{}144c\plus{}12d\plus{}e\equal{}0\end{cases}$

Now this is all calculation, I can't think of a less time-consuming way to do this.

[url=http://en.wikipedia.org/wiki/Newton_polynomial#Main_idea]Newton interpolation[/url] to the rescue!

I don't get this Newton interpolation thing...can somebody explain?

We can write the quartic we're looking for as:

$ 6\left(\frac {(x \minus{} 7)(x \minus{} 8)(x \minus{} 10)(x \minus{} 12)}{(1 \minus{} 7)(1 \minus{} 8)(1 \minus{} 10)(1 \minus{} 12)}\right) \plus{} 4\left(\frac {(x \minus{} 1)(x \minus{} 8)(x \minus{} 10)(x \minus{} 12)}{(7 \minus{} 1)(7 \minus{} 8)(7 \minus{} 10)(7 \minus{} 12)}\right) \plus{} 4\left(\frac {(x \minus{} 1)(x \minus{} 7)(x \minus{} 10)(x \minus{} 12)}{(8 \minus{} 1)(8 \minus{} 7)(8 \minus{} 10)(8 \minus{} 12)}\right) \plus{} 2\left(\frac {(x \minus{} 1)(x \minus{} 7)(x \minus{} 8)(x \minus{} 12)}{(10 \minus{} 1)(10 \minus{} 7)(10 \minus{} 8)(10 \minus{} 12)}\right) \plus{} 0\left(\frac {(x \minus{} 1)(x \minus{} 7)(x \minus{} 8)(x \minus{} 10)}{(12 \minus{} 1)(12 \minus{} 7)(12 \minus{} 8)(12 \minus{} 10)}\right)$

Do you see why?  (If not, see what happens when you plug in one of the five points given.  4 of the terms cancel because they contain a term like $ (8 \minus{} 8)$ on the top. On the fifth term, the top and bottom of the fraction cancel because there are the exact same terms on the top as on bottom, leaving 1 on the inside and the y coordinate on the outside.)  Since this is a quartic that works and there is only one quartic that works, we just need to expand it and collect the terms to find the polynomial being sought.  I will leave this to you.

(I think this is what t0rajir0u was talking about, but I'm not completely sure.)

EDIT: By the way, Dr. Graubner, I believe you left out a 4 in the numerator of $ e$: mathematica gives $ \frac{1444}{99}$.  But the rest of it checks out with mathematica.

hello, yes you have right $ e\equal{}\frac{1444}{99}$.
Sonnhard.

[quote="tenniskidperson3"](I think this is what t0rajir0u was talking about, but I'm not completely sure.)[/quote]
Nope; that's Lagrange interpolation.  Newton interpolation works like this:  write

$ y \equal{} a(x \minus{} 7)(x \minus{} 8)(x \minus{} 10)(x \minus{} 12) \plus{} b(x \minus{} 8)(x \minus{} 10)(x \minus{} 12) \plus{} c(x \minus{} 10)(x \minus{} 12) \plus{} d(x \minus{} 12) \plus{} e$.

The initial conditions then give the [i]triangular[/i] system

$ 4158a \minus{} 693b \plus{} 99c \minus{} 11d \plus{} e \equal{} 6$
$ \minus{}15b \plus{} 15c \minus{} 5d \plus{} e \equal{} 4$
$ 8c \minus{} 4d \plus{} e \equal{} 4$
$ \minus{}2d \plus{} e \equal{} 2$
$ e \equal{} 0$.

What Newton interpolation basically lets you do is perform row-reduction on the Vandermonde matrix you get by posing the interpolation naively, whereas Lagrange interpolation just inverts the whole thing, which is good for theoretical purposes but not optimal for recovering the original coefficients.

t0rajir0u, how did you figure out that triangular system?

Edit: Aha, I finally got it.  It wasn't making sense, but I messed around with the numbers.  Thank you though!

Substitute the given conditions into the polynomial I gave as I've written it.  Do you mean how would you think of writing the polynomial that way?  Well, the cop-out answer is "because it leads to a triangular matrix": it really justifies itself, but perhaps the discussion [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=178381]here[/url], [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=211560]here[/url], or [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=237669]here[/url] might help.  This method is sometimes known as the method of [url=http://en.wikipedia.org/wiki/Divided_differences]divided differences[/url].


Given x^3 - 10x^2 + ax + b = 0 and two if its roots are 5 and 2, find a and b.

[hide="Hint for alternate way to do it"]With Vieta's formulas, we know the coefficient of $x^{2}$ is the negative sum of the roots.[/hide]

Well, I suppose I shall actually solve the problem  :roll:

[hide]By Vieta's formulas, we know that the coefficient of $-x^{2}$ is equal to the sum of the roots.  Thus, we have $5+2+r=10\Rightarrow r=3$ and the polynomial is:

$(x-2)(x-5)(x-3)=x^{3}-10x^{2}+31x-30$

$a=31$ and $b=-30$.

[b]NOTE:[/b]  You could also simply use Vieta's formulas again to solve for $a$ and $b$ here again, given the roots.[/hide]

[quote="AstroPhys"]Does Veita's Theorem work for all polynomials, or just third degreed ones?[/quote]

All polynomials

you can jsut have call P(x)=(x-2)(x-5)(x-c) I dont care what c is and P(x) is that third degree polynomial
but once you plug x as 5 or 2, the equation becomes
5a+b=125
2a+b=32

solve.

[url=http://www.artofproblemsolving.com/Wiki/index.php/Vieta%27s_formulas]check the wiki[/url]

It works for all polynomials; otherwise, it wouldn't be very useful  :roll:

:lol: Thanks, it could still be useful if it were limited to third degreed polynomials  :P .


For which values of $ a, b, c, d$ is it the case that the polynomial
$ x^{6}-6x^{5}+ax^{4}+bx^{3}+cx^{2}+dx+1$ has only positive roots?


Firstly, I just want to say that this is not my problem. I found it on someone's blog here on AOPS (alas, I cannot remember who) and I really hope that this is not some form of stealing.

Well, it is just a math problem right? They are meant to be solved.

A teacher wishes to distribute 20 identical pieces of candy among 7 students, so that the number of pieces of candy for any two students differs by at most 2. In how many different ways can the teacher distribute the candy?

Im not sure if my solution is correct. Someone else want to solve it?

[hide="My Solution"]

First, we notice that no students can receive 0 pieces of candy, because if a student does receive 0 pieces, then the restrictions only allow the other students to have 2 pieces of candy each. This will not be enough to reach 20 candies.

Second, we also notice that no students can receive 1 piece of candy, because if a student does receive 1 piece, then the restrictions only allow the other students to have 3 pieces each. Again, this will not be enough.

We can create a small bijection that lets us arrange the candies in different orders, so that each place represents what each student will receive.

We have 4 cases that we must look at:

[b]Case 1[/b]: Exactly 1 student receives 2 candies. It follows that the rest can only receive 3.

Our sequence is $ 3333332$ and there are 7 ways to arrange this.

[b]Case 2[/b]: Exactly 2 students receive 2 candies. It follows that exactly 1 must also receive 4 and 4 receive 3.

Our sequence is $ 3333422$ which can be arranged $ \dfrac{7!}{4!2!} \equal{} 105$

[b]Case 3[/b]: Exactly 3 students receive 2 candies. It follows that exactly 2 must also receive 4 and 2 receive 3.

Our sequence is $ 3342422$ which can be arranged $ \dfrac{7!}{2!2!3!} \equal{} 210$

[b]Case 4[/b]: Exactly 4 students receive 2 candies. It follows that 3 must also receive 4.

Our sequence is $ 4242422$ which can be arranged $ \dfrac{7!}{3!4!} \equal{} 35$

Therefore, there are $ 7 \plus{} 105 \plus{} 210 \plus{} 35 \equal{} 357$ ways?
[/hide]

[quote="Aristocrat"]$ {7 \choose 1} \plus{} {7 \choose 4}{3 \choose 1} \plus{} {7 \choose 2}{5 \choose 2} \plus{} {7 \choose 3} \equal{} 7 \plus{} 105 \plus{} 210 \plus{} 35 \equal{} \boxed{357}$

I think the approach works.[/quote]

How did you solve it Aristocrat?


Let a,b,c be Pythagorean triples in the form:
$ a\equal{}r^2\minus{}s^2$
$ b\equal{}2rs$
$ c\equal{}r^2\minus{}s^2$

Aside from {3,4,5}, what is the next pythagorean triple such that $ a\equal{}b\plus{}1$?

well, assuming the $ a,b$ are interchangeable with $ 2rs$ and $ r^2 \minus{} s^2$, then we can be like since $ 2rs \equal{} b$, then $ r^2 \minus{} s^2 \equal{} b \pm 1\Longleftrightarrow r^2 \minus{} s^2 \equal{} 2rs \pm 1 \Longleftrightarrow (r \minus{} s)^2 \equal{} 2s^2 \pm 1$. From here we just need to solve the pell equation $ x^2 \minus{} 2y^2 \equal{} \pm 1$, where $ r \minus{} s \equal{} x, s \equal{} y$. However, since I do not know how to solve pell equations, I will just say that it is not difficult to brute force the two smallest solutions to this equation as $ (1,1), (3,2)$, (the $ 1,1,$ solution gives $ 3,4,5$) from which we get that $ r \equal{} 2, s \equal{} 5$, which gives the $ 20,21,29$ triple.


geometry
3D geometry
sphere
group theory
abstract algebra
calculus
integration

Hello all,

I'm supposed to describe the universal covering space of $ S^n \vee \mathbb{R} P^n$ for $ n \geq 2$.

Since isomorphism classes of covering spaces correspond bijectively with conjugacy classes of $ \pi_1(S^n \vee \mathbb{R}^n)$, I am looking for the class of covering spaces which correspond to the identity subgroup of this group (since universal covering spaces are simply connected).  By Van Kampen's Theorem, $ \pi_1(S^n \vee \mathbb{R} P^n) \cong \mathbb{Z}/2 \mathbb{Z}$.  There are only two conjugacy in $ \mathbb{Z}/2 \mathbb{Z}$, the whole thing and $ 0$.  The identity map $ S^n \vee \mathbb{R} P^n \rightarrow S^n \vee \mathbb{R} P^n$ is the covering map which corresponds to $ \mathbb{Z}/2 \mathbb{Z}$, so would the covering space be $ S^n \vee S^n$ where the covering map sends one $ S^n$ to the copy of $ S^n$ in $ S^n \vee \mathbb{R} P^n$, the wedge point to the wedge point, and the second copy of $ S^n$ to its image under the quotient map $ S^n \rightarrow \mathbb{R} P^n$?  This seems like it would be a covering space and its simply connected.  This sort of looks right to me; if I'm wrong, could someone point out where I went astray?

how many pre-images has a point in the sphere $ S^n$?

[hide="hint"]the answer is homotopic to $ S^n\vee S^n\vee S^n$[/hide]

Maybe it's enlightening to consider $ {\mathbb S}^2\vee{\mathbb S}^1$ first. There the universal cover is long string with a lamp attached at each integral point. Your example is similar; you can also think about a general description of the universal cover of $ X\vee Y$ in terms of the universal covers of $ X$ and $ Y$.


find all pairs of positive integers x,y
$(x^{2}+y)-(x+y^{2})=10$

Factorise the left hand side:

\[(x-y)(x+y-1)= 10\]

We know that $x-y < x+y-1$, and clearly $x > y$
So two cases

[list][*]$x-y=1$, then it follows that $(x,y) = (6,5)$
[*]$x-y=2$, then $(x,y)=(4,2)$[/list]

These are the only solutions...


trigonometry
algebra
highschoolmath
trig identities

In a triangle $ABC$, $\angle A = 2 \cdot \angle B$. Prove that $a^2 = b (b+c)$.

[quote="Rushil"][b]INMO 1992 Problem 1[/b]

In a triangle $ABC$ , $\angle A = 2 \times \angle B$. Prove that $a^2 = b (b+c)$[/quote]
Try to search for Double Angle Theorem.

$\frac{a}{\sin 2x} = \frac{b}{\sin x} = \frac{c}{\sin 3x}$ $\Longrightarrow$ $c = b(4\cos^2 x - 1)$ and $a = 2b\cos x$.
$b(b + c) = 4b^2\cos^2 x = a^2$

[quote="Andreas"]$\frac{a}{\sin 2x} = \frac{b}{\sin x} = \frac{c}{\sin 3x}$[/quote]

How did you get this one?

[quote="Rushil"]In a triangle $ ABC$, $ \angle A \equal{} 2 \cdot \angle B$. Prove that $ a^2 \equal{} b (b \plus{} c)$.[/quote]
Sorry for awakening an old topic, but here's a neat solution that no-one here mentioned already:
[hide="Solution"]
Consider the circle passing through $ A$ and tangent to $ BC$.  Let $ CA$ meet it again at $ C'$.  Denote
\[ \angle C'AB \equal{} 180 \minus{} 2\alpha\implies \angle CAB \equal{} 2\alpha\implies \angle ABC \equal{} \alpha\equal{} \angle AC'B\implies \angle ABC' \equal{} 180 \minus{} \angle BAC' \minus{} \angle AC'B \equal{} \alpha \equal{} \angle AC'B\implies AB \equal{} AC'
\]
Hence, $ b(b \plus{} c) \equal{} CA(AB \plus{} CA) \equal{} CA(CA \plus{} AC') \equal{} CA\cdot CC' \equal{} BC^2 \equal{} a^2$. [/hide]

Hi,here is a good one ! please try out! :)
[hide]
$ A = 2 B
\Leftrightarrow A-B=B
\Leftrightarrow \sin (A-B) = \sin B
\Leftrightarrow \sin (A-B)\sin (A+B)=\ sin B \ sin C
\Leftrightarrow \sin^{2} A -\sin^{2} B=\sin B\sin C
\Leftrightarrow (2R \sin A)^{2} -(2R \sin B)^{2} = (2R\sin B)(2R\sin C)
\Leftrightarrow a^{2}-b^{2} = bc
\Leftrightarrow a^{2}=b(b+c) $
[/hide]

[hide="Another solution"]

$\angle A=2\angle B\ \implies\ \triangle ABC\sim\triangle DAC$ , where

$D\in BC\ ,\ \angle BAD=\angle DAC\ \implies\ \frac c{DA}=\frac ab=\frac b{CD}$ .

But $CD=\frac {ab}{b+c}\ \implies\ \frac{ab}{b+c}=\frac {b^2}a\ \implies\ \boxed{\ a^2=b\cdot (b+c)\ }$ .
[/hide]

[hide="A pure geometric solution"]Produce $CA$ to $K$ such that $CK=(b+c)$.So, $AK=c=AB$. From that it follows that $\angle AKB=\angle ABK=\angle B$.So, we have $\triangle CBK\sim \triangle CAB\implies \frac{CK}{CB}=\frac{CB}{CA}\implies \frac{(b+c)}{b}=\frac{a}{b}\implies b(b+c)=a^2$[/hide]

Sorry for reviving a very old thread !
We have to prove a^2=b(b+c)
Rearranging the terms we get a/b=(b+c)/a
L.H.S:-
By sine rule, a/b=sinA/sinB=2cosB ...(A=2B)
R.H.S:-
By sine rule, (b+c)/a=(sinB+sinC)/sinA=(2sin2B*cosB)/sin2B=2cosB ...(A=2B)
Hence, L.H.S=R.H.S proved.

[hide=A bit different solution with the use of trigonometry][b]Proof:[/b] By the law of sines,
$\frac{a}{\sin 2x} = \frac{b}{\sin x} = \frac{c}{\sin 3x}=k$ (let)

[b]LHS:[/b] $k^{2}\sin^{2}2x=4k^{2}\sin^{2}x\cos^{2}x$

[b]RHS:[/b] $ksinx( sin3x+ sinx)=ksinx \cdot (2 \sin2x \cdot cosx)$.
Simplifying the RHS, we get $4k^{2}\sin^{2}x\cos^{2}x$. This expression is equal to the LHS. Hence proved.[/hide]

Suppose,$\angle B =b'$ ,$\implies \angle A =2b'$,
and,$\angle C =180-3b'$.

See that,$\frac{C}{\sin 3b'} =\frac{b}{\sin b'}$.

$\implies c=b(3-4\sin ^2 b')$.

$\implies c=b(2\cos2b'+1)$.

$\implies c^2 = bc+2bc \cos 2b'$.

Using cosine rule we have ,

$a^2=b^2+c^2 -2bc \cos 2b'$.

$\implies a^2=b^2 +bc$.

As desired.$\boxed{a^2 = b (b+c)}$

[quote=Rushil]In a triangle $ABC$, $\angle A = 2 \angle B$. Prove that $a^2 = b (b+c)$.[/quote]
$$\frac{a}{\sin 2B} = \frac{b}{\sin B}\Leftrightarrow  \frac{a}{ 2b}=cosB=\frac{c^2+a^2-b^2}{ 2ca}
\Leftrightarrow  a^2 = b (b+c)$$

In a triangle $ABC$, $a^2 = b (b+c)$. Prove that $\angle A = 2\angle B.$

[quote=BestChoice123][quote=mathmanman]That's just the [url=http://www.ies.co.jp/math/java/trig/seigen/seigen.html]law of sines[/url] ;)[/quote]

oops that link failed :([/quote]
[url=https://artofproblemsolving.com/wiki/index.php/Law_of_Sines]Law_of_Sines[/url]

We know that,
$a/sinA=b/sinB=c/sinC$
$a/sin2x=b /sinx=c/sin(180^{\circ{}}-3x)$
$a/2cosx=b$
$c/sin3x=b/sinx$
$c/(3sinx-4sin^3x)=b/sinx$
$c/(3-4sin^2x)=b$
$b(b+c)=b^2+bc=\frac{a^2}{4cos^2x}+\frac{a}{2cosx}(3-4sin^2x)\frac{a}{2cosx}=\frac{a^2}{4 cos^2x}4(1-sin^2x)=\frac{a^2}{4 cos^2x}4cos^2x=a^2$.

I was just about the bash thing with trigonometry. But here is a geometric solution I guess.

[i]Solution:[/i] Denote $D$ to be the foot of angle bisector of $\angle BAC$ on side $BC$. Now it is easy to notice that $\angle CAD = \angle ABD$. So we can deduce that $AC$ is tangent to $\odot(ADB)$. Applying [i]Power of a Point[/i] along with [i]Angle Bisector Theorem[/i] we get
\begin{align*}
b^2= \frac{ab}{b+c} \cdot a \\
\implies a^2= b(b+c)
\end{align*}
which proves the desired. $\blacksquare$

Let angle B=x
Then , angle A=2x
By sine rule,
a/ b=sin 2x/ sin x
cos x =a/2b
cos x=c²+a²-b²/2ac
bc²+ba²-b²-a²c=0
(c-b){b(b+c)-a²}=0
If c=b
Then also a²=b(b+c)
If b(b+c)-a²=0
a²=b(b+c)
@ Krishijivi


algebra
polynomial
number theory unsolved
number theory

I have a claim . Please solve it...
$P(x)$ is an irreducible in $Z[x]$ nth degree polynomial , having integer coefficents.İs there a prime number  p such that coefficents of at least one of the polynomials
$P(x)$,$P(x+1)$...,$P(x+p-1)$ can be divided by p , except for leading coefficent, and $p^{2}$ does not divide it's constant term

http://www.math.niu.edu/~rusin/known-math/99/eisenstein

The link has the counterexample

$P(x) = x^{4}-x^{2}+1$


Find all pairs of positive integers $(a, b)$ such that $a - b$ is a prime and $ab$ is a perfect
square.

if $a-b$ is prime then $(a,b)=1$, and since $ab$ is a perfect square it follows that $a=m^2$ and $b=k^2$. But since $a-b=m^2-k^2=(m-k)(m+k)$ is prime we must have $m-k=1$, thus $a=(k+1)^2$ and $b=k^2$. Now $a-b=2k+1$ implies that any odd prime will be "good", so just take $a=(\frac{p+1}{2})^2$ and $b=(\frac{p-1}{2})^2$ where $p$ is an odd prime number.
The thing was typed fast so do not complain ;)

[quote="{x}"]if $a-b$ is prime then $(a,b)=1$

...

The thing was typed fast so do not complain ;)[/quote]

Well... the above statement is not entirely correct. I know it's a detail :)

We can also have $a=t\cdot p$ and $b=(t-1)\cdot p$. But in this case $p^2t(t-1)$ is never a prime, except for $t=1$

[quote="JohnConstantine"]But in this case $p^2t(t-1)$ is never a prime, except for $t=1$[/quote]

Hm? You mean a square, right? :D Details, details, I know ;)


Treething's favorite number times Joe's favorite number is equal to Sam's favorite number times Kevin's favorite number. This product is also equal to Ben's favorite number. If the sum of Treething's number and Joe's number is two greater than the sum of Sam's favorite number and Kevin's favorite number, find Treething's favorite number given that Kevin's favorite number plus Ben's favorite number plus Sam's favorite number is a four-digit number equal to the product of four distinct prime numbers.

Treething's favorite number is greater than Sam's favorite number, which is greater than Ben's favorite number, which is greater than Joe's favorite number.

oh right, they're all integers.

are there any restrictions on the numbers?

I'm going to take a wild guess. Is the answer [hide]epsilon[/hide]

oh, right, all the numbers are integers. guys you should be able to solve this without knowing any math at all if you've spent enough time around me

what does this mean?

[quote]the sum of Sam's favorite number times Kevin's favorite number[/quote]

edited, i meant "and" not "times". my problems have a lot of stupid errors.

Ok using Treething's little hint I'm thinking the number is [hide]69[/hide]

Wow this question is Weird.

Is this answer unique?

i'm around (intended answer)% sure it's unique

It's definitely [hide]69[/hide].Come on, the question boils down to "What is treething's favorite number?"

[quote="bpms"]What is a solution for a problem like this?[/quote]

i think treething just made up random facts

Even with the answer can you find other peoples numbers?


function
complex numbers
complex analysis
complex analysis unsolved

Let $p(z)$ be the Weierstrass function for the lattice $\mathbb{Z}+i\mathbb{Z}$. Show that $p(z)$ is real whenever twice the real part or twice the imaginary part of $z$ is an integer.

i've just had it for my exam; i think the following works: pair off the sum over the lattice summing in pairs of conjugate lattice points, then p(x+0i), p(0+iy) are not hard to see to be real as each summand is. to see that p(x+1/2i) is real, look at 2p(x+1/2i)=p(x+1/2i)+p(x-1/2i), use the same strategy for each of the sums and add them componentwise


function
Functional Analysis
real analysis
real analysis unsolved

if $ X$ is a banach space and $ Y$ be a normed linear space. consider sequence $ (F_n)$ of bounded linear maps from $ X$ to $ Y$ such that for each fixed $ x\epsilon X$, the sequence $ ((F_n)(x))$ is bounded then
(a) for each fixed $ x\epsilon X$ $ ((F_n)(x))$ is convergent in $ Y$
(b)for each foxed $ n\epsilon N$ the set $ \{F_n(x): x \epsilon X \}$ is bounded in $ Y$
(c) the seq $ (\parallel{}F_n\parallel{})$ is bounded in $ \mathbb{R}$
(d) the seq $ (F_n)$ is uniformly bounded on $ X$

(a) is not true.
Take $ X \equal{} Y \equal{} \mathbb R$, $ F_{2k} \equal{} I$ (i.e. $ x\mapsto 0$), $ F_{2k \plus{} 1} \equal{} 0$ (i.e. $ x\mapsto 0$). Obviously $ F_n(x)$ is not convergent in $ Y$ for $ x\neq 0$.
(b) is false because $ F_n(X)$ is a linear subspace
(c) and (d) are equal and follow form the uniform boundedness principle.


Suppose that a 6 x 6 square grid of unit squares (chessboard) is tiled by 1 x 2 rectangles (dominoes). Prove that it can be decomposed into two rectangles, tiled by disjoint subsets of the dominoes.

I don't understand what are " disjoint subsets of the dominoes" :(
and if possible, can you draw some examples pls? :)

the problem asks you to prove that there exists a horizontal or vertical [i]line[/i] which devides the board into two parts s.t it doesnt intersect any of the dominoes...
(note that our table has $ 5+5=10$ [i]lines[/i]!)
[hide="hint"]
prove it using contradiction,prove that if the above statement were wrong then any horizontal or vertical [i]line[/i] should intersect at least $ 2$ dominoes so there must be at least $ 20$ dominoes which is a contradiction...[/hide]


Ce zici de asta, Andrei?Nu-ti atrage atentia?E clar ca se foloseste inductia, dar e greu. :wink:

Inlocuieste 3 cu 2 in stanga. Challenge: gasiti o solutie ce nu depaseste 3 randuri!!!

Mai intai,daca x=y,totul e rezolvat.Presupunem atunci ca $x\geq y$,si membrul din mijloc este egal cu$\frac{x^{n+1}-y^{n+1}}{(x-y)(n+1)}$Folosind inductia,presupunem k propozitia $P_{n}: (\frac{x+y}2)^{n}\leq \frac{x^{n+1}-y^{n+1}}{(x-y)(n+1)}$ este adevarata,ai aratam ca $P_{n+1}$ este de asemenea adevarata.Si am descoperit ca daca urmatoarea afirmatie este adevarata,atunci punctul a este rezolvat:
$xy(x^{n}-y^{n})+\frac{x+y}{n+1}\cdot(x^{n+1}-y^{n+1})\leq x^{n+1}-y^{n+1}(*)$.Sper ca nu am facut vreo greseala de calcul..Acum ma straduiesc sa demonstrez $(*)$.

[quote="harazi"]Inlocuieste 3 cu 2 in stanga. Challenge: [b]gasiti o solutie ce nu depaseste 3 randuri!![/b]![/quote]
  eu am rezolvat-o in 12 randuri! :o

notez numerele cu $A,B,C$. avem de aratat ca $A\leq B\leq C$.
de aici incepe solutia :
notam cu $a_{k}$ ($0\leq k\leq s=[n/2]$). coeficientul lui $x^{k}y^{n-k}$ din dezvoltarea lui $A$. observam ca $a_{k}$ este si coeficientul lui $x^{n-k}y^{k}$. analog $b_{k}$,$c_{k}$. avem $a_{1}<a_{2}<\ldots<a_{s}$, $b_{1}=b_{2}=\ldots=b_{s}$ si $c_{2}=c_{3}=\ldots=c_{s}=0$.
$\sum a_{i}=\sum b_{i}=\sum c_{i}=1$.
notam $\alpha=(a_{1},a_{2},\ldots,a_{s})$, $\beta=(b_{1},b_{2},\ldots,b_{s})$, $\gamma=(c_{1},c_{2},\ldots,c_{s})$.
acum aplicam cebasev pentru a transforma termenii $x^{k}y^{n-k}$ in $x^{r}y^{n-r}$, unde $r<k$, si o facem in exact asa fel astfel incat din vectorul $\alpha$ sa obtinem vectorul $\beta$ apoi vectorul $\gamma$.

fara notatii, ies 3 randuri :)

Oricat de scurt ar fi, nu cred ca e rezolvare elementara de clasa a VII-a, sa zic.

da e foarte corecta :) si se incadreaza in programa de baraj  :P

bine, bine..
[hide="solutia mea pentru a)"]
  Cazul $\ n=2$ se verifica usor ca este adevarat.Sa consideram ca $\ n\geq\ 2$ si presupunem ca pentru $\ n$ inegalitatea de la $\ a)$ este adevarata.Vom demonstra ca si in cazul $\ n+1$ inegalitatea are loc.Trebuie sa demonstram ca $\left(\frac{x+y}{2}\right)^{n+1}\leq\frac{(x^{n}+...+y^{n})(x+y)}{2n+2}\leq\frac{x^{n+1}+...+y^{n+1}}{n+2}$, adica $\ (n+2)(x^{n+1}+...+y^{n+1})\leq\ (2n+2)(x^{n+1}+...+y^{^}{n+1})\Leftrightarrow\ 2(x^{n}y+x^{n-1}y+...+xy^{n-1})\leq\ 2n(x^{n+1}+y^{n+1}$.Aceasta din urma se sparge in $\ x^{n-m}y^{m}+x^{m}y^{n-m}\leq\ x^{n+1}+y^{n+1}$, care e echivalenta cu $\ (x^{n-m}-y^{m-n})(x^{m+1}-y^{m+1})\geq\ 0$.Cu asta demonstratia se incheie.
  Astfel, Andrei, nu mai era nevoie sa mai consideri ca $\ x\geq\ y$.Mai tarziu postez si solutia de la $\ b)$. :)  [/hide]

[hide="solutia mea pentru b)"]
  Pentru a doua inegalitate, se ajunge usurel ca $\ x^{n}y+xy^{n}\leq\ x^{n+1}+y^{n+1}$, adica $\ (x-y)(x^{n}-y^{n})\geq\ 0$.[/hide]