This looks impressive but it's not too difficult :

Prove that if a, b, c are positive reals such that abc = 1 then

(1 + ab)/c + (1 + bc)/a + (1 + ca)/b :ge: 18/(a + b + c).

You'll see, standard inequalities work here ! :)

My solution:

[hide]First, variant of Cauchy-Schwarz:

(a1)^2/x1 + (a2)^2/x2 + ... + (an)^2/xn :ge: (a1 + a2 + ... + an)^2/(x1 + x2 + ... + xn) for reals a1 to an, and positive reals x1 to xn.

From this it is obvious that 1/c^3 + 1/a^3 + 1/b^3 :ge: 9/(a^3 + b^3 + c^3).

We also want prove that ab^2/c^3 + bc^2/a^3 + ca^2/b^3 :ge: 9/(a^3 + b^3 + c^3). Use that variant again, we get

a^(1/2) * b + b^(1/2) * c + c^(1/2) * a :ge: 3, which is easily provable by AM-GM.[/hide]

That's my solution too. It was given in Hong Kong in 2000

Cheers :D


quadratics
parameterization
function
Support
probability
real analysis
real analysis unsolved

I have the following linear-quadratic dynamic control problem that has me stumped. I would be grateful for any kind of help.

Suppose the state, $ \pi$, is given by
$ \pi_{t\plus{}1}\equal{}\lambda y_{t}$
where $ y$ is another state variable and $ \lambda$ is a positive parameter.
$ y$ evolves according to
$ y_{t}\equal{}\minus{}\gamma (\phi i_{t}\plus{}(1\minus{}\phi)i_{t\minus{}1})\plus{}e_{t}$
where $ \gamma$ is a positive parameter, $ \phi$ is a parameter that lies between zero and one, $ i$ is the control variable and $ e$ is an iid random variable with mean zero and some finite variance which can be observed by the policy maker but by no one else.
Suppose the policy maker wishes to minimize the following objective function:
$ V_{t}\equal{}\sum_{s\equal{}t}^{\infty}\beta^{s\minus{}t}L_{s}$
with
$ L_{s}\equal{}\mu_{1}E[\pi_{s}]^{2}\plus{}\theta_{1}Var[\pi_{s}]\plus{}\mu_{2}E[y_{s}]^{2}\plus{}\theta_{2}Var[y_{s}]$
where $ \beta$ is a discount factor that lies strictly between zero and one and the $ \mu$'s and $ \theta$'s are positive parameters and $ E$ and $ Var$ refer to the mathematical expectation and variance, conditional on all information up to and including period $ t\minus{}1$, respectively.
Since this is a linear-quadratic problem, the policy maker's optimal choice of instrument $ i$ will be linear:
$ i_{t}\equal{}\Lambda_{1}\plus{}\Lambda_{2}e_{t}$
where the $ \Lambda$'s are yet to be determined. Because the current state also directly depends on the previous period's control, it must be the case that
$ \Lambda_{1}\equal{}\alpha i_{t\minus{}1}$
with $ \alpha$ yet to be determined.
So the question is:
1. How do I set this problem up properly? (preferably as a dynamic programming problem but using an intertemporal Lagrangean is also okay)
2. What are the solutions for $ \Lambda_{1}$, $ \Lambda_{2}$, and $ \alpha$?
As a final twist, what will the solution be if one assumes that $ \phi$ is an iid random variable with support on $ [0,1]$, a mean between zero and one and a finite variance?
Any help would be greatly appreciated.

[quote]
Suppose the policy maker wishes to minimize the following objective function:
$ V_{t}\equal{}\sum_{s\equal{}t}^{\infty}\beta^{s\minus{}t}L_{s}$
[/quote]
I'm not a specialist in probability, but this phrase seems to raise a red flag: $ V_t$ is a family of functions (indexed by $ t$), not a single function, so which one of them do you want to minimize?

$ V_{t}$ is not a family of functions. It is simply the present discounted value of all current and future objective functions ($ L_{t}, L_{t\plus{}1}, L_{t\plus{}2},$ etc.)

OK, but still $ V_t$ depends on $ t$ and I do not see why it cannot happen that the optimal choice of the policy that minimizes $ V_t$ (for that we need to choose $ i_t, i_{t\plus{}1},\dots$) will fail to minimize $ V_{t\plus{}1}$ :? Perhaps, I'm just missing something obvious, but, as I said, I'm not an expert in such questions, so I'm just trying to figure out what exactly is asked...

[quote="fedja"]OK, but still $ V_t$ depends on $ t$ and I do not see why it cannot happen that the optimal choice of the policy that minimizes $ V_t$ (for that we need to choose $ i_t, i_{t \plus{} 1},\dots$) will fail to minimize $ V_{t \plus{} 1}$ :? Perhaps, I'm just missing something obvious, but, as I said, I'm not an expert in such questions, so I'm just trying to figure out what exactly is asked...[/quote]
The t subscript merely points out that we are considering the problem from the perspective of the current period. $ V$ does NOT depend directly on time. The optimization problem is recursive (which is what dynamic programming is all about), so tomorrow's problem will be the same as today's problem, at least in expectations.
However, maybe I'm being daft and missing your point...

OK, let's forget this for a while and let me ask another stupid question: who prohibits you to put $ i_t\equal{}\minus{}\frac{1\minus{}\phi}\phi i_{t\minus{}1}\plus{}\frac 1{\gamma\phi}e_t$, which makes $ y_t$ and, hence $ \pi_{t\plus{}1}$ identically $ 0$?


Hey I set my "Entry" permission to registered users a long time ago, but it seems it isn't working and people can't contribute to my blog...

Hmm, apparently you currently do not have any contributors listed?

"Registered Users" means any registered user can contribute and hence there is no need for a contributer status. Or atleast, I think it does...  :|

Yeah, I believe entry refers to posting a comment, not contributing itself.

[quote="Anaxerzia"][quote="#H34N1"]That's a misconception. The "contribute" actually means "reply"[/quote]

[quote="Blog Option"]Entry Permission:

Select the default type of permission for your entries. This can be changed each time you post an entry[/quote]

If entry does indeed mean reply, then they should at least change that text.[/quote]

Oh no that means who can view your entry

View means who can view your blog as a whole

ANd there's replies for who can reply

The Entry and Reply permissions in the Blog Options are the [b]defaults[/b] for each time you input an entry. Each particular entry may have different permissions (like you can make a private entry, or you can make guest viewable and comentable entries).

[quote="ccy"]Mmm perhaps having bots count as registered users defeats the purpose...?[/quote]The SEO bots can't post stuff.

[quote="#H34N1"][quote="Valentin Vornicu"][quote="Anaxerzia"]I think  he is saying that one can simply go to the search engine and find the cache of the blog, and anyone can view it.[/quote]If you allow viewing for registered users but not guests. If someone is willing to go to that length to find out what is in someone's blog that is AVAILABLE for registered users, why not simply register an account? :)[/quote]

He might mean that it may be used for malicioius purposes. Like spoofing identities etc[/quote]Anyone can spoof Google bot. Doesn't mean it will be useful to something :)

I've changed the texts for the options so that it's more clear what every option does.


trigonometry
inequalities
vector
inequalities proposed

Given reals $ a_i ,b_i,c_i$ $ i\equal{}1,2...n$
and$ \sum a_{i}^2\equal{}1$ $ \sum b_{i}^2\equal{}1$ $ \sum c_{i}^2\equal{}1$ $ \sum b_ic_i\equal{}0$
Prove $ (\sum a_ib_i)^2\plus{}(\sum a_ic_i)^2 \le 1$

[quote="hxy09"]Given reals $ a_i ,b_i,c_i$ $ i \equal{} 1,2...n$
and$ \sum a_{i}^2 \equal{} 1$ $ \sum b_{i}^2 \equal{} 1$ $ \sum c_{i}^2 \equal{} 1$ $ \sum b_ic_i \equal{} 0$
Prove $ (\sum a_ib_i)^2 \plus{} (\sum a_ic_i)^2 \le 1$[/quote]
Because for $ 0\leq\alpha\leq\frac{\pi}{2},$ $ 0\leq\beta\leq\frac{\pi}{2}$ such that $ \alpha\plus{}\beta\geq\frac{\pi}{2}$ holds: $ \cos^2\alpha\plus{}\cos^2\beta\leq1.$ :maybe:

Can you give details,arqady?Your solution seems to be different from what I knew :)

Let $ \vec a(a_1,a_2,...,a_n),$ $ \vec b(b_1,b_2,...,b_n),$ $ \vec c(c_1,c_2,...,c_n),$ $ \measuredangle(\vec a,\vec b)\equal{}\alpha$ and $ \measuredangle(\vec a,\vec c)\equal{}\beta.$
Hence, $ \measuredangle(\vec b,\vec c)\equal{}90^{\circ}$ and we can assume $ \{\alpha,\beta\}\subset[0^{\circ},90^{\circ}].$
But $ \alpha\plus{}\beta\geq90^{\circ}$ for trihedral corner $ SABC,$ where $ \vec{SA}\equal{}\vec a,$ $ \vec{SB}\equal{}\vec b$ and $ \vec{SC}\equal{}\vec c.$
Thus, we need to prove that $ \cos^2\alpha\plus{}\cos^2\beta\leq1,$ which is obvious.

[quote="arqady"]Let $ \vec a(a_1,a_2,...,a_n),$ $ \vec b(b_1,b_2,...,b_n),$ $ \vec c(c_1,c_2,...,c_n),$ $ \measuredangle(\vec a,\vec b) \equal{} \alpha$ and $ \measuredangle(\vec a,\vec c) \equal{} \beta.$
Hence, $ \measuredangle(\vec b,\vec c) \equal{} 90^{\circ}$ and we can assume $ \{\alpha,\beta\}\subset[0^{\circ},90^{\circ}].$
But $ \alpha \plus{} \beta\geq90^{\circ}$ for trihedral corner $ SABC,$ where $ \vec{SA} \equal{} \vec a,$ $ \vec{SB} \equal{} \vec b$ and $ \vec{SC} \equal{} \vec c.$
Thus, we need to prove that $ \cos^2\alpha \plus{} \cos^2\beta\leq1,$ which is obvious.[/quote]
Very [b]nice[/b] your proof :lol: Thank you very much!

[quote="hxy09"]Given reals $ a_i ,b_i,c_i$ $ i \equal{} 1,2...n$
and$ \sum a_{i}^2 \equal{} 1$ $ \sum b_{i}^2 \equal{} 1$ $ \sum c_{i}^2 \equal{} 1$ $ \sum b_ic_i \equal{} 0$
Prove $ (\sum a_ib_i)^2 \plus{} (\sum a_ic_i)^2 \le 1$[/quote]
This one is similar to a nice one due to Cezar Lupu:

If $ a_i,b_i$ are real numbers such that $ \sum^n_{i\equal{}1}a^2_i\equal{}\sum^n_{i\equal{}1}b^2_i\equal{}1$ and $ \sum^n_{i\equal{}1}a_ib_i\equal{}0,$ then
$ \left( \sum^n_{i\equal{}1}a_i\right)^2\plus{}\left(\sum^n_{i\equal{}1}b_i\right)^2 \le n.$

Both inequalities can be proved by the same way and we only need to use Cauchy-Schwarz (once time only). Can someone figure it out?

[quote="can_hang2007"][quote="hxy09"]Given reals $ a_i ,b_i,c_i$ $ i \equal{} 1,2...n$
and$ \sum a_{i}^2 \equal{} 1$ $ \sum b_{i}^2 \equal{} 1$ $ \sum c_{i}^2 \equal{} 1$ $ \sum b_ic_i \equal{} 0$
Prove $ (\sum a_ib_i)^2 \plus{} (\sum a_ic_i)^2 \le 1$[/quote]
This one is similar to a nice one due to Cezar Lupu:

If $ a_i,b_i$ are real numbers such that $ \sum^n_{i \equal{} 1}a^2_i \equal{} \sum^n_{i \equal{} 1}b^2_i \equal{} 1$ and $ \sum^n_{i \equal{} 1}a_ib_i \equal{} 0,$ then
$ \left( \sum^n_{i \equal{} 1}a_i\right)^2 \plus{} \left(\sum^n_{i \equal{} 1}b_i\right)^2 \le n.$

Both inequalities can be proved by the same way and we only need to use Cauchy-Schwarz (once time only). Can someone figure it out?[/quote]
To see the Cauchy-Schwarz solution of this inequality, view [b]Problem 5.15[/b], page 241 of new book:

Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, [url=http://canhang2007.wordpress.com/2009/12/08/new-book-inequalities-with-beautiful-solutions-v-cirtoaje-v-q-b-can-t-q-anh/][b][i]Inequalities with Beautiful Solutions[/i][/b][/url], GIL publishing house 2009.

[quote="can_hang2007"][quote="hxy09"]Given reals $ a_i ,b_i,c_i$ $ i \equal{} 1,2...n$
and$ \sum a_{i}^2 \equal{} 1$ $ \sum b_{i}^2 \equal{} 1$ $ \sum c_{i}^2 \equal{} 1$ $ \sum b_ic_i \equal{} 0$
Prove $ (\sum a_ib_i)^2 \plus{} (\sum a_ic_i)^2 \le 1$[/quote]
This one is similar to a nice one due to Cezar Lupu:

If $ a_i,b_i$ are real numbers such that $ \sum^n_{i \equal{} 1}a^2_i \equal{} \sum^n_{i \equal{} 1}b^2_i \equal{} 1$ and $ \sum^n_{i \equal{} 1}a_ib_i \equal{} 0,$ then
$ \left( \sum^n_{i \equal{} 1}a_i\right)^2 \plus{} \left(\sum^n_{i \equal{} 1}b_i\right)^2 \le n.$

Both inequalities can be proved by the same way and we only need to use Cauchy-Schwarz (once time only). Can someone figure it out?[/quote]

Generally, for vectors $ \vec a, \vec b, \vec c\in\mathbb R^n$, we can write the 'spherical triangle inequality' $ \alpha \plus{} \beta\ge \gamma$ (as implied by Arqady) as
\[ \cos^2\alpha \plus{} \cos^2\beta \plus{} \cos^2\gamma\le1 \plus{} 2\cos\alpha\cos\beta \cos\gamma.\]
Equivalently
\[ \sum\frac { (\vec a\cdot\vec b)^2}{\vec a^2\vec b^2}\le1 \plus{} 2\frac {(\vec a\cdot\vec b) (\vec b\cdot\vec c) (\vec c\cdot\vec a)}{\vec a^2\vec b^2\vec c^2},\]
or
\[ \sum { {\vec c^2}(\vec a\cdot\vec b)^2}\le \vec a^2\vec b^2\vec c^2 \plus{} 2(\vec a\cdot\vec b) (\vec b\cdot\vec c) (\vec c\cdot\vec a).\]
:lol:

I’m still searching how to prove this directly with C-S. :huh:
 And I don’t think there is anything equivalent for more than 3 vectors.

[quote="can_hang2007"][quote="hxy09"]Given reals $ a_i ,b_i,c_i$ $ i \equal{} 1,2...n$
and$ \sum a_{i}^2 \equal{} 1$ $ \sum b_{i}^2 \equal{} 1$ $ \sum c_{i}^2 \equal{} 1$ $ \sum b_ic_i \equal{} 0$
Prove $ (\sum a_ib_i)^2 \plus{} (\sum a_ic_i)^2 \le 1$[/quote]
Let me post now the solution for this inequality:

Setting $ A \equal{} \sum^n_{i \equal{} 1}a_ib_i$ and $ B \equal{} \sum^n_{i \equal{} 1}a_ic_i.$ We have
$ A^2 \plus{} B^2 \equal{} A\sum^n_{i \equal{} 1}a_ib_i \plus{} B\sum^n_{i \equal{} 1}a_ic_i \equal{} \sum^n_{i \equal{} 1}a_i(Ab_i \plus{} Bc_i).$
By the Cauchy-Schwarz Inequality, we have
$ \begin{aligned} \left[\sum^n_{i \equal{} 1}a_i(Ab_i \plus{} Bc_i)\right]^2 & \le \left(\sum^n_{i \equal{} 1}a^2_i\right)\left[ \sum^n_{i \equal{} 1}(Ab_i \plus{} Bc_i)^2\right] \\
& \equal{} \sum^n_{i \equal{} 1} (A^2b^2_i \plus{} 2ABb_ic_i \plus{} B^2c^2_i) \\
& \equal{} A^2\sum^n_{i \equal{} 1}b^2_i \plus{} 2AB\sum^n_{i \equal{} 1} b_ic_i \plus{} B^2\sum^n_{i \equal{} 1}c^2_i \\
& \equal{} A^2 \plus{} B^2.\end{aligned}$
From this and the above identity, we deduce that
$ (A^2 \plus{} B^2)^2 \le A^2 \plus{} B^2,$
or
$ A^2 \plus{} B^2 \le 1.$
The proof is completed.

[b]Remark. [/b]We can consider this inequality as a generalization of the Romania TST 2007 Inequality.

By the way, in new book:

Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, [url=http://canhang2007.wordpress.com/2009/12/08/new-book-inequalities-with-beautiful-solutions-v-cirtoaje-v-q-b-can-t-q-anh/][b][i]Inequalities with Beautiful Solutions[/i][/b][/url], GIL publishing house 2009

We have presented many nice applications of Cauchy-Schwarz and AM-GM (the Romania TST 2007 is an example). You may find to read the book if you are interested.[/quote]
beautiful!
how to buy your book?

Dear friend,

You may contact GIL publishing house to purchase it.

Regards

[quote="can_hang2007"][quote="hxy09"]I collected three solutions for this nice problem
The first approach is based on an important lemma in complex number(solution form Mou Xiaosheng,who gained perfect marks in 2008 IMO)
The second is from arqady(mentioned in post 4)
The last is using Lagrange multiplier

Could you post your nice solution,dear can_hang2007 :lol:[/quote]
I think you uploaded the solutions from your handnote is ok. But please, at least write the solutions in English. To improve knowledge, we must learn from the proofs of the others. But in this case, it is hard to learn, because the solutions was written in Chinese! And the fact is, I dont know Chinese.[/quote]

Very sorry,my friend :blush: I once thought some Chinese in inequality-problems would not prevent reader from catching the main ideas......I have to apologize to all that have got blocked in my handwriting,and be sorry for creating obstacles for exchanging thoughts or solutions in this section or others,I will try to write solutions in English :)

[quote="hxy09"]Given reals $ a_i ,b_i,c_i$ $ i \equal{} 1,2...n$
and$ \sum a_{i}^2 \equal{} 1$ $ \sum b_{i}^2 \equal{} 1$ $ \sum c_{i}^2 \equal{} 1$ $ \sum b_ic_i \equal{} 0$
Prove $ (\sum a_ib_i)^2 \plus{} (\sum a_ic_i)^2 \le 1$[/quote]
I also have a amazing solution which is similar to can_hang's one:
$ (\sum (a_{i}b_{i}))^{2} \equal{} (\sum b_{i}(a_{i} \plus{} kc_{i}))^{2}\leq \sum b_{i}^{2}\cdot \sum (a_{i} \plus{} kc_{i})^{2} \equal{} 1 \plus{} k^{2} \plus{} 2k\sum a_{i}c_{i}$
so $ (\sum a_ib_i)^2 \plus{} (\sum a_ic_i)^2\leq 1 \plus{} k^{2} \plus{} 2k\sum a_{i}c_{i} \plus{} (\sum a_{i}c_{i})^{2} \equal{} (\sum a_{i}c_{i} \plus{} k)^{2} \plus{} 1 \equal{} 1$ for $ k \equal{} \minus{} \sum a_{i}c_{i}$
so the ineq holds.


ratio
geometry
geometric transformation
reflection
trigonometry

Source: 2004 Texas Team Selection Test

Line $k$ is drawn through the centroid of $\bigtriangleup ABC$. Points $B$ and $C$ are on the opposite side of line $k$ from point $A$. The perpendicular distances from $A$ and $B$ to $k$ are $10$ and $6$ respectively. Find the distance from point $C$ to $k$.

Draw an arbitrary line $k$ and the lines $a, b$ parallel to the line $k$, on the opposite sides of the line $k$ and with perpendicular distances from the line $k$ equal to $d_A = 10$ and $d_B = 6$, respectively. Select arbirary points $A, G, B$ on the parallel lines $a, k, b$, which are not collinear. Draw the lines $AG, BG$. Construct the points $M \in AG, N \n BG$ on the opposite sides of the lines $AG, BG$ on from the point $G$, such that $GM = 2 AG$, $GN = 2 BG$. Draw the lines $AM, BM$ intersecting at a point $C$. The cevians $AM, BN$ cut each other in the ratios $AG : GM = 2 : 1$, $BG : GN = 2 : 1$. Thus the point $G$ is the centroid of the triangle $\triangle ABC$ and the points $M, N$ are the midpoints of the segments $BC, AC$. By construction, the perpendicular distances of the vertices $A, B$ from the line $k$ passing through the centroid $G$ are $d_A = 10$ and $d_B = 6$, respectively.

Keeping the vertex $A$ and the centroid $G$ fixed, move the vertex $B$ on the line $b$. The point $M$ on the line $AG$ remains fixed and the vertex $C$ is a reflection of the vertex $B$ in the fixed midpoint $M$ of the segment $BC$. Thus the locus of point $C$ is a line parallel to the line $b$ and consequently, also to the line $k$. This means that the perpendicular distance $d_C$ of the vertex $C$ from the line $k$ does not change in this process, it is the same for any position of the vertex $B$ on the line $b$. Let the vertex $B$ approach the the intersection of the line $AG \equiv AM$ with the line $b$. The triangle $\triangle ABC$ degenerates into a line segment $BC$, while the perpendicular distances of all the points $A, B, C, M, N, G$ from the line $k$ remain fixed. Let $\phi$ be the angle formed by the fixed lines $AG \equiv AM$ and $a \parallel k \parallel b$. Now we can easily calculate the distances:

$AG \cos \phi = d_A = 10$

$BG \cos \phi = d_B = 6$

$AB \cos \phi = (AG + BG) \cos \phi = d_A + d_B = 16$

$BN \cos \phi = \frac 3 2 BG \cos \phi = \frac 3 2 d_B = 9$

$AN \cos \phi = (AB - BN) \cos \phi = d_A + d_B - \frac 3 2 d_B = d_A - \frac 1 2 d_B = 7$

$AC \cos \phi = 2 AN \cos \phi = 2d_A - d_B  = 14$

$BC \cos \phi = (AB - AC) \cos \phi = d_A + d_B - (2d_A - d_B) = 2 d_B - d_A = 2$

$d_C = CG \cos \phi = (BG - BC) \cos \phi = d_B - (2d_B - d_A) = d_A - d_B = 4$

The perpendicular distance of the vertex $C$ from the line $k$ is $d_C = 4$.

I personally think cosines are ok in the intermediate forum.  Although "SINs" are generally considered evil  :P .

[quote="pestich"]I told you before that all those COSs (and SINs as well) will get you in trouble.[/quote]

Well said Maj. I for one would definitely avoid long trig when possible.

Nice solutions.


induction
inequalities unsolved
inequalities

hi

what is the demonstration of the theorem of AM-GM (by induction)?

thank you

hello,here you will find several proofs
http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means
Sonnhard.

thank you ,
can explain me
why they put
\[ (m\minus{}n)\alpha \equal{}x_{n\plus{}1}\plus{}x_{n\plus{}2}\plus{}....\plus{}x_{m}\]

and I don't understand the last

I wait your answer


Add all the equations and get $\sum_{k=1}^{5}x_{k}=31$.  Using equations four and five, we have $x_{4}=17\wedge x_{5}=65\implies 3x_{4}+2x_{5}=181$.

[quote="Xaero"]Add all the equations up and then divide by 3 to get $\sum_{k=1}^{5}x_{k}=62$. Then using the fourth and fifth equations, we get $x_{4}=-14$ and $x_{5}=34$. Thus, $3(-14)+2(34)=\boxed{026}$[/quote]

you have to divide by 6 because you get..

$6x_{1}+6x_{2}+6x_{3}+6x_{4}+6x_{5}= 186$
proceed with Boxedexe's solution


ARML
vector
geometry
ratio
geometric sequence

Just to quell a potentially confusing situation later on, I would like to call attention to a typo in the contest packet given to each team in their final envelope. It is on Team Event #7. The printed answer is 7, whereas the actual answer is 10. (Don't worry, the actual answer was what was posted in the grading rooms.)

If you look at the printed solution, the discrepancy comes from a missing set of solutions to the equation $ c^2 \plus{} d^2 \equal{} 325$, namely $ 15^2 \plus{} 10^2 \equal{} 325$. This yields 3 more valid displacement vectors from $ A(11,5)$: $ (10,15), (15,10),$ and $ ( \minus{} 10,15)$.

On Tiebreaker 3, the solution listed in the packet seems to assume that region I's area must be smaller than region III's. (Or, equivalently, that the geometric sequence is an increasing one.) It also uses 136 in the first paragraph when it should say 2009.

As stated at UNLV, the answer to the problem is 1476 (for a ratio of $ \frac {p}{q} \equal{} \frac {1}{6}$).

I've been in touch with the head writer for ARML. The version of the contest that will appear on the ARML site in the next few days will have these corrections made.

So coaches, if you were planning on giving students copies of this year's contest, go by the pdf's that are soon to appear on arml.com  :)


The digit-sum of a number is the sum of its digits, so 81 has a digit-sum of $ 8 \plus{} 1 \equal{} 9$. What is the greatest possible two-digit integer with a digit-sum of 11?

You want the greatest 2-digit integer, so make the tens digit 9.  The other digit is the only other digit that is part of the sum, so the ones digit is 11-9=2.  The answer is [b]92.[/b]


Guolong Li (silver @ IMO 2005, honourable mention @ IMO 2006)
Jan Vonk (contestant @ IMO 2006 - jan @ ML)
Christophe Debry (perfect score at Flemish mathematical olympiad 2007, also at ML, forgot his username)
Augustin Lu Anh Khoa
Hoan Phung Bui
Pierre-Alain Jacqmin

[quote="Arne"]Christophe Debry [...] forgot his username[/quote] Kurt Gödel. :)

[quote="Arne"]silver @ IMO 2005, honourable mention @ IMO 2006
[/quote]

Something unusual :huh:  :roll: ,isn't it??

That wouldn't be unusual at all. An olympiad score is not a perfect measurement of ability or accomplishment. It's a small sample, based on a small number of questions, and a limited, finite grading scheme. There are many reasons for a person to see the answer to one problem and not another, and everyone has better days and worse days. There's no reason to expect that the same competitor repeating the competition will have his scores change in only one direction.

[quote="nmtruong1986"]There are two guys (or girls?) whose names are vietnamese names.[/quote]

They both have Vietnamese origins, but were born in Belgium and have Belgian nationality.  And they are both guys...  :P

Dear Arne,

 I wish good luck to your IMO team , to which you have given so much !

 Regards - Babis


Let a,b,c be positive real numbers, $ a\plus{}b\plus{}c\equal{}3$. Prove that:
$ \sum\frac{a\plus{}2b^{2}}{a\plus{}2c^{2}}\ge 3$

[quote="manlio"]Can you please give us a hint, NguyenDungTN?

Thank you very much.[/quote]
[hide="Hint"] By Cauchy-Schwarz inequality, we have:
$ \sum \frac {a + 2b^2}{a + 2c^2} \ge \frac {(a + b + c + 2(a^2 + b^2 + c^2))^2}{\sum (a + 2b^2)(a + 2c^2)}$
So we need to prove that:
${ (a + b + c + 2(a^2 + b^2 + c^2))^2 \ge 3\sum (a + 2b^2)(a + 2c^2)}$
$ \Leftrightarrow (a + b + c)^2 + 4(a + b + c)(a^2 + b^2 + c^2) + 4(a^2 + b^2 + c^2)^2 \ge 3[a^2 + b^2 + c^2 + 2\sum ab(a + b) + 4\sum a^2b^2]$
We have: $ 4(a^2 + b^2 + c^2)^2 \ge 12\sum a^2b^2$
We will prove:
$ (a + b + c)^2 + 4(a + b + c)(a^2 + b^2 + c^2) \ge 3 \sum a^2 + \ 6 \sum ab(a + b)$
$ \Leftrightarrow 2(a^3 + b^3 + c^3) + (ab + bc + ca) \ge a^2 + b^2 + c^2 + \sum ab(a + b)$
$ \Leftrightarrow 6(a^3 + b^3 + c^3) + (a + b + c)(ab + bc + ca) \ge (a + b + c)(a^2 + b^2 + c^2) + 3\sum ab(a + b)$
$ \Leftrightarrow 5(a^3 + b^3 + c^3) + 3abc \ge 3\sum ab(a + b)$
Which is clearly true from AM-GM and Schur inequality:
$ 2(a^3 + b^3 + c^3) \ge 6abc$
and: $ 3(a^3 + b^3 + c^3 + 3abc) \ge 3\sum ab(a + b)$
By same method, we can prove the following result:
[i]Let $ a,b,c > 0,a + b + c = 3, k\ge \frac {2}{3}$. Prove that:
$ \sum \frac {a + kb^2}{a + kc^2} \ge 3$[/i]
[/hide]


geometry
combinatorics proposed
combinatorics

In an $m\times n$ table there is a policeman in cell $(1,1)$, and there is a thief in cell $(i,j)$. A move is going from a cell to a neighbor (each cell has at most four neighbors). Thief makes the first move, then the policeman moves and ... For which $(i,j)$ the policeman can catch the thief?

I believe the policeman can catch the thief precisely when $i,j$ have the same parity. I only have time to post a sketch, but basically, it should be obvious that the policeman can "push the thief in a corner", making the [url=http://en.wikipedia.org/wiki/Taxicab_geometry]taxiab distance[/url] between the two equal to either $1$ or $2$. When $i,j$ have the same parity, the two will end up with the thief in a corner, say $(m,n)$, and the policeman in $(m-1,n-1)$, with the thief to move; clearly, the thief is lost in this situation :). Otherwise, if $i-j$ is odd, then after each of the policeman's moves the taxicab distacne between the two is odd, so the policeman can never catch the thief.

your argument seems to be right in the sense that the policeman can catch th thief iff thief is at (m,n) and policeman is at (m-1.n-1) or similar situations at corners. However consider abcissa of cells only. each move changes it by 1. so $ (m \minus{} i) and (m \minus{} 1 \minus{} 1)$ should have same parity, similarly (n-j) and (n-2) should have same parity. similarly for other corners,
Left upper: i should be even;  n-2 and n-j have same parity=> j is even.
Right Lower: j should be even ;  n-2 and n-i have same parity=> i is even.

You forgot to say that the policeman always wins if either m or n is equal to 1.

When m,n>1 we can colour the board in black and white like a chess board and notice that if they star in different colours the policeman never catches the thief. Using the argument of the taxiab distance we can see that the policeman can always decrease the distance between them, but the thief can increase it unless he is pushed in the corner, wich will happen because the policeman can always decrease its distance, now notice that since the distance is even in the beggining the distance between them will eventually be 2 and it's the thief's move. When he reaches a corner he has to decrease the distance and then since the distance is 1 and it's the policeman's move, we all know what happens next :)

So the iff condition, just like grobber said is the distance is even so it means that i-1+j-1 is even meaning that i+j is even meaning they have the same parity. They don't need to both be even.


inequalities AMC USA(J)MO USAMO trigonometry

[quote="Zhero"]Suppose $ x$, $ y$, and $ z$ are positive real numbers such that $ x \plus{} y \plus{} z \equal{} xyz$. Prove that $ \sqrt {x^2 \plus{} 1} \plus{} \sqrt {y^2 \plus{} 1} \plus{} \sqrt {z^2 \plus{} 1} \leq 2 \plus{} \frac {\sqrt {(x^2 \plus{} 1)(y^2 \plus{} 1)(z^2 \plus{} 1)}}{2}$.

(I believe that this inequality is true, though I'm not [i]entirely[/i] certain.)

EDIT: Mods, can you please move this to Pre-Olympiad (or possibly Olympiad)? My fingers slipped, and I posted it here. Sorry. :/[/quote]

what do you mean that you are not certain whether it is true or not?you created this problem? :huh:

No, this is a rip-off of USAMO 2001 #3. I performed a trig substitution, and messing around a bit, I ended up with this inequality. You may have to end up reversing the work I did:

[hide]e.g., setting $ x \equal{} \tan A$, $ y \equal{} \tan B$, and $ z \equal{} \tan C$, where $ A \plus{} B \plus{} C \equal{} \pi$.

EDIT: and $ A, B, C \leq \frac{\pi}{2}$[/hide]


$g(x)=\dfrac{1}{3}(x^3-4x)$

Determine the number of solutions of the equation:

$||g(x)|-1|=\dfrac{1}{2}$

i believe that it might be 6 solutions, but i could be wrong

[quote="YooSam"]$g(x)=\dfrac{1}{3}(x^3-4x)$

Determine the number of solutions of the equation:

$||g(x)-1|=\dfrac{1}{2}$[/quote]

I also believe that there will be 6 solutions, because you could have 3 solutions without the absolute value, and adding the absolute value would double the amount of solutions, making 6.

if you square it, then you get rid of the absolute value symbol, and that would have the lead term being 6, but the only consideration is if you have double roots, i suppose that you could factor it...

$\left|\dfrac{1}{3}x^3 - \dfrac{4}{3}x -1\right| = \dfrac{1}{2}$
$\left(\dfrac{1}{3}x^3 - \dfrac{4}{3}x -1\right)^2 -\left(\dfrac{1}{2}\right)^2= 0$
$\left(\dfrac{1}{3}x^3 - \dfrac{4}{3}x -\dfrac{3}{2}\right)\left(\dfrac{1}{3}x^3 - \dfrac{4}{3}x -\dfrac{1}{2}\right)=0$
i cannot say from that factorization that there are all unique roots, but from graphing them, i know that they are all unique so yes, the answer is 6 roots

[quote="10000th User"]What type of solutions? Complex or real? Also, why is there 2 || in front of the expression but only 1 | after the expression but before the equal sign?[/quote]

Oops, sorry, there is one more | after g(x). I fixed it already  :blush: . Thanks.

The question asks for real numbers. I think if it includes complex numbers, it will have 12 solutions. I have four equations with third-polynomial.

6 is not a right answer. I think you guys must be confused with only one absolute value sign. There are two as I had fixed it.


geometry
incenter
circumcircle
geometric transformation
reflection
geometry proposed

[color=darkred]Let $ ABC$ be a triangle with the incenter $ I$ and the circumcircle $ w\equal{} C(O,R)$. Denote

the reflection $ M$ of $ B$ w.r.t. the line $ CI$ and the reflection $ N$ of $ C$ w.r.t. the line $ BI$ .

Prove that $ OM^2\plus{}ab\equal{}ON^2\plus{}ac\equal{}a^2\plus{}R^2$ and $ MN\perp OI$ .[/color]

Let $ P$ be the projection of $ O$ onto $ AC$.
Then: $ R^2 \minus{} OM^2 \equal{} CP^2 \minus{} PM^2 \equal{} (CP \minus{} PM) (CP \plus{} PM) \equal{} CM \cdot AM \equal{} a(b\minus{}a)$.
Similar argument gives the first 2 equality.

For the fact that $ MN \perp OI$, we need to show that: $ OM^2 \minus{} ON^2 \equal{} IM^2 \minus{} IN^2$. This in turn is equivalent to showing: $ a(c\minus{}b) \equal{} IB^2 \minus{} IC^2$ (because of reflection, $ IB \equal{} IM$ and $ IC \equal{} IN$).

This is easy for now: $ RHS \equal{} (p\minus{}b)^2 \minus{} (p\minus{}c)^2$ , ...

[b] Comment: [/b] Without the first part, I don't know how I would manage to show that $ MN \perp OI$.

for the second:

Call $ E,\ F$ the projection of $ O$ on $ AB,\,AC$; $ H,\,K$ are the projection of $ I$ on $ AB,\,AC$, then:
\[ \overrightarrow{OI}\cdot\overrightarrow{MN} \equal{} \overrightarrow{OI}\left(\overrightarrow{AN} \minus{} \overrightarrow{AM}\right) \equal{} \overline{FK}\cdot\overline{AN} \minus{} \overline{EH}\cdot\overline{AM} \equal{}
\]

\[ \equal{} \left(\overline{AK} \minus{} \overline{AF}\right)\overline{AN} \minus{} \left(\overline{AH} \minus{} \overline{AE}\right)\overline{AM} \equal{}
\]

\[ \equal{} \left(p \minus{} a \minus{} \frac {b}{2}\right)(b \minus{} a) \minus{} \left(p \minus{} a \minus{} \frac {c}{2}\right)(c \minus{} a) \equal{} 0
\]
Then $ OI\perp MN$.

[b][size=100][color=DarkBlue]LEMMA. - A triangle $ \bigtriangleup ABC$ with circumcircle $ (O)$ is given and let $ E,\ F$ be, the tangency points of its incircle $ (I),$ to $ AC,\ AB,$ respectively. We draw the lines through the points $ B'\equiv AC\cap BI,\ C'\equiv AB\cap CI$ and tangent to $ (I)$ at points $ E',\ F',$ respectively. Prove that the points $ O,\ I,\ K\equiv EF'\cap E'F$ are collinear, where $ O,\ I$ are the circumcenter, incenter respectively, of $ \bigtriangleup ABC.$[/color][/size][/b]

$ \bullet$ In our problem now, it is obvious that the line segments $ B'E',\ C'F',$ pass through the points $ N,\ M$ respectively and the point $ K\equiv EF'\cap E'F,$ is the intersection point of the polars of $ M,\ N,$ with respect to $ (I).$

So, we have that the points $ M,\ N,$ lie on the polar of $ K$ with respect to $ (I)$ and then we conclude that $ MN\perp IK.$

Hence, based on the above [b][size=100]Lemma[/size][/b], we conclude that $ MN\perp OI$ and the proof is completed.

Kostas Vittas.

PS. I will post here later, the proof of the [b][size=100]Lemma[/size][/b] I have in mind.

[quote="mathangel"]I 'm sorry but what we want is $ MN \perp OI$, not $ EF \perp OI$. I sketched the figure and found that $ EF \not\perp OI$. I guess something 's missing in your solution.[/quote]ù

clearly I wanted to say $ MN \perp OI$ as fallow clearly by the first step...

[quote="vittasko"][b][size=100][color=DarkBlue]LEMMA. - A triangle $ \bigtriangleup ABC$ with circumcircle $ (O)$ is given and let $ E,\ F$ be, the tangency points of its incircle $ (I),$ to $ AC,\ AB,$ respectively. We draw the lines through the points $ B'\equiv AC\cap BI,\ C'\equiv AB\cap CI$ and tangent to $ (I)$ at points $ E',\ F',$ respectively. Prove that the points $ O,\ I,\ K\equiv EF'\cap E'F$ are collinear, where $ O,\ I$ are the circumcenter, incenter respectively, of $ \bigtriangleup ABC.$[/color][/size][/b][/quote]
[b][size=100]PROOF.[/size][/b] - Let be the points $ P\equiv BC\cap C'F',$ $ Q\equiv BC\cap B'E',$ $ M\equiv CC'\cap AP$ and $ N\equiv BB'\cap AQ.$

The point $ M$ is the midpoint of the segment $ AP$ and applying the [b][size=100]Newton theorem[/size][/b], in the circumscribed quadrilateral $ ACPC',$
we have that the line segment $ EF'$ passes through the point $ M.$

Similarly, the line segment $ E'F$ passes through the point $ N,$ as the midpoint of the segment $ AQ$ and so, we conclude that $ \boxed{MN\parallel BC}$ $ ,(1)$

Let be the points $ X\equiv CC'\cap OF',\ Z\equiv BB'\cap OE'$ and we will prove that $ E'F'\parallel BC$ and $ XZ\parallel BC.$

It is easy to show that $ IF'\parallel OC$ $ ,(2),$   from $ \angle C'IF' \equal{} \angle C'IF \equal{} \angle C'CS \equal{} \angle C'CO \equal{} \frac {\angle A \minus{} \angle B}{2},$

where $ S$ is the orthogonal projection of $ C$ on $ AB$ and similarly, we have that $ IE'\parallel OB$ $ ,(3)$

From $ (2),$ $ (3)$ and because of $ IE' \equal{} IF',\ OB \equal{} OC,$ we conclude that $ \boxed{E'F'\parallel BC}$ $ ,(4)$

From $ (2)$ $ \Longrightarrow$ $ \frac {XI}{XC} \equal{} \frac {IF'}{CO}$ $ ,(5)$

From $ (3)$ $ \Longrightarrow$ $ \frac {ZI}{ZB} \equal{} \frac {IE'}{BO}$ $ ,(6)$

From $ (5),$ $ (6)$ $ \Longrightarrow$ $ \frac {XI}{XC} \equal{} \frac {ZI}{ZB}$ $ \Longrightarrow$ $ \boxed{XZ\parallel BC}$ $ ,(7)$

From $ (1),$ $ (4),$ $ (7)$ $ \Longrightarrow$ $ XZ\parallel MN\parallel E'F'$ $ ,(8)$ and then, we have that the triangles $ \bigtriangleup MF'X,\ \bigtriangleup NE'Z,$ are perspective ( at infinity ).

Hence, based on the [b][size=100]Desarques theorem[/size][/b], we conclude that the points $ O\equiv F'X\cap E'Z,\ I\equiv MX\cap NZ,\ K\equiv MF'\cap NE',$ are collinear and the proof is completed.

Kostas Vittas.

[quote="limes123"]Very nice solution. Could you please show the proof of Newton's theorem? I can't find it...[/quote]
The follows proof is coming form the past and it is due to [b][size=100]L. Anne[/size][/b] ( see the schema [b][size=100]t=232923(a)[/size][/b] and consider $ ACPC',$ as an arbitrary circuscribed quadrilateral ).

$ \bullet$  Let be the point $ M\equiv CC'\cap EF'$ and we see that the triangles $ \bigtriangleup MEC,\ \bigtriangleup MF'C'$ have $ \angle EMC \equal{} \angle F'MC'$ $ ,(1)$ and $ \angle MEC \equal{} 180^{o} \minus{} \angle MEA \equal{} 180^{o} \minus{} \angle MF'C'$ $ ,(2)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ \frac {MC}{MC'} \equal{} \frac {EC}{F'C'}$ $ ,(3)$

We denote as $ D,$ the tangency point of the circle $ (I)$ to $ BC$ and let be the point $ M'\equiv CC'\cap DF.$

By the same way as before, from the triangles $ \bigtriangleup MDC,\ \bigtriangleup MFC',$ we have that $ \frac {M'C}{M'C'} \equal{} \frac {DC}{FC'}$ $ ,(4)$

From $ (3),$ $ (4)$ $ \Longrightarrow$ $ \frac {MC}{MC'} \equal{} \frac {M'C}{M'C'}$ $ \Longrightarrow$ $ M'\equiv M$ $ ,(5)$

From $ (5)$ we conclude that the point $ M\equiv EF'\cap DF,$ as the intersection point of the lines connecting the tangency points of $ (I),$ to the opposite side-segments of $ ACPC',$ lies on its diagonal $ CC'.$

By the same way, we can prove that the point $ M,$ lies also on the diagonal $ AP$ of $ ACPC'$ and so, we conclude that the line segments $ AP,\ CC',\ EF',\ DF,$ are concurrent at one point and we are done.

Kostas Vittas.

Dear Kostas and Mathlinkers,
your nice problem (and proof) comes from Charles Ange Laisant.
Can some one precise the reference?
Sincerely
Jean-Louis


calculus
integration
topology
trigonometry
complex numbers
complex analysis

I want to compute the integral of $\exp(\frac{1}{z})$ along a line from $1$ to $i$ followed by a line from $i$ to $-1$. Can someone help me finding a suitable homotopic path?

I don't think anything that simple will do here- we really need an antiderivative, and that antiderivative isn't an elementary function.

We're probably best off using a semicircle, since the best handle we have on $e^{\frac1x}$ is its Laurent series at zero.

But it is an exercise, and as a hint is given that I should use homotopy invariance of the path integral.

If we take a semicircle, we get $\int_0^\pi\exp(\cos{\phi} - i\sin{\phi})\cdot(-\sin{\phi} + i\cos{\phi})\,d\phi$, but this is also not elementary integrable...

Here's what the semicircle really gets us:
$e^{\frac1z}=\sum_{n=0}^\infty\frac1{z^n}\cdot\frac1{n!}$, so the desired integral is
$\int_0^\pi \sum_{n=0}^\infty e^{-int}\cdot\frac1{n!}\cdot ie^{it}\,dt= \sum_{n=0}^\infty\frac{i}{n!}\int_0^\pi e^{i(1-n)t}\,dt$ $=\sum_{n\neq1}\frac{i}{n!}\cdot\left[\frac{i}{n-1}e^{i(1-n)t}\right]_0^\pi+i\pi =i\pi+\sum_{n\text{ is even}}\frac2{(n-1)n!}$

That last sum is still nasty, but at least we've got something.

A numerical evaluation yields $-0.97165951887903052781 + i \pi$, so perhaps $-1 + i\pi$?

It's not 1. It's probably not expressible in closed form. The imaginary part is exactly $\pi$, but the real part is a lost cause for exact methods.


In a 5 by 5 grid, each of the 25 small squares measures 2 cm by 2 cm and is shaded. Five unshaded circles are then placed on top of the grid as shown. The area of the visible shaded region can be written in the form $ A\minus{}B\pi$ square cm. What is the value $ A\plus{}B$?

[asy]for(int i = 0; i < 5; ++i)
{
	for(int j = 0; j < 5; ++j)
	{
	filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,gray,linewidth(2));
	}
}

filldraw(circle((2.5,.5),.5),white,linewidth(2));
filldraw(circle((4.5,2.5),.5),white,linewidth(2));
filldraw(circle((2.5,4.5),.5),white,linewidth(2));
filldraw(circle((.5,2.5),.5),white,linewidth(2));

filldraw(circle((2.5,2.5),1.5),white,linewidth(2));[/asy]

[quote="GameBot"]In a 5 by 5 grid, each of the 25 small squares measures 2 cm by 2 cm and is shaded. Five unshaded circles are then placed on top of the grid as shown. The area of the visible shaded region can be written in the form $ A \minus{} B\pi$ square cm. What is the value $ A \plus{} B$?

[asy]for(int i = 0; i < 5; ++i)
{
	for(int j = 0; j < 5; ++j)
	{
	filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,gray,linewidth(2));
	}
}

filldraw(circle((2.5,.5),.5),white,linewidth(2));
filldraw(circle((4.5,2.5),.5),white,linewidth(2));
filldraw(circle((2.5,4.5),.5),white,linewidth(2));
filldraw(circle((.5,2.5),.5),white,linewidth(2));

filldraw(circle((2.5,2.5),1.5),white,linewidth(2));[/asy][/quote]

This is simply $ 100\minus{}9\pi\minus{}\pi\minus{}\pi\minus{}\pi\minus{}\pi\equal{}100\minus{}13\pi$

$ 100\plus{}13\equal{}113$.

I asked this because the picture didn't show up on FTW.


Problem 2
If $ x,y,z$ are positive real numbers with $ x,y,z<2$ and $ x^2\plus{}y^2\plus{}z^2\equal{}3$ prove that $ \frac{3}{2}<\frac{1\plus{}y^{2}}{x\plus{}2}\plus{}\frac{1\plus{}z^{2}}{y\plus{}2}\plus{}\frac{1\plus{}x^{2}}{z\plus{}2}<3$

[quote="balan razvan"]I've saw that you've posted 4 problems...these were from a contest?If yes...what contest?  :D[/quote]

From the Greek mathematical Olympiad 2008 (junior level :P )


Prove that if you multiply both sides of an inequality by -1, you have to reverse the sign.

$ a>b$

subtract $ a\plus{}b$ on both sides

$ \minus{}a<\minus{}b$

But now you must prove that when subtracting a+b, you must reverse  the sign.  :P

Nothing of the sort.  Subtracting $ a \plus{} b$ from both sides of the given inequality gives $ \minus{}b > \minus{}a$, which is equivalent to the desired statement.  And I hope you understand why subtracting $ a \plus{} b$ from both sides of an inequality does [b]not[/b] change the sign of the inequality.

The interpretation of this result is geometric:  plot $ a, b$ on the number line.  $ a > b$ means $ a$ is to the right of $ b$.  When you plot $ \minus{}a, \minus{}b$ instead, you'll be flipping $ a, b$ across the point $ 0$, which means that $ \minus{}a$ is now to the left of $ \minus{}b$.  This is easiest to see when $ a, b > 0$ but actually applies either way.

Suppose $ a>b$ and $ \minus{}a \geq \minus{}b$ for some $ (a,b)$. Adding, we get $ a\plus{}(\minus{}a) > b\plus{}(\minus{}b) \iff 0 > 0$, contradiction.

can't you do casework with this, it may seem dumb however,

rtp if a > b then -a < -b

case 1
b>0, a>0
case 2
b<0, a<0
case 3
b<0, a<0

Casework is not really necessary and obscures what's really going on.  $ a > b \Leftrightarrow a \plus{} c > b \plus{} c$ for all $ c$, so we can pick $ c$ sufficiently large so that $ a, b$ are both positive.  But as you have seen, picking $ c \equal{} \minus{}(a \plus{} b)$ is much more effective.  We can also pick $ c \equal{} \minus{} \frac{a \plus{} b}{2}$.

Whoops! Sorry about that earlier one, guys. My bad... :blush:


floor function
function
pigeonhole principle
quadratics
modular arithmetic
algebra
domain

[quote="Peter"]Let $A$ be the set of positive integers of the form $a^2 +2b^2$, where $a$ and $b$ are integers and $b \neq 0$. Show that if $p$ is a prime number and $p^2 \in A$, then $p \in A$.[/quote]

I was about to slay this using either the Thue theorem or properties of $\mathbb{Z}\left[\sqrt{-2}\right]$, as it is absolutely straightforward both of these ways, but then I noticed the "Romania 1997" in the source. Nice problem, I must say, would not some annoying casework mutilate the writeup.

[color=blue][b]Lemma 1.[/b] If $w$ is an integer, and if two coprime positive integers $x$ and $y$ are such that $xy$ is a $w$-th power of a positive integer, then both $x$ and $y$ are $w$-th powers of positive integers.[/color]

[i]Proof of Lemma 1.[/i] Let $x=p_1^{a_1}p_2^{a_2}...p_k^{a_k}$ be the prime factorization of $x$, where all $p_i$ are distinct primes and all $a_i$ are positive integers for $i\in\left\{1;\;2;\;...;\;k\right\}$. Let $y=q_1^{b_1}q_2^{b_2}...q_l^{b_l}$ be the prime factorization of $y$, where all $q_j$ are distinct primes and all $b_j$ are positive integers for $j\in\left\{1;\;2;\;...;\l\right\}$. Then, $p_i\neq q_j$ for every $i\in\left\{1;\;2;\;...;\;k\right\}$ and every $j\in\left\{1;\;2;\;...;\l\right\}$ (else, the numbers $x$ and $y$ were not coprime!), so that the primes $p_1$, $p_2$, ..., $p_k$, $q_1$, $q_2$, ..., $q_l$ are all distinct. Hence, the number $xy=p_1^{a_1}p_2^{a_2}...p_k^{a_k}\cdot q_1^{b_1}q_2^{b_2}...q_l^{b_l}$ has the prime factorization $xy=p_1^{a_1}p_2^{a_2}...p_k^{a_k}q_1^{b_1}q_2^{b_2}...q_l^{b_l}$ (this sounds stupid now, doesn't it?). But since $xy$ is a $w$-th power of a positive integer, each prime factor in the prime factorization of $xy$ must appear with an exponent divisible by $w$; hence, all integers $a_1$, $a_2$, ..., $a_k$, $b_1$, $b_2$, ..., $b_l$ are divisible by $w$. Therefore, the numbers $x=p_1^{a_1}p_2^{a_2}...p_k^{a_k}$ and $y=q_1^{b_1}q_2^{b_2}...q_l^{b_l}$ are $w$-th powers of positive integers (since each prime factor in their prime factorizations appears with an exponent divisible by $w$). Lemma 1 is proven.

Now let's solve the problem:

Since $p^2\in A$, there exist two integers $u$ and $v$ such that $p^2=u^2+2v^2$ and $v\neq 0$. We can WLOG assume that $u$ and $v$ are both nonnegative (else, we can substitute $-u$ for $u$ or $-v$ for $v$ without changing much). Since $v\neq 0$, the nonnegativity of $v$ yields that $v$ is actually positive.

The equation $p^2=u^2+2v^2$ becomes $p^2-u^2=2v^2$, what rewrites as $\left(p-u\right)\left(p+u\right)=2v^2$. Thus, $2\mid\left(p-u\right)\left(p+u\right)$. Since $2$ is prime, this yields that $2$ divides at least one of the integers $p-u$ and $p+u$. But if $2$ divides one of these integers $p-u$ and $p+u$, then $2$ also divides the other one, because $2$ divides the difference $\left(p+u\right)-\left(p-u\right)=2u$ of these integers. Thus, $2$ divides both integers $p-u$ and $p+u$. Hence, $\frac{p-u}{2}$ and $\frac{p+u}{2}$ are integers.

We have $p>u$, since $p^2=u^2+2v^2>u^2$ what is because $v$ is positive.

The prime $p$ is odd. This is because else we would have $p=2$, so that $4=2^2=p^2=u^2+2v^2$, and thus $4\geq 2v^2$, so that $2\geq v^2$, and thus $v\leq\sqrt2$, so that $v=1$ (since $v$ is a positive integer, and the only positive integer $\leq\sqrt2$ is $1$), so that $4=u^2+2v^2=u^2+2\cdot 1^2=u^2+2$ yields $2=u^2$, what is impossible since $u$ is an integer and $2$ is not a square.

Since $p$ is odd, $p^2$ must also be odd. Since $p^2=u^2+2v^2$, this yields that $u^2+2v^2$ is odd, so that $u^2$ is odd, so that $u$ is odd. Thus, $u\neq 0$. Since we know that $u$ is nonnegative, it hence follows that $u$ is positive. We also know that $v$ is positive.

Assume that the two integers $\frac{p-u}{2}$ and $\frac{p+u}{2}$ have a common prime divisor $q$. Then, $q$ also divides $\frac{p-u}{2}+\frac{p+u}{2}=p$. Thus, $q$ is a prime divisor of $p$. Since $p$ is a prime, and thus the only prime divisor of $p$ is $p$ itself, this yields $q=p$. Hence, $p$ is a common prime divisor of $\frac{p-u}{2}$ and $\frac{p+u}{2}$. Therefore, $p$ also divides $\frac{p+u}{2}-\frac{p-u}{2}=u$. Since $u$ is positive, this yields $u\geq p$. This contradicts with $p>u$. Hence, our assumption that the two integers $\frac{p-u}{2}$ and $\frac{p+u}{2}$ have a common prime divisor was wrong. Hence, the two integers $\frac{p-u}{2}$ and $\frac{p+u}{2}$ are coprime.

Now, we divide the equation $\left(p-u\right)\left(p+u\right)=2v^2$ by $4$ and obtain $\frac{p-u}{2}\cdot\frac{p+u}{2}=\frac{v^2}{2}$. Since $\frac{p-u}{2}$ and $\frac{p+u}{2}$ are integers, this yields that $\frac{v^2}{2}$ is an integer, so that $v^2$ is even, so that $v$ is even, and thus $v=2V$ for some integer $V$. Since $v$ is positive, it follows that $V$ is positive as well.

Thus, $\frac{p-u}{2}\cdot\frac{p+u}{2}=\frac{v^2}{2}=\frac{\left(2V\right)^2}{2}=2V^2$. Hence, $2\mid\frac{p-u}{2}\cdot\frac{p+u}{2}$. Since $2$ is a prime, it thus follows that $2$ divides at least one of the numbers $\frac{p-u}{2}$ and $\frac{p+u}{2}$. We will only deal with the case when $2$ divides $\frac{p-u}{2}$; the case when $2$ divides $\frac{p+u}{2}$ is analogous and left to the reader.

Since $2$ divides $\frac{p-u}{2}$, the number $\frac{p-u}{2}/2=\frac{p-u}{4}$ is an integer; besides, it is positive (since $p>u$). Now, dividing the equation $\frac{p-u}{2}\cdot\frac{p+u}{2}=2V^2$ by $2$, we get $\frac{p-u}{4}\cdot\frac{p+u}{2}=V^2$.

Now, $\frac{p-u}{4}$ and $\frac{p+u}{2}$ are two coprime positive integers (in fact, we know that they are positive integers, and they are coprime because $\frac{p-u}{2}$ and $\frac{p+u}{2}$ are coprime). Since $\frac{p-u}{4}\cdot\frac{p+u}{2}=V^2$ is the $2$-nd power of a positive integer, Lemma 1 thus yields that both $\frac{p-u}{4}$ and $\frac{p+u}{2}$ are $2$-nd powers of positive integers. Hence, there exist positive integers $\lambda$ and $\mu$ such that $\frac{p-u}{4}=\lambda^2$ and $\frac{p+u}{2}=\mu^2$.

Thus, $p=\frac{p+u}{2}+2\cdot\frac{p-u}{4}=\mu^2+2\lambda^2$. Since $\lambda\neq 0$ (because $\lambda$ is positive), this yields $p\in A$, and the problem is solved.

  Darij

That's nice and systematic solution darij grinberg! (7 out of 7 points  :smile: ).

[quote="Peter"]Let $ A$ be the set of positive integers of the form $ a^{2}\plus{}2b^{2}$, where $ a$ and $ b$ are integers and $ b\neq 0$. Show that if $ p$ is a prime number and $ p^{2}\in A$, then $ p\in A$.[/quote]

 We assume that $ p > 2$. We use \textsc{Thue's Lemma}.

 [b]Thue's Lemma[/b]  Let $ p$ be a prime number and let $ \alpha$ be an integer with $ \alpha\not\equiv 0\; (mod\; p)$. Then, the congruence  $ \alpha x\equiv y\; (mod\; p)$ admits a solution $ (x, y)$ with $ \vert x\vert ,\vert y\vert\in\left( 0,\sqrt{p}\right)$.

 [b]Proof[/b]
 Let $ \mathcal{S}\equal{}\{ (i,j)\;\vert\; i,j\in\{ 0,\cdots,\lfloor\sqrt{p}\rfloor\}\,\}$ and $ \mathcal{T}\equal{}\{0,\cdots, p\minus{}1\}$. The key idea is to introduce the function $ f :\mathcal{S}\rightarrow\mathcal{T}$ defined by
\[ f(i,j)\equal{}\mathbf{r}_{p}\left(\alpha i\minus{}j\right),\]
where $ \mathbf{r}_{p}(n)$ denotes the remainder in the division of $ n$ by $ p$. Since $ \left\vert\mathcal{S}\right\vert \equal{}\left(\lfloor\sqrt{p}\rfloor\plus{}1\right)^{2}> p \equal{}\left\vert\mathcal{T}\right\vert$, by [b]Pigeonhole Principle[/b], $ f$ is not injective. In other words, we have
\[ f\left(i_{1},j_{1}\right)\equal{}f\left(i_{2},j_{2}\right)\]
or
\[ \alpha i_{1}\minus{}j_{1}\equiv\alpha i_{2}\minus{}j_{2}\; (mod\; p)\]
or
\[ \alpha\left( i_{1}\minus{}i_{2}\right)\equiv j_{1}\minus{}j_{2}\; (mod\; p)\]
for some \textit{distinct} elements $ \left(i_{1}, j_{1}\right)$ and $ \left(i_{2}, j_{2}\right)$ in $ \mathcal{S}$. Setting $ i \equal{} i_{1}\minus{}i_{2}$
 and $ j \equal{} j_{1}\minus{}j_{2}$, it becomes
\[ \alpha i\equiv j\; (mod\; p).\]
Since $ i_{1},j_{1}, i_{2}, j_{2}\in\{ 0,\cdots,\lfloor\sqrt{p}\rfloor\}$, we see that $ \vert i\vert \equal{}\vert i_{1}\minus{}i_{2}\vert <\sqrt{p}$ and  $ \vert j\vert \equal{}\vert j_{1}\minus{}j_{2}\vert <\sqrt{p}$. It remains to check that both $ i$ and $ j$ are nonzero.

 [b]Step 1.[/b] $ i\neq 0$: Assume to the contrary that $ i$ is zero. Then the congruence $ \alpha i\equiv j\; (mod\; p)$ implies that $ 0\equiv j\; (mod\; p)$. Since $ \vert j\vert <\sqrt{p}< p$, this means that $ j$ is also zero. In other words, we get $ i_{1}\equal{} i_{2}$ and $ j_{1}\equal{} j_{2}$. This is a contradiction because $ \left(i_{1}, j_{1}\right)$  and $ \left(i_{2}, j_{2}\right)$ are distinct.

 [b]Step 2.[/b] $ j\neq 0$: Assume to the contrary that $ j$ is zero. It follows that $ \alpha i\equiv j\equiv 0\; (mod\; p)$. Since $ \alpha\not\equiv 0\; (mod\; p)$ and since $ p$ is prime, this implies that $ i\equiv 0\; (mod\; p)$. Since $ \vert i\vert <\sqrt{p}< p$, this means that $ i$ is also zero. In other words, we get $ i_{1}\equal{} i_{2}$ and $ j_{1}\equal{} j_{2}$. This is a contradiction because $ \left(i_{1}, j_{1}\right)$ and $ \left(i_{2}, j_{2}\right)$ are distinct.

  Now, suppose that $ p^{2}\in A$. Write $ p^{2}\equal{} a^{2}\plus{}2b^{2}$ for some positive integers $ a$ and $ b$.  Reading the equation modulo $ p$ yields  $ a^{2}\equiv\minus{}2 b^{2}\; (mod\; p)$.  Multiplying the inverse of $ b$ modulo $ p$,\footnote{  It's clear that $ b < p$ so that $ b$ is not divisible $ p$.} we get
\[ \left( a b^{\minus{}1}\right)^{2}\equiv\minus{}2\; (mod\; p).\]
Taking $ \alpha\equiv a b^{\minus{}1}\; (mod\; p)$ in Thue's Lemma, we can find a pair $ (m,n)$ of integers satisfying that $ n\equiv\alpha m\; (mod\; p)$ and $ 0 <\vert m\vert ,\vert n\vert <\sqrt{p}$. After squaring the congruence, we have
\[ n^{2}\equiv{\alpha}^{2}{m}^{2}\equiv\minus{}2m^{2}\; (mod\; p).\]
In other words, $ n^{2}\plus{}2m^{2}$ is divisible by $ p$. Since  $ 0 <\vert m\vert ,\vert n\vert <\sqrt{p}$ or  since $ 0 < n^{2}\plus{}2m^{2}< 3p$, this means that $ n^{2}\plus{}2m^{2}$ is $ p$ or $ 2p$. In case when $ p \equal{} n^{2}\plus{}2m^{2}$, it is clear that $ p\in A$. In case when $ 2p \equal{} n^{2}\plus{}2m^{2}$, we see that $ n$ is even so that $ p \equal{} m^{2}\plus{}2\left(\frac{n}{2}\right)^{2}\in A$.

$ p^{2} \equal{} a^{2} \plus{} 2b^{2}$ , has solutions $ a , b$ in integers .  We have to prove that $ p$ can be represented in such a form .

$ p^{2} \equal{} a^{2} \plus{} 2b^{2} \equiv 0 \bmod{p}$

$ \implies a^{2} \equiv \minus{} 2b^{2} \bmod{p}$

This implies $ \minus{} 2b^{2}$ is a quadratic residue $ \bmod{p}$ , so $ \left(\frac { \minus{} 2b^{2}}{p} \right) \equal{} 1$

$ \left(\frac { \minus{} 2b^{2}}{p} \right) \equal{} \left(\frac { \minus{} 2}{p} \right)\left(\frac {b^{2}}{p} \right) \equal{} 1$

$ \implies \left(\frac { \minus{} 2}{p} \right) \equal{} 1$ ,thus $ \minus{} 2$ is a quadratic residue modulo $ p$ .

[b]Lemma [/b]
$ \minus{} 2$ is a quadratic residue modulo $ p$ if and only if $ p \equiv 1\bmod{8}$ or $ p \equiv 3 \bmod{8}$

The proof is easy by considering that
$ \left(\frac { \minus{} 2}{p} \right) \equal{} \left(\frac {2}{p} \right)\left(\frac { \minus{} 1}{p} \right)$ and $ \left(\frac {2}{p} \right) \equal{} ( \minus{} 1)^{\frac {p^{2} \minus{} 1 }{8}}$

Thus from the lemma it follows that $ p \equal{} 8k \plus{} 1$ or $ p \equal{} 8k \plus{} 3$ ,and we know that any prime $ p$ of the form $ 8k \plus{} 1$ or $ 8k \plus{} 3$ can be represented in the form $ x^{2} \plus{} 2y^{2}$ where $ x$ and $ y$ are positive integers .

Thus the proof follows .

Sorry if the proof is wrong ,or resembles that of Darij :blush:

[quote="darij grinberg"]It's not wrong, but have fun with proving the fact that

[quote="srinath.r"]any prime $ p$ of the form $ 8k \plus{} 1$ or $ 8k \plus{} 3$ can be represented in the form $ x^{2} \plus{} 2y^{2}$ where $ x$ and $ y$ are positive integers[/quote]

- it is about ten times as difficult as the original problem. The formula for $ \left(\frac 2 p\right)$ that you are using is not quite a piece of cake either.

  darij[/quote]

You are 100% right , couldnt think of a solution like yours :blush:  ,yours was really superb and detailed.

This is the $\mathbb{Z}[\sqrt{-2}]$ method that darij mentioned.

$a^2+2b^2=p^2$ is given.  We work in $\mathbb{Z}[\sqrt{-2}]$ which is a Unique Factorization Domain.  We arrive at \[(a+\sqrt{-2}b)(a-\sqrt{-2}b)=p^2\]

Notice that $\gcd(a+b\sqrt{-2}, a-b\sqrt{-2})|2a, 2\sqrt{-2}b$.  However since $p$ is prime $\gcd(a,b)=1$ therefore $\gcd(a+b\sqrt{-2}, a-b\sqrt{-2})|2\sqrt{-2}=\left(\sqrt{-2}\right)^3$.  Notice that $\sqrt{-2}|\gcd(a+b\sqrt{-2}, a-b\sqrt{-2})$ implies that \[\frac{a+b\sqrt{-2}}{\sqrt{-2}}\in \mathbb{Z}[\sqrt{-2}]\]
However, this implies that $a\equiv 0\pmod{2}$ which contradicts $p$ being prime.  Therefore \[\gcd(a+b\sqrt{-2}, a-b\sqrt{-2})=1\]

Since $(a+\sqrt{-2}b)(a-\sqrt{-2}b)$ this implies that $a+b\sqrt{-2}=uz^2, a-b\sqrt{-2}=\bar{u}\bar{z}^2$ where $u$ is a unit in $\mathbb{Z}[\sqrt{-2}]$.  However, the units are $\pm 1$ because $\text{norm}\left(x+y\sqrt{-2}\right)=x^2+2y^2=1\iff x=\pm 1$.  Since $-1=i^2$ we can write this as $\begin{cases} a+b\sqrt{-2}=z^2\\ a-b\sqrt{-2}=\bar{z}^2 \end{cases}$

By multiplying the two equations above we arrive at $\left(z\bar{z}\right)^2=p^2\implies z\bar{z}=p$.  Write $z$ in the form $m+n\sqrt{-2}$ gives us \[p=\left(m+n\sqrt{-2}\right)\left(m-n\sqrt{-2}\right)=m^2+2n^2\]

Is $ \mathbb{Z}[\sqrt{-2}] $ really unique factorization domain? I've heard about $ \mathbb{Z}[i] $ and $ \mathbb{Z}[\omega] $, but i thought that there are problems in the rest. Where could I found out something more about it? Thank you and sorry if my question is stupid :-)

A completely different approach (based on Euler's method to find all rational solutions of some Diophantine equations).

Write $a^2 + 2b^2 = p^2$, with $a,b \in \mathbb{N}$, $b \neq 0$, as $X^2 + 2Y^2 = 1$, where $X=a/p$, $Y=b/p$. We will find all rational solutions for this equation. The trivial solution $(X,Y) = (1,0)$ hints to substituting $X = 1-A$, $Y = B$, and so $A^2 + 2B^2 - 2A = 0$. Since we look for rational solutions, take $B=\lambda A$, where $\lambda = d/c \in \mathbb{Q}^*$, $\gcd(c,d)=1$ ($\lambda = 0$ leads to $X=-1$, $b/p = Y = 0$, which is anyways forbidden ($b\neq 0$)).
Then $A((1+2\lambda^2)A - 2) = 0$; The case $A=0$ leads to $X=1$, $b/p = Y = 0$, which is anyways forbidden ($b\neq 0$), so assume $A\neq 0$. Then $A = \dfrac {2} {1+2\lambda^2}$,  $B = \dfrac {2\lambda} {1+2\lambda^2}$, thus $X = \dfrac {2\lambda^2 - 1} {1+2\lambda^2} = \dfrac {2d^2 - c^2} {c^2+2d^2}$,  $Y = \dfrac {2\lambda} {1+2\lambda^2} = \dfrac {2cd} {c^2+2d^2}$. Finally, this yields $(2d^2 - c^2)^2 + 2(2cd)^2 = (c^2+2d^2)^2$, this being the complete parameterized set of solutions for $a^2 + 2b^2 = p^2$ and $\gcd(a,b)=1$. But this is precisely the case, since $p$ being prime we cannot have $\gcd(a,b)>1$ (it will then have to be equal to $p$, and that forces $b=0$, forbidden).
We have thus painlessly obtained $p = c^2 + 2d^2$, of the required representation.

@Radar:  Yes there are actually quite a lot of Euclidean domains which all can be found [url=http://en.wikipedia.org/wiki/Euclidean_domain#Norm-Euclidean_fields]here[/url].

A nice article online on this topic is [url=http://math.uga.edu/~pete/4400quadrings.pdf]here[/url].  Private message me your email address and I can share some notes from a class I took, but please do not post them publicly (the teacher is OK with private sharing of the notes).


Say there is a 3-4-5 right triangle.  If this triangle's acute angles are cut, AD being

one, CE being another line, with CE cutting AB into segments of 2 and 1; AD cutting

BC into segments of 3 and 1... the masses of A and C respectively being 1 and 1, all

units of measurement being equal... what is the mass of B?

$2\neq3...\text{this SHOULD be B.}$

1):idea:
2):?:
3):lol:

I'm confused by the faces.. are you joking?  Well if not, you can't just arbitrarily put 1 on both A and C, it wouldn't balance.  If A was 3 and C was 2 however, B could be 6 and everything is good.


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I'm currently in 11th grade and I've been in this youth orchestra for 2 years now (this year would've been the third). The youth orchestra is completely an extracuriccular activity and is not associated with my school in any way. You have to pay a fee to be in the orchestra as well as audition.

I just quit the orchestra with only three more rehersals to go before the concert. My parents were really upset and basically tried to force it down my throat to stay in it but I refused. The thing is that it wasn't enjoyable at all anymore. I felt so out of place at rehersals and I really had little time to practice. One more thing was that on the nights when I had rehersal I would basically be up until around 2 or 3 in the morning trying to get homework done.

In school, I'll challenge myself by taking the most challenging courses. I don't enjoy most some classes but I stick with it. Outside of school though, everything I do I enjoy. And that's the truth. Community service, after school clubs, whatever; I'll find something that I enjoy and I'll do it. But my mom in particular was saying that it was extremely disrepectful to quit and that it would reflect badly on me if any of the directors of the orchestra "said anything to the colleges that I want to apply to."

I know it would've looked better if I satyed in the orchestra but I'd be going to school the next day looking like a zombie and I'm not going to force myself to suffer through it for that. How bad does this look?

It doesn't look [i]bad[/i]. People have their priorities. Clearly, they will not walk off with the impression that you are a highly committed orchestra member, but that's OK, because you're not. At the same time, dabbling with music won't go completely to waste -- it'll make your profile look a bit more colorful.

Keep in mind that just participating in orchestra most likely won't blow admissions out of the water anyway. Qualifying for a district or state level orchestra, on the other hand, could do that.

So what do you do now? Easy -- whatever you want! Of course, you will want to show that you did something with your time. If orchestra doesn't fit into your schedule well, then try to find room for something that interests you more, hopefully with more suitable or flexible hours.

College admissions people are smart.  They don't just look at an application and tally up your extracurriculars to decide if they want to admit you.  If you're making good use if your time, you're making good use of your time, and part of that involves knowing your [b]priorities[/b] and being willing to say no.  A friend of mine quit orchestra halfway through high school and it didn't hurt her application at all - she's currently a sophomore at Brown.  She had more important things to do with her time, and so, it seems, do you!

[quote="Blkmage"]I just quit the orchestra with only three more rehersals to go before the concert. My parents were really upset and basically tried to force it down my throat to stay in it but I refused. The thing is that it wasn't enjoyable at all anymore. I felt so out of place at rehersals and I really had little time to practice. One more thing was that on the nights when I had rehersal I would basically be up until around 2 or 3 in the morning trying to get homework done.
[/quote]
I have had a nearly parallel experience, but with me, my problem was Chinese school.
The Chinese school began revising their books to have [b]more[/b] homework, but at the same time, school was also giving me more homework.
There was no way i could do all that homework (and still do other activities such as marticipate in Math Club and play the piano), so i decided to quit Chinese school. Of course, for me, my parents were more in support of me than they were against me.
Sometimes, school is just a decision between something academic (math, science) versus something extracurricular (notably, orchestra). No matter how impartant the extracurricular activity may seem, it must always be second in line to more important academic things. In your case, i think your decision is justified by many motives. And, quitting orchestra may be a demonstration of you "knowing your priorities". Besides, the other activities you participate in more than outweigh the fact that you quit orchestra.

Don't let the "well-rounded" cult of college admissions fool you -- out here in the real world, the most successful and happy people are those who find something they really like to do, and get *very* good at it.  This is the exact opposite of the "good at everything, great at nothing" that college admissions seems to encourage.

Find what you want to do, and get very good at it, and you should be fine.  If a school doesn't accept you because you have n-1 extracurriculars instead of n, then that's going to be a school of tools that you'd probably rather not attend anyway.

[quote="rrusczyk"]This is the exact opposite of the "good at everything, great at nothing" that college admissions seems to encourage.[/quote]
I'm not sure that's a fair assessment.  College admissions officers like to say that they're looking for both "well-rounded" and "well-lopsided" people - it makes for a more diverse campus and a more interesting intellectual atmosphere overall.  As for the cultivation of well-rounded people in the first place, I'll be a little more philosophical and posit that even if they have a late start, engaging in a variety of fulfilling experiences in high school and college allows well-rounded people to choose a specialty in a very well-informed way later on in life.

Indeed, this is what they say (I'm less convinced that this is how they operate in general) -- but how do students behave?  How many students choose to be "well-lopsided" as a college admissions strategy?  How many choose to be "well-rounded"?  (Most of the "well-lopsided" students I see tend to do so out of love of subject, which is a good reason, though a few do it due to pushy parents, which is a bad reason.  I can't think of any choosing this strategy due to college admissions.)

I do agree with you strongly on the importance of experimenting with a lot of intellectual pursuits in high school and college, in the hopes of finding something you're passionate about.  I'm just not convinced the "6 APs this year, 6 APs next year, and 14 extracurriculars" is a good way to do it.  This approach doesn't allow much time to indulge in those things that you find very enjoyable and important -- this sort of focus seems more valuable to me than those last 3 clubs or those last 2 APs.

I'd like to add another argument in favor of well-roundedness*-- a lot of major advances in science and technology come from using methods and data from one field to solve problems in another, seemingly unrelated field.  This kind of work takes a deep understanding of multiple fields-- even if you know one of them very, very well, you won't get very far if you don't have a good idea of how the others function.  Knowing a lot about a lot of different things will help you make important connections, and the more deeply you understand all of them the more effective you will be.

*Note that I don't think cramming in a bunch of superficial things counts as well-roundedness.  It's not about how many things you do, but how many things you do very well.

"Well-roundedness" and "specialization" are indicators of completely different things, and should not be viewed as opposites at all.

When they say they want well-rounded people, they really mean that they want people who exhibit balance and a wide range of interests. These people are better suited to a liberal education, and are likely to gain more from a diverse environment with lots of interesting things to learn and do.

As for specialization, going out of your way to focus on certain fields or activities shows your particular interests and passions. These "mountains" of interest and skills will likely be a basis for what you do with your education, as well as for how you contribute to the diverse and rich environment of the college.

So the point is, they want open-minded people who are interested in trying and learning new things, and also bring to the table certain interests and passions that they can share with others. When most people in the environment exhibit these qualities, it should result in a sort of intellectual symbiosis and lead to mutual personal growth.

[quote="Osud"]
*Note that I don't think cramming in a bunch of superficial things counts as well-roundedness.  It's not about how many things you do, but how many things you do very well.[/quote]

I think the key here is the "very well".  I don't define "5 on an AP test" as doing things very well.  I think the number of things one can do "very well" is very small, and much smaller than the number of extracurriculars, etc., that I typically see on a punch-ticket college application.  Mainly, I bet we agree on who we'd want to hire or have lunch with, but would define our terminology differently.  I don't think the "well-roundedness", as the term is typically used for college applications, is very valuable (that is, do everything a little better than OK).  I think well-roundedness as you are defining it is (that is, do some things, but not necessarily everything, very well).

As for balance and a wide range of interests, those people would be great for dinner conversation, but in most careers, they'll get crushed by those who have higher focus.

I don't understand why being focused prevents a person from having a wide range of interests, or vice versa.

This is because of the law of diminishing returns -- it doesn't take much to entertain an interest, even to a degree of seriousness. For example, suppose someone is heavy into mathematics, but they are interested in keeping up with current events. Provided they don't screw off and start watching random videos on CNN.com, they can entertain that interest without much time commitment. If they had spent that small additional amount of time on studying mathematics, they would have gained, comparatively, very little.

This is especially true earlier on in life. I don't think committing to, say, a sport, or reading literature, at all affects how someone performs in, say, a math competition, despite the time investment. In fact, from my experience, I experienced the largest rate of improvement in my understanding of mathematics over the years I was rowing (which is about 4 hours a day, 6 days a week), and I can attribute that to a healthier and more balanced lifestyle, which made me happier, more relaxed, and more focused.

Finally, sets of skills acquired through different interests have not only a large union but also a large intersection. For example, there are very few cerebral activities that will not improve your competition math performance. Art, reading, writing, chess, music, science projects, engineering projects, public speaking, and debate, to name a few, all share skills with competition mathematics.

[quote]Mainly, I bet we agree on who we'd want to hire or have lunch with, but would define our terminology differently. [/quote]

I think we also have somewhat different perspectives, coming from different fields.  There will always be an abundance of jobs which require math and strong technical skills, and a large pool of talent.  In my field (historical linguistics), if you want to get hired, you [i]have[/i] to be able to specialize in multiple areas.  There aren't many historical linguistics positions available, so specializing in multiple areas means that you would potentially be employable by more than one kind of department.  Plus, there isn't enough student demand even at a large research university to teach classes only on your specialty, so having a greater breadth of research interests means that you can teach a wide enough range of courses to be useful to the department.

[quote]Finally, sets of skills acquired through different interests have not only a large union but also a large intersection. For example, there are very few cerebral activities that will not improve your competition math performance. Art, reading, writing, chess, music, science projects, engineering projects, public speaking, and debate, to name a few, all share skills with competition mathematics.[/quote]

I would also argue that the range of things you are able to tackle and do well is a good indicator of how good your training has been-- that is, how well it has prepared you to solve problems instead of simply putting a mass of facts in your head.  Here is an essay along those lines:  [url]http://community.middlebury.edu/~harris/Humanities/MyToilet.html[/url]

Again, I don't think we're arguing for very different things here-- that it's good to be able to handle (and enjoy) a wide range of things as an outgrowth of skills and abilities you can apply broadly, and that it's not good to go on a bunch of wild tangents, none of which you can completely master.

[quote="jmerry"]I agree completely with Richard. Extracurriculars didn't matter at all for me, because I was in at Berkeley based entirely on grades and test scores. That was my first choice, and the whole thing minimized stress.[/quote]

Ya but its harder to get in now, u cant just go on grades and scores

Sure you can. It just doesn't work for most top tier schools, with few exceptions, most notably Dartmouth.

At the time I was admitted, Berkeley admitted half its freshman class using only a publicly available formula based on grades and test scores. In response to an affirmative action ban, they switched to "comprehensive review" for all applications. I don't think this changed very many of their actual decisions; those who were automatically admitted before now get a glance at the rest of the application.
Even if you're applying to a place that cares about ECs, grades and test scores matter much more.

[quote="TZF"]I don't understand why being focused prevents a person from having a wide range of interests, or vice versa..[/quote]

Sure -- but a very wide range of interests will reduce how focused you can be, simply because of time.  It's an optimization problem that varies from individual to individual.  I think for most people, spending all your time on activity X is not as good as spending 70% of your time on activity X and 30% on other things.  But 70% on X and 30% on others is better than 20% on X and 80% on others.

I suspect we're talking past each other at this point, and that we don't disagree that much.  I think the current received wisdom on "well-roundedness" places too much emphasis on spending a moderate amount of time on many irrelevant things.  I'll also note that most of the very successful people I know are not "well-rounded" in the traditional college application sense, but are extremely interesting and very good at a handful of things relevant to their interests.

I had another thought on this. Rather than asking what EC focus will help your chances, ask this question:

What will your major be?

If you can answer that question now, years before you actually go to college, you should focus on your chosen field. That leads naturally to a "well-lopsided" profile. If you're undecided, you should be dabbling in a range of interests in search of something you want to stick with, which will lead to a "well-rounded" profile. Still, this is always subject to change, because you might decide on one particular interest- and when that happens, you focus on it.

Around here, we should have an unusually high population in the first category, mostly planning on mathematics. I was certainly one of those.

I talked to my guidance counselor today and I'm getting little nervous b/c I know being in a youth orchestra is really a EC that sets me apart but having quit, she said that really hurts my chances. I do a decent amount of after school stuff, I'm really involved in music (I have qualified for district/county level orchestras and, living out on Long Island, I've been chosen to play in the Long Island String Festival multiple times). As far as ECs go that aren't associated with school, I honestly don't do that much.

I volunteer to help my parents' friend filing papers for eldercare and I was thinking of starting peer-tutoring. I'm enrolled in a science research program (through the school) and am currently in collaboration with a college professor in working on a research project. I know there are a ton of people out there who are much more involved than I am so I'm just nervous that I won't stand out enough. I can do my best to explain that I was prioritizing my time and being like a zombie the morning after rehersals just wasn't good but I don't know if I'll even get an interview to begin with. Also, my parents were pretty disappointed too but the thing is that I really am trying my hardest this year and I'm content with what I've done so far. I was looking at Carnegie Mellon as a possible target school and both parents and guidance counselor were making a big fuss about it. Opinions?

Don't listen to her idea that it will hurt your chances, because it won't. If you write on your CV that you've been invited to play in district/county level orchestras and have played in the Long Island String Festival under activities, then it will be obvious that you are serious about your instrument.

What is this youth orchestra thing? Who plays in it? Where does it perform? How is it chosen? Now it's up to you to decide whether you even want to bother to include it in your resume or not, since you already have a strong case for yourself as a good musician. Unless it's superb, it might not even be worth mentioning that you did it for half of high school, given that you have all these other great things.


What is the remainder when $ 1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} \dots \plus{} 9 \plus{} 10$ is
divided by 3?

1+2+3+4+5+6+7+8+9+10

10(11)/2

5*11

55 is the sum.

55 is 1 mod 3, and the remainder when divided by 3 is $ \boxed{1}$


[quote="mdk"]Consider a set of 2n points in space, n > 1. Suppose they are joined by at least n^2 + 1 segments. Show that at least one triangle is formed. Show that for each n it is possible to have 2n points joined by n2 segments without forming a triangle.[/quote]

Am I missing something.
Counterexample: Say you have $ (0,0,0),(1,0,0),(2,0,0),(3,0,0)$ and all the segments between them. Then no matter how you draw them you get no triangle.

Maybe you mean to say:
"Given $ 2n,n>1$ points in space such that no three are [b]collinear[/b] then blah blah blah ... "


I think we should discuss some International Chemistry Olympiad Problems, National Olympiad Problem etc. like the others. I can find certain IChO Problems not on the internet but I have to translate them because they are in another language (so they won't be the same as the originals). Also you can find Australian Chemistry Olympiad Problems on the internet.

We can use these problems to practice certain concepts or teach/learn these concepts.

What do you think?

That is a very good idea. Let's post some problems - and since you had this idea, it should be you to start to post.

Great. That goes back to the problem that we don't have a moderator and we can't organize our materials.


linear algebra
matrix
analytic geometry
LaTeX

Need help with the following question:

Find T ( [ x1 x2 x3 x4 ] ) if T : |R4 -> |R2 is a linear transformation satisfying
T ( [ 1 1 0 0 ] ) = [ 2 -1 ]
T ( [ 0 1 0 0 ] ) = [ -2 3 ]
T ( [ 0 1 0 2 ] ) = [ 1 0 ]
T ( [ 0 0 1 0 ] ) = [ 0 -1 ]

Note: the numbers to the right of x, such as- x1, x2 are subscripts. |R is real space.

rewritten for neatness

Find a linear transformation $ T \in \mathcal{L}(\mathbb{R}^4,\mathbb{R}^2)$ satisfying

$ \begin{array}{cccc}T\begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}
&
T\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}
&
T\begin{pmatrix} 0 \\ 1 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}
&
T\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ -1 \end{pmatrix}
\end{array}$

but this is pretty straight forward, even done using the standard basis for $ \mathbb{R}^4$ you can calculate that

$ \begin{array}{cccc}
T\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 4 \\ -4 \end{pmatrix}
&
T\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}
&
T\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ -1 \end{pmatrix}
&
T\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -3/2 \\ 3/2 \end{pmatrix}
\end{array}$

so that gives us that

$ \mathcal{M}_{T}(\mathcal{B}_{\mathbb{R}^4},\mathcal{B}_{\mathbb{R}^2}) =
\begin{pmatrix} 4 & -2 & 0 & -3/2 \\
-4 & 3 & -1 & 3/2
\end{pmatrix}$

where ${ \mathcal{B}_{\mathbb{R}^4}}$ and $ \mathcal{B}_{\mathbb{R}^2}$ is the standard basis for ${ \mathbb{R}^4}$ and ${ \mathbb{R}^2}$ respectively.

Could you please elaborate on how to achieve the [4  -4], [-2  3] and such?

$ T$ acts linearly so
$ T\begin{pmatrix} 1 \\
0 \\
0 \\
0 \end{pmatrix} \equal{}
T(\begin{pmatrix} 1 \\
1 \\
0 \\
0 \end{pmatrix} \minus{}
\begin{pmatrix} 0 \\
1 \\
0 \\
0 \end{pmatrix}) \equal{}
T(\begin{pmatrix} 1 \\
1 \\
0 \\
0 \end{pmatrix}) \minus{}
T(\begin{pmatrix} 0 \\
1 \\
0 \\
0 \end{pmatrix}) \equal{}
\begin{pmatrix} 2 \\ \minus{}1 \end{pmatrix} \minus{}
\begin{pmatrix}\minus{}2 \\ 3 \end{pmatrix}
\equal{} \begin{pmatrix} 4 \\ \minus{}4 \end{pmatrix}$

and $ T \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} \equal{} \begin{pmatrix} \minus{}2 \\3 \end{pmatrix}$  from your definition of the problem, the rest are achieved similarly. you may be slightly confused by the notiation. The notation $ \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix}$ in the standard basis for $ \mathbb{R}^4$ is analagous to the coordinates $ (x_1,x_2,x_3,x_4)$, it's written in the first way when dealing with Linear Algebra.

Thank you shallow. Really appreciate it.

BTW, is there a tool for writing the matrix notations in this forum or do you just write the code?

no tools, just code. its not so bad though, latex is pretty easy to learn and has a lot of shortcut commands for making things like matrices (single column or otherwise).

good luck, i recommend learning it early especially if you will be going for a masters or phd. even as an undergrad, i think the professors/graders appreciate the neatness.

heres a good place to start http://www.mathlinks.ro/LaTeX/AoPS_L_About.php


The probability of seeing a shooting star in the first 15 mins is the same as the probability of seeing it in any given 15 minute period.

I don't see how you got 20% though?

[hide="Solution"]$ (1 \minus{} P_{15 mins})^{\frac {1}{15}} \equal{} (1 \minus{} P_{60 mins})^{\frac {1}{60}}$

$ \iff P_{15 mins} \equal{} 1 \minus{} \sqrt [4]{1 \minus{} \frac {4}{5}} \equal{} 1 \minus{} \frac {1}{\sqrt [4]{5}} \approx \boxed{0.33126 \ldots }$

That is, to avoid using a strange variation of PIE logic, you can use the probability of not seeing a shooting star.
[/hide]

[quote="baudhack"]If somebody has the book available, the question is:

"On Saturday night, Becky again goes stargazing. This time, conditions are better, and there's an 80% chance that she will see a shooting star in any given hour. We assume that the probability of seeing a shooting star is uniform for the entire hour. What is the probability that Becky will see a shooting star in the first 15 minutes?"

Since we know that there is a uniform probability, what does the possibility of having more than one star have to do with the probability of a star appearing the first quarter hour?

Even if there were just one star, could we in fact conclude that there was a 20% probability of it appearing the first 15 minutes or is this point irrelevant?[/quote]
NO!!!! sorry, but that's a very common mistake. by that reasoning, wouldn't the probability of seeing a star in two hours be 160%?
for the problem
[hide]divide the hour into intervals of 4 each 15 min long. the probability that you don't see a star in an hour is 20%. now, the only case where you don't see a star is when in each interval you don't. let's denote the probability that you do see a star in 15 min is p. now the probability that you don't see one is 1-p.
so you get an equation: (1-p)^4=.2
solve for p and you have your answer. [/hide]

Thank you for the kind explanations.

I realize that it is not 1/4(80%), but was more curious of the explanation offered by D. Patrick about why it is flawed. After some thinking about his explanation, the reasoning of having multiple stars became clear. I appreciate your additions very much.


inequalities calculus trigonometry vector geometry function integration

As the title suggests this is very easy but i wanted to see ur proofs.

So

Prove that $ (\sqrt3)cos A \plus{} sin A <\equal{} 2$

Looking forward to creative solutions

P.S. Calculus is not allowed.

[hide="standard technique"]$ \text{L.H.S.} \equal{} 2\cdot (\frac{\sqrt{3}}{2} \cos A \plus{} \frac{1}{2} \sin A) \equal{} 2 \sin (A\plus{}\frac{\pi}{3}) \leq 2\cdot 1 \equal{} 2$.
[/hide]

[hide="Cauchy-Schwarz"]By the Cauchy-Schwarz inequality, $ (\sqrt{3}\cos{A}\plus{}\sin{A})^2\le(3\plus{}1)(\cos^2{A}\plus{}\sin^2{A})\Rightarrow|\sqrt{3}\cos{A}\plus{}\sin{A}|\le2.$[/hide]

By the way, that was me accidentally logged in on my brother's account.

[quote="befuddlers"]Well i got the cauchy schwatz method on my own and my friend showed me timwu's method.

Any other methods?[/quote]
[hide]Here's another one:
The inequality is equivalent to: $ (sinA \minus{} 1/2)^2 \plus{} (cosA \minus{}$sqrt{3}/2$ )^2\geq 0$, which is obviously true![/hide]

Don't know if I am eligible to post here but my solution:
$ \sqrt3cosA\plus{}sinA \equal{} 2 (\frac{\sqrt{3}}{2} \cos A \plus{} \frac{1}{2} \sin A) \equal{} 2 \sin (A\plus{}\frac{\pi}{3}) \leq 2$

@Agr_94_Math:
I don't think that any of us mind you posting here.
Generally, when you post a solution, you should hide it so that others can still solve the problem. (Although I think that this was an easy enough problem to skip that, but I'm not sure).

Coincidentally, that solution is almost exactly like timwu's solution. (See the 2nd post).

Xantos C. Guin I am not from Georgia. That is why I had a doubt.
Yes hereafter I will hide my solution and sorry tht I posted the same solution as  post no. 2 but i dont see other's solutions when hidden before I get the problem.
THanks anyway.

[hide="vectors"]$ \text{LHS} \equal{} <\sqrt{3},1> \cdot <\cos A, \sin A> \equal{} 2 \cos \theta \leq 2$[/hide]

I know this is a bit off topic, but what exactly are you finding when you use the dot product?  Like I know the cross product gives a normal vector to the two given vectors you are crossing.

You are finding how parallel are the vectors.

Note that $ \vec{a} \cdot \vec{b} \equal{} \parallel{}\vec{a}\parallel{} \parallel{}\vec{b}\parallel{} \cos{(\theta)} \Longleftrightarrow \cos{(\theta)} \equal{} \dfrac{\vec{a} \cdot \vec{b}}{\parallel{}\vec{a}\parallel{} \parallel{}\vec{b}\parallel{}}$.

If $ \vec{a} \parallel \vec{b}$, then $ \theta \equal{} 0^\circ$, and $ \dfrac{\vec{a} \cdot \vec{b}}{\parallel{}\vec{a}\parallel{} \parallel{}\vec{b}\parallel{}} \equal{} 1$.

If $ \vec{a} \parallel \minus{} \vec{b}$ (anti parallel), then $ \theta \equal{} 180^\circ$, and $ \dfrac{\vec{a} \cdot \vec{b}}{\parallel{}\vec{a}\parallel{} \parallel{}\vec{b}\parallel{}} \equal{} \minus{} 1$.

If $ \vec{a} \perp \vec{b}$, then $ \theta \equal{} 90^\circ$, and  $ \dfrac{\vec{a} \cdot \vec{b}}{\parallel{}\vec{a}\parallel{} \parallel{}\vec{b}\parallel{}} \equal{} 0$.

So, as  $ \dfrac{\vec{a} \cdot \vec{b}}{\parallel{}\vec{a}\parallel{} \parallel{}\vec{b}\parallel{}}$. represents how parallel the vectors are with $ 1$ being perfectly parallel, $ \minus{} 1$ being anti-parallel, and $ 0$ being perpendicular.

I'm sure that there are several other interpretations of the dot product.

On a side note, dot product gives a natural way to express Cauchy-Schwarz Inequality as almost a definition: the cosine of the angle formed by two vectors, of any dimension, has magnitude not greater than 1. This, however, requires proving that the dot product formula for any dimension first, which seems sometimes is taken as a definition(?).

This approach to Cauchy-Schwarz is unsatisfactory, mostly because the theorem holds for more general inner product spaces.  While you can develope the geometric tools to justify this approach, Cauchy-Schwarz is sufficiently simple that it is preferable to use Cauchy-Schwarz to help justify the geomtery, rather than the other way around.

First, some definitions of inner product spaces.  For the sake of simplicity in the following exposition, I will restrict our attention to [i]real[/i] inner product spaces.

An [i]inner product space[/i] over $ \mathbb{R}$ is an $ \mathbb{R}$-vector space $ V$ with a map $ p : V \times V \to \mathbb{R}$ satisfying the conditions listed below for all $ \mathbf{a,b,c} \in V$ and any scalar $ r \in \mathbb{R}$.  For convenience, we will denote $ f(\mathbf{a,b})$ as $ \mathbf{ \langle a,b \rangle }$, or simply $ \mathbf{ a \cdot b}$.
[list]
[*] $ \mathbf{ \langle a, b \rangle \equal{} \overline{ \langle b, a \rangle }}$.
[*] $ \mathbf{ \langle a \plus{} b , c \rangle \equal{} \langle a,c \rangle \plus{} \langle b, c \rangle}$
[*] $ \mathbf{ \langle ra , b \rangle \equal{} r \cdot \langle a,b \rangle}$
[*] $ \mathbf{ \langle a, a \rangle } \ge 0$, with equality only if $ \mathbf{a\equal{}0}$.[/list]
For simplicity, we will abbreviate $ \mathbf{\langle a, a \rangle \equal{} a \cdot a}$ as $ \mathbf{a}^2$, by abuse of notation.  Beware of this notation, though; it does not generalize to higher powers, and it does not distribute over the inner product!  It is merely a convenient shorthand.

It is easy to verify that the dot product as we know it is an inner product, but there are others possible.  For instance, we can let $ V$ be the set of bounded continuous functions mapping $ [0,1]$ into $ \mathbb{R}$; the mapping
\[ (f,g) \mapsto \int\limits_{0}^1 f(t)g(t)dt \]
is then an inner product.

In general, Cauchy-Schwarz says that for any $ \mathbf{a,b} \in V$,
\[ \mathbf{(a \cdot b)}^2 \le \mathbf{(a \cdot a) (b \cdot b)} , \]
with equality if and only if there exist $ x,y$ in $ \mathbb{C}$, not both zero, such that
\[ \mathbf{a} x \plus{} \mathbf{b} y \equal{} 0 , \]
i.e., if and only if $ \mathbf{a}$ and $ \mathbf{b}$ are linearly independent.

To prove this theorem, we consider the binary quadratic form
\[ (\mathbf{a}x \plus{} \mathbf{b}y )^2 \equal{} \mathbf{a}^2 x^2 \plus{} 2 (\mathbf{a\cdot b} ) xy \plus{} \mathbf{b}^2 y^2 . \]
By our fourth axiom for real inner product spaces, this quadratic form is always greater than or equal to zero, with equality if and only if $ \mathbf{a}$ and $ \mathbf{b}$ are linearly dependent.  Thus its discriminant,
\[ 4(\mathbf{a\cdot b})^2 \minus{} 4 \mathbf{a}^2 \mathbf{b}^2 , \]
is less than or equal to zero, with equality if and only if $ \mathbf{a}$ and $ \mathbf{b}$ are linearly dependent.  This is equivalent to the desired inequality.  $ \blacksquare$