geometry
trapezoid
circumcircle
Euler
ratio
power of a point
radical axis

A trapezium $ ABCD$ with $ AD\parallel BC$ is given and let $ P$ be, the intersection point of its diagonals $ AC,$ $ BD.$ We draw the circumcircles $ (O),$ $ (O'),$ of the triangles $ \bigtriangleup QAC,$ $ \bigtriangleup QBD,$ where $ Q\equiv AB\cap CD,$ which re-intersect each other at one point so be it $ M.$ Prove that the line segment $ QM,$ is isogonal conjugate of $ QP,$ with respect to the angle $ \angle AQD.$

Kostas Vittas.

Here' my solution.
Just follow the ray QM and see what it hits... Ehrrr... :wacko:

Let the circumcircles (O), (O') of the $ \triangle QAC,\ \triangle QBD$ meet DA again at E, F. Let FM, EM meet (O), (O') again at U, V. From the circles (O'), (O),

$ \measuredangle VBQ \mod \pi \equal{} \measuredangle VMQ \mod \pi \equal{} \measuredangle EMQ \mod \pi \equal{} \measuredangle EAQ \mod \pi$

It follows that $ VB \parallel DA$ and similarly, $ UC \parallel DA.$ Thus U, B, C, V are collinear, $ UV \equiv BC \parallel DA.$ From the circles (O'), (O), respectively,

$ \measuredangle DVE \mod \pi \equal{} \measuredangle DVM \mod \pi \equal{} \measuredangle DQM \mod \pi \equal{} \measuredangle CQM \mod \pi$
$ \measuredangle CQM \mod \pi \equal{} \measuredangle CEM \mod \pi \equal{} \measuredangle CEV \mod \pi$

It follows that $ EC \parallel DV$ and similarly, $ FB \parallel AU.$ Let (R) be circumcircle of the $ \triangle QAD.$ FB, EC are antiparallel to QD, QA WRT the $ \angle QDA,\ \angle QAD,$ hence $ FB \perp RA,\ EB \perp RD.$ Consequently, $ DV \perp RD,\ AU \perp RA$ are tangents of (R) at D, A. Since $ UV \parallel DA,$ it follows that $ DV \equal{} AU.$ Let the tangents DV, AU of (R) meet at X. Then $ XD \cdot XV \equal{} XA \cdot XU,$ powers of (O'), (O) to X are equal, X is on the radical axis QP of (O'), (O), and QP is the Q-symmedian of the $ \triangle QDA.$ Since $ DA \parallel BC,$ QP cuts the trapezoid bases DA, BC at ther midpoints, and QP is the Q-median of the $ \triangle QDA,$ isogonal to its Q-symmedian QM with respect to the $ \angle AQD.$

It is easy to show $ QP$ is median of $ AQD$ by using ceva theorem. We have $ \angle MAB\equal{}\angle MCD,  \angle MBA\equal{}\angle MDC$.So triangle $ MAB$ and $ MCD$ are similar. Let $ X,Y$ are projection of $ M$ on $ AB,CD$. Since $ MX: MY\equal{}AB: CD\equal{}QA: QD$, line $ QM$ is symedian of $ AQD$. Hence we are done.

[quote="Leonhard Euler"]It is easy to show $ QP$ is median of $ AQD$ by using ceva theorem. We have $ \angle MAB \equal{} \angle MCD, \angle MBA \equal{} \angle MDC$.So triangle $ MAB$ and $ MCD$ are similar. Let $ X,Y$ are projection of $ M$ on $ AB,CD$. Since $ MX: MY \equal{} AB: CD \equal{} QA: QD$, line $ QM$ is symedian of $ AQD$. Hence we are done.[/quote]

Great solution Leonard Euler ,  :)  . Congratulations

Thank you dear [b][size=100]Leonard Euler[/size][/b], for your interest and nice solution, I think the simplest one.

It is exactly what I wanted, because the proof I had in mind, based on the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=175356]Bisection in a trapezium[/url], although it is a synthetic solution however, it is a complicated way to solve this problem.

Best regards, kostas Vittas.

Another way:

(O), (O') meet DA again at E, F.  Powers of D, A to (O), (O') are in the ratio $ \frac {DA \cdot DE}{AD \cdot AF} \equal{} \frac {DC \cdot DQ}{AB \cdot AQ} \equal{} \frac {DQ^2}{AQ^2}$ $ \Longrightarrow \frac {DE}{AF} \equal{} \frac {DQ^2}{AQ^2}.$ The radical axis QM of (O), (O') cuts DA at K, $ KF \cdot KD \equal{} KA \cdot KE\ \Longrightarrow$ $ \frac {KD}{KA} \equal{} \frac {KE}{KF} \equal{} \frac {DE}{AF} \equal{} \frac {DQ^2}{AQ^2},$ $ QK \equiv QM$ is the Q-symmedian of the $ \triangle QAD.$ No angle chase. :lol:

[quote="armpist"]Any true mathematics problem-solver will not start solving
  a problem as is, but will try to modify it so as to fit into one
  already solved. Probably out of two reasons: being lazy and
  show some respect for problem solvers of the past (mostly the
  French for Modern Triangle Geometry, Napoleon directed a lot
  of money into this kind of research after a geometric theorem
  was named after him ).

  In the case of this problem it will remain equivalent if the top parallel side
  of trapezoid is reduced to 0. The circles on the diagonals then become
  conjugated circles and we all know that they intersect on the symmedian.

  Thank you.
  M.T.[/quote]

I am sorry, but that is not a solution. No true problem solver would solve a problem by proving a limiting case only.
As for Bonaparte, he directed far more money into waging uprovoked wars.

Now I think that all of the above solutions (including mine 'non-solution') are not as good
   as could be.  We all proved that the other line is a symmedian and that is a good enough
    reason for it to be an isogonal line to a median.

    A pure geometer like Darij G. (and I usually ask myself  'But how would Darij G. solve
    a particular problem?' ) would have approached it differently, and I'm convinced in
    it more and more.  Just read his cannonical paper on the subject of isogonal conjugation.

    He would prove that one line is isogonally conjugated to another without going
    into specifics of what each line is, he would use only the properties of conjugation
    itself. Darij?

   Thank you.
   M.T.

  To Yetti:
  If looking at one limiting case does not make a solution, then what could be said of some
  'solutions' based on sending some line to infinity? How many limiting cases is that?
  One or two, or infinity with stops along the way?

  At least moving a line along a reasonable distance leaves it a straight line, while
  sending it to infinity ...   So many lines were sent there that there must be a forest
  or them.  I suggest to put stop to this waste and re-use a line or two that are already
  there,  that will be a significant contribution to slow global entropi-cal warming,
  would it not?

[quote="armpist"]Now I think that all of the above solutions (including mine 'non-solution') are not as good
   as could be.  We all proved that the other line is a symmedian and that is a good enough
    reason for it to be an isogonal line to a median.
[/quote]

Let's look at your non-solution. You have to show (1) that the circle through D tangent to AB at Q and the circle through A tangent CD at Q meet on the Q-symmedian at some point N and (2) that Q, N, M are collinear. Yo did not do any of this.

To do (1), you would have to repeat the last solution with powers of A, D to the tangent circles. To do (2), you would have to repeat the last solution with circles (O), (O'), which gives immediate solution of the problem. The tangent circles turn out to be unnecessary and you would not have done anything different in any case.

[quote="armpist"]
  To Yetti:
  If looking at one limiting case does not make a solution, then what could be said of some
  'solutions' based on sending some line to infinity? How many limiting cases is that?
  One or two, or infinity with stops along the way?

  At least moving a line along a reasonable distance leaves it a straight line, while
  sending it to infinity ...   So many lines were sent there that there must be a forest
  or them.  I suggest to put stop to this waste and re-use a line or two that are already
  there,  that will be a significant contribution to slow global entropi-cal warming,
  would it not?[/quote]

More than one general case has the same limiting case $ \Longrightarrow$ the transformation $ \text{general case} \longrightarrow \text{limiting case}$ does not have an inverse. Central projection of a plane to another plane has an inverse $ \Longrightarrow$ the projected case is not a limiting case.

As we all know... Ehrrr... As far as I know, you are the only person, who ever projected two different lines to infinity at the same time. I can even give a link if you insist. Apparently, you have been busy ever since.

For any questions concerning global warming, entropy, etc., the [url=http://www.mathlinks.ro/index.php?f=405]Introductory Physics[/url] or [url=http://www.mathlinks.ro/index.php?f=332]Advanced Physics[/url] forums are places to go.

[hide][quote="armpist"]Hello Yetti

  About that line to infinity business.
  Who told you that I send two lines to infinity at the same time? Not true.
  I did not specify timing. I can tell you now I sent them like entire Air Force of some
  third world country: the first I sent at time t, and the next 5 minutes later.

  I guess the lines handpicked and trained by you for this mission to Infinity
  travel there in a proud loneliness, but dont they pull the lines in their
  immediate vicinity along?  Or do they leave strict orders for the
  rest to freeze and wait for the glorious return with a proof in hand
  which is no doubt better than two proofs by someone else?
  Do they promise to return soon?
  So you see, lately I try to keep myself well away from the mirky
  business of sending lines to infinity. I'm also unsure of and do not trust
  the unnatural irrationality, as you well aware by now.

 Friendly,
 M.T.[/quote][/hide]

Locking this thread.


[quote="aadil"]during a super collision ,as in the case of neutrons what is the external force that provides an increase in velocity after collision[/quote]

[b]aadil[/b], you're evidently new here. Allow me to point out that your question is more appropriate for the [b]Advanced Physics[/b] forum. [b]Introductory Physics[/b] is just what it sounds like -- a place for posting problems in basic mechanics, heat, optics, acoustics, and electricity and magnetism. No calculus required, or allowed, for that matter.

Nevertheless, in a neutron-neutron collision there is no change in speed due to conservation of momentum. There are collision experiments, which produce fast neutrons as a product of the collision, but these are the result of short-lived nuclear fusion reactions.


I just don't understand how generating functions can be used to solve questions like in how many ways can w pennies, x nickels, y dimes, and z quarters be used to form a cents? I've come across this problem  many times, but don't understand how generating functions tie into it. Help!

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=23096]Do you mean generating functionology?[/url]

Find the coefficient of $x^a$ in

$(1+x+x^2+\ldots)(1+x^5+x^{10}+\ldots)(1+x^{10}+\ldots)(1+x^{25}+\ldots)$


algorithm
number theory
Euclidean algorithm

If a and b are relatively coprime, find all possible values of $\gcd(a+b,a^{2}+b^{2}).$

This reminds me of an old Thomas Jefferson practice problem.

[hide="Hint"]
Use the [url=http://www-math.cudenver.edu/~wcherowi/courses/m5410/exeucalg.html]Euclidean algorithm[/url].
[/hide]

Would you mind posting the full answer please? I am still lost... thanks


On the GBML meet today there was this one problem where there was the expression "0.2222...base 9" in it and I couldn't get it, how would you get the fractional form?

What does a decimal mean, exactly?  [hide].a base b = a/b.  .0c base b = c/b2.  And so on.  [/hide]Can you do it now?

also, tare or someone else who was there today, could you post some of the other questions?

Unfortunately we can not bring home the questions. However many of the past problems (the easier ones) are in the GBML Problem Sets so if you look in there you'll be able to have a handful of the problems.

Nope :)

that's pretty weird...

Back to your question.  Maybe you're intimidated because it's in a base different from 10.  So what if it's just .222...   base 10?  How do you find the fractional form of that?  Use what Joel said, and you also have to know how to sum an infinite series.

yeah that last four words is the problem I'm facing :lol:

Another way to think about it: What is 0.999... base 10?  What is 0.888... base 9?

The first one is 1 and the second is 1 also?

Right!  So now can you figure out 0.2222... base 9?

*bangs head on table until brain comes out* Nope! :lol: If I had to guess I'd say 1/3...

You know that 0.888... base 9 is 1.  So what does that make 0.444... base 9?  And so what does that make 0.222... base 9?

Repeating decimals .nnnnnnnnnnn... of base b can be re-written as n/(b-1).

There is a simple proof. since .(b-1)(b-1)(b-1)(b-1)(b-1)(b-1)... is the same as 1, if we devide this by (b-1) we get .11111... or 1/(b-1) now multiply by n and we get .nnnnn...=n/(b-1)

You could also use this method they teach in Algebra 1, before you ever hear the words "geometric sequence," although the motivation behind this is almost equivalent to using geometric series....

if x = 0.222...9,
  9x = 2.222...9,
so 8x = (2.222... - 0.222....) = 2, so x = 2/8 = 1/4.

Oooooooh...:oooooooh: I got it now, thanks for all of your patience (which must've been a whole lot :lol:)


calculus
integration
function
calculus computations

I've been stuck on this problem for some time now, so I'll be grateful for any hints  :)

Let $ f$ be a continuous non-negative function on $ [0,\infty)$ with a bounded primitve-function there (i.e. there exists an M such that $ |\int_{0}^{x}f(x)dx|<M$ for all $ x \in [0,\infty)$). Prove that the improper integral $ \int_{0}^{\infty}f(x)dx$ converges.

Thanks in advance!
Perrin

[quote="Kent Merryfield"]$ g(x) \equal{} \int_0^xf(t)\,dt$ is nondecreasing.

Since it is nondecreasing, it is subject to the monotone series (monotone function) alternative: either it converges or it is unbounded and tends to infinity.[/quote]

doh! Somehow, I always manage to overlook the simplest solutions - it's frustrating :oops:

Thank you very much for your help!
Perrin


esto akolu8ia $a_{n}$ me $a_{1}=1$ kai
 $a_{n}=n(a_{n-1}+1)$ gia ka8e $n\geq 2$
Na ipologisete to

 $K=\prod(1+\frac{1}{a_{i}})$    (apo i=1 mexri apiro)

 :D

edit::sorry.. :maybe: i8ela na to stilo stin algebra..
8a to valo +eki..doste lisi sto topik tis algebras...

Π=1/n! (απο το 1εως το απειρο )=e


quadratics
Vieta
absolute value
algebra
quadratic formula

$ x^2\plus{}3x\minus{}K\equal{}0$ has roots $ x\equal{}a,b$, and $ x^2\plus{}3x\minus{}10\equal{}0$ has roots $ x\equal{}c,d$.  $ |a\minus{}b|\equal{}2|c\minus{}d|$.  Compute $ K$.

[quote="Octahedr0n"]$ x^2 \plus{} 3x \minus{} K \equal{} 0$ has roots $ x \equal{} a,b$, and $ x^2 \plus{} 3x \minus{} 10 \equal{} 0$ has roots $ x \equal{} c,d$.  $ |a \minus{} b| \equal{} 2|c \minus{} d|$.  Compute $ K$.[/quote]NOTE: Are sure you copied those coefficients correctly? With the problem as written, the solution is not textbook pretty.
[u]Hint:[/u]
STEP 1:
Calculate $ c$ and $ d$. You could start by finding the roots of $ x^2 \plus{} 3x \minus{} 10 \equal{} 0$. Do not worry about which root you call $ c$ and which root you call $ d$ for now. Solve the equation by whatever method works for you. Personally, I'd try to solve it by factoring, if possible.
STEP 2:
Calculate $ |c \minus{} d|$ and $ |a \minus{} b| \equal{} 2|c \minus{} d|$. It does not matter which root you call $ c$ and which root you call $ d$, because the absolute value of the difference is the same for both choices.
STEP 3:
Calculate $ a$ and $ b$. Since it does not matter which root is $ a$ and which root is $ b$ either, assume that $ a \ge b$. That makes $ a \minus{} b \equal{} |a \minus{} b|$ to simplify your problem.
STEP 4:
Calculate $ a$ and $ b$. A popular theorem talks about the roots of a quadratic equation and how they are related to the coefficients. Applied to the case of $ x^2 \plus{} 3x \minus{} K \equal{} 0$, it tells you that $ a \plus{} b \equal{} \minus{} 3$ and that $ ab \equal{} K$. Knowing $ a \minus{} b$ from steps 1-3 above, and knowing that $ a \plus{} b \equal{} \minus{} 3$, can you calculate
$ a$ and $ b$?
STEP5:
Calculate $ K$. As stated in STEP 4, $ ab \equal{} K$.

[quote="KMST"]NOTE: Are sure you copied those coefficients correctly? With the problem as written, the solution is not textbook pretty.[/quote]

Yes.

[quote="KMST"]STEP 1:
Calculate $ c$ and $ d$. You could start by finding the roots of $ x^2 \plus{} 3x \minus{} 10 \equal{} 0$. Do not worry about which root you call $ c$ and which root you call $ d$ for now. Solve the equation by whatever method works for you. Personally, I'd try to solve it by factoring, if possible.[/quote]

$ (x \plus{} 5)(x \minus{} 2) \equal{} 0$, so $ x \equal{} \minus{} 5$ and $ x \equal{} 2$.

[quote="KMST"]STEP 2:
Calculate $ |c \minus{} d|$ and $ |a \minus{} b| \equal{} 2|c \minus{} d|$. It does not matter which root you call $ c$ and which root you call $ d$, because the absolute value of the difference is the same for both choices.[/quote]

$ |a \minus{} b| \equal{} 2| \minus{} 5 \minus{} 2| \Rightarrow |a \minus{} b| \equal{} 14$

[quote="KMST"]STEP 3:
Calculate $ a$ and $ b$. Since it does not matter which root is $ a$ and which root is $ b$ either, assume that $ a \ge b$. That makes $ a \minus{} b \equal{} |a \minus{} b|$ to simplify your problem.[/quote]

How do I calculate $ a$ and $ b$ when there is a $ K$ in my equation?

Thanks!

[size=75][EDIT - Fixed Step 0.][/size]

[quote="Octahedr0n"]
How do I calculate $ a$ and $ b$ when there is a $ K$ in my equation?

Thanks![/quote]

There isn't. Assuming $ a \ge b$, $ a \minus{} b \equal{} 14$. From Vieta's formula $ a \plus{} b \equal{} \minus{} 3$. Solve the system.

Anyway, if I had it my way... (edited)
[hide]
$ \left| \frac { \minus{} 3 \plus{} \sqrt {9 \plus{} 4K}}{2} \minus{} \frac { \minus{} 3 \minus{} \sqrt {9 \plus{} 4K}}{2} \right| \equal{} 14 \implies \sqrt {9 \plus{} 4K} \equal{} 14$ which gives $ K \equal{} \frac { 14^2 \minus{}9 }{4} \equal{} \frac {187}{4}$.
[/hide]

The first sentence of each step describes what you're doing in that step.  Thus, "Calculate $ a$ and $ b$" is an instruction immediately followed by a description of the steps necessary to do it.

KMST also left off Step 0: don't make silly computational errors.  Specifically, $ x^2 \plus{} 3x \minus{} 10 \neq (x \plus{} 5)(x \minus{} 3)$.

[hide="nonrootfinding"]$ a \plus{} b \equal{} c \plus{} d \equal{} \minus{} 3$

$ a^2 \plus{} b^2 \plus{} 2ab \equal{} c^2 \plus{} d^2 \plus{} 2cd$

$ a^2 \plus{} b^2 \minus{} 2ab \equal{} 4(c^2 \plus{} d^2 \minus{} 2cd) \equal{} 4(a^2 \plus{} b^2 \plus{} 2ab \minus{} 4cd) \equal{} 4(a^2 \plus{} b^2 \plus{} 2ab \plus{} 40)$

$ (a \minus{} b)^2 \equal{} 4[(a \plus{} b)^2 \plus{} 40] \equal{} 4(9 \plus{} 40) \equal{} 4*49$

Take $ a\ge b$;

$ a \minus{} b \equal{} 2*7 \equal{} 14$
$ a \plus{} b \equal{} \minus{} 3$
$ a \equal{} 11/2$; $ b \equal{} \minus{} 17/2$

$ K \equal{} \minus{} ab \equal{} 11*17/4$[/hide]

[quote="kohjhsd"]Anyway, if I had it my way...
$ \left| \frac { \minus{} 3 \plus{} \sqrt {9 \minus{} 4K}}{2} \minus{} \frac { \minus{} 3 \minus{} \sqrt {9 \minus{} 4K}}{2} \right| \equal{} 14 \implies \sqrt {9 \minus{} 4K} \equal{} 14$ which gives $ K \equal{} \frac {9 \minus{} 14^2}{4} \equal{} \minus{} \frac {187}{4}$.[/quote]

How did you get $ \left| \frac { \minus{} 3 \plus{} \sqrt {9 \minus{} 4K}}{2} \minus{} \frac { \minus{} 3 \minus{} \sqrt {9 \minus{} 4K}}{2} \right| \equal{}\sqrt {9 \minus{} 4K}$?  Also the answer is positive $ \frac{187}{4}$.  Why would that be?  Thanks!

[quote="Octahedr0n"]How did you get $ \left| \frac { \minus{} 3 \plus{} \sqrt {9 \minus{} 4K}}{2} \minus{} \frac { \minus{} 3 \minus{} \sqrt {9 \minus{} 4K}}{2} \right| \equal{} \sqrt {9 \minus{} 4K}$?  Also the answer is positive $ \frac {187}{4}$.  Why would that be?  Thanks![/quote]

Oops, I mixed up the signs in the quadratic formula (thought it was $ \plus{} K$)...
It should be
$ \left| \frac { \minus{} 3 \plus{} \sqrt {9 \plus{} 4K}}{2} \minus{} \frac { \minus{} 3 \minus{} \sqrt {9 \plus{} 4K}}{2} \right| \equal{} \sqrt {9 \plus{} 4K}$.
The two fractions are the two roots of the first equation (from the quadratic formula), and I am taking their difference.


analytic geometry
graphing lines
slope
trigonometry

Given that x^2+y^2=14x+6y+6, what is the largest possible value that 3x+3y can have?

[hide]The equation is one of a circle, with radius 8, centered at $(7,3)$
We want to maximise 3(x+y), or simply maximise (x+y) (the constant coefficient means nothing in this case.)

Because X and Y are on a circle, this problem is asking for where the line $y = x$ intersects with the circle, and thus we solve:
$y = 3+\sqrt{-x^{2}+14x+15}$
$y = x$
From this, we find that
$x = 3+\sqrt{-x^{2}+14x+15}$,
then we can do some algebra (I'll leave that up to you)
and find that $x = \frac{10 \pm \sqrt{112}}{2}$
We only want the positive solution (due to the nature of the problem)
By the equation $y = x$, we see that y is the same, and thus the maximum possible value is
$3 (2 \cdot \frac{10+\sqrt{112}}{2})$
$=\boxed{30+12\sqrt{7}}$
[/hide]

[quote="DiscreetFourierTransform"][hide]Because X and Y are on a circle, this problem is asking for where the line $y = x$ intersects with the circle,
[/hide][/quote]

can you explain that? why is this intersection necessarily the maximum value of $x+y$?

Yes it's not neccessarily the line x+y=0, I think you're supposed to draw the two lines tangent to the circle in the form x+y=c, then the lower value is the minimum, the higher is the maximum. I think this problem is more meant for intermediate.

I'm not sure about the answer 73.  I keep getting something lower. :huh:

There are a few ways to solve this problem:

[hide="Method 1"]
We know that the maximal point will be the tangency point of $(x-7)^{2}+(y-3)^{2}=64$ with $x+y=c$. So, you can substitute $y=c-x$ into the circle's equation and [u]solve for $y$[/u]. Then, the tangency point will be when the discriminant is equal to zero, i.e. $B^{2}=4AC$ in the resulting quadratic. Solve for $c$ and you will get two values; the greater value is the maximal value of $x+y$, so just multiply by 3 to get the final answer[/hide]

[hide=" Method 2"]
Again, we know that the maximal point will be the tangency point of the circle with a line of slope $-1$ (since $y=c-x$); so we know that the line from the center of the circle to the tangency point will be $1$. (The opposite reciprocal of $-1$ is $1$). So we can set up a system; one equation being the slope formula of this radius, and the other will be the distance formula from the center to the tangency point (equal to $8$). Solve as you please, with the tangency point as $(x,y)$.[/hide]

[hide="Method 3"]
Shift the circle so that the center is at the origin. Then the point will be $(x^{'}, y^{'})=(4\sqrt2, 4\sqrt2)$. Shift back.[/hide]

I'm sure there's more... some involving Cauchy's Inequality...   :P

I may have made a computational error, but 73 isn't the maximum I get... [hide]We know that $(x-7)^{2}+(y-3)^{2}= 64$. By Cauchy-Schwarz: \[64\cdot 2 = \left((x-7)^{2}+(y-3)^{2}\right)\left(1^{2}+1^{2}\right) \geq \left((x-7)+(y-3)\right)^{2}= \left(x+y-10\right)^{2}\]Therefore, $8\sqrt{2}\geq x+y-10$, i.e. $3x+3y \leq 24\sqrt{2}+30$.[/hide]

[hide]
you should post this in the real analysis forum to see what kind of solutions they've got  :D [/hide]

[b]Kurt Gödel[/b]'s answer is right: Another method is to parametrize the circle

[hide="thus"]$(x-7)^{2}+(y-3)^{2}=8^{2}\iff x=7+8\cos\varphi\land y=3+8\sin\varphi$, giving $3(x+y)=30+24(\sin\varphi+\cos\varphi)=30+24\sqrt{2}\sin\left(\varphi+{\pi\over 4}\right)\leqslant 30+24\sqrt{2}$[/hide]

Okay, people the question really stated 3x+4y, not 3x+3y.

If you do 3x+3y, then all y'alls answers are right.

If you do it the way's you been doing you get 73.


help me out , thanks :)
and do give complete solutions

is the second question related with spring constant

[hide="My attempt"]
considering the situation when both block and ring stops for first time
we have(using Energy conservation)
$ mga tan{\theta} \equal{} Mga(sec{\theta} \minus{} 1)$
here $ {\theta}$ is the angle made by string with hrizontal,

so,$ \frac {sec{\theta} \minus{} 1}{tan{\theta}} \equal{} \frac {m}{M}$
so,$ \frac {1 \minus{} cos{\theta}}{sin{\theta}} \equal{} \frac {m}{M}$
squaring,we get,
$ \frac {{1 \minus{} cos{\theta}}^2}{1 \minus{} sin^2{\theta}} \equal{} \frac {m^2}{M^2}$

rearranging,we get,
$ \frac {1 \minus{} cos{\theta}}{1 \plus{} cos{\theta}} \equal{} \frac {m^2}{M^2}$

solving,we get,$ cos{\theta} \equal{} \frac {M^2 \minus{} m^2}{M^2 \plus{} m^2}$

then $ tan{\theta} \equal{} \frac {2Mm}{M^2 \minus{} m^2}$

so distance fallen$ \equal{} atan{\theta} \equal{} a\cdot\frac {2Mm}{M^2 \minus{} m^2} \equal{} \frac {2Mma}{M^2 \minus{} m^2}$
as required :)
will try other after coming back from class :P though i am unable to understand the picture
[hide]You are also going to have tough time deciphering my diagram  :rotfl: [/hide]

[/hide]

@skand , good work ,
but
this is where i was stuck ->
i took ->

[img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/2/a/f/2af261fde3d2d0173d2836c91d3756f2f03f34be.gif[/img]

any reasons to oppose the above?

is the ans to second ques 10 cm??i m making a rough attempt
i kno how to do it-tell me where r u stuck?? :)

no , tell me , i tried series parallel something ,but leave it , i am dead sure its a conceptual qn , tell me how u proceed , even if in a rough way , i'll get it :)

I suppose when $ m$ stops,$ M$ also stops

Thats all I can think  :)

[quote="skand"]
so distance fallen$ \equal{} atan{\theta} \equal{} a\cdot\frac {2Mm}{M^2 \minus{} m^2} \equal{} \frac {2Mma}{M^2 \minus{} m^2}$
[/quote]i haven't read it whole but i m getting answer as $ \frac {2m^2a}{M^2 \minus{} m^2}$

2nd ques
i think it shud be 5 cm
see a net force of 50N will act on the block while the tension is 0 and extension in each string is 15 cm
 :)
after that ,i m not sure if this is correct: :maybe:
2(500)x =50
-> x = 5cm :huh:

hey i said that earlier isn't it???
just use kx instead of 2kx in last eqn


How many whole numbers between 1 and 1000 do [b]not[/b] contain the digit 1?

$ \textbf{(A)}\ 512 \qquad \textbf{(B)}\ 648 \qquad \textbf{(C)}\ 720 \qquad \textbf{(D)}\ 728 \qquad \textbf{(E)}\ 800$

[hide="Solution"]
Each digit in the number can be $ 0,2,3,4,\dots,9$, for a total of $ 9^3\equal{}729$.  However, we cannot have the number $ 000$, so we subtract one to get $ 728$, or $ \boxed{\textbf{(D)}}$.
[/hide]

[quote="jupiter68"]Are you sure that (D) is a valid answer, because I don't think that 029,039,049,063,061,etc. are three digit numbers?[/quote]
Where does it say $3$ digit numbers?

The problem asks for whole numbers between $1$ and $1000$ and $029$, $039$, etc. all fit into that category.

There are $9\cdot 9\cdot 9$ numbers. 0 doesn't count so $728$

You can also visualize this problem a bit.

[hide="Another Solution"]Imagine a $10\times10\times10$ cube. This is split into $1000$ unit cubes. Each unit cube corresponds to a number between $000$ and $999$. All the numbers on one face, which we will call A, contains the numbers with a $1$ in the hundreds place, all the numbers on a face adjacent to face A, which we will call B, contains the numbers with a $1$ in the tens place, and all the numbers on a face adjacent to both faces A and B, which we call C, contains the numbers with a $1$ in the ones place.

The edge of the big cube that faces A and B share have a $1$ in both the hundreds and tens place, the edge of the big cube that faces A and C share have a $1$ in the hundreds and ones place, and the edge of the big cube that faces B and C share have a $1$ in the tens and ones place. The unit cube common to all three of these faces will be $111$. So the numbers that are not on any of these three faces do not contain the digit $1$, which is basically the $10\times10\times10$ cube with one layer scraped off, which is a $9\times9\times9$ cube.

So there are $9\times9\times9=729$ numbers without the digit $1$ between $000$ and $999$. But we are looking between $1$ and $1000$, so we exclude $0$, which doesn't have a $1$, to get $728$, and include $1000$, which has a $1$ so it doesn't affect our count. So the answer is $\boxed{728}$ numbers.[/hide]

[hide=Practicing Using Computer]
In[90]:=
prolist = IntegerDigits[Range[1, 1000]];

In[93]:= oh = Select[prolist, FreeQ[#, 1] &];

In[94]:= Length@oh

Out[94]= 728

[/hide]

[Hide=Another probably slower solution]Count what you don't want which gives you 272 numbers and then subtract from 1000. You get $$\boxed{728}$$ in the end. I'm pretty sure it is called complementary counting, but I'm not sure.[/Hide]

[quote=rhydon516][quote=dolphinday]inclusive or exclusive
meaning do 1 and 1000 count as numbers[/quote]

Does it matter?[/quote]

No. They both have a 1.


Find all the integers n>1 for which $x^{25}\equiv x$ (mod n) is satisfied for every x integer. Enjoy! :D

(x^24) *(x - 1) =0modn.

n=2 is obvious.  n also cannot be odd.  For any other n, x can be the first prime greater than n such that x-1 is not a multiple of n.  Thus, n=2 should be the only answer.

Umm, you factored that wrong... $x^{25}-x = x(x^{24}-1)$...

Anyway, $x^{25} \equiv x \mod{n} \Rightarrow x^{24} \equiv 1 \mod{n}$ (assume $(x,n) = 1$ for now) which means by Euler's theorem $\phi(n) | 24$. Since we are assuming $(x,n) = 1$, $n$ must be prime so that the GCD is right for all $x$. Thus we have $n = 2, 3, 5, 7, 13$ as possible answers.

I'm not sure how to find all the composite solutions, but I know $6$ is one...

First lets say p is a prime that divides n...then
$x^[24]=1(mod p)$ for $x=1,2,...,p-1$  and for $p>2$ then p is a divisor of $2^{24}-1$ which easy can be factorized like: $2^{24}-1=(2^6-1)(2^6+1)(2^4+1)(2^8-2^4+1)=(9)(7)(5)(13)(17)(241)$ and its easy to prove that the only primes that can are 3, 5, 7, 13 (checking $3^{24}-1$) plus 2 that is obvious...and also see that if $p$ and $q$ are primes that fulfill that then $pq$ also.

And that n is free of squares because lets say $p^2$ divides n then $p^{25}-p=p(p^{24}-1)$ that only has one p factor so contradiction.

Then the n's are the products of that 5 primes. that makes 31 answers


x,y,z are positive reals such that xyz=1. If 1/x + 1/y +1/z >= x+y+z
Prove that 1/x^k +1/y^k +1/z^k >= x^k +y^k + z^k for every positive integer k.

$ \sum\frac{1}{x}\geq\sum x\Leftrightarrow(x\minus{}1)(y\minus{}1)(z\minus{}1)\leq 0$
$ x^k\minus{}1\equal{}(x\minus{}1)A(A>0);y^k\minus{}1\equal{}(y\minus{}1)B(B>0);z^k\minus{}1\equal{}(z\minus{}1)C(C>0)$
$ \Rightarrow\prod(x^k\minus{}1)\equal{}\prod(x\minus{}1)ABC\leq 0\Rightarrow$ineq is solved


inequalities
geometry
inequalities unsolved

Prove that for any triagle $ABC$ the following inequality holds:
 \[a^3\cdot b^4\cdot c^5 \geq \frac{25\cdot \sqrt 5\cdot 6 \cdot 2S}{27}\]
where $a=BC,b=AC,c=AB$,and $S$ denote the area of the triagle $ABC$
[color=red]Edit[/color]:
 \[a^3\cdot b^4\cdot c^5 \geq \frac{25\cdot \sqrt 5 \cdot (2S)^6}{27}\]

Are you sure it's true?
At least it's not always true for a=b=c.

what if ABC is an equilateral triangle where its sides tend to 0?

Sorry for my bad hand-writting!This problem should as follow:
\[a^3\cdot b^4\cdot c^5 \geq \frac{25\cdot \sqrt 5 \cdot (2S)^6}{27}\]


analytic geometry
geometry
rotation
invariant
geometric transformation

A straight line  has x-intercept 'a' and y-intercept 'b'. The axes are rotated through some angle about the origin. Now the intercepts in the new coordinates are 'p' and q respectively.

Show that
$1/p^{2}+1/q^{2}=1/a^{2}+1/b^{2}$

[hide="possibly hint"]rotating the axes one way is equivalent to rotating the line the other way, and the perpendicular distance from the origin to the line is invariant during the rotation
[/hide]

say we have right triangle ABC with <C=90, then we have that d(C, AB)=2[ABC]/c=ab/sqrt(a^2+b^2)=1/( 1/a^2+1^b^2)^.5
basically what this gives us is that the LHS and RHS are ( d( origin, line) )^-2 of the original line and the rotated line

so the rotation does not change the distance from the origin to the line [rotation preserves distance after all]...so proven


geometry
cyclic quadrilateral
quadrilateral
concurrency
angle bisector
IMO
IMO 2003

Let $ABCD$ be a cyclic quadrilateral. Let $P$, $Q$, $R$ be the feet of the perpendiculars from $D$ to the lines $BC$, $CA$, $AB$, respectively. Show that $PQ=QR$ if and only if the bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$.

in two cyclic quadrilatrals APRD & DRCQ ,AD & CD are diameters respectively and  we have:
                      RQ/sin(<RCQ)=CD/2 or RQ/sin(<BCA)=CD/2
                      PR/sin(<PAR)=AD/2 or PR/sin(<BAC)=AD/2
By dividing:
                      CD/AD= sin(<BAC)/sin(<BCA)=BC/AB
Since CD/AD=BC/AB thus bisectors of <CBA and <ADC intersect AC in the same poin.
S.Ebadollahi

i must admit that this problem was quite easy by imo standards.my solution is based on a standard trick using pedal triangles and projections which is damn easy and pretty similar to the proof  posted here.

here is an interesting thing about this problem and other 2 IMO problems in the past! it's about the so called Pedal Triangle Trick!

this was posted on a forum by: fritue2000 and I thought it would be interesting to post it here also!

"The previous two IMO problems solved by the Pedal Triangle Trick.
(1996 IMO) Let P be a point inside the triangle ABC such that angle
APB - angle ACB = angle APC - angle ABC. Let D, E be the incenters of
triangles APB, APC respectively. Show that AP, BD, CE meet at a
point.

As I know, at the jury meeting at IMO 1996, there is a discussion
for this problem becasuse its solution is very similar to problem
2 of the 34th IMO. The techniques include 'inversion' and 'PTT', etc.
For example, http://home1.pacific.net.sg/~slwee/imo96/imo96op.htm )

(1993 IMO) Let D be a point inside the acute-angled triangle ABC such
that angle ADB = angle ACB + 90 degrees, and AC*BD = AD*BC.

(a) Calculate the ratio AB*CD/(AC*BD).

The well-known 'Pedal Triangle Trick' is "For any point D, let
X, Y, Z be feet of the altitudes from D to AB, BC and CA. Then,
YZ = (DA*BC)/2r, etc, where r is the circumradius of ABC."
The proof is very easy, since D, A, Y, Z lies on a circle with
diameter DA, by the law of sines, YZ = DA sin A = DA*(BC/2r).
(2003 IMO) Given is a cyclic quadrilateral ABCD and let P, Q, R
be feet of the altitudes from D to AB, BC and CA respectively.

Prove that if PR = RQ then the interior angle bisectors of the
angles <ABC and <ADC are concurrent on AC.
Solution) By PTT, PR=RQ implies (DA*BC)/2r = (DC/AB)/2r, so,
CD/DA=BC/AB implies the results. q.e.d.

The most noticeable thing is in the above solution, we did not
used the condtion 'ABCD is cyclic'. And as an IMO problem, it
is not so intersting because all three problems 1993, 1996,
2003 solved by exactly the same TWO ways (PTT and inversion).
Well, the PTT is really well-known, for example, it appears
at "Geometry Revisited" by Coexter. It also used for a proof
of Ptolemy's Theorem.

In a cyclic qudrilateral ABCD, the three points are on the
simson line, we have PR+RQ=RP, with the same notation of 2003
imo. Then, (DA*BC)/2r + (DC*AB)/2r = (DB*AC)/2r or
DA*BC + DC*AB = DB*AC. "

A bit of standard angle chasing shows that triangles DPR and DBC are similar and so are DQR and DBA. Thus PR=QR gives DC/CB = DR/RP = DR/RQ = DA/AB, and the result follows from the angle bisector theorem.

Yeah, I had something like that too, was quite easy :D solved in about 30 minutes (which is extremely few for me)

Lemma: Let $ABCD$ be a cyclic quadrilateral. Let $DR$ and $DP$ be the altitudes from $D$ to $AB$ and $DC$, respectively. Let $RP$ intersect $AC$ at $Q$. Then $\angle AQD$ is a right angle.

Proof of Lemma:

$\angle DRB = \angle DPB = \pi / 2$, so $DRPB$ is cyclic. Then $\angle DRP = \angle DBP$. From cyclic quadrilateral $ABCD$, $\angle DBP = \angle DAP$. $\angle DRP = \angle DAP$, so $RAQD$ is cyclic. Thus $\angle AQD = \pi - \angle ARD = \pi / 2$.

Proof:

Let the bisector of $\angle ABC$ meet $AC$ at S. Let the bisector of $\angle ADC$ meet $AC$ at T.
By the Lemma, $R,Q,P$ are collinear.
Let $\angle DRQ = \alpha$. $\angle APD = \pi - \alpha - \angle RDP = \pi - \alpha - (\pi - \angle ABC) = \angle ABC - \alpha$.
$\frac{RQ}{\sin \angle RDQ} = \frac{DQ}{\sin \alpha}$, $\frac{PQ}{\sin \angle PDQ} = \frac{DQ}{\sin (\angle ABC - \alpha)}$.
$\frac{RQ}{PQ} \cdot \frac{\sin PDQ}{\sin RDQ} = \frac{sin(\angle ABC - \alpha)}{\sin \alpha}$.
$\angle RDQ = \pi - \angle RAQ = \angle BAC$. Also $\angle PDQ = \angle BCA$.
$\frac{RQ}{PQ} \cdot \frac{\sin \angle BCA}{\sin \angle BAC} = \frac{\sin (\angle ABC - \alpha)}{\sin \alpha}$.
$\frac{RQ}{PQ} \cdot \frac{AB}{BC} = \frac{\sin \angle QCD}{\sin \angle QAD} = \frac{AD}{DC}$.
So $RQ=PQ$ if and only if $\frac{AB}{BC} = \frac{AD}{DC}$.

By the Angle Bisector Theorem,
$\frac{AS}{SC} = \frac{AB}{BC}$ and $\frac{AT}{TC} = \frac{AD}{DC}$.
So $\frac{AB}{BC} = \frac{AD}{DC}$ if and only if $\frac{AS}{SC} = \frac{AT}{TC}$.
This can happen if and only if $S=T$.
So $PQ=QR$ if and only if the angle bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$. QED.

Yay I exploded this problem with like 20 cyclic quadrilaterals.

Okay first well-know fact: Simson's Line, so we know P, Q, R are collinear.

i'll go with the only if direction
Let X be the intersection of the angle bisectors. Extend BX, DX to Y, Z, where Y and Z lies on the circumcircle of ABCD. Y, Z lies on the perpendicular bisector of AC, hence YZ is a diameter, and it goes through midpoint M of AC. angle AMY = 90, angle ZDY = 90 because YZ is a diameter, so MYDX is cyclic, so angle ZYB = MDX = ADB. DX bisects ADC hence ADM = BDC = BAC = PDQ.
angle QAD=QPD. Hence triangle PQD is similar to triangle AMD, and triangle ADC is simlar to PDR. DM is a median hence DQ is a median => Q is the midpoint of PR as desired.

If direction:
Define X to be on AC such that DX bisects ADC, DX intersect the circumcircle at Z, M is the midpoint of AC, Y is the intersection of MZ and the circumcircle. triangle ADC is simlar to PDR. DM is a median and DQ is a median, so triangle PQD is similar to triangle AMD, angle ADM = PDQ=BAC = BDC. By definition XD bisects ADC hence XD bisects MDB so MDX = XDQ. QMY=90, YZ is still diameter, MYDX is cyclic, ZYX = MDX = ZDB. Now let BX intersect the circumcircle at Y'. XY'Z = XYZ, now Y' is unique(consider the locus of points Y' such that XY'Z = XYZ, which is a circle, and Y' lies on the cirucmcircle of ABCD. Two circle intersects in two points, one of which is Z, hence there's only one such point left and thus it's unique), hence Y = Y', thus B, X, Y are collinear. Y lies on the perpendicular bisector of AC so YA=YC, thus angle ABY = YBC.

QED

The following will be useful: Let $ \triangle ABC$ be a triangle, $ D$ an arbitrary point and $ X$, $ Y$, $ Z$ the pedal points wrt the sides $ BC$, $ CA$, $ AB$. Then $ YZ\equal{} \frac{AD \cdot BC}{2R}$, where $ R$ denotes the circrumradius of $ \triangle ABC$. That's very easy to prove, I won't write it down now  :) .
Using the lemma shown above, we have $ PQ\equal{} \frac{CD \cdot AB}{2R}$ and $ QR\equal{} \frac{AD \cdot BC}{2R}$. Hence, $ PQ \equal{}QR \Leftrightarrow \frac{AB}{BC} \equal{} \frac{AD}{DC}$. Let $ S$ denote the intersection of the angle bisectors of $ \angle ABC$ and $ \angle CDA$. If $ S \in AC$, then $ \frac{AB}{BC} \equal{} \frac{AS}{SC} \equal{} \frac{AD}{DC}$. If $ PQ\equal{}QR$, then $ \frac{AS_1}{S_1C} \equal{} \frac{AB}{BC} \equal{} \frac{AD}{DC} \equal{} \frac{AS_2}{S_2C}$, hence $ S_2\equal{}S_1\equal{}S$, where $ S_1$ and $ S_2$ denote the intersections of the angle bisectors of angles $ \angle ABC$ and $ \angle CDA$ with the side $ AC$. Thus, the problem is proved.

Hmmm I came up with this, which uses a bit more of Simson line properties.

Let the angle bisector of $ \angle{ABC}$ intersect $ AC$ at $ N$, let the midpoint of $ AC$ be $ M$, let $ E$ and $ F$ be points diametrically opposite on the circumcircle of $ ABCD$ such that $ EF$ is the perpendicular bisector of $ AC$, with $ E$ on the same side of $ AC$ as $ B$. Let $ B'$ be the point diametrically opposite to $ B$.

$ FDQN$ is cyclic, $ FDMN$ is a rectangle (cyclic), so $ \angle{FND} = \angle{FQD} = \angle{MDQ}$.
$ \angle{FND} = \angle{FBD} + \angle{BFE}$
$ \angle{B'BF} = \angle{BFE}$, since both are diameters.
Thus $ \angle{B'BD} = \angle{BNE} = \angle{MDQ}$.

$ P,Q,R$ are collinear and are the Simson line of $ D$. $ AC$ is the Simson line of $ B'$. By a well known property of Simson lines, $ \angle{PQA} = \frac12\overarc{B'D}$, but $ \frac12\overarc{B'D} = \angle{B'BD} = \angle{BNE} = \angle{MDQ}$ and $ \angle{PQA} = \angle{PDA} = \angle{CDR}$, since $ PDR$ is directly similar to $ ADC$. Thus we have a spiral similarity with center $ D$ mapping $ A,M,C$ to $ P,Q,R$ respectively, and since $ M$ is the midpoint of $ AC$, $ Q$ is the midpoint of $ PR$ so we are done.

The argument should work in reverse as well, with some tweaking here and there.

Use that a quadrilateral is harmonic if and only if the diagonals are symmedians(easy to prove with polarity and crossed ratio).
Now, the problem is equivalent to this:
$\triangle{ABC}$,$D$ is on the circumcircle of $\triangle{ABC}$(on the arc$BC$). Denote $X,Y,Z$ the projections of $D$ onto $AB,BC,CA$. $XY=YZ$if and only if $D$ is on the symmedian with respect to $A$.
Proof: See that $BXDY$ and $DYZC$ are inscribed in circles with diameter$BD$,respectively$CD$. We have from Simson's line that $X-Y-Z$are collinear. In $\triangle{XBY}$(aplying law of sines) $XY=BDsinB$, and in $\triangle{YZC}$ we have$YZ=DCsinC$. So, $XY=YZ$ if and only if $\frac{BD}{CD}=\frac{sinC}{sinB}$, but it happens only for symmedian(note that $\frac{sinBAM}{sinCAM}=\frac{sinB}{sinC}$, where $M$ is the midpoint of $BC$.

hey guys)
I solved this problem in about 10 minutes, but I think there is some thing wrong because I didn't use the fact that $ABCD$ is cyclic.
Can it be so?

A different approach:

To begin with, it is clear that triangles $\bigtriangleup  DRP$ and $\bigtriangleup  DAC$ similar. (This follows by a very simple angle chasing, as in the above posts). This also gives us the fact that  triangles $\bigtriangleup  DAR$ and $\bigtriangleup  DPC$ are similar, therefore $\frac{DA}{DC}=\frac{RA}{PC} (1)$

Also, points $P,Q,R$ are collinear (Simson's line)

It remains to prove the equivalence $PQ=QR \Leftrightarrow \frac{BA}{BC}=\frac{DA}{DC}$

Let's focus on triangle $\bigtriangleup RBP$ . Points $A,Q,C $ are collinear and lie on the lines $BR,RP,PB$ respectively. Applying Menelaus' Theorem, we have:
$\frac{AR}{AB}\cdot \frac{CB}{CP}\cdot \frac{QP}{QR}=1$

Therefore, $QP=QR\Leftrightarrow \frac{QP}{QR}=1\Leftrightarrow \frac{AB}{BC}=\frac{RA}{PC}\Leftrightarrow \frac{AB}{BC}= \frac{DA}{DC}$,

since $\frac{DA}{DC}=\frac{RA}{PC}$ by relation (1). The conclusion follows immediately.

Nick

[quote="nickthegreek"]A different approach:[/quote]
Your solution is actually identical to hatchguy's. Take a look:

[quote="hatchguy"]The concurrent with $AC$ bisectors condition is equivalent to $\frac{AB}{BC}=\frac{AD}{DC}$ $=> \frac{AD*BC}{AB*DC} =1$

Since $\angle DAB+ \angle DCB= 180$  and $\angle DCP+ \angle DCB= 180$ we have $\angle DAB= \angle DCP$.

Therefore we have that $ARD$ is similar to $CPD$ and therefore $\frac{AR}{PC}=\frac{AD}{DC}$.  (1)

By Menelaus theorem for transversal $CQA$ in triangle $BRP$ we obtain:

$\frac{AB*RQ*PC}{AR*PQ*BC}=1$.  Therefore $\frac{RQ}{QP}=\frac{AR*BC}{AB*PC}=\frac{AD*BC}{AB*DC}$  (because of (1)).  The conclusion is now obvious.[/quote]

Note that $ARQD$ and $CPDQ$ are cyclic from the right angles. Using Ptolemy's Theorem and some trig, we get that $PQ=CD \sin(\angle BCA)$ and $QR= AD \sin(\angle BAC)$. Using LoS on triangle $ABC$ we get that $PQ=QR \iff \frac{AB}{BC}=\frac{AD}{DC}$, however this is equivalent to the angle bisectors being concurrent by the angle bisector theorem.

...what was I doing with that projective mess last year

Notice that $\frac{RQ}{\sin \angle BAC} = AD$ and $\frac{PQ}{\sin \angle ACB} = CD$ by the Law of Sines, so $$\frac{CD}{AD} = \frac{PQ}{QR} \cdot \frac{\sin \angle BAC}{\sin \angle BCA} = \frac{PQ}{QR} \cdot \frac{BC}{AB}.$$ If $\frac{BC}{AB} = \frac{CD}{AD}$, then the bisectors clearly concur, so we have the equivalence.

Yet another approach with complex numbers:

The feet of the altitudes are
\begin{align*}
p&= \frac 12\left(d+b+c-\frac{bc}d\right) \\
q &= \frac 12\left(d+c+a-\frac{ac}d\right) \\
r &= \frac 12\left(d+a+b-\frac{ab}d\right).
\end{align*}Thus$$PQ = QR \iff |b-a|\left|\frac cd - 1\right| = |b-c|\left|\frac ad - 1 \right|.$$But the other condition is$$\frac{AB}{BC} = \frac{AD}{DC} \iff \frac{|a-b|}{|b-c|} = \frac{|a-d|}{|d-c|},$$which is clearly equivalent.

Clearly the bisectors of $\angle ABC$ and $\angle ADC$ concur on $AC$ iff $\frac{BA}{BC}=\frac{DA}{DC}$, which means that $ABCD$ is harmonic and thus $\overline{BD}$ is a symmedian of $\triangle ADC$. Now, by considering cyclic quadrilaterals $ADQR$ and $CDQP$, we have
$$PR=CD\sin \angle QCP \text{ and } PQ=AD\sin QAR \implies \frac{PQ}{QR}=\frac{AD \sin \angle CDB}{CD \sin \angle ADB}=\frac{AD \sin \angle ADB'}{CD \sin \angle CDB'}=\frac{AB'}{CB'},$$
where $B'$ is a point on $\overline{AC}$ such that $\overline{DB}$ and $\overline{DB'}$ are isogonal.
On the other hand, if $M$ is the midpoint of $\overline{AC}$, by the Ratio Lemma we have
$$1=\frac{AM}{CM}=\frac{AD \sin \angle ADM}{CD \sin \angle CDM}.$$
so $PQ=QR$ iff $B'=M$, which occurs by definition only when $\overline{DB}$ is the symmedian. $\blacksquare$

[b]Remark:[/b] I'm not entirely sure why I used ratio lemma because $\frac{\sin \angle CAB}{\sin \angle ACB}=\frac{BC}{BA}$ from the Law of Sines on $\triangle ABC$. Maybe because I came into the problem thinking something like ratio lemma was going to be useful (see USAJMO 2015/5)

... or if we want to be stupid we just use the same complex number formula three times and get the answer.

WLOG let $(ABCD)$ be the unit circle. Then we compute
\begin{align*}
p&=\frac{1}{2}\left(b+c+d-\frac{bc}{d}\right),\\
q&=\frac{1}{2}\left(a+c+d-\frac{ac}{d}\right),\\
r&=\frac{1}{2}\left(a+b+d-\frac{ab}{d}\right).
\end{align*}
Then
\begin{align*}
p-q&=\frac{1}{2}\left(\frac{c}{d}-1\right)(a-b),\\
r-q&=\frac{1}{2}\left(\frac{a}{d}-1\right)(c-b).
\end{align*}
Thus
$$PQ=QR \iff |p-q|=|r-q| \iff \frac{|c-d|}{|a-d|}=\frac{|c-b|}{|a-b|} \iff \frac{CD}{AD}=\frac{CB}{AB},$$
which clearly is equivalent to the concurrence on the angle bisectors. $\blacksquare$

2nd to last G1...

Note that the angle bisectors concur condition is just $ABCD$ harmonic, which is in turn equivalent to $DB$ being a symmedian in $\triangle ADC$. This is then equivalent to $\angle ADM=\angle CDB$ and note that $\angle CDB=\angle CAB=\angle RDQ$, so we just need to show that $\angle ADM=\angle RDQ$ if and only if $PQ=QR$.

Since $ARDQ$ and $CPQD$ are both cyclic, there is a spiral similarity centered at $D$ that sends $AC$ to $RP$. This also sends $M$ to the midpoint of $PR$. Thus, we have that $\angle ADM=\angle RDQ$ if and only if $Q$ is the midpoint of $P$, so we are done.

marabot solve - the little projective geo i know is actually useful?

We will prove the if direction (only if is simply the reverse of this). The condition is equivalent to $ABCD$ being harmonic by the angle bisector theorem. Let $P_{\infty}$ be the point of infinity along line $PQR$ (which we know is collinear by Simson lines), $X = (ABC) \cap BP_{\infty}$, and $Y = BX \cap AC$. It is well known that $X,Q,D$ are collinear. Now we have \[ -1 = (A,C; B,D) = (C,A; D,B) \stackrel B= (C,A; Q,Y)  \stackrel X= (P, R; Q, P_\infty)\] which implies $Q$ is the midpoint of $PR$, simply reverse this argument for the other direction to get the desired.

Very angry that this took two hours. By Ptolemy's on the cyclic quads we get PQ*CD+PC*QD=PD*QC, hence PQ*CD=CD^2(cosCDPsinCDQ-sinCDPcosCDQ) by trig. This simplifies to PQ=CDsinBDP=CDsinBCA and analogously QR=ADsinBAC. Then law of sines on ABC shows that PQ=QR is equivalent to AB/BC=AD/DC, which, if we let one of the angle bisectors intersect AC at some point, it yields that being one of the angle bisectors makes it both. $\blacksquare$

We use complex numbers. Situate $(ABCD)$ to be the unit circle, where lowercase letters denote the complex number representing the corresponding uppercase letters. It is easy to see by complex foot that $p = \frac{1}{2}(b+c+d-bc\overline{d}),$ $q = \frac{1}{2}(c+a+d-ca\overline{d}),$ and $r = \frac{1}{2}(a+b+d-ab\overline{d}).$ Taking a homothety of scale $2$ centered at the origin (this won't affect $\frac{PQ}{QR}$) and noting that $\overline{d} = \frac{1}{d},$ we get the new points $$p' = b+c+d-\frac{bc}{d},$$ $$q' = c+a+d-\frac{ca}{d},$$ and $$r' = a+b+d-\frac{ab}{d}.$$ This means that $$p' - q' = b - a + \frac{ca}{d} - \frac{bc}{d} = (b-a) - \frac{c}{d}(b-a) = (b-a)\left(1 - \frac{c}{d}\right) = \frac{(b-a)(d-c)}{d}$$ and similarly $$q' - r' = \frac{(b-c)(d-a)}{a}.$$ Taking magnitudes, we get $$|p'-q'| = \frac{|b-a||d-c|}{|d|} = |b-a||c-d| = AB \cdot CD$$ and similarly $$|q' - r'| = |b-c||d-a| = BC \cdot DA.$$ We know that $PQ = QR$ if and only if $P'Q' = Q'R',$ which is equivalent to these magnitudes being equal. Thus we must have $$AB \cdot CD = BC \cdot DA.$$ We can rearrange this expression to $$\frac{AB}{BC} = \frac{AD}{DC}.$$ Hence if $K$ is the foot of the angle bisector of $\angle ABC$ onto $AC,$ then the Angle Bisector Theorem yields $$\frac{AK}{KC} = \frac{AB}{BC} = \frac{AD}{DC}.$$ This implies that $K$ is the foot of the angle bisector of $\angle ADC$ onto $AC$ as well. Therefore, if $PQ = QR,$ then the angle bisectors of $\angle ABC$ and $\angle ADC$ intersect on $AC.$ Because all our steps are reversible, the reverse statement holds as well, and we are done.

By the angle bisector theorem, the second condition is equivalent to $ABCD$ being harmonic.

Now, for the first condition, note that $D$ is the Miquel point of quadrilateral $ABRQ$. Thus, it must be the center of the spiral similarity which takes $\overline{PR}$ to $\overline{AC}$; let $M$ be the image of $Q$ under this spiral similarity. Then since $P$, $Q$, and $R$ are collinear by Simson lines, the $PQ = QR$ condition is equivalent to $M$ being the midpoint of $\overline{AC}$. But
$$\measuredangle ADB = ACB = \measuredangle QCR = \measuredangle QDR = \measuredangle MDC,$$
so this in turn is equivalent to $DB$ being the $D$-symmedian in $\triangle DAC$. However, this, too happens iff $ABCD$ is harmonic, so we are done.

Here's a basic projective approach: Let E,F be the intersection points of the line through B parallel to PR and BQ, each with the circumcircle, respectively. Notice from cyclic quad CDQP and PQR being a simson line, we get $QDC=180-QPC=BPQ=EBC=EDC$, which implies that EQD is a line. Now, $$-1=(PR;QP_{\infty})=^B(CA;FE)=^Q(AC;BD)$$ harmonic, as desired. $\blacksquare$

Note that $PQR$ is a Simson Line. Let $PD$ intersect $(ABCD)$ again at $E$. Note that $AE \parallel PQ$ (this is well known, just an angle chase).

Now, let the point at infinity of lines $PQ$ and $AE$ be $\infty$. Then
\[ (P,R; Q, \infty) \stackrel{A}{=} (P, B; C, (AE \cap BC)) \stackrel{E}{=} (D, B; C, A) \]
which implies that $Q$ is the midpoint of $PR$ if and only if $ABCD$ is harmonic, if and only if the bisectors of $\angle ABC$ and $\angle ADC$ intersect on $AC$. $\blacksquare$

Let $E$ be the intersection of the bisector of $\angle ADC$ and $AC$.The complete diagram is shown below.
[img width = 30%, height = 30%]https://i.imgur.com/hcGi7fc.png[/img]
We know that, $P$, $Q$ and $R$ are colinear by Simson's Line. We have the following quadrilaterals cyclic :
\[\angle ARD=\angle AQD=90^{\circ}\implies AQDR\ cyclic\]
\[\angle BRD=\angle BPD=90^{\circ}\implies BPDR\ cyclic\]
\[\angle CQD=\angle CPD=90^{\circ}\implies CDQP\ cyclic\]
Angle chasing gives us, $\angle DQR=\angle DAR=\angle BCD$ and $\angle QRD = \angle PRD=\angle PBD=\angle CBD$

Which implies, $\triangle QRD\sim \triangle BCD\implies \frac{QR}{DQ}=\frac{BC}{CD}$
Again, $\angle DBA=\angle DBR=\angle DPR=\angle DPQ$ and $\angle BAD=180^{\circ}-\angle DAR = 180^{\circ}-\angle DQR=\angle PQD$

Which implies, $\triangle ABD\sim \triangle PQD\implies \frac{PQ}{DQ}=\frac{AB}{AD}$

Now, suppose $E'$ is the intersection of the $AC$ and bisector of $\angle ABC$.
By Angle Bisector Theorem, we know that, $\frac{AD}{CD}=\frac{AE}{EC}$ and $\frac{AB}{BC}=\frac{AE'}{E'C}$.

So, The angle bisectors of $\angle ABC$ and $\angle ADC $ concur if and only if $E=E'$
\[\frac{AE}{EC}=\frac{AE'}{E'C}\iff \frac{AD}{CD}=\frac{AB}{BC}\iff\frac{AB}{AD}=\frac{BC}{CD}\iff\frac{PQ}{DQ}=\frac{QR}{DQ}\iff PQ=QR \]
as desired. $\Box$

here's a fourth projective solution to this i guess

Note that $\overline{PQR}$ is the Simson line of $D$ with respect to triangle $ABC$. Hence for $E = \overline{DP} \cap (ABCD)$, it is parallel to $\overline{AE}$. Then $$(AC;BD) \stackrel P=(FB;EC) \stackrel A= (PR; Q\infty_{PQR})$$ which implies the result.

Notice that the concurrent angle condition is identical to the condition that $ABCD$ is a harmonic quadrilateral. Observe now that \[\frac{PQ}{QD} = \frac{\sin \angle PDQ}{\sin \angle QPD} = \frac{\sin \angle BAC}{\sin \angle DAC},\] and \[\frac{QR}{QD} = \frac{\sin \angle QDR}{\sin \angle DRQ} = \frac{\sin \angle ACB}{\sin \angle DBA}. \] Hence, \[\frac{PQ}{QR} = \frac{\sin \angle BAC}{\sin \angle DAC} \cdot \frac{\sin \angle DBA}{\sin \angle ACB} = \frac{BC}{BA} \cdot \frac{DA}{DC} = 1 \iff ABCD \text{ is harmonic,}\] as desired. $\blacksquare$


I have a problem, I am trying to find the equation that produces this output.

x= 0     y = 100;
x=0.5    y = 66 & 1/3;
x=1      y = 50;
x= 1.5   y = 40;
x=2      y=33 & 1/3;
x=2.5    y= (approx.)28.58
x=3       y= 25;
x=4       y=20;
x=5    y=16 & 2/3;
x=6    y= (approx.) 14.29;
x=7    y=12.5;
x=8    y = (approx.) 11.12;
x=9    y = 10;

Can someone find the equation that produces these values and also solve x=10 y= ? ?

Thanks and goodluck!

Well $\frac{100}{x+1}$ works, but I guessed it... can somebody find an actual solution? (and for x=10, it would be 100/11 = 9.0909090909...)


[quote="shobber"]We are taking the Simple Harmonic Motion...
I want to know how many ways there are to sketch the graph of the its path??[/quote]

The number of ways to sketch the path is limited only by your ingenuity I should think; however, the equation of motion for a simple harmonic oscillator is a 2nd order ordinary differential equation.

$m\ddot{x}+kx=0$

The solutions of DE's of this sort are guaranteed to be unique. Meaning that if you find a solution to the differential equation, that solution is [u]the[/u] solution.

This implies that while

$A\cos\omega t +B\sin\omega t \text{ and } C_1e^{i\omega t}+C_2e^{-i\omega t}$ don't appear identical, they are mathematically the same.

Also noting that $\cos(\omega t -\phi) = \cos\phi\cos\omega t + \sin\phi\sin\omega t$, then if you let

$C=\sqrt{A^2+B^2}\text{ then } C(\frac{A}{\sqrt{A^2+B^2}}\cos\omega t +\frac{B}{\sqrt{A^2+B^2}}\sin\omega t) = A\cos\omega t +B\sin\omega t\\
\\
\text{Then let }\cos\phi = \frac{A}{\sqrt{A^2+B^2}}\text{ and let }\sin\phi = \frac{B}{\sqrt{A^2+B^2}}\\
\\ \Rightarrow A\cos\omega t +B\sin\omega t = C(\cos\phi\cos\omega t +\sin\phi\sin\omega t)\\
\\
\Rightarrow  A\cos\omega t +B\sin\omega t = C\cos(\omega t -\phi)\text{, where }\phi = \arctan(\frac{B}{A})$


We've got rectangle $ABCD$, and points $E,F,G,H$ on the sides $AB,BC,CD,AD$ respectively, so that $EG||BC$ and $FH||AB$, and the intersection point of $EG$ and $FH$ is $O$. $AC$ and $BD$ intersect at $O_1$, $EF$ and $BO$ at $O_2$.  $AO$ and $EH$ intersect at $R_1$, $DF$ and $CH$ at $R_2$. $FG$ and $CO$ intersect at $Q_1$, $AG$ and $ED$ at $Q_2$. Prove that the lines $O_1O_2$, $R_1R_2$, $Q_1Q_2$ have a common point.

This is my problem and I have very short solution to it just watch to the center of mass of $EOFCDA$ and you will see that all that lines pass through it  :) .


Observe the following algebraic proof. A^2 implies A to the second power or squared.

Given:

A and B are real numbers;
A = B

A = B
A^2 = AB
A^2 - B^2 = AB - B^2
(A + B)(A - B) = B(A - B)
Dividing both sides by (A - B) yields
(A + B) = B
Since A = B, substituting B for A yields
B + B = B
2B = B
Dividing both sides by B yields
2 = 1

Since we know that 2 does not equal 1, clearly state the mistake made in the above algebraic proof.

[quote="shobber"][quote="GoBraves"][quote="shobber"]indeed...
[hide]
$A-B=0$ so we can divide.[/hide][/quote]

can or cant?[/quote]
can not...[/quote]

just making sure...you wrote can

to can or not to can (Shakespere)
just reminded me of that :P


probability
inequalities
analytic geometry
geometry
Alcumus
triangle inequality

Hi,

I am having trouble understanding the solution to Problem 10.16 from Introduction to Counting and Probability.

Problem: A line segment is broken at two random points along its length. What is the probability that the three new segments form a triangle.

Solution (From the Solutions Manual):

Three a,b,c can be the triangle is and only if they satisfy the Triangle Inequality in every possible way; that is, if a+b>c, b+c>a, and c+a>b. [b]Suppose that our line segment is the unit interval, and we are breaking it at point x and y with 0<x<y<1. Then the three resulting segments will form a triangle if and only if x < 1/2, y > 1/2, and y-x < 1/2. So the "possible" region is the triangle with vertices (0,0), (0,1), and (1,1), and the "successful" region is as shaded in the diagram on the right. Therefore, the probability of success is $ \frac {1/8}{1/2} \equal{} \frac {1}{4}.$ [/b]

I do not quite understand things starting from the bolded ones. Why does it have to be 0<x<y<1, why can't x equal y? I do not understand how the "resulting segments will form a triangle if and only if x < 1/2, y > 1/2, and y-x < 1/2" is dervied. Where does the x < 1/2, y > 1/2, and the y-x < 1/2 come from? And why is possible region a "triangle with vertices (0,0), (0,1), and (1,1)"?

For the last part, as far as I understand that the x axis on the coordinate plane could represent the value of x, and the y axis could be the value of y. But why is the possible region a triangle? Why not a square of side 1?

Thank you for the help.

I think this is one of those chapters that I haven't grasped quite that well...

[quote="limac"]Suppose that our line segment is the unit interval, and we are breaking it at point x and y with 0<x<y<1. Then the three resulting segments will form a triangle if and only if x < 1/2, y > 1/2, and y-x < 1/2. [/quote]

So this segment is just the stick from the origin to (1,0). By the triangle inequality, no length can be greater than/equal to 1/2 (otherwise it won't be a triangle). This includes the first segment (the part between origin and X), the second (between X and Y), and the third (Y and (1,0)). So for teh first segment to be under 1/2, X<1/2, and similarly for the third segment, Y>1/2. Then since Y>X, then Y-X can be no greater than 1/2 itself.

If X=Y, then that would leave 2 segments wouldn't it? That's not a triangle...

[quote]So the "possible" region is the triangle with vertices (0,0), (0,1), and (1,1), and the "successful" region is as shaded in the diagram on the right. Therefore, the probability of success is $ \frac {1/8}{1/2} \equal{} \frac {1}{4}.$ [/quote]

After graphing out the the inequalities for the first and third, you find that this region is a square bounded by the origin, (1/2,0), (0,1/2), and (1/2, 1/2). (Of course, X and Y are positive) Then the second inequality is a line going through (0,1/2) and (1/2,0), shaded down. Therefore, this final region is the union of all this, which is just a right triangle bounded by origin, (0,1/2), and (1/2,0). Since 0<X<Y<1, then the total region (shaded and unshaded) is a triangle with vertices at origin, (1,1), and (0,1). And from here you calculate the area and find the probability.

Sorry for not making a diagram.

[hide="Solution"][hide="Beginning"]We can assign any random length to the segment since it won't change the answer, so we'll say the length is 1, or we could say it's the interval $ [0,1]$ and it's broken at the points $ x$ and $ y.$

The total possibilities for $ x$ and $ y$ lie in the triangle below:
[asy]size(150);
import graph;
draw("$y$",(0,1.25)--(0,-0.25),NW);
draw("$x$",(1.25,0)--(-0.25,0),SE);
dot("$(0,1)$",(0,1),NW,black);
dot("$(1,1)$",(1,1),NE,black);
dot("$(0,0)$",(0,0),SE,black);
draw((0,0)--(1,1)--(0,1)--cycle);[/asy]
[hide="Reason for triangle"]
If it was a square, then the pair $ (0.6797,0.346)$ would be counted twice: one as $ (0.6797,0.346)$ and once as $ (0.346,0.6797).$ So the solution set is half of a square.[/hide]

The side lengths of the triangle are $ x, y - x,$ and $ 1 - y.$ Notice that they sum to 1.

Recall the triangle inequality. Using it all three times,
\begin{align*}x + (y - x) & > 1 - y \\
x + (1 - y) & > y - x \\
(1 - y) + (y - x) & > x.\end{align*}
[/hide]

[hide="Part 1"]Looking at the first inequality, we can reduce it to $ y > \frac {1}{2}.$ Graphing it, (I used a dashed line to indicate it wasn't part of the solution set)

[asy]size(150);
import graph;
draw("$y$",(0,1.25)--(0,-0.25),NW);
draw("$x$",(1.25,0)--(-0.25,0),SE);
dot("$(0,1)$",(0,1),NW,black);
dot("$(1,1)$",(1,1),NE,black);
dot("$(0,0)$",(0,0),SE,black);
draw((0,0)--(1,1)--(0,1)--cycle);
draw((0,0.5)--(0.5,0.5),dashed);
filldraw((0,0.511)--(0,1)--(1,1)--(0.5,0.511)--cycle, lightred);[/asy][/hide]

[hide="Part 2"]
Now, we look at the second inequality. It can be reduced to $ y < x + \frac {1}{2}.$ The graph is shown below:
[asy]size(150);
import graph;
draw("$y$",(0,1.25)--(0,-0.25),NW);
draw("$x$",(1.25,0)--(-0.25,0),SE);
dot("$(0,1)$",(0,1),NW,black);
dot("$(1,1)$",(1,1),NE,black);
dot("$(0,0)$",(0,0),SE,black);
draw((0,0)--(1,1)--(0,1)--cycle);
draw((0,0.5)--(0.5,1),dashed);
filldraw((0,0)--(0,0.491)--(0.509,1)--(1,1)--cycle,lightblue);[/asy][/hide]

[hide="Part 3"]
Finally, we look at the last inequality. It simplifies to $ x < \frac {1}{2}.$ When we graph it, it looks like this:
[asy]size(150);
import graph;
draw("$y$",(0,1.25)--(0,-0.25),NW);
draw("$x$",(1.25,0)--(-0.25,0),SE);
dot("$(0,1)$",(0,1),NW,black);
dot("$(1,1)$",(1,1),NE,black);
dot("$(0,0)$",(0,0),SE,black);
draw((0,0)--(1,1)--(0,1)--cycle);
draw((0.5,1)--(0.5,0.5),dashed);
filldraw((0,0)--(0,1)--(0.495,1)--(0.495,0.5)--cycle,lightgreen);[/asy][/hide]

[hide="Finally"]
Consider the intersection of all those shaded areas. That would be
[asy]size(150);
import graph;
draw("$y$",(0,1.25)--(0,-0.25),NW);
draw("$x$",(1.25,0)--(-0.25,0),SE);
dot("$(0,1)$",(0,1),NW,black);
dot("$(1,1)$",(1,1),NE,black);
dot("$(0,0)$",(0,0),SE,black);
draw((0,0)--(1,1)--(0,1)--cycle);
draw((0,0.5)--(0.5,0.5)--(0.5,1)--cycle,dashed);
filldraw((0.05)--(0.495,0.495)--(0.495,0.995)--cycle,gray);[/asy]
(For some reason, I can't fill that triangle).

The successful region is the triangle in the middle, which is $ \boxed{\frac {1}{4}}$ of the whole triangle.[/hide][/hide]

[quote="limac"]Thank you for the prompt reply dragon96. So, one of the basic ideas about Geometric probability using areas is graphing the information in a way such one axis relates to one of the entities and the other axis relates to another of the entity. And each of the axis is basically the total region of the entity occurring in the form of a line. So the axes don't both necessarily have to start from the same quantity. For example, the x axis starts with 0, and the y axis from 2, or something similar?

EDIT: Maybe it is just me, but I think that the solutions in chapter 10 are a little too brief. So maybe clarifying those a little more would help for future editions? Again...maybe I am being too stupid to not understand those.[/quote]

Thank you. However, I just realized some of my coordinates were wrong (graphing issues), so RoFlLoLcOpT's solution would be much more accurate. The concept is the same, and as you said, graphing out inequalities and finding the area of a shaded region vs the entire region is a basic way to solve some geometric probability problems.

You can have areas that are from x < 3 and 2 < y< 4, but the thing you care about is the area of the given region. And not understanding is a natural thing. Perhaps, Alcumus would be a nice place to train for probability.

Thank you very much for your time RoFlLoLcOpT. That was a very well detailed. dragon96, yes I will practice that more on Alcumus. Thank you for that.

Following you guys' advise, the later Challenge seem more doable using the "conceptual" part that I probably missed in the book.

Actually I have another question...(The Solutions Manual might really consider a bit more added explanation)

The Problem is:

A point P is chosen randomly inside square ABCD. What is the probability that triangle ABP has a greater area that each of triangles BCP, CDP, and DAP?

Answer from the Solutions Manual:

Let O be the center of the square. Since ABP, BCP, CDP, and DAP all have the same base, ABP has the greatest area if and only if P is farther from side AB than from all the other sides. In other words, P must be in triangle OCD. So the answer is 1/4.

This explanation makes sense. But I not quite sure how P must only be in the triangular region OCD other than intuition. But that intuition wouldn't really help me if I were to do this on my own. So how come the P is farthest away from A only the mentioned region?

Thank you

It's a square so all regions are symmetrical. If we were to put $ P$ in triangle $ BOC$, the area of triangle $ APD$ would be greater than $ APB$. Likewise, it is the same for all other triangles.

I hope you understand this.

I am not sure i still understand this. I am not sure why the region must be triangular though.

I'm not sure, but I think if you have a second question, it is best you put this in a new thread. Thanks.

So since ABCD is a square, then the area of ABP is the height times base/2 which is the perpendicular distance from P to AB multipled by AB/2. But in ABP, BCP, CDP, and DAP, the base/2 will remain constant as the side length of a square is always same as the other 3 sides.

So let this square be A=(0,2), B=(2,2), C=(2,0), and D is the origin. Then for the AB height to be longer than the CD height, P must be below the line y=1. For the AB height to be longer than the BC height, the BC height cannot go past the line y=x. And similarly, for the DA height, it can't go past the line y=-x+2. Therefore, this area is a triangle from origin, (1,1), and (2,0), thus having an area of 1/4 of the square. Sorry if I was unclear.

That really helps explain it better. Now the reasoning makes sense.

Thank you very much :)

(l'll post a separate thread if I have another question ;))


advanced fields
advanced fields unsolved

prove or disprove

1.  F=$ \sum_{i \equal{} 1}^{n}\oplus R_i$ is a free R-mod.$ f\in F$, $ Rf$ is a sub modular of F, and $ F \equal{} Rf\oplus M$,than there exist K that $ Rf\oplus K\cong R$.
2.  further more, $ K_i M_i$ are all R-mod.$ K_1\oplus K_2\cong M_1\oplus M_2$ and $ K_1\cong M'$ where M' is a sub R-mod of $ M_1$, then there exist $ K_3$ that $ K_1\oplus K_3\cong M_1$

i think P2 is false

[quote="mathiq"]prove or disprove
1.  F=$ \sum_{i \equal{} 1}^{n}\oplus R_i$ is a free R-mod.$ f\in F$, $ Rf$ is a sub modular of F, and $ F \equal{} Rf\oplus M$,than there exist K that $ Rf\oplus K\cong R$.[/quote] Rf is projective such that the short exact sequence 0 -> K -> R -> Rf -> 0 splits.


If the lenghts of a right triangel are 3 and 4, what is the least possible length of the third side?

this was the first question in a high school mathleague. i was thinking of seeing responses to this question, but im too impatient. oh the irony. well for most, the answer would appear to be 5. since 3, 4, 5 is the prbly the most famous pythagorean triple. but here's where my tip comes in. DONT RUSH AND ASSUME TOO MUCH ON A PROBLEM! this could easaly have been the first question on the sprint round in mathcounts. and you wouldnt have given it a second look (most likely). so if you think for just a few seconds more you would think that a:^2:+3:^2:=4:^2: instead of the most popular equation 3:^2:+4:^2:=5:^2:. solving for a in the first equation, you get a= :sqrt: 7, which is smaller than 5, and is the least possible lenght of the third side.

P.S.
if you were wondering why a:^2:+4:^2:=3:^2: doesnt work. its because then the third side would equal
i :sqrt: 7, and the hypotenuse has to be longer than the rest of the sides

a:^2:+4:^2:=3:^2: doesn't work because the hypotenuse is shorter than the leg, thus it can't form a right triangle.

That's the explanation I was thinking of, though what Syntax said works too.
Interesting story, Syntax. Thanks.

indeed. its just to stop some stupid mistakes that may occur.

[quote="moogra"]least possible would be (4^2-3^2)^1/2 which would be 7^1/2[/quote]

Why are you always posting in old topics? All you probably want is to have more posts under your name.

But there are many other problems not posted 3 years ago. :D

Anyway, you could simply do the problem and check your answer. :)

I thought there was a rule that you can't post in something that hasn't been posted in for at least 2 months.  Is that still in effect.

[quote="moogra"]no doing old problems you know. It improves mathematical ability  :rotfl:[/quote]

It also clutters the board and can be qualified as spam. Do them in your head, don't post.

[quote="moogra"]no doing old problems you know. It improves mathematical ability  :rotfl:[/quote]

Your're making the moderators lock these topics. This stuff were talking about is considered spam. DO THEM IN YOUR HEAD!!!!!! :mad:  :mad:

Stop posting in old topics. Leave Syntax Error, MysticTerminator, Tare, and the rest back in history. Don't corrupt the memory of the good old days. ;) (Or something like that!)

Seriously, though, it is generally advised not to revive topics that aren't even from the past year unless there is a REALLY good reason. More than 2 months is frowned upon.


function calculus derivative integration algebra polynomial LaTeX

We consider functions

$ f_{n}(x)\equal{}[(x^{2}\minus{}1)^{n}]^{(n)}$, for all n naturals nenuls .($ u^{(n)}(x)$ means derived function from order n for function $ u : R\to R$ in point $ x$).

a)If   $ f_{n}(x)\equal{}a_{n}x^{n}\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}...\plus{}a_{0}$  with $ a_{i}$ reals,  find the coefficient $ a_{n}$.

b)Using Rolle's theorem, prove that equation $ f_{n}(x)\equal{}0$ has n reals disctinct zeros (root) in the interval $ (\minus{}1,1)$.

c)If $ g : R\to R$, is a function derived from n times, with derrivative from n order  continuous, then prove

                                   $ \int_{\minus{}1}^{1}f_{n}(x)\cdot g(x)dx\equal{}(\minus{}1)^{n}\int_{\minus{}1}^{1}(x^{2}\minus{}1)^{n}\cdot g^{(n)}(x)dx$

d)Prove that    $ \int_{\minus{}1}^{1}f_{n}(x)\cdot h(x)dx\equal{}0$  for any polynomial function $ h : R\to R$ for  rank $ h\leq n\minus{}1$.

e)Prove that function $ f : R\to R, f(x)\equal{}(C_{n}^{0})^{2}x^{n}\plus{}(C_{n}^{1})^{2}x^{n\minus{}1}\plus{}...\plus{}(C_{n}^{n})^{2}$, has n reals disctinct zeros (root) in the interval $ (\minus{}\infty,0)$.

Special thanks for helpful in LATEX  (for Carcul and N.T.TUAN)

Some relevant notation and language: "$ f$ has $ n$ continuous derivatives on the interval $ I$" can be written as "$ f\in C^{n}(I)$" or "$ f$ is $ C^{n}$ on $ I$".
Also, the right English is "for all natural numbers $ n$" in your first sentence.
Polynomials have "degree", not "rank".

Many variations on this have shown up before- search for Legendre polynomials to see some.


conics
ellipse
trigonometry
analytic geometry
geometry
geometric transformation
reflection

The tangent to the ellipse $ \frac{x^2}{a^2} \plus{} \frac{y^2}{b^2} \equal{} 1$ obviously $ a>b$ at $ P(\alpha)$ cuts its circle in $ A$ and $ B$. If $ AB$ subtends a right angle at the centre $ C$ of the ellipse, show that the eccentricity, $ e$ of the ellipse is: $ \frac {1}{\sqrt {1 \plus{} \sin^2 {\alpha}}}$

Should not the excentricity be $ e \equal{} \frac{1}{\sqrt{1 \plus{} \sin \alpha}}$.

[quote="yetti"]Should not the excentricity be $ e \equal{} \frac {1}{\sqrt {1 \plus{} \sin \alpha}}$.[/quote]

The question says [b]eccentricity[/b] $ e \equal{} \frac {1}{\sqrt {1 \plus{} \sin^2 \alpha}}$

[quote="yetti"]Could you please define the angle $ \alpha$ ? $ C$ is the coordinate origin and the ellipse center. The x-axis is the ellipse major axis. I assume $ \alpha$ is angle between the line $ PC$ and the ellipse major axis, but do not know for sure.[/quote]
Hello
If we make an auxillary circle of the ellipse of radius a of eq. $ x^2 \plus{} y^2 \equal{} a^2$
Now any point in circle can be taken in parametric form as $ P(a \cos\theta,a \sin\theta)$

where $ \theta$ is the angle that $ OP$makes with horizontal.
If we draw a perpendicular $ PQ$ on x-axis from $ P$.
the coordinates of point where the perpendicular meets the ellipse is $ (a \cos\theta,b \sin\theta)$
Depending on what value $ \theta$ takes in the range $ [0.2\pi)$ or $ ( \minus{} \pi,\pi],$ the parametric form $ (a \cos\theta,b \sin\theta)$ gives us different point on the circumference on ellipse.
The point $ (a \cos\theta,b \sin\theta)$ in sometimes simply referred as the point $ \theta$

[quote="yetti"]Thanks, kabi.

Let $ UV$ be the ellipse major axis and $ \mathcal C$ is the ellipse pedal circle with center $ C$ and diameter $ UV.$ Perpendicular to $ UV$ through $ P$ cuts the major axis $ UV$ at $ N$ and the pedal circle $ \mathcal C$ at $ Q$ on the same side of $ UV$ as $ P.$ Denote $ \alpha \equal{} \angle QCV$ and $ \omega \equal{} \angle PCV.$

Perpendiculars to the ellipse tangent $ AB$ at $ A, B$ go through the ellipse foci $ E, F.$ Let $ F'$ be reflection of $ F$ in the ellipse tangent $ AB.$ Then $ EF'$ cuts the ellipse tangent $ AB$ at the tangency point $ P.$ $ \triangle ABC$ is isosceles right, $ \angle ABC \equal{} 45^\circ.$ $ CB \parallel EF'$ is midline of the $ \triangle EFF'$ $ \Longrightarrow$ $ \angle PF'F \equal{} \angle CBF \equal{} 90^\circ \minus{} \angle ABC \equal{} 45^\circ.$ Since the $ \triangle PF'F$ is isosceles with $ PF \equal{} PF'$ and $ \angle PF'F \equal{} 45^\circ,$ $ \angle FPF' \equal{} 90^\circ$ and $ FP \perp EP.$ Since $ C$ is the midpoint of $ EF,$ it follows that $ CP \equal{} CE \equal{} CF$ and $ \angle PEF \equal{} \frac {_1}{^2} \angle PCF \equal{} \frac {_{\omega}}{^2}.$ The ellipse excentricity is then

$ e \equal{} \frac {EF}{UV} \equal{} \frac {EF}{EP \plus{} FP} \equal{} \frac {1}{\cos \frac {\omega }{2} \plus{} \sin \frac {\omega }{2}} \equal{} \frac {1}{\sqrt {1 \plus{} 2\cos \frac {\omega }{2} \sin \frac {\omega }{2}}} \equal{} \frac {1}{\sqrt {1 \plus{} \sin \omega }}.$

On the other hand, from the right $ \triangle CNP, \triangle CNQ,$

$ e \equal{} \frac {CP}{CQ} \equal{} \frac {\cos \alpha}{\cos \omega } \equal{} \sqrt {\frac {1 \minus{} \sin^2 \alpha}{1 \minus{} \sin^2 \omega }}.$

Combining, $ \sin^2 \alpha \equal{} \sin\omega$ and $ e \equal{} \frac {1}{\sqrt {1 \plus{} \sin \omega }} \equal{} \frac {1}{\sqrt {1 \plus{} \sin^2 \alpha}}.$[/quote]

Oh....pretty good.
What is carthage?


[quote="Smartguy"]
How many of you guys do math team?
Is it your priority as extracurriculars or do you guys go when you have time?

How many of you do sports and math team?

Thanks[/quote]

can you guys answer all 3 questions?
thanks

I do math team. I think that it is very high in things I want to do mostly because it is fun. I also swim, run, play tennis and a few others.

sorry, im afraid i didn't clarify this:
What i meant by doing sports was joining your school's sports team

[quote="Smartguy"](unfortunately, i dont know which forum this should go in)

How many of you guys do math team?
Is it your priority as extracurriculars or do you guys go when you have time?

How many of you do sports and math team?
[/quote]

1.) Yes I am in my school's National Math Honor Society.
2.) It is my priority to do it as an extracurricular for college as I'm in high school.
3.) I am planning on joining JV tennis during the spring (Varsity will just kill me during the height of AP). But I do know a bunch of people who are in the math society as well as varsity badminton.

The high school I went to for 11th and 12th grade didn't have a math team per se.

But in 10th grade, I essentially ran the math team, which took a lot of time, and was also doing crew, which meant I got home at 6:30 on weekdays. While it didn't leave me with a ton of free time on weekdays, it was still quite doable.

You will probably find that a substantial commitment to sports and/or exercise will increase your productivity, both out of improved health/concentration and out of necessity.

I would highly recommend that you do both, unless there is no way to avoid a time conflict.

I would have to say that this depends a lot on the individual.  (At my high school, at least) Sports seem to take up a tremendous amount of time for most students who participate in them.  I would estimate that the average person on a sports team goes to bed after 1:00 AM every night, which is not healthy and greatly reduces productivity.  However, if you have no problem finishing most of your homework during school and don't think you'll have to spend more than about 90 minutes a night on homework, I would recommend doing a sport.

I am a member of our "math club," but it was just recently started by my dad, so it isn't very big, yet.  I have plenty of spare time, so I go all of the time.

I do not play any sports, but I have started working out more recently (for the health benefits, etc) and I play piano.

1)  I am on the "math team".
2)  Sometimes, it is...
3)  I play tennis.

if you do studying, then you'll won't healthy. If you only do exercise, then you'll be stupid and doomed to talk nonsense.

1.) I am on Math Team.
2.) Yes.
3.) I don't take any other sports, nor plan to.

@Poincare: I'd rather be unhealthy and smart than healthy and dumb.

[quote="Smartguy"](unfortunately, i dont know which forum this should go in)

How many of you guys do math team?
Is it your priority as extracurriculars or do you guys go when you have time?

How many of you do sports and math team?

Thanks[/quote]
I'm on the math team
I always go to the competitions, not meeting b/c i'm in middle school on the high school team, and the hs meets at lunch
both sports & math team

[quote="math154"]Um...health is the most important quality for a non-suicidal person in my opinion. :wink:

Anyway,
1) I'm on the math team I guess.
2) Most of the time.
3) I mainly play tennis. Probably will join the HS team in HS.[/quote]

I agree, but currently due to the amount of homework I have, I would be up till midnight everyday trying to finish it if I took a sport.

I mean more like exercise as often as possible, not necessarily on a sports team.

[quote="ernie"]lol why does everyone play tennis?[/quote]
Tennis is like the asian sport.
uno. I am on the math team
dos. It's my main priority, I guess.
tres. I'm not very good at sports, but I was thinking about joining a team next year.

I run cross country... in middle school.

WHen you get to high school, cross country and track are about the only events that there aren't tryouts for.


Every 5 months, Hal has to replace the batteries in his calculator. He changed them the first time in May. In what month will they be changed the 25th time?

Since the 1st time he changed the batteries was already given, he has $ 25\minus{}1\equal{}24$ battery changes left until a net total of $ 25$ battery changes. Thus, the 25th change will occur in $ \fbox{May}$ as well, since $ 24$ is a multiple of $ 12$, the number of months in a year.


$\sum a^3\geq \frac {1}{9}(a+b+c)^3\geq \frac {1}{9}(a+b+c)^2\left(\frac {1}{a}+\frac {1}{b}+\frac {1}{c}\right)\geq a+b+c$

the first inequality is power-mean, the second is the condition, the third is cauchy

[quote="mathservant"]a,b,c are positive real numbers and a+b+c>=1/a+1/b+1/c
Prove that  a^3+b^3+c^3>=a+b+c[/quote]

\begin{proof}
For $a+b+c >=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$,
\begin{equation}\label{01}
  (a+b+c)^2 >= (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) >= 9.
\end{equation}
And with the Power Mean Inequality, we get
\begin{equation}\label{02}
\left(\frac{a^3+b^3+c^3}{3}\right)^{\frac{1}{3}}>=\frac{a+b+c}{3},
\end{equation}
The inequality \eqref{02} is equivalent to
\begin{equation}\label{03}
a^3+b^3+c^3>=\frac{(a+b+c)^3}{9}.
\end{equation}
In order to prove the origin inequality, we only need  prove the inequality as follows.
\begin{equation}\label{04}
\frac{(a+b+c)^3}{9}>=a+b+c
\end{equation}
The inequality \eqref{04} is exactly the inequality $(a+b+c)^2 >=9$. Thus, the proof of the origin inequality is completed.
\end{proof}

nice problem and nice solutions.and $Yu-Dong Wu$ i think you should learn a little more about latex.
for this problem,I think i have posted the general case.
see  [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35944]www.mathlinks.ro/Forum/viewtopic.php?t=35944[/url]
I think it still true for $a+b+c \geq 3$ :)

Can you post your solution "regular format"  Yu-Dong Wu?Thank you.

[quote="mathservant"]a,b,c are positive real numbers and a+b+c>=1/a+1/b+1/c
Prove that  a^3+b^3+c^3>=a+b+c[/quote]

For $a+b+c>=\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$,  we have

$(a+b+c)^2 >= (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) >= 9$.

And with the Power--Mean Inequality, we get

$(\frac{a^3+b^3+c^3}{3})^{\frac{1}{3}}>=\frac{a+b+c}{3}$,

The above inequality is equivalent to

$a^3+b^3+c^3>=\frac{(a+b+c)^3}{9}$.

In order to prove the origin inequality, we only need prove the inequality as follows.

$\frac{(a+b+c)^3}{9}>=a+b+c $

The inequality  is exactly the inequality

$(a+b+c)^2 >= 9$.

Thus, the proof of the origin inequality is completed.

This one is trivial. You just have to use a^3+b^3+c^3>=((a+b+c)^3)/9 and AM-GM or AM-HM. :D

Sorry I haven't seen Blahblahblah solution, but I will give you an inequality I prposed at the jbmo test selection:

      if a,b,c>0 and a+b+c>=1/a+1/b+1/c then a=b+c>=3/abc.

Not hard. Let $x=bc$,$y=ca$,$z=ab$. Then we have $xy+yz+zx \ge x+y+z$ and we hv to prove $xy+yz+zx \ge 3$
From the given condition,
$xy+yz+yz \ge x+y+z \ge \sqrt{3(xy+yz+zx}$ which gives the desired inequality.


prove that there are infinitely many solutions for the equation $ a^2 \plus{} b^2 \plus{}c^2 \equal{} 3abc$
where a,b,c are integers

if $ a,b,c$ is a solution so is $ b,c,3bc \minus{} 2a$ (we write triplets in increasing order)
"i hope i haven't made a mistake "

This is very simple.
Just assume wlog that x<y<z and as gaurav pointed out, obtain a quadratic in x and and get a root and so on.

yes sorry :(  it is $ 3bc \minus{} a$ not $ 3bc \minus{} 2a$


Bobbo starts swimming at 2 ft/s across a 100 foot wide river with a current of 5 ft/s. Bobbo doesn't know that there is a waterfall 175 feet from where he entered the river. He realizes his predicament midway across the river. What is the minimum speed that Bobb must increase to make it to the other side of the river safely?

Midway across the river is 50 ft.

So by then he wouldve alreayd traveled 125 ft up stream because (distance across river)/(distance up stream)=(rate of Bob)/(rate of current), so since 2/5=50/x, x=125

So he only has 50 ft left to go before waterfall
he also has 50 ft left to get across the river

so using same equation

50/50=y/5, where y is Bob's new rate, y=5ft/s

so he increases from 2ft/s to new speed of 5ft/s


linear algebra
matrix
inequalities
vector
linear algebra unsolved

Can anyone prove this?

Let A be a complex square matrix with $ \|A\| \equal{} 1$ ( any induced matrix norm). Then for any $ a>0$

$ \|A/a \minus{} I\| < \rho(A/a \minus{} I) \plus{} 1/a$, whre $ \rho$ indicates the spectral radius of the matrix.[/b]

What if we have $ A$ upper tringular and the norm is the $ 1$-norm ($ \|\cdot\|_1$).
Is it easier now to say something?
Thanks

The positive answer to the first question would imply that $ \|B\plus{}I\|\ge \|B\|$ for every nilpotent $ B$. It would be a beautiful inequality if it were true but, unfortunately, it isn't. Consider $ V\equal{}\mathbb R^2$ with the vector norm $ |(x,y)|\equal{}\max\{2|x\minus{}y|,|x\plus{}2y|\}$ and the operator $ B: (x,y)\mapsto(3y,0)$. Then $ \|B\|\equal{}3$ and $ \|B\plus{}I\|\equal{}\frac 52$. If it sounds a bit misterious, just notice that the vector $ v\equal{}(0,1)$ of norm $ 2$ on which the norm of $ B$ is attained looks inside the ball of radius $ 6$ when shifted to $ Bv\equal{}(3,0)$.


superior algebra
superior algebra unsolved

[quote="mehdi cherif"]let $ E$ a group such that $ CardE \equal{} n$.  and  $ A\subset E$ and  $ B\subset E$

and $ F \equal{} \{ X\subset E/Card(X\cap A) \equal{} Card(X\cap B) \equal{} 1 \}$
find $ CardF$
 :)[/quote]
we take $ (A',B',C,D)\equal{}(A\minus{}B,B\minus{}C,A\cap B,E\minus{}A\cup B)$
if $ C\equal{}\emptyset$ then $ \boxed{card(F)\equal{}2^{card(D)}card(A)card(B)}$
if $ C\neq \emptyset$
 if $ X\in F$ then:
1) $ card(X\cap C)\equal{}1$ , there are $ 2^{card(D)}card(C)$ possibilité.
2) $ card(X\cap C)\equal{}0$ , there are $ 2^{card(D)}card(A')card(B')$ possibilité.
so $ \boxed{card(F)\equal{}2^{D}(card(C)\plus{}card(A')card(B'))}$


For this problem I used generating functions.   Could some people post some alternate solutions?

Consider the sequence of numbers which begins 6,3,... and has the property that each
successive term in the sequence is equal to the average of the two terms immediately
preceding it. Find the sum of the first 1000 terms of this sequence, accurate to ten decimal places.

(you can use a calculator ;-) ).

Hint: Let $S_n$ be the nth term of the sequence.  Find $S_n - 4$.

if you look at the first differences between the terms, they form an alternating geometric sequence.  so sum the first 999 terms of $3-1.5+.75-.375...$, then take 6-that.


Hi there.

I've encountered some difficulty with cyclic quadrilaterals in my work and I'm looking for some guidance.

The question is:  ABCD is a cyclic quadrilateral.  If angle EDC = 5x-6, angle C = 7x and angle B = 3x+28, find the measure of angle ADC.

My understanding of cyclic quadrilaterals is that the opposing angles will add up to 180 degrees, much like a triangle would only triangles aren't my issue.  I've not encountered a question regarding cyclic quadrilaterals that use variables.

Any aid would be greatly appreciated.
My lust for knowledge has hit a road block.
 :wallbash:

$\angle EDC$ - where did that come from?  What is point E?  Or is this just supposed to be $\angle BDC$?

I wish I could add a diagram.  I was just reviewing the question and thinking that.

I'll try to describe it.  In the circle is the quadrilateral ABCD, AB on the bottom two points DC on the top two.  Above line AD another segment extends straight up to point E.  Point E is outside the circle.  Angle EDC is the top part of the quadrilateral.

Does this help somewhat?

mspaint + http://www.imageshack.us

=

you can make a picture, upload it, and then put it in img tags

[code]
[img]www.imageshack.us/awlrekdsfj34jrt0843ujrefimageurl[/img]
[/code]

Wonderful idea.
 :)

[url=http://img246.imageshack.us/my.php?image=12342nj1.png][img]http://img246.imageshack.us/img246/8076/12342nj1.th.png[/img][/url]

[hide]Opposite angles must sum to 180.  We have $m\angle ADC = 180-(5x-6)$.  We also have that $3x+28+m\angle ADC = 180$.  By substitution we have
$2x+28+180-5x+6 = 180$
$\implies 34-2x = 0 \implies x = 17$
Then $m\angle ADC = 180-(5\cdot17-6) = 101$.[/hide]

Oh I see now.
Thank you so much for your help.
:D

Though, some of your numbers I don't follow.  When dividing 34 by the -2, wouldn't the answer become -17 and then that is used in the equation?  Then the answer would become significantly more then 180 degrees though...

Oh I see where I've been confused.

 :blush:

Thank you.

[quote="Silis"]
Though, some of your numbers I don't follow.  When dividing 34 by the -2, wouldn't the answer become -17
[/quote]

$34-2x = 0 \implies 34 = 2x \implies x = \frac{34}{2}\implies x = 17$


inequalities
function
inequalities unsolved

[quote="ngoclam1088"]Let $ x,y,z > 0$ and $ x \plus{} y \plus{} z \leq 1$. Prove that:
$ \sqrt {x^2 \plus{} \frac {1}{x^2}} \plus{} \sqrt {y^2 \plus{} \frac {1}{y^2}} \plus{} \sqrt {z^2 \plus{} \frac {1}{z^2}} \geq \sqrt {82}$[/quote]

\[ \sum {\sqrt {x^2  \plus{} \frac{1}{{x^2 }}} }  \ge \sqrt {\left( {\sum x } \right)^2  \plus{} \left( {\sum {\frac{1}{x}} } \right)^2 }  \ge \sqrt {\left( {\sum x } \right)^2  \plus{} \frac{{81}}{{\left( {\sum x } \right)^2 }}}
\]
\[ \equal{} \sqrt {\left( {\sum x } \right)^2  \plus{} \frac{1}{{\left( {\sum x } \right)^2 }} \plus{} \frac{{80}}{{\left( {\sum x } \right)^2 }}}  \ge \sqrt {82}
\]
That's all. It's old.

$ \sum\sqrt{(1\plus{}81)(x^2\plus{}\frac{1}{x^2})}\geq\sum x\plus{}9\sum\frac{1}{x}\equal{}9\sum(9x\plus{}\frac{1}{x})\minus{}80\sum x\geq 82$
We are done :)

karis could you please explain further? Sorry, I am unable to understand.

First is CS and second is by $ 9x\plus{}\frac{1}{x}\geq 6$ and the condition.

[quote="ngoclam1088"]Let $ x,y,z > 0$ and $ x \plus{} y \plus{} z \leq 1$. Prove that:
$ \sqrt {x^2 \plus{} \frac {1}{x^2}} \plus{} \sqrt {y^2 \plus{} \frac {1}{y^2}} \plus{} \sqrt {z^2 \plus{} \frac {1}{z^2}} \geq \sqrt {82}$[/quote]
The general one isn't really much harder:
Let $ x_1,x_2,...,x_n \le 1$.
Then $ \sqrt[m]{x_1^m \plus{}\frac{1}{x_1^m}} \plus{} ... \plus{}\sqrt[m]{x_n^m \plus{}\frac{1}{x_n^m}} \ge \sqrt[m]{n^{2m}\plus{}1}$
for all $ n,m \in \mathbb{N}$ and $ x_1,...,x_n \ge 0$.

I'm sure that there are much nicer solutions, but that's my solution with Jensen.
The function $ f(x)\equal{}(x^m\plus{}x^{\minus{}m})^\frac{1}{m}$ is convex and strictly decreasing on the intervall $ (0,1)$. Thus, Jensen implies
$ \sum_{cyc}f(x_1) \geq nf\left(\frac{\sum_{cyc}x_1}{n}\right) \geq nf(\frac{1}{n}) \equal{} \sqrt[m]{n^{2m}\plus{}1}$


Anyone ever been there?

[url=http://www.battleon.com]www.battleon.com[/url]
very addicting somehow.. its still sort of beta ish.. you guys should try it if you dont.. its completely free..

I've got an account on it, it'e pretty addicting...
so it's a beta version?, you a guardian?

That game gets completly BORING you just keep clicking the attack button.

Lol.Battleon rox. iT is so addicting. Thats why its good. IM lvl 58. heh.

Hey if your not gonna play battle on anymore ill but ur guy lvl 61 gurdien for full addy addy t chest and 20k  and liek 1900 steel arrows

jeff can i have ur character then???? if i can or u lvl 61 guardien for my deal contact me with kaconley@gmail.nom or kaneallen144 on aim

[quote="Jeffreywang"][quote="RunescapeFan123"]yea i play battleon for some time and how it's getting boring
Im also a gurdain lvl 61 but i dont have the blade thingly
if you want my account i could give it to you for runescape money muhahahah anyway good game[/quote]
How much money?
jk, I don't play Battleon anymore, it got boring.[/quote]

hey ill trade my rs acc if some1 makes me a gardiun on battleon the rs acc is lvl 67  email me at polley_simon@yahoo.co.uk if u want my rs acc

runescape 73 and 83 and 87 and 93
three of em are friend's accts...but we share

and err this thing: level 28? lol it's so boring after like 10 min tho

Can i use your account name just for one name please because i really want to be a guardian u can email the account at rep009er09@hotmail.com i swear to god that i will not give your info away

[quote="RunescapeFan123"]yea i play battleon for some time and how it's getting boring
Im also a gurdain lvl 61 but i dont have the blade thingly
if you want my account i could give it to you for runescape money muhahahah anyway good game[/quote]

HOW MUCH U WANT IN RUNESCAPE? TELL ME AT haider1234567@hotmail.com  :blush:  :|


inequalities
combinatorics unsolved
combinatorics

Prove that by dividing 67 integers 1-67 into 4 groups randomly (empty group is allowed), there must be at least one group in which there are two integers such that the sum of these two integers also belongs to the same group.

It follows from the fact that [url=http://mathworld.wolfram.com/SchurNumber.html]Schur number[/url] $S(4)=44$ and $67>44$.

Can you explain it in a combinatorial way?


LaTeX
combinatorics unsolved
combinatorics

All points in a $100 \times 100$ array are colored in one of four colors red, green, blue or yellow in such a way that there are $25$ points of each color in each row and in any column. Prove that there are two rows and two columns such that their four intersection points are all in different colors.

Suppose it is not possible to find an intersection with points of four different colors (*).

let $ l_{1}$ and $ l_{2}$ be any two lines of the grid. For a point $ A$ on $ l_{1}$, we name corresponding point of $ A$ the point of $ l_{2}$ on the same column as $ A$.

We first show that there are at least 25 points on $ l_{1}$ which are of the same color as their corresponding points on $ l_{2}$.(1)

To do this we define a graph with four vertices $ R$, $ B$, $ Y$ and $ G$ (each vertex is one color). one vertex $ a$ is connected to another vertex $ b$ with an edge if one can find two corresponding points (one on $ l_{1}$, the other on $ l_{2}$) of colors $ a$ and $ b$.

If one vertex has degree $ 0$, it implies that all the 25 points on $ l_{1}$ of the color considered have corresponding points of the same color, so property (1) is true.

Otherwise, the $ 4$ vertices have degrees $ \geq 1$. If $ e$ is the number of edges of the graph we have : $ deg(R) \plus{} deg(B) \plus{} deg(Y) \plus{} deg(G) \equal{} 2e$.
- if $ e \equal{} 2$ then the $ 4$ vertices have degrees $ \equal{} 1$, but this is not possible (we would have for example $ R$ connected to $ V$ and $ Y$ connected to $ G$, which would give an intersection of 4 points of different colors, but this is in contradiction with (*)).
- if $ e\geq 4$ we also get a contradiction with (*).
- if $ e \equal{} 3$, then one vertex has degree $ \equal{} 3$ and the three others have degree $ \equal{} 1$.Suppose that $ R$ has degree $ \equal{} 3$. $ R$ is connected to $ B$,$ Y$ and $ G$, but :
$ B$ is not connected to $ Y$,
$ B$ is not connected to $ G$,
$ Y$ is not connected to $ G$. (If only I knew how to draw a graph with latex)
This implies that :
- a blue point on $ l_{1}$ has its corresponding points on $ l_{2}$ either red or blue; let $ p$ be the number of blue points on $ l_{2}$ corresponding to a blue point on $ l_{1}$ (the number of red points on $ l_{2}$ corresponding to a blue point on $ l_{1}$ is $ 25 \minus{} p$)
- a green point on $ l_{1}$ has its corresponding points on $ l_{2}$ either red or green; let $ q$ be the number of green points on $ l_{2}$ corresponding to a green point on $ l_{1}$ (the number of red points on $ l_{2}$ corresponding to a green point on $ l_{1}$ is $ 25 \minus{} q$)
- a yellow point on $ l_{1}$ has its corresponding points on $ l_{2}$ either red or yellow; let $ r$ be the number of yellow points on $ l_{2}$ corresponding to a yellow point on $ l_{1}$ (the number of red points on $ l_{2}$ corresponding to a yellow point on $ l_{1}$ is $ 25 \minus{} r$)
As there are 25 red points on $ l_{2}$, we have :
$ 25 \minus{} p \plus{} 25 \minus{} q \plus{} 25 \minus{} r\leq 25$
$ p \plus{} q \plus{} r\geq 50\geq 25$
So property (1) is true.
In conclusion we have proved that for any given two lines $ l_{1}$ and $ l_{2}$ there are at least 25 monocolor vertical pairs, ie pairs $ \{a_{i},b_{i}\}$ such that :
$ a_{i}\in l_{1}$
$ b_{i}\in l_{2}$
$ a_{i}$ and $ b_{i}$ are on the same column
$ a_{i}$ and $ b_{i}$ are of the same color.

As there are $ \binom{100}{2}$ ways to choose two lines out of the 100 lines of the grid, there are at least: $ 25\times \binom{100}{2}$ monocolor vertical pairs.

But :
on each column there are: $ 4\times \binom{25}{2}$ monocolor vertical pairs (first choose the color, then the positions of $ a_{i}$ and $ b_{i}$). Therefore, in the whole grid there are:  $ 100\times 4\times \binom{25}{2}$ monoclor pairs.
As $ 25\times \binom{100}{2} > 100\times 4\times \binom{25}{2}$, we get a contradiction.

So it is possible to find an intersection with four different colors.

[quote="Xell"](*).

your solution is perfect xell
i have a question where can i find a lot of combinatorics problems by their solution or which books are good in combinatorics(not very easy combinatorics)

thank you :wink:

Great Solution!

So can you now help M.A about this one:
http://www.mathlinks.ro/Forum/viewtopic.php?t=196972 :D

[quote="Xell"]Thanks for your kind comment. Actually I think here is a really good place to find problems and solutions. By the way, where did you find this problem ?[/quote]

I found this problem in the book (A PATH IN COMBINATORICS By Dr Titu Andreescu)

this problem is from russia 2000:

let the colors used be $ A,B,C,D$.we call an unordered pair of squares [i]sanguine[/i] if the two squares lie in the same row and are of different colors.every row gives rise to $ 6.25^2$ sanguine pairs (given by $ \binom{4}{2}$ possible pairs of colors and $ 25$ squares of each color). thus, summing over all the rows,there is a total of $ 100.6.25^2$ sanguine pairs.on the other hand,each such pair is simply the intersection of one row with a pair of distinct columns.because there are $ \binom{100}{2}\equal{}\frac{100.99}2$ pairs of columns,some pair of columns contains at least:

$ \frac{100.6.25^2}{100.\frac{99}2}\equal{}\frac{2.6.25^2}{99}>\frac{12.25^2}{4.25}\equal{}75$

sanguine pairs.thus,some two fixed columns form sanguine pairs in at least $ 76$ rows.we henceforth ignore all other rows and columns;we may as well assume that we have only a $ 76\times 2$ board colored in four colors,in which each row contains two different colors and no color occurs more than $ 25$ times in each column.
for each row,consider the pair of colors it contains.if the pairs $ \{A,B\}$ and $ \{C,D\}$ each occure in some row,we are done;likewise for $ \{A,C\},\{B,D\}$ and $ \{A,D\},\{B,C\}$.thus,suppose that at most one pair of colors from each of these three sets occurs;we now seek a contradiction.we have only two possibilities,up to a possible relabelling of colors: either $ \{A,B\}.\{A,C\},\{A,D\}$ are the only pairs that occur,or $ \{A,B\},\{A,C\},\{B,C\}$ are.in the first case,each of the $ 76$ rows contains a square of color $ A$,implying that one column has more than $ 25$ squares of color $ A$,a contradiction.in the second case,each column can contain only the letters $ A,B,C$.there can only be $ 25$ squares of each color $ A,B,C$ in each column,for a total of at most $ 150$ squares,but there are $ 152$ squares in total,a contradiction.this completes the proof.


Prove that:
1/2^(n-1)<=x^2n+(1-x^2)^n<=1, for x in [-1,1] and n is natural!

bye~

Okay the trick is in the fact that x is in [-1,1]
here is a hint x=sint ... try and solve it now...

:D


Let ABCD be a parallelogram, P and Q be points on the side BC such that AB = BP. P is between B and Q and angle BPA + angle PQA + angle BCA = angle BAD. If AP = squareroot2 (PQ), prove that PQ = QC.

It is not hard... you quickly see that we must have $AQP$ ~ $CAP$ so $PQ \cdot PC = AP^2 = 2PQ^2$ and the result follows.