trigonometry
geometry
trig identities
Law of Sines

Hi all,
I'm not sure how difficult or easy this is, what I do know is that I am unable to find an answer.

Scenario: Let's have any triangle, picture here is just an example, ABC. you know the size of AB, AC and the angle Beta (90 degrees in this case). Alfa doesn't change (of course).

Challenge: find a formula to calculate the size of AD, AE or AF (D,E and F are shown as examples only) by entering the BAD, BAE or BAF degrees. (any number from 1 to 89 in this case).

It has to be possible !!!!

Law of sines.  We are given $ AB$ and your angle alpha is complementary to $ \angle B$ ($ m\angle B \equal{} 90^\circ \minus{} \alpha$).  Lets call your beta $ \angle A$, and call the point on $ \overline{BC}$ $ D$, so we are looking for $ AD$.  But $ m\angle D \plus{} m\angle A \plus{} m\angle B \equal{} 180^\circ \iff m\angle D \equal{} 180^\circ \minus{} m\angle A \minus{} m\angle B$.  Then, by the law of sines, $ \frac {\sin D}{AB} \equal{} \frac {\sin B}{AD} \iff AD \equal{} \frac {AB \sin B}{\sin D}$, and we are done because all the quantities on the RHS are known.

Hi Facis,

Thank you for your help on this. I got it !!

You have made a positive contribution to my state of mind.


Given acute triangle ABC. Let D, E be the two moving points on AB, AC such that the perpendicular projection of ED onto BC is equal to (1/2).BC. Prove that the circumcirlce of triangle ADE goes through a fixed point.
Posted: Mon Jan 12, 2009 1:40 pm
By Mashimaru in Geometry Unsolved Problem section

Yes, I see it in Geom Unsoved section with yetti's proof. And no reason for posting this problem here again. The topic should be locked!

Continue discussion at [url]http://www.mathlinks.ro/viewtopic.php?t=250794[/url].


I think this might have been posted by some other Mathpathers, but I didn't quite catch it.

A woman has 5 stones and, using those stones and an ordinary 2 pan balance, can determine the weight of a rock weighing a whole number of pounds, up to 242 pounds.

What are the weights of the rocks.

I can only figure it up to 5 rocks that can determine the weight of rocks up to 100 something.

Well, if 5 rocks have weights 1, 3, 9, 27, and 81, they can weigh anything up to 121 pounds. 242 is double of 121, but how do we get there?


Δίνεται το πολυώνυμο $ P(x)\equal{}ax^3\plus{}bx^2\plus{}cx\plus{}d$ για το οποίο ισχύει $ |P(x)|\leq 1$ για κάθε $ x$ με $ |x|\leq 1$. Να αποδείξετε ότι

$ |a|\plus{}|b|\plus{}|c|\plus{}|d|\leq 7$.

Αλέξανδρος

Ωραιο προβλημα. Εδειξα το ζητουμενο με την επιπλεον υποθεση οτι το P παιρνει τις τιμες -1 και 1 στο εσωτερικο του (-1,1).
Μηπως θελει χρηση των τυπων Vieta? Επισης για $ x\equal{}0$ ειναι $ |d|<\equal{}1$. Αυτο δεν βελτιωνεται,ετσι δεν ειναι?

mipws tha borouse kapoios pou tn exei lysi na kanei post tn lysi tou gia na dw tn methodo antimetwpisis tetoiwn provlimatwn?

Θα δώσω την κατευθυντήρια οδό ... :wink:

Το πρόβλημα ανάγεται στο να εκφράσουμε τα αθροίσματα $ a\plus{}b\plus{}c\plus{}d$ , $ a\plus{}b\minus{}c\plus{}d$ , $ a\plus{}b\plus{}c\minus{}d$ , $ a\plus{}b\minus{}c\minus{}d$
ως άθροισμα ή διαφορά τιμών του πολυωνύμου .

Είναι $ a\plus{}b\plus{}c\plus{}d\equal{}P(1)$  , $ a\plus{}b\plus{}c\minus{}d\equal{}P(1)\minus{}2P(0)$

$ a\plus{}b\minus{}c\plus{}d\equal{}\frac{4}{3}P(1)\minus{}\frac{1}{3}P(\minus{}1)\minus{}\frac{8}{3}P(\frac{1}{2})\plus{}\frac{8}{3}P(\minus{}\frac{1}{2})$

kai $ a\plus{}b\minus{}c\minus{}d\equal{}\frac{5}{3}P(1)\minus{}4P(\frac{1}{2})\plus{}\frac{4}{3}P(\minus{}\frac{1}{2})$

 :)


This is little math-related problem that I got it from my friend.

There is 1.5mL of pure water in container A. There is one hole on the left and one hole on the right. From the left hole, I input 100% methanol at a rate of 20nL/min into container A. Assume methanol and water mixed well immediately. Then the mixed solution is getting out from container A thru the right hole at a rate of 20nL/min also. So how long it will take the 100%-water in container A to turn into 100%-methanol????

Thanks so much for the help

Never. 100% is a limit that can be approached at infinity, not earlier.

Right..that makes sense....

But what if I wanna calculate how long it takes to reach 99% of methanol??

What would be the formula? And how to come up with it???

Sounds like kinetic problem with differential equation to solve.


A box contains exactly five chips, three red and two white. Chips
are randomly removed one at a time without replacement until all the
red chips are drawn or all the white chips are drawn. What is the
probability that the last chip drawn is white?

Every rearrangement of $ RRRWW$ gives a unique representation of the drawing.

The last chip drawn can only be white if the last letter in the sequence is $ R$. (Why?)

Thus, the probability is simply $ \boxed{\frac35}$, since there are three $ R$'s and two $ W$'s.

This should also make sense intuitively: if the number of white chips is less, shouldn't they be more likely to be drawn out sooner?

But since the last two letters in the sequence have to be $ RW$, wouldn't the answer be $ \frac35 \times \frac12 \times \frac23 \times \frac{3!}{2} \times \frac12$, or $ \boxed{\frac3{10}}$?

Why does it have to be RW? Isn't RRWW a valid solution?

math154 is correct (i looked at the official solution).

Can someone explain to me why you want to find the possibilities that end with red? Wouldn't you want to find the possibilities that end with white? Wouldn't also make the problem make since? Is it just me or are all of amc solutions not making since...., such as the 2011 AMC 12A solution for elite members. If you plug in there "solutions" for the problems, you get the number of elite members to be 45, whereas the resulting number should be 15. However if you just solve the equation, you get the answer 21, which if plugged in into the equation gets you the answer, 15, which is the resulting number.

I'm not sure what the elite members problem you're refering to is, but an explanation for this problem:
[hide="I hope this is an OK explanation"]
Note that if we end the 5-letter sequence with $R$, then both of the white chips have already been chosen (which is what we want), and not all of the reds have been chosen (also what we want). Likewise, if we choose $W$ to be last, all three of the red chips would have already been chosen (not part of the desired probability), which would have ended the game already. Therefore, the probability of choosing a $R$ last is $\boxed{\frac{3}{5}}$. I apologize if this is confusing.[/hide]

[hide=Solution] You can list all of the ways that you choose a white piece last
$WW$
$RWW$
$RRWW$
$WRW$
$WRRW$
$RWRW$
We then find that each of the choices has $1/10$ probability, so $6 * 1/10$ is $3/5$. You could also use complementary counting to find all possible ways  the solution ends with a red and then subtract from 1.

[url=https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_21]See 2016 AMC 8/21.[/url]

Here's my approach to it:

We have that there is 5!/(2!*3!) = 10 total unique arrangements.

We also have that if the first chip has to be white, there are 4!/(2! * 2!) = 6 unique arrangements.
By symmetry, if the last chip is has to be white, then there also has to be 6 unique arrangements for the last chip being white.

Therefore the probability is 6/10 = 3/5

47 questions later I finally mastered probability - think about it
[hide=sol]
we want the last chip to be red so that the chip before last (the one that's last drawn) is white
so $\boxed{\frac{3}{5}}$
[/hide]


limit
logarithms
function
real analysis
real analysis unsolved

[color=darkblue]Prove that for any $n\in N\ ,\ n>1$ the equation $\sqrt [n]{x+1}-\sqrt [n] x=\frac{1}{n}$ has only one zero $x_{n}\in (0,1)\ .$ Ascertain L=$\lim_{n\to\infty}x_{n}\ .$
[hide="Answer."]$L=\frac{1}{e-1}\ .$[/hide][/color]

It's easy to see that $x \mapsto \sqrt [n]{x+1}-\sqrt [n] x$ is strictly decreasing. Since $f(0)=1$ and $f(1) = \sqrt[n] 2-1 < \frac1{n}$ (by Bernoulli), $x_{n}$ exists and is unique in $(0,1)$.

Now write the relation in the form $\frac{(x+1)^{1/n}-x^{1/n}}{1/n}= 1$ and let $\ell_{1}= \limsup_{n \to \infty}x_{n}, \, \ell_{2}= \liminf_{n \to \infty}x_{n}$.

We note that $\lim_{t \to 0^{+}}\frac{(a+1)^{t}-a^{t}}{t}= \lim_{t \to 0^{+}}a^{t}\cdot \frac{\left( \frac{a+1}{a}\right)^{t}-1}{t}= \ln \left( \frac{a+1}{a}\right) = \ln \left( 1+\frac1{a}\right)$.

There is a subsequence $x_{n_{k}}$ of $x_{n}$ which tends to $\ell_{1}$. If $\ln \left( 1+\frac1{\ell_{1}}\right)$ were $\neq 1$, then, by using some estimates, we would get that the relation $\frac{(x+1)^{1/n}-x^{1/n}}{1/n}= 1$ would eventually be untrue. Thus, $\ln \left( 1+\frac1{\ell_{1}}\right) = 1$.
Similarly, $\ln \left( 1+\frac1{\ell_{2}}\right) = 1$.

Therefore, $\ell_{1}= \ell_{2}= \frac{1}{e-1}= L$.

It seems too short to be correct. :huh:

Hi,
Here is a possible approach of this very nice problem:
let $f[0,1]\rightarrow R$ given by $f(x)=\sqrt[n]{x+1}-\sqrt[n]{x}-\frac{1}{n}$.
A calculation shows that:
$f'(x)=\frac{\sqrt[n]{x^{n-1}}-\sqrt[n]{(x+1)^{n-1}}}{n\sqrt[n]{x^{n-1}(x+1)^{n-1}}}<0$. Thus, $f$ is a decreasing function on the interval $[0,1]$. Another calculation shows that:
$f(0)=1=\frac{1}{n}>0$ and $f(1)=\sqrt[n]{2}-1-\frac{1}{n}<0$. Thus there is a unique solution of the equation $f(x)=0$ on the interval $[0,1]$.

let $a_{n}=\frac{n^{n}}{(n+1)^{n}-n^{n}}$. A calculation shows that
$f(a_{n})=\frac{1}{\sqrt[n]{(n+1)^{n}-n^{n}}}-\frac{1}{n}<0$, since $2<\frac{(n+1)^{n}}{n^{n}}$.
Let $b_{n}=\frac{n^{n+1}}{(n+1)^{n+1}-n^{n+1}}$. Another calculation shows that $f(b_{n})>0$, therefore $x_{n}\rightarrow \frac{1}{e-1}$.
Now:
find $\lim n(x_{n}-\frac{1}{e-1})=$


I can not understand the line denoted by red line.
1.  I think it should be the subsequences of $ (x_n)_{n\geq 1}$: $ k>K$ such that $ f^{n_k}(x)>x^*$.
2. $ |f^n(x)\minus{}f^{n\plus{}1}(x)|<\delta$ and $ f^n(x)>x^*$ can not have same N, because $ n_0$ is the smallest such integer.

Thanks

Oh, my! Why don't you find a better book to study from? :roll: This one seems so full of stupid lapses. Of course, formally you are right: the reason for $ n_0$ being the least may be just its being $ N\plus{}1$. Fortunately, you can go beyond $ N$, find $ m>N$ such that $ f^m(x)$ is close to $ \xi_1<x^*$ first and take the least $ \color{red}n\ge m$ such that $ f^n(x)>x^*$ afterwards.

Thanks, the book has many errors.
So it is terrible to study alone.


calculus
integration
logarithms
function
calculus computations

What about it? It obviously converges and I doubt very much it can be expressed in terms of some classical constants: it would be almost the same as to find the antiderivative of $e^{-t}/t^{2}$, which is not an elementary function.

you can't do any fancy integration tools? how about
$\int_{1}^{\infty}\frac{e^{-t}+e^{-3t}+e^{-5t}}{t}dt$

fedja's explanations apply to that with equal force. Nothing nice will happen with it.

Here's a problem for which there might be hope - note that it is vital that the interval be "all" of something, in this case all of $[0,\infty),$ and that the coefficients be arranged so as to have it converge near $0.$

$\int_{0}^{\infty}\frac{e^{-t}-2e^{-3t}+e^{-5t}}{t^{2}}\,dt$

It seems to me that Kent's integral equals:

$\ln \frac{5^{5}}{3^{6}}$.

To see this we let

$I(a)=\int_{0}^{\infty}\frac{e^{-at}-2e^{-3at}+e^{-5at}}{t^{2}}dt$.
Then a calculations shows that:

$I'(a)=5\int_{0}^{\infty}\frac{e^{-3at}-e^{-5at}}{t}dt-\int_{0}^{\infty}\frac{e^{-at}-e^{-3at}}{t}dt=\ln \frac{5^{5}}{3^{6}}=b$.
Integration yields:

$I(a)=ba+d$. Now is $a=0$ we see that the $d=0$, hence

$I(1)=b=\ln \frac{5^{5}}{3^{6}}$.

wow...I understand these general methods you guys are using but how do you just know when the method works? is there an outline of preferred methods to use for particular functions?

Checking didi's work by adopting a different method:

Use the fact that $\frac{1}{t^{2}}=\int_{0}^{\infty}xe^{-tx}\,dx:$

$\int_{0}^{\infty}\frac{e^{-t}-2e^{-3t}+e^{-5t}}{t^{2}}\,dt$

$=\int_{0}^{\infty}\int_{0}^{\infty}xe^{-tx}\left(e^{-t}-2e^{-3t}+e^{-5t}\right)\,dx\,dt$

$=\int_{0}^{\infty}x\int_{0}^{\infty}e^{-(1+x)t}-2e^{-(3+x)t}+e^{-(5+x)t}\,dt\,dx$

$=\int_{0}^{\infty}x\left(\frac{1}{1+x}-\frac{2}{3+x}+\frac{1}{5+x}\right)\,dx$

Now use that $\frac{x}{1+x}=1-\frac{1}{1+x},\ \frac{x}{3+x}=1-\frac{3}{3+x},\ \frac{x}{5+x}= 1-\frac{5}{5+x}:$

$=\int_{0}^{\infty}\left(1-\frac{1}{1+x}\right)-2\left(1-\frac{3}{3+x}\right)+\left(1-\frac{5}{5+x}\right)\,dx$

$=\int_{0}^{\infty}-\frac{1}{1+x}+\frac{6}{3+x}-\frac{5}{5+x}\,dx$

$=\left.\ln\left(\frac{(3+x)^{6}}{(1+x)(5+x)^{5}}\right) \right|_{0}^{\infty}=0-\ln\left(\frac{3^{6}}{1\cdot5^{5}}\right)$

$=\ln\left(\frac{5^{5}}{3^{6}}\right).$

So it checks.

One note on justifying the exchange of integrals:

$e^{-t}-2e^{-3t}+e^{-5t}=e^{-t}(1-e^{-2t})^{2}\ge 0.$

Since the integrand is nonnegative, the interchange is justified by Tonelli's Theorem; the absolute integrability is then a consequence of the succeeding calculations rather than something we needed to establish beforehand.


search
number theory
prime numbers
algebra
system of equations

Solve the system of equations

$ x^2 \plus{} yu \equal{} (x \plus{} u)^n$

and

$ x^2 \plus{} yz \equal{} u^4$,

where $ x,y,z$ are prime numbers, and $ u,n$ positive integers

We search for all $ (x,y,z,u,n)$:

$ x^2+yz=u^4\Leftrightarrow yz=u^4-x^2=(u^2+x)(u^2-x) \Rightarrow$ so $ u\ge 2$ and we have 3 cases:
[hide="Case 1"]
$ \{\begin{array}{c} yz=u^2+x \\ 1=u^2-x\end{array} \ \Rightarrow x=u^2-1 \Leftrightarrow x=u+1, 1=u-1 \Rightarrow u=2, x=3$ so $ yz=2^2+3=7$, no solutions.
[/hide]
For the second a third case $ (u,x)=1$ other way $ y$ or $ z$ wouldn't be prime.
[hide="Case 2"]
$ \{\begin{array}{c} y=u^2+x \\ z=u^2-x\end{array}\ \Rightarrow x^2+(u^2+x)u=(u+x)^n$ but for $ n\ge 3, (x+u)^n\ge x^2+(u^2+x)u$ so $ n\le 2$
[hide="n=2"]
$ x^2+(u^2+x)u=x^2+u^2+2ux \Leftrightarrow x=u(u-1) \Rightarrow x=u, 1=u-1 \Leftrightarrow u=2=x$, again no solutions.
[/hide]
[hide="n=1"]
$ x+u=x^2+u^3+ux$, but $ x^2>x$ and $ u^3>u$, no solutions.
[/hide]
[/hide]
[hide="Case 3"]
$ \{\begin{array}{c} y=u^2-x \\ z=u^2+x\end{array}\$ as in Case 2 $ n\le 2$.
[hide="n=2"]
$ x^2\plus{}(u^2\minus{}x)u\equal{}x^2\plus{}u^2\plus{}2ux \Leftrightarrow 3x\equal{}u(u\minus{}1) \Rightarrow 3\equal{}u, x\equal{}u\minus{}1 \Leftrightarrow x\equal{}2$ so $ y\equal{}3^2\minus{}2\equal{}7$ and $ z\equal{}3^2\plus{}2\equal{}11$, $ (2,7,11,3,2)$ is a solution.
[/hide]
[hide="n=1"]
$ x^2\plus{}u^3\equal{}ux\plus{}x\plus{}u$, if $ x\ge u\plus{}1 \Rightarrow x(u\plus{}1)\plus{}u\le x^2\plus{}u<x^2\plus{}u^3$.
If $ x<u$: $ 0<x^2\minus{}x\equal{}ux\plus{}u\minus{}u^3<u^2\plus{}u\minus{}u \Rightarrow 0<1\plus{}u\minus{}u^2\equal{}\frac{5}{4}\minus{}(u\minus{}\frac{1}{2})^2 \Rightarrow u\equal{}1$, no solutions.
[/hide]
[/hide]
Sumarazing, there is only one solution $ (2,7,11,3,2)$.

 :)


algorithm
algebra
polynomial
number theory
greatest common divisor

Find the GCD of $x^{n}-1$ and $x^{m}-1, m,n \in \mathbb{Z^{+}}$.

Has been posted several times, also recently. [hide="Once more"]Let $d = (x^{n}-1, x^{m}-1)$.  Then $x^{n}\equiv x^{m}\equiv 1 \bmod d$.  By Bezout's Theorem $\exists x, y \in \mathbb{Z}: nx+my = (n, m)$.  Hence

$x^{nx+my}= x^{(n, m)}\equiv 1 \bmod d \implies d | x^{(n, m)}-1$

Easily verified that $x^{(n, m)}-1 | d$; ergo the two are the same up to units.[/hide]

Wow, that's Bezout's? I thought it was Euclid  :blush:

The actual statement is known as Bezout's Theorem or Bezout's Lemma, although a nice way to prove it is by using Euclid's Algorithm (an algorithm, and so neither a theorem nor a lemma).


There are 3 pipes, one small, one medium, and one large.  They are all used to fill a pool.  If all 3 are used at the same time then it takes 3 hours to fill the entire pool.  It takes the largest pipe 1/3 the time of the smallest pipe to fill the pool.  It takes the medium twice the time of the large pipe.  How long does it take the smallest pipe to fill the entire pool.

The answer is 16.5.  I can't seem to get that.  Maybe the textbook is wrong?  Iam sooooo confused.

                                                      Help?

This post is a repeat of a previous one by the same user that has already been answered. May I sugest it be closed?

yea, mods could you please lock this please?


induction
number theory
relatively prime

[quote="nateharman1234"]In a proof is it alright to assume that it is true for x and then use contradiction to prove that implies x +1 is true?[/quote]

Do you got an example??? Well, I'm not sure you can, but if you have an example it would be better

I was going to use it to shorten my proof that consecutive Fibonacci numbers are relatively prime, but I'm not sure if I can use both in one proof.

[quote="nateharman1234"]In a proof is it alright to assume that it is true for x and then use contradiction to prove that implies x +1 is true?[/quote]

sure, as long as you do the base case.

Really short proof: prove $F_n^2-F_{n+1}F_{n-1}=(-1)^n$.

Remember two integers $a,b$ are relatively prime if and only if there exist integers $x,y$ such that $ax+by=1$.


geometry
incenter
circumcircle
power of a point
radical axis
geometry unsolved

[quote="marko avila"]In a non-equilateral triangle $ABC$, the angle bisectors of the angles $CAB$ and $ABC$ meet the opposite sides $BC$ and $CA$ at two points $K$ and $L$ respectively. Let $I$, $O$, $H$ be the incenter, the circumcenter, and the orthocenter of the triangle $ABC$, respectively. Show that the following two statements are equivalent:

(a) The line $KL$ is tangent to the circumcircles of triangles $ALI$, $BHI$, and $BKI$.

(b) The points $A$, $B$, $K$, $L$, $O$ lie on one circle.[/quote]

Using customary notation: Let the (internal) bisectors of the angles $\angle A,\ \angle B$ of a triangle $\triangle ABC$ meet the opposite sides at $K$ and $L.$ Let $O,\ I$ and $H$ be the circumcenter, incenter and orthocenter of triangle $\triangle ABC,$ respectively. Prove that $A, B, K, L, O$ are on the same circle if and only if $KL$ is tangent to the circumcircles of $\triangle BHI,\ \triangle ALI,\ \triangle BKI.$

[color=blue][b]1.[/b][/color] $KL$ is tangent to the circumcircles of $\triangle ALI,\ \triangle BKI$ iff $\triangle ABC$ is isosceles with $\angle A = \angle B.$

Let $P,\ Q$ be the circumcenters of $\triangle ALI$ and $\triangle BKI.$

$\angle AIL = \frac{\angle A+\angle B}{2},\ \ \angle APL = 2 \angle AIL = \angle A+\angle B\ \ \Longrightarrow$

$\angle PLA = \frac \pi 2-\frac{\angle APL}{2}= \frac{\angle C}{2}= \angle ICA$ and similarly, $\angle QKB = \frac{\angle C}{2}= \angle ICB.$ It follows that $PL \parallel IC \parallel QK.$ $KL$ is tangent to the circles $(P),\ (Q)$ iff $PL \perp KL,\ QK \perp KL \Longleftrightarrow\ CI \perp KL\ \Longleftrightarrow\ \triangle LKC$ is isosceles with $CL = CK.$ Let $a = CB,\ b = CA,\ c = AB.$

$CL = \frac{ab}{c+a},\ \ CK = \frac{ab}{b+c}\ \Longrightarrow$

$CL = CK\ \Longleftrightarrow\ c+a = b+c\ \Longleftrightarrow\ a = b\ \Longleftrightarrow\ \angle A = \angle B$

[color=blue][b]2.[/b][/color] $KL$ is tangent the circumcircles of $\triangle BHI,\ \triangle BKI$ iff $B, K, H, I$ are concyclic (i.e., iff these 2 circumcircles are identical).

Let $(Q')$ be the circumcircle of $\triangle BHI$ and $K'$ the tangency point of $KL$ with $(Q').$ $BI \equiv BL$ is the radical axis of the circles $(Q), (Q'),$ hence $LK = LK',$ which means that $K' \equiv K,\ (Q') \equiv (Q).$

[color=blue][b]3.[/b][/color] Assume $A, B, K, L, O$ are concyclic. Then $\frac{\angle A}{2}= \angle LAK = \angle LBK = \frac{\angle B}{2},\ \angle A = \angle B.$ From (1), $KL$ is tangent to the circumcircles of $\triangle ALI,\ \triangle BKI.$ Moreover, $2 \angle C = \angle AOB = \angle AKB = \angle ALB = \pi-\frac{3}{2}\angle A.$ Since $2 \angle A+\angle C = \pi,$ we get $\angle C = \frac{\pi}{5},\ \angle A = \angle B = \frac{2 \pi}{5}.$ $\angle BHI \equiv \angle BHC = \angle A+\angle C$ and $\angle BKI = \angle BKA = \pi-\frac{3}{2}\angle A,$ so that $\angle BHI+\angle BKI = \pi-\angle A+\angle C = \pi-\frac \pi 5+\frac \pi 5 = \pi$ and $B, K, H, I$ are concyclic, which means that $KL$ is also tangent to the circumcircle of $\triangle BHI.$ Conversely, assume that $KL$ is tangent to the circumcircles of $\triangle BHI,\ \triangle ALI,\ \triangle BKI.$ From (1), $\angle A = \angle B$ and from (2), $B, K, H, I$ are concyclic. This means that $\angle BHI+\angle BKI = \pi.$ But $\angle BHI \equiv \angle BHC = \angle A+\angle C$ and $\angle BKI = \pi-\frac{3}{2}\angle A,$ so that $\pi-\frac{\angle A}{2}+\angle C = \pi,$ which again yields $\angle C = \frac \pi 5,\ \angle A = \frac{2 \pi}{5}.$ Then $\angle AOB = 2 \angle C = \frac{2 \pi}{5}= \pi-\frac{3}{2}\angle A = \angle AKB = \angle ALB,$ so that $A, B, K, L, O$ are concyclic.


function
limit
inequalities
LaTeX
real analysis
real analysis unsolved

Let $f:[0,+\infty)\rightarrow R$ be a function with the following caractheristics:
1. $\exists$ M such that $\mid\mbox{f}(x)\mid\leq\mbox{M} x,\forall x\in[0,+\infty)$
2.$f(x)+f(y)\leq\mbox{f}(x+y),\forall x,y\in[0,+\infty)$
Prove that:
3.$\exists\mbox{a,b}\in R$ so that $a=\lim_{x\rightarrow 0}\frac{f(x)}{x}$ and $b=\lim_{x\rightarrow+\infty}\frac{f(x)}{x}$
4.$\forall x\in [0,+\infty)$ the following inequalities hold $ax\leq\mbox{f(x)}\leq\mbox{b}x$

Sorry about that. I was experiencing some difficulties with Latex.

I have corrected some $\LaTeX$ related things:
[quote="vlad"]Let $f:[0,+\infty)\rightarrow \mathbb R$ be a function with the following properties:
1. $\exists\ M$ such that $|f(x)|\leq Mx\ \forall x\in[0,+\infty)$;
2.$f(x)+f(y)\leq f(x+y)\ \forall x,y\in[0,+\infty)$.
Prove that:
3.$\exists\,a,b\in\mathbb R$ so that $a=\lim_{x\rightarrow 0}\frac{f(x)}{x}$ and $b=\lim_{x\rightarrow+\infty}\frac{f(x)}{x}$;
4.$\forall x\in [0,+\infty)$ the following inequalities hold $ax\leq f(x)\leq bx$.[/quote]

The conclusion immediately follows if we consider $a=\inf \frac{f(x)}{x}$ and $b=\sup \frac{f(x)}{x}$ ;)

This is a classical oral problem for X entrance exam.

harazi could you please post a solution?
thanks

I have one principle: do not post solution before the others think a little bit on it. It is not fair.

Ok, put $a=inf_{x>0}{\frac{f(x)}{x}}$. Of course we have $f(0)=0$ and $ f(x)\geq ax$ for all x. Next, take $eps>0$ and $x_0$ such that $f(x_0)<(a+eps)x_0$. For all $x<x_0$ we have $(a+eps)x_0>f(x_0)=f(x [x_0/x]+r_x)\geq [x_0/x]f(x)-x M>(x_0/x-1)f(x)-Mx$, from where you conclude that $a$ is the limit of $f(x)/x$. The same for the limit at infinity.
   OBS: edited after Myth observed some idiot mistakes I've made. I was thinking to infinity when proving a limit in à!

He-he. You must be mixed up some things...  :)

Is this also wrong? Oh, my god, I should take a break probably.

You wrote $f(x_0)>(a-eps)x_0$ :) Though $a=\inf$. And some other things are also mixed. ;)

You're right, sorry, I will try to correct the proof.


Putting this ellipse equation in fraction form:

$ \frac {x^2}{49} \plus{} \frac {y^2}{36} \equal{} 1$

This basically is an ellipse with $ a \equal{} 7$ and $ b \equal{} 6$, so it's an ellipse with a horizontal major axis, thus the foci are on the x-axis.
The distance the foci are from the center (origin), $ c$, is basically $ c \equal{} \sqrt {a^2 \minus{} b^2} \equal{} \sqrt {7^2 \minus{} 6^2} \equal{} \sqrt {49 \minus{} 36} \equal{} \sqrt {13}$. so the foci of this ellipse are $ ( \minus{} \sqrt {13},0)$ and $ (\sqrt {13},0)$.

[color=indigo] :blush: i realized that as soon as I had posted...lolz..
can someone help me with this next problem? i'll write it as soon as i have a volunteer... :ninja: [/color]

Well, anyone here can volunteer to solve problems in this forum's threads. I'll be happy to help if I can.
What's the next problem, if you're posting it soon?

YAY!!!!! ur onf.... the next problem is to find the center, the length of the major axis, and the length of the minor axis for this nex equattions... $ (x\minus{}3)^2/9 \plus{} (y\minus{}6)^2/16\equal{}1$

Basically this equation is in the form $ \frac{(x\minus{}h)^2}{b^2}\plus{}\frac{(y\minus{}k)^2}{a^2}\equal{}1$, where $ (h,k)$ is the center and $ a>b$. $ 2a$ is the major axis and $ 2b$ is the minor axis.

Plugging in the corresponding values into that equation form (with h=3, k=6, $ a^2\equal{}16$, $ b^2\equal{}9$, a=4, and b=3), we have:

Center: (3,6)
Length of major axis: 2(4)=8
Length of minor axis: 2(3)=6

And that's all. :)

i just passed a quiz using the formula u gave me...thanks :D  :lol:


quadratics
function
algebra unsolved
algebra

Find all integer numbers $x$ and $y$ satisfying:
$x^2-y^2+xy+1=0$

Treat this equation as a one-variable quadratic function of $x$ and look at $\sqrt \Delta$ - it must be an integer, you get easy to solve another equation.


inequalities
Cauchy Inequality
inequalities proposed

1) Let $ x$ and $ y$  are non-negative numbers. Prove that:
\[ (x^3 \plus{} 5)(y^3 \plus{} 5)\geq18(x \plus{} y)
\]
2) Let $ x,$ $ y$ and $ z$ are non-negative numbers. Prove that:
\[ \frac{(x^3\plus{}2)(y^3\plus{}2)(z^3\plus{}2)}{27}\geq\sqrt{\frac{(x^2\plus{}y^2)(x^2\plus{}z^2)(y^2\plus{}z^2)}{8}}\]

For the first one .Apply Holder inequality we have:
    $ (x^3\plus{}1\plus{}1\plus{}1\plus{}1\plus{}1)(y^3\plus{}1\plus{}1\plus{}1\plus{}1\plus{}1)(1\plus{}1\plus{}1\plus{}1\plus{}1\plus{}1) \geq (x\plus{}y\plus{}4)^3$
And now apply Cauchy inequality we have $ (x\plus{}y\plus{}4)^3\equal{}(x\plus{}y\plus{}2\plus{}2)^3 \geq 27.2^2.(x\plus{}y)$
Then ,easily we get QED

For the second.We only need to prove that $ \frac{(x^3\plus{}2)(y^3\plus{}2)}{9} \geq \frac{x^2\plus{}y^2}{2}$
Case 1: If $ (x^2\minus{}1)(y^2\minus{}1) \geq 0$ then LHS-RHS = $ 2(2x^3\plus{}1\minus{}3x^2) \plus{} 2(2y^3\plus{}1\minus{}3y^2) \plus{} 2 (x^3y^3\plus{}2 \minus{}3xy) \plus{}
\plus{} 3(x^2\minus{}1)(y^2\minus{}1) \geq 0$
Case 2 :If $ 0 \geq (x^2\minus{}1)(y^2\minus{}1)$ .supposed that $ x\geq 1\geq y \geq 0$
Consider $ f(x)\equal{} \frac{y^3\plus{}2}{9}.(x^3\plus{}2) \minus{} \frac{x^2}{2} \minus{}\frac{y^2}{2}$
    $ f'(x)\equal{}x(\frac{x(y^3\plus{}2)}{3}\minus{}1)$
Then we have the minimum value of $ f(x)$ is $ f(\frac{3}{y^3\plus{}2}) \equal{}g(y)\equal{} \frac{2}{9}.(y^3\plus{}2) \minus{} \frac{3}{2(y^3\plus{}2)^2}\minus{}
\minus{} \frac{y^2}{2}$
 $ g'(y) \equal{} \frac{2}{3}.y^2 \minus{} y \plus{} \frac{9y^2}{(y^2\plus{}2)^3}$
By Cauchy inequality we have $ (y^2\plus{}2)^3 \geq 27y^2$ then we get $ 0 \geq y^2\minus{}y \geq g'(y)$
because $ 1\geq y \geq 0$
Since then $ g(y) \geq g(1)\equal{}0$.QED


modular arithmetic
algorithm
number theory
Euclidean algorithm
number theory open

Theorem: If $ n$ is odd prime, then $ x^n\plus{}y^n\equal{}z^n$ have no solution $ x,y,z$ such that $ (x, y)\equal{}(y, z)\equal{}(z, x)\equal{}1, 2|y, |x|>1,
|z|>1.$
Proof: Let $ a$ meet $ z^a\equiv\minus{}1, (\minus{}y)^a\equiv x\minus{}1, \pmod {x^2}.$
In case of $ a\equal{}bn:$
If $ 2|b,$ then $ z^a\minus{}y^a\equiv \minus{}x, \pmod {x^2}$ which is contradictory;
If $ 2\nmid b,$ then $ z^a\minus{}y^a\equiv x\minus{}2, \pmod {x^2}$ which is contradictory.
In case of $ n\nmid a:$ Since we have $ x>z\minus{}y, (z, y)\equal{}1,$ according to Euclidean algorithm, if $ (m, n)\equal{}1,$ we get $ (\frac{z^m\minus{}y^m}{z\minus{}y}\frac{z^n\minus{}y^n}{z\minus{}y})\equal{}1.$
If $ 2|a,$ then $ x|(\frac{z^a\minus{}y^a}{z\minus{}y},\frac{z^n\minus{}y^n}{z\minus{}y})$ which is contradictory.
In case of $ 2\nmid b:$ Let $ (x, z\minus{}y)\equal{}t.$ Since $ x|z^a\plus{}y^a,$ hence $ t|(x,2y^a).$ It follows that $ t\equal{}1.$ Then we verify $ |z\minus{}y|\equal{}1.$
Let $ u^n\plus{}v^n\equal{}w^n, 2|v.$
In case of $ z\equal{}w, y\equal{}v, x\equal{}u,$ we have $ |w\minus{}v|\equal{}1.$
In case of $ z\equal{}u, y\equal{}\minus{}v, x\equal{}w,$ we have $ |u\plus{}v|\equal{}1.$
There is no solution of $ |w\minus{}v|\equal{}|u\plus{}v|\equal{}1$ which is contradictory.

I assumed a lemma in the first line of the proof above:

Every $ a^x\equiv c, b^x\equiv d, \pmod p$ has a positive integer $ x$ solution.

[quote="Taoli"]Every $ a^x\equiv c, b^x\equiv d, \pmod p$ has a positive integer $ x$ solution.[/quote]
This isn't true even for one equation.

[quote="Taoli"]Let $ a$ meet $ z^a\equiv \minus{} c, ( \minus{} y)^a\equiv x \minus{} c, \pmod {x^2}, 0<c<|x|, (c, x)\equal{}1.$[/quote]Theorem: If $ n$ is odd prime, then $ x^n \plus{} y^n \equal{} z^n$ have no solution $ x,y,z$ such that $ (x, y) \equal{} (y, z) \equal{} (z, x) \equal{} 1, 2|y, |x| > 1, |z| > 1.$
Proof: Let $ a$ meet $ z^a\equiv \minus{} c, ( \minus{} y)^a\equiv x \minus{} c, \pmod {x^2}, 0<c<|x|, (c, x)\equal{}1.$
In case of $ a \equal{} bn:$
If $ 2|b,$ then $ z^a \minus{} y^a\equiv \minus{} x, \pmod {x^2}$ which is contradictory;
If $ 2\nmid b,$ then $ z^a \minus{} y^a\equiv x \minus{} 2c, \pmod {x^2}$ which is contradictory.
In case of $ n\nmid a:$ Since we have $ x > z \minus{} y, (z, y) \equal{} 1,$ according to Euclidean algorithm, if $ (m, n) \equal{} 1,$ we get $ (\frac {z^m \minus{} y^m}{z \minus{} y}\frac {z^n \minus{} y^n}{z \minus{} y}) \equal{} 1.$
If $ 2|a,$ then $ x|(\frac {z^a \minus{} y^a}{z \minus{} y},\frac {z^n \minus{} y^n}{z \minus{} y})$ which is contradictory.
In case of $ 2\nmid b:$ Let $ (x, z \minus{} y) \equal{} t.$ Since $ x|z^a \plus{} y^a,$ hence $ t|(x,2y^a).$ It follows that $ t \equal{} 1.$ Then we verify $ |z \minus{} y| \equal{} 1.$
Let $ u^n \plus{} v^n \equal{} w^n, 2|v.$
In case of $ z \equal{} w, y \equal{} v, x \equal{} u,$ we have $ |w \minus{} v| \equal{} 1.$
In case of $ z \equal{} u, y \equal{} \minus{} v, x \equal{} w,$ we have $ |u \plus{} v| \equal{} 1.$
There is no solution of $ |w \minus{} v| \equal{} |u \plus{} v| \equal{} 1$ which is contradictory.

[quote="Taoli"]I assumed a lemma in the first line of the proof above:

Every $ a^x\equiv y, b^x\equiv p \plus{} y, \pmod {p^2}, (a, b) \equal{} 1, (y(p \plus{} y), p) \equal{} 1$ has positive integer $ x, y$ solution such that $ 0 < y < p$.[/quote]

It's still not even true that $ a^x \equiv y \bmod p$ always has a solution.  Take $ a \equal{} 1, y \neq 1$.

Since in the theorem it assumed that $ (x, y)\equal{}1, 2|y, |x| > 1, |z| > 1,$, hence it follows that $ |a| > 1, |b| > 1$ in the lemma.

Any other counterexample?

You should start to try to check your "theorems" for counterexamples yourself. For this one, it's very easy, especially in comparison to the stuff you talk about.

lemma:

Every $ a^x\equiv y, b^x\equiv p \plus{} y, \pmod {p^2}, |a|>1, |b|>1,$
$ (a, b)\equal{}(a, p)\equal{}(b,p)\equal{}1,(y(p \plus{} y), p) \equal{} 1$ has positive integer
 $ x$ and integer $ y$ solution such that $ 0 < |y| < p$.

Stop talking about $ b$; you still haven't resolved anything with $ a$.  Take $ a \equal{} p \plus{} 1$, and if you're going to whine about that counterexample take $ a \equal{} p \minus{} 1$ with $ y$ equal to anything other than $ \pm 1$ and $ p > 3$.

More generally, $ a^x \bmod p$ can take $ d$ different values for any divisor $ d$ of $ p\minus{}1$ depending on $ a$.  For any $ a$ other than the primitive roots $ \bmod p$, of which there are $ p\minus{}1 \minus{} \varphi(p\minus{}1)$, $ a^x \equiv y \bmod p$ where $ y$ is a primitive root is impossible.

[quote="Taoli"][quote="Taoli"]Let $ a$ meet $ z^a\equiv \minus{} c, ( \minus{} y)^a\equiv x \minus{} c, \pmod {x^2}, 0 < c < |x|, (c, x) \equal{} 1.$[/quote]Theorem: If $ n$ is odd prime, then $ x^n \plus{} y^n \equal{} z^n$ have no solution $ x,y,z$ such that $ (x, y) \equal{} (y, z) \equal{} (z, x) \equal{} 1, 2|y, |x| > 1, |z| > 1.$
Proof: Let $ a, p, k$ meet $ p|x, p>1, a>3,z^a\equiv \minus{} c, ( \minus{} y)^a\equiv x^{a\minus{}1} \minus{} c, \pmod {x^k}, k>a, 0 < c < |x|, (c, x) \equal{} 1.$
In case of $ a \equal{} bn:$
If $ 2|b,$ then $ z^a \minus{} y^a\equiv \minus{} x^{a\minus{}1}, \pmod {x^k}$ which is contradictory;
If $ 2\nmid b,$ then $ z^a \minus{} y^a\equiv x^{a\minus{}1} \minus{} 2c, \pmod {x^k}$ which is contradictory.
In case of $ n\nmid a:$ Since we have $ x > z \minus{} y, (z, y) \equal{} 1,$ according to Euclidean algorithm, if $ (m, n) \equal{} 1,$ we get $ (\frac {z^m \minus{} y^m}{z \minus{} y},\frac {z^n \minus{} y^n}{z \minus{} y}) \equal{} 1.$
If $ 2|a,$ then $ x|(\frac {z^a \minus{} y^a}{z \minus{} y},\frac {z^n \minus{} y^n}{z \minus{} y})$ which is contradictory.
In case of $ 2\nmid b:$ Let $ (x, z \minus{} y) \equal{} t.$ Since $ x|z^a \plus{} y^a,$ hence $ t|(x,2y^a).$ It follows that $ t \equal{} 1.$ Then we verify $ |z \minus{} y| \equal{} 1.$
Let $ u^n \plus{} v^n \equal{} w^n, 2|v.$
In case of $ z \equal{} w, y \equal{} v, x \equal{} u,$ we have $ |w \minus{} v| \equal{} 1.$
In case of $ z \equal{} u, y \equal{} \minus{} v, x \equal{} w,$ we have $ |u \plus{} v| \equal{} 1.$
There is no solution of $ |w \minus{} v| \equal{} |u \plus{} v| \equal{} 1$ which is contradictory.[/quote]
Otherwise let $ a\equal{}n\plus{}1, d\equiv \minus{}z^a, \pmod {x^a}.$ Since $ x^n \plus{} y^n \equal{} z^n,$ hence $ (\minus{}y)^a\equal{}y(z^n\minus{}x^n)\equiv y(\frac {\minus{}d}{z}\minus{}x^{a\minus{}1}), \pmod {x^a}$. It follows that $ c\equal{}\frac {\minus{}d}{z}\equiv \minus{}z^{a\minus{}1}, (\minus{}y)^{a\minus{}1}\equal{}\minus{}c\plus{}x^{a\minus{}1}, \pmod {x^a}.$.

Theorem: If $ n$ is odd prime, then $ x^n \plus{} y^n \equal{} z^n$ have no solution $ x,y,z$ such that $ (x, y) \equal{} (y, z) \equal{} (z, x) \equal{} 1, 2|y, |x| > 1.$
Proof: For any prime $ p|x.$ Let $ y^{p \minus{} 1} \equiv z^{p \minus{} 1}\equiv 1, \pmod p.$ In case of $ n\nmid p \minus{} 1:$ Since we have $ x > z \minus{} y, (z, y) \equal{} 1,$ according to Euclidean algorithm, if $ (m, n) \equal{} 1,$ we get $ (\frac {z^m \minus{} y^m}{z \minus{} y},\frac {z^n \minus{} y^n}{z \minus{} y}) \equal{} 1.$ Then $ p|(\frac {z^{p\minus{}1} \minus{} y^{p\minus{}1}}{z \minus{} y},\frac {z^n \minus{} y^n}{z \minus{} y})$ which is contradictory. Then $ n|p \minus{} 1.$ Hence $ x\equiv1, \pmod n$. Let $ u^n \plus{} v^n \equal{} w^n, 2|v.$ In case of $ z \equal{} w, y \equal{} v, x \equal{} u,$ we have $ u\equiv1, \pmod n.$
In case of $ z \equal{} \minus{} w, y \equal{} \minus{} v, x \equal{} \minus{} u,$ we have $ \minus{} u\equiv1, \pmod n.$ Then $ 0\equiv2, \pmod n,$ which is contradictory.


How many ways can we make 20 dollars using only dimes, quarters, and half dollars? (note: there is not a limited amount of coins to be used, you can use as many dimes/quarters/Half dollars as you want, but they need to add up to 20 dollars.)

I can figure out how to "brute force" it, but i would like to know an easier way to do this.

Wait, you say dimes, quarters, and half dollars and then dimes/quarters/nickels.
Is it nickels or half dollars?

I meant half dollars.

oops

I get the feeling that this is really basic and I just can't remember the formula.

there really isn't a formula for doing this, and it's not that simple to boot.
i guess the best bet you have is using generating functions.

This problem is supposed to be doable in about 5 minutes (according to the people who organize our local math league problems.) Judging by how easy the other problems are, I wouldn't expect they want us to use "generating functions". I suppose they just want it to be brute forced. Oh well..

Also, how do I delete a thread on this forum, I accidentally posted a thread when I meant to post a message.

Generating functions are most definitely not required for this problem :wink:

[hide="Hint for simplification by divisibility"]We want the number of solutions to:
Let a be the number of dimes, b be the number of quarters, and c be the number of half-dollars

10a+25b+50c = 2000, which, upon division by 5, gives 2a+5b+10c = 400

Note that the "a" and "c" terms are even, and the RHS, 400 is even, so 5b must be even, and hence b must be even. Let b = 2d.

2a+10(c+d) = 400

Also, since 10(c+d) is divisible by 10 and 400 is divisible by 10, 2a must be divisible by 10, and a must be divisible by 5. Let a = 5e

10(c+d+e) = 400, [b]c+d+e = 40[/b], where c, d, and e are nonnegative integers.

Now does it seem more doable?  :wink: The equation in the form above is fairly quick to solve, but even if you can't figure out how immediately, it is still much faster to do out, since there are some pretty blatant patterns.
[/hide]

[hide="Continued into full solution"]
c+d+e=40, so c+d=40-e

For any number k, the solutions to c+d=k range from (c,d) = (k,0), (k-1, 1) ... (1, k-1), (0, k), so there are k+1 solutions.

Thus, there are (40-e)+1 = 41-e solutions in (c,d) to c+d=40-e

e can be any integer from 0 to 40:

e=0 gives 41 solutions in (c,d)
e=1 gives 40 solutions in (c,d)
...
e=40 gives 1 solution in (c,d)

So, the answer is 1+2+3+...+41 = (41*42)/(2) = 861 solutions to c+d+e=40 for nonnegative integers.

Since each (c,d,e) corresponds to exactly one (a,b,c), there are [b]861[/b] solution to 10a+25b+50c = 2000.

[/hide]

Oh, I sort of understand it now.

Did you set a=5e so that you could factor out the 10?

I never would have thought of doing it that way.


Jim is hosting a dinner party.Including himself there will be 8 ppl.Jim is going to prepare a seating plan around his circular table.If noah cannot sit beside Amanda or Ryan,how many different seating palns can jim make.

hint: the end result doesn't depend on what order you seat the people in. try this order: seat noah first, then amanda and ryan, then the remaining people


geometry
geometric transformation
reflection
conics
Euler
circumcircle
analytic geometry

In the scalene triangle $ \triangle ABC,$ let $ \triangle H_aH_bH_c$ be its orthic triangle and $ \triangle G_aG_bG_c$ the pedal triangle of its centroid $ G.$ $ \triangle A_0B_0C_0$ is its anticomplementary triangle, whose vertices are the reflections of $ A,B,C$ about the midpoints of $BC,CA,AB$ and  $ A',B',C'$ are the othogonal projections of $ A_0,B_0,C_0$ onto $ BC,CA,AB.$ Show that centroids of $ \triangle ABC,$ $\triangle H_aH_bH_c,$  $\triangle G_aG_bG_c,$ $\triangle A'B'C'$ lie on a parallel line to the [url=http://www.xtec.cat/~qcastell/ttw/ttweng/definicions/d_Brocard_r.html]Brocard axis[/url] of $ \triangle ABC.$

The problem is a consequence of a more general configuration

[b]Lemma.[/b] Let $ \triangle ABC$ be a triangle and $ P$ a variable point on a conic $ \mathcal{H}$ in its plane. Then locus of the centroid of the pedal triangle of $ P$ WRT $ \triangle ABC$ is another conic. When $ P$ moves on a line (degenerate conic) the locus is another line and if this line coincides with the Euler line of $ \triangle ABC,$ then  the locus is parallel to its Brocard axis.

that can be said：
P is an any point on eular line of △ABC，the pedal triangle of P is △DEF，then the locus of centroid of △DEF is a line perpendicular to the Brocard axis of △ABC
there is a little not complicated way to sovle it，use vector.
and what's your way?

Lym, I get you, that was indeed the first result that I observed (see the lemma above). However, I think you're confusing the Brocard axis, which passes through the circumcenter and symmedian point, with the Brocard line, which passes through the two Brocard points of ABC. By the way, I do not have a synthetic proof for the lemma.

I'll prove in the upper lemma the case when $ P$ moves on a line, which I think is the most fruitful result. The case when $ P$ moves on a fixed general conic $ \mathcal{H}$ can be approached with  the same method.

[b]Lemma:[/b] Let $ P$ be a varible point on a fixed line $ \ell$ on the plane of $ \triangle ABC.$ Let $ \triangle XYZ$ be the pedal triangle of $ P$ with respect to $ \triangle ABC.$ Then the locus of the centroid $ G'$ of $ \triangle XYZ$ is another line.

Define the orthogonal reference xy where the sidelines of $ \triangle ABC$ are given by the equations $ ax \plus{} by \plus{} c \equal{} 0,$  $dx \plus{} ey \plus{} f \equal{} 0$ and $ gx \plus{} hy \plus{} k \equal{} 0$ and w.l.o.g asumme that $ \ell$ is identical to the y-axis $ \Longrightarrow$ $ P(0,\varrho).$

Coordinates of the orthogonal projections  $ X,Y,Z$ of $ P$ on $ BC,CA,AB$ are given by

$ X \left ( \frac { \minus{} ac \minus{} ab\varrho}{a^2 \plus{} b^2} , \frac {a^2\varrho \minus{} bc}{a^2 \plus{} b^2} \right)$

$ Y \left (\frac { \minus{} df \minus{} de\varrho}{d^2 \plus{} e^2} , \frac {d^2\varrho \minus{} ef}{d^2 \plus{} e^2} \right )$

$Z \left (\frac { \minus{} gk \minus{} gh\varrho}{g^2 \plus{} h^2} , \frac {g^2\varrho \minus{} hk}{g^2 \plus{} h^2} \right )$

Coordinates of the centroid $ G'(\overline{x},\overline{y})$ of $ \triangle XYZ$ are then

$ \overline{x} \equal{} \minus{} \frac {\varrho}{3} \left(\frac {ab}{a^2 \plus{} b^2} \plus{} \frac {de}{d^2 \plus{} e^2} \plus{} \frac {gh}{g^2 \plus{} h^2} \right ) \minus{} \frac {1}{3} \left (\frac {ac}{a^2 \plus{} b^2} \minus{} \frac {df}{d^2 \plus{} e^2} \minus{} \frac {gk}{g^2 \plus{} h^2} \right )$

$ \overline{y} \equal{} \frac {\varrho}{3} \left(\frac {a^2}{a^2 \plus{} b^2} \plus{} \frac {d^2}{d^2 \plus{} e^2} \plus{} \frac {g^2}{g^2 \plus{} h^2} \right ) \minus{} \frac {1}{3} \left (\frac {bc}{a^2 \plus{} b^2} \plus{} \frac {ef}{d^2 \plus{} e^2} \plus{} \frac {hk}{g^2 \plus{} h^2} \right )$

Locus of $ G'(\overline{x},\overline{y})$ clearly represents a parametric equation of a line.
_____________________________________________________________

Using the lemma in the proposed problem, where $ G,G_1,G_2,G_3$ are the centroids of the pedal triangles of the circumcenter $ O,$ the orthocenter $ H,$ the centroid $ G$ and the De Longchamps point $ X_{20},$ all lying on the Euler line of $ \triangle ABC,$ we get that $ G,G_1,G_2,G_3$ are collinear. Thus it only remains to show that this line is parallel to the Brocard axis of $ \triangle ABC.$

Now, let us use barycentric coordinates WRT $ \triangle ABC.$ Brocard axis passing through the symmedian point $ K \ (a^2: b^2: c^2)$ and the circumcenter $ O \ (a^2S_A: b^2S_B: c^2S_C)$ has infinite point  $T_{\infty}$ with coordinates

$(a^2(c^4 \plus{} b^4 \minus{} a^2c^2 \minus{} a^2b^2): b^2(a^4 \plus{} c^4 \minus{} a^2b^2 \minus{} b^2c^2): c^2(a^4 \plus{} b^4 \minus{} b^2c^2 \minus{} a^2c^2))$

On the other hand, coordinates of the orthogonal projections of $ G(1: 1: 1)$ on $ BC,AC,AB$ are

$ G_a  (0: a^2 \plus{} S_C: a^2 \plus{} S_B)  , \ G_b  (b^2 \plus{} S_C: 0: b^2 \plus{} S_A)  , \ G_c (c^2 \plus{} S_B: c^2 \plus{} S_A: 0)$

Therefore, the centroid $ G_2$ of $ \triangle G_aG_bG_c$ has coordinates

$ G_2 \left ( \frac {b^2 \plus{} S_C}{b^2} \plus{} \frac {c^2 \plus{} S_B}{c^2}: \frac {a^2 \plus{} S_C}{a^2} \plus{} \frac {c^2 \plus{} S_A}{c^2}: \frac {a^2 \plus{} S_B}{a^2} \plus{} \frac {b^2 \plus{} S_A}{b^2} \right )$

These coordinates can be re-expressed as

$(a^2(c^2S_C \plus{} b^2S_B \plus{} 2b^2c^2): b^2(a^2S_A \plus{} c^2S_C \plus{} 2a^2c^2): c^2(a^2S_A \plus{} b^2S_B \plus{} 2a^2b^2))$

The infinite point $R_{\infty}$ of the line $ GG_2$ is then

$(a^2(c^4 \plus{} b^4 \minus{} a^2c^2 \minus{} a^2b^2): b^2(a^4 \plus{} c^4 \minus{} a^2b^2 \minus{} b^2c^2): c^2(a^4 \plus{} b^4 \minus{} b^2c^2 \minus{} a^2c^2))$

$ \Longrightarrow$ $ T_{\infty} \equiv R_{\infty},$ hence the line passing trough $ G,G_1,G_2,G_3$ is parallel to $ OK.$

Is OK the Brocard axis，cause I'm still think a pair Brocard points are the Brocard axis，OH，I got an elementary mistaken

“P is an any point on eular line of △ABC，the pedal triangle of P is △DEF，then the locus of centroid of △DEF is a line perpendicular to the Brocard axis of △ABC ”
Dear luis，aboat this，I have got a geometric method，next I will post here， I'm sovling a three circle problem

The point [b]G[/b] with respect △GaGbGc is calld [b]X1[/b]，[b]G[/b] is a new point aboat [b]△GaGbGc [/b]，[b]X2[/b] is the isogonal conjugate point of [b]X1[/b] with respect to [b]△GaGbGc[/b]
 [b]X1，X2[/b] found by chinese expert [b]Ye zhonghao[/b]，let [b]G' （G2）[/b]be the centroid of [b]△GaGbGc[/b]，[b]F1，F2 [/b]are the fermat points of [b]△GaGbGc[/b]， [b]S1，S2 [/b]are the isodynamic points of [b]△GaGbGc[/b]，Then ：
[b]X1X2//OK[/b]（OK belong △ABC），[b]G'（G2）[/b] is the midpoint of [b]X1X2[/b]，[b]F1、F2、X1、X2 [/b]are on a circle，[b]S1、S2、X1、X2 [/b]are also on a circle ，and [b]lester circle [/b]
[b]X1（G），X2 [/b]are also the focal points of [b]steiner ellipse[/b] aboat [b]△GaGbGc [/b]，[b]F1F2X1X2 ，S1S2X1X2[/b] both are Harmonic Quadrilateral
all of this is new Conclusion

tell you a good information，I have solved lester circle with good geometric method，
only “[b]F1、F2、X1、X2 [/b]，[b]S1、S2、X1、X2 [/b] are on a circle” is not solved with geometric method

Denote $ L$ the symmedian point, $ A_1,B_1,C_1$ the midpoints of $ BC,CA,AB; A_2,B_2,C_2$ the midpoints of $ AH_a,BH_b,CH_c$, respectively.
We have $ \vec{GG_3} \equal{} \frac {1}{3}(\vec{G_aA_1} \plus{} \vec{G_bB_1} \plus{} \vec{G_cC_1}) \equal{} \frac {1}{9}(\vec{H_aA_1} \plus{} \vec{H_BB_1} \plus{} \vec{H_cC_1})$
$ \equal{} \frac {2}{9}.(\vec{A_1A_2} \plus{} \vec{B_1B_2} \plus{} \vec{C_1C_2})$
$ \equal{} \frac {2}{9}.\sum_{cyc}(\vec{B_1C_1}.(\frac {A_1B_2}{A_1C_1} \minus{} \frac {A_1C_2}{A_1B_1})$
$ \equal{} \frac { \minus{} 1}{9}.\sum_{cyc}.(\vec{BC}.(\frac {CH_b}{b} \minus{} \frac {BH_c}{c})) \equal{} \frac { \minus{} 1}{9}.\sum_{cyc}.(\vec{BC}.(\frac {a\cos C}{b} \minus{} \frac {a\cos B}{c}) (1)$
We know that $ L(a^2,b^2,c^2); O(\sin2A,\sin2B,\sin2C)$ then $ \vec{OL}$ has the direction: $ \vec{BC}.(b^2.\sin2C \minus{} c^2.\sin2B) \plus{} \vec{CA}.(c^2.\sin2A \minus{} a^2.\sin2C) \plus{} \vec{AB}.(a^2.\sin2B \minus{} b^2.\sin2A) (2)$
From $ (1)$ and $ (2)$ it's easy to show that $ \vec{GG_3}//\vec{OL}$.
For an arbitrary point $ P$ which lies on Euler's line of trangle $ ABC$, denote $ A_pB_pC_p$ the pedal triangle of $ P, G_p$ the centroid of triangle $ A_pB_pC_p$.Since $ \frac {A_1G_a}{A_1A_p} \equal{} \frac {B_1G_b}{B_1B_p} \equal{} \frac {C_1G_c}{C_1C_p} \equal{} \frac {OG}{OP}$ We get $ \vec{GG_p} \equal{} \frac {1}{3}(\vec{A_1A_p} \plus{} \vec{B_1B_p} \plus{} \vec{C_1C_p}) \equal{} k(\vec{A_1G_a} \plus{} \vec{B_1G_b} \plus{} \vec{C_1G_c})$. Therefore $ G,G_3,G_p$ are collinear. So the locus of the centroid of the pedal triangle of a variable point on Euler line is a line through $ G$ and parallel to $ OL$.

[b]Remark:[/b] [b](1):[/b]  From this problem we get an interesting lemma: Let $ X$ and $ Y$ be two isotomic conjugate points wrt $ \Delta ABC; X_aX_bX_c$ the cevian triangle of $ X$. Denote $ \vec{u}$ so that $ \vec{u} \equal{} \vec{AX_a} \plus{} \vec{BX_b} \plus{} \vec{CX_c}$Then $ \vec{XY}//\vec{u}$.

[b](2): [/b]I see a beautiful contaction of this problem and a property of [b]Fontene[/b]'s theorem:
Given triangle $ ABC$ with its circumcenter $ O$, its centroid $ G$. Let $ M_aM_bM_c,G_aG_bG_c$ be the pedal triangles of $ G$ and $ O$ wrt $ \Delta ABC. (G_aG_bG_c)$ intesects the Nine-point circle of triangle $ ABC$ at two points. Then:
a, One point of the intersections (we call $ X$) is the Anti-Steiner point of $ OG$ wrt $ \Delta M_aM_bM_c$.
b, Another point of the intersections is called $ Y$. Then two Steiner lines $ l_1,l_2$ of $ Y$ wrt $ \Delta M_aM_bM_c$ and $ \Delta G_aG_bG_c$ are parallel.
c, The locus of the centroid of the pedal triangle of a variable point $ P$ on $ OG$ wrt $ \Delta ABC$ is a line which bisects the distance of two lines $ l_1,l_2$.

[b]It's another way to show this problem[/b]  :)

This problem could be solved easily by using the E.R.I.Q Theorem.([url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=97493[/url])
I will return later with more detail.(sorry I'm busy now)

After quite some time, I think I've found a nice synthetic solution to this problem and a couple of additional things like: $G_1$ is the anticomplement of $G$ WRT $\triangle G_aG_bG_c$ and  $G$ is the midpoint of $G_1G_3.$ But first of all let us introduce some general lemmata.

[b]Lemma 1:[/b] In $\triangle ABC$ with circumcircle $(O),$  $M$ is the midpoint of $BC$ and $N,L$ denote the orthogonal projections of $M$ on $AB,AC.$ If $E$ is the midpoint of $NL,$ then $ME \parallel AO.$

[b]Proof: [/b] Let $D$ be the intersection of the tangents of $(O)$ through $B,C$ and let $U$ be the midpoint of $DM.$ If $C'$ denotes the projection of $C$ onto $AB,$ from $\angle BAC=\angle BCD,$ it follows that $\frac{_{CA}}{^{CD}}=\frac{_{CC'}}{^{DM}}=\frac{_{MN}}{^{MU}}$ $\Longrightarrow$ $\triangle ACD \sim \triangle NMU$ (s.a.s). Thereby $\angle UNM=\angle CAD=\angle NLM$ $\Longrightarrow$ $UN$ is tangent to $\odot(MNL)$ through $N.$ Likewise, $UL$ is tangent to $\odot(MNL)$ through $L$ $\Longrightarrow$ $UM \equiv UO$ is the M-symmedian of $\triangle MNL,$ thus $\angle LME=\angle OMN.$ Consequently, $\angle EMC=90^{\circ}-|\angle C-\angle B|,$ which is precisely the measure of the angle between $AO$ and $BC$ $\Longrightarrow$ $AO \parallel ME.$
_________________________________________________________________

[b]Lemma 2:[/b] $(M,M') , (N,N')$ and $(L,L')$ are three pairs of points on the sidelines $BC,$ $CA,$ $AB$ of $\triangle ABC,$ respectively. $D,E,F$ denote the midpoints of $MM',NN'$ and $LL'.$ Then centroid $G$ of $\triangle DEF$ is the midpoint of the segment connecting the centroids $U,U'$ of $\triangle MNL$ and $\triangle M'N'L'.$

[b]Proof:[/b] Let $\delta(P)$ denote the directed distance from point $P$ to the sideline $BC.$ Then

$3\delta(U)=\delta(N)+\delta(L) \ , \ 3\delta(U')=\delta(N')+\delta(L')$

$\Longrightarrow \delta(U)+\delta(U')=\frac{_1}{^3}[\delta(N)+\delta(N')+\delta(L)+\delta(L')]=\frac{_2}{^3}[\delta(E)+\delta(F)]=2\delta(G)$

By similar reasoning we deduce that $G$ is the midpoint of $\overline{UU'}.$
_________________________________________________________________

[b]Lemma 3:[/b] Mittenpunkt $M,$ Spieker point $S$ and Bevan point $B_e$ of any scalene $\triangle ABC$ are collinear.

[b]Proof:[/b] Let $(I_a),(I_b),(I_c)$ be the excircles of $\triangle ABC$ against vertices $A,B,C.$ Incircle $(I)$ and excircle $(I_a)$ are tangent to $BC$ through $X,Y$ and let $D,E,F$ be the midpoints of $BC,$ $CA,$ $AB.$ A-Gergonne ray $AX$ cuts $(I_a)$ at the antipode $Y'$ of $Y$ WRT $(I_a).$ Since $D$ is also the midpoint of $XY,$ it follows that $DI_a$ is the Y-midline of $\triangle XYY'$ $\Longrightarrow$ $DI_a \parallel AX,$ i.e. $DI_a$ is the complement of $AX.$  In other words, tangency points $X_1,X_2,X_3$ of the Spieker circle $(S)$ with $EF,$ $FD,$ $DE$ lie on $I_aD,$ $I_bE,$ $I_cF.$ On the other hand, perpendicular lines dropped from $I_a,I_b,I_c$ to $BC,CA,AB$ concur at the Bevan point $B_e,$ i.e. circumcenter of $\triangle I_aI_bI_c.$ From the parallel radii $B_eI_a \parallel SX_1,$ $B_eI_b \parallel SX_2$ and $B_eI_c \parallel SX_3,$ it follows that Mittenpunkt $M \equiv I_aD \cap I_bE \cap I_cF$ is the insimilicenter of $(S)$ and $\odot(I_aI_bI_c).$
_________________________________________________________________

Back to the problem, let $M_a,M_b,M_c$ be the midpoints of $BC,CA,AB$ and $D_a,D_b,D_c$ the midpoints of segments $G_bG_c,G_cG_a,G_aG_b.$ Since $\triangle GG_bG_c$ and the pedal triangle of $M_a$ WRT $\triangle ABC$ are homothetic, their G- and Ma- medians are parallel. From Lemma 1, we deduce then $GD_a \parallel AO \ (*)$ where $O$ is the circumcenter of $\triangle ABC.$

On the other hand, let $M_1$ be the midpoint of $H_bH_c.$ Hence $M_aM_1$ is the perpendicular bisector of $\overline{H_bH_c},$ thus

$\frac{H_aG_1}{G_1M_1}=\frac{AG}{GM_a}=\frac{H_aG_a}{G_aM_a}=2 \Longrightarrow G_aG_1 \parallel M_aM_1 \parallel AO$

Together with $(*),$ we have $ G_aG_1 \parallel D_aG.$

Similarly, $D_bG \parallel G_bG_1$ and $D_cG \parallel G_cG_1$ $\Longrightarrow$ $G_1$ is the anticomplement of $G$ with respect to $\triangle G_aG_bG_c,$ i.e. $G,G_1,G_2$ are collinear, such that $\overline{G_2G}:\overline{G_2G_1}=-1:2.$ Since $A',B',C'$ are the reflections of $H_a,H_b,H_c$ about $M_a,M_b,M_c,$ from Lemma 2 we deduce that centroids $G,G_1,G_3$ are collinear such that $\overline{GG_1}=-\overline{GG_3}.$

Circumcenter $O$ and symmedian point $K$ of $\triangle ABC$ become Bevan point and  Mittenpunkt of $\triangle H_aH_bH_c.$ From Lemma 3, it follows that Spieker point $S'$ of $\triangle H_aH_bH_c$ lies on $OK.$ Further, if $\triangle ABC$ is acute, then $H$ becomes the incenter of $\triangle H_aH_bH_c$ $\Longrightarrow$ $HG_1$ cuts Brocard axis $OK$ at $S'.$ Now, because of $\frac{_{HG}}{^{HO}}=\frac{_{HG_1}}{^{HS'}}=\frac{_{2}}{^{3}},$ we conclude that $GG_1$ is parallel to $OS' \equiv OK$ $\Longrightarrow$ Centroids $G,$ $G_1,$ $G_2,$ $G_3$ lie on the image $\ell$ of the Brocard axis $OK$ under the homothety $(H,\frac{_2}{^3}).$

Dear Mathlinkers,
a synthetic proof concerning this nice result can be seen in the article “Du  théorème  d'Ernesto  Cesàro à une  droite  parallèle  à  l'axe de  Brocard’’ that has been put on my website with some developpments.

http://perso.orange.fr/jl.ayme    vol. 6, p. 13, 20, 22, 23.

Sincerely
Jean-Louis


Hi all,  can anybody prove the chebychev's Inequality?  :(

Rearrangement can be used to show it, i believe. But i forgot what chebyshev explicitly is, so I can't give further details.

Let the indices be cyclic mod $n$ and $a_i\geq a_j$ for $i\geq j$ and $b_i\geq b_j$ for $i\geq j$.

Then

$n \sum a_ib_i\geq \sum a_ib_i+\sum a_ib_{i+1}+\sum a_ib_{i+2}+\ldots+\sum a_ib_{i+n-1}=\left(\sum a_i\right)\left(\sum b_i\right)$

where the middle step follows easily from rearrangment.

A way to prove Chebyshev's inequality without using the rearrangement inequality is to consider the sum
\[\sum_{i = 1}^n \sum_{j = 1}^n (a_i - a_j) (b_i - b_j).\]
Each term of the sum is nonnegative, and so the sum is too.  But when you expand each product and then sum, you will get Chebyshev's.

It's worth noting that that method is very similar to the method used to prove rearrangment.  In particular, say that $x_1\geq x_2$ and $y_1\geq y_2$.  Then

$(x_1-x_2)(y_1-y_2)\geq 0\Leftrightarrow x_1y_1+y_2x_2\geq x_1y_2+x_2y_1$.

This means that for any permutation $b_{\sigma(i)}$ we can 'smooth'

$\sum a_ib_{\sigma(i)}$

by taking terms pairwise and swapping so they are similarly sorted.  Since there are only finitely many pairs, this process must stop, and we eventually get the rearrangement inequality.

[b]blahblahblah wrote[/b]

[quote]Let the indices be cyclic mod n and $a_i\geq a_j for i\geq j$ and $b_i\geq b_j for i\geq j$.

Then

$n \sum a_ib_i\geq \sum a_ib_i+\sum a_ib_{i+1}+\sum a_ib_{i+2}+\ldots+\sum a_ib_{i+n-1}=\left(\sum a_i\right)\left(\sum b_i\right)$

where the middle step follows easily from rearrangment.[/quote]

did u mean $\sum a_ib_{i+1}  =  a_1b_2+a_2b_3+a_3b_4 ....a_{n-1}b_n$ ?

Yes, but the indices are cyclic mod $n$, so we actually get the final term $a_nb_{n+1}=a_nb_1$


I'm afraid I don't follow.  Is the addition being done in base 10 or in base 2?

[quote="cincodemayo5590"]Define $f(x) = \sum_{k=1}^{x}k_{2}$ for $x$ in base-10, where $k_{2}$ is the base-2 representation of the number $k$ (originally in base ten). Note that the output for $f(x)$ will be in base-2.

Is there an explicit formula for the ratio $\frac{ f(2x)}{f(x)}$, when converted back into base-10?[/quote]

adding numbers in a different number base does not change the sum...
the sum is just (1+2+...+2x)/(1+2+...+x)

maybe you mean something else...i think you mean something along the lines of: if you take the base 2 representations and add them as if they were base 10 numbers... so 8=1000_2, and you would be adding like 1000+1000=2000_{10}

Sorry! I did mean for the addition to be in base-10.

So, for example the ratio for $x=5$ would be
 $\frac{(1+10+11+100+101+...+1100)_{10}}{(1+10+11+100+101)_{10}}=\left( \frac{3455}{223}\right)_{10}$.

I hope this makes sense...  :P

Your numerator here is going up to 20, instead of up to 10 - which is it: 2x or 4x?

Oh sorry, an extra zero got in there somehow. I'll edit it now.


geometry
trigonometry
analytic geometry
graphing lines
slope
rectangle
ratio

There is a big right triangle ABC, A being the top angle, B being the bottom left angle,

and C being the bottom right angle, with sides 16 on the vertical leg, and 18 on the

horizontal leg.  B is the right angle, and AC is the hypotenuse.  From angle C to the

side opposite, draw line CD, with CD cutting AB into two different parts, 10 being the

top segment, 6 being the bottom segment.  Now, draw line AE from angle A.  E should

cut line BC into smaller parts, the left segment being 8, the right segment being 10.

The intersection of lines CD and AE is designated as F,  thereby cutting triangle ABC

into 4 parts.  One of these parts is a quadrilateral, BDFE.  Find the area of the

quadrilateral.

 :fool:  :fool:  :fool:

[quote="mgao"]...and C being the bottom right angle...B is the right angle, [/quote]
Bad wording but, here is my  method using [hide="trignometry"]
1)Using Pythag' AC= $2 \sqrt{145}$
2) draw a perpendicular bisector from F to AC. making each side;
$\sqrt{145}$
3) Angle DCB is $\tan {x} = \frac{6}{18} \rightarrow x \approx 18.43$
4) Angle ACB is $\tan {y} = \frac{16}{18} \rightarrow y \approx 41.63$
5) Subtract x from y to get Angle ACD = 23.
6) To find CF: $\cos {23} = \frac{\sqrt{145}}{CF} \rightarrow \text {CF} \approx 13.08$
7) Angle ABE: $\tan a = \frac{8}{16} \rightarrow a \approx 26.57$
8) Angle BAC $\tan b = \frac{18}{16} \rightarrow b \approx 48.37$
9) Angle EAC = b-a = 21.8
10) AF: $\cos 21.8 = \frac{\sqrt{145}}{AF} \rightarrow AF \approx 12.94$
11) We draw a line form D to E splitting the quad into two triangles:
      DE= 10 (Using Pythag. Th.)
12) $DF = CD - CF = 6 \sqrt{10} - 13.08 \approx 5.89$
13) $FE = AE - AF = 8 \sqrt{5} - 12.94 \approx 4.95$
14) Area of Triangle DBE is 24
15) Angle AFG = 68.2 , Angle CFG = 67 (By Angle Sum Theorem)
16) Angle DFE is 135.2 (Vert. Angles are congruent)
17) Triangle DFE ; $\frac{1}{2} a b \sin C = \frac{1}{2} \times 5.89 \times 4.95 \times \sin 135.2 = \approx 10.27$
18) $10.27 + 24 = 34.27$ [/hide]

maybe i overlooked something... but r u right in saying that the perpendicular from F to AC bisects AC... or... just divides it into two segments?

I approached this problem a little differently.

[hide]

Put the triangle on an XY coordinate system.

Set side $AB$ along the Y-axis and side $BC$ along the X-axis.

The next thing to do is to find the equation for $AE$ and $CD$ using the general equation $y = mx + b$ (point-slope equation for a straight line).

for line $AE$: slope $m = -2$

if $x = 0$ then $y = 16$
 $16 = (-2)(0) + b$ and $b = 16$

therefore the equation for $AE$ is $y = -2x + 16$

for line $CD$: slope $m = - \frac{1}{3}$

if $x = 0$ then $y = 6$ so
$6 = (- \frac{1}{3})(0) + b$ and $b = 6$

therefore the equation for $CD$ is $y = - \frac{1}{3}x + 6$

From these two equations, we can find the coordinates of point $F$.

By solving for $x$ and $y$ we find $F$ at the coordinates $(6, 4)$.

We can now divide the quadrilateral $BEFD$ into three figures: two right triangles and one rectangle.

By drawing a perpendicular from $F$ to $AB$ and $BE$ we form the triangles and the rectangle.

The area of the rectangle would be $(6)(4)$ or $24$

One of the right triangles would have legs of lengths of $6$ and $2$. The area of that triangle would be $6$.

The other right triangle would have legs of lengths $4$ and $2$. The area of that triangle would be $4$.

The total area of the quadrilateral $BEFD$ would be $24 + 6 + 4$ or $34$.

[/hide]

[hide]
by taking ritchjp's approach, you have the coordinates of all 4 vertices:
$\\ (0,0), (8,0), (6,4), (0,6)$
applying the shoelace theorem is a quick way to finish

[/hide]

Nice coordinate solution.

I feel like brute forcing this just for fun. :rotfl:

Here's a diagram: (I will explain how I got those lengths)
 [img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=738[/img]

We want to find the area of quadrilateral $BEPD$. Because I wasn't able to find any helpful similar triangles, we will find the lengths of $DP$, $PE$, and $BP$, and apply Heron's Formula twice: on $\triangle BPE$ and on $\triangle BPD$.

First, we draw $BF$ such that it is concurrent with the other two cevians. Now, we can effectively apply Mass Points. Some simple Mass Point principles yield $w_A=6$, $w_B=10$, and $w_C=8$.

Now, $w_F=w_B+w_C=18$, $w_D=w_A+w_B=16$, and $w_F=w_A+w_C=14$. Therefore, $AP: PE=3: 1$ and $CP: DP=2: 1:$. By the pythagorean theorem, $CD=6\sqrt{10}\implies PD=2\sqrt{10}, CP=4\sqrt{10}$ and $AE=8\sqrt{5}\implies AP=6\sqrt{5}, PE=2\sqrt{5}$.

Also by the Pythagorean Theorem, $2\sqrt{145}$, and by mass points, $AF: FC=4: 3$. Thus, $AF=\frac87\sqrt{145}$ and $CF=\frac67\sqrt{145}$. Our final use of mass points is to find that $BP: FP=7: 5$.

We now apply stewart's theorem on triangle $ABC$ to find $BF$. This states that $(AB^2)(CF)+(BC^2)(AF)=(BF^2)(AC)+(AF)(FC)(AC)$. Solving this mess on a calculator yields $BF\approx 12.36189$. (I stored exact value on calculator) Using the fact that $BP: FP=7: 5$, $BP\approx 7.21110$ (I stored this exact value on my calculator too)

Applying Heron's formula, $[\triangle BDP]=18$, and $[\triangle BPE]=16$. Therefore, the area of $BEFD$ is $16+18=34$.

There must be a better geometric solution to this. Oh well, brute forcing is an important skill too.  :P

[hide="Here is a extension. Try prove it !"] [color=darkred][b]Extension.[/b] Given are the triangle $ABC$ and the points: $M\in (AC)$, $\frac{MA}{MC}=m$; $N\in AB$, $\frac{NA}{NB}=n$. Denote the intersection $P\in BM\cap CN$ and the area $S\equiv [ABC]$ of the triangle $ABC$. Prove that\[ \boxed {\ [AMPN]=\frac{mn(m+n+2)}{(m+1)(n+1)(m+n+1)}\cdot S\ge \frac{4mn}{(m+n+1)(m+n+2)}\cdot S\ }  \] [b]Remark.[/b] If $m+n=2$ then $[AMPN]\ge \frac{mn}{3}\cdot S$ (a old and well-known problem).

[b]The our particular case.[/b] For $m=\frac{6}{10}=\frac 35$, $n=\frac{8}{10}=\frac 45$ and $S=\frac 12\cdot (6+10)\cdot (8+10)=144$ you obtain $[AMPN]=\frac{\frac 35\cdot \frac 45\cdot \left(\frac 75+2\right)}{\frac 85\cdot \frac 95\cdot \left(\frac 75+1\right)}\cdot 144=\frac{17}{72}\cdot 144=\boxed {\ 34\ }$
(the [b]4everwise[/b]'s or [b]Fireforce[/b]'s or [b]Ritchjp[/b]'s answer).

[b]Remark.[/b] If $AM=x\cdot AC$ and $AN=y\cdot AB$ then $\boxed {[AMPN]=\frac{xy(2-x-y)}{1-xy}\cdot S}\ \ (\ m: =\frac{x}{1-x}\ ,\ n: =\frac{y}{1-y}\ )\ .$[/color][/hide]

I would use mass points for this problem.
(I know 4everwise used mass points, but mine seems to be simpler :P )

[hide="My solution"]
Using mass points, we get $AF: FE=3: 1$ and $DF: FC=1: 2$.

[hide="masses"]
$B=10$, $A=6$, $C=8$
$D=16$, $E=18$, $F=24$
[/hide]

So $\triangle AFD=\frac13 \triangle ADC=\frac13 \left(\frac12\cdot 10\cdot 18\right) =30$

Hence $[BDFE]=\triangle ABC-\triangle AFD-\triangle AEC=144-30-80=\boxed{34}$
[/hide]
I believe we can use the same method (or Ceva's etc.) to prove the generalized formula, too.

[quote="frt"]I would use mass points for this problem.
(I know 4everwise used mass points, but mine seems to be simpler :P )

[hide="My solution"]
Using mass points, we get $AF: FE=3: 1$ and $DF: FC=1: 2$.

[hide="masses"]
$B=10$, $A=6$, $C=8$
$D=16$, $E=18$, $F=24$
[/hide]

So $\triangle AFD=\frac13 \triangle ADC=\frac13 \left(\frac12\cdot 10\cdot 18\right) =30$

Hence $[BDFE]=\triangle ABC-\triangle AFD-\triangle AEC=144-30-80=\boxed{34}$
[/hide][/quote]Lol, my solution to this problem was terrible. :rotfl:

:wallbash_red: :wallbash_red:

What are mass points?
How did you get the geometric diagrams?

[quote="mgao"]
How did you get the geometric diagrams?[/quote]

He drew it with his hand.

Mass points: http://mathcircle.berkeley.edu/mpgeo/mpgeo.html

Particularly useful when you're looking for ratios or lines are concurrent.

[quote="mgao"]How can you [i]draw[/i] something and enter it into the computer?
You'd better be joking.[/quote]I use geometer's sketchpad.

[quote="Virgil Nicula"][b]Extension.[/b] Given are the triangle $ABC$ and the points: $M\in (AC)$, $\frac{MA}{MC}=m$; $N\in AB$, $\frac{NA}{NB}=n$. Denote the intersection $P\in BM\cap CN$ and the area $S\equiv [ABC]$ of the triangle $ABC$. Prove that\[ \boxed {\ [AMPN]=\frac{mn(m+n+2)}{(m+1)(n+1)(m+n+1)}\cdot S\ge \frac{4mn}{(m+n+1)(m+n+2)}\cdot S\ }  \][/quote]Is this still open?
$\frac{mn(m+n+2)}{(m+1)(n+1)(m+n+1)}\cdot S\ge \frac{4mn}{(m+n+1)(m+n+2)}\cdot S$
$\frac{m+n+2}{(m+1)(n+1)}\ge \frac{4}{m+n+2}$
$m^2+2mn+4m+n^2+4n+4=4mn+4m+4n+4$
$(m-n)^2\ge 0$ $\blacksquare$

Let $MC=a, MA=b, NA=c. NB=d$
$m=\frac{b}{a}$, $n=\frac{c}{d}$
Use mass points to find $\frac{BP}{PM}=\frac{(a+b)d}{ac}$, $\frac{NP}{PC}=\frac{bd}{a(c+d)}$
$S=[AMPN]+[BMC]+[BPN]$
$[BMC]=S\left(\frac{a}{a+b}\right)=S\left(\frac{1}{m+1}\right)$
$[BPN]=S\left(\frac{d}{c+d}\right)\left(\frac{bd}{a(c+d)+bd}\right)=S\left(\frac{1}{n+1}\right)\left(\frac{m}{m+n+1}\right)$
$S=[AMPN]+S\left(\frac{1}{m+1}\right)+S\left(\frac{1}{n+1}\right)\left(\frac{m}{m+n+1}\right)$
$[AMPN]=\frac{2mn+m^2n+mn^2}{(m+1)(n+1)(m+n+1)}$
$[AMPN]=\frac{mn(m+n+2)}{(m+1)(n+1)(m+n+1)}$ $\blacksquare$

can u download the sketchpad?

if so, give link pleez.

oh ya, i used the bad english on purpose.  seems easier to do it in this type of wtng.

I dont believe u can get geometer scatchpad for free, but can recomend three free geom programs [color=red]C.A.R. [/color] mathsrv.ku-eichstaett.de/ MGF/homes/grothman/java/zirkel/
[color=green]wingeom[/color] math.exeter.edu/rparris/wingeom.html or
[color=cyan]KSEG[/color] http://www.mit.edu/~ibaran/kseg.html

I prefer C.A.R. but u can choose


hello

can some one explain this in details thank u

a- How many bit strings of length seven either begin with two 0’s or end with three 1’s?
	]
b- How many bit strings of length 10 contain either five consecutive 0’s or five consecutive

[hide="A"]If the string begins with two 0's, all we have to do is find how many strings of length 5 there are. Since each bit can only be a 0 or 1, there are $2^{5}=32$ strings that begin with two 0's.

There are $2^{4}=16$ strings that end in three 1's.

Since the problem says "either" instead of "and/or," we have to subtract the strings that both begin in two 0's and end in 3 1's. There are $2^{2}=4$ such strings.

$32+16-4=\boxed{44}$[/hide]

[hide="B"]5 consecutive bits can be located in $6$ different positions in a string of length 10. In each location, the other bits can be arranged in $2^{5}=32$ ways. The same goes for 5 consecutive 1's. Also, there are two strings in which both 5 consecutive 0's and 1's appear (0000011111 and 1111100000).

$32\cdot6\cdot2-2=\boxed{382}$[/hide]

For $A$, you have to subtract $8$ since if you subtract only $4$, you have subtracted $4$ for the twice counted strings.

You overcounted in $B$ aswell. For example you counted $0000000000$ 6 times.

thanx for replying

for question b)

i got 222
112+112-2=222 not 382
32+5*16=112
could check it a gain

thanx

for b i get

[hide]
110
[/hide]

Am I undercounting??

Oops...

[hide="A"]$\boxed{40}$[/hide]
B now seems a lot more complicated. :ninja:


If $ m$ is an odd number prove that equation $ t^{2}p^{2m}\plus{}2tp^{m}\plus{}p^{m}\plus{}1\equal{}x^{m}$ does not have an integer solutions.

What's about $ x \equal{} 1$, $ t \equal{} p \equal{} 0$ and any odd integer $ m$ ?


Slove the system of equations
$ (1\plus{}x)(1\plus{}x^2)(1\plus{}x^4)\equal{}1\plus{}y^7$
$ (1\plus{}y)(1\plus{}y^2)(1\plus{}y^4)\equal{}1\plus{}x^7$

I am lazy, so I won't write whole solution.
[hide="Partial solution"]
In cases 1-4 $ x\neq y$
1) $ xy\equal{}0$. As system is symmetric, $ x\equal{}0$, and $ y\equal{}0$
2) $ x>0$, $ y<0$. Just observe first equation- no solutions
3) $ x>0$, $ y>0$. As $ x\neq y$, we may assume that $ x>y$. Then LHS in first equation is greater than RHS in first equation- no solutions
4) $ x<0$, $ y<0$- I won't write it: [hide="Hint"] Multiply first equation with $ 1\minus{}x$ and second with $ 1\minus{}y$ [/hide].
5) $ x\equal{}y$- easy, solution $ x\equal{}y\equal{}\minus{}1$
[/hide]

[quote=chien than]Slove the system of equations $$ \left\{\begin{matrix}
 (1+x)(1+x^2)(1+x^4)=1+y^7\\
\\(1+y)(1+y^2)(1+y^4)=1+x^7
\end{matrix}\right.$$
[/quote]
Any one?


quadratics
number theory unsolved
number theory

[quote="tim1234133"]Show that either there are infinitely many composite numbers of the form $2^{2^{n}}+1$ or $6^{2^{n}}+1$ (or both).[/quote]

One open question is whether or not there are infinitely many composite Fermat primes. Now it seems to me that you cannot show that, however, you can say the following: if there is only a finite amount of Fermat composites, then it must be the case that there are infinitly many of the form $6^{2^{n}}+1$. Am I having the right reasoning?

Kevin's hint was that this should be possible using some properties of quadratic residues.

Then again, he wears trousers that look like a chessboard :D.

Actually tbh, I might buy a pair myself.

Have you solved it? (If so, is it really easy, because all the other questions on that side of the sheet were pretty easy, and even with his hint I can't seem to be able to do it...)

I am pretty sure it's not really easy, but probably really short. I am yet to find a solution.

Ummm ok let's try this... assume not, then there exists N such that for any x > N, both $2^{2^{x}}+1$ and $6^{2^{x}}+1$ are prime. For m > n > N, $6^{2^{m}}+1 \equiv 3^{2^{m}}+1 \mod 2^{2^{n}}+1$. Now if in some case 3 is not a quadratic residue mod $2^{2^{n}}+1$, then taking $m = 2^{n-1}$ gives us $6^{2^{m}}+1 \equiv 0 \mod 2^{2^{n}}+1$, a contradiction. However, 3 is never a quadratic residue mod $2^{2^{n}}+1$ by quadratic reciprocity, since $\left( \frac{2^{2^{n}}+1}{3}\right) =-1$, and $\left( \frac{2^{2^{n}}+1}{3}\right) \left( \frac{3}{2^{2^{n}}+1}\right) = 1$.


If $ m$ is the lebesgue measure on $ \mathbb{R}$ and $ \mu$ is a Borel measure, is it that true that if for all closed sets $ E$ with $ m(E) \equal{} 0$ we have $ \mu(E) \equal{} 0$, then $ \mu$ is absolutely continuous w.r.t. to $ m$?

Are we guaranteed to have $ |\mu|(E) \equal{} \sup \{|\mu|(F) | F \subset E \text{ compact}\}$?

I am confused by the question. Does the sigma-algebra on which mu is defined necessarily contains all the Lebesgue measurable sets of measure zero, for some reason I don't see?

Otherwise, how can we compare these measures? Or is it implicit that we are to compare them relative to the Borel sigma-algebra?

Yes, we are viewing the Lebesgue measure on the Borel sets. But also, we could always complete $ \mu$ to be defined on the Lebesgue sets and it doesn't make any difference for this question.

I seem to remember that all Borel measures on the real line are inner regular, but I can't find such a result in any book I have. I also think all Borel measures are metric outer, but I don't have such a result either. I don't really care about the proofs, I just want to know if it's right

I would like to add to the conditions that $ \mu$ is actually a complex measure, so  $ |\mu|(\mathbb{R}) < \infty$.

I'll think about it after my exam tomorrow!  :P

[quote="Kalle"]
I seem to remember that all Borel measures on the real line are inner regular, but I can't find such a result in any book I have. .[/quote]
All (sigma) finite measures certainly are. As to arbitrary measures, you can try something crazy like defining your measure to be $ 0$ if you can cover your set by countably many compact sets of Lebesgue measure $ 0$ and $ \plus{}\infty$ otherwise. Here you are clearly in trouble because there are sets of Lebesgue measure $ 0$ and of second category.


A rectangle has a perimeter of 64 inches and each side has an integer length. How many non-congruent rectangles meet these criteria?

Let x be the length, y be the width.

$ 2x\plus{}2y\equal{}64$
$ x\plus{}y\equal{}32$

so x can be anywhere from 31 to 1, inclusive. so there are $ \boxed{31}$ rectangles.

$ x\equal{}5,y\equal{}27$ and $ x\equal{}27,y\equal{}5$ are the same rectangles. That means $ x$ can only range from $ 1$ to $ 15,$ thus giving $ \boxed{15}$ rectangles.

Can't you have x=16 and y=16. That counts as one also right? So there are 16.


Solve this problem

Let m be a positive integer.
 Prove that if \[ {\rm 2}^{{\rm m + 1}} + 1 \] divides\[ 3^{2^m } + 1 \], then \[ {\rm 2}^{{\rm m + 1}} + 1 \] is prime.


Can someone telll what exactly is teh chinese remainder theorem with the help of a simple problem?(Don't gimme the link to mathworld)

O.K,I found a good link for this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=65217]go here[/url]

There was this example of chinese remainder theorem  in ACOPS.......
$x\equiv3(mod11)$
$x\equiv5(mod6)$
Find $x$

What they did....

(11,6)=1
so $6z+11y=1$
z=2,y=-1,i.e they found values satisfying the smallest linear combination.
Then they directly did....

$(5)(11)(-1)+(3)(6)(2)=-19$---------$(1)$
So,
$x\equiv-19(mod66)$------$(2)$
$x=47$

Now my doubt is that why did they do $(1)$ and $(2)$?What exactly do they mean?

Please someone clear this doubt up for me bcos I don't want to mug this method,I want to understand it.This seems like a fast way to do such problems,so  if  some  kind  person helps me ,I'tll be of great help. :)


Color $ \{ 1, 2, ... 2006 \}$ with three colors.  Show that there exist $ x, y$ with the same color such that $ |x \minus{} y|$ is a square.

Well, $ 50 < x \in \mathcal{A}, x \plus{} 9 \in \mathcal{B} \Longrightarrow x \minus{} 16 \in \mathcal{C} \Longrightarrow x \minus{} 7 \in \mathcal{A} \Longrightarrow x \minus{} 7 \cdot 7 \in \mathcal{A}$?


geometry
analytic geometry
ratio
calculus
geometry solved

AD and BE are two cevians of point X inside triangle ABC.
    Maximize the area of DXE.

    Maj. Pestich

Very good question! I fear, however, that there is no elegant solution. Still, there is an elementary one:

We denote by |PQR| the area of an arbitrary triangle PQR, and we call M the midpoint of the side AB of triangle ABC. Then, I claim that

$\frac{\left|DXE\right|}{\left|ABC\right|}\leq\frac{5\sqrt5-11}{2}$,

and equality holds if and only if the point X has the homogeneous barycentric coordinates $\left( \frac{\sqrt{5}+1}{2}:\frac{\sqrt{5}+1}{2}:1\right)$ with respect to triangle ABC (this is actually the point lying on the median CM of triangle ABC and dividing it in the ratio $\frac{CX}{XM}=\sqrt5+1$).

[i]Proof.[/i] Let X ( x : y : z ) be the homogeneous barycentric coordinates of the point X with respect to triangle ABC. Then, the cevian feet D and E of the point X have the homogeneous barycentric coordinates D ( 0 : y : z ) and E ( x : 0 : z ), respectively. Transforming into normalized barycentric coordinates, and applying Proposition 1 of [1], paragraph 1, we get

$\frac{\left|DXE\right|}{\left|ABC\right|}=\det\left(\begin{array}{ccc}\frac{x}{x+y+z}&\frac{y}{x+y+z}&\frac{z}{x+y+z}\\0&\frac{y}{y+z}&\frac{z}{y+z}\\\frac{x}{z+x}&0&\frac{z}{z+x}\end{array}\right)=\frac{xyz}{\left(x+y+z\right)\left(y+z\right)\left(z+x\right)}$.

Now, after some easy algebra ;) , we have

$\frac{\left( x+y+z\right) \left( y+z\right) \left( z+x\right)}{xyz}=\frac{5\sqrt{5}+11}{2}+\frac{8z\left( x+y+z\right) ^{2}\left( x-y\right)^{2}+xy\left( 2z+\left( 3+\sqrt{5}\right) \left( x+y\right) \right) \left(4z+\left( 1-\sqrt{5}\right) \left( x+y\right) \right) ^{2}}{8xyz\left(x+y\right) ^{2}}$.

Thus,

$\frac{\left( x+y+z\right) \left( y+z\right) \left( z+x\right)}{xyz}\geq\frac{5\sqrt{5}+11}{2}$.

Taking the reciprocal, we get

$\frac{xyz}{\left(x+y+z\right)\left(y+z\right)\left(z+x\right)}\leq\frac{2}{5\sqrt5+11}=\frac{5\sqrt5-11}{2}$.

Thus,

$\frac{\left|DXE\right|}{\left|ABC\right|}=\frac{xyz}{\left(x+y+z\right)\left(y+z\right)\left(z+x\right)}\leq\frac{5\sqrt5-11}{2}$.

Proof complete.

Actually, the crucial formula of the proof,

$\frac{\left( x+y+z\right) \left( y+z\right) \left( z+x\right)}{xyz}=\frac{5\sqrt{5}+11}{2}+\frac{8z\left( x+y+z\right) ^{2}\left( x-y\right)^{2}+xy\left( 2z+\left( 3+\sqrt{5}\right) \left( x+y\right) \right) \left(4z+\left( 1-\sqrt{5}\right) \left( x+y\right) \right) ^{2}}{8xyz\left(x+y\right) ^{2}}$,

was found using mixing variables and one-variable calculus; the proof is very ugly, but I don't think there is a better one, since the number $\frac{5\sqrt5-11}{2}$ doesn't seem to have any spectacular geometric sense.

[b]References[/b]

[b][1][/b] [url=http://www.math.fau.edu/yiu/barycentric.pdf]Paul Yiu, [i]The uses of homogeneous barycentric coordinates in plane euclidean geometry[/i][/url].

  Darij


Hello,
  The Music season is on in full swing here in Chennai. Being an ardent rasika, I would like to make a few observations.It has been a long time since I heard a Ragam Tanam Pallavi or a detailed exposition of Kambhoji,Thodi,Kalyani,Kharaharapriya and such evergreen ragas.Due to the springing up of innumerable sabhas,the quality has undoubtedly suffered. Artistes have started experimenting with even Papanasam Sivan's compositions as the main piece. Kutcheris are reduced to two hour exercises with brisk swaras, short alapanas and neravals are done just because they have to be done.
And a sad development is that there are certain leading singers who rely on non-Carnatic pieces like Abhangs,Bhajans and so on to draw crowds. Have they forgotten that the Trinity's compositions themselves require more than a lifetime to master? Can they be called kutcheris at all? Many so-called rasikas claim that they are going to such kutcheris when they attend such pseudo-kutcheris.Huge crowds mean nothing. Carnatic music must not be tailored for the listeners. Were there not crowds for Semmangudi, Ariyakudi, Madurai Mani Iyer and other greats? I get to hear classical pieces like Enduku Peddala, Rama Nee Samana, Sri Subhramanyaya Namaste and so on only in kutcheris by veterans like Lalgudi, TN Krishnan, Madurai TN Seshagopalan, MSG and others. Where are the contemporary musicians?Why are they straying like this?Time will tell.
 My blog http://sathej-musings.blogspot.com/ contains some reviews.
Sathej

yeah that is what i meant but i didn't understand if it because i live somewhere else :maybe:

Hello,
    Well, the discussion is about Classical South Indian Music called Carnatic Music. There is a 'Kutcheri' (concert) season during December every year in Chennai, India. It is extremely popular as many people from abroad too visit Chennai for these kutcheris. Hence, the assumption was made that many people in the Forum here would be aware. Anyway sorry for just going on about things which were oblivious to most. I thought Indians might it find interesting and so might people from other countries who follow Carnatic Music.
Sathej

Oh no it is not a problem...it always help to learn something-about absolutely anything! :lol:


Michigan Math Prize Competition Part II 2001.

I am thinking of a number between 0 and 15. You get to ask me
seven yes-or-no questions, and I am allowed to lie at most once in answering your questions. What seven questions can you ask that will always allow you to determine the number? Note: You need to come up with seven questions that are independent
of the answers that are received. (You are not allowed to say, "If the answer to question 1 is yes, then question 2 is XXX; but if the answer to question 1 is no, then question 2 is YYY.")

does this include 0 and 15?  and if so, is 0 considered composite?

The set includes 0 and 15... As for if 0 is composite, I'm not quite sure.

Think about what a composite number means, then note that zero has technically infintely many factors....

err

im not sure either...
hmm

[color=cyan]There is absolutely no reason it could matter if 0 is composite.  Any question can merely be phrased, "Is is a member of the set {_____}?" so that, if you wanted to ask "Is it composite?", you could just ask "Is it 4, 6, 8, 9, 10, 12, 14, or 15?" or include 0 at your own whim.

This is very much like the project I did for my Intel paper this (last)year.  In that case, it was a dynamic situtation, so that you could pick your question after the previous one has been answered.   The dynamic version is known as "Ulam's Game," and a lot of work on it and related problems has been done by Joel Spencer, a mathematician at NYU.  I could talk a lot more about it, if anyone cares to listen.[/color]

seems like a question solved by force, too bad I don't have the time... hope someone gets this question by tonight. I am eager to see the answer.

[color=cyan]Hmm . . . lets say that anyone who is interested in hearing my mini-lecture series on my Intel project private message me, so I don't post something quite that long on a forum where it doesn't really belong and distract from the question being asked.

Hehe.  Mini-lecture series.  I crack myself up.  Shut up, Joel.[/color]

this is not a brute force question and has no complicated solution. mmpc part 2 is 100 minutes long and has 5 questions which are rarely take longer than 10 min. the questions on this exam often have quite stupid answers. im guessing this one youre supposed to ask about the number in base 2 - about the digits (it will have at most 4 digits, the person can lie once, so you can ask each question but one twice and therefore find the number). this is all i can come up with right now. someone tell me if it makes any sense at all.

[color=cyan]I thought about that . . . but the fact that you can't ask one of the questions twice seems problematic to me.  I was thinking maybe one question with 8 numbers, 2 with 4, and 4 with 2?  [/color]

Could I ask if the number was prime or composite?  Maybe thats kind of dependent...

What if we ask about whether each digit in base 2 is 1 and then ask about the sum of the 8s, 4s, and 2s (mod 2), the sum of the 8s, 2s, and 1s (mod 2), and then sum of the 4s and the 1s (mod 2) or something like that? Might that work? I think that's a fairly standard theme in these cryptological problems.

[color=cyan]But think about it -- how would that matter?  We could just as well be playing with the letters A to P, for which primeness would be rather difficult to define, and the game would still be equally as winable.  The specific names we give the numbers, in this case 0 to 15, and the properties of those names (whether being a composite or a vowel) don't affect whether or not you can win.[/color]

[color=cyan]As for what Simon said, asking about the interactions could be interesting . . . I would exhaust the simpler cases first, however.[/color]

If we do the thing Simon says.. (lol) Where can we catch the lie?  Are we supposed to use the facts from the last questions we asked?

The point is that with my method, we're essentially asking about each bit three times, and so we can catch a lie if one of the statements about a bit doesn't agree with the other two.

well, if the guy had lied, there'll obviously be a glitch in the arithmetic somewhere, since we asked him whether there was a 1 in each of the four places. Since we ask him about each digit in the binary representation thrice, we change whatever doesn't agree with the other two.

Actually, I'm not quite sure what you mean by the sum of the 8's, 4's, and 2's, etc. Do you mean the sum of the numbers in the 1st, 2nd and 3rd places of the binary representation, or something different?

Wow, you guys post fast. I'm sorry,  I didn't see your post when I replied.

By 8s digit, I mean the digit to the left, etc. (That's similar to the 1000s digit in base 10, but we're working in base 2, so it's the 8s digits.)

[color=cyan]Actually, I misunderstood what Simon said.  Simon, your method isn't fundamentally different from the other method, is it?  At least, not the first half.  Each of the first 4 questions could be stated as just a set of 8 numbers.  I'm not sure if you can get enough information asking about such large groups.

Also, what exactly would your question about the sum of digits be?  Because you only get 2 answer choices.  So each answer might be too general if it's about the sum of 3 numbers.

On a totally unrelated note, are either of you two from Michigan Ethan Street?[/color]

Well, you already know the digits (with a possible lying error), so you can ask him whether the sum is what they sum up to when you add the digits he gave you. For example, let's take 13. Binary representation is 1101. Let's assume he lied on the first half, for specificity on the 2nd digit. Okay, now we think the number is 1001. We ask him the whether the sum of the 8's 4's and 2's is a 1, which is what we think it should be. He can't lie anymore, so he says no. We get suspicious.(jk). Going on, we ask him if the sum of the 8's 2's and 1's is 2. He says yes. We ask him if the sum of the 4's and 1's is a 1. and he says no. We have two discrepancies, so he must have lied in the first part. (if he lied in the second part, we would only have 1 discrepancy.) We thus fix the number that has both the discrepancies, which is the 2nd digit. We thus get the correct number (since there are only 2 possibilities for each number).

Can you ask nonmathematical question? If so, you could ask "Are you going to answer 'no' to this question?" as your first question...I'll shut up now. :)

[color=cyan]Remember though, this is not a dynamic problem: you have to have your questions picked out in advance, [i]before[/i] he starts answering them.[/color]

Why nonmathmatical questions tare? like what?

Well the answer to my question ("Are you going to answer 'no' to this question?") would always be a lie, so it would make the problem a lot easier. (You'll have six questions but the answers would all be the truth)

:lol: That is a good question to force someone to lie!  Did you get this from Smullyan?  He always says wierd things like that.  I bet you probably could, and maybe the people would give BONUS points for that clever idea....

YES! Finally I say something smart! (Compared to average AoPSers, I'm a "stupid dum-dum head" (copyrighted Cartoon Network))
And I thought up of it myself! Boy am I proud! (of myself :) )

Ok! I think we have this one solved... Time for a new problem.

[color=cyan]Wait -- no one has actually posted a set of 7 questions that could necessarily narrow down 16 numbers to just 1.  It's safe to assume that Tare's idea, while it might make the job easy, is not the solution the creators of the problem had in mind.  Either way, the problem isn't solved until anyone posts those questions.  I'm still waiting.[/color]

And or complexzetas.. that seems to work perfectly fine for me..

[color=cyan]MysticTerminator's answer, as I said before, only demonstratably works in a dynamic situation: he picked his last 3 questions only after the first 4 had been answered.  Thus, it does not satisfy the question.  ComplexZeta never outlined a set of 7 questions -- he gave a basic idea of what the questions might be like, but he did not list seven questions.  This question won't be solved until someone lists 7 questions, and then demostrates that no matter what, those questions will be able to figure out what the number is or where the lie takes place.  No one has done that, as far as I can see.[/color]

Isn't the solution...

What is the sum of the first two digits.
What is the sum of the second and third digit.
What is the sum of the third and fourth digit.
What is the sum of the second and fourth digit.
What is the sum of the first and third digit.
What is the sum of the first three digits.
What is the sum of the last three digits.

that's not yes/no

sigh
I keep forgetting it can't be dynamic.
sigh

AHH? Crap! I forgot that it had to be yes or no...  :evil:  :evil:  :evil:  :evil:

I was thinking, and I discovered that my original solution was not too astray... SPOILER

[hide]Binary only consists of two digits, 0 and 1.  We can just change the questions to ask for single digits.  Then, we can ask if it is one.  Once we have that, we can go on to ask about the sums...  When we have the sum of three numbers, we can ask the question, "Is it odd?"  That will tell us whether the number is three and one or two and zero.  From here, we can perform use system of equations to uncover his lies.

I checked the answer and here it is...

1. Is p equal to 1?

2. Is q equal to 1?

3. Is r equal to 1?

4. Is s equal to 1?

5. Is p + q + r odd?

6. Is p + q + s odd?

7. Is q + r + s odd?[/hide]


Hello,

I finished writing another Mock AMC couple days ago. I wonder if anyone is interested in taking it. If so, please post here so I can schedule the time that will be okay for most people.

I will take it. Thank you for doing this.

i'd like to take it too. thanks a lot, silverfalcon :D

I'm also interested in taking it.  I'd offer some times when I'm available (I assume you wanted us to do this) but my schedule is confusing and constantly changing, so I'll just attempt to go with whatever date and time you choose.

I'm thinking this Saturday at 6:00 PM. How does this sound? I can't do much on weekdays starting next week since I'll be having double soccer practices (2 hour morning and another 2 hour evening so total 4 hours every mon-fri). So, I want to get this done before next week. Is Saturday 6:00 PM (EST) okay?

Actually, that time [i]should[/i] work very well for me.  Coincidentally, I have morning/afternoon soccer practices coming up on the weekdays as well, so a weekend AMC sounds great to me. :)

sure...
i'll probably fail..
why can't it be AMC 10? :P

Saturday at 6 (3 PM on the west coast  :P ) sounds fine!

he is not even ONLINE. which to me it implies he might have forgotten about it. ;)

i came all the way to the library for this... :mad:

*shrugs*  Give him a few minutes, maybe something's holding him up (or his clock at home could be offset from ours).

i thought he would be posting them unless he worked on the test with someone else.

[quote="no_one_2000"]Did they do that on the last one?  I don't remember that. lol[/quote]

Yea, they've did that on most of the Mock AMC's I've done, including the last one.

No, he is working alone on this one I believe. Maybe he has been held back and is requesting someone else to post it???

i sacrifice my lunch for this...
does he said anything in this thread before???? :huh:  :huh:  :huh:

You know, Silverfalcon never actually confirmed that he would be posting it today, did he?  He just asked if it was ok...maybe he changed his mind?

Guys lets just be patient.  :|

No he goes be EST. He lives in my state.

Hm i'm not familair with the PM system. Does it stay in outbox till the recipient comes online?

I'll apologize for not posting the problems at current time.

I came 5 minutes late because I was working on AIME problem  :P .

Hahaha, I just love how the worrying and complaining began right at 6:00 P.M. No room for error guys!


[quote="orangedude654"]limit as x approaches infinity of   ((x^3)-x+1)/((2x^2)-(2x^3))[/list][/code][/quote]

$\lim_{x\to \infty}\frac{x^{3}-x+1}{2x^{2}-2x^{3}}$

Divide numerator and denominator by $x^{3}$ to get,

$\lim_{x\to \infty}\frac{1-\frac{1}{x^{2}}+\frac{1}{x^{3}}}{\frac{2}{x}-2}= \frac{1-0+0}{0-2}$


Bisectors of angles A,B and C of a triangle ABC intersect its circumcircle at D,E and F respectively.Prove that the angles of the triangle DEF are $ 90^o \minus{} \frac {1}{2}A$, $ 90^o \minus{} \frac {1}{2}B$ and $ 90^o \minus{} \frac {1}{2}C$.

[hide]
I'll prove for angle D. You can try for the other angles.

$ \widehat{FDE}\equal{}\widehat{FDA}\plus{}\widehat{ADE}\equal{}\widehat{FCA}\plus{}\widehat{ABE}\equal{}\frac{C}{2}\plus{}\frac{B}{2}\equal{}\frac{180\minus{}A}{2}\equal{}90\minus{}\frac{A}{2}$
[/hide]


geometry
parallelogram
geometry proposed

Let $ABC$ and $XYZ$ be two triangles. Define
\[A_1=BC\cap ZX, A_2=BC\cap XY,\]\[B_1=CA\cap XY, B_2=CA\cap YZ,\]\[C_1=AB\cap YZ, C_2=AB\cap ZX.\]
Hereby, the abbreviation $g\cap h$ means the point of intersection of two lines $g$ and $h$.

Prove that $\frac{C_1C_2}{AB}=\frac{A_1A_2}{BC}=\frac{B_1B_2}{CA}$ holds if and only if $\frac{A_1C_2}{XZ}=\frac{C_1B_2}{ZY}=\frac{B_1A_2}{YX}$.

Okay, I will prove that $\frac{A_1A_2}{BC} = \frac{B_1B_2}{CA} = \frac{C_1C_2}{AB}$ implies $\frac{C_1B_2}{YZ} = \frac{A_1C_2}{ZX} = \frac{B_1A_2}{XY}$. Then, the converse follows by analogy.

Let P be a point such that the quadrilateral $C_2A_1A_2P$ is a parallelogram. Then, $C_2P = A_1A_2$ and $PA_2 = C_2A_1$ with directed segments (segments on parallel lines are directed equally). Moreover, $C_2P \parallel A_1A_2$ and $PA_2 \parallel C_2A_1$; in other words, $C_2P \parallel BC$ and $PA_2 \parallel ZX$.

Now, $\frac{C_2P}{BC}=\frac{A_1A_2}{BC}=\frac{C_1C_2}{AB}=\frac{C_2C_1}{BA}$; on the other hand, $C_2P \parallel BC$ and $C_2C_1 \parallel BA$. This entails that triangles $C_2PC_1$ and BCA are homothetic, so that $PC_1 \parallel CA$, and $\frac{PC_1}{CA}=\frac{C_1C_2}{AB}$, i. e. $\frac{PC_1}{CA}=\frac{B_1B_2}{CA}$, and $PC_1 = B_1B_2$. From $PC_1 \parallel CA$, we have also $PC_1 \parallel B_1B_2$.

Hence, the quadrilateral $PB_1B_2C_1$ is a parallelogram, and we get $PB_1 \parallel C_1B_2$ and $PB_1 = C_1B_2$. We rewrite $PB_1 \parallel C_1B_2$ as $PB_1 \parallel ZY$. On the other hand, $PA_2 \parallel ZX$, as we have seen before. Furthermore, it is obvious that $B_1A_2 \parallel YX$. Thus, the triangles $PB_1A_2$ and ZYX are homothetic, and we get $\frac{PB_1}{ZY}=\frac{B_1A_2}{YX}=\frac{PA_2}{ZX}$, or, in other words, $\frac{C_1B_2}{ZY}=\frac{B_1A_2}{YX}=\frac{C_2A_1}{ZX}$. Rewriting this, we get $\frac{C_1B_2}{YZ}=\frac{A_1C_2}{ZX}=\frac{B_1A_2}{XY}$.

This proof is not really new, it is just the generalization of a similar proof for a problem proposed by Spain for the IMO 1985 long/shortlist:

[quote]Let $A_1A_2$, $B_1B_2$, $C_1C_2$ be three equal segments on the sides of an equilateral triangle. Prove that in the triangle formed by the lines $B_2C_1$, $C_2A_1$, $A_2B_1$, the segments $B_2C_1$, $C_2A_1$, $A_2B_1$ are proportional to the sides in which they are contained.[/quote]

Of course, this problem is not only a special case of sam-n's result, but it is also equivalent to it by affine transformations.

  Darij


On the replies i've seen of this kind of topic, people normally say study and practice old tests. What years should we study and are there any links to sites that have the tests? What years have the highest state dificulty?

Also how many hours would you normally study a day to get to national?

as for difficulty i thought 2007 was hard and i heard other people also say it was difficult.

hm.  try looking here.  this topic has been posted a gazillion times and if only people would just scroll up and look!  :D

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133189[/url]

does anyone where to get state tests  b4  2000

I don't think they have any on the internet but you should ask your coach.

[quote="xpmath"]Yeah, seeing as we have a while, you should do all of them. Doing 1 a day isn't that hard.
It's not about how long you do, it's about how you do it. If you live in a moderately easy state, I think doing a state a day and reviewing questions you get wrong is pretty good.[/quote]

well the problem with that, is most of us only have what is onthe net (2001-2007)

so we either have to get more, or do less tests.


you mean x,y,z in the first condition, right?


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a group of order $ p^{a}m$, where $ p$ is a prime that doesn't divide $ m$. Assume that $ P$ is a subgroup of $ G$ of order $ p^a$ and $ N$ is a normal subgroup of $ G$ of order $ p^{b}n$, where $ p$ doesn't divide $ n$. Prove that $ | P \cap N | \equal{} p^b$ and $ | PN / N | \equal{} p^{a\minus{}b}$.

Clearly, due to the formula $ |PN| \equal{} \frac{|P\parallel{}N|}{|P \cap N|}$, I only need to prove one of these equalities. Any ideas?

We know that $ \left|PN\right| \equal{} \frac {\left|P\right|\cdot\left|N\right|}{\left|P\cap N\right|}$ because $ N$ is normal in $ G$. But $ \left|PN\right|\mid\left|G\right|$, since $ PN$ is a subgroup of $ G$. Thus, $ \frac {\left|P\right|\cdot\left|N\right|}{\left|P\cap N\right|}\mid\left|G\right|$, so that $ \left|P\right|\cdot\left|N\right|\mid\left|P\cap N\right|\cdot\left|G\right|$. Since $ \left|P\right| \equal{} p^a$, $ \left|N\right| \equal{} p^bn$ and $ \left|G\right| \equal{} p^am$, this becomes $ p^a\cdot p^bn\mid\left|P\cap N\right|\cdot p^am$, what simplifies to $ p^bn\mid\left|P\cap N\right|\cdot m$. Hence, $ p^b\mid\left|P\cap N\right|\cdot m$. Since $ m$ is coprime to $ p$ (because $ p$ is prime, and $ m$ is not divisible by $ p$), this yields $ p^b\mid\left|P\cap N\right|$.

On the other hand, $ \left|N\right|\mid\left|G\right|$ (since $ N$ is a subgroup of $ G$), so that $ p^bn\mid p^am$, so that $ p^b\mid p^am$, and thus $ p^b\mid p^a$ (since $ m$ is coprime to $ p$). Thus, $ b\leq a$.

Obviously, $ P\cap N$ is a subgroup of $ P$ and therefore satisfies $ \left|P\cap N\right|\mid\left|P\right| \equal{} p^a$. Also, $ P\cap N$ is a subgroup of $ N$ and therefore satisfies $ \left|P\cap N\right|\mid\left|N\right| \equal{} p^bn$.

The relations $ \left|P\cap N\right|\mid p^a$ and $ \left|P\cap N\right|\mid p^bn$ yield $ \left|P\cap N\right|\mid\gcd\left(p^a,p^bn\right) \equal{} p^b$ (in fact, $ \gcd\left(p^a,p^bn\right) \equal{} p^b$ because $ b\leq a$ and $ n$ is not divisible by the prime $ p$). Hence, altogether, $ \left|P\cap N\right|\mid p^b$ and $ p^b\mid\left|P\cap N\right|$. Thus, $ \left|P\cap N\right| \equal{} p^b$. Consequently, $ \left|PN\slash N\right| \equal{} \left|PN\right|\slash\left|N\right| \equal{} \frac {\left|P\right|\cdot\left|N\right|}{\left|P\cap N\right|}\slash\left|N\right| \equal{} \frac {\left|P\right|}{\left|P\cap N\right|} \equal{} \frac {p^a}{p^b} \equal{} p^{a \minus{} b}$.

[b]EDIT:[/b] This problem was also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=179918 .

  darij

A shorter proof:

Take a $ p$-sylow subgroup $ Q'$ of $ N$. It must be a subgroup of some $ p$-sylow subgroup $ P'$ of $ G$. However, all $ p$-sylow subgroups are conjugate. Thus $ P \equal{} xP'x^{ \minus{} 1}$ for some $ x$. Since $ N$ is normal, $ Q \equal{} xQ'x^{ \minus{} 1}$ is also a $ p$-sylow subgroup of $ N$, and is contained in $ P$. Thus

$ Q \cap N \subset P \cap N$

But since $ Q$ is a $ p$-sylow of $ N$, $ |Q \cap N| \equal{} p^b$. Thus $ |P \cap N| \geq p^b$. On the other hand, $ |P \cap N| \leq p^b$ since $ P \cap N$ is a $ p$-group contained in $ N$. This establishes the desired result.


calculus
integration
function
trigonometry
logarithms
calculus computations

Find the differentiable function $f(x)$ such that $f(x)=-\int_{0}^{x}f(t)\tan t\ dt+\int_{0}^{x}\tan (t-x)\ dt\ \left(|x|<\frac{\pi}{2}\right).$

[hide="i think this is right"]
first note that $\int_{0}^{x}\tan(t-x)dt=\int_{0}^{x}\tan(t)dt$
so we now have
$f(x)=-\int_{0}^{x}(f(t)\tan(t)+\tan(t))dt$
differentiating both sides leads us to
$f'(x)=-f(x)\tan(x)+\tan(x)$ we simplify this to

$\frac{f'(x)}{1-f(x)}=\tan(x)$ now we integrate and have

$-\ln\left(1-f(x)\right)=\ln\left(\sec(x)\right)+C$

which leads to

$1-f(x)=A\sec(x)$ or

$f(x)=1-A\sec(x)$ where $A=e^{C}$
[/hide]

maokid, on the first line of your solution, the RHS should be $-\int_{0}^{x}\tan t \,dt.$ The rest looks fine.

thanks for pointing that out [b]FieryHydra[/b] and [b]blahblahblah[/b]

using what you two mentions i got that the
[hide="answer is"]
$f(x)=\cos(x)-1$
[/hide]