Prove that there is no function $ f : (0, \infty) \to (0, \infty)$ such that $f(x+y) \geq f(x) + yf(f(x))$ for all $x, y \in (0, \infty)$

This has already been discussed several times. Clearly the function is increasing. Take $x=1$. You'll find that $f$ goes to infinity if $x$ goes to infinity. Next, take $y=1$ and observe that $ f(x+1)-f(x)$ goes also to infinity for large $x$. With the help of Cesaro-Stolz theorem (or directly) show that this implies $\frac{f(x)}{x}$ foes to infinity. Show then that for large $x$ you have $ f(x+1)>f(f(x))>f(x+1)$.


Thank you:

1.	Find the mean, median and the mode of the following data set:

5 15 9 22 67 42 2 72 81 53 6 70 41 9 42 23

2 5 9 9 15 22 23 41 42 42 53 67 70 72 81

Mean = 559/16 = 34.9375 = 35

Median = 23 + 41 = 64/2 = 32

Mode = 9 and 42

2.	Using the weighted approach, calculate the average number of home runs per game hit by the Sox.

Weighted Mean = (3x2 + 2x5 + 1x6 + 0x7)/20 = 6 + 10 + 6 + 0/20 = 22/20 = 1.1 home runs/game hit by the Sox.

4.	When a pair of dice is rolled, the total will range from 2 (1, 1) to 12 (6,6).  It is a fact that some numbers will occur more frequently than others as the dice are rolled over and over.

a.	Why will some numbers come up more frequently than others?

	The odds are higher that certain numbers will come up, more than others.  When two dice are rolled, any number between 2 and 12 can come up.  Some numbers appear more often than others.  For example there are six different ways to roll a 7, but only one way to roll a 2 or a 12.  The number 7 has a better than 16% chance of coming up on each roll, whereas there is less than 3% chance of rolling a 2 or 12.  Probabilities for other numbers are as follows: 6s and 8s under 14%, 5s and 9s nearly 11%, 4s and 10s almost 8%, 3s and 11s slightly less than 6%.  7 is the most frequently rolled number.

b.	What are the total (x,y) outcomes that can occur?

The sum of two dice:

      1	2	3	4	5	6
    +-------------------------------------------------------------
  1 | 2	3	4	5	6	7
  2 | 3	4	5	6	7	8
  3 | 4	5	6	7	8	9
  4 | 5	6	7	8	9	10
  5 | 6	7	8	9	10	11
  6 | 7	8	9	10	11	12

There are 36 different combinations for the
sum of the two dice in the table.

Probability is the number of times something happens divided by the number of times it could happen. Take the sum of 7. It appears in the table six times, but there are 36 spots it could have been in, so the probability is the number of time it happened (6) divided by the number of times it could happen (36), or a probability of 6/36. To find probabilities for different sums, find the number of times a sum appears divided by the number of spots where it could have appeared (36).

c.	How might you then estimate the percentage of the time a particular
	number will come up if the numbers thrown be determined?

To find probabilities for different sums, find the number of times a sum appears divided by the number of spots where it could have appeared (36).

d.	Once these percentages have been calculated, how might the mean value of all the numbers thrown be determined?
You would add all of the percentages together and divide by 36 as the N value is 36.

e.	Name the very famous distribution that describes the frequency of the number of times a number comes up in a series of dice rolls.
Central limit theorems are a set of weak-convergence results in probability theory. Intuitively, they all express the fact that any sum of many independent identically distributed random variables will tend to be distributed according to a particular "attractor distribution". The most important and famous result is simply called The Central Limit Theorem which states that if the summed variables have a finite variance then they will be approximately normally distributed. Since many real processes yield distributions with finite variance, this explains the ubiquity of the normal distribution.

this looks right enough.. i only skimmed through it though.

Ditto.

Can't see any glaring mistakes...but it's always best to check the work yourself. After all, you don't get Mathlinks in a exam now do you?  :P

Yes, you should always take a look at your own work and try to prove yourself wrong. If you can't, you're good to go. If you can, try to fix your mistakes.

Yes. That way, you won't rely on others to help you check your work. You can do as well just by yourself.

[quote="José"][quote="Mark1"][quote="José"]A lot of homework here[/quote]

That is why I respectfully asked to check it.  I did not ask anyone to do it for me.

:)[/quote]

As other poster have already said, is always better to look yourself at your mistakes....[/quote]Yes José we know, there is no need to rephrase again as this tends to spam the forums.

Mark1: #1 is missing a 6 from the list. #4e is OK but you can give a simpler answer if you want, instead of bringing the core of statistical theory to answer a small question.

[quote="10000th User"]Mark1: #1 is missing a 6 from the list. #4e is OK but you can give a simpler answer if you want, instead of bringing the core of statistical theory to answer a small question.[/quote]

Thanks 10,000.  Good eye.  Good thing I checked my work in Excel based upon the top list.

Thanks for the review.

 :)


Could someone explain to me an easy way to do this problem?

How many of the integers between 1 and 1000, inclusive, can be expressed as the difference of the squares of two nonnegative integers?

Thanks

Suppose $n=a^2-b^2=(a-b)(a+b)$. Numbers $a-b$ and $a+b$ have the same parity. It implies that $n$  is odd OR $4\,|\,n$. And vice versa, if $n=2k+1$ then $n=(k+1)^2-k^2$ and if $n=4k$ then $n=(k+1)^2-(k-1)^2$.
So we are to calculate number of $n=2k+1$ and $n=4k$ from 1, 2, ..., 1000. It is 500+250=750.

[quote="SnowStorm"]hmm i didn't follow that at all

I get the difference of squares and then that both are even or odd, but What is 4|n ?[/quote]

4|n means that n is divisible by 4.

ok, so n = 2k + 1 means n is odd and n = 4k means n is divisble by 4
So the whole idea is just the odd numbers and multiples of 4 between 1000?

I don't understand where you come up with all the k stuff, maybe im just an idiot but could you explain that to me better?

think of the factorization of a^2-b^2

choosing consecutive squares will allow you to create any odd number
for example
(2-1)(2+1)=3
(3-2)(3+2)=5
etc

to get even numbers notice that all squares leave a remainder of 1 or 0 when divided by 4. Since you are subtracting the squares there is no way you can get numbers that leave a remainder of 2 when dividing by 4.

myth was showing the general way to get odd numbers or numbers divisible by 4.

consecutive squares give you the odds, squares that differ by 2 will give you the numbers divisible by 4.

so the answer is odd numbers+numbers divisible by 4=750


In which row of the Pascal Triangle is the sum of the digits $2147483648$?
1 - row 0
1 1 - row 1
1 2 1 - row 2
1 3 3 1 - row 3
.............

It's pretty easy once you find a pattern...

[quote="kstan013"][hide]That big number is 2^30, so it's the 30th row[/hide][/quote]
[hide]Actually, it's $2^{31}$, so it's the $\boxed{31}$st row.[/hide]
You're right, Quatto.

Wait, shouldn't it be the 31st row?
[hide] For the 0th row, we have the sum is 2^0.  For the first row the sum is 2^1.  For the second row, the sum is 2^2.  This means that for the nth row, the sum is 2^n.  In this case n=31, so it is the 31st row. [/hide]


[hide]values of a, b and c are 1, 2 and 3....[/hide]

[hide]i got this answer by trial method.... :rotfl: [/hide]

perhaps we can construct a polynomial that has those conditions
$f(z)=z^3-xz^2+bz-x$

$s_2+(-x)(x)+2(b)=0$
$s_2=x^2-2b$

$s_3+(-x)(x^2-2b)+b(x)+3(-x)=0$
$x^2-x^3+3bx-3x=0$
$x-x^2+3b-3=0$
$b=\frac{x^2-x+3}{3}$

$f(z)=z^3-xz^2+\frac{x^2-x+3}{3}z-x$

(note this ommits the case where x=0)

not sure what to do now...

a+b+c = abc = x
a = -b-c
(-b-c)bc = x
-b^2c-c^2b = x
x^2 = (-b^2c-c^2b)^2
=b^4c^2+2b^3c^3+c^4b^2
--------- Now solve for a^3+b^3+c^3
(-b-c)^3+b^3+c^3 =-b^3-2b^2c+-bc^2-b^2c-2bc^2-c^3+b^3+c^3
Set them equal to each other....
-b^3-2b^2c+-bc^2-b^2c-2bc^2-c^3+b^3+c^3 = b^4c^2+2b^3c^3+c^4b^2

And i'll leave you to solve... (I guess I don't know what I'm doing) good night...
If you can solve for b or c you will then be able to solve for all of them, maybe
 :?

[quote="riddler"]shady, is it integral values?
[hide="if it is..."]
Then the equation which I got to is:
$(a+b+c)^2\cdot(a+b+c-1)=3(a+b)(b+c)(c+a)$
And find factors from there... that satisfy all integral values of a,b,c
[/hide][/quote]
How can we just substitute and see?Isn't there any toher way at all?


geometry
parallelogram
geometry unsolved

Point $ P$ is taken inside a right angle $ KLM$. A circle $ S_1$ with center $ O_1$ is tangent to the rays $ LK,LP$ of angle $ KLP$ at $ A,D$ respectively. A circle $ S_2$ with center $ O_2$ is tangent to the rays of angle $ MLP$, touching $ LP$ at $ B$. Suppose $ A,B,O_1$ are collinear. Let $ O_2D,KL$ meet at $ C$. Prove that $ BC$ bisects angle $ ABD$.

[quote="28121941"]In the number 5 of Kvant 1994, the following outlined solution is published:
The quadrilateral O_1DO_2B is a paralelogram (O_1D and O_2B are perpendicular to LP), and if E is the point in which circle O_2 touches LM, the quadrilateral CLEO_2 is a rectangle. The triangles DCL and BDO_1 are equal, and therefore the angles DCB and DBC are equal.[/quote]

 Probably , something in the solution is different from the wording. For example - in my figure - $ O_1D$ is parallel to $ O_2B$ but these radii are not equal, so the four points do not form a parallelogram.
 If you can see the problem in the original wording , please correct me !May be I have missunderstood the problem.

 Babis

I am trying of solve the problem in another way; I will check the statement of the problem to see if both circles is supposed to be of the same radius. I just translated the sketched solution of Kvant, in which no pictures are included.

I have checked the original version of the problem in Kvant, 1994, issue 5, and in fact both circles is supposed to be of the same radius.

So, when the radii are not equal, does the problem still hold?

I think the picture by Stergiou provides a clear example that the problem does not work if the radiuses are different. Maybe it is possible to think a bit more about the conditions for that.


geometry
circumcircle
quadrilateral
symmetry
IMO Shortlist
homothety

There is given a convex quadrilateral $ ABCD$. Prove that there exists a point $ P$ inside the quadrilateral such that
\[
\angle PAB \plus{} \angle PDC \equal{} \angle PBC \plus{} \angle PAD \equal{} \angle PCD \plus{} \angle PBA \equal{} \angle PDA \plus{} \angle PCB = 90^{\circ}
\]
if and only if the diagonals $ AC$ and $ BD$ are perpendicular.

[i]Proposed by Dusan Djukic, Serbia[/i]

Assume $ AC \perp BD$ and let $ E \equiv AC \cap BD$ be the diagonal intersection. Let $ F$ be intersection, other than $ E,$ of circumcircles $ \odot(ABE), \odot(CDE)$ with diameters $ AB, CD.$  Let $ P$ be intersection, other than $ F,$ of circumcircles $ \odot(FBC), \odot(FDA).$ Then

$ \angle FCA \equal{} \angle FCE \equal{} \angle FDE \equal{} \angle FDB$ and $ \angle FAC \equal{} \angle FAE \equal{} \angle FBE \equal{} \angle FBD$

$ \Longrightarrow$ $ \triangle FAC \sim \triangle FBD$ are similar $ \Longrightarrow$ right $ \triangle FAB \sim \triangle FCD$ are similar $ \Longrightarrow$

$ \angle FAB \plus{} \angle FDC \equal{} \angle FBA \plus{} \angle FCD \equal{} 90^\circ$ $ \Longrightarrow$ $ \angle PAB \plus{} \angle PDC \equal{} \angle PBA \plus{} \angle PCD \equal{} 90^\circ.$

$ \angle PBC \plus{} \angle PAD \equal{} \angle PFC \plus{} \angle PFD \equal{} \angle CFD \equal{} 90^\circ$ and $ \angle PCB \plus{} \angle PDA \equal{} \angle PFB \plus{} \angle PFA \equal{} \angle AFB \equal{} 90^\circ$

$ \Longrightarrow$ $ P$ is the desired point. $ \square$
____________________________________________

Assume a point $ P$ with the desired properties exists. Let perpendiculars $ a, b, c, d$ to $ PA, PB, PC, PD$ at $ A, B, C, D$ pairwise intersect at $ X \equiv d \cap a, Y \equiv a \cap b,\ Z \equiv b \cap c, T \equiv c \cap d.$ Cyclic quadrilaterals $ PDXA \sim ZCPB$ are similar $ \Longrightarrow$ $ XYZT$ is cyclic, let $ (O)$ be its circumcircle. Let $ U, V$ be poles of $ XZ, YT$ WRT $ (O).$ Then $ \angle ZUX \equal{} 180^\circ \minus{} 2 \angle XYZ$ and from similar right $ \triangle PDX \sim \triangle ZCP,$ $ \angle XPZ \equal{} \angle DPC \plus{} 90^\circ \equal{} \angle XYZ \plus{} 90^\circ$ $ \Longrightarrow$ $ P$ is on circle $ (U)$ with center $ U$ and radius $ UX \equal{} UZ.$ Similarly, $ P$ is on circle $ (V)$ with center $ V$ and radius $ VY \equal{} VT.$ $ XZ, YT$ are pairwise radical axes of $ (O), (U)$ and $ (O), (V)$ $ \Longrightarrow$ diagonal intersection $ W \equiv XZ \cap YT$ is radical center of $ (O), (U), (V)$ and $ PW$ is radical axis of $ (U), (V).$ Since $ (U), (V)$ are both perpendicular to $ (O),$ the circumcenter $ O \in PW.$ $ B \in YZ, D \in XT$ are pedals of $ P.$ Let $ B' \in YZ, D' \in XT$ be pedals of $ O$ and let $ B'' \in YZ, D'' \in XT$ be pedals of $ W.$ From collinearity of $ P, O, W$ and from

$ \frac {YB'}{XD'} \equal{} \frac {YB''}{XD''} \equal{} \frac {YZ}{XT}$ $ \Longrightarrow$ $ \frac {YB}{XD} \equal{} \frac {YZ}{XT}.$

Let $ XY, ZT$ intersect at $ K$ and let $ (K), (P)$ be circles with centers $ K, P$ and equal radii $ KP \equal{} PK.$ Since polar of $ K$ goes through $ W,$ $ K$ is on polar $ UV$ of $ W$ $ \Longrightarrow$ circles $ (U), (V), (K)$ are coaxal and $ (K) \perp (O).$ Using sine theorem for $ \triangle UXK, \triangle VYK$ with $ \angle KXU \equal{} \angle KYV$ and $ \angle UKX \equal{} 180^\circ \minus{} \angle VKY,$

$ \frac {UP}{VP} \equal{} \frac {UX}{VY} \equal{} \frac {UK}{VK}$

$ \Longrightarrow$ $ PK$ bisects $ \angle UPV$ and circle $ (K)$ bisects angle formed by circles $ (U), (V).$ Inversion with center $ P$ and power $ PK^2$ takes circles $ (U), (V), (K)$ into concurrent lines $ u, v, k,$ such that $ k$ bisects $ \angle (u, v).$ Circle $ (O)$ perpendicular to $ (U), (V), (K)$ goes to a circle $ (O')$ centered at the concurrency point $ O'$ of $ u, v, k.$ Since $ (K) \cong (P),$ $ k$ is the perpendicular bisector of $ PK.$ Cyclic quadrilateral $ XYZT$ goes to rectangle $ X'Y'Z'T'$ with diagonal lines $ u \equiv X'Z', v \equiv Y'T'$ inscribed in $ (O')$ and with midline $ k \perp (X'Y' \parallel T'Z').$ By symmetry, $ \triangle T'PX' \cong \triangle Z'KY'$ are congruent $ \Longrightarrow$ $ \triangle XPT \sim \triangle Z'KY'$ are similar. On the other hand, $ \triangle ABC$ is pedal triangle of $ \triangle ZKY$ WRT $ P$ and $ \triangle Z'KY'$ is inversion image of $ \triangle ZKY$ WRT $ P$ $ \Longrightarrow$ $ \triangle ABC \sim \triangle Z'KY'$ are also similar (well known and posted before). As a result, $ \triangle XPT \sim \triangle ABC$ are similar. In exactly the same way, $ \triangle YPZ \sim \triangle ADC$ are also similar. Let $ \mathcal S_1: \triangle XPT \to \triangle ABC,$  $ \mathcal S_2: \triangle YPZ \to \triangle ADC$ be the corresponding similarity transformations. Due to $ \frac {XD}{XT} \equal{} \frac {YB}{YZ},$ these similarities map $ \mathcal S_1(D) \equiv \mathcal S_2(B)$ to the same point $ E \in AC.$ Since $ PD \perp XT, PB \perp YZ,$ it follows that $ BE \perp AC, DE \perp AC$ $ \Longrightarrow$ $ E \in BD$ is diagonal intersection of $ ABCD$ and $ BD \perp AC.$ $ \square$

Nice solution, Vladimir!

I have a different proof, but I'm quite in a rush; so, I'll be a little "detail-less".

We have the following results:

[b]Theorem 1[/b]. Let $ ABCD$ be a quadrilateral and let $ P \equal{} AC \cap BD$, $ E \equal{} AD \cap BC$, and $ F \equal{} AB \cap CD$. Then, $ \text{isog}_{ABE}{P} \equal{} \text{isog}_{CDE}{P} \equal{} \text{isog}_{ADF}{P} \equal{} \text{isog}_{BCF}{P}$ if and only if $ AC \perp BD$ (here $ \text{isog}_{XYZ}{P}$ means the isogonal conjugate of $ P$ with respect to triangle $ XYZ$).

[b]Theorem 2[/b]. Let $ ABCD$ be a quadrilateral and let $ P \equal{} AC \cap BD$. Denote by $ U1,\ V1,\ W1,\ T1$ the projections of $ P$ on $ AB,\ BC,\ CD,\ DA$ and by $ U2,\ V2,\ W2,\ T2$ the intersections of $ PU1,\ PV1,\ PW1,\ PT1$ with $ CD,\ DA,\ AB,\ BC,$ respectively. Then,$ ABCD$ is orthodiagonal if and only if $ U1,\ V1,\ W1,\ T1,\ U2,\ V2,\ W2,\ T2$ are all concyclic.

The second is just a simple angle chase. Now by Theorem 2, and from the fact that two points have the same pedal circles wrt. a given triangle if and only if they are isogonally conjugated wrt. that triangle, we get Theorem 1. Now let $ Q \equal{} \text{isog}_{ABE}{P} \equal{} \text{isog}_{CDE}{P} \equal{} \text{isog}_{ADF}{P} \equal{} \text{isog}_{BCF}{P}$ and we shall see that this point satisfies the desired property.

Proof that $ P$ with desired properties exists $ \Longrightarrow$ $ AC \perp BD$ is actually even simpler than the reverse, just by an angle chase. Since $ ABCD$ is convex and the angles undirected, $ P$ must be inside of $ ABCD,$ otherwise some of the angles in the $ 90^\circ$ sums would be obtuse.

Let $ F$ be intersection, other than $ P,$ of circumcircles $ \odot(PBC), \odot(PDA).$ Let $ E$ be intersection, other than $ F,$ of $ \odot(FAB), \odot(FCD).$ WLOG, let $ BFPC, AFPD$ follow on the corresponding circumcircles in this order. (Otherwise, relabel $ ABCD$ as $ DCBA.$)

$ \angle AFB \equal{} 360^\circ \minus{} (\angle PFA \plus{} \angle PFB) \equal{} 360^\circ \minus{} (180^\circ \minus{} \angle PDA \plus{} 180^\circ \minus{} \angle PCB) \equal{}$
$ \equal{} \angle PDA \plus{} \angle PCB \equal{} 90^\circ.$

$ \angle CFD \equal{} \angle PFC \plus{} \angle PFD \equal{} \angle PCD \plus{} \angle PDA \equal{} 90^\circ.$

It follows that $ \angle AEB \equal{} \angle AFB \equal{} 90^\circ$ and $ \angle CED \equal{} \angle CFD \equal{} 90^\circ.$ The case $ P \equiv F$ (when $ BC \parallel DA$) does not present a problem, because then the angle chase can be repeated using the common tangent $ PF$ of the circles $ \odot(PBC), \odot(PDA).$ Assume that $ AEFB, DEFC$ follow on the corresponding circumcircles in this order. (Proof for the other order $ AFEB, DFEC$ is the same.)

$ \angle DEA \equal{} 360^\circ \minus{} (\angle FED \plus{} \angle FEA) \equal{} 360^\circ \minus{} (180^\circ \minus{} \angle FCD \plus{} 180^\circ \minus{} \angle FBA) \equal{}$
$ \equal{} \angle FCP \plus{} \angle PCD \plus{} \angle PBA \minus{} \angle FBP \equal{} \angle PCD \plus{} \angle PBA \equal{} 90^\circ.$

$ \angle BEC \equal{} \angle FEB \plus{} \angle FEC \equal{} \angle FAB \plus{} \angle FDC \equal{}$
$ \equal{} \angle PAB \minus{} \angle PAF \plus{} \angle PDC \plus{} \angle PDF \equal{} \angle PAB \plus{} \angle PDC \equal{} 90^\circ.$

Again, the case $ E \equiv F$ (when $ AB \parallel CD$) does not present a problem, because then the angle chase can be repeated using the common tangent $ EF$ of the circles $ \odot(FAB), \odot(FCD).$ As a result, $ \angle AEB \equal{} \angle BEC \equal{} \angle CED \equal{} \angle DEA \equal{} 90^\circ.$ This means that $ E \equiv AC \cap BD$ is diagonal intersection and $ AC \perp BD.$ $ \square$

Suppose $ P$ exists and let $ M$ be the Miquel point for the complete quadrilateral formed by the lines $ AB, BC, CD, DA$. Perform an inversion of center $ M$. Let $ Q \equal{} AB \cap CD, R \equal{} BC \cap DA$. The circumcircles of $ \triangle QAB$ and $ \triangle PAB$ are orthogonal by the problem condition, thus the image of the circle $ (PAB)$ is a circle of diameter $ A'B'$. Analogous reasoning holds for $ (PBC), (PCA)$, etc. therefore $ P'$ is the intersection of the perpendicular $ A'C'$ and $ B'D'$.

But $ \angle AMB \equal{} \angle CMB \Rightarrow$ the angle bisectors of $ \angle AMC$ and $ \angle BMD$ coincide $ \Rightarrow \angle (AC, BD) \equal{} \angle (B'D', A'C')$. And conversely.

[b]Existence of $P$ $\implies AC\perp BD$:[/b] We show that $AB^2+CD^2=BC^2+AD^2$, which implies that $AC\perp BD$.
Let $PA = a, PB = b, PC = c, PD = d$, and $\angle PDA = \theta_1, \angle PAD = \theta_2$, so $\angle PCB = \frac{\pi}{2}-\theta_1, \angle PBC = \frac{\pi}{2}-\theta_2$. Note that all the mentioned angles above are acute ($P$ is inside the quadrilateral). More importantly, the $\sin$ and $\cos$ of all these angles are positive.

Note that $\angle APD+\angle BPC = \pi, \angle APB + \angle CPD = \pi$. So by the Law of Cosines,

\[AD^2+BC^2=a^2+b^2+c^2+d^2-2(ad-bc)\cos{\angle APD}\] and \[AB^2+CD^2=a^2+b^2+c^2+d^2-2(ab-cd)\cos{\angle APB}\]. So $AB^2+CD^2=AD^2+BC^2 \Longleftrightarrow (ad-bc)\cos{\angle APD} = (ab-cd)\cos{\angle APB}$.

We proceed to compute $\cos{\angle APD}$. By the Law of Sines, \[\frac{a}{d}\cdot \sin{\theta_2} = \sin{\theta_1}\]\[\frac{b}{c}\cdot {\cos{\theta_2}}=\cos{\theta_1}\]. Squaring the 2 above equations, and summing, we get that \[\frac{a^2}{d^2}\cdot\sin^2{\theta_2}+\frac{b^2}{c^2}\cdot \left(1-\sin^2{\theta_2}\right)  = 1 \implies \sin{\theta_2} = d \cdot \sqrt{\frac{c^2-b^2}{a^2c^2-b^2d^2}} \implies \cos{\theta_2} = c \cdot \sqrt{\frac{a^2-d^2}{a^2c^2-b^2d^2}}\] (the sine and cosine are both positive). Now we easily get that \[\sin{\theta_1} = a \cdot \sqrt{\frac{c^2-b^2}{a^2c^2-b^2d^2}} \hspace{1cm} \cos{\theta_1} = b \cdot \sqrt{\frac{a^2-d^2}{a^2c^2-b^2d^2}}\].

Now note that \[\cos{\angle APD} = -\cos({\theta_1+\theta_2}) = -(\cos{\theta_1}\cos{\theta_2}-\sin{\theta_1}\sin{\theta_2}) \\ = -\frac{a^2bc-bcd^2}{a^2c^2-b^2d^2}+\frac{ac^2d-ab^2d}{a^2c^2-b^2d^2} \\= -\frac{(ac+bd)(ab-cd)}{(ac+bd)(ac-bd)}\\ =  -\frac{ab-cd}{ac-bd} \\ \implies (ad-bc)\cos{\angle APD} = -\frac{(ab-cd)(ad-bc)}{ac-bd}\]. Since this is symmetric, $(ad-bc)\cos{\angle APD} = (ab-cd)\cos{\angle APB}$.

Note that I still have dumb cases to deal with, like $a = d, b = c$, but those are silly.
-----------------------------------------------------------------------------------------------------------------------------------------------------------------
[b]$AC \perp BD \implies$ existence of $P$:[/b] Let $AC \cap BD = O$. We show that the isogonals of $AO, BO, CO, DO$, with respect to $\angle BAD, \angle ABC, \angle BCD, \angle CDA$, concur at our desired point $P$. By simple angle chasing, it is obvious that this $P$ satisfies the properties. Let the isogonals of $AO, BO, CO, DO$ be $%Error. "fancy" is a bad command.
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{l}_d$. Let $\angle({line_1, line_2})$ denote the angle between 2 lines (positive or negative). Finally, let $\angle BAC, \angle CAD, \angle ACB, \angle ACD$ be $\theta_1, \theta_2, \theta_3, \theta_4$ respectively.

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{l}_c) = \theta_4, \angle(%Error. "fancy" is a bad command.
{l}_c, AC) = \theta_3-\theta_4$.

So \[\frac{\sin(\theta_1-\theta_2)}{\sin{\theta_2}}\cdot\frac{\sin(\frac{\pi}{2} - \theta_3)}{\sin(\frac{\pi}{2}-\theta_1)}\cdot{\frac{\sin{\theta_4}}{\sin(\theta_3-\theta_4)} = \frac{\sin{\theta_1}\cos{\theta_2}-\sin{\theta_2}\cos{\theta_1}}{\sin{\theta_2}}\cdot\frac{\cos{\theta_3}}{\cos{\theta_1}}\cdot{\frac{\sin{\theta_4}}{\sin{\theta_3}\cos{\theta_4}-\sin{\theta_4}\cos{\theta_3}}}}\] \[\\ = \frac{\frac{OB}{AB}\cdot\frac{OA}{AD}-\frac{OD}{AD}\cdot\frac{OA}{AB}}{\frac{OD}{AD}}\cdot\frac{\frac{OC}{BC}}{\frac{OA}{AB}}\cdot{\frac{\frac{OD}{DC}}{\frac{OB}{BC}\cdot\frac{OC}{CD}-\frac{OD}{DC}\cdot\frac{OC}{BC}} = 1}\] (after elementary calculation). So the result follows.

If $AC\perp BD$ lets consider $P$ to be the isogonal conjugate point of $E=AC\cap BD$ wrt $RAB$ ($R=BC\cap AD$). Now let's consider the orthocenter $X$ of $CDA$. We have $BCX\sim BPA$($\angle CXB=\angle CAD=\angle BAP$ and $\angle PBA=\angle XBC$) so $BX\cdot BP=BA\cdot BC$ Now from this we have $BAX\sim BPC$ giving $\angle DCA=\angle BXA=\angle BCP$ similarly we have that $\angle PDA=\angle BDC$ and it's easy to check that $P$ satisfies all the conditions.
Now assume such $P$ exists. if again $R=AD\cap BC$ we consider $K,M$ which are isogonal conjugates of $P$ wrt $RAB$ and $RCD$. The condition for angles gives that lines $AK,KB,CM,DM$ form a rectangle now note that if $AK,BK$ cut $BC,AD$ at $C',D'$ we have that $RCD$ and $RC'D'$ are homothethic(because $\angle KRA=\angle PRB=\angle MRA$ meaning that $K,M,R$ are collinear). But using the if part(on $ABC'D'$ we know that $P$ is the isogonal conjugate of $K$ wrt triangle $AC'D'$ but now since $P$ is the isogonal conjugate of $M$ wrt $ACD$ we have that the homothethy that sends $ACD$ to $AC'D'$ (and $M$ to $K$) sends $P$ to $P$ this gives $M=K$ which means $AC\perp BD$

Well known: In triangle ABC if P and Q are isogonal conjugates, the midpoint of PQ is the center of the circle that passes through the feet of the perpendiculars from P and Q to the sides.

Let AB and CD cut at F, and AD and BC cut at E.

Assume the diagonals are perpendicular at point X. From X draw the perpendiculars to the sides, and these 4 points are concyclic (easy to see by anglechase). Define P to be the reflection of X across the center of this circle. Then by the well-known theorem, P and X are isogonal conjugates in triangles ABE, DCE, FAD, FBC and the angle condition is immediately obtained.

Now assume the angle condition is true, then do exactly the same but switching P and X.

Nice problem!

[i]Part 1. :- Let $AC \perp BD$[/i]

[i]Proof[/i] Indeed, let $AC,BD$ meet at point $R$, lines $AB,CD$ and $AD,BC$ meet at points $Q,P$ respectively. Let the reflections of $P$ in $AB,BC,CD,DA$ be $X,Y,Z,T$ respectively. By some old USAMO problem, whose link I shall post later on, we know that $X,Y,Z,T$ are con cyclic. Let $S$ be the centre of this circle. It is evident that $S$ is the isogonal conjugate of $R$ in triangles $PAB,PDC,QAD,QBC$. Thus, we easily see that our result is satisfied by point $S$

[i]Part 2. :- Let such a point $X$ exist[/i]

[i]Proof[/i] Indeed, doing as previously, we let the reflections of our point $S$ be $X,Y,Z,T$. Then, by our angle condition, these points are infact con cyclic. Thus, if we let $R'$ be the centre, then it is the isogonal conjugate of $S$ in all these four triangles. Therefore, we see that $AB,BC,CD,DA$ sub tend the same right angle at $R'$ and so $R'=R$ which means that $AC \perp BD$.

Here's an overkill solution:

[hide = Solution]
Suppose $P$ exists; then, the angle condition implies $\angle APD + \angle BPC = 180^{\circ}$, so by a well-known result, $P$ has an isogonal conjugate $Q$. However, then we have $\angle AQB = \angle BQC = \angle CQD = \angle DQA = 90^{\circ}$, which is only possible if $ABCD$ has perpendicular diagonals.

Now suppose that $ABCD$ has perpendicular diagonals, and let $Q$ be their intersection. Then $\angle AQD + \angle BQC = 180^{\circ}$, so $Q$ has an isogonal conjugate $P$, which is in fact the point $P$ we want.
[/hide]

Solution from [i]Twitch Solves ISL[/i]:

The problem is killed by quoting the following theorem:
[b][color=red]Theorem:[/color][/b]  	If $X$ is a point such that 	$\angle AXD + \angle BXC = 180^{\circ}$, 	then the isogonal conjugate of $X$ exists.
[i]Proof.[/i]  	[Proof, for completeness only] 	It's enough for the projections of $X$ 	to the sides to be cyclic, 	by considering the six-point circle. 	Let them be $X_1$, $X_2$, $X_3$, $X_4$. 	\begin{align*} 		\measuredangle XX_4X_1 &= \measuredangle XAX_1 \\ 		&\vdots 	\end{align*} 	and this means $\measuredangle X_1X_2X_3 = \measuredangle X_1X_4X_3$. $\blacksquare$

Now in one direction, if the diagonals are perpendicular and meet at $Q$, then its isogonal conjugate exists, and is seen to have the desired property.

Conversely, given such a point $P$, it has an isogonal conjugate $Q$ which satisfies $\angle AQB = \angle BQC = \angle CQD = \angle DQA = 90^{\circ}$, which implies that $Q$ is the perpendicular intersection of the diagonals.

lol wut.

By the angle condition $\angle APD+\angle CPB=180$, so $P$ has an isogonal conjugate $P'$. However we then have

$$\angle AP'D=\angle DP'C=\angle CP'B=\angle BP'A=90$$

So the diagonals are perpendicular. The reverse condition is true by taking the isogonal conjugate of $AC\cap BD$. $\square$

If $AC\perp BD$ point $E=AC\cap BD$ has isogonal conjugate $Q$ in $ABCD,$ and clearly $P=Q$ works.
If $P$ exists, then $\angle APB+\angle CPD=\pi,$ so $P$ has isogonal conjugate $F$ in $ABCD$ and $$\angle AFB=\angle BFC=\angle CFD=\angle DFA=\frac{\pi}{2}\implies F=E\implies AC\perp BD.$$

Let $E=AC\cap BD$, since $\angle AEB+\angle CED=180^{\circ}$ there exists an isogonal conjugate $E'$ which obviously satisfies the conditions.

(I guess if you wanted to prove the isogonal conjugate lemma you could draw the (cyclic) pedal quadrilaterals of $E$ and $E'$ and then do some PoP, etc.)

Perhaps I should be a bit more specific when proving constructions using geometric continuity.
-----
Note that the angle condition is equivalent to there existing a pedal rectangle. Realize that after any horizontal or vertical affine transform, orthodiagonal quadrilaterals remain orthodiagonal. To prove that the angle condition implies that the diagonals are perpendicular, we will show that given the square with vertices $(\pm 1, \pm 1)$ and any point $(a, b)$ inside the square, the quadrilateral induced by the lines through $V$ and perpendicular to the segment through $V$ and $(a, b)$ for each $V \in \{(\pm 1, \pm 1)\}$ is orthodiagonal. This is easy to see by Cartesian coordinate bash. To prove that the orthodiagonal condition implies the angle condition, consider some orthodiagonal quadrilateral $ABCD$. By geometric continuity[hide=*]We can use this to prove that three linearly moving lines $\ell_1$, $\ell_2$, and $\ell_3$ eventually concur. The technique is actually somewhat simple: find two distinct positions and fix an $x$-axis perpendicular to none of the lines at any time so that in one position, the intersection of $\ell_1 \cap \ell_2$ has a smaller $x$-coordinate than $\ell_2 \cap \ell_3$, and in the other one $\ell_1 \cap \ell_2$ has a larger $x$-coordinate that $\ell_2 \cap \ell_3$. Continuity clearly guarantees an eventual concurrency.[/hide], it is easy to see that there exists a rectangle with two sides parallel to each diagonal of $ABCD$ such that it is a pedal quadrilateral of $ABCD$. Thus we're done.

Observe that $P$ has a pedal circle. We will prove a known lemma:
$\textbf{Claim.}$ If $P$ has a pedal circle in a convex $n$ gon then it also has an isogonal conjugate.
$\textit{Proof.}$ Reflect $P$ over the circumcenter of its pedal circle, and then this point is an isogonal conjugate of $P$ in all the triangles formed by $3$ adjacent sides. $\square$

Take the isogonal conjugate of $P$, call it $Q$, and the conditions rewrite to being $\angle AQB = \angle BQC = \angle CQD = \angle DQA = 90^\circ.$ Thus $B-Q-D$ and $A-Q-C$, and hence $AC \perp BD$.

For the converse, it is known that the intersection of the diagonals has a pedal circle. Thus take the isogonal conjugate to finish.

Let $Q = AC \cap BD$. For the forwards direction, note $Q$ has an isogonal conjugate, and this point satisfies the conditions of our desired $P$. For the backwards direction, note $P$ has an isogonal conjugate, which can be determined to be $Q$. $\blacksquare$


Considering the equation X²-dY²=1, where d,x,y are positive integers. Show that the number of solutions of this equation is infinite.

It's a well known fact about Pell's equation.

Not true when $d$ is a square.

http://mathworld.wolfram.com/PellEquation.html


A standard die is weighted so that the likelihood of any side facing up after a toss is proportional to the number of dots on that face. What is the probability of the side with 5 dots being face up after a single toss? Express your answer as a common fraction.

The ratio of the probabilities of getting 1, 2, 3, 4, 5, or 6 is, respectively, 1:2:3:4:5:6. Adding all of these values up to get 21 gives us the denominator of the fraction. The numerator is 5 (since the it is equal to the value of the rolled number), so the probability of rolling a 5 on this die is [b]$ 5/21$[/b].

there are the #'s

1,2,3,4,5,6

with the ratio of

1:2:3:4:5:6;=

then we add all these together and get

21

the numerator MUST be 5 because thats what the problem said

so we get the fraction 5/21

so our answer is 5/21 or $\boxed{5/21}$

Click here: [hide]JK! Click here: [hide]Gotcha there! Click here: (I promise the solution is there!)[hide]So..since each number is directly proportional to its probability of being rolled, let's call this ratio between the two $k$.
After that, we know that the probability of rolling a $1$ is $1$$k$, or $k$, the probability of rolling a $2$ is $2$$k$, and so on.
Then we can set up an equation for $k$!
We can say-because the sum of all the probabilities in a given outcome is 1- that we can make the equation below:
$k$ $+$ $2$$k$ $+$ $3$$k$ $+$ $4$$k$ $+$ $5$$k$ $+$ $6$$k$ $=$ $1$.
Solving for $k$, we get $21$$k$ = $1$, so $k$ $=$ $\frac{1}{21}$.
Now, we want to know the probability for rolling a $5$, we want to solve for $5$$k$, which is $5\cdot\frac{1}{21}$. And that equals $\frac{5}{21}$.
So our answer is $\boxed{5/21}$.[/hide][/hide][/hide]


integration
complex analysis
complex analysis unsolved

I really have a hard time with conformal mappings onto polygons. This is one of the questions I'm stuck. Hope someone can help.
Let $ F(z)\equal{}\int_0^z z^{\minus{}\beta_1} (1\minus{}z)^{\minus{}\beta_2} dz$ with $ 0<\beta_1, \beta_2<1$ and $ 1<\beta_1 \plus{} \beta_2 <2$. Prove that $ F$ maps the upper half-plane to a triangle whose vertices are the images of 0,1, and $ \infty$ with angles $ \alpha_1\pi , \alpha_2\pi, \alpha_3\pi$ where $ \alpha_i \plus{}\beta_i \equal{}1$ and $ \beta_1 \plus{}\beta_2 \plus{}\beta_3\equal{}2$ and the length of the side of triangle opposite angle $ \alpha_i \pi$ is $ \frac{sin(\alpha_i \pi)}{\pi} \Gamma(\alpha_1) \Gamma(\alpha_2) \Gamma(\alpha_3)$
What happen if $ \beta_1 \plus{}\beta_2\equal{}1$? If $ 0<\beta_1 \plus{} \beta_2 <1$?

Think of what happens on the boundary first, i.e., honestly go along the real line and ask yourself in which direction the antiderivative moves (you need to understand what is the right branch of the power to choose, of course). Once you know that the mapping sends the boundary of the upper half-plane to the boundary of the triangle, there is a great theorem that tells you that under certain easy to verify conditions, the interior is conformally mapped to the interior. ;)


Please Condense:

6(In 3+ In x) + 1/4 In 3

In the [b]n[/b]atural [b]l[/b]ogarithm, the "l" is a lowercase "L", not a capital "I"  :)

Sorry about that!
Thank you very much for your answer


A man, Walter Sparrow, becomes obsessed with a novel that he believes was written about him. As his obsession increases, more and more similarities seem to arise. The acting is great, the camera shots are great, and the story is great.

The link between this movie and maths is the number $ \boxed{23}$. Some meanings of $ \boxed{23}$ can be found on [url=http://en.wikipedia.org/wiki/23_(number)]wikipedia[/url].

[quote="PowerOfPi"]I was born on the 23rd day of a month... I think I watched part of that movie a couple years ago.[/quote]

LOL as was I =)

On an incompletely related note:
[youtube]iNIu3d0CcJY[/youtube]


Is this the correct statement?

[quote="TRAN THAI HUNG"]Caculate $\ M= 2x^{3}+2x^{2}+1$ with ${x=\frac{1}{3}\sqrt[3]{\frac{{23+\sqrt{513}}}{4}}+\sqrt[3]{\frac{{23-\sqrt{513}}}{4}}}-1$[/quote]

The statement now is right.You can see in the first post.

As M4RIO said, the two terms are mutually reciprocal, hence for shortness denote $a=\sqrt[3]{{\frac{{23+\sqrt{513}}}{4}}}$, giving ${1\over a}=\sqrt[3]{{\frac{{23-\sqrt{513}}}{4}}}$

Now $x={1\over 3}(a+{1\over a}-1)$. Put $y=a+{1\over a}$, giving $x={{y-1}\over 3}$. Substituting that into the given expression and rearranging, we get

$M=\frac{2y^{3}-6y+31}{27}=\frac{2y(y^{2}-3)+31}{27}$

Now we have

\begin{eqnarray*}2y(y^{2}-3) &=& 2\left(a+{1\over a}\right)\left(a^{2}+2+{1\over a^{2}}-3\right)\\ &=& 2\left(a+{1\over a}\right)\left(a^{2}-1+{1\over a^{2}}\right)\\ &=& 2\left(a^{3}+{1\over a^{3}}\right)\\ &=& 2\left(\frac{{23+\sqrt{513}}}{4}+\frac{{23-\sqrt{513}}}{4}\right)\\ &=& 2\cdot{23\over 2}\\ &=& 23\end{eqnarray*}

Therefore $M=\frac{23+31}{27}=2$

[quote="TRAN THAI HUNG"][quote="M4RI0"]Does anyone have a solution :?:[/quote]My teacher say that the problem was wrong.I have edited it.[/quote]
 :mad:  Do you have any idea how much time I had to spend on that wrong problem?  :huh:


floor function
limit
arithmetic sequence
number theory unsolved
number theory

Does exist a non-constant infinite arithmetic sequence, such that all it's terms have the form $a^b$. Both $a$ and $b$ are integer numbers greater than $2$.

No, the set of powers contains gaps of arbitrary length (their asymptotic density is also $0$!) as already posted in http://www.mathlinks.ro/Forum/viewtopic.php?t=77183 .
Every infinite sequence would reach such a gap after some steps.

[quote="ZetaX"]their asymptotic density is also $0$!.[/quote]
Can you give some reference/papers ? Seems a quite interesting fact :)

I hope a short post suffices ;)

See my (overkill) post in http://www.mathlinks.ro/Forum/viewtopic.php?t=62176 :
"There are at most $1+\sqrt[2]{n}+\sqrt[3]{n}+\sqrt[4]{n}+...+\sqrt[\left\lfloor log_2(n)\right\rfloor]{n} \leq 1+ \sqrt n log_2(n)$ 'true' powers $m^k , k\geq 2$ in the set $\{1,2,...,n\}$."

And since $\lim_{n \to \infty} \frac{ 1+ \sqrt n log_2(n)}{n} =0$, the density is $0$.

I'm sorry, I actually tought you were speaking of something different.
Something more close to "an asymptotic estitmate of the size of the $n$-th gap (being $o(n)$)".
Anyway no matter what I tough it didn't make much sense :(

[quote="{x}"]Anyway no matter what I tough it didn't make much sense :([/quote]
I don't think that it doesn't make sense, it's an interesting question. When you find something about the gaps, please tell me :)

[quote="dondigo"]I have a different solution:
Assume, that exist such sequence:
$c,c+r,c+2r,c+3r,\cdots$.
Led $d=gcd(c,r) \quad c=c'd, r=r'd$
Now our sequence looks like that:
$dc',d(c'+r'), d(c'+2r'),d(c'+3r'),\cdots$. $gcd(c',r')=1$. By Dirichlet Theorem sequence $c', c'+r', c'+2r',\cdots$ contains infinitely many primes.
So exist a term of this sequence, which is the prime number, that does not divide $d$. Let this prime number be $p_{k+1}$
So certain term of our first sequence has the form $dp_{k+1}=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}p_{k+1}$
Obviously such number cannot be power of natural number.Contradiction[/quote]
This is a more reasonable solution considering that the question came from Baltic MO, in other words, nothing advanced beyond elementary mathematics. And yea, ZetaX's solution is something I completely did not know before.


prove that there are no four positive integers w, x, y, z s.t.

$x^{2}=10w-1$
$y^{2}=13w-1$
$z^{2}=85w-1$

i dont see u use t anywhere but the problem as written can be solved when viewed mod 5.
the square residues mod 5 are $\{0,1,4,4,1\}\Longrightarrow\{0,1,4\}$ so we have

$x^{2}\equiv-1\mod(5)$
$y^{2}+1\equiv 13w\mod(5)$ and
$z^{2}\equiv-1\mod(5)$

 since -1 is not a residue mod 5 there are no integer values that work for x and y. however there are solutions for y (idk how to find them) but w/ trial an error i found (2,5) for (w,y).

o woops silly me how did i forget that $4\equiv-1$ mod 5  :rotfl:

[hide="Sketch"]Just consider $10w-1, 13w-1, 85w-1$ mod 4, and notice that they can't all be perfect squares.[/hide]


any other solution than the one in the book?
I thinks it's pretty amazing the connection between NT and geometry(as shown in the solution of the book)  :)

Well, basically, 180 = 3x3x4x5

So, if n is not divisible by 3, 180/n is.

But what if $n=9, 18,\dots, 9k,\dots$? :) And the point is not to calculate, but to construct.

[i]Solution:[/i] We are given that $\gcd(3,n)=1$,i.e.,
\[an+3b=1,a,b\in\mathbb{Z}. \]
Multiply both sides by $\frac{180^{\circ}}{n}$:
\[180^{\circ}a+\frac{180^{\circ}}{n}(3b)=\frac{180^{\circ}}{n}. \]
Division by 3 gives
\[60^{\circ}a+\frac{180^{\circ}}{n}b=\frac{60^{\circ}}{n}\]
Then solve the linear diophantine given $n$, and then copy angles.  Creating a $60^{\circ}$ is simple while having the centre of a circle.

correct!  :)  that's the solution in the book. any other solution?

That's the only solution I could think of, but certainly there are more solutions. :)


What is the sum of the two least prime numbers greater than 100?

The next two smallest primes are 101 and 103 so 101+103=204.

And we know that they are prime because that they can't be divisible by prime numbers less than their square roots.

I got this in a mem game, and everyone got 101+103=204


geometry
incenter
circumcircle
geometric transformation
homothety
ratio
trapezoid

$A,B,C$ are on circle $\mathcal C$. $I$ is incenter of $ABC$ , $D$ is midpoint of arc $BAC$. $W$ is a circle that is tangent to $AB$ and $AC$ and tangent to $\mathcal C$ at $P$. ($W$ is in $\mathcal C$)
Prove that $P$ and $I$ and $D$ are on a line.

Here are a few observations which I won't prove for now:

(1) If (W) touches AB and AC in S, T respectively then I is the midpoint of ST.

(2) PS and PT are the bisectors of angles <APB, <APC respectively.

By using these we can prove what we want in the following manner:

PA is the symmedian corresponding to P in triangle PST, because it's the intersection of the tangents through S and T to the circumcircle of PST. This means that <APS=<IPT=><APT+<IPT=<SPT and because <APT=<TPC=><IPT+<TPC=<SPT=><IPC=<SPT and in the same manner we show that <IPB=<SPT=<IPC=> I is on the bisector of angle <BPC, which is equivalent to what we wanted to show at the beginning.

Here's the proof of (2):

Let S' and T' be the points where PS and PT cut the circumcircle of ABC (respectively). The circle (C) is obtained from (W) by a homothety of center P and ratio PT'/PT=PS'/PS, so the tangent in T' to (C) corresponds to the tangent to (W) in T (which is AC) by this homothety, so they're parallel, so T' is the midpoint of the arc AC. In the same manner we show that S' is the midpoint of the arc AB and the conclusion follows.

The "standard" proof for (1) uses Casey's thm and the transversal theorem (at least that's what it's called in Romanian). I'll post it in a little while if I fail to find another (nicer, non-computational) one.

Let A' be the foot of the bisector of <A in triangle ABC.

From Casey's thm we get BS*AC+CT*AB=AS*BC (#). On the other hand, from the transversal thm we know that I is on ST iff (BS/SA)*A'C+(CT/TA)*A'B=(A'I/IA)*BC iff b*BS/SA+c*CT/TA=a which is true from (#) by division with AS=AT: AC*BS/AS+AB*CT/AT=BC (a, b, c are the sides of the triangle ABC, BC, CA, AB respectively).

I guess it's not that long or ugly :).

I can't find references to your Casey's theorem or transversal theorem...

Could you please post the above two theorems formally?

Casey's theorem (there is a much more general form, using the notion of cycle instead of that of circle, but I won't post that, because I might make some mistakes):

Let C be a circle and $C_i$ with i from 1 to 4 circles internally tangent to C, arranged around C in the increasing order of the indexes (1->2->3->4). Take $C_iC_j$ be the length of the external common tangent of $C_i$ and $C_j$. Then the analogous of Ptolemy's thm takes place: $C_1C_2\cdot C_3C_4+C_1C_4\cdot C_2C_3=C_1C_3\cdot C_2C_4$.

The transversal theorem:

ABC is a triangle, D is a point on BC, P is a point on AD, M and N are points on AB and AC respectively. Then the line MN passes through P iff DC*BM/MA+BD*CN/NA=BC*DP/PA, where all the 2-letter combinations represent oriented segments.

Sorry, but i don't understand how you can apply Casey's Theorom to this situation? please explain ...

You apply it for the circle W and the degenerate circles A, B, C. You can apply it for 2 circles and 2 points, 3 circles and a point ans other such cases. A point is nothing more than a degenerate circle.

Let us to present an alternative proof of the collinearity of the points $ P,\ I,\ D,$ based on the proof of the fact [b][size=100](1)[/size][/b], mentioned by [b][size=100]grobber[/size][/b] in the above post #2#, which appeared in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=46801]Mixtilinear incircle[/url] of [b][size=100]Leon[/size][/b].

$ \bullet$ We denote the points $ B'\equiv (O)\cap BI,\ C'\equiv (O)\cap CI,$ where $ (O)$ is the circumcircle of the given triangle $ \bigtriangleup ABC$ and it is easy to show that $ B'C'\parallel ST,$ from $ \angle B'XC \equal{} \angle B'C'C \plus{} \angle ACC' \equal{} \frac {\angle B}{2} \plus{} \frac {\angle C}{2} \equal{} 90^{o} \minus{} \frac {\angle A}{2} \equal{} \angle ATS,$ where $ X\equiv AC\cap B'C'.$

So, because of $ IS \equal{} IT,$ from the trapezium $ B'C'ST,$ we have that the point $ Q\equiv B'C'\cap PI,$ is the midpoint of the segment $ B'C'$ and then, we conclude that $ OQ\perp B'C',$ where $ O$ is the circumcenter of $ \bigtriangleup ABC.$

$ \bullet$ It is easy to show that $ OQ \equal{} \frac {A'I}{2}$ $ ,(1)$ where $ A'\equiv (O)\cap AI,$ because of the incenter $ I$ of $ \bigtriangleup ABC,$ is the orthocenter of the triangle $ \bigtriangleup A'B'C'$ as well.

So, from $ (1)$ and because of $ OQ\parallel IA',$ we conclude that the line segment $ IQ,$ passes through the point $ D\equiv (O)\cap A'O,$ as the symmetric point of $ A'$ with respect to $ O$ and then ( from $ OA'\perp BC$ ), as the midpoint of the arc $ BAC$ of $ (O).$

Hence, we conclude that the points $ P,\ I,\ D,$ are collinear and the proof is completed.

Kostas Vittas.

Dear Mathlinkers,
this result comes from Lauvernay (1892).
See: http://perso.orange.fr/jl.ayme   vol. 4  A new mixtilinear incircle adventure I, p. 17.
Sincerely
Jean-Louis

$\sqrt{bc}$-invert, which sends $T$ to the excentral touchpoint $D$, $I$ to the $A$-excenter $I_A$, and $T$ to the intersection point of the $A$-external bisector and $\overline{BC}$. It suffices to show that $A, D, I_A, T$ are concyclic now, but this is obvious.

Label the touch points to $AB$, $AC$, and $(ABC)$ as $K$, $L$, and $T$, and the midpoints of arcs $AB$ and $AC$ as $X$ and $Y$. Homothety tells us $TKX$ and $TLY$ collinear. Pascal on $BACXTY$ then tells us that $KIL$ collinear, and noting $\triangle AKI \cong \triangle ALI$ implies $IK = IL$.

Thus $TI$ and $TA$ are a median and symmedian, respectively, of $\triangle TKL$, so $\angle KTA = \angle ITL = \frac C2$. As a result, it's clear that $TI$ will pass through the top point of $(ABC)$. $\blacksquare$


geometry
3D geometry
pyramid
inequalities
analytic geometry

$SO$ is height of pyramid $SABC$, where $\measuredangle ASB=\measuredangle ASC=\measuredangle BSC=90^{\circ}.$
Let $SO=h,$ $AO=a,$ $BO=b,$ $CO=c.$
Prove that $h\leq\frac{a+b+c}{3}.$

[quote="cincodemayo5590"] Because each $a,b,c$ make a right triangle with the height $h$ and are the hypotenuses of these triangles, this means that $h\leq a$, $h\leq b$, $h\leq c$ as a hypotenuse of a triangle [/quote]
This is wrong. ;)

[hide="A Possible Solution"]Imagine a cross-section, perpendicular to ASB and passing through C. Then, the resulting triangle is a right triangle, where S is the right angle. The maximum value of h occurs when the two legs of the triangle are equal.

We can see, that the maximum possible value of h occurs when SA=SB=SC or when a=b=c. Then, <SAO = <SBO = <SCO = 45 deg. and a=b=c=h. So, h = (a+b+c)/3.

For any other a,b,c h will be smaller so h < (a+b+c)/3.[/hide]

[quote="Altheman"][quote="cincodemayo5590"][hide="Not exactly a formal proof, but..."] Because each $a,b,c$ make a right triangle with the height $h$ and are the hypotenuses of these triangles, this means that $h\leq a$, $h\leq b$, $h\leq c$ as a hypotenuse of a triangle is always greater than or (in this case, if $a$, $b$, or $c$ are the height) equal to either leg. Combining inequalities, we get $3h\leq a+b+c$, or $h\leq \frac{a+b+c}{3}$. [/hide][/quote]

a,b,c are not the lateral edges (considering the base as ABC), but they are the distances from the foot of the perpendicular to A,B, and C[/quote]
oh... hah. I thought it was the lateral edges. Yay reading the problem correctly!  :D

[quote="dyaka"]Imagine a cross-section, perpendicular to ASB and passing through C. Then,  triangle the resulting is a right triangle, where S is the right angle.[/quote]
The resulting is CS.;)
[quote="dyaka"]
We can see, that the maximum possible value of h occurs when SA=SB=SC[/quote]
Why?
 [quote="dyaka"]Then, <SAO = <SBO = <SCO = 45 deg. and a=b=c=h. So, h = (a+b+c)/3.[/quote]
This is obviously wrong:
If $a=b=c=h$ then $SA=SB=SC=h\sqrt2$ and $AB=AC=BC=2h.$
Hence, $O\in AB,$ $O\in AC$ and $O\in BC.$  :D
This problem is still unsolved. ;)

I used other notations: Take a coordinate system s.t. $S(0,0,0)$, $A(a,0,0)$ etc. My $a$ is different from the original $a$.

Well, it seems that it suffices to prove that \[ \sum \sqrt{a^2 - \frac{1}{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} \geq 3 \cdot \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}}  \]

(Computations  :D )

Next, \[ \, \Longleftrightarrow \, \, \, \sum a \sqrt{ \frac{1}{b^2} + \frac{1}{c^2} \right) \geq 3  }\]
\[ \Longleftrightarrow \, \, \, \sum \frac{a}{bc} \cdot \sqrt{ b^2 + c^2 } \geq 3  \]
\[ \Longleftrightarrow \, \, \, \sum a^2 \cdot \sqrt{ b^2 + c^2 } \geq 3abc  \]
Since $b^2+c^2 \geq 2bc$, let's prove that \[ \sum a^2 \cdot \sqrt{2bc} \geq 3abc  \]
\[ \Longleftrightarrow \, \, \, \sqrt 2 \sum \sqrt{a^3} \geq 3 \sqrt{abc}  \]
Also, $\sum \sqrt{a^3} \geq 3 \sqrt{abc}$, so it's done. (even a stronger statement) Or I'm terribly wrong?

It can be proven that $h=\frac{1}{\sqrt{{1\over a^2}+{1\over b^2}+{1\over c^2}}}$ (for example analytically, if we put the coordinate origin at the vertex of the pyramid, find the equation of the plane which the base belongs to, and then find the distance of the origin from that plane - or more tediously, but synthetically, if we find the sides of the base by Pythagoras', then the base area by Heron's, and then calculate the volume two-ways - once using the base area and $h$ and then using one lateral side and the remaining perpendicular edge). Then we have to prove

${\sqrt{{1\over a^2}+{1\over b^2}+{1\over c^2}}\geqslant \frac{3}{a+b+c}}$

which is self-evident by QM-HM: Put ${1\over a}=\alpha, {1\over b}=\beta, {1\over c}=\gamma$. Then $\sqrt{\alpha^2+\beta^2+\gamma^2}\geqslant \sqrt{\frac{\alpha^2+\beta^2+\gamma^2}{3}}\geqslant \frac{3}{{1\over\alpha}+{1\over\beta}+{1\over\gamma}}$ (Moreover, it seems that the inequality is strict, because of the left-hand part.)

Farenhajt, you made the same mistake as I did at first. $OA$ means the distance from $A$ to the foot of the perpendicular from $S$.

Also, here's how I'd phrase that problem:

[color=blue]Let $OABC$ be a pyramid such that $\measuredangle AOB = \measuredangle BOC = \measuredangle COA = 90^\circ$. Also, let $H \in (ABC)$ such that $OH \perp (ABC)$. Prove that \[ \frac{HA+HB+HC}{3} \geq OH .  \][/color]

Is my solution good, or not?

Oh, you're right  :blush:

It seems that your's is good, but for the misread problem (the same goes for mine).

My congratulations, perfect_radio!
I meant the following:
$AO+BO>AB.$ Hence, $a+b>\sqrt{2h^2+a^2+b^2}\Leftrightarrow h^2<ab.$
Similarly, $h^2<ac$ and $h^2<bc.$ Hence, $h<\sqrt[3]{abc}\leq\frac{a+b+c}{3}.$ :)

Indeed, that's very nice. Can you make that trick work with $3 \sqrt 2$ instead of $3$?


function
calculus
integration
geometry
geometric transformation
limit
real analysis

Again I need for some help : :wallbash_red:

Show that for $ f\in L^1(\mathbb{R}^n)$ and $ g\in L^1(\mathbb{R}^n)\cap L^{\infty}(\mathbb{R}^n)$
\[ f\ast g\in C(\mathbb{R}^n)\cap L^1(\mathbb{R}^n).\]

Basically, there are two separate claims being made here: that $ f*g$ is in $ L^1$ and that $ f*g$ is continuous.

For the first claim, that $ f*g\in L^1,$ we don't need to know that $ g$ is bounded; it's enough that both functions are integrable. This is the classic application of Fubini/Tonelli that occurs in most textbooks as soon as those theorems about interchanging order of integration have been presented. In fact, we learn that $ \|f*g\|_1\le \|f\|_1\|g\|_1.$

Now, the second claim is that $ f*g$ is continuous, and this time, we need the hypothesis that $ g\in L^{\infty}.$

Start with this lemma: Define the translation operator $ \tau_t$ by $ \tau_t f(x)=f(x-t).$ (This is for any ${ t\in \mathbb{R}^n}.$) Then $ \lim_{t\to 0}\|\tau_tf-f\|_1=0$ for any $ f\in L^1.$

The proof of the lemma proceeds via the approximation of $ f$ by a "nice" function. There are several variants of the proof depending on what you use for the nice functions. Step functions could be used, or continuous compactly supported functions could be used.

Now, we're going to use that lemma to prove your theorem, and we'll use that lemma for $ f,$ the function that isn't necessarily bounded. We'll also use that for any $ x,$ $ |u*v(x)|\le \|u\|_1\|v\|_{\infty}.$

Is that enough hint yet?


[quote="zserf"]My teammate and I rode them for a total of four or five hours. It was way too much fun. We convinced Wyoming to join us, then later I heard the Utah coach say her kids were riding the elevator.

Also after the banquet when 200+ people were waiting for the elevator me and a different member ran down the escalator and got in on the lobby, and we were the first ones up.
Some questions:

Did you see us
A.) at all?
B.) while juggling?
C.) when we were holding up signs pretending to be other states?
D.) when I was holding a newspaper mask over my face?
E.) when we were playing cards?
F.) when we were staring at the back of the elevator
G.) were you the one that also turned around?
H.) did we compliment you?
I.) were you the kids from Wyoming that rode the elevator with us?[/quote]
I saw the juggling, it was pretty good.

dan tobias had no part in this?

:rotfl:

jorian

Because you have a captive audience that's constantly changing and comes to you. It sounds stupid but it was really fun. Really, Really fun. plus you get to mess with peoples minds.

[quote="DarkKnight"]Dan Tobias is from Jorian's state and apparently he messed around with the elevator last year at nationals. Anyway, it's interesting that people always play with the elevators each year. What's really the point?[/quote]

Well, I'm like writing this document on why not to trust Dan Tobias, and these reasons are only helping (actually true). Anyways, I'm surprised math kids in the elevators don't like press certain buttons and do interesting functions with them...

There really is no point by the way.

Anyone know how he placed btw?

I remember playing elevator tag at state once with my school's team.  You basically need two or three elevators, not too many floors, and some teams.  You ride the elevators and try to catch the other team.  That was fun!

[quote]My teammate and I rode them for a total of four or five hours. It was way too much fun. We convinced Wyoming to join us, then later I heard the Utah coach say her kids were riding the elevator.

Also after the banquet when 200+ people were waiting for the elevator me and a different member ran down the escalator and got in on the lobby, and we were the first ones up.
Some questions:

Did you see us
A.) at all?
B.) while juggling?
C.) when we were holding up signs pretending to be other states?
D.) when I was holding a newspaper mask over my face?
E.) when we were playing cards?
F.) when we were staring at the back of the elevator
G.) were you the one that also turned around?
H.) did we compliment you?
I.) were you the kids from Wyoming that rode the elevator with us?[/quote]

Are you from Alaska?  I think I saw A, B, E, F, and H.

[quote="frost13"]We briefly met Dan Tobias.  He pronounced your name Horian which caught us off guard Jorian.  Is the J an H sound?[/quote]
Did he mention Jorian because you mentioned to Dan that there was someone on AOPS who worshipped him? Or did he bring Jorian up himself?

I saw B, but not in the elevator.

BTW I saw Dan Tobias in the elevator with his parents.  He wasn't messing around.

Oddly enough, it seems Arkansas was trying to pull something similar.  They offered me a normal state pin covered with a leaf- Limited Edition  :D

Coach frost beat me to the chase, leetspeak, I was going to send a pm later on...

[quote="13375P34K43V312"][quote="frost13"]We briefly met Dan Tobias.  He pronounced your name Horian which caught us off guard Jorian.  Is the J an H sound?[/quote]

 :rotfl:  :rotfl:  :rotfl:

oh man I did NOT actually think you were serious when you wanted to ask that[/quote]

Hey Carl, is it true that you took 22 minutes in that maze?

 :rotfl:  :rotfl:

Someone wrote "Carl Lian" "22 Minutes, Lots of Help" in the Wall of Shame board.

:rotfl:  :rotfl:  :rotfl:

Yeah he did!!!  It was quite fun to watch!

I wrote that(he was right there) did it still say that he tied for 13th?(he wrote that part)

[quote="frost13"]We briefly met Dan Tobias.  He pronounced your name Horian which caught us off guard Jorian.  Is the J an H sound?[/quote]

no, he's just annoying like that

(how did he just randomly tell you my name?)

it's pronounced your-ee-ahn (not with pauses though)

he probably didn't fool around cause his parents were there...

btw my name is dutch

jorian

Ugh, are you SERIOSLY this obsessed with Dan? You mention him in every other post, and are now weiting about him. I'd normally say something like this jokingly, but I'm serious.

huh "weiting about him"

i don't like the kid

jorian

p.s. he's just funny to discuss cause he's obnoxious and cocky

jorian, it's getting to the point that you are just asking us questions about dan that you probably already know the answer to or can easily find out the question to, just to mention him in every post

[quote="jhredsox"]huh "weiting about him"
[/quote]

How old are we?

To be honest, I don't know what word you meant by that either

umm...double the amount in ur sig.

when have i EVER said ANYTHING good about dan?

jorian

p.s. backup from klebian

and i think this is ridiculous guys

He mentioned writing about him before. writing. r is right next to the e. This is an entirely childish arguement, and I wouldn't have said any more had I not been correcting my own error.

Ok, I would like to end this saying that that was a joke about the writing thing (that might have thrown people off).

I actually don't have much affiliation AT ALL with dan. I don't see him on a regular day.

the j guy

To answer all your questions...

[quote="kstan013"]I saw all but CGH. I was the one who rode with you for a little bit from Wisconsin. Which one are you, zserf?[/quote]

I was the one with the longer hair, I probably wasn't the one juggling

[quote="uldivad9"]I only saw you juggling. How many people on your team juggle?[/quote]

Two of us juggle knew how to before nationals, and we taught the other two. Weather they keep it up or not is to be seen.

[quote="guyinPA"]Juggler was Very talented -- do you juggle more hours or practice math problems more?[/quote]

It varies, I don't go out of my way to do math a ton, but sometimes I get a problem I have to solve. My juggling amount varies also, but I have juggling balls now so I'll probably juggle more.

P.S. Can we stop the argument about whether or not Jorian is obsessed with Dan. It's pointless and off topic.


function
real analysis
real analysis unsolved

Let $ D \equal{} \{j/2^n \in [0,1]: n \equal{} 1, \cdots, j \equal{} 0, \cdots, 2^n\}$. Let $ f: D \rightarrow \mathbb{R}$ be the function satisfying the following property: there is a sequence $ \epsilon_n > 0$ with $ \sum_{n\equal{}1}^{\infty} \epsilon_n < \infty$ such that $ \vert f(\frac{j\minus{}1}{2^n}) \minus{} f(\frac{j}{2^n}) \vert < \epsilon_n$ for any $ n > 0$ and $ j \equal{} 1, \cdots, 2^n$.

Prove that, for any $ t_1, t_2 \in A$ such that $ \vert t_1 \minus{} t_2 \vert < \frac{1}{2^N}$, $ \vert f(t_1) \minus{} f(t_2) \vert \leq 2 \sum_N^{\infty} \epsilon_n$.

Still asking for solutions to homework questions?


inequalities
calculus
integration
logarithms
real analysis
real analysis unsolved

For positive reals x,y prove that
$\int^{x}_{0}e^{t}\ dt\int^{y}_{0}e^{t}\ dt\geq (\int^{\sqrt{xy}}_{0}e^{t}\ dt)^{2}$

Isn't it straightforward?

We have that:
(i) $\int_{0}^{x}e^{t}\, dt = \sqrt{\frac{x}{y}}\cdot \int_{0}^{\sqrt{xy}}e^{s \cdot \sqrt{\frac{x}{y}}}\, ds$;
(ii) $\int_{0}^{x}e^{t}\, dt = \sqrt{\frac{y}{x}}\cdot \int_{0}^{\sqrt{xy}}e^{s \cdot \sqrt{\frac{y}{x}}}\, ds$.
Now, by Cauchy and $ AM-GM$,
\begin{eqnarray*}{LHS &=& ( \int_{0}^{\sqrt{xy}}\exp ( s \cdot \sqrt{\frac{x}{y}}) \, ds ) \cdot ( \int_{0}^{\sqrt{xy}}\exp ( s \cdot \sqrt{\frac{y}{x}}) \, ds ) \\ \ &\geq& ( \int_{0}^{\sqrt{xy}}\sqrt{\exp ( s \cdot \sqrt{\frac{x}{y}}+s \cdot \sqrt{\frac{y}{x}}})}\, ds )^{2}\\ \ &\geq& RHS .\end{eqnarray*}

Easy generalization : if $f(t) > 0$ is    increasing and $\ln (f(t))$ is convexe , then for $x,y > 0$ :
$\int^{x}_{0}f(t)\ dt\int^{y}_{0}f(t)\ dt\geq (\int^{\sqrt{xy}}_{0}f(t)\ dt)^{2}$
:cool:


USACO has announced the schedule for their 2004-2005 internet contests. The first round will be the first week-end in Nov. The most noticeable change from last year is that there are now 4 divisions, so everyone from novice to expert prgrammer can compete. Languages include C/C++, Pascal, and Java. They have also recently expanded the interactive training pages to support all 3 languages.

USACO is is a great program! The contests and training material are all free. Anyone in the world can enter, even if your school is not interested. In addition to the opportunity for programming glory, you have a chance to win a free training camp in Colorado Springs.

http://www.usaco.org

[quote="h_s_potter2002"]You're already registered to take the contests if you're registered for the training pages, right?[/quote]

You'll automatically get notification of the contests if you are registered for the training pages. There's no separate advance registration for the contests. You just log in during the competition time window and go. Good luck!

[quote="themonster"]During the competition time frame, do we just log into the the training gate, or do we need to go to some other webpage?  Also, is the solution submission similar to that of the training page?[/quote]

you need to log into the contest gateway. The link is always included in the notices about the contests.

Yes, submission procedure is similar to the training page, although you won't get any feedback on how your program has performed (except for the sample input) until after the contest is over.

[quote="texas137"]The most noticeable change from last year is that there are now 4 divisions, so everyone from novice to expert prgrammer can compete.[/quote]

Wow, there's four of them now?  Didn't they just add the blue division last year?

Anyway, this is great news!  Everybody will be competing with programmers of approximately the same level!

[quote="countersixte"]Anyway, this is great news!  Everybody will be competing with programmers of approximately the same level![/quote]

That's right. So enter! And try to get your friends to enter! The bottom division is an hour long contest intended for teachers to give to their beginning programming students during CS classes. So tell teachers at your school.


function
analytic geometry
graphing lines
slope
calculus
derivative
algebra

I graphed the function $ x^x \equal{} y$ because I was bored. Although this function grows horrendously fast, I was surprised when the curve sloped [i]downward[/i] before rocketing out beyond my calculator's screen.

The strange shape of the curve led to a question: what is the minimum possible value of $ y$ for $ x > 0$?

P.S. It might help to know that $ x \approx 0.3679$ and $ y \approx 0.6922$. I figured this out with brute force.

Well, if it helps, the minimum is reached at $ x\equal{}\frac{1}{e}$. As for why, that is a discussion much too advanced for classroom math.

Well, let's look at derivatives:
We want to look at critical points: where the derivative is zero, and the endpoints of the interval.

Let's assume $ x \ge 0$ for this, because if it's not, we have the potential for complex numbers at non-integer values of $ x$, so restricting the domain makes sense here.

The endpoints: $ \lim_{x \to 0}$ and $ \lim_{x \to \infty}$
The limit in this first case is $ \lim_{x \to 0} = 1$
The limit in the infinite case is ${ \lim_{x \to \infty} = \infty}$
So we conjecture that the function has a minimum on its domain not achieved at the endpoints (from inspection).

The minimum must occur at a place where the first derivative is zero, or where
$ x^x ( 1 + \ln x) = 0$
This can happen when either $ x^x = 0$ (never) or $ (1 + \ln x) = 0$
$ \ln x = -1$
$ x = e^{-1}$.

To show this is a minimum, take the second derivative, and show that it's positive.
This is $ x^{x-1} + x^x (1 + \ln x)^2$ from logarithmic differentiation.
At $ x = e^{-1}$, this expression is equal to $ e^{1 - \frac{1}{e}}$, or
$ \frac{e}{\sqrt[e]{e}}$ which is positive.


function
logarithms
real analysis
real analysis unsolved

$P$ the set of prime numbers.
Compute an equivalent of:
$f(s)=\sum_{p \in P} \frac{1}{p^s}$ when $s$ tends to $1^+$

Wow! It was given this year? As far as I know it is not an easy problem.

not this year.( And there were no hints  :o )

[quote="harazi"]As far as I know it is not an easy problem.[/quote]
Hmmm... What's so difficult about it? :? Of course, one has to know Euler's formula $\prod_{p\in P} \frac 1{1-p^{-s}}=\sum_{n\in\mathbb N} n^{-s}$ ($s>1$) but from here on everything is completely straightforward: taking $\ln$ of both sides, we get $\sum_{p\in P} [p^{-s}+O(p^{-2s})]=\ln\left(\frac 1{s-1}+O(1)\right)$ and, finally, $\sum_{p\in P} p^{-s}=\ln\frac 1{s-1}+O(1)$.

Absolutely right, Fedja, I was wrong. The difficult problem I knew was to give an asymptotic developpement with 2 terms for the function.


quadratics conics parabola calculus function

bugmenot, it's like this:
 as a convention, we operate exponents first, then multiplication or division and finally
addition or subtraction. so
 $-5^2$ is simply $-(5^2)$
which is $-25$

however, $(-5)^2$ = $+25$
madatmath

[quote="bugmenot"]Which is it?

I know when I went to school, it was 25, but now it is -25 (at least that is what my wife, whom is a 7th and 8th grade teacher says the correct answer is now).

Why is this?[/quote]
The difference is that $-5^2$ is the negative of the square of 5. However, $(-5)^2$ is -5 times -5.

It depends whether you mean $\ -(5^2)$ or $\ (-5)^2$ which equal $\ -25$ and $\ 25$ respectively.

By order of operations, it is -(5^2) and is therefore -25 :D

wait... how can it be -25?  that would make [i]i[/i] useless and disprove the rule that a quadratic with a not equal to 0 is always a parabola   :?

this is confusing.  order of operations is a good argument but it has to be +25

thatd be pretty intersting if for all reals $a$ that $a=-a$ :D

unfortunately, it doesnt work like that.

$-25 = \sqrt{-25} \cdot \sqrt{-25} = i\sqrt{25} \cdot i\sqrt{25} = -25 \neq 25$

I guess this is really confusing. but without parenthesis -a^2 is negative unless a is imaginary.

:ninja: In a calculus, first we always do with the parenthesis, then we do with the powers,next we do with the multiplications and divisions, last we do with the subtractions additions.
So that: -5^2=-25

Math is fun, math is great, life is boring if we don't have math.

[quote="Ignite168"]It doesn't matter though as both will be right as -25 = 25

-25 = \sqrt{-25} * \sqrt{-25} = \sqrt{-25 * -25} = \sqrt{625} = 25 ;)[/quote]

$-25=25 ????$ Come on!

$\sqrt{a}\sqrt{b}=\sqrt{ab}$ only works if at least one is positive, according to my Pre-Calc teacher.

The heart of the issue is syntax, not logic.  Syntax is entirely a matter of convention.

In any case, you won't see an expression like $-5^{2}$ on, say, a test or in a mathematical paper.  Most mathematicians are much more precise with their language.

I believe it depends like this:

-(5)^2 or -5^2 cause -5 is a number not -(5). -_-

PERSON ABOVE IS ABSOLUTELY CORRECT!!!!!
-5 * -5= 25
but -(5*5)=25
for those who don't know why...

The real reason for this is as follows

$-25^{2}$
this doesn't have the coefficient for the negative
....$-1*25^{2}$

what i say to myself is the only time a negative number squared that equals a positive is when a number is subsituted for $x$ or when it's in $()$

for example

negative    $-7^{2}$
positive     $(-7)^{2}$
positive     $x^{2}$ ......even if x is negative
-jorian

[quote="Nplusnequals"]Think of it this way:

$x$ is like a box.

And by the fundamental theorem of algebra,

$x^{2}= 25$

must have two solutions, and it does: $\pm 5$[/quote]

true but it is more complicated...for example...there is a negative sign

[quote="bibobeo"]I'm not an English speaker, but I'll try.
Actually, I don't uinderstand why you argue about this problem.
It seems very easy, doesn't it?

-5^2 = -25, obviously.[/quote]

thats what we were saying some idiots thought we were wrong

btw how can it be 25% and 74% 25 +74 =99

[quote="Ihatepie"]

btw how can it be 25% and 74% 25 +74 =99[/quote]

Actually, it is 25.64% abd 74.36%

[quote="t0rajir0u"][quote="7h3.D3m0n.117"]that doesn't make sense..i thought order of opporations never said anything about negative signs so therefore the answer should be

$(-5)^{2}= 25$[/quote]

Adding a negative sign before a number is equivalent to multiplying it by $-1$, and squaring (exponentiation) takes precedence over multiplication.[/quote]

You do realise this means -1*-1 = --1 and there is no such thing as --1. Also, - does not mean -1*some number.  :wink:

[quote="i_like_pie"][quote="Necrowarrio0"][quote="t0rajir0u"]Adding a negative sign before a number is equivalent to multiplying it by $-1$, and squaring (exponentiation) takes precedence over multiplication.[/quote]

You do realise this means -1*-1 = --1 and there is no such thing as --1. Also, - does not mean -1*some number.  :wink:[/quote]
There is no such thing as $--1,$ but did he say there was? He said add the negative sign before the number, so if there already is a negative sign, make it positive.

And yes, a negative sign [b]does[/b] mean multiplying the number by $-1.$[/quote]

yeah so since exponents go before multiplication it is -5^2=-25

[quote="bugmenot"]Which is it?

I know when I went to school, it was 25, but now it is -25 (at least that is what my wife, whom is a 7th and 8th grade teacher says the correct answer is now).

Why is this?[/quote]

I don't understand what you are asking.

If you mean $(-5)^{2}$, then it is $25$. If you mean $-(5^{2})$, then it is $-25$.

I don't see why a simple problem needs to go through 3 pages of explanation.

Chances are that if you have a question, the answer is already here.


Is it possible to construct the triangle if the measure of angle A, the length of the bisector of angle A and the length of the median from the vertex A  to the opposite side are given?

kcefle

[quote="probability1.01"]What does it mean if you are given an angle A?  Does that mean you have two lines which intersect with that angle?[/quote]

Yes. Maybe I used the wrong word "measure" but of course we are not given the numerical value of angle A in degrees, grads etc. Lets say we are given two "rays" originating from A and therefore the angle BAC of the looked for triangle .

kcefle


symmetry
combinatorics unsolved
combinatorics

Show that the number of $3-$element subsets $\{ a , b, c \}$ of $\{ 1 , 2, 3, \ldots, 63 \}$ with $a+b +c < 95$ is less than the number of those with $a + b +c \geq 95.$

this post contained something that doesn't really help, so i took it off.

Can you please elaborate your solution? I haven't quite caught your 'catch' ?  :lol:  :D

well what i wrote is not certainly helpful in that regard directly but what you can do is for each set $\{a,b,c\}$ such that $a+b+c< 95$ consider the set $\{64-a,64-b,64-c\}$ and note that this set has a sum $>192-95=97$. And since you also have the sets of size $96$, the result follows.
is it clear now?

Since the average of all sets is 96 and we can use symmetry (eg. the set (1,2,3) can correspond to the set (63,62,61) and the set (1,2,4) can correspond to the set (63,62,60)) we see that for every set that takes on a value of 6+x there is a set that takes on a value of 186 - x. So the number of sets that have a value of 10 is equal to the number of sets that 182 and the number of sets that have a value of 11 is equal to the number of sets that have a value of 181. From this we see that the number of sets that have a sum of between 6 and 95 is equal the number of sets that have a sum of between 97 and 186. The conclusion easily follows as we subtract the number of sets with a sum of 95 from the former and add that to the latter and all the number of sets with a sum of 96 to the latter.

This result can be generalised as follows:
The number of 3 element subsets (a,b,c) of (1,2,...,n) with
a+b+c < $n+ (n+1)/2$ is less than the no. of subsets with a+b+c > or = $n + (n+1)/2$
when n is odd. A similar result is true when n is even. In fact, we can generalise to k element subsets but we need to consider separately the cases when n= 1, 2, ... k-2, 0 mod k-1


function
calculus
derivative
set theory
inequalities unsolved
inequalities

i attach a paper for your refernce, i think the proof of therom 1.1 is wrong, especially in the last step, the reason is not enought for its correction. Do you think so ????????????

Without all the Set theory Mumbo Jumbo, he essentially just abused the fact that if $ f'(x) > 0$ on some interval $ [a,b]$, then it's minimum will occur at $ x \equal{} a$.

We first consider a function of $ x_1$, so since it's first partial derivative is greater than 0, it will achieve it's minimum when it equals the minimal possible value it can attain, $ c$. Then just do the same for $ x_2$, $ x_3$, ... , $ x_{n \minus{} 1}$.

Everything else he said was just to justify that first sentence I wrote for multivariable functions, which is obvious, since we are only considering one of the variables, so when considering the derivative, all other variables are treated as constants.

Dear rofler:
The auother  said f'([b]x[/b])> 0  on D_i (i=1,2,---,n-1) ,no on [a,b]^(n-1). Please read the paper exactly. :)


Hi.  I am running for vice-prez of CTC.  Do you think it's a good idea and will it stop the tourney mania?  This is just a discussion thread.  No spamming.

NIce double thread!

I believe that CTC was the dumbest idea ever, and its creator should be wiped off the planet!

That was sarcasm.

=)
we all know it wuz
=)
i think CTC would be good
but if we allow mewto to run for prez, then he'll definitely win
i think mewto should be dictator and he gets that position for life
then someone else could run for prez
like me!

i dont think ctc is a good idea

wuts the point? if people want to make a BUNCh of tourneys who cares?

none of them ever finish, which makes it pointless

I'm neutral.

At CTC thread, I've posted the "dominoes effect" CTC provoked on this forum. So, I think that mewto55555 is smart, but evil. But, I'm neutral :|

its a BAD idea

i joined anyway
 :rotfl:  :rotfl:  :rotfl:

Smart but evil...Good description of me! :D  :D  :D

[quote="isabella2296"]I'm undecided.

On a good note - it's great that CTC attempts to control the tournies, since there are so many.

On a bad note - I think the tourney fuss will die down soon, and after that CTC will be completely unnecessary.[/quote]

Actually, the tourneys may not continue.  CTC is just trying to stop the tourney mania before it starts up again.

[quote="vallon22"]maybe we can set a limit to how many tournies can be going at any one time
anyone who makes one that exceeds that amount, well nobody joins, and that tourny dies down[/quote]

Yeah... but you think [b]everyone[/b] will listen and/or care?  I think not.

[quote="vallon22"]it's a start
i doubt everyone will follow it, but hopefully most pplz will
also, we can set a limit to how many tounies one person can join, and if anyone exceeds that, he'll get disqualified from all the tournies he/she is signed up in[/quote]

That might work.  However, it may cause more chaos.

[quote="vallon22"]it may or may not,
it depends
we'll see[/quote]

And that's the chaos I'm talking about.  Some guy trying to stop this from happening. [color=white][size=59]idiot.[/size][/color]

[quote="AIME15"]Instead of ignoring rule-breakers, wy not ignore CTC?

[color=blue]* AIME15 turns away from mewto55555[/color][/quote]

Are you trying to cause more tourneys to occur?  If so, then we dont really want you to influence other people.  Just because you don't like it doesn't mean you have to change a whole slew of people's mindsets.  Thank you. (or not.)

[quote="isabella2296"]I really think that we shouldn't limit tourneys in the first place. Who cares if we have tons of them? Does it really matter if most of them don't get finished?[/quote]

What's the point of a tourney?

[quote="mewto55555"]Why don't they keep the Wimbledon and the Australian Open at the same time?[/quote]

I pulled this from another thread.  Now why don't you answer that question?

[quote="iin77"]
Tournies are worthless in the first place.

Just play FTW.

[/quote]

Meh.

Tournies are fun. Who cares if they're worthless?

[quote="iin77"]CTC is worthless.

Tournies are worthless in the first place.

Just play FTW.

People get overly excited when something new starts, then after a couple weeks they don't care any more.

Oh, and the answer to that question, ernie, is:
1.) During June, it's winter in Australia.  During January, it's winter in England.
2.) Tennis is really, really taxing.  They would have to choose one tournament or the other to play in.

As FTW has nothing to do with the seasons and is not taxing, this question proves nothing in regards to the need for the CTC.

I win.[/quote]

It wasn't my question, however, it was mewto's.

they're only fun if they FINISH (cough cough)

[quote="vallon22"]doesn't have to be
you can play pplz you might not have played otherwise[/quote]

It adds to the experience, however, when you finish a tourney.  More games means more chances to play new people.

i'm not sure, but i think that's basically what i said....

I don't think CTC is a good idea. iin77 wins.

Now it seems that EVERYONE is joining CTC...

fishy, good point

i joined!

im not gonna do anything thouh

lol
ya, join just for the sake of it
that's what most pplz do in tournies, BTW

This CTC idea has extinguished my last remaining bit of faith in AoPS members...


Hi everybody,

I am using TeXnicCenter and MikTeX on a WinXP platform. Currently I am trying to work with the package PSTricks and I read somewhere that I should use ps4pdf then. But even this very easy example does not work:

[code]
\documentclass[pdftex,a4paper]{scrreprt}
\usepackage[T1]{fontenc}
\usepackage{graphicx}
\usepackage{pstricks}
\usepackage{pst-plot}
\usepackage{ps4pdf}

\PSforPDF{%
\usepackage{pstricks}
\usepackage{pst-plot}%
}

\begin{document}

\PSforPDF{%
\begin{pspicture}(-5,-5)(5,5)
\rput(0,0){\psovalbox{graphic}}
\end{pspicture}%
}

\end{document}
[/code]

I get two warnings:
[quote]
Package ps4pdf Warning: Graphic container `test-pics.pdf´ not found.
Package ps4pdf Warning: `test-pics.pdf´ is no more valid.
[/quote]

The output pdf only shows:
[quote]
Can’t show picture #1 because test-pics.pdf not found. Create it from test.dvi using dvips and ps2pdf!
[/quote]

Can anybody help me?
Thanks,
greetz,
Sabine

I cannot help it wonder why would you need to attach a pdf file in a .tex file  :?

Why not add directly pictures?

Hi Valentin,

thanks for you answer!

I'll try to explain my problem a bit more. I am writing my master thesis and it is the first time I use Latex. I will have to create some diagrams such as flow charts, gant charts etc and this is why I want to use PSTricks.
The task of ps4pdf is to create from each of the commands of PSTricks a page in a pdf document and insert it at the correct position. This is the file test-pics.pdf that should be created but somehow it is not.
In TeXnicCenter I use the option LaTeX=>PDF and without the ps4pdf I get the error "undefined control sequence" several times.

As I said it is the first time I work with LaTeX so maybe I am doing it too complicated or misunderstood somthing... But I do not know how I could create the diagrams I need and add them as pictures.

If you just want to insert pictures (.png) you can use the
\includegraphics[height=xx in, width=xx in]{picture.png}
command when xx is the numbers, and "in" is the measure unit. You can leave those out.

[quote="Biene227"]Problem is: I do not know how I could create such diagrams (flowcharts etc.). Do you know a good program to do that? It seems for windows there are no free programs...[/quote]But how do you make them a .pdf file then ?!

[quote="Biene227"]Problem is: I do not know how I could create such diagrams (flowcharts etc.). Do you know a good program to do that? It seems for windows there are no free programs...[/quote]
The xypic package for LaTeX is probably pretty good for flow charts.

You'll have to search on the web to find documentation and installation instructions.

Thanks, the xypic package is indeed a good solution! I also did a few charts with WinFig (the windows version of XFig) which is also quite nice!

About PSTricks I figured out that when I leave the ps4pdf stuff and choose LaTeX=>PS, I can then open the .ps file with gsview and convert it to pdf. So, in a bit more complicated way, I finally get what I wanted  :)

Thanks very much for your help!!!

[quote="ComplexZeta"]What if you do LaTeX => PS => PDF? (That's the bottom item in that pull-down menu.)[/quote]

My code is now:
[code]\documentclass[a4paper]{scrreprt}
\usepackage[T1]{fontenc}
\usepackage{pstricks}
\usepackage{pst-plot}

\begin{document}

\begin{pspicture}(-5,-5)(5,5)
\rput(0,0){\psovalbox{graphic}}
\end{pspicture}

\end{document}[/code]
i.e. I left the pdflatex option in the document class and all the ps4pdf stuff.
When I choose LaTeX=>PS=>PDF, it does not create a pdf file. Pushing F5 will yield the error message that the file could not be found.
With the pdflatex option, I get a warning, that the global option is not used.

So I will have to use LaTeX=>PS and convert the .ps file to .pdf.


algebra
polynomial
superior algebra
superior algebra unsolved

We have generalized Legendre symbol for finite field $ F_q$ with $ q$ odd:

$ \left(\frac {x}{F_q}\right)$, which equals $ 1$ if $ t^2 \equal{} x$ has a solution $ t\in F_q^*$

 equals $ \minus{} 1$ if $ t^2 \equal{} x$ has no solution $ t\in F_q$

and equals $ 0$ if $ x\equal{}0$.

Show that:

$ \left(\frac {x}{F_q}\right) \equal{} x^{{q \minus{} 1}/2}$

You mean $ x^{(q\minus{}1)/2}$?  Well, if $ y \equal{} x^{(q\minus{}1)/2}$, what do you know about $ y^2$?  So what are the possible values for $ y$?

Can you show that $ y\equal{}1$ if $ x\equal{}t^2$?  How many $ x$ satisfy this condition?  How many roots are there of the polynomial $ X^{(q\minus{}1)/2} \minus{} 1$?

Let us know which of these questions you're having trouble with.

Ok, I think it could be like this:

If $ x=0$ then it is ok. Let $ x\neq 0$.

If $ \left(\frac{x}{F_{q}}\right) =1$ then it is ok: $ x=t^2$ for some $ t$, and $ x^{(q-1)/2}=t^{q-1}=1$.

If $ \left(\frac{x}{F_{q}}\right) =- 1$ then assuming $ x^{(q-1)/2}=1$  ( $ x^{(q-1)/2}$ can only be $ 1$ or $ -1$ )

lets write $ x=g^j$, where g is a generator of $ F_q^*$.

We have ${ g^{j(q-1)/2}=1}$ then $ j(q-1)/2=k(q-1)$ for some $ k$. That means $ j=2k$ and we have a contradiction

with $ \left(\frac{x}{F_{q}}\right) =- 1$.

Correct ?


Let $a,b,c$ be positive real numbers and
\[(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16.\]
Find the minimum and maximum of
\[P=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}.\]

We can replace 16 by $k$ but I like the number $16$.

[quote="Ji Chen"]Thanks for your gift.
If $x,y,z$ be positive real numbers, let $P=\frac{x}{y}+\frac{y}{z}+\frac{z}{x},Q=(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$, then
$Q\geq 16\Rightarrow P\geq \frac{13-\sqrt{5}}{2}$ and $Q\leq 16\Rightarrow P\leq \frac{13+\sqrt{5}}{2}$.[/quote]
Can you prove it? Thank you!

Yes, Jichen. Good Job! It's also my result. When my brother gave me this one, it takes me hours to solve. You seems to solve it so fast.


for any triangle prove that
$ 1/a^2 \plus{} 1/b^2 \plus{} 1/c^2\leq \frac{1}{4r^2}$

$ \frac {1}{a^{2}} + \frac {1}{b^{2}} + \frac {1}{c^{2}}\leq \frac {1}{4r^{2}}$

$ a = x + y,b = y + z,c = x + z$  $ \implies$

$ (x + y + z)r = \sqrt {(x + y + z)xyz}$

$ \implies$  $ r^{2} = \frac {xyz}{x + y + z}$

The rest is easy with Muirhead's inequality (but my solution is very long)

But $ \frac {1}{(x + y)^{2}}\leq \frac {1}{4xy}$

$ \implies$  $ \sum_{cyc}{\frac {1}{(x + y)^{2}}\leq \sum _{cyc}{\frac {1}{4xy}} = \frac {x + y + z}{4xyz} = \frac {1}{4r^{2}}}$  :D

omg i was expecting a solution with geometrical tricks :S with substitution it becomes really easy :) thanks


I don't understand . Can you post your solution again ?  :(  :(


Evaluate $ {2^{\log_{4} 108}}/{2^{\log_{4} 3}}$. (Intermediate Algebra, pp. 433)

Please state any laws of logarithms you're using because I just learnt them.

:roll: how did you do that ? can you explain please
i dont think you are right in every step, so the answer might be wrong but could you explain please?

Thanks.

Isn't it right?

Maybe you would like to review the [url=http://www.artofproblemsolving.com/Wiki/index.php/Logarithms#Logarithmic_Properties]properties of logarithms[/url][/url]

Typically, this kind of problem has nice integer answers.

Anyway, the question asks you to evaluate, so you must give the numerical value anyway.


Find the solution to $ x|x| \equal{} 2x\plus{}1$ which has the smallest value.

I brute forced this question, to get $ \boxed{\minus{}1}$.

Is there an easier way to do it without guess and check? And is it necessarily faster?

Rearranging, we have $ \minus{} 1 \equal{} x(2 \minus{} |x|)$. It is fairly easy to see that $ x \equal{} \pm1$ (since $ \minus{} 1(1) \equal{} \minus{} 1$ and negative reciprocals other than that can't be made with the expression). We then note that for $ \pm1$, $ 2 \minus{} |x|$ will be positive because of the absolute value. So $ x \equal{} \boxed{ \minus{} 1}$.

Not sure if this is faster ...

I think, for this problem, there are two cases. One is where the absolute value thingy is positive, in which case you can replace it with x^2. The other is if it is negative. You can replace it with -x^2. Thus, you get a piece-wise function.
We all know what happens if x is zero.
If x>0, you have x^2=2x+1
If x<0, you have -x^2=2x+1
For x >0, you have -x^2+2x+1=0, or x^2-2x-1=0
R.I.P... You can't factor this. That doesn't matter, though, because you want the smallest value! :)
Anyway, for x<0,
You get x^2+2x+1=0 This factors out to (x+1)(x+1)=0 The only solution is -1, which fits the first requirement, so it is our answer. This is just a more general solution, and it applies to graphing these types of things.


1. Find all solutions to $ log x\equal{}ln x$.

2. Find all solutions to $ log x\equal{}e^{x}$.

3. Find all solutions to $ sin x\equal{}sin^{\minus{}1} x$.

[hide="Number 1"]The obvious solution is $ x\equal{}1$. Other than that, some algebra yields:
\[ \log{x}\equal{}\ln{x}\quad\longrightarrow\quad\log{x}\equal{}\frac{\log{x}}{\log{e}}\quad\longrightarrow\quad\log{e}\equal{}1\]Since this isn't true, the only solution is the aforementioned one, $ x\equal{}1$. This can be confirmed graphically.[/hide]

[hide="Number 2"]Simply looking at the graphs of the two functions tells us that there is no solution since they do not intersect anywhere.[/hide]

[hide="Number 1"]$ \frac{\ln x}{\ln 10}\equal{}\ln x$
$ (\ln x)\left(\frac{1}{\ln 10}\minus{}1\right)\equal{}0$
$ \ln x\equal{}0\implies \boxed{x\equal{}1}$.[/hide]

[hide="Number 2"]No solutions; graph it.[/hide]

[hide="Number 3"]$ \sin\sin x\equal{}x\implies \boxed{x\equal{}0}$. That's the only solution.[/hide]

Wow? There actually aren't any other solutions!?!?!? I was thinking there was some weird special value out there that coincidentally satisfies both equations...


trigonometry
geometry
circumcircle
function
inequalities
trapezoid
triangle inequality

1.given a triangular EBD,C lies in ED,A lies in EB,∠DAB=∠DCB=90°，cosB+cosD=1,prove:the bisector of ∠BAC and the bisector of ∠DCA meets in BD.
2.(hard)ABCDE lies in line L,AC/CE=BC/CD.F is a point which doesn't belong to L,connect FC,G lies in FC,and satisfies ∠FAC=∠AGD,∠FEC=∠EGB.prove:∠FAC=∠FEC

one of my fellow students told me a nice solution to the second problem
any one solved it?have a try

[quote="bye"]1.given a triangular EBD,C lies in ED,A lies in EB,∠DAB=∠DCB=90°，cosB+cosD=1,prove:the bisector of ∠BAC and the bisector of ∠DCA meets in BD.
[/quote]
Wrong!

(1) $ \cos B \plus{} \cos D \equal{} 1$ $ \Longrightarrow$ $ \angle B, \angle D$ are acute. From $ BC \perp DC, DA \perp BA$ $ \Longrightarrow$ $ BD$ is antiparallel of $ AC$ WRT $ \angle CEA$ and $ \overline{AB} \plus{} \overline{DC} \equal{} \overline{BD} (\cos B \plus{} \cos D) \equal{} \overline{BD}.$

Let internal bisectors of $ \angle CAB, \angle DCA$ meet at $ J,$ the E-excenter of $ \triangle EAC.$ Let circumcircle $ (K)$ of $ \triangle JCA$ cut the lines $ EA \equiv EB, EC \equiv ED$ again at $ A_0, C_0.$ $ A_0C_0 \parallel BD,$ both antiparallel to $ AC$ WRT $ \angle CEA.$ Let parallel to $ A_0C_0 \parallel BD$ through $ J$ cut $ EB, ED$ again at $ B_0, D_0.$ $ \angle A_0B_0D_0 \equal{} \angle B$ and $ \angle JA_0B_0 \equal{} \angle JCA \equal{} \frac{_1}{^2} (180^\circ \minus{} \angle B)$ $ \Longrightarrow$ $ \triangle JA_0B_0$ is isosceles with $ \overline{A_0B_0} \equal{} \overline{B_0J}.$ Similarly, $ \triangle JC_0D_0$ is isosceles with $ \overline{JD_0} \equal{} \overline{D_0C_0},$ so that $ \overline{A_0B_0} \plus{} \overline{D_0C_0} \equal{} \overline{B_0D_0}.$ Since $ K \in EJ,$ $ \overline{A_0A} \equal{} \overline{C_0C},$ and $ \overline{AB_0} \plus{} \overline{D_0C} \equal{} \overline{B_0D_0}$ as well.

For arbitrary parallel $ B'D' \parallel BD$ (where $ B' \in EB, D' \in ED$), both $ \overline{AB'} \plus{} \overline{D'C}$ and $ \overline{B'D'}$ are linear functions of $ \overline{AB'}.$ For $ B' \equiv A_0,$ $ D' \equiv C_0,$ $ \overline{AA_0} \plus{} \overline{C_0C} \equal{} 0 < \overline{A_0C_0}.$ For $ B' \rightarrow \infty,$ $ D' \rightarrow \infty,$ $ \frac{\overline{AB'} \plus{} \overline{D'C}} {\overline{EB'} \plus{} \overline{D'E}} \longrightarrow 1 > \frac{\overline{B'D'}}{\overline{EB'} \plus{} \overline{D'E}} \equal{} \text{const}$ by triangle inequality $ \Longrightarrow$ $ \overline{AB'} \plus{} \overline{D'C} > \overline{B'D'}.$ Therefore, equality occurs for unique $ B' \equiv B \equiv B_0,$ hence $ BD \equiv B_0D_0$ and $ J \in BD.$

[quote="yetti"](1) $ \cos B \plus{} \cos D \equal{} 1$ $ \Longrightarrow$ $ \angle B, \angle D$ are acute. From $ BC \perp DC, DA \perp BA$ $ \Longrightarrow$ $ BD$ is antiparallel of $ AC$ WRT $ \angle CEA$ and $ \overline{AB} \plus{} \overline{DC} \equal{} \overline{BD} (\cos B \plus{} \cos D) \equal{} \overline{BD}.$

Let internal bisectors of $ \angle CAB, \angle DCA$ meet at $ J,$ the E-excenter of $ \triangle EAC.$ Let circumcircle $ (K)$ of $ \triangle JCA$ cut the lines $ EA \equiv EB, EC \equiv ED$ again at $ A_0, C_0.$ $ A_0C_0 \parallel BD,$ both antiparallel to $ AC$ WRT $ \angle CEA.$ Let parallel to $ A_0C_0 \parallel BD$ through $ J$ cut $ EB, ED$ again at $ B_0, D_0.$ $ \angle A_0B_0D_0 \equal{} \angle B$ and $ \angle JA_0B_0 \equal{} \angle JCA \equal{} \frac {_1}{^2} (180^\circ \minus{} \angle B)$ $ \Longrightarrow$ $ \triangle JA_0B_0$ is isosceles with $ \overline{A_0B_0} \equal{} \overline{B_0J}.$ Similarly, $ \triangle JC_0D_0$ is isosceles with $ \overline{JD_0} \equal{} \overline{D_0C_0},$ so that $ \overline{A_0B_0} \plus{} \overline{D_0C_0} \equal{} \overline{B_0D_0}.$ Since $ K \in EJ,$ $ \overline{A_0A} \equal{} \overline{C_0C},$ and $ \overline{AB_0} \plus{} \overline{D_0C} \equal{} \overline{B_0D_0}$ as well.

For arbitrary parallel $ B'D' \parallel BD$ (where $ B' \in EB, D' \in ED$), both $ \overline{AB'} \plus{} \overline{D'C}$ and $ \overline{B'D'}$ are linear functions of $ \overline{AB'}.$ For $ B' \equiv A_0,$ $ D' \equiv C_0,$ $ \overline{AA_0} \plus{} \overline{C_0C} \equal{} 0 < \overline{A_0C_0}.$ For $ B' \rightarrow \infty,$ $ D' \rightarrow \infty,$ $ \frac {\overline{AB'} \plus{} \overline{D'C}} {\overline{EB'} \plus{} \overline{D'E}} \longrightarrow 1 > \frac {\overline{B'D'}}{\overline{EB'} \plus{} \overline{D'E}} \equal{} \text{const}$ by triangle inequality $ \Longrightarrow$ $ \overline{AB'} \plus{} \overline{D'C} > \overline{B'D'}.$ Therefore, equality occurs for unique $ B' \equiv B \equiv B_0,$ hence $ BD \equiv B_0D_0$ and $ J \in BD.$[/quote]
Not good proof! Is there exists a nice one?

There is nothing wrong with the proof, perhaps it is too elementary. You really should try a different approach yourself, instead of just saying that you do not like it.

$ BC \perp DC,$ $ DA \perp BA$ $ \Longrightarrow$ $ ABDC$ is cyclic, with a circumcircle $ (O).$
$ \cos B \plus{} \cos D \equal{} 1$ $ \Longrightarrow$ $ \overline{AB} \plus{} \overline{DC} \equal{} \overline{BD}.$
Let $ J$ be intersection of the internal bisectors of $ \angle CAB, \angle DCA.$
Let $ P \in BD$ be such that $ \overline{BP} \equal{} \overline{AB},$ then also $ \overline {PD} \equal{} \overline{DC}$ $ \Longrightarrow$ $ \triangle PAB, \triangle PCD$ are both isosceles.
Let $ AJ, CJ, AP, CP$ cut $ (O)$ again at $ U, V, X, Y,$ respectively.
$ \angle UAB \equal{} \frac {_1}{^2}(180^\circ \minus{} \angle BDC) \equal{} \frac {_1}{^2}(180^\circ \minus{} \angle PDC) \equal{} \angle DCP \equal{} \angle DCY$ $ \Longrightarrow$ $ UB \equal{} DY$
$ \Longrightarrow$ $ BYUD$ is isosceles trapezoid and $ UY \parallel BD.$ Similarly, $ VX \parallel BD$ $ \Longrightarrow$ $ UY \parallel VX.$
By Pascal theorem for cyclic hexagon $ AVYUXC,$
the intersections $ J \equiv AU \cap CV,$ $ P \equiv AX \cap CY,$ $ Q \equiv UY \cap VX$ are collinear.
Since $ UY \parallel VX,$ $ Q$ is at infinity and $ JP \parallel UY \parallel VX \parallel BD.$
Since $ P \in BD,$ it follows that $ J \in BD$ as well.

$ BDCA$ is a cyclic quadrilateral (or let's call it semicyclic ?), with BD as the circumcircles diameter, and the problem essentially asks to prove that $ cos B + cos D = 1$ is a sufficient condition to the center, $ O$ , of the circunference arc inscribed to the open polygonal  $ BACD$ , be situated in  $ BD$   ($ O\in BD$).
Maybe it would be interesting and simpler (not "elementary" in yettis sense...) to show that the condition is also necessary.
We can easily find $ \overline{AC} = - \overline{BD}. cos ( B + D )$, for instance , using Ptolomeus Theorem.
 So we can write that twice the quadrilateral area, $ S$ is  \[ 2S = {{\overline{BD}^2\over2}  [ sin B . cos B + sin D . cos D - cos ( B + D ) . sin ( B + D )] =\overline{BD}^2 . sin ( B + D ) . sin B . sin D   \;\;(*)}\]
In order to  $ O\in BD$ , we must have the same area to be given by
$ 2S =  \overline{BD}. [ cos B + cos D - cos (B + D )].r$
where the circunferences arc radius  is $ r = \overline{BD}.{{sin B . sin D} \over { sin B + sin D }}$

So the last expression for 2S becomes \[ 2S = \overline{BD}^2. [cos B + cos D - cos ( B + D ) ]. {{sin B . sin D}\over{ 2. sin {{B+D}\over2} . cos { {B - D}\over2}}} (**)\]
Equating (*) (slightly modified) to (**) and simplifying,

\[ cos B + cos D - 1 + 2 . {sin}^2 {{B+D}\over 2} = sin ( B + D ) . 2 . sin {{B+D}\over2} . cos {{B-D}\over2}\]

Rearranging and multiplying by $ 2.cos^2 {{B+D}\over 2}$ we get

\[ ( cos B + cos D - 1).[1 + cos ( B + D)] = ( cos B + cos D - 1) . sin^2 ( B + D )\]

whose unique geometric valid solution is $ cos B + cos D - 1 = 0 \,\,$  ; the other algebraic ones leads to $ B + D = \pi$ and $ B + D = 0 \;\;$.


geometry
number theory
superior algebra
superior algebra unsolved

let p be a prime with $p=3$ mod $4$
suppose a is in $\mathbb{F}_{p}$ with a nonzero

how many solutions does the equation in  $\mathbb{F}_{p}$ have :

$y^2=x(x-a)(x+a)$

The answer should be p

I was confused where to post this, do you regard this as algebra or number theory?
It is related apparently to finding rational triangles of certain area

Please help, I am very interested

apparently proving this is the same as proving

in $\mathbb{F}_{p}$ there are p-3 solutions of $u^2+a=v^2=w^2-a$

does that help?

If you disregard the solutions $(0,0),(\pm a,0)$ and you identify the solutions with the same $y^2$, you're left with trying to show that $\left(\frac{x(x-a)(x+a)}p\right)=1\ (*)$ has $\frac{p-3}2$ solutions different from $0,\pm a$. This is obvious, since $(*)$ holds iff $\left(\frac{(-x)(-x-a)(-x+a)}p\right)=-1$, so you can partition the elements of $\mathbb F_p\setminus\{0,\pm a\}$ into pairs of elements $x,-x$, and in each pair exactly one satisfies $(*)$.

Am I missing something here? :?

no
this is nice
thank you
do you know more about this, for other primes

 :blush: it was so easy??


Given an integer polynomial in x with degree three such that x^3 coefficient is 1, prove that this polynomial is not a square for infinitely many positive integer values of x

[quote="fourierseries"]Consider $x^3$

This takes on infinitely many square values.[/quote]

Yes, but it takes on infinitely NON-SQUARE values too.

Basically you need to prove that there exist infinitely many $x \in \mathbb{Z}$ such that $p(x) = x^3 + ax^2 + bx + c$ is not a perfect square, you see?