Find all positive integers $ a$ , such that $ 0 \le a \le 20$ , and $ a^5\plus{}9$ is a muliple of  $ 5$.

Nope, that's it.  (This problem doesn't really belong in Pre-Olympiad.)

[quote="mathemagician1729"]at a glance i found 1 11 6 and 16 theres probably more though...[/quote]
Have you a proof please?

All you have to do is check the values 0,1,2,3,4. For the ones of these that work, adding any multiple of 5 will give another valid value of a.

One way to see that a=5k+1 works is if you notice that $ a^5\plus{}9 \equal{} 10\plus{}(a\minus{}1)(a^4\plus{}a^3\plus{}a^2\plus{}a\plus{}1)$ is divisible by $ 5$ if $ a\minus{}1$ is divisible by 5.

The reverse direction is [url=http://en.wikipedia.org/wiki/Fermat%27s_little_theorem]Fermat's little theorem[/url] for $ p \equal{} 5$.

anyway here is my solution (simple) :

we can easily check that $ a^5 \minus{} a$ is divisible by $ 5$,

so we have:

$ a^5 \minus{} a \equal{} 5b$

and

$ a^5 \plus{} 9 \equal{} 5c$

so

$ a \equal{} 5(c \minus{} b \minus{} 1) \minus{} 4 \equal{} 5k \minus{} 4$

$ 0 \le 5k \minus{} 4 \le 20;$

$ k \equal{} 1,2,3,4$ then the possible values of $ a$ are$ : 1,6,11,16.$  :)


calculus limit function trigonometry derivative inequalities conics

[color=blue]Problem 1[/color] 2005 Kokushikan University/Engineering

Find the limit : $ \lim_{x \rightarrow \minus{}4} \frac{x^3\plus{}4x^2\plus{}2x\plus{}8}{x^2\plus{}x\minus{}12}$

[quote="kunny"][color=blue]Problem 1[/color] 2005 Kokushikan University/Engineering

Find the limit : $ \lim_{x \rightarrow \minus{} 4} \frac {x^3 \plus{} 4x^2 \plus{} 2x \plus{} 8}{x^2 \plus{} x \minus{} 12}$[/quote]
$ \lim_{x \rightarrow \minus{} 4} \frac {x^3 \plus{} 4x^2 \plus{} 2x \plus{} 8}{x^2 \plus{} x \minus{} 12} \equal{} \lim_{x \rightarrow \minus{} 4} \frac {x^2 \plus{} 2}{x \minus{} 3} \equal{} \frac {18}{\minus{}7}$

That's correct. :lol:

[color=blue]Problem 2[/color] 2005 Kwansei University/Enginnering

For a function $ f(x)$, express $ \lim_{h \rightarrow 0} \frac{f(a\plus{}3h)\minus{}f(a)}{h}$ in terms of $ f'(a)$.

[quote="kunny"]That's correct. :lol:

[color=blue]Problem 2[/color] 2005 Kwansei University/Enginnering

For a function $ f(x)$, express $ \lim_{h \rightarrow 0} \frac {f(a \plus{} 3h) \minus{} f(a)}{h}$ in terms of $ f'(a)$.[/quote]
 :lol:

$ \lim_{h \rightarrow 0} \frac {f(a \plus{} 3h) \minus{} f(a)}{h} \equal{} \lim_{3h \rightarrow 0} 3.\frac {f(a \plus{} 3h) \minus{} f(a)}{3h}\equal{}3f'(a)$

That's correct.  :lol:

[color=blue]Problem 3[/color] 2005 Syonan Institute of Technology

Find the values of $ a,\ b$ such that both two curves of $ y\equal{}x^3\plus{}ax$ and $ y\equal{}bx^2\plus{}c$ pass the point $ (\minus{}1,0)$ and have a common tangent line at the point.

[quote="kunny"]That's correct.  :lol:

[color=blue]Problem 3[/color] 2005 Syonan Institute of Technology

Find the values of $ a, b, c$ such that both two curves of $ y_{1} \equal{} x^3 \plus{} ax$ and $ y_{2} \equal{} bx^2 \plus{} c$ pass the point $ ( \minus{} 1,0)$ and have a common tangent line at the point.[/quote] (Edited slightly for my own purposes)

At $ (\minus{}1,0)$ we have $ y_{1} \equal{} 0 \equal{} (\minus{}1)^{3}\plus{}a(\minus{}1) \equal{} \minus{}1\minus{}a$ hence $ a\equal{}\minus{}1$.
We also have $ y_{2} \equal{} 0 \equal{} b(\minus{}1)^{2}\plus{}c \equal{} b\plus{}c$ hence $ b\plus{}c\equal{}0$.

The tangents are equal so we have $ y'_{1} \equal{} \frac{\mathrm{d}y_{1}}{\mathrm{d}x} \equal{} 3x^{2}\plus{}a \equal{} 3x^{2}\minus{}1$ and $ y'_{2} \equal{} \frac{\mathrm{d}y_{2}}{\mathrm{d}x} \equal{} 2bx$. At the point $ (\minus{}1,0)$, then, we have:
$ y'_{1} \equal{} 3(\minus{}1)^2\minus{}1 \equal{} 2$ and $ y'_{2} \equal{} 2b(\minus{}1) \equal{} \minus{}2b$ and this $ \equal{}2$, from $ y'_{1}$ so $ b\equal{}\minus{}1$ which means that we have:
$ a\equal{}\minus{}1, b\equal{}\minus{}1, c\equal{}1$.

That's correct. :lol:

[color=blue]Problem 4[/color] 1987 Yamagata University/Agiculture, Enginerring

For the function $ f(x) \equal{} 2x^3 \minus{} 3x^2 \minus{} 36x$,

(1) Examine of the variation of $ f(x)$ and draw the graph of $ y \equal{} f(x)$.

(2) Find the number of the distinct real solution for the cubic equation :$ 2x^3 \minus{} 3x^2 \minus{} 36x \minus{} a \equal{} 0$.

(3) Find the range of $ a$ such that the cubic equation $ 2x^3 \minus{} 3x^2 \minus{} 36x \minus{} a \equal{} 0$ has distinct two positive roots and a negative root.

Differentiating and setting to zero, I get a local maximum at $ (\minus{}2, 44)$ and a local minimum at $ (3, \minus{} 81)$. This and the fact that the graph passes through the origin gives the shape of the curve.

The number of distinct real roots is $ 1, 2$ or $ 3$, depending on the value of $ a$.

For two distinct positive roots and a negative root, we need $ \minus{} 81 < a < 0$.

Your answer is incorrect from the beginning. :(

Hi kunny. I have edited my previous post which included one typo ($ (2, 44)$ instead of $ (\minus{}2, 44)$) and one miscalculation ($ (3, \minus{}66)$ instead of $ (3, \minus{}81)$) and perhaps some other errors.

(2) This simply states find the number of values for $ a$ such the $ f(x) \equal{} a$. Since the range of $ f(x)$ is all real numbers (from the graph in (1) it's visible) there are infinitely many values for $ a$.

Please read the context of question 2 carefully.

It seems to me that there is one real negative solution when $ a < \minus{} 81$, two distinct real solutions (one negative, one positive) when $ a \equal{} \minus{} 81$, three distinct real solutions (one negative, two positive) when $ \minus{} 81 < a < 44$, two distinct real solutions (one negative, one positive) when $ a \equal{} 44$ and one real positive solution when $ a > 44$.

At last the problem has finished. :lol:

[color=blue]Problem 5[/color] 1991 Himeji Institute of Technology

Determine the range of $ \theta \ (0\leq \theta <2\pi)$ such that the cubic equation in $ x$ : $ 4x^3\minus{}6x^2\minus{}24x\plus{}13\plus{}54\sin \theta \equal{}0$ has distinct three real roots.

[quote="kunny"]That's correct. :lol:

Here is my solution:

The converse of Vieta, $ a,\ b,\ c\in x^3 \minus{} 3x \minus{} k \equal{} 0$ where $ k \equal{} abc$.

$ a < b < c\Longleftrightarrow y \equal{} x^3 \minus{} 3x\ \text{has distinct three real roots with}\ y \equal{} k\ \cdots[*]$.

Now let $ f(x) \equal{} x^3 \minus{} 3x$, we have $ f'(x) \equal{} 3(x \plus{} 1)(x \minus{} 1)$, yielding local maximum:$ f( \minus{} 1) \equal{} 2$ and local minimum:$ f(1) \equal{} \minus{} 2$. Thus $ [*]\Longleftrightarrow \minus{} 2 < k < 2$. On the basis of the condition, let the cubic function $ C: y \equal{} f(x)$ is tangent to the line $ l: y \equal{} k$ at $ x \equal{} \minus{} 1$ and intersects at $ x \equal{} \beta$, by Vieta gives $ \minus{} 1 \minus{} 1 \plus{} \beta \equal{} 0\Longleftrightarrow \beta \equal{} 2$, analogously let $ C$ is tangent to $ l$ at $ x \equal{} 1$ and intersects at $ x \equal{} \alpha$, by the symmetry of $ C$ with respect to the origin, we have $ \alpha \equal{} \minus{} 2$. Let $ l$ move with $ [*]$, the ranges of $ a,\ b,\ c$ are $ \alpha < a < \minus{} 1$ and $ \minus{} 1 < b < 1$ and $ 1 < c < \beta$, yielding $ \minus{} 2 < a < \minus{} 1 < b < 1 < c < 2.$ Q.E.D.

Oh, I have just found the another approach, Flavour of ''Tchevichev'' comes out. :)[/quote]

thx for sharing ideas. nice solution!

Thank you! :lol:

[url=http://www.youtube.com/watch?v=Gj9hqDXuDnk&feature=player_embedded]Have a nice Break 1[/url]

[url=http://www.youtube.com/watch?v=rRvtsDUHvyo&feature=player_embedded#t=42]Have a nice Break 2[/url]

[url=http://www.youtube.com/watch?v=fMMXtJ2KZ-w&feature=player_embedded]Have a nice Break 3[/url]

[quote="kunny"][color=blue]Problem 11[/color] 1984 Tokyo Univesity/Humanities

For a real constant $ t$, consider the function $ f(x) \equal{} (x^2 \minus{} 3x \plus{} 2)(x \minus{} t)$. If we denote the roots of $ f'(x) \equal{} 0$ by $ \alpha ,\ \beta\ (\alpha < \beta)$ , we can regard them as functions in $ t$. Find the maximum and minimum values of the function in $ t: \ |t \minus{} \alpha| \plus{} |t \minus{} \beta|$ in the range of $ 1\leq t\leq 3$.[/quote]

Hi Kunny,

Edit: For some reason not adequately explained, I thought the expression was $ f(x) \equal{} \frac{(x^2 \minus{} 3x \plus{} 2)}{(x \minus{} t)}$

Ah well, back to the brain transplant room.

[quote="kunny"][color=blue]Problem 11[/color] 1984 Tokyo Univesity/Humanities

For a real constant $ t$, consider the function $ f(x) \equal{} (x^2 \minus{} 3x \plus{} 2)(x \minus{} t)$. If we denote the roots of $ f'(x) \equal{} 0$ by $ \alpha ,\ \beta\ (\alpha < \beta)$ , we can regard them as functions in $ t$. Find the maximum and minimum values of the function in $ t: \ |t \minus{} \alpha| \plus{} |t \minus{} \beta|$ in the range of $ 1\leq t\leq 3$.[/quote]

[color=blue]
my solution:

Case: 1

Let $ \ t\ge 2$ and $ \ g(t)\equal{}|t \minus{} \alpha| \plus{} |t \minus{} \beta|$

As $ \ f(x)\equal{}(x\minus{}1)(x\minus{}2)(x\minus{}t)$, By applying Rolle's theorem we get $ 1<\alpha<2\le\beta\le t$

$ \implies g(t)\equal{}2t\minus{}(\alpha\plus{}\beta)$

Also $ f'(x)\equal{} 3x^2\minus{}2(t\plus{}3)x\plus{}2\plus{}3t\implies \alpha\plus{}\beta\equal{}\frac {2(t\plus{}3)}{3}$  and $ \ \alpha\cdot\beta\equal{}\frac {2\plus{}3t}{3}$

$ \implies g(t)\equal{}2t \minus{} \frac {2(t\plus{}3)}{3}\equal{}\frac {2}{3}(2t\minus{}3)$

$ \implies \frac{2}{3}\le g(t)\le 2$ for $ \ t\in [2,3]$

Case: 2

Let $ \ t< 2$ and $ \ g(t)\equal{}|t \minus{} \alpha| \plus{} |t \minus{} \beta|$

As $ \ f(x)\equal{}(x\minus{}1)(x\minus{}t)(x\minus{}2)$, By applying Rolle's theorem we get $ 1\le \alpha\le t<\beta< 2$

$ \implies g(t)\equal{}\beta\minus{}\alpha\equal{}\sqrt {(\alpha\plus{}\beta)^2\minus{}4\alpha\cdot\beta}$

$ \implies g(t)\equal{}\sqrt {\left(\frac {2(t\plus{}3)}{3}\right)^2\minus{}\frac {4(2\plus{}3t)}{3}}\equal{}\frac {2}{3}\sqrt {t^2\minus{}3t\plus{}3}$

$ \implies g(t)\equal{} \frac {2}{3}\sqrt {\left(t\minus{}\frac {3}{2}\right)^2\plus{}\frac {3}{4}}$

$ \implies \frac{\sqrt {3}}{3}\le g(t)\le \frac {2}{3}$ for $ \ t\in [1,2)$

Hence $ g(t)\in \left[\frac{\sqrt {3}}{3},\  2\right]$ for $ \ t\in [1,3]$
[/color]

[quote="kunny"]Perfect! :lol:

[color=blue]Problem 12[/color] 1998 Tokyo University/Humanities

Let $ a\neq 0$ be real number. Find the value of $ a$ such that the difference of local maximum and local minimum values of

the function $ f(x) \equal{} (3x^2 \minus{} 4)\left(x \minus{} a \plus{} \frac {1}{a}\right)$ is minimal.[/quote]

we get that the critical numbers are $ p \equal{} \frac {2a}{3}$, $ q \equal{} \minus{} \frac {2}{3a}$.

it suffices to find $ a$ such that $ t\equal{}|f(p) \minus{} f(q)| \equal{} \frac {4}{9}\left|a^3 \plus{} \frac {1}{a^3} \plus{} 3a \plus{} \frac {3}{a}\right|$ is minimal.

by AM-GM, $ t \ge \frac{32}{9}$, equality occurs if $ a\equal{}\pm 1$.

am I missing anything? :trampoline:

That's correct. :lol:

[color=blue]Problem 13[/color] 1981 Kyoto University/Science

In $ x\minus{}y$ coordinate plane suppose a line passing through the point $ (4,\ 5)$ intersects the parabola $ y\equal{}\frac{1}{4}x^2$ at two points $ P,\ Q$.

Find the minimum value of the length of line segment $ PQ$.

[quote="kunny"]Perfect! :lol:

[color=blue]Problem 12[/color] 1998 Tokyo University/Humanities

Let $ a\neq 0$ be real number. Find the value of $ a$ such that the difference of local maximum and local minimum values of

the function $ f(x) \equal{} (3x^2 \minus{} 4)\left(x \minus{} a \plus{} \frac {1}{a}\right)$ is minimal.[/quote]
[color=blue]
My Solution:

$ f(x)\equal{} 3x^3\minus{}3\left(a\minus{}\frac {1}{a}\right)x^2\minus{}4x\plus{}4\left(a\minus{}\frac {1}{a}\right)$

$ \implies f'(x)\equal{}9x^2\minus{}6\left(a\minus{}\frac {1}{a}\right)x\minus{}4$

Let $ \ x_1< x_2$ be the roots of $ \ f'(x)\equal{}0$

$ \implies x_1 \plus{} x_2 \equal{} \frac {2}{3}\left(a\minus{}\frac {1}{a}\right)$  and  $ \ x_1x_2\equal{}\minus{}\frac {4}{9}$

$ \implies x_2 \minus{} x_1 \equal{} \frac {2}{3}\left|a\plus{}\frac {1}{a}\right|$

By sign scheme of $ \ f'(x)$ we can conclude that at  $ \ x_1,\  x_2$  we have local maximum and local minimum respectively.

$ \implies f(x_1)\minus{}f(x_2)\equal{} \left[3(x_1^3\minus{}x_2^3)\minus{}3\left(a\minus{}\frac {1}{a}\right)(x_1^2\minus{}x_2^2)\minus{}4(x_1\minus{}x_2)\right]$

$ \implies f(x_1)\minus{}f(x_2)\equal{} (x_2\minus{}x_1)\left[\minus{}3(x_1^2\plus{}x_2^2\plus{}x_1x_2)\plus{}3\left(a\minus{}\frac {1}{a}\right)(x_1\plus{}x_2)\plus{}4\right]$

$ \implies f(x_1)\minus{}f(x_2)\equal{} \frac {2}{3}\left|a\plus{}\frac {1}{a}\right|\left[\minus{}3(x_1\plus{}x_2)^2\plus{}3x_1x_2\plus{}2\left(a\minus{}\frac {1}{a}\right)^2\plus{}4\right]$

$ \implies f(x_1)\minus{}f(x_2)\equal{} \frac {2}{3}\left|a\plus{}\frac {1}{a}\right|\left[\minus{} \frac {4}{3}\left(a\minus{}\frac {1}{a}\right)^2\minus{}\frac{4}{3}\plus{}2\left(a\minus{}\frac {1}{a}\right)^2\plus{}4\right]$

$ \implies f(x_1)\minus{}f(x_2)\equal{} \frac {2}{3}\left|a\plus{}\frac {1}{a}\right|\left[ \frac {2}{3}\left(a\minus{}\frac {1}{a}\right)^2\plus{}\frac{8}{3}\right]$

$ \implies f(x_1)\minus{}f(x_2)\equal{} \frac {4}{9}\left|a\plus{}\frac {1}{a}\right|\left(a\plus{}\frac {1}{a}\right)^2\equal{}\frac {4}{9}\left|a\plus{}\frac {1}{a}\right|^3$

As  $ \  \left|a\plus{}\frac {1}{a}\right|\ge 2$  and equality for  $ \ a \equal{} \pm 1\ ,$

$ \implies f(x_1)\minus{}f(x_2)\ge\frac {32}{9}$  and equality for  $ \ a \equal{} \pm 1\ ,$

[/color]

[quote="kunny"]That's correct. :lol:

[color=blue]Problem 13[/color] 1981 Kyoto University/Science

In $ x \minus{} y$ coordinate plane suppose a line passing through the point $ (4,\ 5)$ intersects the parabola $ y \equal{} \frac {1}{4}x^2$ at two points $ P,\ Q$.

Find the minimum value of the length of line segment $ PQ$.[/quote]

[color=blue]
My solution:

Consider the parametric form of the straight line $ \  PQ$

$ x\equal{} 4 \plus{}r\cos \theta ,\ y \equal{} 5\plus{}r \sin \theta$

By solving the line with the curve we get,

$ r^2 \cos^2\theta \plus{}4 (2\cos\theta \minus{}\sin\theta)r\minus{}4\equal{}0$

$ \implies r_1 \plus{} r_2 \equal{} \frac {4(\sin\theta\minus{}2\cos\theta)}{\cos^2\theta}$ and  $ \ r_1r_2\equal{}\minus{}\frac {4}{\cos^2\theta}$

$ \implies PQ\equal{}|r_1\minus{}r_2|\equal{}\sqrt{(r_1\plus{}r_2)^2\minus{}r_1r_2}\equal{}4\sqrt {(\tan\theta\minus{}1)^4\plus{}4}$

$ \implies PQ \ge 8$ equality for $ \ \tan\theta\equal{}1\implies \theta \equal{} \frac {\pi}{4}$

$ \boxed{\boxed{\text {Minimum} \ PQ\equal{} 8}}$

[/color]

[quote="makar"][quote="kunny"]Perfect! :lol:

[color=blue]Problem 12[/color] 1998 Tokyo University/Humanities

Let $ a\neq 0$ be real number. Find the value of $ a$ such that the difference of local maximum and local minimum values of

the function $ f(x) \equal{} (3x^2 \minus{} 4)\left(x \minus{} a \plus{} \frac {1}{a}\right)$ is minimal.[/quote]
[color=blue]
My Solution:

$ f(x) \equal{} 3x^3 \minus{} 3\left(a \minus{} \frac {1}{a}\right)x^2 \minus{} 4x \plus{} 4\left(a \minus{} \frac {1}{a}\right)$

$ \implies f'(x) \equal{} 9x^2 \minus{} 6\left(a \minus{} \frac {1}{a}\right)x \minus{} 4$

Let $ \ x_1 < x_2$ be the roots of $ \ f'(x) \equal{} 0$

$ \implies x_1 \plus{} x_2 \equal{} \frac {2}{3}\left(a \minus{} \frac {1}{a}\right)$  and  $ \ x_1x_2 \equal{} \minus{} \frac {4}{9}$

$ \implies x_2 \minus{} x_1 \equal{} \frac {2}{3}\left|a \plus{} \frac {1}{a}\right|$

By sign scheme of $ \ f'(x)$ we can conclude that at  $ \ x_1,\ x_2$  we have local maximum and local minimum respectively.

$ \implies f(x_1) \minus{} f(x_2) \equal{} \left[3(x_1^3 \minus{} x_2^3) \minus{} 3\left(a \minus{} \frac {1}{a}\right)(x_1^2 \minus{} x_2^2) \minus{} 4(x_1 \minus{} x_2)\right]$

$ \implies f(x_1) \minus{} f(x_2) \equal{} (x_2 \minus{} x_1)\left[ \minus{} 3(x_1^2 \plus{} x_2^2 \plus{} x_1x_2) \plus{} 3\left(a \minus{} \frac {1}{a}\right)(x_1 \plus{} x_2) \plus{} 4\right]$

$ \implies f(x_1) \minus{} f(x_2) \equal{} \frac {2}{3}\left|a \plus{} \frac {1}{a}\right|\left[ \minus{} 3(x_1 \plus{} x_2)^2 \plus{} 3x_1x_2 \plus{} 2\left(a \minus{} \frac {1}{a}\right)^2 \plus{} 4\right]$

$ \implies f(x_1) \minus{} f(x_2) \equal{} \frac {2}{3}\left|a \plus{} \frac {1}{a}\right|\left[ \minus{} \frac {4}{3}\left(a \minus{} \frac {1}{a}\right)^2 \minus{} \frac {4}{3} \plus{} 2\left(a \minus{} \frac {1}{a}\right)^2 \plus{} 4\right]$

$ \implies f(x_1) \minus{} f(x_2) \equal{} \frac {2}{3}\left|a \plus{} \frac {1}{a}\right|\left[ \frac {2}{3}\left(a \minus{} \frac {1}{a}\right)^2 \plus{} \frac {8}{3}\right]$

$ \implies f(x_1) \minus{} f(x_2) \equal{} \frac {4}{9}\left|a \plus{} \frac {1}{a}\right|\left(a \plus{} \frac {1}{a}\right)^2 \equal{} \frac {4}{9}\left|a \plus{} \frac {1}{a}\right|^3$

As  $ \ \left|a \plus{} \frac {1}{a}\right|\ge 2$  and equality for  $ \ a \equal{} \pm 1\ ,$

$ \implies f(x_1) \minus{} f(x_2)\ge\frac {32}{9}$  and equality for  $ \ a \equal{} \pm 1\ ,$

[/color][/quote]

That's correct. Impressive Solution! :lol:

Here is my solution 1:

Let $ f(x) \equal{} (3x^2 \minus{} 4)(x \minus{} b)\ ,\ b \equal{} a \minus{} \frac {1}{a}$, we have $ f'(x) \equal{} 9x^2 \minus{} 6bx \minus{} 4$.

$ \therefore f'(x) \equal{} 0\Longleftrightarrow x \equal{} \frac {b\pm \sqrt {b^2 \plus{} 4}}{3}: \equal{} \alpha ,\ \beta \ (\alpha < \beta)$.

$ f(x) \equal{} f'(x)\left(\frac 13 x \minus{} \frac19 b\right) \minus{} \frac 23 (b^2 \plus{} 4)x \plus{} \frac {32}{9}b$. Since $ f'(\alpha) \equal{} f'(\beta) \equal{} 0$, we get the difference of local extrema, $ f(\beta) \minus{} f(\alpha) \equal{} \frac 49 (b^2 \plus{} 4)^{\frac 32}\geq \frac {32}{9}$.
The equality holds when $ b \equal{} 0\Longleftrightarrow a \minus{} \frac {1}{a} \equal{} 1$, yielding $ a \equal{} \boxed{\pm 1}$.

[quote="kunny"][hide][quote="makar"][quote="kunny"]Perfect! :lol:

[color=blue]Problem 12[/color] 1998 Tokyo University/Humanities

Let $ a\neq 0$ be real number. Find the value of $ a$ such that the difference of local maximum and local minimum values of

the function $ f(x) \equal{} (3x^2 \minus{} 4)\left(x \minus{} a \plus{} \frac {1}{a}\right)$ is minimal.[/quote]
[color=blue]
My Solution:

$ f(x) \equal{} 3x^3 \minus{} 3\left(a \minus{} \frac {1}{a}\right)x^2 \minus{} 4x \plus{} 4\left(a \minus{} \frac {1}{a}\right)$

$ \implies f'(x) \equal{} 9x^2 \minus{} 6\left(a \minus{} \frac {1}{a}\right)x \minus{} 4$

Let $ \ x_1 < x_2$ be the roots of $ \ f'(x) \equal{} 0$

$ \implies x_1 \plus{} x_2 \equal{} \frac {2}{3}\left(a \minus{} \frac {1}{a}\right)$  and  $ \ x_1x_2 \equal{} \minus{} \frac {4}{9}$

$ \implies x_2 \minus{} x_1 \equal{} \frac {2}{3}\left|a \plus{} \frac {1}{a}\right|$

By sign scheme of $ \ f'(x)$ we can conclude that at  $ \ x_1,\ x_2$  we have local maximum and local minimum respectively.

$ \implies f(x_1) \minus{} f(x_2) \equal{} \left[3(x_1^3 \minus{} x_2^3) \minus{} 3\left(a \minus{} \frac {1}{a}\right)(x_1^2 \minus{} x_2^2) \minus{} 4(x_1 \minus{} x_2)\right]$

$ \implies f(x_1) \minus{} f(x_2) \equal{} (x_2 \minus{} x_1)\left[ \minus{} 3(x_1^2 \plus{} x_2^2 \plus{} x_1x_2) \plus{} 3\left(a \minus{} \frac {1}{a}\right)(x_1 \plus{} x_2) \plus{} 4\right]$

$ \implies f(x_1) \minus{} f(x_2) \equal{} \frac {2}{3}\left|a \plus{} \frac {1}{a}\right|\left[ \minus{} 3(x_1 \plus{} x_2)^2 \plus{} 3x_1x_2 \plus{} 2\left(a \minus{} \frac {1}{a}\right)^2 \plus{} 4\right]$

$ \implies f(x_1) \minus{} f(x_2) \equal{} \frac {2}{3}\left|a \plus{} \frac {1}{a}\right|\left[ \minus{} \frac {4}{3}\left(a \minus{} \frac {1}{a}\right)^2 \minus{} \frac {4}{3} \plus{} 2\left(a \minus{} \frac {1}{a}\right)^2 \plus{} 4\right]$

$ \implies f(x_1) \minus{} f(x_2) \equal{} \frac {2}{3}\left|a \plus{} \frac {1}{a}\right|\left[ \frac {2}{3}\left(a \minus{} \frac {1}{a}\right)^2 \plus{} \frac {8}{3}\right]$

$ \implies f(x_1) \minus{} f(x_2) \equal{} \frac {4}{9}\left|a \plus{} \frac {1}{a}\right|\left(a \plus{} \frac {1}{a}\right)^2 \equal{} \frac {4}{9}\left|a \plus{} \frac {1}{a}\right|^3$

As  $ \ \left|a \plus{} \frac {1}{a}\right|\ge 2$  and equality for  $ \ a \equal{} \pm 1\ ,$

$ \implies f(x_1) \minus{} f(x_2)\ge\frac {32}{9}$  and equality for  $ \ a \equal{} \pm 1\ ,$

[/color][/quote][/hide]

That's correct. :lol:

Here is my solution 1:

Let $ f(x) \equal{} (3x^2 \minus{} 4)(x \minus{} b)\ ,\ b \equal{} a \minus{} \frac {1}{a}$, we have $ f'(x) \equal{} 9x^2 \minus{} 6bx \minus{} 4$.

$ \therefore f'(x) \equal{} 0\Longleftrightarrow x \equal{} \frac {b\pm \sqrt {b^2 \plus{} 4}}{3}: \equal{} \alpha ,\ \beta \ (\alpha < \beta)$.

$ f(x) \equal{} f'(x)\left(\frac 13 x \minus{} \frac19 b\right) \minus{} \frac 23 (b^2 \plus{} 4)x \plus{} \frac {32}{9}b$. Since $ f'(\alpha) \equal{} f'(\beta) \equal{} 0$, we get the difference of local extrema, $ f(\beta) \minus{} f(\alpha) \equal{} \frac 49 (b^2 \plus{} 4)^{\frac 32}\geq \frac {32}{9}$. The equality holds when $ b \equal{} 0\Longleftrightarrow a \minus{} \frac {1}{a} \equal{} 1$, yielding $ a \equal{} \boxed{\pm 1}$.

My Solution 2:[/quote]

Your solution is very good! But what about alternate:
One More thing what about[b] solution to problem 13[/b]

[quote="makar"][quote="kunny"]That's correct. :lol:

[color=blue]Problem 13[/color] 1981 Kyoto University/Science

In $ x \minus{} y$ coordinate plane suppose a line passing through the point $ (4,\ 5)$ intersects the parabola $ y \equal{} \frac {1}{4}x^2$ at two points $ P,\ Q$.

Find the minimum value of the length of line segment $ PQ$.[/quote]

[color=blue]
My solution:

Consider the parametric form of the straight line $ \ PQ$

$ x \equal{} 4 \plus{} r\cos \theta ,\ y \equal{} 5 \plus{} r \sin \theta$

By solving the line with the curve we get,

$ r^2 \cos^2\theta \plus{} 4 (2\cos\theta \minus{} \sin\theta)r \minus{} 4 \equal{} 0$

$ \implies r_1 \plus{} r_2 \equal{} \frac {4(\sin\theta \minus{} 2\cos\theta)}{\cos^2\theta}$ and  $ \ r_1r_2 \equal{} \minus{} \frac {4}{\cos^2\theta}$

$ \implies PQ \equal{} |r_1 \minus{} r_2| \equal{} \sqrt {(r_1 \plus{} r_2)^2 \minus{} r_1r_2} \equal{} 4\sqrt {(\tan\theta \minus{} 1)^4 \plus{} 4}$

$ \implies PQ \ge 8$ equality for $ \ \tan\theta \equal{} 1\implies \theta \equal{} \frac {\pi}{4}$

$ \boxed{\boxed{\text {Minimum} \ PQ \equal{} 8}}$

[/color][/quote]

That's correct. :lol:

Here is my Solution:

Let the equation of the line $ PQ: y \equal{} m(x \minus{} 4) \plus{} 5$, by $ \frac 14 x^2 \equal{} m(x \minus{} 4) \plus{} 5\Longleftrightarrow x^2 \minus{} 4mx \plus{} 4(4m \minus{} 5) \equal{} 0$ we have the

$ x$ coordinates of the intersection points of the line and the parabola as $ x \equal{} 2m\pm 2\sqrt {m^2 \minus{} 4m \plus{} 5}: \equal{} \alpha ,\ \beta (\alpha <\beta).$

$ \therefore \overline{PQ} \equal{} \sqrt {1 \plus{} m^2}(\beta \minus{} \alpha) \equal{}\sqrt{1\plus{}m^2}\cdot 4\sqrt{m^2\minus{}4m\plus{}5} \equal{}4\sqrt {(m \minus{} 1)^4 \plus{} 4}\geq \boxed{8}$ with equality $ m \equal{} 1$.

[quote="makar"][/quote][quote="kunny"][hide][quote="makar"][quote="kunny"]Perfect! :lol:

[color=blue]Problem 12[/color] 1998 Tokyo University/Humanities

Let $ a\neq 0$ be real number. Find the value of $ a$ such that the difference of local maximum and local minimum values of

the function $ f(x) = (3x^2 - 4)\left(x - a + \frac {1}{a}\right)$ is minimal.[/quote]
[color=blue]
My Solution:

$ f(x) = 3x^3 - 3\left(a - \frac {1}{a}\right)x^2 - 4x + 4\left(a - \frac {1}{a}\right)$

$ \implies f'(x) = 9x^2 - 6\left(a - \frac {1}{a}\right)x - 4$

Let $ \ x_1 < x_2$ be the roots of $ \ f'(x) = 0$

$ \implies x_1 + x_2 = \frac {2}{3}\left(a - \frac {1}{a}\right)$  and  $ \ x_1x_2 = - \frac {4}{9}$

$ \implies x_2 - x_1 = \frac {2}{3}\left|a + \frac {1}{a}\right|$

By sign scheme of $ \ f'(x)$ we can conclude that at  $ \ x_1,\ x_2$  we have local maximum and local minimum respectively.

$ \implies f(x_1) - f(x_2) = \left[3(x_1^3 - x_2^3) - 3\left(a - \frac {1}{a}\right)(x_1^2 - x_2^2) - 4(x_1 - x_2)\right]$

$ \implies f(x_1) - f(x_2) = (x_2 - x_1)\left[ - 3(x_1^2 + x_2^2 + x_1x_2) + 3\left(a - \frac {1}{a}\right)(x_1 + x_2) + 4\right]$

$ \implies f(x_1) - f(x_2) = \frac {2}{3}\left|a + \frac {1}{a}\right|\left[ - 3(x_1 + x_2)^2 + 3x_1x_2 + 2\left(a - \frac {1}{a}\right)^2 + 4\right]$

$ \implies f(x_1) - f(x_2) = \frac {2}{3}\left|a + \frac {1}{a}\right|\left[ - \frac {4}{3}\left(a - \frac {1}{a}\right)^2 - \frac {4}{3} + 2\left(a - \frac {1}{a}\right)^2 + 4\right]$

$ \implies f(x_1) - f(x_2) = \frac {2}{3}\left|a + \frac {1}{a}\right|\left[ \frac {2}{3}\left(a - \frac {1}{a}\right)^2 + \frac {8}{3}\right]$

$ \implies f(x_1) - f(x_2) = \frac {4}{9}\left|a + \frac {1}{a}\right|\left(a + \frac {1}{a}\right)^2 = \frac {4}{9}\left|a + \frac {1}{a}\right|^3$

As  $ \ \left|a + \frac {1}{a}\right|\ge 2$  and equality for  $ \ a = \pm 1\ ,$

$ \implies f(x_1) - f(x_2)\ge\frac {32}{9}$  and equality for  $ \ a = \pm 1\ ,$

[/color][/quote][/hide]

That's correct. :lol:

Here is my solution 1:

Let $ f(x) = (3x^2 - 4)(x - b)\ ,\ b = a - \frac {1}{a}$, we have $ f'(x) = 9x^2 - 6bx - 4$.

$ \therefore f'(x) = 0\Longleftrightarrow x = \frac {b\pm \sqrt {b^2 + 4}}{3}: = \alpha ,\ \beta \ (\alpha < \beta)$.

$ f(x) = f'(x)\left(\frac 13 x - \frac19 b\right) - \frac 23 (b^2 + 4)x + \frac {32}{9}b$. Since $ f'(\alpha) = f'(\beta) = 0$, we get the difference of local extrema, $ f(\beta) - f(\alpha) = \frac 49 (b^2 + 4)^{\frac 32}\geq \frac {32}{9}$. The equality holds when $ b = 0\Longleftrightarrow a - \frac {1}{a} = 1$, yielding $ a = \boxed{\pm 1}$.

My Solution 2:[/quote][quote="makar"]

Your solution is very good! But what about alternate:
One More thing what about[b] solution to problem 13[/b][/quote]

Thanks. :)

My Solution 2:

$ f(x) = (3x^2 - 4)\left(x - a + \frac {1}{a}\right)\Longrightarrow f'(x) = 9x^2 - 6\left(a - \frac {1}{a}\right)x - 4 = 0$

$ \Longleftrightarrow x = - \frac {2}{3a},\ \frac {2}{3}a\ (a\neq 0)$. Let $ p = - \frac {2}{3a},\ q = \frac {2}{3}a\ (a\neq 0)$,

the difference of the local extrema is $ \boxed{\boxed{|f(q) - f(p)| = \left|\int_p^q f'(x)\ dx\right|}}$

$ = 9\left|\int_p^q (x - p)(x - q)\ dx \right| = \frac {3}{2}|q - p|^3 = \frac {4}{9}\left|a + \frac {1}{a}\right|^3$, by $ a\cdot \frac {1}{a} = 1 > 0$, $ a$ and $ \frac {1}{a}$ have same sign,
so we have $ \left|a + \frac {1}{a}\right| = |a| + \frac {1}{|a|}\geq 2\sqrt {|a|\cdot \frac {1}{|a|}} = 2$ with equality $ |a| = \frac {1}{|a|}\Longleftrightarrow |a| = 1$.

[quote="kunny"][hide][/quote][quote="makar"][/quote][quote="kunny"][quote="makar"][quote="kunny"]Perfect! :lol:

[color=blue]Problem 12[/color] 1998 Tokyo University/Humanities

Let $ a\neq 0$ be real number. Find the value of $ a$ such that the difference of local maximum and local minimum values of

the function $ f(x) = (3x^2 - 4)\left(x - a + \frac {1}{a}\right)$ is minimal.[/quote]
[color=blue]
My Solution:

$ f(x) = 3x^3 - 3\left(a - \frac {1}{a}\right)x^2 - 4x + 4\left(a - \frac {1}{a}\right)$

$ \implies f'(x) = 9x^2 - 6\left(a - \frac {1}{a}\right)x - 4$

Let $ \ x_1 < x_2$ be the roots of $ \ f'(x) = 0$

$ \implies x_1 + x_2 = \frac {2}{3}\left(a - \frac {1}{a}\right)$  and  $ \ x_1x_2 = - \frac {4}{9}$

$ \implies x_2 - x_1 = \frac {2}{3}\left|a + \frac {1}{a}\right|$

By sign scheme of $ \ f'(x)$ we can conclude that at  $ \ x_1,\ x_2$  we have local maximum and local minimum respectively.

$ \implies f(x_1) - f(x_2) = \left[3(x_1^3 - x_2^3) - 3\left(a - \frac {1}{a}\right)(x_1^2 - x_2^2) - 4(x_1 - x_2)\right]$

$ \implies f(x_1) - f(x_2) = (x_2 - x_1)\left[ - 3(x_1^2 + x_2^2 + x_1x_2) + 3\left(a - \frac {1}{a}\right)(x_1 + x_2) + 4\right]$

$ \implies f(x_1) - f(x_2) = \frac {2}{3}\left|a + \frac {1}{a}\right|\left[ - 3(x_1 + x_2)^2 + 3x_1x_2 + 2\left(a - \frac {1}{a}\right)^2 + 4\right]$

$ \implies f(x_1) - f(x_2) = \frac {2}{3}\left|a + \frac {1}{a}\right|\left[ - \frac {4}{3}\left(a - \frac {1}{a}\right)^2 - \frac {4}{3} + 2\left(a - \frac {1}{a}\right)^2 + 4\right]$

$ \implies f(x_1) - f(x_2) = \frac {2}{3}\left|a + \frac {1}{a}\right|\left[ \frac {2}{3}\left(a - \frac {1}{a}\right)^2 + \frac {8}{3}\right]$

$ \implies f(x_1) - f(x_2) = \frac {4}{9}\left|a + \frac {1}{a}\right|\left(a + \frac {1}{a}\right)^2 = \frac {4}{9}\left|a + \frac {1}{a}\right|^3$

As  $ \ \left|a + \frac {1}{a}\right|\ge 2$  and equality for  $ \ a = \pm 1\ ,$

$ \implies f(x_1) - f(x_2)\ge\frac {32}{9}$  and equality for  $ \ a = \pm 1\ ,$

[/color][/quote]

That's correct. :lol:

Here is my solution 1:

Let $ f(x) = (3x^2 - 4)(x - b)\ ,\ b = a - \frac {1}{a}$, we have $ f'(x) = 9x^2 - 6bx - 4$.

$ \therefore f'(x) = 0\Longleftrightarrow x = \frac {b\pm \sqrt {b^2 + 4}}{3}: = \alpha ,\ \beta \ (\alpha < \beta)$.

$ f(x) = f'(x)\left(\frac 13 x - \frac19 b\right) - \frac 23 (b^2 + 4)x + \frac {32}{9}b$. Since $ f'(\alpha) = f'(\beta) = 0$, we get the difference of local extrema, $ f(\beta) - f(\alpha) = \frac 49 (b^2 + 4)^{\frac 32}\geq \frac {32}{9}$. The equality holds when $ b = 0\Longleftrightarrow a - \frac {1}{a} = 1$, yielding $ a = \boxed{\pm 1}$.

My Solution 2:[/quote][quote="makar"]

Your solution is very good! But what about alternate:
One More thing what about[b] solution to problem 13[/b][/quote][quote="kunny"]
[/hide]

Thanks. :)

My Solution 2:

$ f(x) = (3x^2 - 4)\left(x - a + \frac {1}{a}\right)\Longrightarrow f'(x) = 9x^2 - 6\left(a - \frac {1}{a}\right)x - 4 = 0$

$ \Longleftrightarrow x = - \frac {2}{3a},\ \frac {2}{3}a\ (a\neq 0)$. Let $ p = - \frac {2}{3a},\ q = \frac {2}{3}a\ (a\neq 0)$,

the difference of the local extrema is $ \boxed{\boxed{|f(q) - f(p)| = \left|\int_p^q f'(x)\ dx\right|}}$

$ = 9\left|\int_p^q (x - p)(x - q)\ dx \right| = \frac {3}{2}|q - p|^3 = \frac {4}{9}\left|a + \frac {1}{a}\right|^3$, by $ a\cdot \frac {1}{a} = 1 > 0$, $ a$ and $ \frac {1}{a}$ have same sign,
so we have $ \left|a + \frac {1}{a}\right| = |a| + \frac {1}{|a|}\geq 2\sqrt {|a|\cdot \frac {1}{|a|}} = 2$ with equality $ |a| = \frac {1}{|a|}\Longleftrightarrow |a| = 1$.[/quote]

In my opinion, your alternate is best among all solutions.I liked it much,specially the boxed expression!!


linear algebra
matrix
inequalities
algebra
system of equations
college contests

[b]Problem 3[/b]

Let $ A$ be an $ n \times n$ square matrix with integer entries. Suppose that $ p^{2}A^{p^{2}} \equal{} q^{2}A^{q^{2}} \plus{} r^{2}I_{n}$ for some positive integers $ p, q, r$ where $ r$ is odd and $ p^{2} \equal{} q^{2} \plus{} r^{2}$. Prove that $ |detA| \equal{} 1$.
(Here $ I_{n}$ means the $ n \times n$ identity matrix.)

We will prove slightly stronger results.

[b]Claim 1: All eigenvalues $ \lambda$ of $ A$ fulfill $ \lambda^{q^2}\equal{}\lambda^{r^2}\equal{}\lambda^{p^2}\equal{}1$ and $ \det A\equal{}1$.[/b]

[b]Claim 2: $ A$ splits, i.e. it can be diagonalized (and we do not need $ A\in M_n(\mathbb{Z})$)[/b]

[i]Proof of claim 1[/i]: Let $ \lambda$ be an eigenvalue of $ A$, then $ p^2\lambda^{p^2}\equal{}r^2\plus{}q^2\lambda^{q^2}$ and therefore $ \lambda\not\equal{}0$ and $ \det A\not\equal{}0$ hold. Moreover, we have $ p^2|\lambda|^{p^2}\equal{}|r^2\plus{}q^2\lambda^{q^2}|\leq r^2\plus{}q^2|\lambda|^{q^2}$, which implies $ 0\leq r^2(1\minus{}|\lambda|^{p^2})\plus{}q^2|\lambda|^{q^2}(1\minus{}|\lambda|^{r^2})$ and therefore $ |\lambda|\leq 1$. If $ |\lambda|\equal{}1$ is fulfilled, the inequalities turn into equalities, which is only possible for $ \lambda^{q^2}\equal{}1$ and then we get $ \lambda^{r^2}\equal{}\lambda^{p^2}\equal{}1$, too. This leaves the other possibility $ 0<|\lambda|<1$. If such an eigenvalue occurs, we get $ |\det A|<1$. Since $ A\in M_n(\mathbb{Z})$, we get the contradiction $ \det A\equal{}0$. It remains to verify the statement about the determinant. Complex eigenvalues come in pairs $ (\lambda,\bar{\lambda})$ which contribute the factor $ \lambda\bar{\lambda}\equal{}1$ to the determinant of $ A$. Real eigenvalues fulfill $ \lambda^{r^2}\equal{}1$, i.e. $ \lambda\equal{}1$, since $ r$ is odd.

[i]Proof of claim 2 (without using claim 1):[/i] We can assume $ A$ in Jordan normal form. The equation is only fulfilled, if it is true for each Jordan block. Let $ J_t(\lambda)\equal{}\lambda I_t\plus{}N$ be a Jordan block of size $ t$ for the eigenvalue $ \lambda$. Here $ N$ is the standard nilpotent matrix and remember that $ N^t\equal{}0$. Moreover, the powers $ I_t,N,\ldots,N^{t\minus{}1}$ are non-zero matrices, which have their entries in different (off-)diagonals, i.e. $ \sum_{k\equal{}0}^{t\minus{}1}\alpha_kN^k\equal{}0\Longleftrightarrow\forall k: \alpha_k\equal{}0$. Now, direct calculation gives

$ p^2\sum_{k\equal{}0}^{\min\{p^2,t\minus{}1\}}\left(p^2\atop k\right)N^k\lambda^{p^2\minus{}k}\equal{}r^2I_t\plus{}q^2\sum_{k\equal{}0}^{\min\{q^2,t\minus{}1\}}\left(q^2\atop k\right)N^k\lambda^{q^2\minus{}k}$

Since $ p^2>q^2>0$, we must have $ q^2\geq t\minus{}1$ and and according to our preparations this requires $ p^2\lambda^{p^2}\equal{}r^2\plus{}q^2\lambda^{q^2}$ and for all $ k\equal{}1,\ldots,t\minus{}1$ the identities $ \;p^2\left(p^2\atop k\right)\lambda^{p^2\minus{}k}\equal{}q^2\left(q^2\atop k\right)\lambda^{q^2\minus{}k}$. Since $ \lambda\not\equal{}0$ due to the first identity, this system of equations is equivalent to $ \forall k\equal{}1,\ldots,t\minus{}1: \;p^2\left(p^2\atop k\right)\lambda^{r^2}\equal{}q^2\left(q^2\atop k\right)$, which implies $ t\equal{}1,2$. If $ t\equal{}2$ holds, we get $ 0<\lambda^{r^2}\equal{}\frac{q^4}{p^4}<1$. Substitution into the first identity yields $ \lambda^{q^2}\equal{}\minus{}\frac{p^2}{q^2}<\minus{}1$, a contradiction. Therefore, we must have $ t\equal{}1$ and we are finished.


geometry
circumcircle
trigonometry
geometry unsolved

Let $ \triangle ABC$ be triangle and R,r be its circumradius,inradius. O is circumcenter of $ \triangle ABC$. If $ \triangle ABC$ is acute triangle, then $ d(O,AB)\plus{}d(O,BC)\plus{}d(O,CA) \equal{} R\plus{}r$. If $ \angle{A}$ is obtuse,
$ \minus{}d(O,BC)\plus{}d(O,CA)\plus{}d(O,AB) \equal{} R\plus{}r$. How can we express this theorem not dividing by two case?

[quote="elmagodelosnumeros"]Does $ O$ mean the circumcenter of $ ABC$?[/quote]
Yes, I've edited it

[quote="Leonhard Euler"][color=darkred][b][u]The Carnot's relation.[/u][/b] Let $ \triangle ABC$ be triangle with the circumcircle $ C(O,R)$ and the incircle $ C(I,r)$. Denote the middlepoints $ D$, $ E$, $ F$ of the sides $ [BC]$, $ [CA]$, $ [AB]$ respectively and $ OD = x$, $ OE = y$ $ OF = z$. Prove that : if $ \triangle ABC$ is acute, then $ x+y+z = R+r$ ; if $ A > 90^{\circ}$, then $ -x+y+z = R+r$ ; if $ A = 90^{\circ}$, then $ y+z = R+r$. How can we express this theorem not dividing by three case?[/color][/quote]
[color=darkblue] [b]Proof of the Carnot's relation.[/b] Denote $ a+b+c = 2p$. Generally, for example $ x = R\cdot |\cos A|$, i.e. $ x = R\cdot\{\begin{array}{ccc}\cos A &\mathrm{if}& A\le 90^{\circ}\\ \ -\cos A &\mathrm{if}& A > 90^{\circ}\end{array}$.

Therefore, the general [b]Carnot's relation[/b] is $ \boxed{R\cdot(\cos A+\cos B+\cos C) = R+r}$ which in the case $ A > 90^{\circ}$ becomes $ -x+y+z = R+r$.

Suppose w.l.o.g. that $ A\ge B$ and $ A\ge C$. Apply the [b]Ptolemeu's relation[/b] to the cyclical quadrilaterals $ AEOF$, $ BFOD$ and $ CDOE$ :

$ \{\begin{array}{cccccc}1\blacktriangleright & A\le 90 &\implies &\{\begin{array}{ccccccc}aR & = & & & cy &+& bz\\ bR & = & cx & & &+& az\\ cR & = & bx &+& ay & &\\\ 2pr & = & ax &+& by &+& cz\end{array}\| &\implies & (a+b+c)(R+r) = (a+b+c)(x+y+z)\\ \\ 2\blacktriangleright & A > 90^{\circ}&\implies &\{\begin{array}{ccccccc}aR & = & & & cy &+& bz\\ bR & = &-cx & & &+& az\\ cR & = &-bx &+& ay & &\\\ 2pr & = &-ax &+& by &+& cz\end{array}\| &\implies & (a+b+c)(R+r) = (a+b+c)(-x+y+z)\end{array}\|$ $ \implies$

$ x\cdot\mathrm{sign}(90^{\circ}-A\right)+y+z = R+r$. Generally, $ \boxed{\ x\cdot\mathrm{sign}(90^{\circ}-A)+y\cdot\mathrm{sign}(90^{\circ}-B)+z\cdot\mathrm{sign}(90^{\circ}-C) = R+r\ }$.

[b]Remark.[/b] If $ A = 90^{\circ}$, then the [b]Carnot's relation[/b] becomes $ y+z = R+r$. But $ \{\begin{array}{c}2y = b\\ \ 2z = c\\ \ 2R = a\end{array}\|\implies$ $ r = p-a$.

[u]With trogonometry[/u], $ -1+\sum\cos A =$ $ (\cos B+\cos C)-(1-\cos A) =$ $ 2\cos\frac{B-C}{2}\sin\frac{A}{2}-2\sin^{2}\frac{A}{2}=$

$ 2\sin\frac{A}{2}(\cos\frac{B-C}{2}-\cos\frac{B+C}{2}) =$ $ 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}=$ $ \frac{4(p-a)(p-b)(p-c)}{abc}=$ $ \frac{4pr^{2}}{4Rpr}=\frac{r}{R}$ $ \implies$ $ \boxed{\sum\cos A = 1+\frac{r}{R}}$.

[u]Without trigonometry[/u], only in the case of the acute $ \triangle ABC$, suppose that we know the remarcable following relations :

the [b]arrow[/b] $ s_{A}$ [b]of the arc $ BC$ which doesn't include the vertex $ A$[/b] is $ s_{A}=\frac{r_{a}-r}{2}= R\cdot (1-\cos A)$ a.s.o. ; $ r_{a}+r_{b}+r_{c}= 4R+r$.

Thus, $ \boxed{R\cdot\sum\cos A}=\sum [R-R\cdot (1-\cos A)] =$ $ \sum (R-s_{a})=$ $ 3R-\frac{1}{2}\cdot\sum(r_{a}-r) =$ $ 3R-\frac{1}{2}\cdot [(4R+r)-3r]=$ $ \boxed{R+r}$.[/color]


limit
integration
calculus
calculus computations

This is a limit related to one of the limits we have discussed on the forum.

Find:

$\lim_{n\rightarrow \infty}\int_{0}^{\infty}\frac{1+x}{1+x^{2}}\frac{1+x^{2}}{1+x^{4}}\cdots \frac{1+x^{n}}{1+x^{2n}}dx$.

We have even
\[\lim_{n\to\infty}\int_{0}^{1}\prod_{k=1}^{n}\frac{1+x^{k}}{1+x^{2k}}\,dx=\infty\]
This is almost obvious

I got that  the limit is

$\int_{0}^{1}\prod_{n=1}^{\infty}\frac{1+x^{n}}{1+x^{2n}}dx$.

You are right.
The product is not  integrable.

An observation:

If $f_{n}(x)=\prod_{k=1}^{n}\frac{1+x^{k}}{1+x^{2k}}$, then the sequence does not converge uniformly on $[0,1]$.


Hola gente, necesito demostrar el siguiente enunciado, espero me ayuden

[color=blue]Sea $ABC$ un triángulo cualquiera, Sea $O_{c}$ el excirculo q se opone a $C$, $O_{b}$ el excirculo q se opone a $B$.
$O_{c}$ y $O_{b}$ son tangentes a la recta $BC$ en $J$ y $F$, respectivamente, tangentes a la recta $AB$ en $Z$ y $G$, respectivamente, tangentes a $AC$ en $L$ y $Y$, respectivamente, y sea $l$ la cuarta tangente comun a $O_{c}$ y $O_{b}$.
Las rectas $FG$ y $JL$ se intersectan en $U$, las rectas $ZG$ y $YL$ intersectan a $l$ en $M$ y $N$ respectivamente.
Muestre q el circuncentro del triangulo $AMN$ se encuentra sobre la recta $AU$.
[/color]

Probaremos que $ U$ está en la recta de la altura $ AH$ de $ \triangle ABC.$ Así que supongamos $ U \equiv{} JL \cap AH $ y $ U' \equiv{} TG \cap AH.$ Hay que probar que $ U$ se confunde con $ U'.$ Aplicando Teorema de Menelao a la transversal $ \overline{UAH}$ en  $ \triangle CLJ$ se tiene:

$\frac {UA}{UH} \cdot \frac {JH}{JC} \cdot \frac {CL}{AL} \equal{} 1 \ \Longrightarrow \ \frac {UA}{UH} \equal{} \frac {AL}{JH}$

Por teorema de Menelao a la transversal $ \overline{U'AH}$ en $ \triangle BGF$ se obtiene  $\frac {U'A}{U'H} \equal{} \frac {AG}{FH}$

Por otro lado de la semejanza $ \triangle O_bGA \sim \triangle O_cZA$ se tiene $ \frac {O_bA}{O_cA} \equal{} \frac {AG}{AL}$

Las proyecciones ortogonales de $ O_b,O_c,A$ sobre  la recta $ BC$ son $ F,J,H.$ Entonces por teorema de Thales

$ \frac {O_bA}{O_cA} \equal{} \frac {FH}{JH} \Longrightarrow \ \frac {FH}{JH} \equal{} \frac {AG}{AL} \Longrightarrow \frac {UA}{UH} \equal{} \frac {UA'}{U'H} \Longrightarrow \ U \equiv U'$

Es claro que $ \triangle AMN$ es simétrico a  $ \triangle ABC$ con respecto a la recta $ O_bO_c,$ por ende la altura $ AH$ de $ \triangle ABC$ es isogonal de la  A-Altura del $ \triangle AMN$ con respecto al ángulo comun $ \angle BAC.$ A saber, ésta isogonal $ AU$ contiene el circuncentro de $ \triangle AMN.$


Could you post some resources for mathcounts practice? thanks.


How would you distinguish between acetone and cyclohexanone:

1) in the laboratory?

2) at home?

Wild guesses

1. Boiling Point (cyclohexanone should have a higher boiling point)

2. Solubility in water (acetone should be more soluble in water)

Both of you are right, although the iodoform test is much more simpler than finding boiling points.

Probably at home we can distinguish it by seeing the evaporation.
Acetone evaporates at room temperature itself whereas cyclohexanone does not. :wink:


I propose two problemes

probleme 1

                                     [b] find all functions Z->Z such that for all X,Y,T belongs to Z*
                                         F( X^2 + Y^2 -5T^2 ) = X^2 + F( Y^2 ) +5F( -T^2 )[/b]
probleme 2

                                     [b]find all integers n for wich these exist an integer k such that
                                                           (2n)!/n!=(2k+1)*2^2005[/b]

it's a really good solution

mine is about writing n in base 2 and using legendary formula i proved that the highest power of two in (2n)!/n! is n so we get n=2005.

ok now try to solve the first probleme!


I clicked on the link, and I don't quite understand what you're trying to do....are we supposed to use a technique that me@home did, or are you just trying to give us some inequalities to do?


group theory
abstract algebra
superior algebra
superior algebra unsolved

what is the order of normalizer of Sylow p-subgroup of simple groups A_n?


[quote="ernie"]I'm taking 10B at Fairfield, for my first time.  Anyone else?[/quote]

Does your school offer it at least? :lol:

Mine doesn't.

[quote="Ihatepie"]Did you just talk to yourself? :D

Anyways, I'm also taking it there. My school only offers the 12.

What grade are you in?[/quote]

I'm in 8th grade.  Which kid were you? :D :D :D

I was the Indian kid all the way in the back (behind the kid that looked more Indian than me).  Were you one of the two in the front-right corner?

I was the guy on the far left who was in the middle row. They guy to my right was wearing glasses and twitching the entire time. I'm Japanese-Brazilian (little Armenian) and I'm pretty tall at around 5-8. I don't remember looking right so I guess I don't know who you are. Oo

I was somewhat short (5-1).  Sigh, I think I saw you. :P


integration
inequalities
real analysis
real analysis unsolved

Set
\[ \mathcal E(x) \equal{} \begin{cases} \frac 1{2\pi} \log|x| & \text{if } n \equal{} 2, \\
\minus{} \frac 1{n(n \minus{} 2)\omega_n} \frac 1{|x|^{n \minus{} 2}} & \text{if } n > 2, \end{cases}
\]
and for any $ f \in L^1(\Omega)$, consider the Newtonian potential:
\[ (Nf)(x) \equal{} \int_\Omega \mathcal E(x \minus{} y)f(y) dy.
\]
(It is known and is easily verified that $ N \colon L^p(\Omega) \to L^p(\Omega)$).

1) Given $ p > n$, show that $ N \colon L^p(\Omega) \to C^1(\overline\Omega)$.
2) More precisely, if $ p \equal{} \frac n{1 \minus{} \alpha}$, $ 0 < \alpha < 1$, then prove that $ N \colon L^p(\Omega) \to C^{1,\alpha}(\overline\Omega)$ is a bounded linear operator (that $ N$ maps $ L^p(\Omega)$ into $ C^{1,\alpha}(\overline\Omega)$ is a part of the statement).

[b]P.S.[/b] If you want, prove (or disprove) 1) only. I just want to make sure that the standard assumption that $ p \equal{} \infty$ can be weakened.

Denote \[ \omega _N  \equal{} {\text{ volume of unit ball in }} \mathbb{R}^N  \equal{} \frac{{2\pi ^{\frac{N}
{2}} }}
{{N\Gamma \left( {\frac{N}
{2}} \right)}}.
\]Then $ \mathcal E(x)$ is the normalized fundamental solution of Laplace's equation, that is $ \minus{}\Delta u \equal{} 0$. By simple computation, one has
\[ \begin{gathered}
  D_i \mathcal E\left( {x \minus{} y} \right) \equal{} \frac{1}
{{N\omega _N }}\left( {x_i  \minus{} y_i } \right)\frac{1}
{{\left| {x \minus{} y} \right|^N }} \hfill \\
  D_{ij} \mathcal E\left( {x \minus{} y} \right) \equal{} \frac{1}
{{N\omega _N }}\left( {\left| {x \minus{} y} \right|^2 \delta _{ij}  \minus{} N\left( {x_i  \minus{} y_i } \right)\left( {x_j  \minus{} y_j } \right)} \right)\frac{1}
{{\left| {x \minus{} y} \right|^{N \plus{} 2} }}. \hfill \\
\end{gathered}
\]In order to prove (1), the following inequalities are usefull
\[ \begin{gathered}
  \left| {D_i \mathcal E\left( {x \minus{} y} \right)} \right| \leqslant \frac{1}
{{N\omega _N }}\frac{1}
{{\left| {x \minus{} y} \right|^{N \minus{} 1} }} \hfill \\
  \left| {D_{ij} \mathcal E\left( {x \minus{} y} \right)} \right| \leqslant \frac{1}
{{\omega _N }}\frac{1}
{{\left| {x \minus{} y} \right|^N }} \hfill \\
  \left| {D^\beta  \mathcal E\left( {x \minus{} y} \right)} \right| \leqslant C\left( {N,\left| \beta  \right|} \right)\frac{1}
{{\left| {x \minus{} y} \right|^{N \plus{} \left| \beta  \right| \minus{} 2} }}. \hfill \\
\end{gathered}
\]$ (Nf)(x)$ is called Newtonian potential of density $ f$. It is found in a book by Gilbarg and Trudinger that

[list]
$ \star$ If $ f$ is bounded and integrable in $ \Omega$ then $ Nf \in C^1(\mathbb R^N)$.
$ \star$ If $ f$ is bounded and locally Holder continuous with exponential $ \alpha \leq 1$ in $ \Omega$ then $ Nf \in C^2(\mathbb R^N)$.
[/list]

So, how Newtonian potential is important? The reason is that the solvability of the following problems are equivalent
\[ \boxed{\begin{gathered} {\text{Laplace}} \hfill \\
\Delta u \equal{} 0 \hfill \\
\end{gathered} } \Longleftrightarrow \boxed{\begin{gathered} {\text{Poisson}} \hfill \\
\Delta u \equal{} f \hfill \\
\end{gathered} }
\]
Indeed, set $ v \equal{} u \minus{} Nf$ then the problem $ \Delta u \equal{} f$ in $ \Omega$ together with $ u \equal{} \varphi$ on $ \partial \Omega$ is equivalent to the problem $ \Delta v \equal{} 0$ in $ \Omega$ with $ v \equal{} \varphi \minus{} Nf$ on $ \partial \Omega$. The choice of $ Nf$ is that for the weak density potential $ f$ is given, the high differentiable level of $ Nf$ will be obtained.

You misunderstood the problem. I know that the statements you put near the "*" sign are true. Look at your first "*". You demand that $ f \in L^\infty(\Omega)$. I say it is [i]too strong[/i], claiming [i]it is sufficient[/i] for $ f$ to belong to $ L^p(\Omega)$ with $ p > n$ only. Moreover, I claim that [i]not only[/i] $ Nf \in C^1(\mathbf R^n)$ but [i]also[/i] $ Nf \in C^{1,\alpha}(\mathbf R^n)$ for $ \alpha$ specified in 2). Yet, I claim that $ N \colon L^p(\Omega) \to C^{1,\alpha}(\overline\Omega)$ is a [i]bounded linear operator[/i]. You read the topic inattentively.

[quote="volodja"]You misunderstood the problem.[/quote]

Oops, I did not give any solutions here, just fundamential background in brief.


induction
extremal principle
number theory proposed
number theory

Find all positive integers $ n$ such that $ 2^{n\minus{}1} \equal{} \minus{}1 (mod n)$

Let p be a prime divisor of n. We see that p must be odd. $ p|n|2^{kp\minus{}1}\plus{}1$ and by FLT $ p|2^{p\minus{}1}\minus{}1$ => $ p|2^{p\minus{}1}(2^{(k\minus{}1)p}\plus{}1)\rightarrow p|2^{(k\minus{}1)p}\plus{}1$ => $ p|2^{p\minus{}1}(2^{(k\minus{}2)p\plus{}1}\plus{}1)\rightarrow p|2^{(k\minus{}2)p\plus{}1}\plus{}1$ etc... => $ p|2^{0\cdot p\plus{}1}\plus{}1\rightarrow p\equal{}3$ or $ p|2^{0\cdot p}\plus{}1$ - contradiction. So $ n\equal{}3^k$ is the only solution, and it's true for every natural k (which can be proven easily by induction).

[quote="limes123"]$ p|2^{p \minus{} 1}(2^{(k \minus{} 1)p} \plus{} 1)\rightarrow p|2^{(k \minus{} 1)p} \plus{} 1$ => $ p|2^{p \minus{} 1}(2^{(k \minus{} 2)p \plus{} 1} \plus{} 1)\rightarrow p|2^{(k \minus{} 2)p \plus{} 1} \plus{} 1$ etc... => $ p|2^{0\cdot p \plus{} 1} \plus{} 1\rightarrow p \equal{} 3$ or $ p|2^{0\cdot p} \plus{} 1$[/quote]
This step is wrong.
$ 2^{kp \minus{} 1} \equiv 2^{k(p \minus{} 1) \plus{} k \minus{} 1} \equiv 2^{k \minus{} 1} \bmod p$, and we don't nescesarilly have $ k \minus{} 1 \equiv 1 \bmod p \minus{} 1$..

(And $ n \equal{} 3^k$ doesn't hold for $ k\equal{}1$ or $ k\equal{}2$)

[quote="matex (L)(L)(L)"]Find all positive integers $ n$ such that $ 2^{n \minus{} 1} \equal{} \minus{} 1 (mod n)$[/quote]
[hide="Solution"]
Let $ p_i$ be the prime factors of $ n$, when $ i$ ranges from $ 1$ to $ x$.  Consider the numbers $ p_i \minus{} 1$.  Let $ p_y \minus{} 1$ be the one that divides the smallest power of $ 2$ (if there are more, then just consider any one of them).  Let $ p_y \equal{} p$, so let $ 2^l\parallel{}(p \minus{} 1)$.  Now, we have that $ 2^{n \minus{} 1}\equiv \minus{} 1\bmod p$.  Since $ n$ is the product of product of primes congruent to $ 1\bmod 2^l$, we have that $ n\equiv 1\bmod 2^l$, so $ n \minus{} 1\equiv 0\bmod 2^l$.  Letting $ k$ be the order $ 2\bmod p$, we have that $ n \minus{} 1$ must be a multiple of $ \frac {k}{2}$, but not of $ k$.  We have that $ k$ is a factor of $ p \minus{} 1$, so the power of two that fully divides $ k$ is at most $ 2^l$.  Yet, $ 2^l|(n \minus{} 1)$, so if $ \frac {k}{2}|(n \minus{} 1)$, then $ k|(n \minus{} 1)$, which is a contradiction.  We conclude that $ n$ has no prime factors, so it is $ n\equal{}1$ is the only solution. [/hide]

[quote="The QuattoMaster 6000"][quote="matex (L)(L)(L)"]Find all positive integers $ n$ such that $ 2^{n \minus{} 1} \equal{} \minus{} 1 (mod n)$[/quote]
[hide="Solution"]
Let $ p_i$ be the prime factors of $ n$, when $ i$ ranges from $ 1$ to $ x$.  Consider the numbers $ p_i \minus{} 1$.  Let $ p_y \minus{} 1$ be the one that divides the smallest power of $ 2$ (if there are more, then just consider any one of them).  Let $ p_y \equal{} p$, so let $ 2^l\parallel{}(p \minus{} 1)$.  Now, we have that $ 2^{n \minus{} 1}\equiv \minus{} 1\bmod p$.  Since $ n$ is the product of product of primes congruent to $ 1\bmod 2^l$, we have that $ n\equiv 1\bmod 2^l$, so $ n \minus{} 1\equiv 0\bmod 2^l$.  Letting $ k$ be the order $ 2\bmod p$, we have that $ n \minus{} 1$ must be a multiple of $ \frac {k}{2}$, but not of $ k$.  We have that $ k$ is a factor of $ p \minus{} 1$, so the power of two that fully divides $ k$ is at most $ 2^l$.  Yet, $ 2^l|(n \minus{} 1)$, so if $ \frac {k}{2}|(n \minus{} 1)$, then $ k|(n \minus{} 1)$, which is a contradiction.  We conclude that $ n$ has no prime factors, so it is $ n \equal{} 1$ is the only solution. [/hide][/quote]
The exact same solution as I came up with in school today. (Except i used induction instead of extremal principle, but they are the same in this case).
Do you know a generalisation or a related problem, cause I can't seem to make up a good problem which is related to this one. (i.e. uses the same technique)


A sequence is generated in which each term is 5 less than the previous term. If the 20th term is -32, what is the 15th?

My way. {If you have any questions feel free to ask.}

First, find the common difference. Which is given to you, as -5.
∴ Let the common difference equal d. {d=-5}

∴ A20= A15+5d.
It is given A20= -32

∴ -32= A15+5d
∴ -32= A15+5(-5)
∴ -32= A15-25
∴ -7= A15.

By the way, ∴ means 'Therefore'.

[quote="vallon22"]well, since we know the 20th term, and we wanna find the 15th term, it would probably be easier to count back

the 20th termis -32, so the 19th term would be -27, 18th is -22, 17th is -17, 16th is -12, and 15 is -7

our answer is -7

edit, beaten
but forte's way is more formula-like
mine is what you would do is you forgot the formula =) :rotfl:[/quote]

Haha, damn right! Formulas! ♥ ^^  :P

wait where did the d come from? and a20 etc.....

PiLuvah, in an arithmetic sequence, the first term is generally called $ a$ and the common difference is $ d$.

So the sequence looks like this:

$ a, a\plus{}d, a\plus{}2d, a\plus{}3d, ...$

[quote="Textangle"][quote="Forte"]First, find the common difference. Which is given to you, as -5.
∴ Let the common difference equal d. {d=-5} [/quote]

d is the common difference, or -5, which means that for every term in the sequence, or for every d, you subtract 5.

Anyways, $ 5( \minus{} 5) \equal{} 25$, $ \minus{} 32 \minus{} ( \minus{} 25) \equal{} \minus{} 7$

As for A20, I think it means the 20th term (subscripts people! :) )[/quote]

Yeah, thats right. I'm sorry I didn't clarify that.  :P


Let $x_{1},x_{2}, ..., x_{n}$ be positive real numbers such that \[x_{1}+x_{2}+...+x_{n}=1\]
Prove that \[\frac{a^{x_{1}-x_{2}}}{x_{1}+x_{2}}+\frac{a^{x_{2}-x_{3}}}{x_{2}+x_{3}}+...+\frac{a^{x_{n}-x_{1}}}{x_{n}+x_{1}}\geq \frac{n^{2}}{2}\]
holds true for every positive real number $a$. Determine also when the equality holds.

One Liner:

[hide]

where the indices are taken modulo , and the solution follows.

[/hide]


See http://www.mathlinks.ro/Forum/viewtopic.php?t=4344 .

  darij

Let the reflection of $E$ across $AB,BC,CD,DA$ are $P,Q,R,S$ and let $EP,EQ,ER,ES$ intersect $AB,BC,CD,DA$ at $K,L,M,N$.clearly $AKEN$ is cyclic,and $EK=KP$,$EN=NS$,so $KN\parallel PS$,hence $\angle KAE=\angle KNE=\angle PSE$,similarly $\angle EDM=\angle RSE$,$\angle KBE=\angle PQE,\angle ECM=\angle RQE$,but $\angle EAK+\angle EBK+\angle ECM+\angle EDM=180$ because $\angle AEB=90$,hence $\angle PSR+\angle PQR=180$.

Let's say that the  reflections intersects the sides $BA$,$AD$,$DC$ and $CB$ at the points $A"$,$B"$,$C"$ and $D"$, respectively. And let's name the points of the reflections $A'$,$B'$,$C'$ and $D'$ across the sides $BA$,$AD$,$DC$ and $CB$ , respectively. Since the quadrilaterals $A'B'C'D'$ and $A"B"C"D"$ have parallel sides, they are similar. So, we must to show that the quadrilateral $A"B"C"D"$ is cyclic: from the cyclic quadrilaterals $AA"EB"$,$DB"EC"$,$C"ED"C$ and $D"BA"E$ we have $\angle B"AE=\angle B"A"E $, $\angle B"DE=\angle B"C"E$,$\angle D"C"E=\angle D"CE$ and $\angle D"BE=\angle D"A"E$, respectively. And since $\angle B"AE+\angle B"DE=90^\circ$ and $\angle D"BE +\angle D"CE=90^\circ$. From equal angles we get $\angle B"A"E+\angle D"A"E+\angle D"C"E+\angle B"C"E=180^\circ$, as desired.

[quote=aayush-srivastava]homothety(dilation)centred at E with ratio $-2$ brings the reflections back to the feet of perpendiculars simplifying the solution[/quote]

Shouldn't the ratio be $\frac{1}{2}$? Can anyone explain reason of ratio being $-2$?

[hide=Proof]
Let $P, Q, R, S$ be the foot of perpendicular from $E$ onto $AB, BC, CD, DA$, respectively.

[b]Claim 1:[/b] $\{P, E, A, S\}, \{P, B, E, Q\}, \{Q, E, R, C\}, \{R, B, S, E\}$ are concyclic.
\[\angle EPA  = \tfrac12\pi = \pi - \tfrac12\pi = \pi - \angle ESA\]
Thus, $P, E, A, S$ are concyclic.
Similarly, for other sets of points.

[b]Claim 2:[/b] $P, Q, R, S$ are concyclic.
\begin{align*}
\angle SPQ &= \angle SPE + \angle EPQ = \angle SAE + \angle EDQ\\
&= \angle BAE + \angle EDC = (\tfrac12\pi - \angle ABE) + (\tfrac12\pi - \angle ECD)\\
&= \pi - \angle SBE - \angle ECQ = \pi - \angle SRE - \angle ERQ\\
&= \pi - \angle SRQ
\end{align*}

[b]Main Proof:[/b]
Let $W, X, Y, Z$ be the reflections of $E$ in $AB, BC, CD, AD$, respectively.
Consider the homothety at $E$ with factor $2$, $PQRS \to WXYZ$.
Since, $PQRS$ is concyclic, $WXYZ$ are concyclic.

And we are done.
[/hide]

Let the reflections of $E$ over $AB,BC,CD,DA$ respectively be  $P,S,R,Q$ respectively.

Let $\omega_A,\omega_B,\omega_C,\omega_D$ be the circles centered at $A,B,C,D$ with radius $EA,EB,EC,ED$.

We see that, $P=\omega_A\cap\omega_B, Q=\omega_A\cap\omega_D, R=\omega_C,\omega_D$ and $S=\omega_A\cap\omega_C$.

Now By an inversion $\Psi$ centered at $E$ with an arbitary radius, (let the circle be $\omega$) we see that $\omega_A\mapsto$ Radical Axis of $\omega$ and $\omega_A$. Similarly for others. Let $\omega_A,\omega_B,\omega_C,\omega_D\mapsto \omega_A',\omega_B',\omega_C',\omega_D'$ respectively .

Now as $P=\omega\cap\omega_A$. Hence, $P\mapsto\omega_A'\cap\omega_B'$. Similarly for others.

If $P,Q,R,S\mapsto P',Q',R',S'$ then $P'Q'R'S'$ is a rectangle which is actually a cyclic quadrilateral. Now again inverting back we get that $PQRS$ is a cyclic quadrilateral. Hence, proved. $\blacksquare$.

Take a homothety centered at $E$ with scale factor $\frac{1}{2}$, which brings the reflection of $E$ across each side of $ABCD$ to the projection from $E$ onto that side. Because dilations preserve circles, it suffices to prove that these four projections are concyclic, so let the projection from $E$ onto $AD, AB, BC$, and $CD$ be $W, X, Y$, and $Z$, respectively.

Because $\angle{EXB}=\angle{EYB}=90$, we have that $EXBY$ is cyclic and thus $\angle{EXY}=\angle{EBY}$, and by right angles, $\angle{EBY}=\angle{CEY}$. Because $\angle{EZC}=\angle{EYC}=90$, we have that $EYCZ$ is cyclic as well, so thus $\angle{CEY}=\angle{YZC}$, and therefore $\angle{EXY}=\angle{YZC}$. Now we do basically the same angle chase on the "other side": we know $\angle{AWE}=\angle{AXE}=90$, so $\angle{WXE}=\angle{EAW}$, and by right angles, $\angle{EAW}=\angle{WED}$. Because $\angle{DWE}=\angle{EZD}=90$, we have that $WEZD$ is cyclic as well, so thus $\angle{WED}=\angle{WZD}$, and therefore $\angle{WXE}=\angle{WZD}$.  Hence, $\angle{WXY}=\angle{WXE}+\angle{EXY}=\angle{WZD}+\angle{YZC}=180-\angle{WZY}$, so $\angle{WXY}+\angle{WZY}=180$, so thus $WXYZ$ is cyclic, and we are done.

let E1 E2 E3 E4 be the reflectons E. It is req to prove that E1 E2 E3 E4 are cocyclic
define the intersections of E and E1 E2 E3 E4 with the sides of the quadrilateral as P1 P2 P3 P4
consider a circular inversion centered at E with arbitary radius.
it is obvious that P1^ P2^ P3^ P4^ are cocyclic.
So P1 P2 P3 P4 must also be concyclic.
moreover we notice a homothety centered at E which maps P1 P2 P3 P4 to E1 E2 E3 E4 resp.
So E1 E2 E3 E4 must be concyclic as well (as desired)

let $E,F,G,H$ be foot of altitudes from $E$ to sides. obviously $EFGH$ is cyclic. with scaling with center of $E$ and $k=2$ we are done.

Let $X,Y,Z,W$ denote the foot of the perpendicular from $AB,AD,DC,BC$ respectively. Let the reflection of $E$ over $AB,AD,DC,CB$ be $X’,Y’,Z’,W’$. $X’Y’Z’W’$ are $XYZW$ merely dilations of each other, so by homothety, it suffices to prove that $XWZY$ is cyclic. Noting that $AXEY, BWEX,WCZE, YEZD$ are all cyclic,
$$\angle XWE + \angle XYE = \angle XBE + \angle XAE = 90$$
$$\angle EYZ + \angle EWZ = \angle EDZ + \angle ECZ = 90$$
Hence, $\angle XWE + \angle XYE + \angle EYZ + \angle EWZ = \angle XYZ + \angle XWZ = 180$, which implies that $XYZW$ is cyclic, as desired.

$\angle AEB + \angle CED = \pi \implies E$ has isogonal conjugate ($E'$) in $ABCD$ and it's well-known that $E'$ is a center of circle in question.

[quote=MithsApprentice]Let $\, ABCD \,$ be a convex quadrilateral such that diagonals $\, AC \,$ and $\, BD \,$ intersect at right angles, and let $\, E \,$ be their intersection.  Prove that the reflections of $\, E \,$ across $\, AB, \, BC, \, CD, \, DA \,$ are concyclic.[/quote]

[Hide=My Inversion Solution]
The points $X,Y,Z,T$ are reflections of $\, E \,$ across $\, AB, \, BC, \, CD, \, DA \,$.
$ BX=BE=BT\implies \omega_1$is center $B$.(radius-$BE$)
Then
$ \omega_1,\omega_2,\omega_3,\omega_4$

Inversion $\Phi$ around $E$.

Then

$\Phi(\omega_1)\parallel \Phi(\omega_3),\Phi(\omega_2)\parallel \Phi(\omega_4).$

Then

$\Phi(X)\Phi(Y)\Phi(Z)\Phi(T)-$rectangle $\implies X,Y,Z,T-$cyclic.
[/hide]

Let the feet of the altitudes from $E$ to $AB$, $BC$, $CD$, and $DA$ be $E_A$, $E_B$, $E_C$, and $E_D$, respectively. There is a homothety centered at $E$ with factor $2$ sending $E_AE_BE_CE_D$ to the quadrilateral with vertices of the reflections of $E$ over $AB$, $BC$, $CD$, and $DA$, so we just need to prove that $E_AE_BE_CE_D$ is cyclic. We see that $AE_AEE_D$, $BE_BEE_A$, $CE_CEE_B$, and $DE_DEE_C$ are cyclic, because $\angle AE_AE=\angle BE_BE=\angle CE_CE=\angle DE_DE=90^\circ$. Therefore, $\angle EE_AE_B=\angle EBC$, $\angle EE_CE_B=\angle ECB$, $\angle EE_AE_D=\angle DAE$, and $\angle EE_CE_D =\angle ADE$. Therefore, $$\angle E_BE_AE_D+\angle E_BE_CE_D=\angle EBC+\angle ECB+\angle DAE+\angle ADE=180^\circ$$ so $E_AE_BE_CE_D$ is cyclic.

Let $A_0, B_0, C_0, D_0$ be the reflections of $E$ across $AB$, $BC$, $CD$, and $DA$, respectively. Let $A_1, B_1, C_1, D_1$ denote the feet of the altitudes from $E$ to $AB, BC, CD, DA$, respectively.

Note that the quadrilateral $A_1B_1C_1D_1$ is the result after applying a homothety to $A_0B_0C_0D_0$ with scale factor $\frac{1}{2}$, centered at $E$. Thus, $A_0B_0C_0D_0$ is cyclic iff $A_1B_1C_1D_1$, so it suffices to show that $A_1B_1C_1D_1$ is cyclic.

We have \begin{align*}
\angle AD_1E  \\
= \angle AA_1 E  \\
= \angle DC_1E  \\
= \angle D D_1 E  \\
= \angle C C_1 E  \\
= \angle CB_1 E  \\
= \angle B B_1 E  \\
= \angle B A_1 E \\
=90^{\circ}, \\
\end{align*}so the quadrilaterals $AD_1EA_1, DC_1ED_1, CB_1EC_1$, and $BA_1EB_1$ are all cyclic.

Thus, we get \begin{align*}
\angle D_1A_1E = \angle D_1 AE \\
\angle D_1C_1 E = \angle D_1 D E \\
\angle B_1 C_1 E = \angle B_1 C E \\
\angle B_1 A_1 E = \angle B_1 B E \\
\end{align*}.

Using this information, we deduce, \begin{align*}
\angle B_1A_1D_1 + \angle B_1C_1D_1 \\
= \angle B_1A_1 E + \angle D_1 A_1 E + \angle D_1 C_1 E + \angle B_1 C_1 E \\
=\angle B_1 B E + \angle D_1 AE + \angle D_1 D E + \angle B_1 C E \\
 = (\angle B_1 B E + \angle B_1 C E ) + (\angle D_1 AE+ \angle D_1 D E) \\
= (\angle CBE + \angle BCE) + (\angle DAE + \angle ADE) \\
= 90^{\circ} + 90^{\circ}  \\
= 180^{\circ}, \\
\end{align*}so $A_1B_1C_1D_1$ is cyclic.

If we take a homothety centered at $E$ to the reflections of $E$ over the sides with ratio $1/2$, we will get the feet of the altitudes from $E$ to the sides. If we prove this is concylic, then the reflections will be as well by rules of homothety. Now let $F,G,H,I$ be the feet of the perpendiculars from $E$ to $AB,BC,CD,DA$ respectively. Notice that $HEID$ is cyclic so $\angle HIE = \angle HDE = \angle CDE$. Similarly, $\angle FIE = \angle FAE = \angle BAE.$ Furthermore, $\angle FGE = \angle FBE = \angle ABE$ and $\angle HGE = \angle HCE = \angle DCE$. But $$\angle HGF + \angle HIF = \angle HIE + \angle IEF + \angle FGE + \angle EGH = \angle CDE + \angle BAE + \angle DCE + \angle ABE = 180,$$ so $FGHI$ is cyclic and we are done.

Wait can someone please check this. This was too easy for USAMO P2 :o  I see people using directed angles but is that necessary?

Label the reflected points $E_1, E_2, E_3, E_4$ over $AB, BC, CD, DA$ respectively. Consider a homothety centered at $E$ with ratio $\frac{1}{2}$. This brings the points $E_1, E_2, E_3, E_4$ on the sides of the quadrilateral $ABCD$. Notice that $EE_1 \perp AB$, $EE_2 \perp BC$, $EE_3 \perp CD$, $EE_4 \perp DA$ by reflection properties. Thus we see that $$E_1BE_2E$$ $$E_2CE_3E$$ $$E_3DE_4E$$ $$E_4AE_1E$$ are all cyclic. So by cyclic condition we know that $$\angle{E_2E_1E} = \angle{E_2BE} = 90 - \angle{E_2CE} = 90 - \angle{E_2E_3E}.$$ Similarly,  $$\angle{E_4E_1E} = \angle{E_4AE} = 90 - \angle{E_4DE} = 90 - \angle{E_4E_3E}.$$ Thus we see that $\angle{E_4E_1E_2} + \angle{E_4E_3E_2} = 180$, which means $E_1E_2E_3E_4$ is cyclic. Since homothety preserves circles, we know that the quadrilateral in the problem is cyclic.

Here's a solution that stupidly does not use homotheties and is pure angle chase.

[hide=Sol]
Let $W, X, Y, Z$ denote the reflections over $AB$, $BC$, $CD$, and $DA$ respectively. We have that $$\measuredangle ZWX = \measuredangle ZWA + \measuredangle AWB + \measuredangle BWX$$ Since $AW = AE = AZ$, $A$ is the center of $(WEZ)$ and $\measuredangle ZWA = 90^{\circ} + \measuredangle ZEW$. Similarily, $\measuredangle BWX = 90^{\circ} + \measuredangle WEX$. As $\measuredangle AWB = \measuredangle BEA = 90^{\circ}$, it follows that $$\measuredangle ZWX = 90^{\circ} + \measuredangle ZEW + \measuredangle WEX = 90^{\circ} + \measuredangle ZEX$$ It follows from a similar process that \begin{align*}\measuredangle ZYX = \measuredangle ZYD + \measuredangle DYC + \measuredangle CYX &= \\ (90^{\circ} + \measuredangle ZEY) + (90^{\circ}) + (90^{\circ} + \measuredangle YEX) &= 90^{\circ} + \measuredangle ZEX\end{align*} Thus, $\measuredangle ZWX = \measuredangle ZYX$ and $W, X, Y, Z$ are concyclic.
[/hide]

Dilate the quadrilateral with scale factor 1/2 so that they end up on the sides of ABCD. Then we have <KFE+<GFE+<KHE+<GHE=<EAK+<EBG+<ECG+<EDK=180 degrees. $\blacksquare$

[quote=Taco12]what
-----
Invert about a circle centered at $E$ with radius $1$. It suffices to show $W^*X^*Y^*Z^*$ is cyclic, but this is obvious as it is a rectangle. $\blacksquare$[/quote]

ay cmon man why u using heavy artillery use [i]elementary[/i] techniques please. :)

@below what in the tomfoolery bullsh-- is that

[quote=huashiliao2020]Dilate the quadrilateral with scale factor 1/2 so that they end up on the sides of ABCD. Then we have <KFE+<GFE+<KHE+<GHE=<EAK+<EBG+<ECG+<EDK=180 degrees. $\blacksquare$

[quote=Taco12]what
-----
Invert about a circle centered at $E$ with radius $1$. It suffices to show $W^*X^*Y^*Z^*$ is cyclic, but this is obvious as it is a rectangle. $\blacksquare$[/quote]

ay cmon man why u using heavy artillery use [i]elementary[/i] techniques please. :)[/quote]

Now, with the addition of homographies and moving points, inversion is elementary lol

Let $F$, $J$, $H$, and $I$ be the feet of $E$ on $BC$,$BA$,$AD$, and $DC$
Because $90+90=180$, $BFEJ$, $FEIC$, $HEID$, and $AJEH$ are cyclic
Let $\angle JFE=a$ and $\angle EFI = b$
$\angle IFC=90-b$
$\angle IEC=90-b$
$\angle DEI=180-90-(90-b)=b$
$\angle IHD = b$
$\angle FHI = 180-90-b=90-b$

$\angle BFJ = 90-a$
$\angle BEJ = 90-a$
$\angle JEA = 180-90-(90-a)=a$
$\angle JHA = a$
$\angle JHE = 90-a$

$\angle JHI = \angle FHI + \angle JHE = 180-a-b = 180 - \angle JFI$ so $JFIH$ is cyclic
The reflections of $E$ is quadrilateral $JFIH$ scaled x2, so its also cyclic

Let $E_\ell$ denote the projection of $E$ onto line $\ell$. Clearly $AE_{DA}EE_{AB}$, $BE_{AB}EE_{BC}$, $CE_{BC}EE_{CD}$, and $DE_{CD}E_{DA}$ are all cyclic, meaning
\begin{align*}
    \angle E_{AB}E_{BC}E_{CD}&=\angle EE_{BC}E_{AB}+\angle EE_{BC}E_{CD}\\
    &=\angle EBE_{AB}+\angle ECE_{CD}\\
    &=(90^\circ-\angle EAE_{AB})+(90^\circ-\angle EDE_{CD})\\
    &=180^\circ-(\angle EAE_{AB}+\angle EDE_{CD})\\
    &=180^\circ-(\angle EE_{DA}E_{AB}+\angle EE_{DA}E_{CD})\\
    &=180^\circ-\angle E_{AB}E_{DA}E_{CD}.
\end{align*}
So $E_{AB}E_{BC}E_{CD}E_{DA}$ is cyclic as well, and taking a homothety at $E$ with scale factor $2$ finishes. $\square$

[hide=Easy problem][b]Solution:[/b]Reflect point $E$ over sides $AB$,$BC$,$CD$,$DA$ respectively to say, $E_1,E_2,E_3,E_4$
Let, $EE_1 \cap AB=P,$
    $EE_2 \cap BC=Q,$
    $EE_3 \cap CD=R,$
    $EE_4 \cap AD=S,$
[b]Claim:[/b] Quadrilateral $PQRS$ is cyclic.
[b]Proof:[/b] Notice that quadrilaterals $APES,PBQS,CRES,RDSE$ are all cyclic (opposite angles are $90$).We use directed angles ( modulo $\pi$).\
We have $\measuredangle PQB=\measuredangle PEB=\measuredangle BAE=\measuredangle PAE=\measuredangle PSE$. Also $\measuredangle EQP=90-\measuredangle PQB= 90-\measuredangle PSQ$
Doing similar angle chasing in the other side, we get $\measuredangle RQS =90-\measuredangle QSR$.
Now we have $\measuredangle PSR+\measuredangle RQP=90+90=180$.
Our claim is thus proved.

Now consider a homothety centered at $E$ with a scale factor $2$. This send quadrilateral $PQRS$ to quadrilateral $E_1E_2E_3E_4$. Thus $E_1E_2E_3E_4$ is a cyclic quadrilateral as well.Our proof is thus complete.[/hide]

Denote the feet from $E$ to the sides of $ABCD$ as $K$, $L$, $M$, and $N$. Noting the clear cyclicities, we invert at $E$ with arbitrary radius. Then $K^*L^*M^*N^*$ is a (cyclic) rectangle, so $KLMN$ must also be cyclic. We finish with a homothety of scale factor 2. $\blacksquare$

Indeed a good design in the diagram is observable  :)  .
[b]Solution :[/b] Let the reflections of $E$ across $AB,BC,CD,DA$ be $X,Y,Z,W$ and let $EX \cap AB=P, EY \cap BC=Q, EZ \cap CD=R, EW \cap DA=S$. So we have $PE \perp AB$ and $ES \perp AD$ so we get $APES$ as cyclic quadrilateral. Similarly we have $EPBQ,EQCR,ERDS$ as cyclic quadrilaterals. Now we shall move on with a claim first .

[b]Claim 1 : [/b] $PQRS$ is a cyclic quadrilateral.

[b]Proof : [/b]Observe that due to those cyclic quadrilaterals we have $$\angle{SPQ}+\angle{SRQ}=(\angle{SPE}+\angle{EPQ})+(\angle{SRE}+\angle{ERQ})=(\angle{SAE}+\angle{QBE})+(\angle{SDE}+\angle{ECQ})$$ $$=(\angle{SAE}+\angle{SDE})+(\angle{QBE}+\angle{ECQ})=\frac{\pi}{2}+\frac{\pi}{2}=\pi.$$

[b]Claim 2 : [/b] $SP||XW,PQ||XY,QR||YZ,RS||ZW$

[b]Proof : [/b] Since $X$ is the reflection of $E$ across $AB$ and $XE \cap AB=P$ so we have $PX=PE$, similarly we have $SW=SE$ so by Midpoint Theorem we get $PS||XW$ . Similarly we have $PQ||XY,QR||YZ,RS||ZW$.

 [b]Finish : [/b] So we have $\angle{PSE}=\angle{XWE}$ and $\angle{RSE}=\angle{ZWE}$ . Adding we get $$\angle{PSR}\angle{XWZ}$$ and simlarly we have $$\angle{PQR}=\angle{XYZ}$$. Adding again we get $$\angle{XYZ}+\angle{XWZ}=\angle{PQR}+\angle{PSR}=\pi$$ since $PQRS$ is cyclic. So we are done !

[hide= Inversion Solution]Let, $W$,$X$,$Y$,$Z$ be the reflections of $E$ over the sides $AB$,$BC$,$CD$, and $AD$ respectively.
Define circle $\omega_A$ as the circle centered at $A$ and with radius $AE$. Similarly define $\omega_B$, $\omega_C$ and $\omega_D$ respectively. It is clear that $W$ is the point of intersection of the circles $\omega_A$ and $\omega_B$. Similar results hold for $X$,$Y$ and $Z$.
Now invert about $E$ with an arbitrary radius. It is clear that ${\omega_A}*$ is a line parallel to $BD$ and perpendicular to $AE$. Again similar results hold for the other inverted clines. Also $W^*$ is the intersection of ${\omega_A}^*$ and ${\omega_B}^*$. Similarly define $X^*$, $Y^*$ and $Z^*$.
It is clear that the inverted quadrilateral of $WXYZ$ is cyclic (it is a rectangle). Thus $WXYZ$ must also be cyclic.
Our proof is thus complete. [/hide]
I found homothety on this problem way easier than inversion lol  :blush:


algorithm
number theory unsolved
number theory

$(a_{n},b_{n})$ where $a_{n}=n+[\frac{\sqrt{5}-1}{2}n]$   $b_{n}=2n+[\frac{\sqrt{5}-1}{2}n]$  is a solution. But I still don't know how to get this.

Since such sequences are |arbitrary|, there is no way to find any rule.

You are theorically right ZetaX, but in such problems there is an "obvious" pattern to discover.
In this particular case, the algorithm is $a_{n}$ = least number not already used, and $b_{n}=a_{n}+n$.

Pierre.

In another hand, if you see that the sequences $(a_{n})$ and $(b_{n})$ divide the set of the positive integers in two disjoint sets, you may think to Beatty's sequences, so that $a_{n}= [xn]$ and $b_{n}=[yn]$ for all $n$, for some positive irrational numbers $x,y$, such that $y=x+1$ (see above) and $\frac{1}{x}+\frac{1}{y}= 1$. This leads $x = \frac{\sqrt 5+1}2$ as desired.

Pierre.


articles
geometry
geometric transformation

I'd like to know if any of you ever came across works of other mathematicians that never got translated to english ? I'm sure there's many but which ones in particular stuck you ? and how important do you think it is to know other languages in the world of maths especially ?

Given that I was born and educated in Russia, it is easy to guess that have I read (and am still reading) plenty of such stuff. The Russian math. journals were only partially translated into English during the Soviet Union times and vice versa, so it was not unusual that some results were "rediscovered" across the Atalntic a few times (an American colleague of mine told me once that his worst nightmare was proving some brilliant theorem, publishing it, and then discovering that it was used as a lemma in some article in an obscure Russian journal over 15 years ago; the stories with the roles exchanged were, probably, not uncommon either).

English (or, at least, broken English) is gradually becoming a standard for the scientific communication (just as Latin was in the 18-19th centuries), so the most important stuff is either written in English or translated immediately. Still, I benefit quite a bit from my knowing Russian and being able to read mathematical papers in French. Perhaps, some day we'll all want to learn some basic Chinese too (though, probably, not tomorrow).

In general, you can live and thrive in the mathematical world with English only today but it never hurts to know an additional language.

I wish my Chinese was good enough to read Chinese math articles, I use Chinese math books, but my mom has to help me with the translation


Let $ a,b,c > 0$ . Prove that
 $ \frac{a^3}{b}\plus{}\frac{b^3}{c}\plus{}\frac{c^3}{a}\geq\frac{3(a^4\plus{}b^4\plus{}c^4)}{a^2\plus{}b^2\plus{}c^2}$

[quote="PHONGVAN"]Let $ a,b,c > 0$ . Prove that
 $ \frac {a^3}{b} \plus{} \frac {b^3}{c} \plus{} \frac {c^3}{a}\geq\frac {3(a^4 \plus{} b^4 \plus{} c^4)}{a^2 \plus{} b^2 \plus{} c^2}$[/quote]
By the Cauchy Schwarz we have
\[ \frac{{a^3 }}{b} \plus{} \frac{{b^3 }}{c} \plus{} \frac{{c^3 }}{a} \ge \frac{{\left( {a^2  \plus{} b^2  \plus{} c^2 } \right)^2 }}{{ab \plus{} bc \plus{} ca}}
\]
Hence it suffices us to show that
\[ \left( {a^2  \plus{} b^2  \plus{} c^2 } \right)^3  \ge 3\left( {ab \plus{} bc \plus{} ca} \right)\left( {a^4  \plus{} b^4  \plus{} c^4 } \right)
\]
Assume $ a\plus{}b\plus{}c\equal{}1$ and denote $ ab \plus{} bc \plus{} ca \equal{} \frac{{1 \minus{} q^2 }}{3},r \equal{} abc$ then the inequality becomes
\[ \left( {1 \plus{} 2q^2 } \right)^3  \ge \left( {1 \minus{} q^2 } \right)\left( {6q^4  \minus{} 24q^2  \plus{} 108r \minus{} 3} \right)
\]
Because we have $ r \le \frac{{\left( {1 \minus{} q} \right)^2 \left( {1 \plus{} 2q} \right)}}{{27}}$ so we just need to prove
\[ \left( {1 \plus{} 2q^2 } \right)^3  \ge \left( {1 \minus{} q^2 } \right)\left( {6q^4  \plus{} 8q^3  \minus{} 36q^2  \plus{} 1} \right)
\]
\[ \Leftrightarrow q^2 \left( {14q^4  \plus{} 8q^3  \minus{} 30q^2  \minus{} 8q \plus{} 43} \right) \ge 0
\]
Obviously true, equality holds if and only if $ a\equal{}b\equal{}c$, :).

[quote="nguoivn"][quote="Honey_S"]
Hence it suffices us to show that
$ \left( {a^2 \plus{} b^2 \plus{} c^2 } \right)^3 \ge 3\left( {ab \plus{} bc \plus{} ca} \right)\left( {a^4 \plus{} b^4 \plus{} c^4 } \right)$
[/quote]
It isn't trues, my friend  :wink:[/quote]
Thank you, nguoivn. I see my mistake,  :oops: .

I got it to $ (a^4\plus{}b^4\plus{}c^4)(a^2\plus{}b^2\plus{}c^2) \ge 3(a^5b\plus{}b^5c\plus{}c^5a)$. Is this true?

[quote="dgreenb801"]I got it to $ (a^4 \plus{} b^4 \plus{} c^4)(a^2 \plus{} b^2 \plus{} c^2) \ge 3(a^5b \plus{} b^5c \plus{} c^5a)$. Is this true?[/quote]
It's wrong. Try $ a\equal{}1.3,$ $ b\equal{}1$ and $ c\equal{}0.$

Felix, he means as c gets infinitely close to 0, for example 0.0001.

[quote="dduclam"][quote="PHONGVAN"]Let $ a,b,c > 0$ . Prove that
 $ \frac {a^3}{b} \plus{} \frac {b^3}{c} \plus{} \frac {c^3}{a}\geq\frac {3(a^4 \plus{} b^4 \plus{} c^4)}{a^2 \plus{} b^2 \plus{} c^2}$[/quote]

I think us to be employable S.O.S method, But seem as plenty complexity.  :mad:[/quote]
It's wrong! Try $ a\equal{}5,$ $ b\equal{}3$ and $ c\equal{}2.$ :wink:

when i was trying... i just multiplied it out and got the reverse for the given inequality by muirhead ....where could have this gone wrong??


Let a,b,c be positive numbers such that :a+b+c=1.Prove that:
a^2/(a^2+b) + b^2/(b^2+c) + c^2/(c^2+a) ≥ 3/4.

[quote="kunny"][quote="quykhtn-qa1"]Let a,b,c be positive numbers such that :$ a \plus{} b \plus{} c \equal{} 1$.Prove that:
$ \frac {a^2}{a^2 \plus{} b} \plus{} \frac {b^2}{b^2 \plus{} c} \plus{} \frac {c^2}{c^2 \plus{} a} \geq \frac 34$.[/quote][/quote]
After $ \sum_{cyc}\frac {a^2}{a^2 \plus{} b}\equal{}\sum_{cyc}\frac {a^4}{a^4 \plus{} a^2b}\geq\frac{(a^2\plus{}b^2\plus{}c^2)^2}{\sum(a^4\plus{}a^2b)}$ use the Vasc's inequality:
$ (a^2\plus{}b^2\plus{}c^2)^2\geq3(a^3b\plus{}b^3c\plus{}c^3a).$ :wink:

[quote="Dumel"]I also used Vasc's inequality and CS. my solution:
$ \sum_{cyc}^{} \frac {a^2}{a^2 \plus{} b} \equal{} \sum_{cyc}^{} \frac {a^2}{a^2 \plus{} b \sum_{}^{} a} \equal{} \sum_{cyc}^{} \frac {a^2}{a^2 \plus{} b^2 \plus{} ba \plus{} bc} \ge \frac {( \sum_{}^{} a^2)^2}{\sum_{}^{} a^4 \plus{} \sum_{}^{} a^2b^2 \plus{} abc \sum_{}^{} a \plus{} \sum_{cyc}^{}a^3b } \ge \frac {( \sum_{}^{} a^2)^2}{\sum_{}^{} a^4 \plus{} \sum_{}^{} a^2b^2 \plus{} abc \sum_{}^{} a \plus{} \frac {1}{3}( \sum_{}^{} a^2)^2 } \ge \frac {3}{4} \Leftrightarrow \sum_{}^{} a^2b^2 \ge abc \sum_{}^{} a$
it's obviously true by Muirhead or AM-GM[/quote]
It is similary arqady's solution. :oops:


If $x^2+y^2+z^2=kxzy$ has infinate interger solutions in $(x,y,z)$, find all possible values for k. (constant)

Can't think of any off the topic of my head, but definitely k != 0 :P .

That does not provide infinate solutions!


Is this true for a three digit number? If it is, prove it.

To find the square of a 3-digit number ending in 5, take all of the digits exept the last 5. Add one to that number, and then square it. After that, subtract the number you squared from the square number. Then add 25 at the end.

For example, let's solve 135^2
take the 13 and add one to it, making it 14.
squaring it makes it 196.
Then we subtract that number we just squared (14) from the squared number we just obtained (196) and we get 182.
adding 25 at the end of that gives us 18225.

I think it works for all numbers ending in 5.

I seriously didn't see your blog before I posted this, i wonder how I posted this out of nowhere and then we ended up on the same topic.
I thought about this while lying on the floor bored  :)

My blog is certainly not the first place this has been discussed.  It's been a problem solving trick since long before I was born and many people have discovered it over time.  I just thought you might enjoy reading the article since it goes into the algebraic reasoning behind it.

[quote="236factorial"]I guess it isn't polite to ask when you were born, but I did prove that rule shortly after I explored it. Are there any similar tricks like this??[/quote]

There are many tricks of arithmetic, some of which are more useful and some of which are more artificial than others.  Just knowing how to expand expressions like $(x+1)^2$ might help you learn to square numbers like 81 in your head.  If you generalize that idea and practice using it, you can learn how to perform a great deal of mental arithmetic.

An understanding of algebraic expressions lies at the heart of many arithmetic tricks.

I think that one of the most useful factoring tricks is

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

This is an actual countdown problem (i think from chapter though...)

The sum of two numbers is 50 and their difference is 12. What is the difference of their squares?

quickly the answer is 600 (50 * 12). of course you can do this the long way and set up
x + x + 12 = 50
x = 19
x + 12 = 31
31^2 - 19^2 is 600. (961 - 361)

[quote="MCrawford"]You might enjoy reading through [url=http://www.artofproblemsolving.com/Forum/weblog.php?w=7&month=1&year=2005&day=28]this[/url].[/quote]i enjoyed reading that

if the countdown round ever asked for this and you knew it, you would be the luckiest guy  :)  Of course, if your opponent just happens to know it too...


function
algorithm
algebra
polynomial
algebra proposed

Hi  ,  My math teacher gave me 2 days ago this problem , and I'm stuck :blush:  , so I would like a hint  please  :roll: :

Find all functions from N to N , where N defines the set of natural numbers ,  suc that :
$ f(n) \plus{} f(f(2n))\plus{}f(f(f(4n)))\equal{}7n$

Thanks / :)

[quote="KapioPulsar"][quote="serguei"]Hi  ,  My math teacher gave me 2 days ago this problem , and I'm stuck :blush:  , so I would like a hint  please  :roll: :

Find all functions from N to N , where N defines the set of natural numbers ,  suc that :
$ f(n) \plus{} f(f(2n)) \plus{} f(f(f(4n))) \equal{} 7n$

Thanks / :)[/quote] i found the fruction f(n)=n  :oops:[/quote]

Can u please give me a hint  :)

[quote="serguei"]Can u please give me a hint  :)[/quote]

it seems nobody has hint for the moment. There is a trivial solution $ f(n)\equal{}n$, as KapioPulsar said you and I dont think you need any hint to show this.

The question is to know if there are some other solutions (my opinion is that we have infinitely many solutions for this equation) and what are they.

[quote="serguei"]I tought that he showed that $ f( n ) \equal{} n$ is the only solution,   dear pco can you give me other solutions  than $ Id_{N}$ ? Thanks  :)[/quote]

I dont have explicit examples but it seems that it is possible to build such functions step by step :

for each value, choose the fewest number which dors not lead to a contradiction :

$ f(1)\equal{}1$
$ f(2)\equal{}1$
$ f(3)\equal{}1$
$ f(4)\equal{}5$
$ f(5)\equal{}4$
$ f(6)\equal{}1$
$ f(7)\equal{}1$
$ f(8)\equal{}8$
$ f(9)\equal{}1$
$ f(10)\equal{}1$
$ f(11)\equal{}1$
$ f(12)\equal{}13$
$ f(13)\equal{}14$
$ f(14)\equal{}19$
$ f(15)\equal{}1$
$ f(16)\equal{}14$
....
This is not a proof since I'm not quite sure I can continue this algorithm up to $ \plus{}\infty$ but it explains why I said that I think there is infinitely many solutions.

Dont hesitate, if we dont find any proof, to post your teacher's solution when you'll get it.

hi PCO
excuse me but I didn't understand your previous post
how do you say that? :?:

easy to prove is that f(0)=0 beacause f:N->N and if we got n<>0 $ \frac{f(n)}{n}\plus{}\frac{f(f(n))}{n}\plus{} \frac{f(f(f(n)))}{n}\equal{}4$ so beacause f(A)=N we takw that f(n)=kn , k is  {1,2,3,4,5,6,7} i am not sure  :blush:

how can we go through this considering that zero isnt contained in the set of naturals?

[quote="pco"][quote="KapioPulsar"]so f(n)=kn. k is {1,2,3,4,5,6,7}.. is the final answer?[/quote]
Did you just verify ?

$ f(n) \equal{} 2n$ is surely not a solution ...
$ f(n) \equal{} 3n$ is surely not a solution ...
...

So nobody found a solution different from $ f(n) \equal{} n$ and nobody up to now was able to show that $ f(n) \equal{} n$ was the unique solution.

We are waiting for serguei telling his teacher that he did not find. We hope he'll ask him the clever answer and kindly post it in this forum.[/quote]if with my way say that beacause for k=2,3,4,5,6,7 it not solutions we get that the only solution is the f(n)=n  :blush:

since f is from N to N f should be polynomial
deg(f) shouldn't be more than 1 because in that case it wont be =7n(because deg(7n)=1)
so deg(f)=1
so f(n)=an+b
and a>=1 becuase if a<1 the equation won't be correct
so a>=1
by substitution in the equation we find:
 i)a + 2a^2 + 4a^3 = 7
ii)3b + 2ab + (a^2).b = 0
if a isnt equal to 1 so a>=2 and then the equation won't be correct so a=1
and by a=1 we find in ii that b=0
so f(n)=n
this the unique f

[quote="irantst"]since f is from N to N f should be polynomial[/quote]
Surely not : there exists infinitely many  functions from $ \mathbb N\to\mathbb N$ which are not  polynomials (for example, $ f(n)\equal{}1\plus{}[2010\sin(3^n)]$ :) )

[quote="irantst"]hi PCO
I got surprised
your answer is absolutely true and I made a mistake
but I want to know that how do you find such difficult f liks f(n)=1+[2010\sin(3^n)]
very nice
bye[/quote]

It was a kind of joke. You can take $ f(n)=1+[|h(n)|]$ where $ h(n)$ is any function from $ \mathbb N\to \mathbb ^R$ (and btw I made a mistake and would have written $ 1+[2010|\sin(3^n)|]$.
So you can choose

$ 1+[|\frac{2012}n|]$

$ 1+[\ln(n)]$

...

oh
I taught that it is true and I said the answer to someone
but thanks


I believe you can get this from [url=http://en.wikipedia.org/wiki/Sharkovskii%27s_theorem]Sharkovskii's theorem[/url], or else the argument behind it.

[quote="bakerbakura"]If $ n$ is a positive integer, $ 2^m$ is the highest power of $ 2$ dividing $ n$, and $ f : \mathbb{R} \rightarrow \mathbb{R}$ is a continuous function such that $ f^n(x) \equal{} x$ for all $ x \in \mathbb{R}$, prove that $ f^{2^m}(x) \equal{} x$ for all $ x \in \mathbb{R}$.[/quote]
Let $ g(x) \equal{} f^{2^m}(x), k \equal{} \frac{n}{2^m}$. Then $ g^k(x) \equal{} x$ and we should prove $ g(x) \equal{} x$ when $ k$ is odd. And we can prove that $ g$ is continuous too.

Because of $ g^k(x) \equal{} x$: $ g$ is bijective. ($ g(a) \equal{} g(b) \Rightarrow g^{k\minus{}1}(g(a)) \equal{} g^{k\minus{}1}(g(b)) \iff a\equal{}b$). Since $ g$ is continuous we see that $ g$ must be monotomic.
So for some $ a$: $ g(a) \equal{} a$.
First assume that $ g$ is increasing.
If $ g(b) > b$ then: $ g^k(b) > g^{k\minus{}1}(b) > ... > g(b) > b$
If $ g(b) < b$ then: $ g^k(b) < g^{k\minus{}1}(b) < ... < g(b) < b$
Hence $ g(b) \equal{} b$.
Now assume that $ g$ is decreasing.
Assume $ g(a) \neq a \forall a \in \mathbb{R}$.
Since $ g$ is continuous, we either have $ g(a) < a$ or $ g(a) > a$ $ \forall a \in \mathbb{R}$. If $ g(a) > a$, then $ g^k(a) > g^{k\minus{}1}(a) > ... > g(a) > a$, which gives a contradiction, and likewise with $ g(a) < a$.
$ g(a) \equal{} a$.
For all $ x > a$: $ g(x) < g(a) \equal{} a$. So $ g(g(x)) > g(g(a))\equal{}a$ and $ g^3(x) < g(a) \equal{} a$, and so on. So $ g^k(x) < g(a) \equal{} a < x$. Contradiction. Hence $ g$ is increasing and $ g(x) \equal{} x \forall x \in \mathbb{R}$


vector
geometry
rectangle
analytic geometry
algebra unsolved
algebra

A 2 x 4 rectangle is divided into 8 squares of side 1. Let M be the set of 15 vertices thus
obtained.
Find all points A in M for which it is possible to arrange the set \{A} into pairs (A1;B1),
(A2;B2), : : :, (A7;B7) such that

A1B1+A2B2+...+AnBn=0 where A1B1,...,AnBn,O are all vectors.

Check this [url=http://www.olimpiade.ro/data/2004_Matematica_Judeteana_Bareme_Clasa%20a%20IX-a_0.pdf]pdf file[/url]

Your pdf file would be useful if it was not in some foreign language!

Bomb

geez, wat's the point of posting here? u get the solutions anyway...

just add 2sum(OB_i) to both sides (vectors), then consider your 14 remaining position vectors in coordinate form. then the sum of x-coordinates is even and the sum of y-coordinates is even. if u put the origin as the 'bottom-left' corner, then (0,1)(2,1)(4,1) are the only satisfactory points which you can remove to enable this.

examples verify these as valid solutions.

- Sam.


AMC
AIME
geometry
ratio
trigonometry
trig identities
Law of Cosines

13.  AB = √30, BC = √15, CA = √6. M is the midpoint of BC. ∠ADB = 90 degrees. Find area ADB/area ABC.

[hide] I found out that the ratio is AD / 2 AM but how can i find

those lengths...[/hide]

For those who enjoy brute forcing, the following way is almost guaranteed to get you an answer with 10 minutes of algebra, which is more than acceptable on the AIME:

[hide]

The area of ABC/ABD = 2AMD/ADB = 2AM/AD.  AM can be found with stewart's.  Then, apply Law of cosines on AMB to find cos(DAB), and use this info to set up another law of cosines / pythagorean system of equations with ADB.  Solve for AD, and we have the answer

I personally am sober from the law of cosines, and I try to avoid it whenever possible, so I'll post the entire solution when i find a more elegant way.

[/hide]


If I intentionally leave an empty line, how do I have it not show up as a bad box?  I have tried putting white text on the line, but that messed things up.

Do the bad boxes affect anything??  :maybe:

I just ignore whatever bad boxes come up.  :lol:

[quote="ehehheehee"]Not really (unless text runs off the page), but I don't like it when I build the file and something like "24 bad boxes" shows up.[/quote]

I agree. It just makes you feel really bad inside.

There could be lots of reasons for the problem so could you give an example of where you get it then we can see what is happening and how to overcome it.

I want to intentionally leave a line blank to show division.

So it do something like this: text text text\\\\

And it says it is a bad box.

When I do something like this: text text text\\a\\

It is not a bad box.

I tried putting white letters (\\\textcolor{white}{text}\\) so it gives the impression of an empty line, but then adobe reader has a pop up when I get to that line that says "unrecognizable token 'gray' was found" (I also have pictures).  Then, the whole page with that line is blank.

\\\\ willgive an underfull \hbox because LaTeX can't left or right-justify a blank line. So what you need to do is add extra vertical space so there is no line with nothing in it. There are various ways to do this - choose whichever you prefer.
Instead of \\\\ use
1. \bigskip \\ or
2. \medskip \\ or
3. \vspace{2ex} \\ or
4. \\*[2ex]

\bigskip and \medskip (and there's also a \smallskip) give preset vertical spacing.

2ex=2 times the height of the letter x. You can have any number you wish or change the units to in, cm, mm or other units.


This is more of an algorithmic doubt than programming....

Prob. 24.1-5 of CLRS

Let $ G \equal{} ( V , E )$  be a weighted, directed graph with weight function $ w : E \longrightarrow R$. Give an $ O(V
E)$ -time algorithm to find, for each vertex $ v \in V$ , the value $ \delta^{*}(v) \equal{} min_{u \in V} \{ \delta(u, v) \}$.

I was thinking of adapting the standard Bellman-Ford algorithm for this problem, with its distance function exchanged with the new $ \delta ^{*} (v)$ function. I think it satisfies the same basic properties as $ d [v]$ / $ \delta (v)$ as in the standard algorithm .......

I would appreciate any ideas on this!!

Thnx

$ \delta (u, v)$  is the weight for the shortest path from vertex $ u$ to vertex $ v$, where each edge is assigned a weight.

So $ \delta^*(v)$ is the distance from $ v$ to the closest vertex, right? Do you want to compute $ \delta^*(v)$ for each vertex or only for some given $ v$? Bellman-Ford should work fine if you mean to use it to compute all the possible distances $ \delta(u,v)$ for fixed $ v$ and then take the minimum for answer. If you want to find $ \delta^*(v)$ for all $ v$ then Floyd-Warshall will be faster.

Now I realized that $ \delta(u,v)$ is not the same as $ \delta(v,u)$... but that doesn't really change anything.

well,as the question is , I think we need to calculate $ \delta^{*} (v)$ for each vertex .......

I agree with you that an all pairs shortest path algorithm (Floyd-Warshall) would work best for our problem (when we need to calculate it for all vertices) , but given the fact that the problem was in a chapter before floyd-warshall was presented, I am motivated to believe that it is possible to achieve an $ O(VE)$ complexity with bellman-ford or another single source shortest path algorithm

Do you think that the 'change' i have proposed to bellman ford will work?

Oh Nice!!!

That pretty much solves the problem.......

any generic method/reason for applying such a trick?!

It's a common trick to try to create a single source or destination.  Note that the given objective function involves the min function, and so does the shortest-path problem.  So the idea is to smuggle the min calculation into the graph itself.

Hi,

i haven't got point of hint presented by Ravi_B. How could I solve the given problem (24.1-5 of CLRS) by adding dummy vertex [i]s[/i]? Yes, i can get single-source (I suppose [i]s[/i] is this single source of the graph). Then I could run Bellman-Ford with modified function [i]Delta(u,v)[/i] . How to do?

Adding dummy vertex s helped me to solve problem 24.1-6 (Suppose that a weighted, directed graph [i]G = (V, E)[/i] has a negative-weight cycle. Give an efficient algorithm to list the vertices of one such cycle. Prove that your algorithm is correct.) Because I got single source which is every vertex reachable from so i could detect negative weight cycle by running Bellman-Ford.

But i can't solve primordial problem of this thread. Can someone help me? I haven't got basic idea  :blush: .

Ahh :wallbash: I got it.. stupid me.. I didn't understand problem correctly. I thought I have to compute min shortest path weight from vertex v of V to any vertex u of V - min (delta(v,u)) for each vertex v of V. But the problem is to compute for each vertex v the shortest path weight to v from any of the
vertices. Then using the dummy vertex s as single-source of Bellman-Ford is really great idea.
I apologize my "spam" posts.

Btw. Is there a algorithm which solves computing for each vertex v the shortest path weight from v to any of the vertices in time better than Floyd-Warshall?

If all edges have positive weight, then running dijkstra with a heap $ V$ times gives a running time of $ O(V^2\log V)$.

Yep... Even i have noticed Johnson's algorithms is asymptotically better than Floyd-Warshall (to be able to be negative edges)....

[quote="Ravi B"][hide="Hint"]Add a dummy vertex $ s$ to the graph.  Add dummy edges from $ s$ to every other vertex; assign each of these dummy edges a weight of 0.[/hide][/quote]

one question this is for every other vertex or for all vertices?

By "every other vertex", I meant "every vertex except $ s$".

i have not got the point of the hint. Would you please give me an introduction in detail .
thanks


[quote="Thales418"]i think the answer is $0.5$ but i cant find a easy solution

this is true?![/quote]
What is your solution? :D

$x_{n}$ incress too much

and using approximations summing a few numbers we get approximatelly $0.5$

Just a Physics Easy Way to do things. Using Approximations.

I just want to know if my reply is rigth or wrong.

Yes, it is $\frac{1}{2}$. Now, please post your solution!

I dont have a solution! i said this

I just use approximations and use the fact that x grows very very fast

:(

sorry

This is more of an algebra than a calculus problem.

From the given recurrence we see that $x_{n+1}= x_{n}^{2}+3x_{n}+1$ (with a variety of proofs), which means that $1+x_{n+1}= (1+x_{n})(2+x_{n})$, or:

$\frac{1}{x_{n+1}}= \frac{1}{(1+x_{n})(2+x_{n})}= \frac{1}{1+x_{n}}-\frac{1}{2+x_{n}}$

or: $\frac{1}{2+x_{n}}= \frac{1}{1+x_{n}}-\frac{1}{1+x_{n+1}}$ (1)

Summing (1) we see that the sum telescopes:

$\sum_{k=0}^{\infty}\frac{1}{2+x_{k}}= \lim_{n}\sum_{k=0}^{n}\frac{1}{2+x_{k}}= \lim_{n}\left( \frac{1}{1+x_{1}}-\frac{1}{1+x_{n+1}}\right)$

which, since $x_{n}\rightarrow+\infty$, clearly equals $\frac{1}{2}\; \Box$

[quote="cmad"]From the given recurrence we see that $x_{n+1}= x_{n}^{2}+3x_{n}+1$[/quote]

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=435103#p435103]here[/url] for a problem related to that first step. It gives a proof as well.

me too  :wink:
[quote="cmad"]

From the given recurrence we see that $x_{n+1}= x_{n}^{2}+3x_{n}+1$ (with a variety of proofs), which means that $1+x_{n+1}= (1+x_{n})(2+x_{n})$, or:

$\frac{1}{x_{n+1}}= \frac{1}{(1+x_{n})(2+x_{n})}= \frac{1}{1+x_{n}}-\frac{1}{2+x_{n}}$

or: $\frac{1}{2+x_{n}}= \frac{1}{1+x_{n}}-\frac{1}{1+x_{n+1}}$ (1)

Summing (1) we see that the sum telescopes:

$\sum_{k=0}^{\infty}\frac{1}{2+x_{k}}= \lim_{n}\sum_{k=0}^{n}\frac{1}{2+x_{k}}= \lim_{n}\left( \frac{1}{1+x_{1}}-\frac{1}{1+x_{n+1}}\right)$

which, since $x_{n}\rightarrow+\infty$, clearly equals $\frac{1}{2}\; \Box$[/quote]


three consecutive positive prime numbers have a sum that is a multiple of 7. what is the least possible sum?

no calc allowed

You pretty much have to do this one out, since there is no formula for the nth prime.


Hi!I'm a new member and I have just started learning Physics...

I've got a question: :?:

What is the meaning of "uniform motion"?
Does it mean "uniform [size=150]speed[/size]"?Or "uniform [size=150]velocity[/size]"?

What we learned in school was that they're both types of uniform motion. Pretty pointless phrase if you ask me...


This is a tough AMC problem:
For how many positive integers n less than or equal to 24 is n! evenly divisible by 1+2+3...+n?

juse use n(n+1)/2. and check each number less than 24 for n.

Rephrase:  When does $ n(n+1) | 2n!$?

Second rephrase:  When does $ n+1 | 2(n-1)!$?

[hide="Answer"] Precisely when $ n+1$ is not prime.  If $ n+1$ is composite, then its factors appear in the expansion of $ (n-1)!$. [/hide]


geometry
circumcircle
geometric transformation
reflection
parallelogram
geometry proposed

Let $ ABCDEFGHILMN$ be a regular dodecagon. Let $ P$ be the meet of $ AF$ and $ DH$, and $ S$ be the circumcircle of $ APH$. Prove that:

i) the radius of $ S$ is the same as the radius of the circumscribed circle of $ A...N$;

ii) the centre of $ S$ belongs to $ HN$

iii) $ PE \equal{} AB$

[hide="outline"]i) Clearly $ AH \equal{} AF$, $ \angle AHD \equal{} \angle DFA$, $ \angle HAF \equal{} \angle HDF \equal{} \angle DHF \equal{} \angle DAF$, and so by ASA triangles $ APH$ and $ ADF$ are congruent and have same circumradius.

ii) Let $ U, V$ be the centers of the dodecagon and $ S$, respectively. Then $ AU \equal{} AV \equal{} UH \equal{} VH$ and $ UV$ is perpendicular to $ AH$. Hence $ AUHV$ is a rhombus. Then $ AU \parallel HV$ and so $ V$ is on $ HN$.[/hide]

i)We have that $ AF//MH$ and $ DH//AM$ so $ APHM$ is a parralellogram so the radius of the circumcircles of $ APH$ and $ AHM$ are equal.
ii)The circumcircle of $ APH$ is the reflection of the circumcircle of $ AMH$ with respect $ AH$ and because the reflection of $ HN$ with respect $ AH$ is $ HB$ we have that the center of $ S$ belongs to $ HN$.
iii)Because of the parallelogram $ APHM$ we have that $ AP\equal{}MH\equal{}BE$ and $ BE//AP$ so $ APEB$ is a parallelogam so $ PE\equal{}AB$.


trigonometry
inequalities unsolved
inequalities

Let $ f(x)\equal{}4x^3\minus{}3x\plus{}1$.Consider all sequences $ \minus{}1\equal{} x_{0}\le x_{1}\le...\le x_{6}\le 1$.
Determine the maximum value of the expression:
$ P\equal{}\sum _{i\equal{}1}^{6}\left/ f(x_{i})\minus{}f(x_{i\plus{}1})\right/$

$ P \equal{} \sum_{i \equal{} 1}^6 |\cos 3\theta _{i \plus{} 1} \minus{} \cos 3\theta _{i}|$.

Kunny I don't know your point.Trigonometry Substitution

Let $ x \equal{} \cos \theta _{i}$, we have $ 4\cos ^ 3 \theta _{i} \minus{} 3\cos \theta _{i} \equal{} cos 3\theta _{i}$.


floor function
trigonometry
function
geometry
geometric transformation
reflection

Is there a closed form for the expression $ \sum_{i \equal{} 0}^{n} ( \minus{} 1)^{\lfloor\frac {i}{2}\rfloor} a^{n \minus{} i}b^i \binom{n}{i}$?
And when $ a^2 \plus{} b^2\equal{}1$, can the expression equal 1?

The answer to the first question is yes: it's $ \frac {(a \plus{} ib)^n \plus{} (a \minus{} ib)^n}{2} \plus{} \frac {(a \plus{} ib)^n \minus{} (a \minus{} ib)^n}{2i}$ where $ i \equal{} \sqrt { \minus{} 1}$.  (The basic idea here is something that I've seen described on the fora as a "roots of unity filter.")

As for the second question, let $ a \plus{} bi \equal{} e^{ix}$.  The sum then becomes $ \cos nx \plus{} \sin nx$.  Now, $ \cos \theta \plus{} \sin \theta \equal{} 1 \implies 1 \plus{} 2 \sin \theta \cos \theta \equal{} 1 \implies \sin 2 \theta \equal{} 0 \implies 2 \theta \equal{} k \pi, k \in \mathbb{Z}$.

Thanks for the fast answers. I am working on a difficult problem (at least in my opinion and in the way I'm trying to prove it), and your solutions help me to progress...

@JBL - I've seen this "roots of unity filter"' or stuff similar to it (for example - a beautiful article about generating functions in "Mathematical Reflections"), but I always forget it, even though it is quite logical.  :blush:

@t0rajir0u - thanks for the result. I think you have a small error actually ($ 2\theta\equal{}4k\pi, (4k\plus{}1)\pi, k \in Z$)

I will notify you when I will solve the problem...  :lol:

My apologies (I forgot that my implications weren't reversible).  Instead, just write $ \cos \theta \plus{} \sin \theta \equal{} 1 \Leftrightarrow \cos \left( \theta \minus{} \frac {\pi}{4} \right) \equal{} \frac {1}{\sqrt {2} } \Leftrightarrow \theta \minus{} \frac {\pi}{4} \equal{} 2 \pi k \pm \frac {\pi}{4}$.