calculus
integration
abstract algebra
calculus open

Hallo everybody,

by a chance I encountered the following kind of problems.

Given a hermitian positive-semidefinite integral kernel $K(x,y)$ find a general expression for $f(x)$ such that:

$\int f^*(x)K(x,y)f(y)\,dx dy=0$.

Particularly intresting is the solution of the problem under an additional conditition:$|f(x)|^2=F(x)$, where $F(x)$ is a non-negative real function.

Is the problem well known? Where can I read on it?

Thank you in advance.

[quote="seamusoboyle"]Er, wrong forum my friend[/quote]

What is the correct one? :)

[quote="dr-er"][quote="joml88"][quote="seamusoboyle"]Theres a calculus one further down the forum, mods dont like it too much when you post calculus in here. They'll probably move it.[/quote]

Exactly :D
[color=red]
Now moved from Intermediate.[/color][/quote]

Thank you! Hope to see some ideas... :)[/quote]

I have solved the "problem". The answer is trivial. The solution is any eigenfunction of the integral kernel with the eigenvalue 0.


geometry
3D geometry
tetrahedron
inequalities
geometry proposed

Prove that $ a,b,c,d > 0$ are the areas of the faces of a tetrahedron if and only if all the inequalities of this type are true :

$ {\begin{array}{ccc} a < b + c + d \\
b < a + c + d \\
c < a + b + d \\
d < a + b + c \end{array}}$ .

Romanian NMO , Shortlist

Call A the vertex opposit to the face a, and equal for B, C and D. Call H the projection of A in a. Denote with $ [XYZ]$ the area of the triangle XYZ. Then we have that $ d>[HBC]$, $ b>[HCD]$, $ c>[HBD]$ cause the altitude is bigger. Then $ a\equal{}[BCD]\equal{}[HBC]\plus{}[HCD]\plus{}[HBD]<b\plus{}c\plus{}d$ and equally for the other.


what is this rating thing you see on the left under my name?

it's the average numbers of stars(1-6)depending on how good people think your post is. By the way, what's your number?


Just wondering if any body has any comment about the new medication that there coming out with to prolong aging up to 50 years. Would you want to live that long why or why not?

It depends what I can do. I basically see no point, as I am the person that likes to do things and cannot 'kick back and relax' especially for another 50 years. There is a limited amount of things you can do at that age, and it will not be the best time ever.
It would be nice to see how the world changes.

My teacher had once said that if she had the opportunity to live much longer, but would be in a hospital bed hooked up with tubes the whole time, she wouldn't take the opportunity. Why live when you're not makin any difference?

i think a lot of ppl that just live to be 90 get pretty bored with their lives. :D

Supposely youare suppose to age slower so for example if you live to 150 you would feel like a 100 year old doses that make a diffrence?

only if the 50 years came extending your childhood rather than just live longer.

lol noneoftheabove.
When I was 6 I used to tell everyone that I'd be playing varsity football when I was 150, so if that medication would do that :lol: .

it compleatly matters about my condition at the age

What is more important than an Elixir of Life is an Elixir of Youth.
I would much rather leave till 90 as a 30 year old and die suddenly. But of course I would be happy to replace 90 in the above statement with 150! :D

By the way, who told you our average dying age is 100 :?:

Our average dying age is 77 :D If we could stay as old as we are now...There are many things good about that (Mathcounts, Statistics...)

I was just giving an example i know less then 10% of people make it to 100.

[quote="bubka"]What is more important than an Elixir of Life is an Elixir of Youth.
I would much rather leave till 90 as a 30 year old and die suddenly. But of course I would be happy to replace 90 in the above statement with 150! :D [/quote]
Which would also replace 30 with 50  :)
I don't think I would want to live to 150. Definitely not forever.
[hide="my reason"] I don't want to outlive everybody.[/hide]

[quote="RiemannZetaFunction"]I personally would like to live a 1000 years or more you would get to see human kind progress. and you have to admit you could learn a ton.[/quote]
Ditto.

you would also see a lot of human degradation at that!

dunno

I don't like the prospect of becoming very slow, with short attention span, or even becoming senile

but if i can still walk around, and do stuff like taking showers etc without help( it must be hard not being able to do private stuff on your own) , and surf all over the internet, watch movies,  and do math, it should not be bad  :oops:

Consider you all that if you live 150 years, maybe you'll be a teen until you're 40... but well, I think it depends on how I am at 100.....

[quote="Vihang"]Mebbe dying(or maybe hibernating) at 80 and waking up after 70 years just to watch human progress is a good idea. ....[/quote]
Make that waking up for several days every 25.314159265 years.

There was a story by H.G. Wells in which a man went to sleep for many centuries and woke up in a new world

I prefer quality... someone can also live through the memories of people


Find all continuous functions $f: \mathbb R \to \mathbb R$ which satisfy
\[ f(f(x))=3f(x)-2x,\quad \forall x \in \mathbb R.\]

[quote="tanpham90"]Find all continuous functions $ f : R \to R$ which is satisfy the requirements :
$ f(f(x)) \equal{} 3f(x) \minus{} 2x$ $ \forall x \in R$[/quote]

$ 1)$ $ f(x) \equal{} x$ is a solution.
So we'll now consider functions for which it exists an $ x_0$ such that $ f(x_0)\neq x_0$

$ 2)$ $ f(x)$ is obviously injective, and so monotonous (since continuous).
If $ f(x_0) > x_0$, then $ f(f(x_0)) \equal{} 3f(x_0) \minus{} 2x_0 > f(x_0)$ and so $ f(x)$ is strictly increasing.
If $ f(x_0) < x_0$, then $ f(f(x_0)) \equal{} 3f(x_0) \minus{} 2x_0 < f(x_0)$ and so $ f(x)$ is strictly increasing.
If $ \lim_{x\rightarrow \plus{} \infty}f(x) \equal{} L$, then, in $ f(f(x)) \equal{} 3f(x) \minus{} 2x$, when $ x\rightarrow \plus{} \infty$, LHS$ \rightarrow f(f(L))$ and RHS$ \rightarrow \minus{} \infty$. So $ \lim_{x\rightarrow \plus{} \infty}f(x) \equal{} \plus{} \infty$
If $ \lim_{x\rightarrow \minus{} \infty}f(x) \equal{} l$, then, in $ f(f(x)) \equal{} 3f(x) \minus{} 2x$, when $ x\rightarrow \minus{} \infty$, LHS$ \rightarrow f(f(l))$ and RHS$ \rightarrow \plus{} \infty$. So $ \lim_{x\rightarrow \minus{} \infty}f(x) \equal{} \minus{} \infty$
So $ f(x)$ is a strictly increasing bijection and $ f(\mathbb{R}) \equal{} \mathbb{R}$

$ 3)$ Using $ f(f(x)) \equal{} 3f(x) \minus{} 2x$, it is easy to establish that $ f(u_{n}) \equal{} u_{n \plus{} 1}$ with $ u_0 \equal{} x_0$, $ u_1 \equal{} f(x_0)$ and $ u_{n \plus{} 2} \equal{} 3u_{n \plus{} 1} \minus{} 2u_n$ for any $ n\geq 0$. Then, since $ u_n \equal{} (2x_0 \minus{} f(x_0)) \plus{} (f(x_0) \minus{} x_0)2^n$, we have $ f(u_n) \equal{} 2u_n \minus{} (2x_0 \minus{} f(x_0))$
Using that $ f(x)$ is a bijection and that $ f(f(x)) \equal{} 3f(x) \minus{} 2x$ implies $ 2f^{ \minus{} 1}(x) \equal{} 3x \minus{} f(x)$, it is possible to show that the previous equalities are true even for $ n < 0$ and so :
$ f(x) \equal{} 2x \minus{} L_0$ with $ L_0 \equal{} 2x_0 \minus{} f(x_0)$ for any $ x\in\{(2x_0 \minus{} f(x_0)) \plus{} (f(x_0) \minus{} x_0)2^n\: \: \forall n\in\mathbb{Z}\}$.
With continuity, we have, when $ n\rightarrow \minus{} \infty$, $ f(L_0) \equal{} L_0$

$ 4)$ Now, consider $ f(x_0) > x_0$. If it exists $ x_1$( Wlog $ x_1 > x_0$) such that $ f(x_1) > x_1$ and $ f(x_1)\neq 2x_1 \minus{} L_0$, then we have $ f(L_0) \equal{} L_0$ and $ f(L_1) \equal{} L_1$.
For any $ L\in(L_0,L_1)$, with continuity, it exist a value $ a$ such that $ f(a) \equal{} 2a \minus{} L$. Applying then the demo of point 3 above starting with $ a$ and $ f(a)$, we conclude $ f(L) \equal{} L$, and so $ f(x) \equal{} x$ $ \forall x\in(L_0,L_1)$ which is in contradiction with $ f(x) \equal{} 2x \minus{} L_0$ for any $ x\in\{(2x_0 \minus{} f(x_0)) \plus{} (f(x_0) \minus{} x_0)2^n\: \: \forall n\in\mathbb{Z}\}$.
So $ f(x_0) > x_0$  implies :
$ f(x)\leq x$ $ \forall x\leq L_0$
$ f(x) \equal{} 2x \minus{} L_0$ $ \forall x\geq L_0$

$ 5)$ Now, consider $ f(x_0) < x_0$. The same demo as above gives :
$ f(x)\geq x$ $ \forall x\geq L_0$
$ f(x) \equal{} 2x \minus{} L_0$ $ \forall x\leq L_0$

And so the solutions :

$ a)$ $ f(x) \equal{} x$
$ b)$ Let $ L\in\mathbb{R}$. $ f(x) \equal{} x$ $ \forall x\leq L$ and $ f(x) \equal{} 2x \minus{} L$ $ \forall x\geq L$
$ c)$ Let $ L\in\mathbb{R}$. $ f(x) \equal{} x$ $ \forall x\geq L$ and $ f(x) \equal{} 2x \minus{} L$ $ \forall x\leq L$
$ d)$ Let $ L_0\leq L_1\in\mathbb{R}$. $ f(x) \equal{} 2x \minus{} L_0$ $ \forall x\leq L_0$, $ f(x) \equal{} x$ $ \forall x\in[L_0,L_1]$ and $ f(x) \equal{} 2x \minus{} L_1$ $ \forall x\geq L_1$

Thank a million , Patrick  :)  !
[quote="pco"]
$ 4)$ Now, consider $ f(x_0) > x_0$. If it exists $ x_1$( Wlog $ x_1 > x_0$) such that $ f(x_1) > x_1$ and $ f(x_1)\neq 2x_1 \minus{} L_0$, then we have $ f(L_0) \equal{} L_0$ and $ f(L_1) \equal{} L_1$.
For any $ L\in(L_0,L_1)$, with continuity, it exist a value $ a$ such that $ f(a) \equal{} 2a \minus{} L$. Applying then the demo of point 3 above starting with $ a$ and $ f(a)$, we conclude $ f(L) \equal{} L$, and so $ f(x) \equal{} x$ $ \forall x\in(L_0,L_1)$ which is in contradiction with $ f(x) \equal{} 2x \minus{} L_0$ for any $ x\in\{(2x_0 \minus{} f(x_0)) \plus{} (f(x_0) \minus{} x_0)2^n\: \: \forall n\in\mathbb{Z}\}$.
So $ f(x_0) > x_0$  implies :
$ f(x)\leq x$ $ \forall x\leq L_0$
$ f(x) \equal{} 2x \minus{} L_0$ $ \forall x\geq L_0$
[/quote]
Can you explain more detail , please ! What is $ L_1$ and why do we have $ f(L_1)\equal{}L_1$

[quote="tanpham90"]Thank a million , Patrick  :)  !
[quote="pco"]
$ 4)$ Now, consider $ f(x_0) > x_0$. If it exists $ x_1$( Wlog $ x_1 > x_0$) such that $ f(x_1) > x_1$ and $ f(x_1)\neq 2x_1 \minus{} L_0$, then we have $ f(L_0) \equal{} L_0$ and $ f(L_1) \equal{} L_1$.
For any $ L\in(L_0,L_1)$, with continuity, it exist a value $ a$ such that $ f(a) \equal{} 2a \minus{} L$. Applying then the demo of point 3 above starting with $ a$ and $ f(a)$, we conclude $ f(L) \equal{} L$, and so $ f(x) \equal{} x$ $ \forall x\in(L_0,L_1)$ which is in contradiction with $ f(x) \equal{} 2x \minus{} L_0$ for any $ x\in\{(2x_0 \minus{} f(x_0)) \plus{} (f(x_0) \minus{} x_0)2^n\: \: \forall n\in\mathbb{Z}\}$.
So $ f(x_0) > x_0$  implies :
$ f(x)\leq x$ $ \forall x\leq L_0$
$ f(x) \equal{} 2x \minus{} L_0$ $ \forall x\geq L_0$
[/quote]
Can you explain more detail , please ! What is $ L_1$ and why do we have $ f(L_1) \equal{} L_1$[/quote]

Sure, it's not very clear. i'm sorry.
let's try in a different manner :

Let $ x_0$ and $ x_1$ such that $ f(x_0)>x_0$ and $ f(x_1)>x_1$

Then we have shown in the point 3 above that :
$ f(x) \equal{} 2x \minus{} L_0$ with $ L_0 \equal{} 2x_0 \minus{} f(x_0)$ for any $ x\in\{(2x_0 \minus{} f(x_0)) \plus{} (f(x_0) \minus{} x_0)2^n\: \: \forall n\in\mathbb{Z}\}$.
$ f(L_0)\equal{}L_0$
And similarly :
$ f(x) \equal{} 2x \minus{} L_1$ with $ L_1 \equal{} 2x_1 \minus{} f(x_1)$ for any $ x\in\{(2x_1 \minus{} f(x_1)) \plus{} (f(x_1) \minus{} x_1)2^n\: \: \forall n\in\mathbb{Z}\}$.
$ f(L_1)\equal{}L_1$

Then, if $ L_0\neq L_1$, WLOG say $ L_0<L_1$, Let $ L\in(L_0,L_1)$
The line $ g(x)\equal{}2x\minus{}L$ is below the curve $ f(x)$ at point $ x\equal{}L_0$ and is above the curve $ f(x)$ at point $ x\equal{}L_1$.
So, using continuity, it exist a value $ a\in(L_0,L_1)$ such that $ f(x)\equal{}g(x)$ and so $ f(a) \equal{} 2a \minus{} L$. Applying then the demo of point 3 above starting with $ a$ and $ f(a)$, we conclude $ f(L) \equal{} L$, and so $ f(x) \equal{} x$ $ \forall x\in(L_0,L_1)$ which is in contradiction with $ f(x) \equal{} 2x \minus{} L_0$ for any $ x\in\{(2x_0 \minus{} f(x_0)) \plus{} (f(x_0) \minus{} x_0)2^n\: \: \forall n\in\mathbb{Z}\}$.
So $ L_0\equal{}L_1$ and :

$ f(x_0) > x_0$  implies :
$ f(x)\leq x$ $ \forall x\leq L_0$
$ f(x) \equal{} 2x \minus{} L_0$ $ \forall x\geq L_0$

And the reminder of the demo.

Is it a bit more clear ?.

Now I understand ! Thank you Patrick   :lol:


probability logarithms trigonometry geometry trapezoid algebra polynomial

[b][size=150]Mock AMC H[/size][/b]

1.	This test has 25 multiple-choice questions.  Only one of the five answers is correct.

2.	A correct response receives 6 points, unanswered questions receive 2 points, and no points are awarded for incorrect answers.
3.	The only aids allowed are scratch paper, graph paper, rule, compass, protractor, erasers, and SAT approved calculators.
4.	Any figures are not necessarily drawn to scale.
5.	You have 75 MINUTES to complete this test.
6.	Please send your answers to me by any of the following methods:

 E-mail: 				jlaurendi@gmail.com
 Private Message:  			joml88
 AOL Instant messenger: 		BaseballJoe2002

 (Remember to include your AoPS screen name)

 Any questions, comments, or misunderstandings during the test may be conveyed to me via any of those three methods.

1.  Rebecca has $7$ different sticks of lengths $1,  \ 3, \ 5, \ 7, \ 8, \ 11$ and $15$.  If she randomly chooses three of them what is the probability that they form a non-degenerate triangle?

(a) $0$
(b) $9/35$
(c) $13/35$
(d) $4/5$
(e) $6/7$

2.  How many even numbers less than $2004$ are divisible by $3$?

(a)	$256$
(b)	$333$
(c)	$334$
(d)	$667$
(e)	$668$

3.  A right triangle has legs of length $x$ and $y$.  Given that $2^x\cdot 2^y=1024$ and $(2^x)^y=16$ find the length of the hypotenuse.

(a) $10$
(b) $4\sqrt{6}$
(c) $8$
(d) $2\sqrt{23}$
(e) None of the above

4.  Find the sum of the real roots of $f(x)=x^3+7x-8=0$.

(a) $-7$
(b) $7$
(c) $8$
(d) $-8$
(e) $1$

5. Twin primes are pairs of primes of the form $(p,p+2)$.  Which one of the following is a possible sum of a pair of twin primes?  For example, one possible sum would be $3+5=8$.

(a) $99$
(b) $122$
(c) $188$
(d) $204$
(e) $218$

6. Jim and David are playing a game.  At the beginning of the game there is a pile with $76$ pennies.  On each turn they take at least one but no more than seven pennies.  The person who picks up the last penny loses.  If Jim goes first how many pennies should he take on his first turn to ensure that he will win?

(a) $1$
(b) $2$
(c) $3$
(d) $4$
(e) $7$

7. The numbers $1$, $3$, $9$, $27$, $81,$ $243,$ $729,$ $2187$ and $6561$ are placed in a $3\times3$ grid such as the one below.  The product of the numbers in any row, column or diagonal is the same.  The value of this product is

\[\begin{tabular}{|c|c|c|} \hline
& & \\
\ \ \ \ \ \ & \ \ \ \ \ \ & \ \ \ \ \ \ \\ \hline
& &\\
& & \\ \hline
& & \\
& & \\ \hline
\end{tabular}\]

(a) $19683$
(b) $531441$
(c) $59049$
(d) $177147$
(e) $729$

 8. The 3-digit integer, $n$, is multiplied by the number formed by reversing the digits of $n$ and the result is $214 875$.  Find the middle digit of $n$.

(a) $1$
(b) $3$
(c) $5$
(d) $7$
(e) $9$

9. Given that $\displaystyle \frac{\log_mx}{\log_nx}=\frac{37}{55}$ and $\displaystyle \frac mn=n^k$ find $k$.

(a) $1/2$
(b) $55/37$
(c) $37$
(d) $1$
(e) $18/37$

10. In how many ways can the letters from $A$ to $H$ be arranged so that the following conditions are met:

(i) $A$ comes before $B$
 (ii) $C$ comes before $D$
(iii) $E$ comes before $F$ and
(iv) $G$ comes before $H$?

(a) $384$
(b) $2520$
(c) $1680$
(d) $40320$
(e) None of the above

11. Evaluate $\tan(2\sin^{-1}(x))$.

(a)	$\displaystyle \frac{2x\sqrt{1-x^2}}{1-2x^2}$

(b)	$\displaystyle \frac{2\sqrt{1-x^2}}{1-2x^2}$

(c)	$\displaystyle \frac{2x\sqrt{1-x^2}}{1+2x^2}$

(d)	$\displaystyle \frac{x\sqrt{1-x^2}}{1-2x^2}$

(e) $\displaystyle \frac{2\sqrt{1+x^2}}{1+2x^2}$

12.  A trapezoid whose legs are both integers has acute base angles such that the sum of their sines is equal to $\displaystyle \frac 13$.  Given that the trapezoid has a height of $20$ calculate the minimum value of the sum of the legs.

(a) $60$
(b) $120$
(c) $240$
(d) $61$
(e) $3661$

13.  Find the coefficient of $x^{18}$ in
\[(1+x)^{20}+x(1+x)^{19}+x^2(1+x)^{18}+\cdots+x^{18}(1+x)^2.\]

(a) $10$
(b) $320$
(c) $1330$
(d) $1140$
(e) None of the above.

14. Alec, Brittany, Cindy, and Derek are planning to run a marathon as a relay team in that order.  The rules mandate that each runner run at least $2$ miles and that handoffs can only be made at the mile markers.  In how many different ways can they run the marathon? (A marathon can be assumed to be 26 miles)

(a) $2600$
(b) $7315$
(c) $5985$
(d) $1540$
(e) $1330$

15. The number $47$ in base $a$ represents the same number as $74$ in base $b$.  It is assumeed that both $a$ and $b$ are positive integers greater than one.  What is the smallest value of $a+b$?

(a) $11$
(b) $13$
(c) $19$
(d) $24$
(e) None of the above.

16. A quadrilateral inscribed in a circle has sides of length $\sqrt{2002}$, $\sqrt{2005}$, $\sqrt{2006}$, and $\sqrt{2009}$ in that order.  The diameter of the circle can be expressed in the form $\sqrt{x}$ where $x$ is a squarefree integer i.e. $x$ does not contain any squares in its prime factorization.  Find the sum of the digits of $x$.

(a) $1$
(b) $4$
(c) $6$
(d) $10$
(e) $11$

17. Suppose that $f(x)$ is polynomial of degree $5$ with leading coefficient of $2001$ and that
\[f(1)=1 , \ \ f(2)=3 , \ \ f(3)=5 , \ \ f(4)=7, \ \ \mathrm{and} \ \ f(5)=9.\]
What is the value of $f(6)$?

(a) $11$
(b) $-296$
(c) $-432$
(d) $121$
(e) None of the above.

18. Let $a_1$, $a_2$, $a_3,\cdots$ be the numbers which can be written as a sum of one or more different powers of $4$ with $a_1<a_2<a_3<\cdots$ e.g. $a_1=4^0=1$, $a_2=4^1=4$, $a_3=4^0+4^1=5$ etc.  Find the value of $a_{70}$.

(a) $216$
(b) $3220$
(c) $3796$
(d) $4116$
(e) None of the above.

19. Two concentric circles have radii $r$ and $R > r$. Three new circles are drawn so
that they are each tangent to the big two circles and tangent to the other two new circles.
Find $\frac{R}{r}.$

(a) $\sqrt{3}$
(b) $3/2$
(c) $2$
(d) $3$
(e) Cannot be determined with the given information.

20. How many primes, $p$, make $13p^4-888$ a positive prime number?

(a) 0
(b) 1
(c) 2
(d) 3
(e) 4 or more

21. The sides of triangle $ABC$ have length $AB=7$, $BC=8$, and $AC=9$.  The distance between the incenter and the orthocenter of $\triangle ABC$ is

(a) $1$
(b) $\sqrt{30}/5$
(c) $\sqrt{2}$
(d) $\sqrt{5}/2$
(e) None of the above.

22. How many triples of natural numbers, $(a,b,c)$ are there such that $a+b+c=111$ and $a$, $b$, and $c$ are in geometric progression?

(a) $2$
(b) $3$
(c) $4$
(d) $5$
(e) None of the above.

23. $O$ is the center of a semicircle with radius $OA$ on the base.  $C$ is a point on the base between $O$ and $A$, and $B$ is a point on the semicircle such that $\overline{BC}$ is perpendicular to $\overline{OA}$.  $BO$ is extended through $O$ to $E$ and $D$ is the intersection of the lines $AE$ and $BC$.  If $OB=1$, $OE=\sqrt{2}$,  $\displaystyle AD=\frac{2\sqrt{5}-\sqrt{10}}{4}$,  and $\displaystyle DE=\frac{2\sqrt{5}+\sqrt{10}}{4}$, then $AB=$

(a) $\sqrt{2}$
(b) $1$
(c) $\sqrt{2-\sqrt{2}}$
(d) $\sqrt{5}$
(e) None of the above.

24. The sum of a certain number of positive integers is $2005$.  The maximum value of the product of these integers is $P$ where $P$ can be expressed in the form $a^b\cdot c^d$ where $a$ and $c$ are prime numbers such that $c>a$.  Find the value of $a+2b+3c+4d$.

(a) $2005$
(b) $2401$
(c) $2617$
(d) $2683$
(e) None of the above.

25. Let $N_b=1_b+2_b+\cdots+100_b$ where the integer, $b$, is greater than $2$.  Compute the number of values of $b$ for which the sum of the squares of the digits of $N_b$ is at most $512$.

(a) $30$
(b) $32$
(c) $16$
(d) $15$
(e) None of the above.

This post is so you will know that the problems are posted.

Great Test!  I woke up 3 minutes before the test and my brain is not still warmed up :)

Thanks alot - that test was very good. I didn't find it as difficult as some of Beta's tests, but there are tons of things I didn't get because I didn't know certain properties. I hope we review all of them in the discussion thread.


I am looking for help to solve a probability question.
Q:  A hand of 5 cards is dealt from a deck of 52 cards. (a) What is the probability that the hand contains at least one ace? (b) What is the proabability that the hand contains exactly one act? (c) What is the probability that the hand contains no aces?
Any help in breaking this down would be appreciated.

[hide=" (a) and (c)"]
Note that, since you either have 0 aces, or 1 or more aces, the probabilities of (a) and (c) must add to 1.
For (c), you can choose 48 out of 52 cards, then 47 out of 51, etc.
So $ P(c)\equal{} \dfrac{48}{52} \cdot \dfrac {47}{51} \cdot \dfrac {46}{50} \cdot \dfrac {45}{49} \cdot \dfrac {44}{48} \equal{} \dfrac {35673}{54145}$
And then $ P(a)\equal{} 1 \minus{} P(c) \equal{} \dfrac{18472}{54145}$

I will do (b) later... Stupid history...
[/hide]

No problem.   :D
[hide="(b)"]
Assume,without loss of generality, that the first card you pick is the ace required. (Multiply the result of this by 5 to cover all the cases of which card is the ace.)
The first card, you have 4 aces to pick. Then you have 48, 47, 46... non-aces to pick. So, the probability that you get one ace total and that this ace is the first card you pick is:
$ \dfrac 4{52} \cdot \dfrac {48}{51} \cdot \dfrac {47}{50} \cdot \dfrac {46}{49} \cdot \dfrac {45}{48} \equal{} \dfrac {3243}{54145}$
Then multiply by 5 to get: $ \dfrac {3243}{10829}$.
[/hide]

[quote="Dee Wells"]Thanks heaps, I was on the wrong track altogether, I was looking at success / unfavourable ratios.[/quote]

This is not at all the wrong track (at least for question (c)).  If you wanted to frame Brut3Forc3's arguments in those terms, you have two choices:

Choice 1: We keep track of the order in which the cards are dealt to us.  This gives a total of $ 52\cdot51\cdot50\cdot49\cdot48$ possible hands (any of 52 cards could be first; then, there are 51 remaining cards, any of which could be second, etc.).  Of these, exactly $ 48\cdot47\cdot46\cdot45\cdot44$ have no aces in them (48 cards could be first (no aces allowed), etc.).  Thus, the probability is $ \frac {48\cdot47\cdot46\cdot45\cdot44}{52\cdot51\cdot50\cdot49\cdot48}$.

Choice 2: We don't keep track of the order in which the cards are dealt to us.  This gives a total of $ \binom{52}{5}$ possible hands (that's a binomial coefficient, meaning the same thing as $ 52C5$ -- I don't know when people learn this notation) and $ \binom{48}{5}$ possible ace-free hands (choosing 5 cards from the 48 non-ace cards).  Thus the probability is $ \frac {\binom{48}{5}}{\binom{52}{5}}$.

Easy exercize: we've done the same question two different ways.  Verify that the two answers are in fact equal.

For question (a), Brut3Forc3's method (so-called "complementary counting") is far-and-away the best way to go.  Once you have more experience with counting problems of this sort, you'll begin to be able to recognize when two questions are complementary and which of the two is easier to solve directly.

Approaching part (b) with our Choice 2:  to get one ace and 4 other cards, we have $ \binom{4}{1}\cdot \binom{48}{4}$ ways.  The total number of hands (denominator in our probability) is the same as it was before.

Followup: having seen two solutions to part (b), can you find a solution to part (a) not relying on complementary counting?


Okay, that is the title of the second chapter I need to do today. So, again, please actually help me learn this stuff, don't just give me the answers. So, here are the problems.  :mad:

 2. List the steps necessary to complete the square of the expression $ x^2\minus{}10x$.

 State whether the Trinomial is a perfect square.

 4. $ x^2\plus{}18x\minus{}9$

 6. $ x^2\plus{}8x\plus{}16$

 Find the value of $ c$ that makes each Trinomial a perfect square.

 8. $ z^2\plus{}18z\plus{}c$

 10. $ p^2\minus{}6p\equal{}0$

 Help me...?

For #2, see #6 and use a similar method.
Btw, #10 confuses me, too. I don't see any c

[hide="4"]
First of all, it is best if you remember that $ (a\plus{}b)^{2} \equal{} a^{2} \plus{} 2ab \plus{} b^{2}$. This can be derived by either FOILing it out or the binomial theorem.

So now, we look at the equation. $ x^2$ is a perfect, square is not, but -9 is not (assuming we are only using real values). Thus, it is not in the forum $ a^2 \plus{} 2ab \plus{} b^2$.

[/hide]

[hide="6"]
Note that 16 is indeed a perfect square (its square root is 4). Now notice that the second term is also 4*2x. Thus, since it is in the forum $ a^2 \plus{} 2ab \plus{} b^2$, it is a perfect square.  [/hide]

[hide="8"]
Let's call c $ b^2$ (since it must be a perfect square). We know that for it to be a perfect square, the second term is 2bz. Solving for 2bz = 18z, we get that b=9. Now plugging it in, c = 81.[/hide]

Thank you guys! That helps a lot. -Prints- I am pretty sure I can do the rest of the problems now. -Verra out-

for all Completeing the square put the binomial in form x^2+bx/a+c/a=0, then go x^2+bx/a=-c/a
now because (x+y)^2= (x^2)+2xy+(y^2)
So, we need to add a y^2 to right hand side to get a perfect square.
but what is this y? this y is equal to b/2a (you already have the x)
so you would add (b/2a)^2 to both sides
Then take the square root and use easy algebra to figure out the values of x.
Make sure you understand that most of these quadratics have 2 answers so make sure you take the + and - of the right hand side.

I hope i explained that clear. I'm not that good at explaining and am trying to learn.


A bug is crawling on the coordinate plane from $(7,11)$ to $(-17,-4)$. The bug travels at constant speed one unit per second everywhere but quadrant II, where it travels at $.5$ units per second. What path should the bug take to complete its journey in minimal time? Generalize!

I just know that locally, say if you move from a region of 1m/s to 2m/s, then to minimize your time your new sine of the angle of *incidence* must be twice the old.

Interestingly, this is the exact way light refracts!   :huh:

So in your example, if you wanted to get from (1,0) to (-2,3), you would find that your new sin is half your old one. That's a bad example; I chose it because you just happen to cross (0,2) (sin of quad1 section =2/root5, and sin of quad2 section = 1/root5).

The reason has to do with the differential calculus property of minimums: specifically, it's the certain angle where changing the point of crossing adds/substracts distances proportional to the "cost" of either area. If you get what I mean.

EDIT: if we call the angle of incidence x, then the rate of change of the distance as we slide the "point of crossing" along the y-axis is equal to d(secx)/d(tanx), which is secxtanx/(secx)^2, or simply sinx. Now the rates of change on either side must cancel each other out. The [b]weighted[/b] rate of change depends on how fast or slow you move, so you end up with something like n1sinx1 = n2sinx2.

[img]http://64.40.110.41/incidence.gif[/img]

For your case, I believe the fastest way is just through the origin. Because whenever you cut across quadrant II, the gains are outweighed by the losses.

In fact, the only way for crossing the ".5 unit/s" zone to be faster is if its shape juts out at a sharper (<60 degree) angle. Why 60? Because sin(60/2)=0.5, and if that were any higher, the other sine would have to be >1!

This still does not provide a conclusive proof though that the optimal path is a straight line to the origin and then a straight line to the other point.

Let $l_{1}$ be the length of path in non-II quadrants and $l_{2}$ the length of path in the II quadrant. Then we have to minimize $l=l_{1}+2l_{2}$. Let the optimal path cross y-axis at $(0,a)$ and x-axis at $(-b,0)$ where $a,b$ are non-negative. Then $l_{1}=\sqrt{7^{2}+(11-a)^{2}}+\sqrt{(17-b)^{2}+4^{2}}, l_{2}=\sqrt{a^{2}+b^{2}}$. By QM-AM, $l\geqslant{1\over\sqrt{2}}(7+|11-a|+|17-b|+4+2(a+b))\geqslant{1\over\sqrt{2}}(39+a+b)$. Hence $l$ is obviously minimized when $a=b=0$.

NOTE: It's easy to show that the optimal path which avoids II quadrant leads through the origin.

EDIT: Farenhajt's solution is much more straightforward.

Given $(x_{1},y_{1})$ in quadrant 1 and $(x_{2},y_{2})$ in quadrant 2, consider the path between them. (see incidence.gif)

Call B the length of the blue segment and R the length of the red segment.
The total time is $R+2B = |x_{1}|\sec\theta_{1}+2|x_{2}|\sec\theta_{2}$. We want to find the y-intercept $Y$ that will minimize this time.

$\frac{\Delta}{\Delta Y}(R+2B) = 0$
$\frac{\Delta}{\Delta Y}|x_{1}|\sec\theta_{1}+\frac{\Delta}{\Delta Y}2|x_{2}|\sec\theta_{2}= 0$
$\frac{\Delta}{-\Delta x_{1}\tan\theta_{1}}|x_{1}|\sec\theta_{1}+\frac{\Delta}{\Delta x_{2}\tan\theta_{2}}2|x_{2}|\sec\theta_{2}= 0$
$2\frac{\Delta\sec\theta_{2}}{\Delta\tan\theta_{2}}= \frac{\Delta\sec\theta_{1}}{\Delta\tan\theta_{1}}$
$2\frac{\sec\theta_{2}\tan\theta_{2}}{\sec^{2}\theta_{2}}= \frac{\sec\theta_{1}\tan\theta_{1}}{\sec^{2}\theta_{1}}$
$2\sin\theta_{2}= \sin\theta_{1}$

Notice that the whole path must obey this property, otherwise it can be made shorter by modifying a section of it as shown by the lighter path in incidence2.gif

Therefore $\theta_{2}$ and $\theta_{3}$ must have sines equal to half of those of $\theta_{1}$ and $\theta_{4}$ respectively. That means that $\theta_{2},\theta_{3}\leq 30^\circ$, which is a contradiction because they are supposed to be complimentary.


Calculating $ \frac{1}{4\cdot7}\plus{}\frac{1}{7\cdot 11}\plus{}\cdots \plus{}\frac{1}{n(n\plus{}1)}$ is simple:

$ \sum \frac{1}{n(n\plus{}3)}\equal{}\sum \frac{A}{n}\plus{}\frac{B}{n\plus{}3}$

From where we find $ A$ and $ B$ and solve the problem. With three terms in the denominator it is a little more time-consuming to find $ A$ , $ B$ , and $ C$:

$ \frac{1}{5\cdot11\cdot17}\plus{}\frac{1}{11\cdot17\cdot23}\plus{}\cdots$

$ \sum \frac{1}{(n\minus{}6)n(n\plus{}6)}\equal{}\sum \frac{A}{n\minus{}6}\plus{}\frac{B}{n}\plus{}\frac{C}{n\plus{}6}$

But with four or more terms in the denominator it is much more time-consuming. Is there any easier method for calculating this without Calculus?

What would be the general term for an expression in the form of:

$ \frac{1}{n(n\plus{}k)(n\plus{}2k)...}$

if this is much harder without calculus, how would we go with four or five terms in the denominator?

[quote="triplebig"] What would be the general term for an expression in the form of:

$ \frac {1}{n(n \plus{} k)(n \plus{} 2k)...}$

if this is much harder without calculus, how would we go with four or five terms in the denominator?[/quote]

Simplest hint : $ \frac{1}{\prod_{i\equal{}0}^{p}(n\plus{}ki)}\equal{}\frac{1}{kp}\frac{(n\plus{}kp)\minus{}n}{\prod_{i\equal{}0}^{p}(n\plus{}ki)}$

Let $ k\in\mathbb{R}$, $ p\in\mathbb{N}^*$ and $ S\equal{}\sum_{n\equal{}1}^{\plus{}\infty}\frac{1}{\prod_{i\equal{}0}^{p}(n\plus{}ki)}$

Then :
$ S\equal{}\frac{1}{kp}\sum_{n\equal{}1}^{\plus{}\infty}\frac{n\plus{}kp \minus{} n}{\prod_{i\equal{}0}^{p}(n\plus{}ki)}$

$ S\equal{}\frac{1}{kp}\sum_{n\equal{}1}^{\plus{}\infty}(\frac{n\plus{}kp}{\prod_{i\equal{}0}^{p}(n\plus{}ki)}\minus{}\frac{n}{\prod_{i\equal{}0}^{p}(n\plus{}ki)})$

$ S\equal{}\frac{1}{kp}\sum_{n\equal{}1}^{\plus{}\infty}(\frac{1}{\prod_{i\equal{}0}^{p\minus{}1}(n\plus{}ki)}\minus{}\frac{1}{\prod_{i\equal{}1}^{p}(n\plus{}ki)})$

$ S\equal{}\frac{1}{kp}\sum_{n\equal{}1}^{\plus{}\infty}(\frac{1}{\prod_{i\equal{}0}^{p\minus{}1}(n\plus{}ki)}\minus{}\frac{1}{\prod_{i\equal{}0}^{p\minus{}1}((n\plus{}k)\plus{}ki)})$

$ S\equal{}\frac{1}{kp}\sum_{n\equal{}1}^{k}\frac{1}{\prod_{i\equal{}0}^{p\minus{}1}(n\plus{}ki)}$

SO :
$ \sum_{n\equal{}1}^{\plus{}\infty}\frac{1}{\prod_{i\equal{}0}^{p}(n\plus{}ki)}$ $ \equal{} \frac{1}{kp}\sum_{n\equal{}1}^{k}\frac{1}{\prod_{i\equal{}0}^{p\minus{}1}(n\plus{}ki)}$

And that's all.

For example :
$ p\equal{}1$, $ k\equal{}3$

$ \sum_{n\equal{}1}^{\plus{}\infty}\frac{1}{\prod_{i\equal{}0}^{1}(n\plus{}3i)}$ $ \equal{} \frac{1}{3}\sum_{n\equal{}1}^{3}\frac{1}{\prod_{i\equal{}0}^{0}(n\plus{}3i)}$

$ \sum_{n\equal{}1}^{\plus{}\infty}\frac{1}{n(n\plus{}3)}$ $ \equal{} \frac{1}{3}\sum_{n\equal{}1}^{3}\frac{1}{n}$ $ \equal{}\frac{1}{3}(1\plus{}\frac{1}{2}\plus{}\frac{1}{3})$ = $ \frac{11}{18}$

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=273420[/url]


Who do you think is the best chess player of all time, or at least of these five...

The Greatest Ever... Topalov lost to Kramnik. These are the most commonly known as the best currently. The players you mentioned are all those of the past when chess theory was not fully developed, compared to modern times. If these played the best now, they would lose to extravagant variations unknown then.

[quote="Temperal"]..you don't seem to be very knowledgeable about chess history. Most of these players are nothing compared to the old nineteenth century greats.

My list:

1) Emmanuel Lasker
2) Jose Casablanca
3) Wilhelm Steinz

Not to mention Alekhine, Blackburne, and several others I've probably forgotten.

You haven't even listed the best modern players: Topalov, Polgar...[/quote]

Yeah, I think Emmanuel Lasker held the title for 27 years. Or was it 17? Right now the champions are holding it for 2 years max  :P

Karpov is better than those old ones will ever be! If any one of these played the best 19th century or earlier player, he would crush them completely. With the computers we have now, the GMs memorize lines that would completely outplay a person 200 years ago. The more modern the player, the more chess is advanced for them to learn, the better they are.

I've actually played Karpov. Too bad it was 3 years ago when I sucked and wasn't as good and perceptive now. Sigh.

In my opinion a good chess player is not afraid to take risks. I really hate playing high level players and they castle within the first 10 moves and not even launching an offensive the entire game. Chess is only exciting to me when both players go all out, sacking pieces, and playing for tempos. Based on my perspective, I believe Mikhail Tal is the most fun to watch play.

Topalov may have lost to Kramnik, but that doesn't mean he's worse.

I truly doubt Karpov could have beaten either Lasker or Casablanca.

Karpov, Kasparov, and Fisher are way better than Lasker are Casablanca. Casablanca sat in his room calculating variations for years, Karpov, Kasparov, and Fisher have computers. Karpov is one of the most drawish players ever, he can also win on slight possessional advantages in the endgame that cream everyone else. Likewise for Fisher/Kasparov minus the drawishness. Go to anyone good at chess, they say Karpov, Kasparov, and Fisher(and less usually Kramnik/Anand). Topalov toppled from his peak when he lost to Kramnik, he wasn't all that good. He is also a jerk and accuses Kramnik of cheating every game, than loses the match even when Kramnik forfeit a game because he was pissed at the TD and he lost an hour on his clock. Karpov, Kasparov, and the rest were undefeatable in their times(except by each-other). Ask people like Raven Sturt(1900 twelve year old), they don't think Casablanca is best do they? Do you personally know WGM Rusudan Goletiani, former US women's champion?

Tal is cool to watch, I totally agree to that, but if he played Karpov in a match it would be:
Karpov/Tal
1/2-1/2
1/2-1/2
1/2-1/2
1-0
1/2-1/2
1/2-1/2
1/2-1/2
1/2-1/2

I've always thought kasparov was the best.
he lost to the computer though

isn't bobby fisher in jail?

[quote]Karpov, Kasparov, and Fisher are way better than Lasker are Casablanca. Casablanca sat in his room calculating variations for years, Karpov, Kasparov, and Fisher have computers.[/quote]

Would you care to explain this? Are you implying that the modern greats can learn quicker because the computer generates better results than humans?

[quote]
Karpov is one of the most drawish players ever, he can also win on slight possessional advantages in the endgame that cream everyone else. Likewise for Fisher/Kasparov minus the drawishness. Go to anyone good at chess, they say Karpov, Kasparov, and Fisher(and less usually Kramnik/Anand). Topalov toppled from his peak when he lost to Kramnik, he wasn't all that good. He is also a jerk and accuses Kramnik of cheating every game, than loses the match even when Kramnik forfeit a game because he was pissed at the TD and he lost an hour on his clock. Karpov, Kasparov, and the rest were undefeatable in their times(except by each-other). Ask people like Raven Sturt(1900 twelve year old), they don't think Casablanca is best do they? Do you personally know WGM Rusudan Goletiani, former US women's champion? [/quote]

What does much of this have to do with anything? Especially those last sentences.

On the relevant parts: Topalov is quite good, despite a slump. Him being a jerk changes nothiing in relevance to your argument.

[quote]
You are all wrong, just admit it. [/quote]

Uh... I have nothing to say to such a rude and incorrect statement.[quote]

Tal is cool to watch, I totally agree to that, but if he played Karpov in a match it would be:
Karpov/Tal
1/2-1/2
1/2-1/2
1/2-1/2
1-0
1/2-1/2
1/2-1/2
1/2-1/2
1/2-1/2

You people just don't know chess about good chess players... I'm not trying to insult all of you, just study more before you argue...[/quote]

Tal is okay, but nothing compared to Lasker nor Casablanca. Again, I have nothing to say to the rude and incorrect last sentence.

Seriously, the old players might not know as many computer-generated positions and techniques, but their foresight and tactics out match the modern players.

Ok, I admit it, maybe my argument is wrong, maybe I just like Karpov, maybe my friend at school typed half of what I said while I wasn't looking, maybe you are all wrong, maybe none of us are right... I have simply stated my argument, if you can overcome it with your own, please do so. But criticizing my argument, when not even writing your own, is just as rude as some of the statements in mine.

Firstly, with the computers, the newer players can simply put in a position and get the best move, people like Casablanca took much longer, giving the new players time to memorize more and longer variations. Humans can find better variations than computers, but only with years of study, and that isn't what a player has when the championship final is tomorrow and you need the best move you can get.

Secondly, those people agree with my argument, not yours. I'm not saying people like Casablanca and Tal are bad(on the contrary, they are quite good), they just cannot compare to what we have now. Over 65 years after Casablanca's death, we have come a long way in the world of chess.

Thirdly, I took credit for all the incorrect statements, but the rude comments were none of my doing, I have edited them out.

A new player has the advantage of openings almost completely unknown and unused in their time. Even when the computer doesn't create them. What is Kasparov found a winning variation vs. one of Casablanca's openings(which he did, I believe), Casablanca doesn't know the opening since he hasn't been studying for the over 65 years since his death, unlike the mixed chess world.

Even if the players of the past were naturally better at chess(which I doubt, though it may be close), they would not stand up to the modern chess world.

PS: Fisher is on the run from the US, hiding in either Iceland or Japan, nobody is quite sure...
PSS: Karpov was champion for 16 years, Kramnik for 7, and Kasparov for 8, I'm not so sure about '2 years max'...

Okay, so that's settled then, we've reached a compromise.  :)

Personally, actually, my choice would be probably a tie between Lasker and Anand... so I chose "other".

I think i'm the best ever...it's so obvious to me :D

[quote="Fraenkel"]Yeah you should definitely add ProtestanT in the poll.
Perhaps even Chuck Norris (but that might be overkill)[/quote]

adding Chuck Norris is pointless. Everyone would vote for him

Wikipedia has an interesting article on comparing the top chess players from different eras:
[url]http://en.wikipedia.org/wiki/Comparing_top_chess_players_throughout_history[/url].

As I have said before, this is a poll of the 5 players called #1 every by the majority of the general populace of the world. aka: More people call these players the best evert than any other players.

Is that Humprhey Bogart in Casbablanca or Ted Casablanca the journalist? Seriously, after having read that I'm pretty sure at least 60% of you on this thread have no idea whatsoever about chess history, so having studied chess history and played competitively until I was 16 (I was 2.2k+ on fide but didn't reach CM or FM and I quit later that year due to lack of interest)  I'll give you a relatively accurate depiction of the "best player" of all time.

First, there is no such thing as "best player of all time". Players are measured by decades and during the romantic era and previously all players were awful, that is with the exception of Philidor, Gioacchino Greco and Ruy Lopez (though Greco and Lopez lived during the 16th and 17th century and very little is known about them, but I'll trust history on this one).

In the 18th century it was all Philidor, he was by far the best player, no one could beat him unless he gave knight odds, and was easily 50 years+ years ahead of his time. The romantic era ensued after this and players like La Bourdonnais became the top dogs, even though their game was vastly inferior to Philidor's.

Howard Staunton then began defeating the romantics and for a short time he was considered the top player. Later Anderssen and Morphy emerged, and both were vastly superior over Staunton (he refused to play Morphy however). Morphy completely crushed his opposition, he even beat Anderssen while terminally ill, with a score of 7-2 I think. Morphy, just like CAPABLANCA, Reshevsky and Fischer was a chess prodigy and the game came very naturally to him. Fischer summed up his genius pretty well when he said something along the lines of "If Morphy was alive today he'd beat any modern GM." Then unfortunately Morphy died of a heart attack while suffering of mental illness at 47 and did not get to play chess against Steinitz for he renounced the game during his last years.

Steinitz then became the first world champion, by defeating Anderssen (a very old Anderssen at that) and Zuckertort, for five years. He had a very defensive style which troubled his contemporaries for they found his defense almost impenetrable. He was indeed a good player but nowhere near Morphy. Then Lasker and Alekhine came, I won't delve much into them since I think they're over-rated (in terms of best players ever) but they were both top players at some point, as can be noted by Alekhine taking the WC title to the grave and Lasker holding it for 27 years.

Then there is Capablanca, he was truly one of a kind. He never studied chess only observed it when he was young, never played for more than a decade, came to U.S. defeated Frank Marshall (considered the best player then) and then made it his career-- completely crushing for the remainder of his years. He went on a 12 year win streak and had the fewest losses of any GM. A computer generated study shows his game contained the fewest positional errors, that is fewest blunders and fewes tactical disadvantageous positions (and he played in the 19th century wowza!). Yes he did lose to Alekhine in a rematch (he took the title from him before), losing the title and never regaining it, but that was an awfully close game and a series of draws ended up frustrating him and threw him off balance. In the end he lost -6 +3 =25 and Alekhine never gave him a chance to redeem himself, for he knew better of course. Alekhine is remebered for responding, several years after defeating Capablanca, after being asked how he had had done it, with the humble response of: "Even now I cannot explain that."

Then came Fischer, Kasparov, Kramnik, Anand, Karpov, Botvinnik, Tal, Petrosian, Topalov, Reshevsky, Flohr, Keres, Pillsbury (played very few games due to his unfortunate early death but crushed tournaments when he did) -- and I'm sure I'm forgeting some important ones since I'm tired... I won't go into details about them since this post is way too long already but will just say they were/are very goooooooooooot.

With that being said here are my top five players of all time (disregarding era, even though it's a huge factor)

1. José Raúl Capablanca
2. Robert James Fischer
3. Paul Morphy
4. Garry Kasparov
5. Vladimir Kramnik/Viswanathan Anand (very very very close so I'll throw them both in)

Best by era:

Ruy Lopez (1500?)
Gioacchino Greco (1600?)
Philidor (until 1795)
Louis-Charles Mahé de La Bourdonnais (until 1840)
Howard Staunton (until around 1860)
Paul Morphy (until 1884)
Harry Nelson Pillsbury (until 1906) The short years before were dominated by Steinitz
José Raúl Capablanca (until 1942)
Mikhail Botvinnik/Tal and Anatoly Karpov (each dominated a small portion of the 20th century)
Robert James Fischer (after Karpov until his exile)
Anatoly Karpov (before losing to Kasparov)
Garry Kasparov  Vladimir Kramnik and Vishy Anand(after Fischer's departure starting with Kasparov who dominated for a long time afterwards, the latter two after 2000 or so, even though Kasparov was always hanging in there)

* I omitted Alekhine and Lasker because I chose Pillsbury instead. He played few tournaments but had stellar results whenever he did, I believe that if he was pitted in games against either Lasker or Alekhine at their prime he would come out on top. However he did not "dominate" those years like Lasker and Alekhine did. Still I think he deserves that spot.

There you have it folks... this was not proof read and written in a rush so there's probably a lot of typos and syntax errors. However, I hope you do get some insight as to qualifying chess players.

P.S: I'll add the top five I consider the best rapid players.

1. R.J Fischer
2. V. Anand
3. J.R. Capablanca
4. P. Morphy
5. A. Morozevich

Look, everyone is either too skeptical or don't pay attention. I was simply looking for humble opinions, I didn't know that many people on AoPS knew all of facts thrown at on this post. If I listed everyone, they would all have 1%, so I listed the best know players who are good. I consider Karpov better than Kasparov, for one Kasparov quit chess, Karpov is playing a tourney now.

Though I do say all(ok, most) of you have set logical claims before me, but as I said before, at the time I did not know anyone on AoPS knew much on chess history, I was hopping to see if [i]anyone[/i] at all did, I seems that there is a large, educated chess community on this site. Any future posts will be better thought over

Sincerely yours,
dbkarp


IF a,b,c are real numbers such that a,b,c  [0,1) prove the following:
   (a+b+c)^2  <= 2(a+b+c)+ac(a+c)+bc(b+c)+ab(a+b)

Good luck :)

You inequality is obvious wrong.let $a=b=c \to 1$
but we can get:$2(a+b+c)+ac(a+c)+bc(b+c)+ab(a+b) \ge 2(a^2+b^2+c^2+ab+bc+ca)$
I hope mine is right :D
Notice:
$a^2b+a \ge a^2+ab$(because it is equal to $a(a-1)(b-1) \ge 0$)
and
write the other 5 inequalities,add them up we have:
$2(a+b+c)+ac(a+c)+bc(b+c)+ab(a+b) \ge 2(a^2+b^2+c^2+ab+bc+ca)$

[quote="crizty"]sorry i made a mistake writtig it now it's right (i hope :? ) i can't see any mistake now in my proof  :)[/quote]
But the inequality I proved is stronger than your "new" inequality. :)


What is the 1997th digit to the right of the decimal point in the decimal expansion of $ \frac17$?

$ \frac17\equal{}0.\overline{142857}$, and since $ 1997\equiv5\pmod6$, our answer is $ \boxed5$.


Prove that if $\overline{a_na_{n-1}...a_0}$be a prime number . Then $a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$is iiredubic Plynomials.

:?:  :?:  ;) There was an answer here.What happend for it :?:

I deleted it because it was wrong.  :P :oops:


Two standard dice are thrown. What is the probability, expressed as a common fraction, that a prime sum is rolled?


calculus
integration
trigonometry
real analysis
real analysis solved

Compute the following integral :
\[ \int \frac{(2+x^2)\sin{\ln(x-1)}+x^2\cos{\ln(x-1)}}{2x^3}\ dx\]
on the interval $(1,+\infty)$.

The integral diverges (weakly) at $\infty$. Are you sure this is the question you wanted to ask?

If the denominator was $x^a$ for some $a>3$, it would converge, although it would still be very hard to compute.


[color=darkblue][b]Prove that in any triangle $ ABC$ ocurs the identity:[/b][/color]
[color=red]
\[ \displaystyle\frac {\cos(A \minus{} \frac {B}{2})}{a\sqrt {b(p \minus{} b)}} \plus{} \displaystyle\frac {\cos(B \minus{} \frac {C}{2})}{b\sqrt {c(p \minus{} c)}} \plus{} \displaystyle\frac {\cos(C \minus{} \frac {A}{2})}{c\sqrt {a(p \minus{} a)}} \equal{} \frac {1}{2\sqrt {rR}}(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})
\]
[/color]

nowone?

[hide="see here"]We have $ \frac{cos(A\minus{}\frac{B}{2})}{cos\frac{B}{2}}\equal{}cos A \plus{}sin A tan \frac{B}{2}\equal{}\frac{b^2\minus{}a^2\plus{}ac}{bc}$.Therefore
$ cos(A\minus{}\frac{B}{2})\equal{}\frac{b^2\minus{}a^2\plus{}ac}{bc} \cdot \frac{\sqrt{p(p\minus{}b)}}{\sqrt{ac}}$ $ \implies$ $ \frac{cos(A\minus{}B/2)}{a\sqrt{b(p\minus{}b)}}\equal{}\frac{\sqrt{p}}{abc\sqrt{abc}}\cdot (b^2\minus{}a^2\plus{}ac)$[/hide]


calculus
integration
calculus computations

Let $f : [0,1] \to (0,1)$ be a (Riemann) integrable function. Prove that the equation \[2x-\int_{0}^{x}f(t) \, dt = 1\] has an unique solution in $[0,1]$.

We need the following lemma:

If $g$ is Riemann integrable on $[a,b]$ and $g(x)>0$ for all $x$ in that interval, then $\int_{a}^{b}g(x)\,dx>0.$

Assuming that:

Let $A=\int_{0}^{1}f(t)\,dt.$ By the lemma, $0<A<1.$

Let $h(x)=2x-\int_{0}^{x}f(t)\,dt.$

Then $h$ is continuous, $h(0)=0$ and $h(1)=2-A>1.$

By the Intermediate Value Theorem, there is at least one solution to $h(x)=1.$

But if $a<b,$ then $h(b)-h(a)=2(b-a)-\int_{a}^{b}f>2(b-a)-\int_{a}^{b}1>b-a.$

Thus, $h$ is strictly increasing and hence the solution to the equation $h(x)=1$ is unique.

Now, about that lemma ...

[quote="Kent Merryfield"]Now, about that lemma ...[/quote]
See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=37648]here[/url]


geometry
geometric inequality
perimeter
Triangle
circumcircle
IMO Shortlist

Let $ABC$ be a triangle with semiperimeter $s$ and inradius $r$. The semicircles with diameters $BC$, $CA$, $AB$ are drawn on the outside of the triangle $ABC$. The circle tangent to all of these three semicircles has radius $t$. Prove that
\[\frac{s}{2}<t\le\frac{s}{2}+\left(1-\frac{\sqrt{3}}{2}\right)r. \]
[i]Alternative formulation.[/i] In a triangle $ABC$, construct circles with diameters $BC$, $CA$, and $AB$, respectively. Construct a circle $w$ externally tangent to these three circles. Let the radius of this circle $w$ be $t$.
Prove: $\frac{s}{2}<t\le\frac{s}{2}+\frac12\left(2-\sqrt3\right)r$, where $r$ is the inradius and $s$ is the semiperimeter of triangle $ABC$.

[i]Proposed by Dirk Laurie, South Africa[/i]

I just wanted to note that on the first German TST 2004 in Rostock, we've got a weaker version of this problem (in fact, slightly simpler because of the $\leq$ instead of the < sign in the left member, but yet nobody gave a complete solution):

Let ABC be a triangle with perimeter 2s and inradius r. Construct the semicircles with diameters BC, CA, AB outwardly (with respect to the triangle). The radius of the circle tangent to these three semicircles will be denoted by t. Prove

$\frac{s}{2}\leq t\leq \frac{s}{2}+\left( 1-\frac{\sqrt3}{2}\right) r$.

I got 2 of 7 points on this problem, I think nobody got more... it is a really sadistic problem.

  Darij

[quote="darij grinberg"]I got 2 of 7 points on this problem, I think nobody got more... it is a really sadistic problem.[/quote]
It's not only "sadistic", it's useless too (not nice) (or at least in my opinion)  :maybe:  :P . Besides the Soddy circles (which I never used at any problem  :blush:  :lol: ) I didn't learn anything out of it.

See also

  http://mathlinks.ro/viewtopic.php?t=4796

I am attaching to this post a PDF file containing a detailed solution of this problem.

  darij

If either of you are interested, this problem also appeared in [url=http://mathlinks.ro/viewtopic.php?t=4796]the 3rd MathLinks Contest (the current edition) in the 2nd Round[/url].  I also saw it come up on the [url=http://science.up.ac.za/samo/s3q2003.pdf]South African Math Olympiad 2003[/url].

Probably darij got 2 points by proving the left part of the inequality like this(or am I wrong??):

[u][b]Notations:[/b][/u]We denote by $\Omega_A,\Omega_B,\Omega_C$ the circles with diameters $BC,CA,AB$ respectively.Let $\Omega$ be the circle externally tangent to these three circles.Let $A'$ be the point of tangency of $\Omega_A$ with $\Omega$ and define $B',C'$ analogously.Let $A",B",C"$ be the midpoints of $BC,CA,AB$ and $O$ the center of $\Omega$.

[u][b]Proof of the first part[/b][/u]:It is clear that the points $(A'A",O),(B',B",O),(C',C",O)$ are collinear.Thus $OC"=t-\frac{c}{2},OA"=t-\frac{a}{2}$.We also have $A"C"=\frac{b}{2}$.So triangle inequality in $\triangle{A"C"O}$ gives

$A"O+C"O > A"C" \Rightarrow (t-\frac{a}{2})+(t-\frac{c}{2}) > \frac{b}{2} \Rightarrow t > \frac{a+b+c}{4}=\frac{s}{2}$

I shall prove that more difficult inequality (the RHS):

Let $ A', B', C' $ the midpoints of segments $ BC, CA, AB $ respectively.  Let the incircle of $ \triangle{A'B'C'} $ touch segments $ B'C', C'A', A'B' $ at $ M, N, P $ respectively. Consider the circle internally tangent to the circles with centers $ A', B', C' $ and radii $ A'N, B'P, C'M $ respectively.  This is called the inner Soddy circle of $ \triangle{A'B'C'} $ and has radius $ t - \frac{s}{2} $.  Letting $ R $ be the circumradius of $ \triangle{ABC} $, it is well-known (and can be easily derived from Descartes' Theorem) that this inner Soddy circle also has radius $ \frac{rs}{2(4R + r + 2s)} $.

So it suffices to show that $ \frac{rs}{2(4R + r + 2s)} \leq \left(1 - \frac{\sqrt{3}}{2}\right)r \Longleftrightarrow \frac{s}{4R + r + 2s} \leq 2 - \sqrt{3} \Longleftrightarrow 4R + r \geq \sqrt{3}s $.  Let the tangent lengths of $ \triangle{ABC} $ be $ x, y, z $ respectively.

Then this inequality is equivalent to $ \frac{(x + y)(y + z)(z + x)}{\sqrt{xyz(x + y + z)}} + \sqrt{\frac{xyz}{x + y + z}} \geq \sqrt{3}(x + y + z) $.

Upon expanding the denominators and applying some simple algebraic manipulations becomes $ (x^2y + x^2z + y^2z + y^2x + z^2x + z^2y + 3xyz)^2 \geq 3xyz(x + y + z)^3 $.  By cancelling similar terms from both sides this becomes:

$ \sum_{sym}x^4y^2 + \sum_{sym}x^3y^3 \geq \sum_{cyc}x^4yz + \sum_{sym}x^3y^2z + 3x^2y^2z^2 $ which is trivial by some applications of Muirhead's inequality.

this is a stupid problem. It takes a lot of skill to write a beautiful geometry problem, even more for a beautiful solution, even more to carry through a stupid and ugly solution, and yet more to write such a stupid, ugly, and terrible and not worth solving problem such as this one. No deep beautiful geometrical results can be applied, and this problem can only be attacked by sufficient reading into algebraic literature and by enough computational fortitude. This problem is an insult to all of the beautiful geometry problems that have made their way onto the IMO.

I think the criticism is rather harsh. At least $LHS$ can be solved somewhat nicely.
Let the midpoints of $AB,BC,CA$ be $F,D,E$, and the centre of new circle be $X$.
From homothety we can determine that:
$XD+\frac{BC}{2}=XE+\frac{AC}{2}=XF+\frac{AB}{2}=t$. WLOG $X$ is not on $DE$, then we have $2t=XD+\frac{BC}{2}+XE+\frac{AC}{2}>DE+\frac{BC}{2}+\frac{AC}{2}=s$.

As for $RHS$, there is not really any synthetic involved, but not too messy computational wise too. Note that $XD,XE,XF$ are of the form $\frac{b+c-a}{4}+k,\frac{a+c-b}{4}+k,\frac{a+b-c}{4}+k$( rewrite as $u+k,v+k,w+k$).

Hence we need to show $XD+\frac{BC}{2}\leq \frac{s}{2}+(1-\frac{\sqrt3}{2})r\Leftrightarrow k\leq (1-\frac{\sqrt3}{2})r$. Draw the circles centered at $D$,$E$,$F$ with radius $u,v,w$, giving us the configuration needed to apply Descartes (the last circle is that centered at $X$ with radius $k$). We get that $k=\frac{uvw}{uv+vw+uw+2\sqrt{uvw(u+v+w)}}$. Further, the inradius of $DEF$ is half of that of $ABC$, so $r=\frac{2\sqrt{uvw(u+v+w)}}{u+v+w}$.

It remains to show that $\frac{uvw}{uv+vw+uw+2\sqrt{uvw(u+v+w)}}\leq (1-\frac{\sqrt3}{2})\frac{2\sqrt{uvw(u+v+w)}}{u+v+w}$
$\Leftrightarrow \sqrt{uvw(u+v+w)}\leq(2-\sqrt3)(uv+vw+uw+2\sqrt{uvw(u+v+w)})$
$\Leftrightarrow (2\sqrt3-3)\sqrt{uvw(u+v+w)}\leq (2-\sqrt3)(uv+vw+uw)$
$\Leftrightarrow (u+v+w)uvw\leq u^2v^2+v^2w^2+u^2w^2$ which is $AM-GM$

I think I found the problem quite nice  :)
[Hide=Solution]
Let $A',B',C'$ denote the mid points of $BC,CA,AB$ respectively. Consider circles $\omega_A = \odot (A',\frac s2 - \frac a2)$. Similarly draw in $\omega_B$ and $\omega_C$.

Note that the centre of the requested circle, call it $O'$, is the center of the Inner Soddy circle $\omega$ of $\triangle A'B'C'$. Suppose $\omega$ has radius $x$. It follows that the circle in question has radius $x+\frac s2$ which readily proves the left part of the inequality because $x>0$.

We will now prove the right part of the inequality. Consider the incircle $\Omega$ of $\triangle A'B'C'$ , centered at $I$ , having a radius of $r'=\frac r2$. We wish to prove $x\leq (2-\sqrt 3)r'$. We will distinguish two cases.

[i]Case 1 :[/i] $O\equiv I$.

Straightforward calculations reveal $x=  (2-\sqrt 3)r'$ , so the bound holds in this case.

[i]Case 2 :[/i] $O \neq I$.

We claim that there exists $X \in OI$ such that,  under the inversion $\psi (X,1)$ , the circles $\psi(\omega)$ and $\psi (\Omega)$ become con-centric and $\psi(\omega)$ lies inside $\psi (\Omega)$.

[i]Proof : [/i] Impose coordinates on the line $OI$ and let the points where $\omega$ and $\Omega$ hit $OI$ be denoted by $i,j,k,\ell$ with $i<j<k<\ell$.

Consider the function given by $$f(x)= \frac {\frac {1}{x-\ell}-\frac {1}{x-k}}{\frac {1}{x-j}-\frac {1}{x-i}}$$
(for a point $X\in OI$.)

Note that we have $\lim_{x\rightarrow \ell^{+}}f(x) = \infty$ and $\lim_{x\rightarrow \infty}f(x) = \frac {\ell-k}{j-i}<1$. This means that by [b]IVT[/b] , we can choose a point $X$ such that $f(x)=1$. Note that we have, in that case $:=$

$$ \frac {1}{x-\ell}-\frac {1}{x-k} =  \frac {1}{x-j}-\frac {1}{x-i} \implies \frac {1}{x-\ell}+\frac {1}{x-i}=\frac {1}{x-j}+\frac {1}{x-k}$$

This means that the inversion $\psi(X,1)$ makes the circles $\omega$ and $\Omega$ concyclic. We also have $:=$

$$ \frac 1{x-i}<\frac 1{x-j}<\frac 1{x-k}<\frac 1{x-\ell} \implies \frac 1{(x-i)(x-\ell)}< \frac 1{(x-j)(x-k)}$$

So $\omega$ lies inside $\Omega$.

We now claim that the ratio of the radius of $\omega$ and $\Omega$ increases after this inversion.

We have $:=$

$$ \frac { \frac 1 {x-k} -  \frac 1 {x-j} } {\frac 1 {x-\ell} -  \frac 1 {x-i} }= \frac {k-j}{\ell-i} \cdot \frac {(x-i)(x-\ell)}{(x-k)(x-j)} >  \frac {k-j}{\ell-i} = \frac x{r'}$$

So since $\omega_A$,$\omega_B$,$\omega_C$ invert to circles tangent to $\psi (\omega)$ and orthogonal to $\psi (\Omega)$ , hence this ratio is exactly $2-\sqrt 3$ as noted in the previous case.

We are done. $\blacksquare$


probability
parameterization
calculus
calculus computations

Lets say $X$~$Bin(5000,0.001)$ and we wish to find $P(X\leq 2)$ .

Since $n=5000$ and $p\to 0$ , so we can use Poisson approximation $X$~$Po(5)$ and obtain

$P(X\leq 2)=P(X=0)+P(X=1)+P(X=2)=e^{-5}(1+5+5^{2}/2)\\=0.124652019...$

What if we approximate it to normal distribution ? Since $n=5000$ is large , isnt it that we can use Central Limit Theorem and hence approximate it to Normal Distribution ?

Unfortunately , when I try to do so i.e

$\overline{X}$~$N(5,\frac{4.995}{5000})$ , the probability turns out to be too small (calculator shows $0$) .. Why this happen ?

Someone please clear my doubt  :(

Let's put it this way: you already know that the Poisson distribution is an excellent approximation for this. The only way that the normal distribution could possibly also be considered a good approximation would be for the normal distribution to be a good approximation to the Poisson distribution. That only happens for large $\lambda$ (using $\lambda$ for the single parameter - the mean - of the Poisson distribution.) In this case, $\lambda=5$ which isn't large.

If you want to use the normal approximation to the binomial, the criterion isn't "$N$ large." Rather is is that $Np$ and $N(1-p)$ are both large. If one of $Np$ or $N(1-p)$ is large but the other is not, then use the Poisson approximation.

--

A side issue on calculations of the binomial probability: One can do this recursively. In what follows, set $q=1-p.$

$P(X=0)=q^{N}$

After that, $P(X=k+1)=P(X=k)\cdot \frac{N-k}{k+1}\cdot \frac{p}{q}.$

Use that in this case:

$P(X=0)=.999^{5000}\approx 0.006721112$

(The Poisson estimate for this was $e^{-5}\approx 0.006737947$.)

$P(X=1)=P(X=0)\cdot\frac{5000}1\cdot\frac1{999}\approx 0.033639199$

$P(X=2)=P(X=1)\cdot\frac{4999}2\cdot\frac1{999}\approx 0.084165343$

$P(X=3)=P(X=2)\cdot\frac{4998}3\cdot\frac1{999}\approx 0.140359822$

For your cumulative probabilty, we get $P(X\le 2)\approx 0.12452565$ Compare that to the $0.12465202$ you get from the Poisson approximation.

This direct recursive computation takes no more time and effort than the Poisson calculation; the only restriction is that we have to be able to compute $q^{N}.$

[quote="Kent Merryfield"]Oh, the standard deviation for this distribution is

$\sqrt{Npq}=\sqrt{4.995}\approx 2.234949664.$

(The standard deviation for the Poisson is $\sqrt{\lambda},$ which in this case would be $\sqrt{5}$.)

So part of your problem is that you were using the wrong standard deviation for your normal distribution. But it still wouldn't be a good approximation.[/quote]

ok thanks for clearing my doubt .

still one more thing about central limiting theorem .

Suppose now we have $X$~$Bin(500,0.2)$ , and we wish to find from a sample of $500$ , lets say , $P(\overline{X}\leq 98)$ .

so $Np=100$ , $Npq=80$ (which i think is large enough)
so we use estimation where $E(X)=100$ , sampling variance $s^{2}=\frac{Var(X)}{N}=0.16$ then by Central Limit Theorem we have

$\overline{X}$~$N(100,0.16)$

and so to find $P(\overline{X}\leq 98)$.

Is this correct ?  :maybe:

You made the same mistake again. What you're calling the sampling variance is not the variance of $X,$ it is the variance of $\frac{X}N.$ And we have $X$ here, not $\frac{X}N.$

Here's another way to do it: note that $\text{Var}(X)=80.$

Now let $Z=\frac{X-100}{\sqrt{80}}.$ $Z$ is then a distribution with mean zero and variance 1.

We approximate $Z$ by $\overline{Z}\sim N(0,1).$

But note that $X$ is discrete and we are trying to approximate it by a continuous distribution. You're better off slicing things halfway between the integers.

$P(X\le 98)=P(X\le 98.5)=P\left(Z\le \frac{98.5-100}{\sqrt{80}}\right) \approx P\left(\overline{Z}\le \frac{98.5-100}{\sqrt{80}}\right)$

I should add that $\text{Binom}(500,0.2)$ is well within the practical range of the recursive direct calculation outlined above. It took me just a 2 or 3 minutes with a spreadsheet to construct the entire cumulative probability distribution.

$P(X\le 98)\approx 0.437710718$

oh gosh , i guess i mess up  :blush:
thanks .

By constructing the entire cumulative probablity from $X=0$ to $X=98$ ? Isnt that takes times ? Though some of them is too small that we can ignore . So if I follow your step , means I have to start from $P(X=98)$ then keep reducing it ? no ?

Ah - one should learn to appreciate just how valuable a mathematical tool a spreadsheet is and just how friendly toward recursion it is. Yes, if you were trying to use a hand calculator, it might take a significant amount of time to construct the whole distribution, but with the spreadsheet (and the power of the "fill down" command), one can do 500 calculations as fast as one can do 10.

The best starting place for the recursion is all the way at the end, with $q^{n}.$ (Or the other end, with $p^{n}$). Yes, the probabilities out there are laughably small, and it doesn't really matter whether you include them in the cumulative probability or not, but it is the best place to start.

The problem with starting in the middle, say with $X=98,$ is that you have to compute $\binom{500}{98}$ and it takes some considerable cleverness to keep that calculation from overflowing your calculator. And that cleverness might well be the equivalent in terms of work of just starting the whole recuresion from the end.


calculus
integration
trigonometry
calculus computations

Find the value of $n\in{\mathbb{N}}$ satisfying the following inequality.

\[\left|\int_0^{\pi} x^2\sin nx\ dx\right|<\frac{99\pi ^ 2}{100n}\]

I think there are three such values: $n = 1, 3, 5$.

Proof: First I will show that $n$ cannot be even. To see this, suppose $n$ is even. Then,

\[ I = \int_0^\pi x^2 \sin(nx) dx = \left. - \frac{x^2\cos(nx)}{n} \right |_0^{\pi} + \frac{1}{n} \int_0^{\pi} x\cos(nx) dx\]

I omit some intermediate steps here. It should be easy to find that $| I | = \pi^2 / n$. But
\[ \frac{\pi^2}{n} < \frac{99\pi^2}{100n}\]
is absurd. Therefore, we need only consider the odd case. Using the same approach, we find that for odd $n$
\[ I = \frac{n^2 \pi^2 - 4 }{n^3}\]
Hence, solving $|I| < 99\pi^2 / 100n$, it should be easy to find that $n = 1, 3, 5$.

Thank you for your reply,skanderbeg.

You are right.

kunny


geometry
trigonometry
MATHCOUNTS
Pythagorean Theorem
area of a triangle
Heron\u0027s formula
trig identities

How do you find the area of a triangle with only the lengths of the legs and hypotenuse?

ummm...is this a joke?

leg x other leg/2 AKA (length)(height)/2

Yes, the area of a right triangle is just $ \frac 12 l_1l_2$ where $ l_1$ and $ l_2$ are the legs. You don't even need to know the hypotenuse if you know the legs. However, if you knew only one leg and the hypotenuse, then the other leg by the Pythagorean theorem is (if $ h$ is the hypotenuse and $ l$ is the known leg)
\[ \sqrt {h^2 \minus{} l^2}
\]
And the area is then $ \frac 12l\sqrt {h^2 \minus{} l^2}$.

The reason being because the area of any triangle is just $ \frac {1}{2}bh$, where $ b$ is the base and $ h$ is the height. So the legs of the right triangle are equal to the base and height(do you see why?)

In general, if you have a triangle and you only know the side lengths, you can use Heron's Formula, which goes like this:

$ A \equal{} \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}$

where $ a$, $ b$, and $ c$ are the lengths of the triangle and $ s$ is the semi perimeter(half the perimeter).

[hide="Proof (warning: contains trig)"]
Let $ a$, $ b$, and $ c$ be the lengths of the sides, and $ A$, $ B$, and $ C$ be the angles opposite them. From the Law of Cosines, we get:

$ \text{cos}(C) \equal{} \frac {a^2 \plus{} b^2 \minus{} c^2}{2ab}$

We then get $ \text{sin}(C) \equal{} \sqrt {1 \minus{} \text{cos}^2(C)} \equal{} \frac {\sqrt {4a^2b^2 \minus{} (a^2 \plus{} b^2 \minus{} c^2)^2}}{2ab}$

The altitude of a triangle with base $ a$ is $ b\text{sin}(C)$ so:

$ A \equal{} \frac {1}{2}ab$

$ \equal{} \frac {1}{2}ab\text{sin}(C)$

$ \equal{} \frac {1}{4}\sqrt {4a^2b^2 \minus{} (a^2 \plus{} b^2 \minus{} c^2)^2}$

$ \equal{} \frac {1}{4}\sqrt {(2ab \minus{} (a^2 \plus{} b^2 \minus{} c^2))(2ab \plus{} (a^2 \plus{} b^2 \minus{} c^2))}$

$ \equal{} \frac {1}{4}\sqrt {(c^2 \minus{} (a \minus{} b)^2)((a \plus{} b)^2 \minus{} c^2)}$

$ \equal{} \frac {1}{4}\sqrt {(c \minus{} (a \minus{} b))(c \plus{} (a \minus{} b))((a \plus{} b) \minus{} c)((a \plus{} b) \plus{} c)}$

$ \equal{} \boxed{\sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}}$

[size=75]Proof taken from Wikipedia.[/size]
[/hide]
[color=darkred][size=75]When possible, try to avoid using trig in the Mathcounts forum.
nebula42[/size][/color]

[quote="nick42"]ummm....basic triangle formulas? $ length*height/2$[/quote]

I think you mean:

$ \frac{\text{base} \times \text{height}}{2}$

Please do not repeat things already mentioned before you. This is considered spam.

[quote="isabella2296"]The reason being because the area of any triangle is just $ \frac {1}{2}bh$, where $ b$ is the base and $ h$ is the height. So the legs of the right triangle are equal to the base and height(do you see why?)

In general, if you have a triangle and you only know the side lengths, you can use Heron's Formula, which goes like this:

$ A \equal{} \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}$

where $ a$, $ b$, and $ c$ are the lengths of the triangle and $ s$ is the semi perimeter(half the perimeter).

[hide="Proof"]
Let $ a$, $ b$, and $ c$ be the lengths of the sides, and $ A$, $ B$, and $ C$ be the angles opposite them. From the Law of Cosines, we get:

$ \text{cos}(C) \equal{} \frac {a^2 \plus{} b^2 \minus{} c^2}{2ab}$

We then get $ \text{sin}(C) \equal{} \sqrt {1 \minus{} \text{cos}^2(C)} \equal{} \frac {\sqrt {4a^2b^2 \minus{} (a^2 \plus{} b^2 \minus{} c^2)^2}}{2ab}$

The altitude of a triangle with base $ a$ is $ b\text{sin}(C)$ so:

$ A \equal{} \frac {1}{2}ab$

$ \equal{} \frac {1}{2}ab\text{sin}(C)$

$ \equal{} \frac {1}{4}\sqrt {4a^2b^2 \minus{} (a^2 \plus{} b^2 \minus{} c^2)^2}$

$ \equal{} \frac {1}{4}\sqrt {(2ab \minus{} (a^2 \plus{} b^2 \minus{} c^2))(2ab \plus{} (a^2 \plus{} b^2 \minus{} c^2))}$

$ \equal{} \frac {1}{4}\sqrt {(c^2 \minus{} (a \minus{} b)^2)((a \plus{} b)^2 \minus{} c^2)}$

$ \equal{} \frac {1}{4}\sqrt {(c \minus{} (a \minus{} b))(c \plus{} (a \minus{} b))((a \plus{} b) \minus{} c)((a \plus{} b) \plus{} c)}$

$ \equal{} \boxed{\sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}}$

[size=84][b]Proof taken from Wikipedia.[\b][/size]
[/hide][/quote]
You just copy and pasted the proof didn't you  :P

Never underestimate isabella2296's powers! :lol:

I don't know, maybe she did, maybe she didn't. It all depends on whether I expect her to know trig or not :D

Um, no, I didn't copy and paste...I wrote it in my own words. Just copied the numbers because I thought it was a good proof.  :)

There were barely no numbers; perhaps you meant you copied the formulas (which is basically the proof) with $ a, b$, and $ c$?

lol, i didn't know you could do trig  :o

This is a proof that some guy on Wikipedia wrote, not me. I know very, very little trig. (But I plan to learn on my own soon.)

multiple the leg by the leg and you get the answer which is the area of a triangle unless it is not a equilateral triangle than you would need the other information. It can be easy but it can be hard according to the triangle.

Please don't repeat solutions, it's spam.

I think this question has been answered sufficiently. Moderators of this forum, please lock this thread.

For a better and much more elegant proof of Heron's, see the link below for a proof by fellow AoPSer and Georgian ARMLer Miles Dillon Edwards, published some time ago in a mathematical journal:

[url]http://www.artofproblemsolving.com/Resources/Papers/Heron.pdf[/url]

Umm... I don't think you need Heron's for this, and that proof is very complex (with all those complex numbers).

I was simply pointing out that when you have a non-right triangle, and the side lengths, you could use Heron's Formula.

Yea Heron's formula would work but in this situation it would just be a lot easier to use the basic area of a triangle formula

they were just giving things that people can do to find out the area of a triangle

in this case, some may be easier, but it's good to know them all


There's a guy in the middle of nowhere. He wants to go to the nearest hotel. There is a road 20 miles away (perpenducularly), heading toward the hotel. That road is a line. The guy is 40 miles away from the hotel.

(So, it's a 30-60-90 triangle)

Finally, the problem

The guy can walk at 3 mph in the grass, and 5 mph on the road. How does he walk the fastest to the hotel?

If you know any calculus, this should be easy.

Plus, can you not do it with a calculator? Just by hand?

What I mean is that the calculator is not used for graphs or problems, etc...

[hide]

Here we go.

I knew it was going to be some angle x above the horizontal and then figured that the equation y=(20/(3cos(x)))+(34.64-(20tan(x)))/5 and then found the minimum which i got as 36.86 degrees and 12.26 hours.

The forumla basically found the distance in miles of each leg then converted it into hours long the trip took then found the minimum of x.

Enjoy

[/hide]

There is a similar problem which was used to test if dogs know calculus.  (Dogs  can run  faster than they can swim, and they have to go retrieve a ball in the water)

BTW: [b]The problem is  same as law  of refraction for light- That is Snells law [/b]-  The ratio of sines of the angles = refractive index = ratio of speeds in different mediums.... for example  sin(x)=3/5 will give the answer)

Anyway coming back to the story of dogs...

Tim Pennings, an Professor of Mathematics  did try this with his dogs and it turned out that his dogs indeed found the shotest path-  (Result -  same as shown in the above post) :)

Read all about it in:[url=www.maa.org/features/elvisdog.pdf]Do Dogs Know Calculus?
[/url]
Enjoy!! :)
[quote]Most calculus students are familiar with the calculus problem of finding the optimal path from A to B. Optimal may mean, for example, minimizing the time of travel, and typically the available paths must transverse two different mediums, involving different rates of speed. This problem comes to mind whenever I take my Welsh Corgi, Elvis, for an outing to Lake Michigan to play fetch with his favorite tennis ball. Standing on the waters edge (See Figure 1) at A, I throw the ball into the water to B. By the look in Elvis eyes and his elevated excitement level, it seems clear that his objective is to retrieve it as quickly as possible rather than, say, to minimize his expenditure of energy..[/quote]

BTW  He does say that:[quote]. we confess that although he made good choices, Elvis does not know calculus. In fact, he has trouble differentiating [/quote]


calculus
integration
derivative
function
logarithms
trigonometry
calculus computations

I'm pretty new to this calculus stuff, but is there a rule for finding integrals for terms that are multiplied/divided together? I know for derivatives there's the product rule, division rule, etc. and finding integrals for terms that are added, you could integrate each one individually. but is integration through substitution and through parts the only way?

There is no general rule for the integral of a product. (Or to quote me alone in my office while grading a test, "THERE'S NO PRODUCT RULE FOR INTEGRALS!!! ARRGH!!!)

Bear in mind that all of our rules for "integration" - meaning antidifferentiation - can only come from taking rules for differentiation and reading them backwards.

The linearity of the derivative does read backwards simply as the linearity of the integral - that is, \[ \int af(x)\plus{}bg(x)\,dx\equal{}a\int f(x)\,dx\plus{}b\int g(x)\,dx.\]
The chain rule reads backwards as integration by substitution. The single most important thing you must know about differentiation is the chain rule, and this carries over to the single most important thing you must know how to do for doing integrals is how to substitute.

Now, can you read the product rule backwards? If we took that literally, we'd have \[ \int u\,dv\plus{}v\,du\equal{}uv\plus{}C,\]
but the problem with that is that we're relatively unlikely to encounter $ u\,dv\plus{}v\,du$ in exactly that form, unless we perform another step to create that form (see integrating factor methods in differential equations). So we subtract part of that off to the other side, thereby creating integration by parts. My capsule description of integration by parts is "the product rule read backwards with an uncancellation," meaning that if you check the result by differentiation, you'll use the product rule and something will cancel.

The quotient rule is just too specific to have much of a chance of producing a useful general rule if read backwards - and the quotient rule is more of a second-level, derived rule that it is a primary rule.

The real lessons to be learned about the process of integrating elementary functions (we'll discuss what that means) are these:
(1) It's not a machine.
(2) It doesn't always work. That is, there are elementary functions with no elementary antiderivatives.

These two points are closely related. Differentiation [b]is[/b] a machine. By "elementary function" I mean functions built out of a defined list of building blocks - powers, exponentials, logarithms, trig functions, inverse trig functions - combined by addition/subtraction/multiplication/division and by composition of functions. You know the elementary derivatives of each building block function and you can handle the combinations by linearity and by the product rule, quotient rule, and chain rule. And it's guaranteed to always work - the derivative of any elementary function is always an elementary function.

But you can't run that machine in reverse, and it doesn't always work. In fact, I'd prefer to think that for an elementary function, [i]not[/i] having an elementary antiderivative is the normal state of affairs; having one is a happy accident.


Every subset of {1,2,3,4,5,6,7} is arranged in descending order, and then all of the numbers in that subset is added, but with alternating signs. For instance, from subset {2}, we got 2, and for subset {1,3,6}={6,3,1}, we got 6-3+1=4. Determine the sum of all numbers gotten that way.

Clarify: when there's at least 2 in the set, does the minus always come first?


analytic geometry
geometry
trapezoid
rectangle
probability
induction
LaTeX

Here are the problems of COMC 2006:

[u][b]Part A[/b][/u]

[b]1.[/b] What is the value of $(1+\frac{1}{2}) (1+\frac{1}{3}) (1+\frac{1}{4}) (1+\frac{1}{5})$?

[b]2.[/b] If $f(2x+1)=(x-12)(x+13)$, what is the value of $f(31)$?

[b]3.[/b] In $\triangle{ABC}$, $M$ is the midpoint of $BC$. If $\angle{ABM}=15^{o}$ and $\angle{AMC}=30^{o}$, what is the size of $\angle{BCA}$?

[b]4.[/b] Determine all solutions $(x,y)$ to the system of equations:
$\frac{4}{x}+\frac{5}{y^{2}}=12$
$\frac{3}{x}+\frac{7}{y^{2}}=22$

[b]5.[/b] In $\triangle{ABC}$, $BC=4$, $AB=x$, $AC=x+2$, and $cos(\angle{BAC})=\frac{x+8}{2x+4}$.
Determine all possible values of $x$.

[b]6.[/b] Determine the number of integers $n$ satisfying [i]all three[/i] of the conditions below:
a. each digit of $n$ is either $1$ or $0$,
b. $n$ is divisible by $6$,
c. $0<n<10^{7}$

[b]7.[/b] Suppose $n$ and $D$ are integers with $n$ positive and $0\le D\le 9$.
Determine $n$ is $\frac{n}{810}=0.9D59D59D5...$

[b]8.[/b] What is the probabiblity that $2$ or more consecutive heads will occur at least once in $10$ tosses of a fair coin?

[u][b]Part B[/b][/u]

[b]1.[/b] Piotr places numbers on a $3$ by $3$ grid using the following rule, called "Piotr's Prinicple":
For any three adjacent numbers in a horizontal, vertical, or diagonal line, the middle number is always the average (mean) of its two neighbours.
(a) Using Piotr's principle, determine the missing numbers in the grid below:
[3]  [  ]  [19]
[8]  [  ]  [   ]
[  ]  [  ]  [   ]
(b) Determine, with justification, the total of the nine numbers when the grid below is completed using Piotr's Principle.
[x]  [  ]  [   ]
[5]  [  ]  [23]
[  ]  [  ]  [   ]
(c) Determine, with justification, the values of $x$ and $y$ when the grid below is completed using Piotr's Principle.
[x]  [7]  [   ]
[9]  [  ]  [ y]
[  ]  [  ]  [20]

[b]2.[/b] The circle $x^{2}+y^{2}=25$ intersects the $x-axis$ at poits $A$ and $B$. The lines $x=11$ intersects $x-axis$ at pt $C$. The point $P$ moves along the line $x=11$ above the $x$-axis and $AP$ intersects the circle at $Q$.
(a) Determine the coordinates of $P$ when $\triangle{AQB}$ has maximum area. Justify your answer.

(b) Determine the coordinates of $P$ when $Q$ is the midpoint of $AP$. Justify your answer.

(c) Determine the coordinates of $P$ when area of $\triangle{AQB}$ is $\frac{1}{4}$ of the area of $\triangle{APC}$. Justify your answer.

[b]3.[/b] (a) The trapezoid $ABCD$ has parallel sides $AB$ and $DC$ of lengths $10$ and $20$, respectively. Also, the length of $AD$ is $6$ and the length of $BC$ is $8$. Determine the area of trapezoid $ABCD$.

(b) $PQRS$ is a rectangle and $T$ is the midpoint of $RS$. The inscribed circles of $\triangle{PTS}$ and $\triangle{RTQ}$ each have radius $3$. The inscribed circle of $\triangle{QPT}$ has radius $4$. Determine the dimensions of rectangle $PQRS$.

[b]4.[/b](a) Determine, with justification, the fraction $\frac{p}{q}$, where $p$ and $q$ are positive integers and $q<100$, that is closest to, but not equal to, $\frac{3}{7}$.

(b) The baseball sum of two rational numbers $\frac{a}{b}$ and $\frac{c}{d}$ is $\frac{a+c}{b+d}$. Starting with the rational numbers $\frac{0}{1}$ and $\frac{1}{1}$ at Stage $0$, the baseball sum of each consecutive pair of rational numbers in a stage is inserted between the pair to arrive at the next stage. Prove:
(i) no rational number will be inserted more than once;
(ii) no inserted fraction is reducible;
(iii) every rational number between $0$ and $1$ will be inserted in the pattern at some stage

Alright guys, let em loose! :surf: (The Solutions)
[hide="Long awaited simple Algebra solution for No.7! :D "]$\frac{n}{810}=0.9D59D59D5...$
let $\frac{n}{810}=x$
so, $x=0.9D59D59D5...$
since the repeating decimal goes in a pattern of 3,
we multiply both sides by 1000.
$1000x=9D5.9D59D5...$
(here's the crux move! :lol: )
since $x=0.9D59D59D5...$, we subtract x from both sides thus giving us
$999x=9D5$
$x=\frac{9D5}{999}$
Now we sub $x=\frac{n}{810}$ back in to get
$n=\frac{(810)(9D5)}{999}$,which we can cancel down to
$n=\frac{(30)(9D5)}{37}$. Since 30 isn't divisble by 37,
9D5 must be. We use logic to find 9D5=925, thats the only
number with the same first and last digit in the 37th
times table :)
therefore,
$n=\frac{(30)(925)}{37}=(30)(25)=750$[/hide]
Hope you find it useful.

That is so much easier then the way i did it..

Actually i got the same equation but i just guessed and check to see what value for d worked

Nicely done

That's exactly what my solution was... except I didn't think that I needed to prove that $\frac{9D5}{999}$ was $x$, I just used prior knowledge... nice job! :D  :D

- NAWAHD WERDNA

Thanks Guys. :lol:
Everybody, begin posting your solutions. :)

Could Somebody PLEASE post a solution to A8... I really need to know! lol...  :help:

- NAWAHD WERDNA

[quote="rem"][hide="Solution to A8"]The total probability is $1$. We will substract the probabilities of all the "bad" cases ie no two heads are consecutive. It's easy to see the number of heads is then at most $5$.
Now let us look at the case when we have $k$ heads with no two consecutive. The number of tails is $10-k$ and there are $10-k+1=11-k$ different places to "put" the heads in and no two heads will be in the same place, because no two are consecutive. So the number of bad outcomes is:
$\sum_{i=0}^{5}{{11-i}\choose{i}}=144$
Then the number of favourable outcomes is $2^{1}0$(total outcomes is $2$ for each toss) minus number of bad outcomes $144$ which is $880$.
Then the reuqired probability is $\frac{880}{1024}=\frac{55}{64}$
[b]ANSWER[/b] : $\frac{55}{64}$[/hide]
Also there was some solution using Fibonacci numbers or something, does anybody know it?[/quote]

Let $f_{n}$ be the number of possible cases not having two consecutive heads. We know that $f_{0}$ is 1 and $f_{1}$ is 2.

Consider n coins. If the first coin is head, the second one is tail. The rest is n-2 coins not having two consecutive heads. Therefore, there are $f_{n-2}$ cases.
If the first coin is tail, the rest is n-1 coins not having two consecutive heads. Therefore, there are $f_{n-1}$ cases.

Therefore,  $f_{n}= f_{n-1}+f_{n-2}$ and $f_{10}=144$ (10th Fibonacci number). So the answer is $\frac{2^{10}-144}{2^{10}}= \frac{55}{64}$.

I'll be around for a few days on this forum. Ask me any questions concerning this competition.

That solution is just so pwnage. :omighty:

Question @David:
When will the results be posted on the web?

To be honest, I don't know. However, I know the chair of this competition(and where his office is). I could ask him.

how did you people do question 4 a) and b)

yes, i'd would like the solution to 4a and 4 b as well, it was such as strange question it seemed. i now know the answer to 4a after making a program to find it, but i still don't understand how to obtain the answer normally. as well, how should we prove 4b? i don't even know how to start that kind of question

I think the solution to 4b) is a little lengthy (if i'm not mistaken) unless if there's some real cool way of doing it

ya Alex Rem. told me the answer and then i made a program and it gave me that answer so its got to be right

I got all of Part A, and in Part B, I got all of #1 (I actually had 1c right on my test), #2, and #3a. On #3b, I messed up by saying the dimensions of the rectangles are $2x*y$, and solving for $x=12, y=9$, and I put down $x*y=12*9$ for the dimensions. I got #4a (shaky proof), but didn't get #4b. lightree, what score can I expect for missing 4b, having 4a kind of shaky, and messing up on 3b? 68-72?

[hide="Solution for B.4(a)"]In order to make the fraction as close as possible to $\frac{3}{7}$ we need to minimize $|\frac{p}{q}-\frac{3}{7}|=| \frac{7p-3q}{7q}|$.
So we need to minimize $7p-3q$ and maximize $7q$.
Now $min(7p-3q)=1$ because it cannot be $0$ because then $\frac{p}{q}=\frac{3}{7}$.

Now if $7p-3q=1$ the trivial solution is $(1,2)$ so the general solution is $(1+3k, 2+7k)$ where $k$ is an integer. Then maximum value of $2+7k<100$ is $93$ ie $k=13, p =40, q=93$. So, $min|\frac{p}{q}-\frac{3}{7}|=\frac{1}{7*93}$

Now if $7p-3q=-1$ the trivial solution is $(2,5)$ so the general solution is $(2+3k, 5+7k)$ where $k$ is an integer. Then maximum value of $5+7k<100$ is $96$ ie $k=13, p =41, q=96$. So, $min|\frac{p}{q}-\frac{3}{7}|=\frac{1}{7*96}$
So taking the better of the two cases $min|\frac{p}{q}-\frac{3}{7}|=\frac{1}{7* 96}$

Now consider the case when $100>q>96$ then $7p-3q\geq 2$ and so $|\frac{p}{q}-\frac{3}{7}|\geq \frac{2}{7q}\geq \frac{2}{7*99}>\frac{1}{7*96}$
In case when $q<96$, $|\frac{p}{q}-\frac{3}{7}|\geq \frac{1}{7*95}>\frac{1}{7*96}$

So, the answer is $p=41, q= 96$.[/hide]

right right... i think it is now time to stop with the discussion of chinese mathematics. Every country has what they are good at. I am chinese, but i dont care how good or how bad my comrades are...[i]i just care about how good [i]I[/i] do in math contests[/i].

So lets all just shut up and be friends...good?

[quote="bryant124"][/quote][quote="Phelpedo"][quote="bryant124"][quote="SM4RT"]I'm pretty sure its hard..because some of the world's best high school students in math cant even get all the questions..[/quote]
well if uve been in china thatll be different...they say a lot of chinese high school students can get a really high score on imo...its just that there are too many ppl there...[/quote]

uh...who says that? But by all means, try your hands at the problems from last year, eh? http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=16&year=2006

Iurie Boreico will probably be the most decored IMO-er ever. He got a silver in 7th grade, a gold in 8th grade, and then two consecutive perfect scores- and he still has more to go![/quote][quote="bryant124"]
the first question was pretty easy...but u do need a lot of experience of doing math exercises to be able to prove it...[/quote]
Numbers 1 (everyone from Canada got 7 on problem 1) and 4 were considerably much easier than other years.  Number 5 was delightful to work on. :)

I agree with "sunchips" lets not worry about other people lets focus on ourselves..btw whats wrong with this site, its not wroking to well on my PC..ya i think we should post more problems on this forum rather then discussing which country is the best..

[quote="SM4RT"]AOPS is up and running again... :)..
[quote=Phelpedo]
Hm, strange. Now what was I talking about? Oh yeah, it's like us have 5 Brian Lawrences.
[/quote]

Why wasn't Brian Lawrence on the US Olympiad Team last year...didnt he get 42 on the IMO the year before![/quote]

He went to RSI, which is a really good Science Institute.

I think the cutoff is 66, because there are 4 povincial plaque winners who did not make the cutoff. if the cutoff is 65, there would be 88 people write CMO, so i think the cutoff is 66.

[quote="mathgeniuse^ln(x)"]How many people would there be if 66 was the cutoff score?[/quote]

69 people got 66 or above, plus the 4 provincial champions who didn't make it, it would be 73 people.

[quote="Kileel"]Didn't the results say something like "the top [b]50[/b] students will be invited to CMO 2007"?[/quote]

It did.  Dang it, thanks for ruining the fun. jk.

Because then 69 would be the cutoff since there would be 44 people with 69+ and the four provincial winners=48

And if 68 were the cutoff there would be 52 people with 68+ and the four provincial winners=56

I really hate myself.

they usually take around 75 people or more. no more than 90 people would be invited. they don't only take 50 people.

wtf..  i think i got like 65.. *cuts wrists*

hmm ya that sux that ur off by one..but o well if ur not in grade 12 then u still got next year...

Wait, did anyone get an email of what the exact cutoff is?


parameterization
advanced fields
advanced fields unsolved

$ \varphi' \varphi'' \equal{} \epsilon^2 \varphi'''$   on  $ [\minus{}1,1]$

$ \varphi(\minus{}1)\equal{}\varphi(1)\equal{}0$, $ \varphi'(\minus{}1)\equal{}1$.

Discuss the solutions.  Start with $ \epsilon\equal{}0$.

(What is an appropriate response to this question in an oral exam?  The question seems to motivate something that I am unfamiliar with.  Can someone enlighten me?)

if ep is zero, i beleive there is no solution...


inequalities
calculus
derivative
function
calculus computations

Show that $ e^x \ge 1 \plus{} x$ without using Maclaurin Series.

They are equal at $ x\equal{}0$. Compare the derivatives for other values of $ x$.

Can you elaborate?

I'm not quite sure I get it.

This was a question on a BC calculus test, after finishing first semester.

In other words: show that $ e^x\minus{}x\minus{}1$ has a minimum at zero.

Even more specifically:  show that $ x \plus{} 1$ is the tangent line to $ e^x$ at $ x \equal{} 0$.

Sorry to be pedantic, but that is not actually more specific*: $ y \equal{} x$ is a tangent line to the curve $ y \equal{} x^3 \plus{} x$ at $ x \equal{} 0$, but it is not true that $ (x^3 \plus{} x) \minus{} x$ has a minimum at $ x \equal{} 0$.

* That is to say, proving what t0r wrote is not sufficient to conclude what jmerry wrote.

Sorry; show that $ x \plus{} 1$ is the tangent line to $ e^x$ at $ x \equal{} 0$ and that $ e^x$ is convex.

[u]Solution.[/u] Consider the function $ f(x)\equal{}e^x\minus{}x$.  Notice that $ f'(x)\equal{}0\iff x\equal{}0$ and that $ f'(x)<0$ and $ f'(x)>0$ in $ (\minus{}\infty,0)$ and $ (0,\infty)$, respectively.  Therefore, $ e^x\geq x\plus{}1$ for all $ x\in\mathbb{R}$. $ \Box$