Given $ a, b, c \ge 0$ satisfy $ a\plus{}b\plus{}c\equal{}2$
Prove that $ (a\plus{}b\minus{}ab)(b\plus{}c\minus{}bc)(c\plus{}a\minus{}ca) \le 1$  :)

[quote="nguoivn"]Given $ a, b, c \ge 0$ satisfy $ a \plus{} b \plus{} c \equal{} 2$
Prove that $ (a \plus{} b \minus{} ab)(b \plus{} c \minus{} bc)(c \plus{} a \minus{} ca) \le 1$  :)[/quote]
For $ ab \plus{} bc \plus{} ca \ge 1,$ applying the AM-GM Inequality, we have
$ \begin{aligned} \sqrt [3]{(a \plus{} b \minus{} ab)(b \plus{} c \minus{} bc)(c \plus{} a \minus{} ca)} & \le \frac {2(a \plus{} b \plus{} c) \minus{} (ab \plus{} bc \plus{} ca)}{3} \\
& \le \frac {2(a \plus{} b \plus{} c) \minus{} 1}{3} \equal{} 1, \end{aligned}$
so the inequality holds.

Let us consider now the case $ ab \plus{} bc \plus{} ca \le 1.$ Assume that $ c$ is the smallest numbers among $ a,b,c.$ From the inequality $ 1 \ge ab \plus{} bc \plus{} ca$ and $ (a \minus{} c)(b \minus{} c) \ge 0,$ we find that
$ 1 \ge ab \plus{} bc \plus{} ca \minus{} (a \minus{} c)(b \minus{} c) \equal{} 2c(a \plus{} b) \minus{} c^2 \equal{} 2c(2 \minus{} c) \minus{} c^2.$
This yields $ 0 \le c \le \frac {1}{3}$ (because $ c < 1$).

Notice now that
$ \begin{aligned} (b \plus{} c \minus{} bc)(c \plus{} a \minus{} ca) & \equal{} [(1 \minus{} a)c \plus{} a][(1 \minus{} b)c \plus{} b] \equal{} (1 \minus{} a \minus{} b \plus{} ab)c^2 \plus{} (a \plus{} b \minus{} 2ab)c \plus{} ab \\
& \equal{} ab(1 \minus{} c)^2 \plus{} (1 \minus{} a \minus{} b)c^2 \plus{} (a \plus{} b)c \equal{} ab(1 \minus{} c)^2 \plus{} (c \minus{} 1)c^2 \plus{} c(2 \minus{} c) \\
& \equal{} ab(1 \minus{} c)^2 \plus{} c(c^2 \minus{} 2c \plus{} 2). \end{aligned}$
Therefore, the inequality can be written as
$ (2 \minus{} c \minus{} ab)[ab(1 \minus{} c)^2 \plus{} c(c^2 \minus{} 2c \plus{} 2)] \le 1,$
or
$ [(2 \minus{} c)(1 \minus{} c)^2 \minus{} ab(1 \minus{} c)^2][ab(1 \minus{} c)^2 \plus{} c(c^2 \minus{} 2c \plus{} 2)] \le (1 \minus{} c)^2.$
After using the AM-GM Inequality, we see that it is enough to check the following inequality
$ (2 \minus{} c)(1 \minus{} c)^2 \plus{} c(c^2 \minus{} 2c \plus{} 2) \le 2(1 \minus{} c).$
This is equivalent to
$ c(1 \minus{} 2c) \ge 0,$
which is true because $ 1 \ge 3c.$
The proof is completed.

[quote="nguoivn"]Given $ a, b, c \ge 0$ satisfy $ a \plus{} b \plus{} c \equal{} 2$
Prove that $ (a \plus{} b \minus{} ab)(b \plus{} c \minus{} bc)(c \plus{} a \minus{} ca) \le 1$  :)[/quote]
My solution.
Assume:$ a \ge b \ge c$
$ \equal{}>b;c \le 1$
Case 1:$ 1 \ge a;b;c$
We have:$ a\plus{}b\minus{}ab \le 1;b\plus{}c\minus{}bc \le 1;c\plus{}a\minus{}ca \le 1$
Case 2:$ a \ge 1 \ge b \ge c$
We have:$ b\plus{}c\minus{}bc \le 1$
And:$ 4(a\plus{}b\minus{}ab)(b\plus{}c\minus{}bc) \le (2\plus{}b\minus{}ab\minus{}bc)^2 \le 4$
Done!

[quote="can_hang2007"]How about the following inequality?

If $ a,b,c$ are nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 2,$ then
$ (a \plus{} b \minus{} ab)(b \plus{} c \minus{} bc)(c \plus{} a \minus{} ca) \le 1 \minus{} abc.$

(There is a one-line solution for this)

[/quote]
Of course, $ RHS \minus{} LHS \equal{} (ab \plus{} bc \plus{} ca \minus{} 1)^2 \plus{} abc(a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \minus{} 1) \ge 0$
Obviously trues with $ a \plus{} b \plus{} c \equal{} 2$    :oops:
By this way, we can prove the stronger:
$ (a \plus{} b \minus{} ab)(b \plus{} c \minus{} bc)(c \plus{} a \minus{} ca) \le 1 \minus{} \frac{4}{3}abc.$

[quote="can_hang2007"][quote="nguoivn"][quote="can_hang2007"]How about the following inequality?

If $ a,b,c$ are nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 2,$ then
$ (a \plus{} b \minus{} ab)(b \plus{} c \minus{} bc)(c \plus{} a \minus{} ca) \le 1 \minus{} abc.$

(There is a one-line solution for this)

[/quote]
Of course, $ RHS \minus{} LHS \equal{} (ab \plus{} bc \plus{} ca \minus{} 1)^2 \plus{} abc(a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \minus{} 1) \ge 0$
Obviously trues with $ a \plus{} b \plus{} c \equal{} 2$    :oops:
[/quote]
Why this is obvious, nguoivn?[/quote]
Because $ a^2\plus{}b^2\plus{}c^2 \ge \frac{(a\plus{}b\plus{}c)^2}{3}\equal{}\frac{4}{3}>1$
So, $ abc(a^2\plus{}b^2\plus{}c^2\plus{}abc\minus{}1) \ge 0$  :)

[quote="can_hang2007"]I meant this, nguoivn. Why do we have:
$ RHS \minus{} LHS \equal{} (ab \plus{} bc \plus{} ca \minus{} 1)^2 \plus{} abc(a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \minus{} 1) \ge 0$?
If my calculation is not very bad, then this identity cannot hold.[/quote]
Sorry, I saw my mistake  :oops:
We have: $ LHS \equal{} (2 \minus{} x)(2 \minus{} y)(2 \minus{} z)$ with $ x \equal{} a \plus{} bc$...
So, $ LHS \equal{} 8 \plus{} 2(xy \plus{} yz \plus{} zx) \minus{} 4(x \plus{} y \plus{} z) \minus{} xyz$
Notice that: $ xyz \equal{} abc(a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \minus{} 3) \plus{} (ab \plus{} bc \plus{} ca)^2$
$ xy \plus{} yz \plus{} zx \equal{} 3(ab \plus{} bc \plus{} ca) \minus{} abc$
$ x \plus{} y \plus{} z \equal{} 2 \plus{} ab \plus{} bc \plus{} ca$
Then, $ LHS \minus{} RHS \equal{} (ab \plus{} bc \plus{} ca \minus{} 1)^2 \plus{} abc(a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \minus{} 4) \ge 0$?  :maybe:

[quote="can_hang2007"]How about the following inequality?

If $ a,b,c$ are nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 2,$ then
$ (a \plus{} b \minus{} ab)(b \plus{} c \minus{} bc)(c \plus{} a \minus{} ca) \le 1 \minus{} abc.$

(There is a one-line solution for this)
[/quote]
Under the condition $ a\plus{}b\plus{}c\equal{}2,$ the following identity holds
$ (a\plus{}b\minus{}ab)(b\plus{}c\minus{}bc)(c\plus{}a\minus{}ca)\equal{}1\minus{}abc\minus{}(1\minus{}a)^2(1\minus{}b)^2(1\minus{}c)^2,$
and from it, we can deduce that
$ (a\plus{}b\minus{}ab)(b\plus{}c\minus{}bc)(c\plus{}a\minus{}ca) \le 1\minus{}abc.$
This is the desired inequality. By the way, the identity also shows that this inequality holds for any [b]real[/b] numbers $ a,b,c$ satisfying $ a\plus{}b\plus{}c\equal{}2.$ Equality occurs for $ a\equal{}1,$ or $ b\equal{}1,$ or $ c\equal{}1.$

[quote=can_hang2007]How about the following inequality?

If $ a,b,c$ are nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 2,$ then
$ (a \plus{} b \minus{} ab)(b \plus{} c \minus{} bc)(c \plus{} a \minus{} ca) \le 1 \minus{} abc.$

(There is a one-line solution for this)

[quote="caubetoanhoc94"][quote="nguoivn"]Given $ a, b, c \ge 0$ satisfy $ a \plus{} b \plus{} c \equal{} 2$
Prove that $ (a \plus{} b \minus{} ab)(b \plus{} c \minus{} bc)(c \plus{} a \minus{} ca) \le 1$  :)[/quote]
My solution.
Assume:$ a \ge b \ge c$
$ \equal{} > b;c \le 1$
Case 1:$ 1 \ge a;b;c$
We have:$ a \plus{} b \minus{} ab \le 1;b \plus{} c \minus{} bc \le 1;c \plus{} a \minus{} ca \le 1$
Case 2:$ a \ge 1 \ge b \ge c$
We have:$ b \plus{} c \minus{} bc \le 1$
And:$ 4(a \plus{} b \minus{} ab)(b \plus{} c \minus{} bc) \le (2 \plus{} b \minus{} ab \minus{} bc)^2 \le 4$
Done![/quote]
In case 2, I think you were wrong. Because if we use these two inequalities, we will obtain that
$ (a \plus{} b \minus{} ab)(b \plus{} c \minus{} bc)^2 \le 1,$
not
$ (a \plus{} b \minus{} ab)(b \plus{} c \minus{} bc)(c \plus{} a \minus{} ca) \le 1.$[/quote]

[img]http://s13.sinaimg.cn/middle/006ptkjAzy77HsucEGw2c&690[/img]

[quote=scpajmb]Let $a,b,c \ge 0$,and $a^2  + b^2  + c^2  = 2$，prove that
$$\left( {a^2  - ab + b^2 } \right)\left( {b^2  - bc + c^2 } \right)\left( {c^2  - ca + a^2 } \right) + abc \le 1$$[/quote]
This inequality is equivalent to $f(w^3)\geq0$, where $f$ is an increasing function and the rest is easy checking.


Triangle ABC has a periemter of 10, and the lengths of its sides are all integers. If a is the length of side BC, what is the difference  between the largets and smallest possible values of a.

[hide="Solution"]
Since all sides are integers, the smallest possible size of A is 2. (It can not be 1. Take for instance the triangle 5-4-1. $4+1 = 5$, forming a straight line instead of a triangle. All other triangles with 1 like 6-3-1 are also impossible. A triangle that could work would be 4.5-4.5-1, but we need integer sides).

The largest possible value of A would mean $B+C > A$ Clearly, if $B+C = 5$, A is 5, and the perimeter is 10. However, it needs to be greater than. The smallest $B+C > 5$ with integer values is $B+C = 6$. Since $A+B+C = 10$, we have $A+6 = 10, \ A = 4$

Using min of A as 2 and max of A as 4, we see that their difference is $\boxed{2}$
[/hide]

[hide]$a+b+c=10$ or $b+c=10-a$

if a is the longest side, we can say that

$b+c>a = 10-a>a = 10>2a = 5>a$

Since $a$ must be an integer its highest value is $4.$

One may assume the lowest value of a should be $1$, but it is not possible

If the highest value of one of the sides is $4$, the highest perimeter possible if we let $a=1$ would be $4+4+1=9$, and we know the perimeter must be 10, so the lowest possible value for $a$ is $2$. The diference is hence $\boxed{4-2= 2}$[/hide]


inequalities
function
inequalities unsolved

x1,x2,....,xn are positive real numbers such that x1+x2+...+xn=1
prove that (x1+1/x1)(x2+1/x2).....(xn+1/xn)≧(n+1/n)^n .
plz help thx

[hide="Solution"]By AM-GM, $ k \equal{} \sqrt [n] {x_1x_2...x_n} \le \frac {1}{n}$
By Holder's inequality, $ LHS \ge (k \plus{} \frac {1}{k})^n$.
Let $ \frac {1}{k} \equal{} m$, then $ m \ge n$. Then $ k \plus{} \frac {1}{k} \equal{} m \plus{} \frac {1}{m}$. As $ m \ge n \ge 2$ and $ x \plus{} \frac {1}{x}$ is an increasing function for $ x \ge 2$, we have $ (m \plus{} \frac {1}{m})^n \ge (n \plus{} \frac {1}{n})^n$, which proves the claim.[/hide]


Is there a relationship between $ log_{a}(x), log_{b}(x), log_{ab}(x)$?

Yes.

If you use change of base formula for base $ 10$, we have $ \frac{\log_{10} x}{\log_{10} a}$, $ \frac{\log_{10} x}{\log_{10} b}$, and $ \frac{\log_{10} x}{\log_{10} (ab)} \rightarrow \frac{\log_{10} x}{\log_{10} a \plus{}\log_{10} b}$.  Now take the reciprocal of the fractions.  These should add together.

$ \frac{1}{\log_a x} \plus{} \frac{1}{\log_b x}\equal{}\frac{1}{\log_{ab} x}$.


vector
function
Functional Analysis
superior algebra
superior algebra unsolved

thinking about an example in functional analysis this question arose:
what is the dual of $ \mathbb{R}^{\infty}$ (real number sequences)?
I was trying to imagine the doble dual of  $ \mathbb{R}^{(\infty)}$ (almost everyterm zero sequences).... is well known that the dual of $ \mathbb{R}^{(\infty)}$ is $ \mathbb{R}^{\infty}$
...well I'll appreciate any help  :)

PS: I mean algebraic dual, without any topological considerations.

If $ B$ is a Hamel basis of real vector space $ V$, then the algebraic dual of $ V$ can be identified with the set $ \mathbb R^B \equal{} \{f\colon B\to\mathbb R\}$. The space $ \mathbb R^{(\infty)}$ (or $ c_{00}$ for me) has a countable Hamel basis, hence its dual is $ \mathbb R^{\infty}$. The latter space does not seem to have a natural Hamel basis, but we can hope to find the cardinality of such a basis (Hamel dimension). There exists a linearly independent subset of $ \mathbb R^{\infty}$ of continuum cardinality (indeed, there are continuum many infinite subsets of $ \mathbb N$ with finite overlap). Using some [url=http://en.wikipedia.org/wiki/Cardinal_number]cardinal arithmetics[/url] that I don't really understand, one may be able to prove that the Hamel dimension of $ \mathbb R^{\infty}$ is exactly the cardinality of continuum. If this is true, then the dual of $ \mathbb R^{\infty}$ is isomorphic to $ \mathbb R^{\mathbb R}$, although not in any natural way.

thanks you gave me a way how to start :D ... here is what I've done:

Let $ X$ be a vector space and $ B$ a basis then $ X'\cong \mathbb{R}^B$. Let  $ P(B)$ the set of not empty parts of $ B$ and for each $ A\in P(B)$ consider $ \chi_A\in \mathbb{R}^B$ the characteristic function of $ A$.

[b]Claim[/b] $ \chi_A,$ $ A\in P(B)$ are linearly independent i.e. $ A_1,\cdots,A_r$ are distinct then $ \alpha_1\chi_{A_1} \plus{} \cdots \plus{} \alpha_r\chi_{A_r} \equal{} 0\Longrightarrow \alpha_1 \equal{} \ldots \equal{} \alpha_r \equal{} 0$

$ r \equal{} 2$ is trivial and in the case $ \alpha_1\chi_{A_1} \plus{} \cdots \plus{} \alpha_r\chi_{A_r} \equal{} 0$ since $ A_i$ are distinct $ \exists i\leq r,\exists a\in A_i,\forall j\neq i, a\notin A_j$ because if we deny it $ \forall i\leq r,\forall a\in A_i,\exists j\neq i, a\in A_j$ in other words each $ A_i\subseteq A_{j_i}$ so we obtain a chain $ A_1\subseteq A_{j_1}\subseteq A_{j_2}\subseteq\ldots$ which must have repeated terms since there are only $ r$ $ A_i$'s and it means that the next $ A_i$ after the repeated term must coincide with it, which is a contradiction.

Evaluating $ a$ in $ \alpha_1\chi_{A_1} \plus{} \cdots \plus{} \alpha_r\chi_{A_r}$ we obtain $ \alpha_i \equal{} 0$ so we can finished by induction.

Finally if $ B$ is infinite, $ \{\chi_A| A\in P(B)\}$ has the same cardinal of the parts of $ B$ which is strictly bigger than the cardinal of $ B$ ([url=http://planetmath.org/?op=getobj&from=objects&id=136]Cantor th.[/url]) so $ dim X'\geq Card \{\chi_A| A\in P(B)\} > dim X$.

That's not quite right- merely having distinct sets is not enough for linear independence. $ A,B,A\cup B,A\cap B$ give a linearly dependent collection of functionals. The characteristic functions of single-element sets span the set of all characteristic functions of finite sets.

The $ \chi_A$ also don't span, unless the field is finite.

Don't try to find a basis for the space of linear functionals- just look at the cardinality of the whole space, and derive the cardinality of a basis from that.

it is quite nonsense to consider the algebraic dual. the algebraic dimension does not reveal anything about the structure. nevertheless ...

there is a theorem of kaplansky: if $ X$ is an infinite set, then $ K^X$ has dimension $ |K^X|$.

in particular: $ \dim_{\mathbb{R}}(\mathbb{R}^{\infty}) = |\mathbb{R}^{\mathbb{N}}| = |({2^\mathbb{N})^{\mathbb{N}}| = |2^{\mathbb{N} \times \mathbb{N}}| = |2^{\mathbb{N}}| = |\mathbb{R}|}$

as a consequence of the theorem above, we have for every infinte-dimensional vector space $ V$ over $ K$ the formula $ dim_K(V^*)=|K|^{dim_K(V)}$.

thus: $ \dim_{\mathbb{R}}(\mathbb{R}^{\infty})^* = |\mathbb{R}^\mathbb{R}| = |2^\mathbb{R}|$.


calculus
probability
blogs
linear algebra

Hello all,

My name is Lefteris and I recently discovered the Math forum here. I just recently obtained my bachelor degree in Computer engineering from my university here in Greece.

I also managed to acquire a certain scholarship issued by the Japanese government for graduate studies in Japan. The only "problem" is the entrance examinations of the university I applied for. They are really math-intensive. I will be taking said exams a year from now, so I have ample time before me.

In my undergrad university the math courses we had were:
"Analysis (Very basic calculus stuff), Discrete Math, Probability and Statistics, and Numerical Analysis. And they were in the first 3 semesters so I have to admit that I need to get re-accustomed to problem solving methods.

The entrance exams of the Japanese university state this:
"The problems will be in the following three fields:
(1) linear algebra; (2)analysis(differential/integral calculus, ordinary differential equations, etc.); and
(3) probability /statistics."

So what I am basically asking is advice for books which tackle the above subjects. Since I will be self-studying what I would need is books with lots of exercises to solve for practise and equally as many provided solutions so that I can self-check the solutions as I go. I need to develop very very good problem solving skills for the above subjects.

If anyone would like to check the level of the problems that the entrance exams would have he can go here:
[url]http://www.i.u-tokyo.ac.jp/edu/entra/examarchive_e.shtml[/url],
 and click on any year(the english version) to see the problems for that year.

I know that I might be asking for a lot but I would be really glad if anyone could provide me with any good advice.

Thanks in advance,
-Lefteris

I think the best person to help you here is AoPS member [url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=3182][b]kunny[/b][/url]. From the posts he's been making, I can assume that he is involved with some Japanese Universities as I see him frequently posting questions from Japanese universites' entrance exams. He is also situated in Japan so I'm sure he would have an idea of how difficult the entrance exams are. I would recommend sending him a private message. (Or getting his attention on his blog if he doesn't reply for a while.)

Thanks for taking the time to answer my question and thank you very much for the suggestion. I will send him a private message immediately.

Nice to meet you, Lefteris.

Can you undestand Japanese?

I consider myself an intermediate in the Japanese language. In no way can I say that I am fluent. I can have a conversation but I am sure I must be sounding kind of strange.

I have taken the 4th, 3rd and 2nd Kyu of The Japanese Language Proficiency exams. Thankfully the math will be in English too so I won't have to worry about not understanding any part of the problem.

As per the opening post of this topic what books would you recommend about the subjects and taking into account my situation? Any other advice since it is about Japan and Japanese universities?

All advice is really welcome. Thanks for taking the time to reply.


trigonometry
quadratics
geometry
perpendicular bisector
geometry unsolved

Let ABC triangle,$ \widehat{ABC} \equal{}75^0$and heigh $ AH\equal{}\frac{1}{2}BC$.Prove that  triangle ABC isosceles at C.

this can be solved by fairly simple trig
let AH=h
now AH/HB = cot 75 = tan 15
so HB = h tan 15
but BC = 2h, so HC = h(2 - tan 15)
using double angle formula for tan, i.e. tan 2x = 2 tan x / [1 - tan^2 x] and x = 15
we get a quadratic in tan 15, and taking positive soln, tan 15 = 2 - sqrt 3
so 2 - tan 15 = sqrt 3
so HB = h sqrt 3
so tan ACH = AH/HC = 1 / sqrt 3
but angle ACH is between 0 and 180 degrees, so it must be 30 degrees
this means angle CAB is 75 degrees, so angle A  = angle B, i.e. the triangle is isosceles at C

Let's call $ D$ the foot of the perpendicular of $ C$ onto $ AB$, and $ E$ the projection of $ D$ onto $ BC$. The triangle $ BCD$ is rightangled at $ D$ and having the angle $ \angle DBC$, $ \implies 4 \cdot DE=BC$ ( 1 ); $ AH$ being half of $ BC$ , with $ (1)$ we get $ D$ the midpoint of $ AB$, i.e. $ CD$ is the perpendicular bisector of $ AB$ and, therefore, the triangle $ \triangle ABC$ is isosceles at $ C$.

Best regards,
sunken rock

[quote="sunken rock"]Let's call $ D$ the foot of the perpendicular of $ C$ onto $ AB$, and $ E$ the projection of $ D$ onto $ BC$. The triangle $ BCD$ is rightangled at $ D$ and having the angle $ \angle DBC$, [color=red][size=150]$ \implies 4 \cdot DE = BC$ ( 1 ); [/size][/color]$ AH$ being half of $ BC$ , with $ (1)$ we get $ D$ the midpoint of $ AB$, i.e. $ CD$ is the perpendicular bisector of $ AB$ and, therefore, the triangle $ \triangle ABC$ is isosceles at $ C$.

Best regards,
sunken rock[/quote]
hi ,
from where u get the relation with red colore???
thanks

[hide]Let $ C'$ be the point such that $ \angle ABC'\equal{}\angle BAC'\equal{}75^{\circ}$, where $ C$ and $ C'$ lie on the same side of line $ AB$. We have $ \angle HAC'\equal{}75^{\circ}\minus{}\angle HAB\equal{}60^{\circ}$. Thus $ 2AH\equal{}AC'\equal{}BC'$. Thus $ C'\equal{}C$. Since $ \triangle ABC'$ is isosceles at $ C'$, then $ \triangle ABC$ is isosceles at $ C$.

Q.E.D. :D[/hide]

[quote="mathz"]Let ABC triangle,$ \widehat{ABC} \equal{} 75^0$and heigh $ AH \equal{} \frac {1}{2}BC$.Prove that  triangle ABC isosceles at C.[/quote]
[img]http://www.mathmontada.net/vb/uploaded/24_1260611216.jpg[/img][/quote]

See the attached drawing: ABC is the subject triangle, with $ \measuredangle BAC\equal{}15^\circ$, AD is symmetrical of AB about AC, hence $ \measuredangle BAD\equal{}30^\circ$. CE is the altitude of the rightangled triangle, DF the altitude of $ \triangle BAD$. $ DF\equal{}\frac{AD}{2}$, while CE is midline of $ \triangle BDF$, done.

Best regards,
sunken rock

[quote="sunken rock"]See the attached drawing: ABC is the subject triangle, with $ \measuredangle BAC = 15^\circ$, AD is symmetrical of AB about AC, hence $ \measuredangle BAD = 30^\circ$. CE is the altitude of the rightangled triangle, DF the altitude of $ \triangle BAD$. $ DF = \frac {AD}{2}$, while CE is midline of $ \triangle BDF$, done.

Best regards,
sunken rock[/quote]
hi sunken rock
according to your solution and graph
how can BCD be triangle??!!!!
[quote][sunken rock"]Let's call $ D$ the foot of the perpendicular of $ C$ onto $ AB$, and $ E$ the projection of $ D$ onto $ BC$. [color=red]The triangle $ BCD$ is rightangled at $ D$ and having the angle $ \angle DBC$[/color], $ \implies 4 \cdot DE = BC$ ( 1 ); $ AH$ being half of $ BC$ , with $ (1)$ we get $ D$ the midpoint of $ AB$, i.e. $ CD$ is the perpendicular bisector of $ AB$ and, therefore, the triangle $ \triangle ABC$ is isosceles at $ C$.[/quote]


[quote="pytheagle"]This is equivalent to $ 1^2 \plus{} 2^2 \plus{} 3^2 \plus{} 4^2 \plus{} 5^2 \plus{} 6^2 \plus{} 7^2 \equal{} \boxed{140}$.

There is $ 1$ way to choose a row of $ 8$ squares on one side and similarly with the other ($ 1\cdot1$).
There are $ 2$ ways to choose a row of $ 7$ squares on one side and similarly on the other ($ 2\cdot2$).
There are $ 3$ ways to choose a row of $ 6$ squares on one side and similarly on the other ($ 3\cdot3$).
There are $ 4$ ways to choose a row of $ 5$ squares on one side and similarly on the other ($ 4\cdot4$).
There are $ 5$ ways to choose a row of $ 4$ squares on one side and similarly on the other ($ 5\cdot5$).
There are $ 6$ ways to choose a row of $ 3$ squares on one side and similarly on the other ($ 6\cdot6$).
There are $ 7$ ways to choose a row of $ 2$ squares on one side and similarly on the other ($ 7\cdot7$).
We cannot do $ 1$ by $ 1$ because that only contains $ 1$ square whereas the minimum is $ 4$.

I hope you understand it.[/quote]

The question said, that there should be 4 black squares inside the square, so no 2x2 squares fit that criteria.

So it is $1^2 \plus{} 2^2 \plus{} 3^2 \plus{} 4^2 \plus{} 5^2 \plus{} 6^2  \equal{} \boxed{91}$.

[quote="anant1993"]The answer to this should be 11 instead. Since the question says "distinct" squares. So all 3*3 squares which are considered are not different, there can be only 2, 1 with 5 black 4 white and the other with 4 black 5 white. Similarly 2 squares for 4*4, 2 for 5*5, 2 for 6*6, 2 for 7*7 and only 1 for 8*8
So answer is 2*5 + 1  = 11[/quote]

You are right, the question can be interpreted that way. It does not specify, what counts as "distinct" squares - distinct by position, or distinct by shape. I will send a note about this.

PS. Alcumus said correct answer to be $\boxed{91}$.

[quote="Python54"]When I tried doing this problem, I did $49+36+25+16+9+4+2+1$ which is equal to 140 but Alcumus said the correct answer was 91.
This was because it said that there must be at least 4 small squares in each of these squares so 1x1s and 2x2s wouldn't count because they don't have 4 squares.

But 2x2 squares have 4 squares! $2\cdot 2=4$.[/quote]
They must contain $4$ [b]black[/b] squares.  All $2\times2$ squares in the figure contains exactly $2$ black and $2$ white squares.

No $1\times1$ squares or $2\times2$ squares contain five black squares. Every square which is $4\times4$ or larger does. However, a $3\times3$ square will only contain 5 black squares if its upper left corner is black. We can choose the upper left corner of a $3\times3$ square in $6\cdot6=36$ ways, but for only half of these squares will the upper left corner be black. Therefore, there are $36/2=18$ $3\times3$ squares containing at least 5 black squares. We can choose the position of the upper left square of a $4\times4$ square in $5\cdot5=25$ ways, so there are 25 $4\times4$ squares. Similarly, there are 16 $5\times5$ squares, 9 $6\times6$ squares, 4 $7\times7$ squares and 1 $8\times8$ square. There are a total of $18+25+16+9+4+1=\boxed{73}$ squares containing at least 5 black squares.
Readings:

However, we only need 4 black squares, not 5.
It might be a different problem where it has 5 black squares . . .

[quote=m.sun][hide][hide][hide]the answer's 73[/hide][/hide][/hide][/quote]

It's 91

[quote=sabbasi][quote="shazi"]I apologize if this sounds like a stupid question, but i'm not getting how the solution got $6\times6$ 3 by 3 squares? Where did the 6 come from in the first place?

Once again, sorry if this is a stupid question, but i need to know.[/quote]

If you consider for example the top left corner of a 3x3 square, then there are 6 positions in the first row, 6 in the second, etc. 6 in the sixth row, so that is how we get 6x6 and in general perfect squares for the number of nxn squares :)[/quote]

But aren't they supposed to have sides on the grid lines?

yeah but he's talking about the square in the top left corner not the sides

[quote=pytheagle]This is equivalent to $ 1^2 \plus{} 2^2 \plus{} 3^2 \plus{} 4^2 \plus{} 5^2 \plus{} 6^2 \plus{} 7^2 \equal{} \boxed{140}$.

There is $ 1$ way to choose a row of $ 8$ squares on one side and similarly with the other ($ 1\cdot1$).
There are $ 2$ ways to choose a row of $ 7$ squares on one side and similarly on the other ($ 2\cdot2$).
There are $ 3$ ways to choose a row of $ 6$ squares on one side and similarly on the other ($ 3\cdot3$).
I thought the answer was 91

There are $ 4$ ways to choose a row of $ 5$ squares on one side and similarly on the other ($ 4\cdot4$).
There are $ 5$ ways to choose a row of $ 4$ squares on one side and similarly on the other ($ 5\cdot5$).
There are $ 6$ ways to choose a row of $ 3$ squares on one side and similarly on the other ($ 6\cdot6$).
There are $ 7$ ways to choose a row of $ 2$ squares on one side and similarly on the other ($ 7\cdot7$).
We cannot do $ 1$ by $ 1$ because that only contains $ 1$ square whereas the minimum is $ 4$.

I hope you understand it.[/quote]


geometry perimeter ratio inradius MATHCOUNTS

[quote="dimak11214"]1) Find at least two ways to get the sum of the first 101 odd numbers.

2) What is the largest decimal thats less than zero?

3) A triangle and square both have the same perimeter, and both have incribed circles in them. Find the ratio of the area of the big circle to the small circle.[/quote]

[hide="3"]If the perimeters are x, the length of the square is $\frac{x}{4}$. so the area of its circle is $\frac{x^2\pi}{64}$. The triangle has length $\frac{x}{3}$ and the inradius is $\frac{\frac{\sqrt{3}x}{6}\frac{x}{6}}{\frac{x}{2}}=\frac{x\sqrt{3}}{18}$ and its area is $\frac{3x^2\pi}{18^2}$. So the ratio is $\frac{\frac{x^2\pi}{64}}{\frac{3x^2\pi}{18^2}}=\frac{27}{16}$[/hide]

[hide="number 1"]
1. Use the formula $n^2$.
$101^2=(100+1)^2=10000+200+1=10201$
2. Pair up the numbers.
There are 50 pairs, each with an average of $(201+1)/2=101$
That total is $101\times50\times2=10100$.
But we didn't count 101 because it didn't have a pair.
$10100+101=10201$[/hide]
[hide="number 2"]
Uhhh...
$-.\overline{0}1$?[/hide]
[hide="number 3"]
Say that the perimeter of each the triangle and circle is $12x$
The side length of the square is $3x$. Thus the radius of the inscribed circle is $1.5x$, and its area is $x^21.5^2\pi=2.25\pi x^2$
The side length of the triangle is $4x$.
To find the inradius, we can use the formula $rs=A$
The area is $\frac{16x^2\sqrt{3}}{4}=4x^2\sqrt{3}$
The semiperimeter is $6x$
So the inradius is $\frac{4x^2\sqrt{3}}{6x}=\frac{2x\sqrt{3}}{3}$
So the area of the circle os $\frac{12x^2}{9}=\frac{4x^2}{3}$
So the ratio is
$\frac{\frac{4x^2}{3}}{2.25\pi x^2}=\frac{7.75\pi}{4}$
But I doubt thats right[/hide]

[hide]1. There are actually 3 ways. First, you can square 101. Secondly, you can simply add all the numbers. Last but not least, you know the 101st odd number is 203. The formula is 1+2+3+...2x-1. In this case, you can square 101.(kind of like number 1, but a little different approach)

2. I don't quite get the question because there is no exact answer. For example, -.00000000000000000003 is a decimal and so is -.0000000001. So theres no answer.

3. I'm too lazy to do this one.[/hide]

2 doesn't have an answer unless you consider the Infinitesimal valid.

It seems like all the mathematical solutions I can think of have been posted, but I can think of a new solution for 1:
[hide="1"]Punch in all the numbers with the addition sign inbetween in a calculator.[/hide]
 :P

Number 2 is definetly strange.

biffanddoc, in number 3, you had the right steps until the end, when you screwed up on your ratio (it was the wrong ratio, and it was calculated incorrectly).

Well there is no answer for Q.2
If a is the answer for Q.2 Then we have a<0
But a<a/2<0
This contrdicts the maximality of a

Also let x = \[ -.\overline{0}1 \]
Thus 10x = \[ -.\overline{0}1 \]
Thus x=10x
Thus x=0
Thus \[ -.\overline{0}1=0 \]

[quote="Vihang"]Thus x=0
Thus \[ -.\overline{0}1=0  \][/quote]

Which is true, by the way. :lol:

Moderator's note:

During MATHCOUNTS competition season, please specify the source of all problems posted to this Forum.

Problem 3 is underspecified.  Some of you have assumed the triangle to be equilateral, but I don't see that in the problem statement.  So, there is not a single valid solution.

I do not understand Problem 2.  Please don't make up non-standard notation, such as $0.\overline{0}1.$  Less experienced students may not understand you are joking about this notation.

Then number three doesn't have an answer, and therefore I am not wrong :wow: :wow:

On number 2, you could also right it as negative one divided by infinity, which is the largest negative number.

$-\frac{1}{\infty}$

Yeah, but anything divided by $\infty$ is $0$, isn't it?

[quote="b-flat"]I don't know if it is by definition, but the way I look at it, it is as close to zero as possible, but not actually zero.[/quote]

It does seem that way, but that's actually defined as 0.

We all agree that the problem is messed up, so can we just cut discussion? This just keeps bumping this topic around.

ok.
$\infty$ is not a number, therefore there is no answer

[hide="1"]
Use the $n^2$ formula and $101\cdot2\cdot50$
You get $10201$[/hide]
Yeah, number 2 and 3 aren't that clear...

number 2:
[hide]well .00000000000000000(0 forever)1 but are we allowed that?[/hide]
number 1:
[hide]1. divide odd number by 2 and add .5. then square the number.
2. Do n(n+1)/2 and subtract the even numbers[/hide]
nubmer 3 is hard

$0.00000(0 forever)1=0.\overline{0}1$
Moogra what you mean to say is $-.\overline{0}1$ is a solution

This was explained in post number 8 in this topic.

This topic is really getting offtrack. Remember to use standard notation, which $.\overline{0}\cdots1$ is not.


How many 6-digit numbers are there such that-:
a)The digits of each number are all from the set $ \{1,2,3,4,5\}$
b)any digit that appears in the number appears at least twice ?
(Example: $ 225252$ is valid while $ 222133$ is not)
[b][weightage 17/100][/b]

Sorry if it isn't exactly!
1) Number consist $ 1$ digit appear $ 6$ times: $ 5$ number
2) Number consist $ 2$ digit $ a,b$ which $ a$ appear $ 4$ times, $ b$ appear $ 2$ times: $ 2.C_5^2.C_6^4 \equal{} 300$ number
3) Number consist $ 2$ digit $ a,b$ which $ a,b$ appear $ 3$ times: $ C_5^2.C_6^3 \equal{} 200$ number
4) Number consist $ 3$ digit $ a,b,c$ which $ a,b,c$ appear $ 2$ times:$ C_5^3.C_6^2.C_4^2 \equal{} 900$ number
Sum: $ 1405$ number

[quote="nktp"]Sorry if it isn't exactly!
1) Number consist $ 1$ digit appear $ 6$ times: $ 5$ number
2) Number consist $ 2$ digit $ a,b$ which $ a$ appear $ 4$ times, $ b$ appear $ 2$ times: $ C_5^2.C_6^4 \equal{} 150$ number
3) Number consist $ 2$ digit $ a,b$ which $ a,b$ appear $ 3$ times: $ C_5^2.C_6^3 \equal{} 200$ number
4) Number consist $ 3$ digit $ a,b,c$ which $ a,b,c$ appear $ 2$ times:$ C_5^3.C_6^2.C_4^2 \equal{} 900$ number
Sum: 1255 number[/quote]
shouldnt the second case be multiplied by 2 ? For onr digit may appear 4 and other twice, or vice versa ?

OK! I too hurry! Case 2 be $ a$ appear $ 4$, $ b$ appear $ 2$ and $ a$ appear $ 2$, $ b$ appear $ 4$, it have $ 300$ number and total is $ 1405$

i did this by another approach,could some1 plz tell whats the error i am making
let there be 6 slots then the first slot can be filled in 5 ways, the next in one way(since the digit is supposed to repeat) the next in 4,1,3,1 So the total possible ways are $ 5\times 4\times 3\times \frac {6!}{2!.2!.2!}$ as there are three pairs of two digits each. This gives $ 5400$ which is too big in my opinion. (There are other cases also but i think i am going wrong here itself)
Could some1 plz help  :maybe:

I thinks if you do by this way then if you select example $ 1$ in first step, $ 2$ in third step and $ 3$ in fifth step is difference with the way that you select example $ 1$ in first step, $ 3$ in third step and $ 2$ in fifth step. It is wrong because $ 2$ this way is same:select $ 3$ number $ 1,2,3$, then change place $ 3$ this number (you count $ \frac {6!}{2!.2!.2!}$ is number of way to change place $ 3$ couple this number)

Call a number $good$ if it satisfies conditions $(a)$ and $(b)$.
Case 1: The number contains only 1 distinct digit.
               No of such $good$ numbers $= 5$
Case 2 : The number consists of exactly 2 distinct digits. By condition $(b)$ there are two possible subcases i.e $(2+4),(4+2)$ and $(3+3)$
Total number of such $good$ numbers
$ = \binom{5}{2}( \frac {6!}{2!4!} + \frac {6!}{3!3!} +\frac {6!}{4!2!} )$
$ = 10 \cdot ( 15+15+20) = 500$
Case 3 : The number has exactly 3 distinct digits. There is only one option i.e $(2+2+2)$
No of such $good$ numbers $= \binom{5}{3} \cdot \frac{6!}{2!2!2!}$
$= 900$
There can't be more than 3 distinct digits as it violates condition $(b)$ .
Total number of $good$ numbers $= 1405$


videos
analytic geometry
geometry
3D geometry

The original thread: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=137926&search_id=1132268238&start=20

It was so funny I had to make a new one.

If you drive rings around your grampa on a motorcycle, you are either an Armagetron addict or a crazy Simpsons fan.

If you stick a flag on your toy tank and shoot lasers out of it, you might be a BZFlag addict.

Let's see...

If you have 3 brothers, and you all coordinate your outfits so each one of you is wearing a different color out of: green, red, blue, and purple, you may be Legend of Zelda: Four Swords addicts.

If you throw colorful plastic balls at animals and hope they disappear inside, you may be a Pokemon addict.

If you think cake is good for you, you may be a Runescape addict.

All I got.

[quote]If you think cake is good for you, you may be a Runescape addict. [/quote]
The cake is a lie...

if you like to throw litle sticks and pretend they are ninga-type-throwing-dagger-things, you may be a Naruto Addict.

You've been playing Halo 3 too long when your grenade finger twitches when you're near large crowds.

You've been playing COD4 too long when your aiming finger twitches at any sign of movement.

You've been playing Halo 3 too long if you're AoPS account reference the game in more than one way ( :blush: ).

you've been playing DDR for too long when your feet twitch whenever your computer mouse makes a quick move up the screen.

you've been reading manga too long when you continuosly type the name of your favorate character.

Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao Hao

ok so i guess 5 hours straight was too long...

You may be a Mario Kart addict if...

You've ever dreamed of turning into a bullet and soaring above everyone else.

Your windshield has ever been sprayed with ink.

You make a sudden evasive movement when you see a banana peel in front of you.

The only castle you've ever heard of is Bowser Castle.

Time for SSBM references...

If...
you greet your friends with a knee
you're M2K
you're NoTA
when asked what game Marth is from, you say,"NO DUH, YOU IDIOT, SMASH"
you try to wavedash in Naruto Clash of Ninjas
you automatically twitch you fingers to l-cancel whenever you land on the ground

etc. etc.

Also (I'll let you try to figure out what game I referencing to),
If...
when you start referring to money as minerals
when their are too many people in your classroom, you tell the teacher to construct additional pylons (big giveaway).

Mario Kart:

If when you have a child you put them on motor bikes and age 2.

I would make more, but I gotta leave.

[quote="erabel"]If you think cake is good for you, you may be a Runescape addict.[/quote]

If you can eat 20 cakes in one minute

Hey. I know a dude who thinks he is Darth Vader. He is hilarious.

In a slurred voice: "Luke! I am your Father"

If you try to be cool by playing Dragonforce on your GH guitar.

If you understand all of this references. You might be a vg addict.

When you avoid rooms full of people due to lag possibilities you might be avg addict.

[quote="hunter34"]If you try to be cool by playing Dragonforce on your GH guitar.[/quote]

someone did that once, i think

If you think when you die in real life you'll respawn in lumbridge you MIGHT be addicted to runescape...

if u suddenly start having a habit of pushing red buttons, u've played "The Red Button" on i-am-bored.com one too many times.


(7-x)/(x+3)-(x+5)/(x+3)=27
 solve for x.

need some help...
pls and thx =]

[hide] Since they have a common denominator,
\[\frac{7-x}{x+3}-\frac{5+x}{x+3}=\frac{2-2x}{x+3}=27\Rightarrow 27{x+3}=2-2x\Rightarrow 27x+81=2-2x\Rightarrow 29x=-79\Rightarrow x=\frac{-79}{29}\]

[quote="April"]:(
$\frac{7-x}{x+3}-\frac{x-5}{x+3}=27\Leftrightarrow 12-2x=27(x+3)\Leftrightarrow x=-\frac{79}{29}$[/quote]

it should just be 12 on the numerator in the second step

-edit- you wrote the problem down wrong :P


function
complex analysis
complex analysis unsolved

Suppose that $ f$ is analytic on the entire plane, and that $ f$ maps 1-1 & onto the complex plane.
Show $ f(z)\equal{}az\plus{}b$ for $ a,b\in \mathbb{C}, a\ne 0$.
(Hint: $ f$ has a pole at $ \infty$)

I strongly recommend taking a look at the last five topics in this subforum next time...


How many people would have to be randomly selected for the probability to be greater than 50% that at least one had a birthday on February 29?(Leap Year)
[hide]1013[/hide]
Solutions, please?

[hide]Probability that someone has a birthday on Feb 29 = $ \frac{1}{1461}$
probability of some other birthday = $ \frac{1460}{1461}$
so the least number of people needed to have a probability of 0.50 such that none of them have a birthday on Feb 29 is:

$ (\frac{1460}{1461})^n = 0.50$

$ n{\log(\frac{1460}{1461}) = {\log0.50}}$

so $ n = 1012$

this means that the probability that at least one of the 1012 has a birthday on Feb 29 is 0.50, so the number needed to give a probability greater than 0.50 will be $ 1013$.[/hide]


what book are you using to work at math?can you recomand me some good books for math cotests?Thank you!

Also books from Dusan Djukic,especially IMO compedium,and Arthur Engels book-"Problem-solving strategies"...


limit
induction
real analysis
real analysis unsolved

You don't really need to find the exact value.

$f: [\dfrac {1}{5},\dfrac {1}{3}] \rightarrow [\dfrac {1}{5},\dfrac {1}{3}]$ defined by $f(x)=\dfrac {1}{4-3x}$ is well defined, continuous and increasing. Whence it has a limit $l$, which must be solution of l = f(l) by continuity. But this equation has only two solutions $1$ and $\dfrac {1}{3}$, whence $l=\dfrac {1}{3}$.

in julien_santini solution I have to prove, that there exists $i$, such that $x_i \in [\dfrac{1}{5},\dfrac{1}{3}]$, right?

Yes, you would.

We can make it easier to work with, since $\dfrac15$ was artificial. For any $x\in [0,\dfrac13]$, $f(x)\in[\dfrac14,\dfrac13]\subset[0,\dfrac13]$ and $f$ is increasing in this range.  For your value of $a_1$, $a_2$ is slightly negative and $a_3$ is in the usable range.


geometry
rectangle
inequalities
analytic geometry
geometry proposed

Let $\{R_i\}_{1\leq i\leq n}$ be a family of disjoint closed rectangular surfaces with total area 4 such that their projections of the $Ox$ axis is an interval. Prove that there exist a triangle with vertices in $\displaystyle \bigcup_{i=1}^n R_i$ which has an area of at least 1.

[Thanks Grobber for the correction]

hi!

just a question about problem 2, Can the rectangles have points in common? and in that case are the common areas considered once for the total area?

:lol:
psykyk

p.s. sorry for posting in the wrong place the last occasion

If I remember correctly, the closed rectangular surfaces had to be disjoint. I think this was mentioned in the text.

another question my friend, and thx for clearing my previous doubts. First the rectangles must be disjoint, and can their projections onto the Ox axis be not disjoint?

Thank you!

psykyk

[quote="psykyk"]i am deadly confused with this problem !!!

It seems that the interval condition is not at all needed... after being puzzled trying to figure out how could i use this condition i found the following:

sinc U Rn (the union)is closed, there is among all triangles with vertices in this region one with maximum area.... then all points are within a triangle that has the first one as median traingle, and it's area is at least 4, so the original one has area at least 1, i'm not sure but with this argument even a strict inequality can be proved.[/quote]
i really agree with you .i think if there exists a set of points with area 4,then exist three points $A,B,C$ in the set such that the area of triangle $ABC$ is at least 1

The corect statement of the problem ends with:
Prove that there exist a triangle with vertices in $\displaystyle \bigcup_{i=1}^n R_i$ which has the area equal to 1.

:)

[hide="Fubini"]
Let $[a, b]$ be the projection of $ \bigcup_{i=1}^n R_i $ onto $Ox$. For $x \in [a, b],$ define $m(x) = \text{min}\{y|(x, y) \in \bigcup_{i=1}^n R_i\}$ and $M(x) = \text{max}\{y|(x, y) \in \bigcup_{i=1}^n R_i\}.$ Since $\bigcup_{i=1}^n R_i$ is included in $\{(x, y)|x \in [a, b], m(x) \le y \le M(x)\}$ and has area of at least $4$, it follows that there exists $x_0 \in [a, b]$ such that \[(b-a)(M(x_0)-m(x_0)) \ge 4.\] Consider the points $A$ and $B$ having coordinates $(x_0, m(x_0))$ and $(x_0, M(x_0))$ respectively, and suppose, for instance, $x_0 \le {{a+b}\over2}.$ Then, for \[x_1 = x_0 + {2\over{M(x_0) - m(x_0)}}\] we have $x_0 < x_1 \le x_0 + {{b-a}\over2} \le b,$ therefore we can find a point $C \in \bigcup_{i=1}^n R_i$ having coordinates $(x_1, y_1)$ such that $\text{area}(ABC) = 1$.
[/hide]


A library has six identical copies of a certain book. At any given time, some of these copies are at the library and some are checked out. How many different ways are there for some of the books to be in the library and the rest to be checked out if at least one book is in the library and at least one is checked out?  (The books should be considered indistinguishable.)

Obviously, we can have $ 1\minus{}5$ books in the library, so the answer is $ \boxed5$.

Yeah, and you don't have to spam and say that.  You can private message them, or something like that, but please don't post on the forum just to say "Nice solution!" or "That makes sense!"

To be specific, there can be 1-5 books in the library and 5-1 books cheked out

From 1 to 5 books can be in the library, and the rest are checked out. Therefore, there are $\boxed{5}$ possibilities.

Oh, I was considering the books distinguishable :(

i tried using stars and bars? so 6 books implies 5 gaps in which to divide the books. you divide them into categories of checked out and checked in. thus, since you only need one gap to divide them into its binary categories--and you have 5 gaps to choose from--you do 5c1


We define a lattice point as a point whose coordinates are integers, zero admitted. Then the number of lattice points on the boundary and inside the region bounded by the $ x$-axis, the line $ x \equal{} 4$, and the parabola $ y \equal{} x^2$ is:

$ \textbf{(A)}\ 24 \qquad
\textbf{(B)}\ 35\qquad
\textbf{(C)}\ 34\qquad
\textbf{(D)}\ 30\qquad
\textbf{(E)}\ \text{not finite}$

[hide]E, not finite; although x = 4 gives a positive boundary, there's an infinite number of points with negative x values that are inside the parabola. The x-axis boundary is irrelevant, as it does not intersect with the parabola y = x^2.[/hide]

I think you pictured the graph wrong. We are looking at the set of points under the parabola between $ x \equal{} 0$ and $ x \equal{} 4$ (including the parabola, the line $ x \equal{} 4$, and $ x \equal{} 0$). So all the lattice points that satisfy the above conditions are non-negative integers.

[hide="My Solution"]
We simply consider the $ y$-coordinate of the point on the parabola at the points $ 0 \le x \le 4$. For $ x \equal{} 0$, $ y \equal{} 0$, so we get 1 point. For $ x \equal{} 1$, $ y \equal{} 1$, so we get 2 points. For $ x \equal{} 2$, $ y \equal{} 4$, so we get $ 5$ points. Notice that these numbers are all one more than a perfect square. So we simply have to calculate the following sum:

$ \sum_{k \equal{} 0}^{4}{(k^2 \plus{} 1)} \equal{} \sum_{k \equal{} 1}^{5}{((k \minus{} 1)^2 \plus{} 1)} \equal{} \sum_{k \equal{} 1}^{5}{(k^2 \minus{} 2k \plus{} 2)} \equal{} \sum_{k \equal{} 1}^{5}{k^2} \minus{} 2\sum_{k \equal{} 1}^{5}{ k} \plus{} \sum_{k \equal{} 1}^{5}{2} \equal{} \frac {5(5 \plus{} 1)(2(5) \plus{} 1)}{6} \minus{} 5(5 \plus{} 1) \plus{} 5(2) \equal{} 55 \minus{} 30 \plus{} 10 \equal{} \boxed{35 \text{ (B)}}.$
[/hide]

I think I want to clear up some confusion here, because I also believe that the answer is $ E$, not finite.

$ x \equal{} 4$ is a vertical line, correct?

Yes, it is. So you end up getting some weird region under the curve.

OK, so apparently, I can't use geogebra.

But I'm not sure where you're getting this weird region from. A parabola in the form $ y \equal{} x^2$ is going to have a vertex at the origin  and will open upwards (and thus $ x \equal{} 0$ is irrelevant). The only bound that $ x \equal{} 4$ will give is a positive bound...the negative bound is still free to go on to infinity.[/img]

There's a unique region that has $ y \equal{} x^2$, $ y \equal{} 0$ and $ x \equal{} 4$ as part of its boundary.  It's in the first quadrant, [i]below[/i] the parabola but [i]above[/i] the $ x$-axis.

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh, ok.

Then Sunehra's solution works, because each positive x value up to 4 will have lattice points equal to its square + 1 for the point on x = 0.


I have been very busy lately and only think that I will get around to solving one problem. It says on the usamts site on the part for submitting solutions that you should send solutions for at least two solutions. However, it says that those are guidelines that we are strongly being urged to follow. Obviously, I would like to send in my one solution.  Is this allowed, or will my work only be graded if I send in more solutions?[/quote]

Hey, try to at least write something for the other problems, to get one or two points on them. There's no reason that you shouldn't be able to write anything about any problem except one of them.


Hi
can there be integral soln for the following eqn?
                 x 2+y  2 =19

[hide]

x^2 + y^2 = 19

(x-y)^2 + 4xy = 19

(x-y)^2 = 3 mod 4

But it is well known that every perfect square is either congruent to 0 or 1 mod 4!  Thus there are no integral solutions.

[/hide]

Quote:(x-y)^2 + 4xy = 19

I think this should be (x-y)^2 + 2xy. However after that it is simple to finish:

[hide](x-y)^2 + 2xy = 19. Clearly, (x-y)^2 must be odd, so (x-y) is odd. Then x,y, cannot both be odd, so one of them is even, say x= 2k. So we have (x-y)^2 +4ky = 19, and now we see that a perfect square cannot be 3 mod 4, so there are no solutions.[/hide]

It has already poseted but -  among x and y , one is even (say 2k) and other is odd (say 2j+1) then x^2+y^2 = 4*k^2+4*j^2+4*j+1   so when divided by 4, remainder is 1 hence no solution possible for 19.

BTW  It is well known that  for  odd prime p  of the form (4n+1)  there is unique solution for x^2+y^2 = p.   For  the form ..4n+3 there is no solution...

May I just point out that in this case it is much easier to check x=0, 1, 2, 3, 4 :) In fact you don't even need 4 if you assume x is the smaller one. Was there any chance the right hand side meant to have another digit?

Good job
the soln i thought was similar
no. of the form (2n)
 2 is 0 mod(4).And no of the form (2n+1) 2 is 1 mod (4)
The following cases arise:
Case 1:if both x and y are even
Then x  2 +y 2  is 0 mod(4)
Case 2:if both x and y are odd:
then x 2 +y 2  is 2 mod(4)
Case 3:if x is even and y is odd:
then x 2 +y 2 is 1 mod (4)
so there is no possibility of getting
x 2 +y 2 as 3 mod(4)

Sriram


counting
distinguishability
geometry
geometric transformation
rotation
reflection

There is an unlimited supply of
congruent equilateral triangles made of colored paper.  Each
triangle is a solid color with the same color on both sides of the
paper. A large equilateral triangle is constructed from four of
these paper triangles as shown.  Two large triangles are considered
distinguishable if it is not possible to place one on the other,
using translations, rotations, and/or reflections, so that their
corresponding small triangles are of the same color.
Given that there are six different colors of triangles from which to
choose, how many distinguishable large equilateral triangles can be
constructed?

[asy]draw((0,0)--(10,0)--(5,8.7)--cycle);
draw((5,0)--(2.5,4.3)--(7.5,4.3)--cycle);[/asy]

I think that its:

6 choices for the middle triangle x 5 for one of the outside x 4 x 3=360?

Well let's do some casework:

All same color-Clearly 6 ways.

All outside same color, center different-6x5=30 ways

One outside and one center same,all else different-6x5c2=60

2 outside and center same,other different-6x5=30

2 outsides same, all else different-6x5x4=120

All different-6x5c3=60

Thus the answer is 6+30+60+30+120+60=306?

Let us first consider only the three small triangles not placed in the center.

The number of ways to construct this is in one-to-one correspondence with the number of solutions to $ a\plus{}b\plus{}c\plus{}d\plus{}e\plus{}f\equal{}3$ where $ a$, $ b$, $ c$, $ d$, $ e$, and $ f$ represent the number of small triangles with each color.

Using the divider method (http://www.artofproblemsolving.com/Forum/viewtopic.php?t=242328) we find that the answer to this is $ \binom{8}{3}$ which is equal to $ 56$.

So, now consider the triangle in the center. It can be one of $ 6$ colors, so there are $ 6$ possibilities for the triangle in the center.

The final answer is $ 56\cdot6$ which is $ 336$. I may be wrong...

Well that differs from mine by 30, so maybe I forgot a possibility.

mewto missed center/one outside same (color A) and others different (color B)

336 is the answer:Explaination:[hide]Because any permutation of the vertices of a large triangle can be obtained by rotation or reflection, the coloring of the large triangle is determined by which set of three colors is used for the corner triangles and the color that is used for the center triangle. If the three corner triangles are the same color, there are six possible sets of colors for them. If exactly two of the corner triangles are the same color, there are $6\cdot5=30$ possible sets of colors. If the three corner triangles are different colors, there are ${6\choose3}=20$ possible sets of colors. Therefore there are $6+30+20=56$ sets of colors for the corner triangles. Because there are six choices for the color of the center triangle, there are $6\cdot56=\boxed{336}$ distinguishable triangles.[/hide]

I first did get 360 but after considering the problem in parts, I got [hide]336[/hide].

[hide=My solution]
$6\cdot\left( \binom{6}{1}+ 2\binom{6}{2}+ \binom{6}{3}\right)=\boxed{336}$
[/hide]

ugh I botched this one up by considering circular permutation. Only later did I realize that allowing reflections renders this useless...

Can someone post a solution explaining what each choose symbolizes? In other words, much more detailed explanation.

help please i don't understand anybody elses solution :(

@above, it's kinda late, but I can help you out if you don't know yet.
Here's my solution.

[hide=Sol]
So first, we might as well find the number of ways to choose the colors for the outer parts. We don't have to do it like this, but it would be simpler.

Lets split this into different cases:

Case 1: All three colors are the same.
There are $\binom{6}{1} = 6$ ways to do this, and we do $6$ choose $1$ because there are $6$ total colors, and we just have to choose $1$ of them since all the colors are the same. Also, we use the combination formula because the order of choosing colors doesn't matter.

Case 2: Two colors are the same, but one of them is different.
There are $ 6 \cdot 5 = 30$ ways to choose the colors here, for there are $6$ choices for the color where two triangles are the same, and $5$ choices for the color that is different.

Case 3: All of the colors are different.
There are $\binom{6}{3} = 20$ ways to choose the colors here, because there are $6$ total colors, and we have to choose $3$ of them.

To conclude this part, there are $20+30+6 = 56$ ways to color the outer triangles.

We are now done with finding the number of ways to color the outer triangles. The number of ways to color the inner triangle is $6$, because there are $6$ different colors to choose from.

Now, we can multiply the two results to get our answer:
$56 \cdot 6 = \boxed{336}$ ways to create a large equilateral triangle.
 [/hide]

Do we need to count for rotation and reflection?  The solution doesn't say that.

[hide]
"If the three corner triangles are different colors, there are ${6\choose3}=20$ possible sets of colors." ==> (6*5*4)/3!

"If exactly two of the corner triangles are the same color, there are $6\cdot5=30$ possible sets of colors."  If applying the same logic as above, shouldn't it be (6*5)/2!?
[/hide]

Thank you for your help.

[hide=Solution]We proceed by casework on the number of colors used.\\[b]Case 1: One color.[/b] There are $6$ ways to choose the color.\\[b]Case 2: $2$ colors.[/b]There are $\tbinom{6}{2}=15$ ways to choose the colors and $6$ ways to arrange them (this can be found by considering which color is the middle).\\[b]Case 3: $3$ colors.[/b] There are $\tbinom{6}{3}=20$ ways to choose the colors and $3(1+2)=9$ ways to arrange them (this can be found by considering which color has two tiles and proceeding by casework whether that color is in the center ).\\[b]Case 4: $4$ colors.[/b] There are $\tbinom{6}{4}=15$ ways to choose the colors, and $4$ ways to choose the middle color, which fixes the large triangle.\\ \\ Our answer is $6+15\cdot 6+20\cdot 9+15\cdot 4=\boxed{336}.$[/hide]

[hide=Overkill]
We use Burnside's Lemma.  In particular, let all the colorings of the triangle be the set being acted upon, and let the group acting on the set be $(\text{identity}, 120^{\circ} \text{rotation}, 240^{\circ} \text{rotation}, \text{reflection across height 1}, \text{reflection across height 2}, \text{reflection across height 2})$.  In the identity, all hold, and so we have $6^4$, in the rotations, the outer three must be the same, giving $6^2$ for each, and in the reflections, two of the three must be the same, giving $6^3$ for each.  Thus, the answer is $\frac{1296+36+36+216+216+216}{6} = 336$[/hide]

[hide=Solution]The inside triangle is independent so we multiply by 6. The answer is

6(number of ways for 3 triangles same color+number of ways for 2 triangles same color+number of ways for no triangles same color)

The number of ways for 3 of the outer triangles to be the same is 6 because 6 different triangles to choose from.

The number of ways for 2 of the outer triangles to be the same is 6x5 because 6 ways to choose which color will be same and 5 ways to choose which color will be different.

The number of ways for no triangles to be same is 6C3 because order doen't matter.
Answer is 6(6+30+20)=336[/hide]


A lottery gives out prizes which are powers of 11. Thus, the powwible prizes are 1, 11, 121, 1331, 14641, and 161051 dollars. They will give out no more than 10 awards ofany particular amount. If the lottery gives out exactly $1,111,111, now many prizes do they award? Please show your work.

the sum of the digits must end with one...

[quote="xenon_nightmare152"][hide]
41
[/hide]
don't pay attention to my previous post.... :oops:[/quote]

Read the last sentence to the problem. I can't see how you got 41.

41 is the correct answer. Would you like to show how you got that solution?

i used guess and check and got [hide]41[/hide]

What about the combinations to get that (solution?) :?

[hide="Solution"]The problem is analgous to the sum of the digits of 1111111 in base 11. $1111111=699881_{11}$ and $6+9+9+8+8+1=41$[/hide]

Thanks. My method was tediously long and not worth explaining.

Just add 11^6 until you can't fit any more in, then add 11^5 until you can't fit any more in, etc.

in response to natmc's comment about 236factorial's method being slow, I think it takes the same time as converting to base 11 because that's basically how you convert to a different base, unless there's some secret methods I dont know about  :)

I was talking about Treethings method.  I highly dought this is a sprint question, so I'd use by Basecon (base conversion) program and find it in 2 seconds, or use the method of just dividing by 11

sorry, not 236factorial, Treething is who I meant.  Isn't that how you convert to base 11 [without using calculator program lol]


arithmetic sequence
combinatorics unsolved
combinatorics

Prove that, for each $ n$, there is $ N$ such that $ A \bigcup B \equal{} \{1, 2, ..., N\} \Rightarrow A$ contains an arithmetic progression of lengh $ n$ or $ B$ contains such a progression.

This is just a special case of [url=https://en.wikipedia.org/wiki/Van_der_Waerden's_theorem]Van der Waerden's Theorem[/url] (but essentially equivalent to it).


Change the position of just one digit to make the following equation correct:

$ 62\minus{}63\equal{}1$

[hide="Answer"]$ 62 \equal{} 63 \minus{} 1$.[/hide]
 :wink:

Nope, that's not right.

The problem says to move [b]a digit[/b], not signs.

Fine, fine, if we play "by the rules" we have to do
[hide="this."]$ 2^6 \minus{} 63 \equal{} 1$.[/hide]


function
geometry
3D geometry
quadratics
rectangle
calculus
calculus computations

What conditions must a and b have if the function is to have a local maximum and a local minimum. The function is: y= x^3 + ax^2 +bx

You have a Cone with the heigh 24 cm and the radie 12 cm. Inside it, there is a cylider. Which is the highest volume that the cylinder can have inside the cone?

You're supposed to use derivate in both problems but I guess you already figured that out :p
Thanks to anyone who can solve these for me. I'm having a test in two days and these are the only problems that I haven't solved.

Thanks again!

The volume of a cylinder is $\pi r^{2}h$, but the variables $r$ and $h$ are constrained by the cylinder being inside the cone. Draw a cross-section of both solids by a vertical plane that passes through their common axis. You will see a rectangle inscribed into an isosceles triangle. From this you should be able to deduce that $h+2r=24$. Then maximize $\pi r^{2}(24-2r)$ over $0\le r\le 12$.

[quote="akech"]For the first one, differentiate it with respect to x and you will get a quadratic equation. Then set this equal to zero to solve for x which you can solve by using quadratic formula. Then decide what a and b are going to be by requiring the discriminant to be either zero or positive.[/quote]

Add to this commemt:

The desired condition is that the sign of $y'$ can change twice; the discriminant >0.

[quote="Swiifty"]What conditions must $a$ and $b$ have if the function is to have a local maximum and a local minimum. The function is: $y= x^{3}+ax^{2}+bx$.[/quote]
$a^{2}>3b$.

The first problem wasn't hard at all when I think of it now... Second one is a bit more problematic but I'll see if I can get it to work now.

[quote="mlok"]From this you should be able to deduce that $h+2r=24$. Then maximize $V=\pi r^{2}h$.[/quote]
$24=2r+h\geq3\sqrt[3]{r^{2}h}=3\sqrt[3]{\frac{V}{\pi}}$,
with equality if and only if $r=h=8$.

I can't figure out how you've drawn the conclusion that h + 2r = 24

[quote="Swiifty"]The first problem wasn't hard at all when I think of it now[/quote]

What is your answer?

[quote="Swiifty"]a^2 > 3b

Ji chen was wrong because he set a^2 > 3b to be equall which it can't be if the equation is to have 2 points where y´ = 0.[/quote]

That's right. :)

But I still don't get the second problem :S
Does the way its been solved here include all cylinders inside a cone? What if the height is 20 cm and radie 5 cm? Does it still apply?


analytic geometry
graphing lines
slope
function
limit
trigonometry
calculus

I haven't really read books about division of 0's, but please see if my explanation is correct.

1.) Why is 1/0, undefined ?

[hide]Suppose $ \frac {1}{0}$ is defined, then $ \frac {1}{0} \equal{} b, b \in R$. Then b can be 0 or any real number.
Case 1:
Suppose b = 0. Then 1/0 = 0. Multiplying both sides by 0, since cancellation works for multiplication, we cancel 0 on the left side of the equation, leaving 0 = 1, which is not true.

Case 2: Suppose $ b \neq 0$. Then 1/0 = b => 0 = 1. [/hide]

Since $ b$ is not a real number, therefore, 1/0 is not defined under the set of real numbers.

2. What is  indeterminate such as 0/0 or 0 raise to 0? What is the difference between undefined and indeterminate?

Well I'm probably not the best person to reply to this but as far as I know... Indeterminate refers to a value that can be [i]made[/i] determinate. You can get information out of it after applying some theorems or clever algebra during the limit process. An undefined value, such as the slope of $ y\equal{}\frac{1}{x^2}$ as x tends towards zero is undefined. Essentially the slopes reciprocal is zero but the concept of something changing infinitely for each instant is... well, absurd.

[quote="t0rajir0u"][quote="guile"][hide]Suppose $ \frac {1}{0}$ is defined, then $ \frac {1}{0} \equal{} b, b \in R$. Then b can be 0 or any real number.
Case 1:
Suppose b = 0. Then 1/0 = 0. Multiplying both sides by 0, since cancellation works for multiplication, we cancel 0 on the left side of the equation, leaving 0 = 1, which is not true.

Case 2: Suppose $ b \neq 0$. Then 1/0 = b => 0 = 1. [/hide][/quote]
This is perfectly fine.  What you have shown is that $ \frac {1}{0}$ is not a real number that satisfies the usual properties of real numbers, which is correct.

[quote="guile"]2. What is  indeterminate such as 0/0 or 0 raise to 0? What is the difference between undefined and indeterminate?[/quote]
The concept of an indeterminate form comes from calculus.  Often we have a function $ f(x)$ with the property that evaluating $ f$ at some point (say, $ 0$) gives $ f(a) \equal{} \frac {0}{0}$ or $ f(a) \equal{} 0^0$ (this is an abuse of notation; what I mean is that evaluating $ f(a)$ requires evaluating an expression of that form), which cannot be meaningfully assigned a value without further information.  On the other hand, it may turn out that $ \lim_{x \to a} f(x)$ exists.

The reason such forms are indeterminate is that the same indeterminate form can correspond to different limits for different functions.  For example, if $ f(x) \equal{} \frac {x^2 \minus{} 9}{x \minus{} 3}$, then $ f(3) \equal{} \frac {0}{0}$ is an indeterminate form and

$ \lim_{x \to 3} f(x) \equal{} 6$.

On the other hand, if $ f(x) \equal{} \frac {\sin x}{x}$, then $ f(0) \equal{} \frac {0}{0}$ is an indeterminate form and

$ \lim_{x \to 0} f(x) \equal{} 1$.[/quote]

but is there a proof, like the one, i've made above, that would suggest that 0/0 or 0 raise cannot be computed?

[quote="guile"]but is there a proof, like the one, i've made above, that would suggest that 0/0 or 0 raise cannot be computed?[/quote]

Not to my knowledge. I think it is taken intuitively; a mathematical axiom if you will. It's quite clear that where $ y \equal{} f(x)$, $ \lim_{c\to 0} y^c \equal{} \lim_{c\to 0} [f(x)]^c \equal{} 1$, but that's just a limit. It's not the actual value of $ [f(0)]^0$, just the value that $ y^c$ approaches from the left and right.

Check this out: \[ \frac {\infty}{\infty} \equal{} \infty\cdot \frac {1}{\infty} \equal{} \infty\cdot 0 \equal{} 0\] \[ \frac{0}{0}\equal{}\frac{\frac{a}{\infty}}{\frac{b}{\infty}}\equal{}\frac{a}{b}\cdot \frac{\infty}{\infty}\equal{}\frac{a}{b}\cdot \infty \cdot \frac{1}{\infty}\equal{}0\]

[quote="guile"]but is there a proof, like the one, i've made above, that would suggest that 0/0 or 0 raise cannot be computed?[/quote]
On the one hand, $ \frac{0}{0} \equal{} \frac{a \cdot 0}{0} \equal{} a \cdot \frac{0}{0}$ for any real $ a$ if we assume the usual properties of fractions; it follows that we must have $ \frac{0}{0} \equal{} 0$.  On the other hand, the usual properties of fractions also tells us that $ \frac{0}{0} \equal{} 1$.

The deeper issue here is the notion of dividing by zero at all.  The definition of the quotient $ \frac{a}{b}$ is in terms of the multiplicative inverse $ b^{\minus{}1} \equal{} \frac{1}{b}$, which does not exist when $ b \equal{} 0$.  Thus $ \frac{a}{0}$ cannot be defined in the real numbers (in fact, over any ring or field) for any $ a$.

The second question involves a similar contradiction between two usual properties and deeper rules.  On the one hand, $ x^0 \equal{} 1$ for any $ x$, so $ 0^0 \equal{} 1$.  On the other hand, $ 0^x \equal{} 0$ for any $ x$, so $ 0^0 \equal{} 0$.

The deeper issue here is how you define exponentiation.  From a calculus standpoint, the preferred definition (because it is immediately defined for positive reals) is

$ a^b \equal{} e^{b \ln a}$.

This expression is not defined when $ a \equal{} 0$; in fact, for most calculus purposes it is not useful to have a non-positive base anyway.  See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=204218&search_id=245007287]the discussion here[/url].  (Threads like this are all over the place.)


A stained glass window is composed of pieces of glass shaped as a semicircle, a rectangle, a square, an equilateral triangle, and two trapezoids. Find the number of square units, to the nearest tenth, of the region labeled T.

[asy]pair A=(0,10), G=(20,10), F=(40,10), E=(50,0), D=(40,-10), C=(30,-10), I=(30,0), J=(40,0), H=G-(0,10), B=(0,-10), K=(10*sqrt(3),0);
draw(A--G--F..E..D--C--B--A);
draw(A--K--B);
draw(K--H--G);
draw(H--I--J);
draw(I--C);
draw(F--D);
label("20",(A+G)/2,N);
label("20",(G+F)/2,N);
label("10",(J+D)/2,W);
label("T", H-(0,5));[/asy]

Ok, to solve this diagram, we'll to find the height of the equilateral triangle.  You'll see why in a minute.

We could draw the height, and we could immediately notice two $ 30$-$ 60$-$ 90$ triangles! (gasp!)

[asy]draw((0,0)--(2,0)--(1,1.732)--cycle);
draw((1,1.732)--(1,0));[/asy]

In a $ 30$-$ 60$-$ 90$ triangle, the longer side (in this case, the height) is always $ s \sqrt{3}$, where $ s$ is the shorter side.

The shorter side in this case would be $ \frac{10}{2}$, or $ 5$.

Thus, the height is $ 5\sqrt{3}$.

We can then subtract $ 5\sqrt{3}$ and $ 10$ to find the "top" base of the trapezoid, which is approximately $ 21.34$.

We can easily see that the length of the "bottom" base is $ 40 \minus{} 10$, or $ 30$.

And finally, our height is the same as the square, so our height is $ 10$.

Use the area of a trapezoid, $ A \equal{} h \cdot \frac{b1 \plus{} b2}{2}$.

Plug in the values, to recieve an area of $ \boxed{256.7}$.

That was a long one...

[quote="ernie"]
The shorter side in this case would be $ \frac{10}{2}$, or $ 5$.
[/quote]
how do you know the diameter is 10?

[quote="ernie"]
Plug in the values, to recieve an area of $ \boxed{256.7}$.
[/quote]

The answer is 213.4 on FTW


A similar problem has been posted here, but it's not quite the same.
Suppose $ a_n \in R$, $ a_n > 0$, and $ \sum a_n$ diverges.

What can be said about these series?

a) $ \sum \frac{a_n}{1 \plus{} na_n}$

b) $ \sum \frac{a_n}{1 \plus{} n^2a_n}$

My guess is that a) diverges too, while b) converges.
But I was unable to prove it so far.
Thanks.

a) Divergent , because of : $ \ \frac {a_n}{1 \plus{} na_n}\geq \min (\frac 1{2n},\frac {a_n}2)$
b) Convergent , because of :  $ \ \frac {a_n}{1 \plus{} n^2a_n} < \frac 1{n^2}$
 :cool:

Actually, I think we missed something here, Diogene.

A series $ b_n$ satisfying: $ b_n \geq \min (\frac 1{2n},\frac {a_n}2)$ can still be convergent.

Example: Construct $ a_n$ as follows:

$ a_n \equal{} 1$, if $ n \equal{} 2^k$ for some natural $ k$.
$ a_n \equal{} \frac{1}{10^n}$, otherwise.

The series $ \sum{a_n}$ obviously diverges, because it has infinitely
many terms $ 1$, and $ a_n > 0, \forall n$.

However, if we take $ b_n \equal{} \min (\frac 1{2n},\frac {a_n}2)$, then $ \sum{b_n}$ converges.

We need a better estimate.

[quote="Lele-Br"]Actually, I think we missed something here, Diogene.

A series $ b_n$ satisfying: $ b_n \geq \min (\frac 1{2n},\frac {a_n}2)$ can still be convergent.

Example: Construct $ a_n$ as follows:

$ a_n \equal{} 1$, if $ n \equal{} 2^k$ for some natural $ k$.
$ a_n \equal{} \frac {1}{10^n}$, otherwise.

The series $ \sum{a_n}$ obviously diverges, because it has infinitely
many terms $ 1$, and $ a_n > 0, \forall n$.

However, if we take $ b_n \equal{} \min (\frac 1{2n},\frac {a_n}2)$, then $ \sum{b_n}$ converges.

We need a better estimate.[/quote]

Sorry , I was in an absent mind :oops:  The serie $ \sum\frac{a_n}{1\plus{}na_n}$ can be divergent or convergent  :!:

Actually in your exemple it's convergent : $ \sum \frac {a_n}{1\plus{}na_n}<  \sum  \frac  1{1\plus{}2^k} \plus{} \sum  \frac 1{10^n\plus{}n}$
 :cool:


Dima has written number $ 1/80!,\,1/81!,\,\dots,1/99!$ on $ 20$ infinite pieces of papers as decimal fractions (the following is written on the last piece: $ \frac {1}{99!} \equal{} 0{,}{00\dots 00}10715\dots$, 155 0-s before 1).  Sasha wants to cut a fragment of $ N$ consecutive digits from one of pieces without the comma. For which maximal $ N$ he may do it so that Dima may not guess, from which piece Sasha has cut his fragment?

[i]A. Golovanov[/i]

The problem asks for the maximum fragment that appears in $ 2$ numbers. It's easy to see that we have a fragment with $ 155$ digits $ 00\dots0010$ that appears in $ \frac{1}{99!}$ and $ \frac{1}{98!}$(without the comma of course). Suppose there exists a fragment with $ 156$ digits. Then there exists $ k_{1},k_{2}\ge 155$ ($ k_{i}$ would be the position of the last digit from the fragment in a number, if we number the digits in the number from $ 0$) , $ 80\leq x,y\leq99$ ($ x,y$ different) such as the following holds:
$ \frac{[10^{k_{1}}\cdot \frac{1}{x!}-10^{k_{2}}\cdot \frac{1}{y!}]}{10^{156}}= z$, where $ z$ is an integer. This implies:
$ z \cdot 10 \leq 10^{k_{1}-155}\cdot \frac{1}{x!}-10^{k_{2}-155}\cdot\frac{1}{y!}< z\cdot 10+\frac{1}{10^{155}}$
Suppose $ x>y$, then $ 99 \ge x \ge 81$. We have:
$ z \cdot 10\cdot x! \leq 10^{k_{1}-155}-10^{k_{2}-155}\cdot(y+1)(y+2) \dots x < z\cdot 10\cdot x!+\frac{x!}{10^{155}}<z\cdot 10\cdot x!+1$ (because $ \frac{99!}{10^{155}}<1$).
The number in the middle is an integer($ k_{1},k_{2}\ge 155$) between two consecutive integers so: $ z \cdot 10\cdot x! = 10^{k_{1}-155}-10^{k_{2}-155}\cdot(y+1)(y+2) \dots x$. This implies $ x|10^{k_{1}-155}$, which is impossible since $ 99 \ge x \ge 81$($ x$ will have always a divisor different from $ 2$ and $ 5$). So the maximum is $ 155$.


Let ABC be a triangle with angle A<C<90<B. Consider the bisectors of exterior angles at A and B each measured from the vertex to the opposite side.Suppose both these line segments are equal to AB.Find angle A

D =AC intersection with external angle B bissector.As AB=BD the triangle ABD is isosceles so pi/2-B/2+B+2A=pi
E=BC intersection wih external angle A bissector.As AE=AB the triangle ABE is isosceles so 2pi-2B+pi/2-A/2=pi
Then we find A=pi/15


Two gear wheels A and B are in contact. One wheel (A) has 36 teeth. The other (B) has 24 teeth. How many times must the smaller wheel turn before the larger wheel completes a revolution?

[hide]

24:36 is the ratio
Simplifing gives you
1:3/2
So the answer is the smaller wheel must rotate one and a half times before the big gear rotates once.[/hide]


geometry
circumcircle
radical axis
geometry proposed

[quote="pohoatza"]Let $ABCD$ be a convex quadrilateral where $O \in AC \cap BD$, and $(AO), (CO)$ are symmedians in triangles $ABD$ , $CBD$. Let $w1$ be a circle tangent in $D$ to the circumcircle of $\triangle{ACD}$, and $w2$ a circle tangent in $B$ to the circumcircle of $\triangle{ABC}$ and denote $X,Y \in w1 \cap w2$. Prove that $XYCA$ is cyclic.[/quote]

First of all, we will prove that $ABCD$ is cyclic. This is obvious because we ll have:
$\frac{CD}{CB}=\frac{AD}{AB}$ ...etc...

So, the circles $ACD$ and $ABC$ are identically same with the circumcircle $C$ of the triangle $ABD$.The common tangents of the circles $w_{1},C$ and $w_{2},C$ intersect at L (fixed point). But then, the point L will be the radical center of the circles $w_{1},w_{2},C$, so the points $X,Y,L$ are collinear and also holds:
$LY\cdot LX=LD^{2}=LC \cdot LA$ and the proof is completed!

P.S.The points $A,O,L$ are collinear because $AO$ is the symmedian (well-known theorem)
Thanks for the nice problem!
Nick Rapanos

[quote="robinson"][quote="Nick Rapanos"]
First of all, we will prove that $ABCD$ is cyclic. This is obvious because we ll have:
$\frac{CD}{CB}=\frac{AD}{AB}$ ...etc...

[/quote] why is it obvious?[/quote]

Ok let me prove it then (although u could think it on yr own...)
[b]If $AO$ is the symmedian of $ADB$ then C lies on the circumcircle of $ADB$.[/b]
[i]Proof[/i]
We want $\frac{CD}{CB}=ct$ and $C\in AC$, so the point $C$ is the intersection point of $AC$ with the apolonious circle of $ADB$.
But it is a well known fact that the symmedian and the apolonious circle intersect on the circumcircle...
(I have two proofs in mind but no time-I ll post them on Monday)

the point $L$ is also the pole of line $BD$ wrt to the circumcircle of $ABCD$, and $ABCD$ is a harmonic quadrilateral, thus $L$ lies on $AC$.


group theory
abstract algebra
superior algebra
superior algebra solved

Let $G$ be a group and $H$ a subgroup of $G$ for which $[H,G']=\{e\}$.
Prove that $[H',G]=\{e\}$.


Let $ ABCD$ be a convex quadrilateral so that $ m(\widehat {ABD}) \equal{} 2\cdot m(\widehat {ADB}) < 60^{\circ}$ ,

$ AB \equal{} AC$ and $ m(\widehat {AEB}) \equal{} 60^{\circ}$ , where $ E\in AC\cap BD$ . Ascertain  $ m(\widehat {ADC})$ .

[hide="Answer."]
$ 30^{\circ}$ .[/hide]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=200985][color=darkred][b]See here[/b][/color][/url] a particular case.

[b]Solution[/b] (similiar with suleymansoyler)
Let $ \angle{ABD} \equal{} 2\alpha, \angle{ADB} \equal{} \alpha$. Let $ T$ outside quadrilateral, such that $ TA \parallel BD$, and $ \angle{ADT} \equal{} \alpha$.Let $ CT \cap BD \equal{} O$ In trapezoid $ ABDT$ we have: $ \angle{ABD} \equal{} \angle{BDT} \equal{} 2\alpha$, so it is isosceles and $ AB \equal{} TD$. Also $ \angle{DAT} \equal{} \angle{TDA} \equal{} \alpha$, so $ AT \equal{} TD \equal{} AB \equal{} AC$. $ \angle{CAD} \equal{} \angle{BEA} \equal{} 60^\circ$, so $ \triangle{ACT}$ - equilateral and $ AB \equal{} AC \equal{} CT \equal{} TD \equal{} AT$. $ \angle{TOD} \equal{} \angle{ATC} \equal{} 60^\circ, \angle{COD} \equal{} 120^\circ$. $ \angle{ATD} \equal{} 180^\circ \minus{} 2\alpha, \angle{CTD} \equal{} 120 \minus{} 2\alpha$. $ \angle{TCD} \equal{} 30 \plus{} \alpha, \angle{CDB} \equal{} 30 \minus{} \alpha. \angle{CDA} \equal{} 30 \minus{} \alpha \plus{} \alpha \equal{} 30^\circ$


The numerator and the denominator of a fraction are integers differing by 16.  Find the fraction if its value is more than 5/9 but less than 4/7.

We can write this problem as 5/9<x/(x+16)<4/7.
so that 0.555...<x/(x+16)<0.57142857...
We know that the fraction is a little over 1/2.  And since the denominator is 16 more than the numerator, we can see that 21/37=0.567567... works.

-interesting_move

Can you do this without using trial and error?  Say, if you didn't have a calculator and you didn't want to have to long-division a bunch of numbers.

well, my solution differs slightly with interesting_move's. I solved for 5/9<x/(x+16) and x/(x+16)<4/7 and found that 20<x<21+1/3, and so x, an integer, is clearly 21. the two solutions yield the same result, but i just like mine better since it requires less guessing, and decided to post it  8-)

Fierytycoon

Yes, we can. 5/9<x/(x+16)<4/7, so (by the first inequality), 5x+80<9x, or x>20. By the second inequality, 4x+64<74, so 3x<64, or x<22. Thus x=21.

5/9 < 4/7 < 1 why we write the unknown as x > 0 and x/(x+16) < 1, so no multiplication by negatives will be made. 5/9 < x/(x+16) < 4/7 gives 20 < x < 21 + 1/3 where all the rational x satisfying the inequality work.

Mm-hmm.  That's the nice, non-guessing way, you three (Fierytycoon, ComplexZeta, and Mindspa).

Well, three answers that show the preferred solution.  I guess that means we're on to another one.  This time harder, preferably.

I made an error, x clearly needs to be an integer otherwise the supposed form x/(x+16) won't work.

Prove that for all integers a,b>=1 sqrt(3) lies inbetween a/b and (a+3b)/(a+b).

This can be proved for a general case sqrt(n), then the righths fraction needs to be altered.

Did anyone bother to consider the possibility when the numerator and denominator are negative?

What the problem is really saying is 5/9 < |x - 16|/x < 4/7.

It turns out there aren't any solutions there, but I don't think that's obvious until you check.

Well first suppose that our fraction a/b < rt3. rt3 equals (3-rt3)/(rt3-1), so we have a(rt3 - 1) < b(3 - rt3), we can do this as we aren't multiplying by any negatives. So a+3b > art3 + brt3, and thus (a+3b)/(a+b) > rt 3.
Similarly we can do the other case, if a/b > rt3 just reverse all the signs in the above process. a/b never can equal rt3 as it is irrational, so we are done.

The same method proves that rt(n) is between a/b and (a+nb)/(a+b).
Nice problem, by the way..


Find all integer roots of the equation : $y^2 = x^3 + (x+4)^2$

[hide]$(0, 4)$?

We must have $x \equiv 0$ $mod4$

Then we divide never ending times the expression by $2$ obtaining a contradiction. [/hide]


calculus
integration
function
search
calculus computations

That's an elliptic integral. Unless $A=-1$, the antiderivative is not an elementary function.

[quote="jmerry"]... the antiderivative is not an elementary function.[/quote]

[quote="Ish"]Is there a method to solve it?[/quote]

Ish, you're not listening. You presented an antiderivative problem (as opposed to an integral, meaning definite integral, problem). For that, "solve" is a somewhat ambiguous term but can be taken to mean "find an elementary function which is this antiderivative." And you've been told that that is impossible.

How do I approximate it closest to the true value?

You are not integrating over a given interval, so there is no value you are looking for.  You are looking for a general solution, not a number. :wink:

lol. sorry im not too familiar with elliptic integrals.
What can I do with them?

A suggestion: before taking a look at such integrals, study and learn easier materials thoroughly.  And for elliptic integrals, search the web (a good source would be http://www.wikipedia.com ).


Suppose a coin collection contains 30 coins which are either nickels, dimes, and/or quarters. How many different combinations of these coins are there whose value is $5?

Is this just a case by problem?

is the group of 30 coins is variable or fixed?

[quote="236factorial"][hide="I think this is true"]
You need at least 14 quarters in each arrangement. [/hide][/quote]
True, using quarters and dimes

[quote="236factorial"]Make an organized list, adjusting the numbers. I've done that so many times it's making me sick.

[hide="I think this is true"]
You need at least 14 quarters in each arrangement. [/hide][/quote]
By make an organized list you mean make a list for each arrangement?

ex: make $69 out of 5's and 2's? answer 6? 4 2's, 6 5's, 14 2's, 5 5's, etc?

Making a table might help, because everything is in straight (hopefully) columns and rows. A random list can confuse you.

[hide="this really helps"]
25q+10d+5n=500
5q+2d+n=100
q+d+n=30
4q+d=70, nickels will follow.
so d=2 mod 4, and none can be negative. its trivial from there[/hide]

[quote="nat mc"]if you can't get it then here:
[hide]minimum of q:
q+d<=30
4q+d<=30+3q
70<=30+3q
40<=3q
q>=13.33333

max of q:
4q+d=70
4q<=70
q<=17.5

Obviously there are only 4 values of q, 14,15,16,17.
sO the answer is 4[/hide][/quote]
Well, I got it, in a different way.

That really was not casework. just 2 equations, 3 unknowns

Im still sketchy on the definition of casework............
I did the algebra in my head a slightly different way, only getting one  of the bounds and guessing the stuff in between....


Why there is no Portuguese Fourm? Where could I use the portuguese?

Back when the forum layout displayed national flags with posters' screen names, I do recall seeing several Brazilian flags. If you could find those posters, they could be your allies. (That's certainly the largest body of Portuguese speakers in the world.)

[quote="Ifai"]Why there is no Portuguese Fourm?[/quote]

Good question. Portuguese is one of the most widely spoken languages of the world (the 7th?), probably as much as Russian, and certainly more than the languages spoken in many subforums here in the National & Local Communities. Anyway, you have here another "ally", since I am Portuguese.

For reference, here is a list of the users of APS that have location in Brazil or Portugal, along with user number and number of posts:

iurion (10331 - 8)
ajrocha (2647 - 0)
F u r u y á (11242 - 30)
Claudio Freitas (29820 - 0)
McCory (8071 - 6)
localixto (11068 – 0)
renato2050 (8760 - 0)
skydiver (19743 - 0)
Brambila (20959 - 76)
Calsaverini (14879 - 0)
josecelio (9800 - 0)
sergio_tadao (3683 - 0)
Musashi (4368 - 0)
Mauro (5822 - 0)
Erkenbrand (10308 - 3)
xlu_henry (10703 - 0)
amunhoz (12379- 0)
DrCarl (16877 - 0)
arnaldom (20222 - 1)
enildus (10522 - 0)
bersantos1 (12488 - 0)
trustno1 (9763 – 3)
brazil (17880 - 4)
Whyzzerd (569 - 0)
thiago_j (4609 - 10)
dsbraz (4388 - 0)
hjbortol (6203 - 3)
edward elric (8122 - 0)
MParaujo (7726 - 0)
jbotelho (9604 - 0)
aheimlich (11150 - 0)
MasterMath (10543 - 0)
Antonio Luiz (10363 - 0)
chdvf (10869 - 0)
IgorCastro (11940 - 30)
Felipe Diniz (11963 - 4)
Kellem (11964 - 0)
mffmartinelli (14685 - 106)
Serginhomello (14729 – 0)
Cako (20666 - 0)
jonfv (21160 - 40)
ricardotb (13499 - 8)
carlito84 (14108 - 0)
lflorencio (11479 - 0)
VascoMoreira (31861 - 1)
rodrigsen (24331 - 1)
Batalha (2191 - 0)
Bajoukx (3688 - 30)
Tiago Guerreiro (17537 – 50)
JC1 (23608 - 9)
joaopedro (24051 - 1)
Afonso Bandeira (31295 - 26)
Isix (3639 - 0)
Otávio_Menezes (20160 - 25)
windtalker (15998 – 1)
odracir (19908 - 1)
Aamodt (10186 - 1)
Oreste (7868 - 6)
Ricardo (9157 - 8)
fav0rit (18197 - 0)
Joao Guerreiro (9437 - 51)
POmPOm (15479 – 0)
mymycat (19380 – 0)
Daniel Hartmann (20195 - 0)
e.lopes (13603 – 341)
Davi Maximo (9739 - 0)
The Papangu (3374 – 0)
Edson Lopes (8622 - 9)
Overshine (9619 - 2)
ramonn (9952 - 0)
Régis Barbosa (15805 - 0)
Jensen Popp (17348 - 0)
Jose A Delgado (18488 - 0)
lnrd (19006 - 0)
Pimentel-ES (7698 - 0)
jceia (19270 – 0)
Fischer° (15870 - 0)
FFP (13823 – 1)
Roligtroll (7920 - 0)
Diogo (10868 – 0)
jcconegundes (18829 - 0)
Lucas Molina (7697 - 17)
Bruno Bonagura (10913 - 95)
lucianne (5397 - 0)
erick_nascimento (14977 - 0)
fskilnik (8714 - 65)
schumi (8770 - 0)
LoboSolitario (12730 - 154)
r mendes (18292 - 0)
Metal (20080 - 1)
niski (20464 - 26)
Levi (23009 – 9)
sauloplay (25712 – 0)
bruninha (29255 - 1)
Versus (30986 - 1)
MarcioCohen (10296 – 1)
gabriel ponce (18149 – 92)
Cloud (16335 – 0)
Fernando Cunha Córes (9830 – 0)
santosguzella (25872 – 160)
Benzadeus (14988 – 0)
eduflavio (19133 – 0)
Fabio_Pastorius (16408 – 0)

There also some users whose name appears to be Portuguese, and there are about more users from Macau. In total, there are about 110 users that speak Portuguese.

Obrigado, Carcul! :D

Eu estudo Português na universidade. Since I think it's not enough for studying Portuguese if I just spend time to speak it on the lessons, I want to meet some people who know Portuguese so that I can learn from them and talk with them. Hence I will PM you later.

Nevertheless, how do you know the list of who know Portuguese? Moveover, I know that most of the people in Macao do not know Portuguese.

Yes , I'm a brazillian too.

Seria ótimo se tivessemos um fórum em português ;)
Qualquer coisa só me avisar ;)

Well, what is the process to be made to create a Portuguese-speaking forum here?

ptBR:
Bem, qual é o processo que se tem que fazer para se criar um sub-fórum de falantes de português por aqui?


topology
function
real analysis
real analysis unsolved

$ U\subset R^n$ is an open set. $ 1\leq m \leq n$.
$ \forall i\equal{}1,...,m$, $ F_i(x) \in C^{(p)}(U,R)$.
Let $ F(x)\equal{}(F_1(x),...,F_m(x))^T$, $ S\equal{}{x \in U|F(x)\equal{}0}$ is not null. Moreover $ \forall x \in S, rankF'(x)\equal{}n\minus{}k$.
Prove or disprove: S is a $ C^{(p)}$ k-dimensional manifold.

I think it was asked and solved by me long ago (search). Amazingly I remember almost nothing about this topic..

The book 'analysis on manifolds' by munkres has a related result which should get you going. i think implicit/inverse function theorem is primarily what you need.

Well, I read your post and think you missed the detail of my problem.

My F(x) is from $ R^n$ to $ R^m$ (and the rank is n-k), your post only provided an answer when m=n-k.


Suppose that i have three equations in three variables in the form of ax + by + cz = 0 and i solve for the common point, that is (x+y+z), can you explain what is its geometric interpretation?

intersection of three planes in 3 space is a point


abstract algebra
group theory
superior algebra
superior algebra unsolved

Suppose $ G\equal{}\{a_1,a_2,\cdots,a_n\}$ is an abelian group of order $ n$ and let $ g\equal{}a_1 a_2 \cdots a_n$. Assume that there is [b]NO[/b] element $ b$ in $ G$ such that $ b\neq e$ but $ b^2\equal{}e$. Prove that $ g\equal{}e$.

Well, since $ b^2 \equal{} e$ implies $ b\equal{}e$, you can write $ G\equal{}\{e, a_1, a_1^{\minus{}1}, \hdots, a_k, a_k^{\minus{}1} \}$ where each element is distinct from it's inverse...

Perhaps more interesting is the following:

Let $ G$ be a finite abelian group.
Prove that the product of all elements in $ G$ equals $ e$, unless there is a unique involution $ i$, in which case the product equals $ i$.


Prove that the following equation has no integer solutions: $x^{5}+31=y^{2}.$

I've got a conceptual proof, so bear with me...

[hide] So basically, we need y^2.  And since it equals x^5+32, that means that it has to be within 32 of x^5, automatically meaning that -17<x<17, since at 17 from 16, squares are more than 32 apart from each other, and thus cannot be reached with the added 32.

Next, negative numbers below and including -2 are eliminated, because -(-2)^5 > 31, and thus y^2 would be negative, and y, imaginary.

-1, 0, 1, and 2 can be ruled out quickly, since 30, 31, 32, and 63 are obviously not squares.  So the new range is 2<x<17.

For these remaining positive values, sqrt(x^5+31)=y, which means that y is between x^5 and x^6 (which, notice, is a square).  However, when you are between 3 and 17 and using the fifth power, things grow extremely fast.  The 31 does not raise the x^5 to x^6, which would be the closest square (this part is probably wrong...), and thus does not work.  x^5 needs to triple, at the very least, from 243 (3^5), and adding 31 won't help much. This pretty much covers everything, except proving how when 2<x<17, the +31 does not land on other squares that are between x^5 and x^6...[/hide]

This is probably riddled with errors, because I am still new to this whole critical thinking stuff, so please correct me if I'm wrong.

[quote="ASquare"][hide] So basically, we need y^2.  And since it equals x^5+32, that means that it has to be within 32 of x^5, automatically meaning that -17<x<17, since at 17 from 16, squares are more than 32 apart from each other, and thus cannot be reached with the added 32.[/hide][/quote]

x^5 isn't a square.

Don't suppose you are iura's friend Armo?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26704

[quote="zscool"]Don't suppose you are iura's friend Armo?

[/quote]
No that's not me. :D  :D


hello, what kind of calculator can you use?
Sonnhard.

[quote="admire9898"][size=200](7^2007+13^10)/(2007^513)[/size]
[size=150]you can use your calculator(no program needed),but no computer allowed.
It's a problem of a math contest which allowed to use calculator
Anybody who can solve this ,just post your method ,thanks![/size]
[b] :) Accuracy by 0．001[/b][/quote]

You can compute first $ \log_{10}(\frac {7^{2007}}{2007^{513}})$ using the calculator ($ 2007\log_{10}(7) \minus{} 513\log_{10}(2007)$). You get $ 1.9049...$ and then calculate $ 10^{this result}$ with the calculator.

You get $ 80.34599...$

Then, you write $ \frac {13^{10}}{2007^{513}} < \frac {(10^2)^{10}}{(10^3)^{513}} < 10^{ \minus{} 1519}$ and so may be ignored with the required accuracy ($ 0.001$)

And the result is $ 80.346$

But this is not an olympiad problem ... :(


AwesomeMath
summer program
probability
videos
MathPath

Between Art of Problem Solving and my own books, I probably spend about 2 to 2.5 hours a day.

Well that's a very vague question.
I could give completely different answers the week before MC state, the week before MC nationals, the middle of the school year, and now.

During the summer it depends on what I'm doing that day, and if I feel like doing something or not.
I'd say about 45 minutes a day as well, if you average all days in the summer.

Less than an hour at all times. :D  It's SO easy.

None at all during the summer (thank god).

[quote="mbabbitt2003"]It's SO easy.[/quote]
Which type of math is so easy? all of them?

Yes. I'm only doing one type, though.  :oops:

Depends on whether I'm stuck on an interesting problem or not... Today, for example, I tried to derive the quartic formula for an hour (but I'm still stuck). Also the AoPS books.

I derived it using Maple! I cheated and it is lost. Please tell me where you are stuck.

Ahhhhhhhhh... First I started with $0=x^4+ax^3+bx^2+cx+d$ and reduced it to $0=y^4+ey^2+fy+g$ by replacing $x=y-\frac a3$. Then I replaced $y=\sqrt[4]{u}-\sqrt[4]{v}$, did some more ugly manipulation, and then replaced $P=\sqrt[4]{u}$ and $Q=\sqrt[4]{v}$ (don't know why). And now I'm stuck with two variables in one equation. :roll:

Don't ask, I was just following a random train of thought.

What's Maple?

[quote="robinhe"][quote="mathgeniuse^ln(x)"]I don't do math at all, except for the practices that I get from my coach and teachers.[/quote]

But doing practices is math ;)

Maybe an hour a day in the summer for me.  Sometimes more, sometimes less.[/quote]

Maybe that is why I said, [size=150]except[/size]

That is like saying:

I got every problem right on the sprint round test, except for 3 or them.

I watch a lot of TV and the only work I do is School Work and if necessary I prepare for Math Competitions  from practices I get from my coaches.

Treething, don't you catch up on sleep at school? I mean, last year I got refreshing naps in my English and Spanish class almost every day  :lol:

Except if I sleep in my class I will automatically be told to wake up, so I don't sleep in class.

How do people sleep in class anyways, won't the teacher notice? I only sleep in Study Hall when I have to. :D

Most people in my classes are taller than me so it works out well for me, especially since the teachers usually sit me in the back because I understand the concepts and whatnot. Plus, the teacher won't notice since I don't snore like someone was in my class once  :lol:

I think all my teachers would notice if someone fell asleep. Except for gym, you could hide in the locker room the entire time and he wouldn't notice, since he just reads his newspaper the whole class. And maybe health class, because we watch a lot of movies  :roll: .

[quote="lotrgreengrapes7926"]I think all my teachers would notice if someone fell asleep. [/quote]
Same here, especially since at my school classes are quite small.

This topic has really digressed...from amount of math in a day to sleeping in class  :roll:

I do about 3 hours a day right now, and yes, on topic please.

1.5 or so hours when there's a class that night (intermediate counting / probability / numbertheory)

I spend 1 hour a day..... for me, just the right amount, because I don't want to start doing it like a routine, just for fun

I spend roughly 1 to 2 hours each day...and the rest of the day on eating, sleeping, video games, and TV. I wish there was no school, only math competitions...*[color=red]s[/color][color=green]i[/color][color=blue]g[/color][color=brown]h[/color]* :D

[quote="Ignite168"][quote="Treething"][quote="SplashD"]That's because he barely sleeps[/quote]

Five hours a day.

(planning to catch up over the summer)[/quote]

You are such a liar, don't act like I don't know you're asleep when I'm trying to message you at 2am and you're not responding.

What did I do today? Instead of going to 2 of my classes, I went to watch the World Cup. During Language Arts, we watched Yo Momma, and during the one class I actually did go to, the teacher put out lots of Sports Illustrated Magazines to read, or you could sign each other's yearbooks, or you could just go to sleep. Productive days :D[/quote]

uh that's cuz it's summer now

Swimsuit Edition?

[quote="Treething"]
Swimsuit Edition?[/quote]

My parents would kill me.

Anyways, I'm about to go to mathpath which is one month of math basically non-stop.

So I do 0-0.5 hours on average everyday.

[quote="robinhe"]

Anyways, I'm about to go to mathpath which is one month of math basically non-stop.

So I do 0-0.5 hours on average everyday.[/quote]

Yeah, I am going to Math Path as well, so my average after MathPath would be around 3 hours a day, opposed to my 0-1 hours a day that i believe I said, even though I hardly do 0-1 hours of day on math.


[quote="cf249"]I am holding a bee in my hand.  What is in my eye?[/quote]

I spent a while considering whether or not this had something to do with "B" and "I"

[hide]Beauty[/hide]

Yes, you're right.

[hide]If you don't know the saying, it's "Beauty is in the eye of the beholder".  Since I am holding a bee, that makes me the "bee-holder", and beauty is in my eye.[/hide]


[quote="Arne"]Use De Moivre's theorem.[/quote]
yes Arne Moivre's theorem it's better then induction

An old trusty induction will do it since

$(x^n+\frac{1}{x^n}) \times (x+\frac{1}{x}) - x^{n-1}+\frac{1}{x^{n-1}} = x^{n+1}+\frac{1}{x^{n+1}}$

[quote="dondigo"]Prove that if $x+\frac{1}{x}=2\cos{\alpha}$ then
$x^n+\frac{1}{x^n}=2\cos{n\alpha}$

$n\in\mathbb{N}$[/quote]

AM-GM gives us

$|x+\frac{1}{x}|\geq 2$

Hence $2|\cos\alpha|\geq 2$ $\implies \cos\alpha =\pm 1$ $\implies \alpha=k\pi$ where $k$ is an integer . And the equality holds when $x=\pm 1$

So when $x=1$ , $\alpha=2k\pi$ and substitute it into the equality we see that it holds . Similarly for $x=-1$ and $\alpha=(2k+1)\pi$ it also holds .

[quote="Arne"]Obviously, $x$ can be a complex number.

[Strange that you didn't notice as your nick is about complex numbers.][/quote]

Guess i forget the complex . Thought that it can be solved without using induction or De Moivre's theorem . Well ... i guess im wrong  :P


function
limit
calculus
calculus computations

[b]20.[/b]  Which of the following statements are true for every function $ f$, defined on the set of all real numbers, such that $ \displaystyle \lim_{x \to 0}{\frac{f(x)}{x}}$ is a real number $ L$ and $ f(0)\equal{}0$?

I.  $ f$ is differentiable at $ 0$.
II.  $ L\equal{}0$
III.  $ \lim_{x \to 0}{f(x)} \equal{} 0$

I eliminated II. rather quickly with the example of $ \frac{sin(x)}{x}$ but I have trouble with the other 2 statements.  Any explanation would be helpful.  Thanks!

For 1) consider the  $ \lim_{x \rightarrow a} \frac {f(x) \minus{} f(a)}{x \minus{} a}$ version of the definition of differentiability at a point $ a$.

For 3) think about limits of the form $ \frac {f(x)}{g(x)}$ where $ g \rightarrow 0$ as $ x \rightarrow 0$, if this limit is to equal $ L$, where does $ f$ have to go?


Give x,y,z are real numbers and we have xyz=1.
Please find the smallest value of A= \sum {a -> b -> c} (a 2 +2b 2 +3ab)/c.

Do you mean a, b and c positive reals and abc=1?
In that case using Cauchy:
a2+2b2+3ab=a2+b2+b2+ab+ab+ab \geq 6a5/6b7/6.
And again using Cauchy we obtain \sum 6a5/6b7/6/c \geq 18 (abc)1/3=18.
And equality holds iff a=b=c=1.

I'm sorry about my mistake and i think you are right.
If i change the problem to :find the smallest value of
A= \sum {a->b->c}(m 2 a 2 +n 2 b 2 +pab)/c.
with ( a,b,c>0,abc=1 and m,n,p are positive real numbers).
I think m,n,p have a condition.

Let x=m2/(m2+n2+p), y=n2/(m2+n2+p) and z=p/(m2+n2+p). We have x+y+z=1.
Then m2a2+n2b2+pab = (m2+n2+p)(xa2+yb2+zab) \geq (m2+n2+p) a2xb2y(ab)z = (m2+n2+p)a2x+zb2y+z.
Applying Cauchy we obtain A \geq 3(m2+n2+p).

A stronger problem :
 \sqrt a ^3 + \sqrt b ^3 + \sqrt c ^3 =3.
Find the smallest value of A=   \sum {a -> b -> c} (a 2 +2*b 2  +3ab)/c.
 :maybe: