# Tag Info

8

This is pretty standard. Since the equation is symmetric in $A,B,C$, it suffices to find all solutions with $A \leq B \leq C$ and then by permutations you get all of them. Then $$ABC = A+B+C \leq C+C+C =3C$$ which implies $AB \leq 3$. Now, since $A \leq B$, there are only three possibilities such that $A B \leq 3$. In each of them $ABC=A+B+C$ becomes a ...

6

The key is the following known observation: If $n=2k+1$ then $$a^{n}+b^n=a^{2k+1}+b^{2k+1}=(a+b)( \mbox{junk} )$$ Thus, for all $n$ odd, $a+b$ divides $a^n+b^n$.

6

Since you have a quartic, there are potentially 4 real solutions. Note that any solution to $x = x^2 - 3x - 2$ is also a solution to the given equation. Hence, $x^2 - 4x - 2$ is a factor of the quartic. Now find the other factor, and solve both quadratics. Alternatively, see this almost 10 year old thread on the Art of Problem Solving Forum, or this ...

5

\begin{align*}x=5:&\;\;\;\;I\;\;\;\;\;\;f(5)+2f\left(\frac15\right)=5\\ x=\frac15:&\;\;\;\;II\;\;\;f\left(\frac15\right)+2f(5)=\frac15\end{align*} Now try $\;I-2\cdot II\;$ .

5

Holder's Inequality gives: $$\left( \frac{a^3}x + \frac{b^3}y + \frac{c^3}z \right)(x + y + z)(1+1+1) \ge (a+b+c)^3$$

5

Here is how I would start: First we set $L := 1370^{685}$ for readability. Two trivial solutions for $y^2=x^3+x+L^2$ are $P_{1,2}=(0,\pm L)$. Using the double formula for elliptic curves given here we get two new points on the curve: $\lambda = \frac{1}{2L} ; x_3=\frac{1}{4L^2} ; y_3=-(L+\frac{1}{8L^3})$ Plugging it in it can be veryfied that ...

4

Suppose there exist $a, b, c, x, y, z > 0$ such that $$a+b+c=4,$$ $$ax+by+cz = xyz,$$ $$x+y+z \leq 4.$$ Observe that we have $$axyz + bxyz + cxyz = 4xyz,$$ so that $$axyz + bxyz + cxyz = 4ax + 4by + 4cz,$$ and hence $$ax(yz-4) + by(xz-4) + cz(xy-4) = 0.$$ Since $x+y+z \leq 4$ and $a, b, c, x, y, z > 0,$ what follow are the inequalities $$xy, yz, xz ... 4 Since 2\cdot 31<63, there are at least three numbers such that a_i\equiv a_j\equiv a_k\pmod{31}. Similarly, there are two numbers such that a_m\equiv a_n\pmod{32}. Of a_i, a_j and a_k, take the ones of same parity. We are done. 3 Note that N=\frac{10^{2008}-1}{9}. One can use the Taylor series of \sqrt{x} at a=10^{2008} to solve this problem. More precisely, let f(a)=\sqrt a,$$f(x)=f(a)+(x-a)f'(a)+\frac{f''(a)}2(x-a)^2+\cdots$$Or$$\sqrt{9N}=10^{1004}-\frac{1}{2\cdot 10^{1004}}-\frac{3}{4\cdot 10^{3\cdot 1004}}+\cdots$$That means$$3\sqrt{N}=10^{1004}-5\cdot ...

3

This is a well-known fact in the theory of Farey series. You have $aq>bp$ and $pd>qc$, so the integers $\delta_1=aq-bp$ and $\delta_2=dp-cq$ are both $\geq 1$. Now $$d\delta_1+b\delta_2=(ad-bc)q=q$$ It follows that $q\geq b+d$.

3

The statement we are to prove is not that we can partition the triangle into $2n + 1$ triangles using the $n + 3$ points as described. Rather, it says that if we do partition the triangle using the $n + 3$ points as described, the result will always be $2n + 1$ triangles. The argument fails because there are partitions of the triangle that can be done using ...

3

I assume that we are talking about real values, so that we may suppose $a_n\ge0$. And as you have already disposed of $a_n>1$, we may assume that $0\le a_n\le1$. So, let your $v_n$ be $\cos^2\theta_n$, that is, $$a_n=\cos^{4028}\theta_n\ .$$ Then $$\cos^2\theta_{n+1}=(2\cos^2\theta_n-1)^2=\cos^2(2\theta_n)\ .$$ The sequence will be periodic if and only ...

3

The answer is 1. The result is known as Goldbach-Euler theorem. See Wikipedia entry for "proof". For rigorous proof, you could consider sum of reciprocals of all perfect powers, $S$. Note that sum equals $$S = \sum_{x \in B}\sum_{n = 2}^{\infty}\frac{1}{x^{n}} = ... 3 Macavity's way is certainly the most elegant. Here is a way to do it just using Cauchy-Schwarz and AM-GM. Use Cauchy-Schwarz inequality to get:$$(x+y+z)\left(\frac{a^3}{x}+\frac{b^3}{y}+\frac{c^3}{z}\right)\geq (a^{3/2}+b^{3/2}+c^{3/2})^2$$So to prove your inequality, it suffices to show that$$ (a^{3/2}+b^{3/2}+c^{3/2})^2 \geq \frac{(a+b+c)^3}{3}$$... 2 Your argument is wrong in several places. There are (n-k+1)^2 possible k\times k squares that need to be considered. If you only consider a few of them, say a nonoverlapping subset, you will only consider \lfloor \frac nk\rfloor^2 such squares. Then indeed, if \lfloor \frac nk\rfloor^2>n, you can be sure that there exists a k\times k that ... 2 a+b\mid a^k+b^k when k is odd. 2 {\bf Hint}\ \ n-m\mid n^k-m^k.\  So \,\ m=-(n\!+\!1),\,\ k\, odd {\rm yields}\ \ 2n\!+\!1\mid n^k\!+(n\!+\!1)^k.\,\  Yours is case \,k=n. Remark \  The above divisibility can be viewed as a special case of the Factor Theorem \, x-y\,\ \mid\,\ f(x)-f(y)\  for any polynomial \,f\, or, by the Polynomial Congruence Theorem \,x\equiv ... 2 In the second solution, when = in \ge is satisfied, a=b=c. But then, abc=17 and a+b=c makes the contradiction. 2 The official solution is wrong. The equality happens when$$\frac{a^2}{17}=\frac1{2a}$$or a^3=\frac{17}2<a^3. Note that in your solution, the equality does not happen either. Edit: I would solve the problem as follows. Let a=x-y,b=y-z then ab(a+b)=17 and we need to minimize ... 2 Write your equation as$$B = \dfrac{AC+1}{C-A}$$Let C = A + x, so this becomes$$ B =A + \dfrac{A^2+1}{x}$$Thus x must divide A^2+1, and B + C = 2A + x + \dfrac{A^2+1}{x}. You'll want a divisor of A^2+1 that is closest to \sqrt{A^2+1} (on one side or the other). 2 Since \dfrac{1}{i} \le \dfrac{1}{p} for p \le i \le q, and 0 < q-(p-1) < q+\frac{1}{2}, and p > p - \frac{1}{2} > 0, we have \displaystyle\sum_{i = p}^{q}\dfrac{1}{i} \le \sum_{i = p}^{q}\dfrac{1}{p} = \dfrac{q-(p-1)}{p} < \dfrac{q+\frac{1}{2}}{p-\frac{1}{2}}, as desired. 2 A sequence exists. Define a_1 = 1, and  a_n recursively as$$ a_n = 1 + 2 \sum_{i=1}^{n-1} i a_i. $$For example, we get  a_2 = 1 + 2 \times (1\times 1) = 3, a_3 = 1 + 2 \times ( 1 \times 1 + 2\times 3) = 15, a_4 = 1 + 2 \times ( 1 \times 1 + 2 \times 3 + 3 \times 15 ) = 105. For the first few terms of m, we have \begin{array}{llll} 1 & ... 1 Given a row r, define the pattern of the row to be the set of indices i such that the ith number in row r is different from the ith number in the first row. Clearly, the first row has pattern \emptyset, and if two rows have different patterns, they must be different. We claim that, given your constraints, if p is a pattern with an even number ... 1 Maybe we need to take a few steps back ( though I imagine this has all been covered in your course). If a is any real number other than 1 and r is a positive integer, then 1 + a + a^{2} +\ldots +a^{r-1} = \frac{a^{r}-1}{a-1}. If you need to verify that, multiply the left side by a-1 and notice that there is a lot of cancellation. If b is another ... 1 I hope that the solution is clear from the picture: \angle PAD=\angle PMB because P lies on the circumcircle of \triangle ACM, similar for \angle ADP=\angle PBM AD=BM(=AM) as \angle A=90^\circ. Therefore \triangle APD and \triangle MPB are congruent. 1 After giving this more thought, I see that the answer is in fact 2, and that this is the correct answer for any right triangle. The diagram provided by Quang Hoang is very well done, and Quang's observation about the congruent triangles is insightful, but I don't see how it settles the issue. Here is my most recent response to the original question: ... 1 Maybe if we write N=\lfloor 10^{2008}\cdot 0.\overline{111}\rfloor=\dfrac{10^{2008}-1}{9}. Then we must have$$ \sqrt{N}\approx\sqrt{N+\tfrac19}=\sqrt{\frac{10^{2008}}{9}}=\frac{10^{1004}}3=10^{1004}\cdot 0.\overline{333}  and with the derivative of the square root function $f(x)=\sqrt x$ being $f'(x)=\frac{1}{2\sqrt x}$ the difference between $\sqrt{N}$ ...

1

$y$ and $z$ are $10$ and $15$, because from $9$, you can move to $100$ via $10$ in $1+T(10)$ steps or via $15$ in $1+T(15)$ steps. Taking the minimum gives $T(9)=1+\min(T(10),T(15))$. Now, we have $yz=10\cdot 15=150$. EDIT: A problem like this is usually solved using Dynamic Programming. In this case, we calculate all $T(i)$ starting at $T(100)$ and going ...

1

As @Michael says, you can count the diagonals in one direction, and note that you can't have both ends of a long diagonal, which then reduces the number by $1$ As for configurations, note that the bishops on black squares and those on white squares are completely independent of each other. So, for example, the black-squared bishops could be on the first and ...

1

You can "unofficially" (well, nobody really keeps track) participate in (mostly short answer, but some proof-based) student-run contests like the NIMO and OMO (see http://internetolympiad.org/), or the (proof-based) Olympiad-style ELMO (see this year's AoPS-version of the contest). Of course, AoPS also has a good repository of past contest problems; in ...

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