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11

Assume that there is some $x_1$ such that $0<x_n<x_{n+1}<1$ then $\{x_n\}$ is monotonically increasing and bounded. Hence converges by monotone convergence theorem. Let $\lim\limits_{n\rightarrow \infty}{x_n}=l$, then obtain $l^2=l \Rightarrow l=0,1$, we can readily omit $0$ and hence $x_n\rightarrow 1$. Proof for uniqueness of $x_1$. Assume ...


6

Hint: try integration by parts. Note that $$ \int \cos(\ln x)\; dx= x\cos(\ln x)+\int \sin(\ln x)\; dx $$ And the same "trick" on the last integral will lead you to the answer. Edit: As said before, integration by parts yields for the second integral $$ \int \sin(\ln x)\; dx= x\sin(\ln x)-\int \cos(\ln x)\; dx $$ So $$ \int \cos(\ln x)\; dx=x\sin(\ln x)+ ...


6

Short answer: it's a lot of practice. Math is like music, it requires tons of practice before you can reach superstar status. The most talented musicians are those who spend thousands of hours practicing their instrument. It's the same for mathematicians. Math and music actually stimulate/rely on similar areas of the brain and I've read that practicing one ...


6

Hint: $$x^3 + bx^2 + cx + d = (x-r_1)(x-r_2)(x-r_3)$$


6

Part 1. Prove that it is impossible to place $10$ tokens. Let's try to build with $10$ tokens. First, note that in each column/row must be at least $1$ token. Lets call column with token in $1$st cell as column $a$; and row with token in $2$nd cell as column $b$. Column $a$ and column $b$ must have at least $2$ tokens. Column $a$ must have cells $1,6$ ...


6

Let $ABCD$ be some tetrahedron with incenter $O$, and let $E$ be the incenter of the tetrahedron $OBCD$. Draw a small sphere around $O$ and select $B'C'D'$ such that this sphere is inscribed in $AB'C'D'$. This is always possible, by selecting three tangent planes through $A$ that enclose the sphere, and intersecting them with a tangent plane behind the ...


5

Case 1 $n=2k$ is even. Then, $a^{2k},b^{2k} \equiv 0,1,4 \pmod{8}$ and therefore both $a,b$ need to be even. Then $a^{2k}, b^{2k}$ are both divisible by $2^{2k}$. Thus $2^{2k}|2008$ which implies that $k=1$. Writing $a=2a', b=2b'$ we have $$a'^2+b'^2=502 \,.$$ Taking this equation modulo 8, there are no solutions. Case 2 $n=2k+1$ is odd. Then ...


5

Let $a,b,c$ be $5/2,5/4,1/4.$ Then there are two solutions of the system giving different values for $x(1-x)$, so that one cannot in general determine it from the equations. The two solutions are $(x,y,z)=(3/2,1/2,0)$ and $(x,y,z)=(1/2,-1/2,0).$ Then note that $x(1-x)$ is $-3/4$ for the first solution, but is $+1/4$ for the second solution. Perhaps the OP ...


4

I am lucky to study with two IMO medalists in the university. I would say that they are almost 50/50 talent & hard work. Their mind generally works a bit faster than mine; nevertheless when discussing problem sets, I notice that they spend almost as much time to solve a problem set as me. But the most important thing that I noticed is that they have ...


4

This is an opinionated question. While true math "prodigies" may exist, I think that hard work and determination outweighs this quality. Also, understand that people don't "practice" to do well in Math Olympiads. People learn and practice math because they enjoy doing it, and through competitions one can evaluate oneself to see how well he has done. True ...


4

This proof is obtained by "working backwards" and picking the most reasonable breadcrumb that can lead us along the backtracked way. As I champion the approach of "working from the front and the back and figuring out what the middle is" as a way of solving problems, let me walk you through how this is done in this scenario. There is a lot to learn, so bear ...


4

Hint: for a given $h$ one has $${f(x + h) - f(x) \over h} = {f(x + h) - f(x + h/2) \over h} + {f(x + h/2) - f(x + h/4) \over h} + ....$$ $$= {1 \over 2}{f(x + h) - f(x + h/2) \over h/2} + {1 \over 4}{f(x + h/2) - f(x + h/4) \over h/4 } + ....$$ You actually need continuity of $f(x)$ already for the above. Now take limits as $h$ goes to zero in the above ...


4

When I worked on this problem back in 2002, showing uniqueness was really easy through the "average of neighbors" observation (albeit on a slanted hexagonal board, instead of the regular chessboard). Proof of uniqueness: Suppose we have 2 solutions $ f(p,q,r)$ and $ g(p,q,r)$. Let $ h(p,q,r) = f(p,q,r) - g(p,q,r)$. Then, we get that $$ 6 h(p,q,r) = ...


3

Achille Hui did the hardest part of the work by discovering the closed formula $\frac{3pqr}{p+q+r}$. The rest is a routine "maximum principle" argument that I explain below. For a positive integer $k$, let $T_k$ be the finite set $\lbrace (p,q,r)\in T | p+q+r=k\rbrace$. For $x=(p,q,r)\in T$, define the neighborhood $N(x)$ of $x$ to be $$ \begin{array}{lcl} ...


3

I will show that 1 is true and therefore 2 is false. Note that since $\Sigma=\{p_2\to p_1, p_3\to p_2,\, \dots\,\}$, we have that $\Sigma\equiv\{\neg p_1\to \neg p_2, \neg p_2\to \neg p_3,\, \dots\,\}$. In particular, we have $\Sigma \equiv \bigcup_{1 \leq n<\omega}\{\neg p_n \to \neg p_{n+1}\}$. (1) Claim: for all $n<\omega$ such that $\Sigma \cup ...


3

The bad and brute-force approach would be like this: You see that $2^{11} > 2008$, so you only have to consider $ 2 \leq n \leq 10$ It will become tedious, but you will probably finish it in about 30-50 min. You can neglect $n = 6$ pretty soon from FLT, since $2008 \equiv 6 \bmod 7$ and $a^6 + b^6 \equiv (0,1) + (0,1) \not\equiv 6 \bmod 7$ Same goes ...


3

Not an answer but a simplification perhaps: Well $$a^n+b^n\equiv 2008\equiv 0 \text{ mod 2}$$ $$\implies a^n\equiv -b^n\equiv b^n \text{ mod 2}$$ $$\implies a\equiv b \text{ mod 2}$$ Thus either both $a$ and $b$ are both odd or they are both even. Now suppose both $a$ and $b$ are even then we can write: $$a=2c \text{ and } b=2d \text{ so that we get:}$$ ...


3

HINT: For $\displaystyle(abcd)_{10}, a+b+c+d=20, a+b=11\iff c+d=9$ $\displaystyle a+b=a+d\iff b=d$ $\displaystyle a=d+3\implies a=b+3, (b+3)+b=11$ Hope you can take it home from here


3

Per light of @Blue's last comment, here's a (very brief) solution: Since $E$ is the midpoint of $E'E''$ and $\angle A=90^\circ$, $\triangle AEE'$ is isosceles at $E$. It follows that $\angle AEE''=2 \angle EAE'$, equivalently $E$ trisects the small arc $AC$. Similarly, $F$ trisects the small arc $AB$ and $D$ trisects the large arc $AC$ (and $AB$). That ...


3

Edit It looks from later comments that you may be interested in log to the base $10$. Whatever base $b$ you are interested in, there is an easily computed constant $a$ such that $\log_b(x)=a\ln x$. So we integrate $\cos(a\ln x)$. We try integration by parts, $u=\cos(a\ln x)$ and $dv=dx$. Then $du=-a\frac{1}{x}\sin(a\ln x)$ and we can take $v=x$. Thus our ...


3

Draw a diagram. For $y\ge0$, the curve $$y(y+1)=(x+1)^2$$ is the upper half of a hyperbola with turning point at $(-1,0)$. One of the asymptotes of this hyperbola is $y=x+\frac{1}{2}$. The inequality $$y(y+1)\le (x+1)^2$$ defines the region below this hyperbola. The hyperbola $$y(y-1)=x^2$$ has turning point $(0,1)$ and is congruent to the first ...


2

If $0 \le y \le 1$ the statement is trivially true, because $y(y-1) \le 0 \le x^2$. So now you need to see what happens when $y > 1$.


2

In your example, we could take $A=\{1,2,3,4\}$ and $B=\{1,2,3,5\}$. Then $$f(\{1,2,3\})=\min\{f(\{1,2,3,4\}),f(\{1,2,3,5\})$$The same holds for all other pairs of sets containing $\{1,2,3\}$. I'll call a function $f$ admissible if $$f(A\cap B)=\min\{f(A), f(B)\}$$ holds for all subset $A,B$ of $\{1,\cdots,n\}$. We will use the shorthand $[n]=\{1,\cdots,n\}$. ...


2

I think you misinterpret repeating pattern. We know that $a_n$ only depends on $a_{n-1}$ and $a_{n-2}$. Now, we start with some given $a_1$ and $a_2$ (modulo $10$, since only the last digit is of interest). There are only $10\cdot 10=100$ possible combinations for a pair $(a_{n},a_{n+1})$. Since the sequence $(a_n)_{n\in \mathbb N}$ is infinite, we know by ...


2

For any integral of the form $\int\frac{dx}{x(1+x^n)}$, it is effective to factor out the $x^n$ in the denominator. Therefore: $\int\frac{dx}{x(1+x^n)}$ $ = \int\frac{dx}{x^{n+1}(1+\frac{1}{x^n})}$ $ = -\int\frac{du}{u}$ with $u = 1+\frac{1}{x^n}$, $du = \frac{-n}{x^{n+1}}{dx}$ $ = -\frac{1}{n}\log({1+\frac{1}{x^n}})+C$ For $n=3$ ...


2

This is not a separate answer, but rather a (too long) comment in response to the proper answer by boywholived. Sure, there exists exactly one such value of $x_1$, but the comment by barak manos still stands: any chance you can tell us what that value of $x_1$ is? Here is a little computer program that does the job: program IMO_1985; procedure find; const ...


2

One solution is the number 7454. The solution can potentially start 29, 38, 47, 56, 65, 74, 83, 92. Since the sum of the first and second digit is the same as the sum of the first and last digit, the possible solutions look like this: 29?9, 38?8, 47?7, 56?6, 65?5, 74?4, 83?3, 92?2. But the constraint that the first digit be 3 more than the last rules out ...


2

Hint: A quadratic function that has its minumum (or extremum) at $x=x_0$ is of the form $y=a(x-x_0)^2 +b$.


2

Just use $P(x)=(x-r_1)(x-r_2)(x-r_3)$. Then compare the coefficients on each side.


2

Wins Alice because she has the possibility of choosing the last digit and is enough to choose something coprime with 10, say 1. Then the remaining number seen as an element of $\mathbb{Z}/10^7\mathbb{Z}$ is an invertible element of that ring. And the group of unit,G, of $\mathbb{Z}/10^7\mathbb{Z}$ has $\phi(10^7)$ elements that is a number coprime with 7. ...



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