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2

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 ...


2

Let $a_1, a_2, a_3, \dots, a_{77}$ be the number of games played in the $nth$ day, such that $a_i < 55 = 132 - 77, i\in [1, 77]$. Also consider the sequence $a_1 + 21, a_2 + 21, \dots ,a_{77} + 21$. The 154 sequences $a_1, a_2, a_3, \dots, a_{77}$ and $a_1 + 21, a_2 + 21, \dots ,a_{77} + 21$ are all less than $76 = 55 + 21$. Thus by pigeon hole ...


0

To distribute 6 marbles in 6 bags we just have to line up the 6 balls and then place 5 markers to indicate the possible distribution. For example using o for marble and | for a bag marker the sequence oooooo can be marked as oo||o|o|oo| indicating 2 in the first bag, none in the second bag, 1 in the third bag, 1 in the fourth bag, 2 in the fifth bag and none ...


0

Coordinatize the triangle on the unit circle via $$A = (\cos\alpha, \sin\alpha)\qquad B = (-1,0) \qquad C = (1,0)$$ Let $T$ be a point of tangency representing generically one of $D$, $E$, $F$.$$T = (\cos\theta,\sin\theta)$$ Define $T_B$ and $T_C$ as points on the tangent line, at distance $t$ from $T$; we can express these as $T \pm t\;T^\perp$, where ...


1

Given, $abc=P$, $a+b+c=S$. Hence, $ab=P/c$ and $a+b=S-c$ or $x^2-(S-c)x+P/c=0$, but the determinant must be a square, so, $(S-c)^2-4P/c=r^2$, for some integer $r$. Let $P/c=k^2$ so that $(S-c)^2=(2k)^2+r^2$ a pythagora's equation with $S-c=l^2+m^2$, $k=lm$ and $r=l^2-m^2$. So the solution of the quadratic, which solves for $a$ and $b$, is $a=l^2$, $b=m^2$, ...


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. ...


2

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


6

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


0

Your problem is a bit unclear to me. In particular the part “Similarly for $E$ on arc $AC$ and $F$ on arc $AB$” could be interpreted to mean that for $E$ we'd intersect with $BA$ and $BC$ to obtain $E'$ and $E''$, or to intersect again with $AB$ and $AC$. The former was my first interpretation of this question, but in that case the claim is false. The latter ...


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 ...


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

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 ...


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) = ...


1

Well, it has at least two prime factors (counting multiplicity) because it's not prime. If they're distinct, you're done, if not then you have the square of a prime, but it's not a square, so you must either have a) another prime factor not equal to the square, in which case that number times one copy of the prime would produce a smaller number ...


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} ...


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 ...


0

According to your definition, you do not even require $\alpha, \beta, \gamma, \text{or } \delta$ to be zeroes. You merely require some interpolating function, not even necessarily polynomial in nature. In the most general sense, you first need to define two things. What space should the function be in? It will, at the very least, need to be a superset ...


1

Let us write the condition as $$ f(x_k) = f_k. $$ Consider the function $$ p_k(x) = \prod_{\ell \ne k} \frac{x-x_\ell}{x_k-x_\ell}. $$ Note that $$ p_k(x_m) = \prod_{k\ne\ell} \frac{x_m-x_\ell}{x_k-x_\ell} = \delta_{km}. $$ So we define $$ f(x) = \sum_k f_k p_k(x), $$ and it is clear that $$ f(x_m) = \sum_k f_k p_k(x_m) = \sum_k f_k ...


0

First, a general fact is that if you want a polynomial in which $a, b, c, $and $d$ are the only zeroes, then it will look like $$P(x)=(x-a)^{n_a}(x-b)^{n_b}(x-c)^{n_c}(x-d)^{n_d}$$ where $n_a, n_b, n_c, n_d$ are positive integers. Second, if you have a function $f$ that satisfies the properties you stated above, note that the function $f+P$ will also ...


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 $n=3$ ...


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 $1,6$ ...


0

I know it isn't using the Euler characteristic, but I couldn't help it. Consider the following 'proof by picture': Start by rounding everything out, so that your original picture looks like a ball with holes drilled out of it. Then just follow the pictures. This is an ambient isotopy of the manifold, so the genus is preserved, and it follows that the ...


0

I believe that you should first break up those faces a bit more, to make them polygonal. That mean, add an edge between any vertex on an inner square to the outer square in a given plane. There are 3 big faces, plus 12 big faces that have been cut in 4, plus 4 faces from the very small rod, 4 from the medium rod, 7 medium faces in the inner cube and 12 more ...


0

I, too, have a hard time with Olympiad/Putnam (like a college version of Olmpiad math) style questions, but have gotten much better with time. The first thing I would suggest is to be patient - problem solving is hard and is about improving your fundamental ability to think. In particular I would suggest you adopt the following "strategies" that I have ...


0

As someone who's also striving for the IMO: Learning how to google will help you a lot. Despite what's usually advertised, math contests' syllabi can be very extensive so you'll often find yourself googling some new technique you heard about in a solution. Also, the following two books are almost unanimously agreed to be must-reads: "The Art and Craft of ...


1

What I'd personally recommend AoPS book series followed by Polynomials by E.J. Barbeau (I loved this book!) and then practice AMC 8-10-12 exams, AIME exams, USAJMO exams, USAMO exams, MOP/IMO exams as well as other math contests such as national qualifying tests for other countries and ARML. Brilliant and AOPS Alcumus are also excellent places to warm ...


1

Define an increasing sequence $s_0,...,s_n\in \{1,...,n\}$ as follows. $s_0 = 1$ and $s_k$ is equal to the number of distinct rows in the $n\times k$ matrix consisting of the first $k$ columns of $M$. It is evident that this is an increasing sequence and if $s_{k-1} = s_k$, then we can delete the $k$-th column without any rows becoming equal. But $s_k ...


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 $y > 1$.


0

Just a first step, i.e. a way to inscribe a rhombus in a convex quadrilateral $ABCD$. Fix the direction of the diagonals of the rhombus, let them be parallel to lines $r$ and $s$, with $r\perp s$. Consider the segments parallel to $r$ having their extrema on $AB$ and $CD$: the center of the rhombus must belong to the locus of midpoints of such segments, ...


0

EDIT: the full algorithm: The full algorithm can be summarized as follow: First check if $N<p$. If it is, then the answer is $N$. Otherwise proceed to next step. Calculate $M=\lfloor\frac{N}{p}\rfloor$. As explained in the explanation part, now we need to find the number of $k$ such that $kp$ is a valid $n$, and that $0\leq k\leq M$. This is equivalent ...


3

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 ...


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 ...


1

Through integration by parts, $$\int \cos(\ln(x))dx $$ Let $u = \cos(\ln(x))$ and $dv = dx$, thus, $$du = -\frac{\sin(ln(x))}{x}, v = x$$ Hence, $$ \int \cos(\ln(x))dx = u \times v - \int du \times v\\ \int \cos(\ln(x))dx = \cos(\ln(x)) \times x + \int \sin(ln(x))dx $$ Since we have another awkward integral on the rhs, integrate it again, this time we ...


1

Let $u = \ln x$, then $x = e^u$, and $dx = e^udu$. Thus $I = \int e^u\cdot \cos udu$. This integral is popular in calculus textbook and you can find an answer.


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)+ ...


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 ...


2

Let $M$ be the midpoint of $CH$. Furthermore, let $P$ be a point on (the extension of) $CH$ such that $SP$ is perpendicular to $CH$. Because $\angle CHS - \angle CSB = 90^{\circ}$, it is straightforward to show that the triangles $\triangle SBC$ and $\triangle SPH$ are similar. Consider spiral homotheties $$B(90^\circ \text{counter ...


1

Construct a graph $G$ with a vertex corresponding to each city, and an edge between each pair of cities for which a flight between the cities exists. Consider any vertex $v \in V(G)$. Let $d_4(v)$ be the number of vertices of distance $4$ from $v$ in $G$. Let $\{v_1, v_2, \dots , v_p\}$ be the set of vertices of distance $1$ from $v$ in $G$. Now for each ...


1

I propose that the question has a typo and that the inequalities in the index of the summation cannot be strictly less than and need to be less than or equal to. Here is why. Consider the case where $n=2$ and $r=3$. Clearly, the left hand side is an application of binomial theorem and we have $a_1^3+3a_1^2a_2+3a_1a_2^2+a_2^3$ The right hand side is (from ...


1

The first two problems ask for Steiner Triple Systems. It is known that a Steiner Triple System exists if and only if $n \equiv 1 \text{ or } 3 \pmod 6$. For $K_7$ we have $7 \equiv 1 \pmod 6$, so one exists. An example is drawn on the Wikipedia page (with triangles are drawn as lines): (see also Fano plane). For $K_{12}$ we have $12 \equiv 0 \pmod ...


1

first of all you should find $A$, $B$ and $C$ with initial $y(-2)=41$, $y(5)=20$ and quadratic function is minimum at $x_{min}=-\frac{-B}{2A}=\frac{B}{2A}$. so you have three equations and three : $41=4A+2B+C$ $20=25A-5B+C$ $2=\frac{B}{2A}$ Solve this for $A$, $B$, $C$ and then you can find $D=y_{min}=y(x_{min})$ or ...


2

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


0

Firs off: notice the race lasts 40 minutes. Note Dexter and Prexter are only simultaneously at the edge of the pool at minute 26. Since Dexter is at the edge after an even amount of minutes and Prexter is at the edge after a multiple of $3.25$ minutes, the only even multiple of $3.25$ under $40$ is $26$. So they are only at the edge at the same time at ...


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 ...


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


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 ...


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 ...


9

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 ...



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