id int64 41 3.22M | problem stringlengths 38 4.44k | solution stringlengths 8 38.6k | problem_vector listlengths 1.02k 1.02k | solution_vector listlengths 1.02k 1.02k | last_modified stringdate 2025-08-11 00:00:00 2025-08-11 00:00:00 |
|---|---|---|---|---|---|
599,349 | Let $P$ be a point in the interior of an acute triangle $ABC$, and let $Q$ be its isogonal conjugate. Denote by $\omega_P$ and $\omega_Q$ the circumcircles of triangles $BPC$ and $BQC$, respectively. Suppose the circle with diameter $\overline{AP}$ intersects $\omega_P$ again at $M$, and line $AM$ intersects $\omega_... | [b][color=red]Claim:[/color][/b] $M$ and $N$ are isogonal conjugates.
[i]Proof.[/i] Redefine $N$ as the isogonal conjugate of $M.$ It suffices to prove $N=(BQC)\cap(AQ).$ First, we claim $CBQN$ is cyclic. Indeed, $$\measuredangle BPC+\measuredangle BQC=\measuredangle BAC+180=\measuredangle BMC+\measuredangle BNC$$ and ... | [
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599,350 | Let $ABCD$ be a cyclic quadrilateral with center $O$.
Suppose the circumcircles of triangles $AOB$ and $COD$ meet again at $G$, while the circumcircles of triangles $AOD$ and $BOC$ meet again at $H$.
Let $\omega_1$ denote the circle passing through $G$ as well as the feet of the perpendiculars from $G$ to $AB$ and $CD$... | First, let $ E = AB \cap CD $ and $ F = BC \cap DA $ and denote the circumcircle of $ ABCD $ by $ \omega $. Let $ O_1, O_2 $ be the centers of $ \omega_1, \omega_2 $ respectively. Let $ M $ be the midpoint of $ GH $.
Consider the inversion about $ \omega $. It is clear that line $ AB $ goes to $ \omega_1 $ and t... | [
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599,351 | In triangle $ABC$ with incenter $I$ and circumcenter $O$, let $A',B',C'$ be the points of tangency of its circumcircle with its $A,B,C$-mixtilinear circles, respectively. Let $\omega_A$ be the circle through $A'$ that is tangent to $AI$ at $I$, and define $\omega_B, \omega_C$ similarly. Prove that $\omega_A,\omega_B,\... | I will fill in some intermediate steps in XmL's post.
Lemma 1: If $ X $, $ Y $ are the tangency points of the $ A $-mixtilinear incircle with $ AB, AC $ respectively then $ I $ is the midpoint of $ XY $.
[i]Proof:[/i] This is a well-known result, and moreover can be proved immediately with a limiting case of Saw... | [
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599,353 | We are given triangles $ABC$ and $DEF$ such that $D\in BC, E\in CA, F\in AB$, $AD\perp EF, BE\perp FD, CF\perp DE$. Let the circumcenter of $DEF$ be $O$, and let the circumcircle of $DEF$ intersect $BC,CA,AB$ again at $R,S,T$ respectively. Prove that the perpendiculars to $BC,CA,AB$ through $D,E,F$ respectively inters... | My solution:
Since the lines through $ A, B, C $ and perpendicular to $ EF, FD, DE $ are concurrent at the orthocenter of $\triangle DEF $ ,
so $ \triangle DEF $ and $ \triangle ABC $ are orthologic $ \Longrightarrow $ the perpendiculars through $ D, E, F $ to $ BC, CA, AB $ are concurrent at $ X $ .
Since $ AD, BE,... | [
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599,355 | Define the Fibanocci sequence recursively by $F_1=1$, $F_2=1$ and $F_{i+2} = F_i + F_{i+1}$ for all $i$. Prove that for all integers $b,c>1$, there exists an integer $n$ such that the sum of the digits of $F_n$ when written in base $b$ is greater than $c$.
[i]Proposed by Ryan Alweiss[/i] | Pretty easy. It is [i]well known[/i] that $F_i$ is strictly periodic mod $n$, to say that there exists large $N$ where
$F_N \equiv 0$ and $F_{N+1} \equiv 1$ (mod $n$). Thus, $F_{N-1} \equiv 1$ (mod $n$), implying that $F_{N-2} \equiv -1$ (mod $n$). Taking $n=b^k$ for arbitrarily large $k$ implies that we can find some ... | [
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599,358 | Let $d$ be a positive integer and let $\varepsilon$ be any positive real. Prove that for all sufficiently large primes $p$ with $\gcd(p-1,d) \neq 1$, there exists an positive integer less than $p^r$ which is not a $d$th power modulo $p$, where $r$ is defined by \[ \log r = \varepsilon - \frac{1}{\gcd(d,p-1)}. \][i]Prop... | I'm not sure whether my solution is correct or not, but I'll write it down anyway.
Let $a$ be the least non-$d$th power modulo $p$. Let $b$ be the least integer such that $ab > p$ and let $ab=p+k$. Since $p$ is a prime, obviously $0 < k < a$. Hence $ab \equiv k$ is a $d$th power and $b$ is a non-$d$th power since $a$... | [
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599,362 | Given positive reals $a,b,c,p,q$ satisfying $abc=1$ and $p \geq q$, prove that \[ p \left(a^2+b^2+c^2\right) + q\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \geq (p+q) (a+b+c). \][i]Proposed by AJ Dennis[/i] | Here is a slightly different solution: By AM-GM, we have $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge 3.$ Therefore, by AM-GM again, we have \[a^2 + b^2 + c^2 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge \left(a^2 + 1\right) + \left(b^2 + 1\right) + \left(c^2 + 1\right) \ge 2(a + b + c).\] In addition, since $1 + 1 + ... | [
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599,363 | Let $a$, $b$, $c$ be positive reals. Prove that \[ \sqrt{\frac{a^2(bc+a^2)}{b^2+c^2}}+\sqrt{\frac{b^2(ca+b^2)}{c^2+a^2}}+\sqrt{\frac{c^2(ab+c^2)}{a^2+b^2}}\ge a+b+c. \][i]Proposed by Robin Park[/i] | Another proof with expansion...
By Holder, \[(\text{LHS})^2\left(\sum a(b^2+c^2)\right)\left(\sum \frac{a}{a^2+bc}\right)\ge (a+b+c)^4.\]It suffices to show that \[\left(\sum_{\text{sym}} a^2b\right)\left(\sum a(b^2+ca)(c^2+ab)\right)\le (a+b+c)^2(a^2+bc)(b^2+ca)(c^2+ab).\]
[hide="The expansion"]We can actually do th... | [
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599,366 | Let $ABC$ be a triangle with circumcenter $O$. Let $P$ be a point inside $ABC$, so let the points $D, E, F$ be on $BC, AC, AB$ respectively so that the Miquel point of $DEF$ with respect to $ABC$ is $P$. Let the reflections of $D, E, F$ over the midpoints of the sides that they lie on be $R, S, T$. Let the Miquel point... | [b]Generalization:[/b]
Let $ D, E, F $ be the point on $ BC, CA, AB $ , respectively .
Let $ K $ be a point and $ X, Y, Z $ be the projection of $ K $ on $ BC, CA, AB $, respectively .
Let $ R, S, T $ be the reflection of $ D, E, F $ in $ X, Y, Z $ , respectively .
Let $ P $ be the Miquel point of $ \{ D, E, F \} $ an... | [
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599,368 | Let $t$ and $n$ be fixed integers each at least $2$. Find the largest positive integer $m$ for which there exists a polynomial $P$, of degree $n$ and with rational coefficients, such that the following property holds: exactly one of \[ \frac{P(k)}{t^k} \text{ and } \frac{P(k)}{t^{k+1}} \] is an integer for each $k = 0,... | We claim that $m=n$ is the desired maximum value. Note that the required condition is just a cute way of saying $t^k \mid P(k)$ and $t^{k+1} \nmid P(k)$ for $k=0, 1, \dots, m$. For construction, take $P(j)=t^j$ for $j=0, 1, \dots, n$ and apply Lagrange Interpolation to get $P$ with rational coefficients.
For the estim... | [
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599,369 | Show that the numerator of \[ \frac{2^{p-1}}{p+1} - \left(\sum_{k = 0}^{p-1}\frac{\binom{p-1}{k}}{(1-kp)^2}\right) \] is a multiple of $p^3$ for any odd prime $p$.
[i]Proposed by Yang Liu[/i] | Nice! Here's my solution: We will use congruences on rationals, i.e. for $a,b \in \mathbb{Z}_p,$ we write $a \equiv b \pmod{p^e}$ iff $p^e$ divides the numerator of $a-b$ when written in simplest terms. Note that, by Binomial Theorem, we get $$\sum_{k=0}^{p-1} \frac{\binom{p-1}{k}}{(1-kp)^2}=\sum_{k=0}^{p-1} \binom{p-1... | [
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599,371 | Let $a,b,c,d,e,f$ be positive real numbers. Given that $def+de+ef+fd=4$, show that \[ ((a+b)de+(b+c)ef+(c+a)fd)^2 \geq\ 12(abde+bcef+cafd). \][i]Proposed by Allen Liu[/i] | I have not yet solved this problem but it may be helpful to realize that we can perform the following parametrization: $ d = \frac{2x}{y + z}, e = \frac{2y}{x + z}, f = \frac{2z}{x + y} $ for some positive real numbers $ x, y, z $ | [
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599,373 | Does there exist a strictly increasing infinite sequence of perfect squares $a_1, a_2, a_3, ...$ such that for all $k\in \mathbb{Z}^+$ we have that $13^k | a_k+1$?
[i]Proposed by Jesse Zhang[/i] | Yes. Note that all we have to do is show that for all positive integers $k$, there exists an integer $x$ such that $13^k | x^2+1$. Prove this using induction. The base case is easy (take $x=5$). If for some $k=l$ we have $13^l | x^2+1$, let $y=x+13^l \cdot z$. Then $13^{l+1} | y^2 + 1 \iff 13^{l+1} | x^2+1+2xz13^l$. If... | [
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599,374 | Find all positive integer bases $b \ge 9$ so that the number
\[ \frac{{\overbrace{11 \cdots 1}^{n-1 \ 1's}0\overbrace{77 \cdots 7}^{n-1\ 7's}8\overbrace{11 \cdots 1}^{n \ 1's}}_b}{3} \]
is a perfect cube in base 10 for all sufficiently large positive integers $n$.
[i]Proposed by Yang Liu[/i] | The number in question when evaluated turns out to be\[ \frac{b^{3n}-b^{2n+1}+7b^{2n}+b^{n+1}-7b^n-1}{3(b-1)}=\frac{(b^n-1)(b^{2n}+b^n(8-b)+1)}{3(b-1)}\]Let the sequence of these numbers be $x_n^3$. We shall show that $b-1$ is a power of $3$. Suppose that this is not the case. Then, $\exists p \in \mathbb{P}$ such that... | [
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-0.00... | 2025-08-11 |
599,376 | Let $n$ be a positive integer. For any $k$, denote by $a_k$ the number of permutations of $\{1,2,\dots,n\}$ with exactly $k$ disjoint cycles. (For example, if $n=3$ then $a_2=3$ since $(1)(23)$, $(2)(31)$, $(3)(12)$ are the only such permutations.) Evaluate
\[ a_n n^n + a_{n-1} n^{n-1} + \dots + a_1 n. \][i]Proposed by... | These $ a_k $ are called Stirling numbers of the first kind. It is not hard to check that they satisfy the following generating function:
\[ \sum_{k = 1}^{n}a_{k}x^k = x(x + 1)(x + 2) \dots (x + n - 1) \]
Letting $ x = n $ in the expression we immediately find that the desired sum is $ \frac{(2n - 1)!}{(n - 1)!} $ | [
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599,379 | $ABCD$ is a cyclic quadrilateral inscribed in the circle $\omega$. Let $AB \cap CD = E$, $AD \cap BC = F$. Let $\omega_1, \omega_2$ be the circumcircles of $AEF, CEF$, respectively. Let $\omega \cap \omega_1 = G$, $\omega \cap \omega_2 = H$. Show that $AC, BD, GH$ are concurrent.
[i]Proposed by Yang Liu[/i] | Nice and simple problem.By the radical axes concurrence theorem we have $AG,CH,EF$ are concurrent at $X$(let).It is well known that the polar of $J$ is $EF$ where $J=AC \cap BD$.From this it follows that the polar of $X$ passes through $J$.Let $AC \cap GH=J'$.Then the polar of $X$ passes through $J'$.Now if $AC$ is not... | [
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1,908,062 | Determine the value of the expression $x^2 + y^2 + z^2$,
if $x + y + z = 13$ , $xyz= 72$ and $\frac1x + \frac1y + \frac1z = \frac34$. | $$\sum_{cyc}\frac{1}{x}=\frac{\sum_{cyc}xy}{xyz}=\frac{\sum_{cyc}xy}{72}=\frac{3}{4}\text{ therefore }\sum_{cyc}xy=54$$
$$\text{ Furthermore }\sum_{cyc}x^2=\left(\sum_{cyc}x\right)^2-2\sum_{cyc}xy=13^2-2\cdot 54=61$$
$$\blacksquare.$$ | [
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1,908,068 | The radius $r$ of a circle with center at the origin is an odd integer.
There is a point ($p^m, q^n$) on the circle, with $p,q$ prime numbers and $m,n$ positive integers.
Determine $r$. | $p^{2m}+q^{2n}=r^2$
$\implies p\ne q$
$\implies p^m=a^2-b^2;q^n=2ab;r=a^2+b^2$ and $(a,b)=1$
$\implies q=2 \implies q^n=2^n=2ab \implies a=2^{n-1},b=1$
$\implies 3|p^m=2^{2(n-1)}-1 \implies p=3$
$\implies 3^m=2^{2(n-1)}-1$
$\implies m=1,n=2$
$\implies r=5$ | [
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615,460 | Let $a_1 \leq a_2 \leq \cdots$ be a non-decreasing sequence of positive integers. A positive integer $n$ is called [i]good[/i] if there is an index $i$ such that $n=\dfrac{i}{a_i}$.
Prove that if $2013$ is [i]good[/i], then so is $20$. | Weird that there is no reply yet, since this problem is rather simple.
[hide="Solution"]If $2013$ is [i]good[/i], then there is an $i$ such that
\[ 2013=\frac{i}{a_i} \Leftrightarrow 2013a_i=i.\]
Thus $2013$ divides $i$.
We'll proceed by contradiction. Assume that $2013$ is [i]good[/i] and $20$ isn't. Then let $i=2013... | [
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615,467 | In Sikinia we only pay with coins that have a value of either $11$ or $12$ Kulotnik. In a burglary in one of Sikinia's banks, $11$ bandits cracked the safe and could get away with $5940$ Kulotnik. They tried to split up the money equally - so that everyone gets the same amount - but it just doesn't worked. After a whil... | Let there were $m$ $11$-valued and $n$ $12$-valued coins in someone's share. On equal sharing, each bandit would get $540$ Kulotnik. We have to prove this does not happen. Consider $11m+12n=540$.
So $m$ must be a multiple of $12$, let $m=12s$. Then $11s+n=45$. The possible solutions to $(m,n)$ are $(0,45),(12,34),(24,2... | [
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1,830,290 | Find all the pairs of real numbers $(x,y)$ that are solutions of the system:
$(x^{2}+y^{2})^{2}-xy(x+y)^{2}=19 $
$| x - y | = 1$ | $$| x - y | = 1$$
$$\implies x^2 + y^2 = 1 + 2xy$$
$$(x-y)^2 = (x+y)^2 - 4xy$$
$$\implies (x+y)^2 = 1 + 4xy$$
$$\text{Let} \quad xy = a$$
$$(x^{2}+y^{2})^{2}-xy(x+y)^{2}=19$$
$$\implies (1 + 2xy)^2 - xy(1 + 4xy) = 19$$
$$\implies (2a + 1)^2 - a(4a + 1) = 19$$
$$\implies a = 6$$
Wlog $x > y$
So we have $xy = 6$ and $x-y... | [
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1,830,291 | Givan the set $S = \{1,2,3,....,n\}$. We want to partition the set $S$ into three subsets $A,B,C$ disjoint (to each other) with $A\cup B\cup C=S$ , such that the sums of their elements $S_{A} S_{B} S_{C}$ to be equal .Examine if this is possible when:
a) $n=2014$
b) $n=2015 $
c) $n=2018$ | We need ${{n(n+1)}\over 2}\equiv 0(3)\iff n\equiv 0,2(3)$ which shows that it is impossible for $n=2014$.
If it's possible for $n$ then also for $n+9$ since we can do $$A\rightarrow A\cup \{n+9,n+4,n+2\},B\rightarrow B\cup \{n+8,n+6,n+1\}, C\rightarrow C\cup \{n+7,n+5,n+3\}$$
For $n=8$ we have $A=\{8,4\},B=\{7,5\},C=\... | [
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577,559 | Find all the polynomials with real coefficients which satisfy $ (x^2-6x+8)P(x)=(x^2+2x)P(x-2)$ for all $x\in \mathbb{R}$. | The relation $ (x^2-6x+8)P(x)=(x^2+2x)P(x-2)$ writes $ (x-2)(x-4)P(x)=x(x+2)P(x-2)$.
For $x=2$ we thus have $P(0) = P(2-2)=0$; for $x=4$ we have $P(2) = P(4-2)=0$; and for $x=-2$ we have $P(-2) =0$. Therefore $P(x) = x(x-2)(x+2)Q(x)$. Plugging in and cancelling equal factors, we are left with $ (x-2)Q(x)=xQ(x-2)$.
Fo... | [
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594,859 | Let $(x_{n}) \ n\geq 1$ be a sequence of real numbers with $x_{1}=1$ satisfying $2x_{n+1}=3x_{n}+\sqrt{5x_{n}^{2}-4}$
a) Prove that the sequence consists only of natural numbers.
b) Check if there are terms of the sequence divisible by $2011$. | [hide="Part (a)"]
We have $2x_2=3x_1+\sqrt{5x_1^2-4}=4$, so $x_2=2$. Now, for all $n\geq 1$, we have $2x_{n+1}=3x_{n}+\sqrt{5x_n^2-4}> 2x_n\implies x_{n+1}>x_n$. Also, $2x_{n+1}=3x_{n}+\sqrt{5x_n^2-4}$ is equivalent to
\[\begin{aligned}
(2x_{n+1}-3x_n)^2-5x_n^2=-4\implies x_{n+1}^2-3x_{n+1}x_n+x_n^2=-1.
\end{aligned}\... | [
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602,483 | Square $ABCD$ is divided into $n^2$ equal small squares by lines parallel to its sides.A spider starts from $A$ and moving only rightward or upwards,tries to reach $C$.Every "movement" of the spider consists of $k$ steps rightward and $m$ steps upwards or $m$ steps rightward and $k$ steps upwards(it can follow any poss... | In that case, the spider chooses a path as follows: first, the spider picks a value of $a$ in $[0,\ell]$. Then the spider chooses one of the $\binom{\ell}{a}$ ways to make $a$ moves of the form $(m,k)$ and $\ell-a$ of the form $(k,m)$. Finally, the spider finds itself with $\ell(m-k)$ remaining steps to make, of whic... | [
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1,902,223 | (i) $ABC$ is a triangle with a right angle at $A$, and $P$ is a point on the hypotenuse $BC$.
The line $AP$ produced beyond $P$ meets the line through $B$ which is perpendicular to $BC$ at $U$.
Prove that $BU = BA$ if, and only if, $CP = CA$.
(ii) $A$ is a point on the semicircle $CB$, and points $X$ and $Y$ are on t... | i) Let $AB=BU$. We will prove that $AC=CP$. We know that $\angle BPU=\angle APC=x$.
Also, $\angle BUP=\angle BAP=y$.
But $x+y=90^{\circ}$.
Also, $\angle PAC=90 ^{\circ}-\angle BAP=90^{\circ}-y=x$.
Therefore, $\angle PAC=\angle APC=x$ and thus, $AC=CP$. | [
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577,613 | There are $100$ students who want to sign up for the class Introduction to Acting. There are three class sections for Introduction to Acting, each of which will fit exactly $20$ students. The $100$ students, including Alex and Zhu, are put in a lottery, and 60 of them are randomly selected to fill up the classes. What ... | [hide="Solution"]
The probability any given student gets into a certain acting class is $20/100 = 1/5$. Then the probability that a second student gets into the same class is $19/99$. There are three ways to pick the acting class Alex and Zhu will both be in. Thus, the probability they get into the same class is $3 * (... | [
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577,615 | Bob writes a random string of $5$ letters, where each letter is either $A, B, C,$ or $D$. The letter in each position is independently chosen, and each of the letters $A, B, C, D$ is chosen with equal probability. Given that there are at least two $A's$ in the string, find the probability that there are at least three ... | [hide=Solution]
First, note that the total number of strings is $4^5=1024,$ the number of strings with no $A$s is $3^5=243,$ the number of strings with exactly one $A$ is $5\cdot3^4= 405,$ and the number of strings with exactly two $A$s is $10\cdot3^3=270.$
Thus, by complementary probability, the desired probability i... | [
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0.058273207396268845... | 2025-08-11 |
577,618 | We have a calculator with two buttons that displays and integer $x$. Pressing the first button replaces $x$ by $\lfloor \frac{x}{2} \rfloor$, and pressing the second button replaces $x$ by $4x+1$. Initially, the calculator displays $0$. How many integers less than or equal to $2014$ can be achieved through a sequence o... | [hide="Counting those sequences (different method)"]Alternatively, we can just split up into cases based on how many 1's there are and then "attach" a 0 to all but the last 1 in each configuration. Then the number of such sequences is \[1+{11\choose 1}+{10\choose 2}+{9\choose 3}+{8\choose 4}+{7\choose 5}+{6\choose 6}=1... | [
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... | 2025-08-11 |
577,665 | Given that $x$ and $y$ are nonzero real numbers such that $x+\frac{1}{y}=10$ and $y+\frac{1}{x}=\frac{5}{12}$, find all possible values of $x$. | [hide=Solution]
From the first equation, we get $$x+\frac{1}{y}=10 \implies x = 10-\frac{1}{y}.$$
Then, substituting this into the second equation and then solving for $y$ yields
\begin{align*}
y + \frac{1}{x}=\frac{5}{12} &\implies y + \frac{1}{10-\frac{1}{y}} = \frac{5}{12} \\
&\implies y + \frac{y}{10y-1} = \frac{5... | [
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0.0017... | 2025-08-11 |
577,667 | Let \[ A = \frac{1}{6}((\log_2(3))^3-(\log_2(6))^3-(\log_2(12))^3+(\log_2(24))^3) \].
Compute $2^A$. | [hide=Solution]
Let $a=\log_2 3.$ Then, we have
\begin{align*}
A &= \frac{1}{6}((\log_2(3))^3-(\log_2(6))^3-(\log_2(12))^3+(\log_2(24))^3) \\
&= \frac{1}{6}((\log_2(3))^3 - (\log_2(3)+1)^3 - (\log_2(3)+2)^3 + (\log_2(3)+3)^3) \\
&= \frac{1}{6}(a^3-(a+1)^3-(a+2)^3+(a+3)^3) \\
&= \frac{1}{6}(a^3-a^3-3a^2-3a-1-a^3-6a^2-12... | [
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577,668 | Let $b$ and $c$ be real numbers and define the polynomial $P(x)=x^2+bx+c$. Suppose that $P(P(1))=P(P(2))=0$, and that $P(1) \neq P(2)$. Find $P(0)$. | [hide=Solution]
Since $P(1)=1+b+c$ and $P(2)=4+2b+c$ are both roots of $P$, we have, from Vieta's, that \[ P(1) + P(2) = -b \implies 4b + 2c = -5 \]and \[ (1+b+c) \cdot (4+2b+c) = c. \]Substituting $ b = - \frac {2c+5}{4} $, we get \[ \left(-\frac{1}{4}+\frac{c}{2}\right) \cdot \left( \frac{3}{2} \right) = c \implies\f... | [
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0.00116786... | 2025-08-11 |
577,669 | Find the sum of all real numbers $x$ such that $5x^4-10x^3+10x^2-5x-11=0$. | Basically, if $\alpha$ is a root then $1-\alpha$ is also a root, so the function is symmetric about $x=\tfrac12$ and the number of real roots is even. Then (I believe?) one can prove that $x^5-(x-1)^5$ is a monotonically increasing function along the interval $[\tfrac12,\infty)$, so there must be only one real root gr... | [
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577,670 | Given $w$ and $z$ are complex numbers such that $|w+z|=1$ and $|w^2+z^2|=14$, find the smallest possible value of $|w^3+z^3|$. Here $| \cdot |$ denotes the absolute value of a complex number, given by $|a+bi|=\sqrt{a^2+b^2}$ whenever $a$ and $b$ are real numbers. | $|w^3+z^3|=|w+z|\cdot |w^2-wz+z^2|=\frac{1}{2}|3(w^2+z^2)-(w+z)^2|$
$\ge \frac{1}{2}|3|w^2+z^2|-|w+z|^2|=\frac{41}{2}.$ | [
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577,733 | Let $O_1$ and $O_2$ be concentric circles with radii 4 and 6, respectively. A chord $AB$ is drawn in $O_1$ with length $2$. Extend $AB$ to intersect $O_2$ in points $C$ and $D$. Find $CD$. | [hide="Solution"]
Let the points be labeled $C,A,B,D$ in that order. Furthermore, let $O$ be the center of both circles and let $M$ be the orthogonal projection of $O$ onto $AB$. Note that by the Pythagorean Theorem we have \begin{align*}MC^2&=OC^2-OM^2=OC^2-(OA^2-OM^2)=6^2-(4^2-1^2)=21\\\implies MC&=\sqrt{21}.\end{a... | [
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0.0034981... | 2025-08-11 |
577,907 | Point $P$ and line $\ell$ are such that the distance from $P$ to $\ell$ is $12$. Given that $T$ is a point on $\ell$ such that $PT = 13$, find the radius of the circle passing through $P$ and tangent to $\ell$ at $T$. | Power of a point :huh: :rotfl: :stretcher:
Sane solution: [hide]Let A be the point on the circle opposite to T so that AOT is the diameter. Let B be the point on l directly below P. Notice that triangles PAT and TBP are similar by angle-side since PB || AT and PAT is a right triangle. Use similar triangle ratios t... | [
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-0.00113094... | 2025-08-11 |
577,909 | $ABC$ is a triangle such that $BC = 10$, $CA = 12$. Let $M$ be the midpoint of side $AC$. Given that $BM$ is parallel to the external bisector of $\angle A$, find area of triangle $ABC$. (Lines $AB$ and $AC$ form two angles, one of which is $\angle BAC$. The external angle bisector of $\angle A$ is the line that bisect... | [hide] Let the extension of AC be Y. (away from BC)
Let the intersection of the external bisector of <A and BC be X.
Let <YAX=x.
Therefore, <AXB=x.
Because AX is parallel to BM, we know that <ABM=<AMB=x, so AM=AB=6.
Using Heron's Formula with the side lengths 6, 10, 12, we have the area=8sqrt(14). [/hide] | [
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577,910 | In quadrilateral $ABCD$, $\angle DAC = 98^{\circ}$, $\angle DBC = 82^\circ$, $\angle BCD = 70^\circ$, and $BC = AD$. Find $\angle ACD.$ | Okay then.
Let $P$ be the intersection of $\overline{AC}$ and $\overline{BD}$. Then $\frac{AD}{\sin{APD}}=\frac{BC}{\sin{BPC}}$. By Law of Sines, the former equals $\frac{DP}{\sin{98^{\circ}}}=\frac{DP}{\sin{82^{\circ}}}$ and the latter equals $\frac{CP}{\sin{82^{\circ}}}$, so $CP=DP$. Thus $\angle{ACD}=\angle{BDC}=18... | [
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577,911 | Let $\mathcal{C}$ be a circle in the $xy$ plane with radius $1$ and center $(0, 0, 0)$, and let $P$ be a point in space with coordinates $(3, 4, 8)$. Find the largest possible radius of a sphere that is contained entirely in the slanted cone with base $\mathcal{C}$ and vertex $P$. | [hide] Note that you can take a cross-section of the cone at the points (3,4,8), (3/5, 4/5, 0), and (-3/5, -4/5, 0).
The inradius of this triangle is equivalent to the longest length radius of the sphere.
We know that the area of this triangle is 8 (height=8, side length=2).
Using A=sr, calculate the side lengths throu... | [
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577,912 | In quadrilateral $ABCD$, we have $AB = 5$, $BC = 6$, $CD = 5$, $DA = 4$, and $\angle ABC = 90^\circ$. Let $AC$ and $BD$ meet at $E$. Compute $\dfrac{BE}{ED}$. | We get that $AC=\sqrt{61}$. Using law of cosines, we get that $\cos{DAC}=\frac{13}{2\sqrt{61}}$, so $\sin{DAC}=\frac{5\sqrt{3}}{2\sqrt{61}}$. Additionally, note that $\sin{CAB}=6/\sqrt{61}$. By using the ratio lemma, we get that the ratio is $\frac{5}{4}*\frac{\frac{6}{\sqrt{61}}}{\frac{5\sqrt{3}}{2\sqrt{61}}}$, which ... | [
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577,913 | Triangle $ABC$ has sides $AB = 14$, $BC = 13$, and $CA = 15$. It is inscribed in circle $\Gamma$, which has center $O$. Let $M$ be the midpoint of $AB$, let $B'$ be the point on $\Gamma$ diametrically opposite $B$, and let $X$ be the intersection of $AO$ and $MB'$. Find the length of $AX$. | Interesting point: $AM = BM$ and $BO = OB'$, so $BM$ and $AO$ are both medians of $ABB'$. Thus, their intersection $X$ is the centroid of said triangle, and $AX = \frac{2}{3}AO = \frac{2}{3}R = \boxed{\frac{65}{12}}$. | [
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577,914 | Let $ABC$ be a triangle with sides $AB = 6$, $BC = 10$, and $CA = 8$. Let $M$ and $N$ be the midpoints of $BA$ and $BC$, respectively. Choose the point $Y$ on ray $CM$ so that the circumcircle of triangle $AMY$ is tangent to $AN$. Find the area of triangle $NAY$. | Like the previous outline, we let $G$ be the centroid. Note that $AN=BC/2=5$, and that $CM=\sqrt{8^2+3^2}=\sqrt{73}$. Now $AG=\frac{2}{3}AN=\frac{10}{3}$, and $GM=\frac{1}{3}CM=\frac{\sqrt{73}}{3}$, so
\[
GY\cdot GM=GA^2\Rightarrow GY=\frac{100}{9}/\frac{\sqrt{73}}{3}=\frac{100}{3\sqrt{73}}
\]
So
\[
\frac{GY}{GM}=\fra... | [
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577,916 | Two circles are said to be [i]orthogonal[/i] if they intersect in two points, and their tangents at either point of intersection are perpendicular. Two circles $\omega_1$ and $\omega_2$ with radii $10$ and $13$, respectively, are externally tangent at point $P$. Another circle $\omega_3$ with radius $2\sqrt2$ passes th... | The conditions imply that $w_3$ is tangent to $C_1C_2$, $C_2C_4$, and $C_4C_1$ where $C_i$ is the center of $w_i$. Suppose $w_3$ is tangent to $C_1C_2$, $C_2C_4$, and $C_4C_1$ at $X$, $Y$, and $Z$, respectively. Then, $C_1X = C_1Z = 10$ and $C_2X = C_2Y = 13$. Let $c = C_4Y = C_4Z$. The radius of $C_3$ is $2\sqrt{2}$, ... | [
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0.04448768496513367,
0.00208485964... | 2025-08-11 |
577,917 | Let $ABC$ be a triangle with $AB = 13$, $BC = 14$, and $CA = 15$. Let $\Gamma$ be the circumcircle of $ABC$, let $O$ be its circumcenter, and let $M$ be the midpoint of minor arc $BC$. Circle $\omega_1$ is internally tangent to $\Gamma$ at $A$, and circle $\omega_2$, centered at $M$, is externally tangent to $\omega_1$... | This was how I solved it during the test:
[hide="Solution"]
By Monge's theorem, $AS$ passes through the insimilicenter of the circumcircle and $\omega_2$. To find the ratio of the radii, it suffices to find its distance from $O$: this motivates us to define $N$ to be the intersection of $AS$ with $OM$, as it is the ins... | [
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0.000... | 2025-08-11 |
577,919 | [5] If four fair six-sided dice are rolled, what is the probability that the lowest number appearing on any die is exactly $3$? | [hide]There is a $P(\text{lowest number is at least 3})=(\frac{2}{3})^4=\frac{16}{81}$ and $(P(\text{lowest number is at least 4})=\frac{1}{2})^4=\frac{1}{16}$. $P(\text{lowest number is exactly 3})=P(\text{lowest number is at least 3})-P(\text{lowest number is at least 4})$, so $P(\text{lowest number is exactly 3})=\f... | [
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-0.... | 2025-08-11 |
577,921 | [5] Find all integers $n$ for which $\frac{n^3+8}{n^2-4}$ is an integer. | As Sonnhard said, this fraction reduces to $n + \frac{4}{n - 2}$.
For those of you who do not understand, here is an explanation:
The terms $n^3$ and $8$, which are found in the numerator, are both perfect cubes.
The sum of two perfect cubes $a^3$ and $b^3$ can be factored into $(a + b)(a^2 - ab + b^2)$.
Similarly... | [
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-0.0034... | 2025-08-11 |
577,923 | [4] Let $x_1,x_2,\ldots,x_{100}$ be defined so that for each $i$, $x_i$ is a (uniformly) random integer between $1$ and $6$ inclusive. Find the expected number of integers in the set $\{x_1,x_1+x_2,\ldots,x_1+x_2+\cdots+x_{100}\}$ that are multiples of $6$. | [hide="What"]For each number in the set, the expected number of $\sum_{j=1}^ix_j$ that are in the set is $\frac{1}{6}$ because we can ignore everything but $x_i$, so by Linearity of Expectation the answer is just $100\left(\frac{1}{6}\right)=\frac{50}{3}$[/hide] | [
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... | 2025-08-11 |
577,924 | [4] Let $ABCDEF$ be a regular hexagon. Let $P$ be the circle inscribed in $\triangle{BDF}$. Find the ratio of the area of circle $P$ to the area of rectangle $ABDE$. | Rectangle $ABDE$ has a width of $s$ and a length of $s\sqrt{3}$. Its area is therefore $s^2\sqrt{3}$.
The radius of the equilateral triangle is equal to $s$, making the radius of the triangle's incircle $\frac{s}{2}$.
Then, the incircle's area is $\frac{s^2\pi}{4}$.
When you combine the areas into the ratio, you g... | [
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0.004867479670... | 2025-08-11 |
577,925 | The Evil League of Evil is plotting to poison the city's water supply. They plan to set out from their headquarters at $(5, 1)$ and put poison in two pipes, one along the line $y=x$ and one along the line $x=7$. However, they need to get the job done quickly before Captain Hammer catches them. What's the shortest dista... | Reflect the line $x = 7$ over the line $y = x$ to produce the line $y = 7$.
The point $(1, 7)$ is a reflection of $(7, 1)$ about $y = x$. By connecting the image point with $(5, 1)$, we have the shortest distance to the point $(1, 7)$, which equals the shortest distance to $(7, 1)$.
But, we have to return back to $(5... | [
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0.004050... | 2025-08-11 |
577,926 | [4] Let $D$ be the set of divisors of $100$. Let $Z$ be the set of integers between $1$ and $100$, inclusive. Mark chooses an element $d$ of $D$ and an element $z$ of $Z$ uniformly at random. What is the probability that $d$ divides $z$? | [hide="No way this works"]A random divisor of $100$ is of the form $2^a5^b$, with $a, b$ uniformly and randomly distributed among $0, 1, 2$. Then the expected number of $z$'s the $2$'s will divide is $\frac{1}{3}\left(100+50+25\right)=\frac{175}{3}$, and the expected number the $5$'s will divide is $\frac{1}{3}(100+20+... | [
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577,930 | Compute the side length of the largest cube contained in the region
\[ \{(x, y, z) : x^2+y^2+z^2 \le 25 \text{ and } x, y \ge 0 \} \]
of three-dimensional space. | [img]https://i.postimg.cc/RVhK0p1d/image.png[/img]
this is the top half of the cube
$\sqrt{x^2+x^2+(\frac{x}2)^2}=\frac{3x}2=5$ so $x=\boxed{\frac{10}3}$ | [
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-... | 2025-08-11 |
578,708 | Compute \[\sum_{k=0}^{100}\left\lfloor\dfrac{2^{100}}{2^{50}+2^k}\right\rfloor.\] (Here, if $x$ is a real number, then $\lfloor x\rfloor$ denotes the largest integer less than or equal to $x$.) | [hide=Solution]Note that $$\frac{2^{100}}{2^{50}+2^k}+\frac{2^{100}}{2^{50}+2^{100-k}}=\frac{2^{100}\cdot 2^{50-k}+2^{100}}{2^{50}+2^{100-k}}=2^{50}$$ Since clearly $\frac{2^{100}}{2^{50}+2^k}$ is not an integer for $k\in \{0,1,2,...,100\}\setminus \{50\}$, we know that the sum of the floor functions, $\left\lfloor\fr... | [
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-0... | 2025-08-11 |
578,715 | Fix a positive real number $c>1$ and positive integer $n$. Initially, a blackboard contains the numbers $1,c,\ldots, c^{n-1}$. Every minute, Bob chooses two numbers $a,b$ on the board and replaces them with $ca+c^2b$. Prove that after $n-1$ minutes, the blackboard contains a single number no less than \[\left(\dfrac... | Let $N$ be the final number; it is sufficient to show \[N^{1/L}\stackrel?\ge\frac{c^{n/L}-1}{c^{1/L}-1}=1^{1/L}+c^{1/L}+\left(c^2\right)^{1/L}+\cdots+\left(c^{n-1}\right)^{1/L}.\]
Thus we only need the following monovariant:
[color=red][b]Claim.[/b][/color] For any $a$, $b$, we have $\left(ca+c^2b\right)^{1/L}\ge ... | [
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0.... | 2025-08-11 |
586,231 | Find a nonzero monic polynomial $P(x)$ with integer coefficients and minimal degree such that $P(1-\sqrt[3]2+\sqrt[3]4)=0$. (A polynomial is called $\textit{monic}$ if its leading coefficient is $1$.) | [hide]Let $S=1-\sqrt[3]{2}+\sqrt[3]{4}$. Note that $\sqrt[3]{2}S=-S+9\Rightarrow 2S^3=(9-S)^3\Rightarrow S^3-9S^2+81S-243=0$. So the number is a root of $P(x)=x^3-9x^2+81x-243$. (Clearly it can't be a root of a quadratic because there are cube roots).[/hide] | [
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-0.0... | 2025-08-11 |
586,233 | Let $ABCD$ be a trapezoid with $AB\parallel CD$ and $\angle D=90^\circ$. Suppose that there is a point $E$ on $CD$ such that $AE=BE$ and that triangles $AED$ and $CEB$ are similar, but not congruent. Given that $\tfrac{CD}{AB}=2014$, find $\tfrac{BC}{AD}$. | [hide]
Let $H$ be the point on $\overline{CD}$ such that $\overline{BH} \perp \overline{CD}$. Let $AB = a$, $AD = x$. We see that
\[\frac{CH + DH}{AB} = \frac{CH + a}{a} = 2014 \Longrightarrow CH = 2013a\] We see that $\triangle AED \sim \triangle CEB \sim \triangle CBH$, so
\[\frac{ED}{AD} = \frac{BH}{CH} \Longr... | [
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0.006118... | 2025-08-11 |
586,367 | Let $f:\mathbb{N}\to\mathbb{N}$ be a function satisfying the following conditions:
(a) $f(1)=1$.
(b) $f(a)\leq f(b)$ whenever $a$ and $b$ are positive integers with $a\leq b$.
(c) $f(2a)=f(a)+1$ for all positive integers $a$.
How many possible values can the $2014$-tuple $(f(1),f(2),\ldots,f(2014))$ take? | EDIT: Not relevant anymore. | [
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... | 2025-08-11 |
586,374 | Let $S=\{-100,-99,-98,\ldots,99,100\}$. Choose a $50$-element subset $T$ of $S$ at random. Find the expected number of elements of the set $\{|x|:x\in T\}$. | Sorry for bumping, but in the above sol it should be $1 - \tfrac{\binom{199}{50}}{\binom{201}{50}}$, everything else's good. Got me confused for half an hour on this problem. | [
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596,747 | Find all real numbers $k$ such that $r^4+kr^3+r^2+4kr+16=0$ is true for exactly one real number $r$. | If $r$ satisfies the original equation, then $r/2$ satisfies the equation $16x^4+8kx^3+4x^2+8kx+16=0$. So this means that $16x^4+8kx^3+4x^2+8kx+16=0$ can also only be true for one real number.
However, it is easy to see that for any real number other than 0, the reciprocal will also satisfy this equation (symmetric c... | [
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-0... | 2025-08-11 |
603,988 | Let $f(n)$ and $g(n)$ be polynomials of degree $2014$ such that $f(n)+(-1)^ng(n)=2^n$ for $n=1,2,\ldots,4030$. Find the coefficient of $x^{2014}$ in $g(x)$. | [hide="Non-Lagrange Solution"]
By the binomial theorem, we have \[f(2x)+g(2x)=4\left(\displaystyle\sum_{n=0}^{2014}3^n\binom{x-1}{n}\right)\] and \[f(2x+1)-g(2x+1)=2\left(\displaystyle\sum_{n=0}^{2014}3^n\binom{x}{n}\right).\]
If $a$ is the leading coefficient of $g(x)$, then subtracting the two equations and comparin... | [
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603,989 | Natalie has a copy of the unit interval $[0,1]$ that is colored white. She also has a black marker, and she colors the interval in the following manner: at each step, she selects a value $x\in [0,1]$ uniformly at random, and
(a) If $x\leq\tfrac12$ she colors the interval $[x,x+\tfrac12]$ with her marker.
(b) If $x>\... | Solved with Elliott Liu.
Note that we can think of this as coloring a circle. Additionally, note that the set of black points and set of white points are both a single continuous region. Thus, we can do states based on how much of the circle is colored. Let $f: \left(\frac12 ,1\right) \to \mathbb{R}$ be the expected n... | [
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603,990 | Let $ABC$ be a triangle with circumcenter $O$, incenter $I$, $\angle B=45^\circ$, and $OI\parallel BC$. Find $\cos\angle C$. | This follows from the two identities \[\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2 = \dfrac{r}{4R}\] and \[\cos A+\cos B + \cos C = 1+\dfrac rR.\] | [
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603,993 | Find all ordered pairs $(a,b)$ of complex numbers with $a^2+b^2\neq 0$, $a+\tfrac{10b}{a^2+b^2}=5$, and $b+\tfrac{10a}{a^2+b^2}=4$. | aren't $(a,b)=(4,2);(1,2) $solutions
let $z=a+ib$
$5+4i=a+ib+\frac{10(bi+a)}{a^2+b^2}$
$=a+ib+\frac{i10(a-ib) }{a^2+b^2}=z+\frac{10i\bar{z}}{|z|^2}=z+\frac{10i}{z}$
$\implies z^2-(5+4i)z+10i=0 \implies z=4+2i,1+2i$
$\implies (a,b)=(4,2);(1,2) $ | [
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578,065 | Let $l$ be the tangent line at the point $(t,\ t^2)\ (0<t<1)$ on the parabola $C: y=x^2$. Denote by $S_1$ the area of the part enclosed by $C,\ l$ and the $x$-axis, denote by $S_2$ of the area of the part enclosed by $C,\ l$ and the line $x=1$. Find the minimum value of $S_1+S_2$. | $S_1+S_2=\int\limits_{0}^{1}x^2dx-S_{\triangle}ABC$, where $A$ is the x-intercept of $l$, $B$ the intersection of $l$ and $x=1$, and $C(1,0)$.
It is easy to compute that $l=2tx-t^2$, and thus $A(\frac{t}{2},0)$ and $B(1,2t-t^2)$.
So $S_1+S_2=\frac{1}{3}-\frac{(1-\frac{t}{2})(2t-t^2)}{2}=\frac{1}{3}-\frac{t^3-4t^2+4t}... | [
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578,067 | Find the triplets of primes $(a,\ b,\ c)$ such that $a-b-8$ and $b-c-8$ are primes. | Since $a-b-8$ and $b-c-8$ are both primes, $a > b > c >= 2$. Clearly $a$ and $b$ are odd. Thus $a-b-8$ is even and a prime, which implies that $a-b-8 = 2$ or $a = b+10$. Consider the cases $c = 2$ and $c$ is odd and > 2.
(i) $c = 2$.
In this case, $a = b+10$, $b$ and $b-c-8 = b-10$ are all odd primes. Under modulo 3, ... | [
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578,091 | A sphere with radius 1 is inscribed in a right circular cone with the radius $r$ of the base.
Let $S$ be the surface area of the cone.
(1) Express $S$ in terms of $r$.
(2) Find the minimum value of $S$. | $S(r)=\frac{2\pi r^4}{r^2-1}\ (r>1)$
$=2\pi\left(r^2+1+\frac{1}{r^2-1}\right)$
$=2\pi\left(r^2-1+\frac{1}{r^2-1}+2\right)$
$\underbrace{\geq}_{AM-GM} 2\pi\left(2\sqrt{(r^2-1)\cdot \frac{1}{r^2-1}}+2\right)$
$=8\pi$
The equality hods when $r^2-1=\frac{1}{r^2-1}\ (r>1)\Longleftrightarrow r=\sqrt{2}.$
$\therefore S_... | [
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578,102 | Let $l$ be the tangent line at the point $P(s,\ t)$ on a circle $C:x^2+y^2=1$. Denote by $m$ the line passing through the point $(1,\ 0)$, parallel to $l$. Let the line $m$ intersects the circle $C$ at $P'$ other than the point $(1,\ 0)$.
Note : if $m$ is the line $x=1$, then $P'$ is considered as $(1,\ 0)$.
Call $T$... | Let $E = (1,0)$.
(1) Since $P'$ is obtained by rotating $(1,0)$ by $2 \angle POE$, we have $s' = s^2-t^2 = 2s^2-1$, $t'=2st$. ($t'$ cannot be expressed as a polynomial of $t$. Actually, it is even not a function of $t$; $t' = \pm 2t\sqrt{1-t^2}$.)
(2) Applying the formula obtained, we have $P_1\left( \frac{1}{2}, \fra... | [
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607,373 | For each positive integer $n$, let $s(n)$ be the sum of the digits of $n$. Find the smallest positive integer $k$ such that
\[s(k) = s(2k) = s(3k) = \cdots = s(2013k) = s(2014k).\] | Indeed, $9999=10^4-1$ is the minimum value of $k$. First, suppose that $k$ have one digite. Then, $s(11k)=2s(k)$, absurd. If $k$ has 2 digits, then $s(101k)=2s(k)$, contradiction. If $k$ has 3 digits, then $s(1001k)=2s(k)$, again a contradiction. Then, we have that $k$ has at least 4 digits.
If $k$ have exactly four d... | [
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607,376 | Find all polynomials $P(x)$ with real coefficients such that $P(2014) = 1$ and, for some integer $c$:
$xP(x-c) = (x - 2014)P(x)$ | Let ${x\choose n}=\frac{x(x-1)\dots (x-n+1)}{n!}$, for each $n\in \mathbb{N}, x \in \mathbb{R}$. First, one must prove that $c \ne 0$. This is easy, because if $c=0$, then $xP(x)=(x-2014)P(x)$, which impies $P(x)=0$, absurd, since $P(2014)=1$. Next, we prove that $c$ must be a divisor of $2014$, and to do this, we can ... | [
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607,378 | $2014$ points are placed on a circumference. On each of the segments with end points on two of the $2014$ points is written a non-negative real number. For any convex polygon with vertices on some of the $2014$ points, the sum of the numbers written on their sides is less or equal than $1$. Find the maximum possible va... | Let's solve the general case for $n$ even. We say that a segment with endpoints on the $n$ points in the circle is a $k$-segment if it separates the circle into two parts, with the part with the minimum quantity of points having $k-1$ points (for example, any side of the $n$-agon is a $1$-segment, and any diameter is a... | [
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607,494 | $N$ coins are placed on a table, $N - 1$ are genuine and have the same weight, and one is fake, with a different weight. Using a two pan balance, the goal is to determine with certainty the fake coin, and whether it is lighter or heavier than a genuine coin. Whenever one can deduce that one or more coins are genuine, t... | This problem seems to be a little confuse, but i think that we must find all $N$ such that, for any wheights of the $N-1$ genuine coins, and for any wheight of the fake coin, one can find the fake coin, and whether it is lighter or heavier than a genuine coin.
First, it's easy to see that $N=2,3$ are not solutions, be... | [
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607,527 | Given a set $X$ and a function $f: X \rightarrow X$, for each $x \in X$ we define $f^1(x)=f(x)$ and, for each $j \ge 1$, $f^{j+1}(x)=f(f^j(x))$. We say that $a \in X$ is a fixed point of $f$ if $f(a)=a$. For each $x \in \mathbb{R}$, let $\pi (x)$ be the quantity of positive primes lesser or equal to $x$.
Given an posi... | (a) given a catracha function $f$ and $k \in [n] := \{1,2, \dots , n\}$, let $a_k \in \mathbb{N}$ be the least positive integer such that $f^{a_k}(k)=k$ (such $a_k$ always exists, since $f^{f(k)}(k)=k$). Now, from $f^{f(k)}(k)=k$, changing $k$ by $f^{a}(k)$, we get $f^{f(f^{a}(k))}(f^{a}(k))=f^{a}(k) \Rightarrow f^{f^{... | [
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1,929,976 | Find the least natural number $n$, which has at least 6 different divisors
$1=d_1<d_2<d_3<d_4<d_5<d_6<...$, for which $d_3+d_4=d_5+6$ and $d_4+d_5=d_6+7$. | Note that d(2) = 2,otherwise d(3),d(4) and d(5) are odd.Also d(3) > 6 ,so d(3) is prime.If d(4) is even ,then d(4)=2d(3) and d(5) is a multiple of 3,implying d(3)=3,contradiction.So d(4) must be prime and d(5) even.It follows that d(5) = 2d(3),from which we get that d(4) = d(3) + 6 and d(6) = 3d(3) - 1.But d(6) is... | [
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1,929,980 | Let $f(x)$ be a polynomial with integer coefficients, for which there exist $a,b\in \mathbb{Z}$ ($a\neq b$), such that $f(a)$ and $f(b)$ are coprime. Prove that there exist infinitely many values for $x$, such that each $f(x)$ is coprime with any other. | If $f(x)=1$ everywhere then we trivially have the claim. So, suppose $f(\cdot)$ is non-constant. Note that for any prime $p$, there is an $n$ such that $p\nmid f(n)$ (otherwise, there would be a prime $p$ with $p\mid f(n)$ for all $n$, contradicting with the fact that $f(a)$ and $f(b)$ are coprime).
Now, let us const... | [
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1,929,985 | If $AG_a,BG_b$, and $CG_c$ are symmedians in $\Delta ABC$ ($G_a\in BC,G_b\in AC,G_c\in AB$), is it possible for $\Delta G_a G_b G_c$ to be equilateral when $\Delta ABC$ is not equilateral? | Yes, take $\angle ACB = 120^{\circ}$ and $\angle BAC = \angle ABC = 30^{\circ}$. | [
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1,932,197 | Prove that for $\forall$ $x,y,z\in \mathbb{R}^+$ the following inequality is true:
$\frac{x}{y+z}+\frac{25y}{z+x}+\frac{4z}{x+y}>2$.
| We can simply use Cauchy-Schwarz inequality:
$$
\frac{x}{y+z}+\frac{25y}{z+x}+\frac{4z}{x+y}=\frac{x^2}{xy+xz}+\frac{25y^2}{yz+xy}+\frac{4z^2}{xz+yz}\ge \frac{(x+5y+2z)^2}{2(xy+yz+zx)}>2,
$$
because $(x+5y+2z)^2=(x^2+25y^2+4z^2)+(10xy+20yz+4xz)>4(xy+yz+zx)$. | [
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1,932,212 | Find all $f: \mathbb{N}\rightarrow \mathbb{N}$, for which
$f(f(n)+m)=n+f(m+2014)$
for $\forall$ $m,n\in \mathbb{N}$.
| Step 1- Injectivity.
Step 2- Substitute $P(m,f(t)-f(m+2014))$ to get $f(f(t)-f(m+2014))=t-m$, for appropriate $t$.
Step 3- Thus order the elements for integers $>2014$, using $k^{th}$ least value principle with induction for integers larger than 2014.
Step 4- $P(k-f(n),n): f(k)-f(k+2014-f(n))=n$, for large enough $k$. ... | [
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1,932,570 | Find all pairs of natural numbers $(m,n)$, for which $m\mid 2^{\varphi(n)} +1$ and $n\mid 2^{\varphi (m)} +1$. | Clearly \( m \) and \( n \) are odd. Suppose that neither of them equals 1, and denote \( \varphi(m) = 2^{m_0} m_1 \) and \( \varphi(n) = 2^{n_0} n_1 \), where \( m_0 \) and \( n_0 \) are non-negative integers, and \( m_1 \) and \( n_1 \) are odd positive integers. Without loss of generality, we may assume that \( m_0 ... | [
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1,932,571 | Does there exist a natural number $n$, for which $n.2^{2^{2014}}-81-n$ is a perfect square? | No. Assume that $n(2^{2^{2014}}-1)=m^2+81$. Note that the left-hand side is divisble by $43$ which is $3 \mod 4$ and hence cannot have $-1$ as a quadratic residue and neither $-81$ too. Done. | [
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1,932,576 | Is it true that for each natural number $n$ there exist a circle, which contains exactly $n$ points with integer coordinates? | Pick the Cartesian point $ P=\left( 1/7, \sqrt 3 \right) . $ Prove that no two lattice points are equidistant from $ P. $ The circle with center $ P $ and radius $ 2 $ contains exactly one lattice point. Now, suppose there is a circle centered at $ P $ containing $ n $ points. Increase its radius until it touches anoth... | [
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1,934,614 | Polly can do the following operations on a quadratic trinomial:
1) Swapping the places of its leading coefficient and constant coefficient (swapping $a_2$ with $a_0$);
2) Substituting (changing) $x$ with $x-m$, where $m$ is an arbitrary real number;
Is it possible for Polly to get $25x^2+5x+2014$ from $6x^2+2x+1996$ wi... | Discriminant is invariant, so no. | [
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1,934,627 | It is known that each two of the 12 competitors, that participated in the finals of the competition “Mathematical duels”, have a common friend among the other 10. Prove that there is one of them that has at least 5 friends among the group. | [b]The official solution.[/b]
Let us denote the competitors by $A_1,A_2,\dots,A_{12}$. For the sake of contradiction, assume every student has at most $4$ friends. First, we prove it's impossible someone to have less than $4$ friends.\\
1) Suppose $A_1$ has only one friend (let it be $A_2$), then $A_1$ and $A_2$ do... | [
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1,934,633 | We define the following sequence: $a_0=a_1=1$, $a_{n+1}=14a_n-a_{n-1}$. Prove that
$2a_n-1$ is a perfect square.
| An alternative finish from this identity: So
\[(a_{n+1}-7a_n)^2=48a_n^2-12=12 \cdot (2a_{n-1}-1)(2a_{n-1}+1).\]
Hence $3(2a_n-1)(2a_n+1)$ is a perfect square for all $n$.
But it is easy to prove by induction that $a_n \equiv 1 \mod 3$ so that this immediately implies that $2a_n-1$ is a perfect square. | [
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1,832,189 | A convex quadrilateral $ABCD$ is inscribed into a circle $\omega$ . Suppose that there is a point $X$ on the segment $AC$ such that the $XB$ and $XD$ tangents to the circle $\omega$ . Tangent of $\omega$ at $C$, intersect $XD$ at $Q$. Let $E$ ($E\ne A$) be the intersection of the line $AQ$ with $\omega$ . Prove that ... | Obviously,$BDCA$ and $CDEA$ are harmonic quadrilateral.So $(B,D;C,A)=-1$ $(C,D,E,A)=-1$. Now,Let $BE \cap AD=M$.We will prove that CQ will pass through $M$.
$BC \cap AD=N$
Now $-1=(C,D;E,A)=B(C,D;E,A)=(N,D;M,A)$
And $-1=(B,D;C,A)=C(B,D;C,A)=(N,D;CC \cap AD, A)$.So $CQ \cap AD= M$. So $AD$,$BE$,$CQ$ are concurrent. | [
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1,834,741 | a) Prove that the equation $2^x + 21^x = y^3$ has no solution in the set of natural numbers.
b) Solve the equation $2^x + 21^y = z^2y$ in the set of non-negative integer numbers. | [b]a)[/b] The Diophantine equation
$(1) \;\; 2^x + 21^x = y^3$
has no solution in natural numbers.
[b]Proof:[/b] Assume equation (1) has a minimal solution $(x,y)=(x_0,y_0)$ ($x_0$ is minimal). We know that $x_0 = 3q + r$, where $q \geq 0$ and $0 \leq r \leq 2$ are two integers. Hence by (1)
$(2) \;\; y_0^3 ... | [
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1,834,749 | The function $f: N \to N_0$ is such that $f (2) = 0, f (3)> 0, f (6042) = 2014$ and $f (m + n)- f (m) - f (n) \in\{0,1\}$ for all $m,n \in N$. Determine $f (2014)$. $N_0=\{0,1,2,...\}$ | $f (6042)- f (2014) - f (4028) \in\{0,1\}$ so let say it is $y$.So $y \in \{0,1\}$.So $f(4028)+f(2014)=2014-y$
Again, $f(4028)-2f(2014)=x\in\{0,1\}$
Substractinf the 2 equations we get $3f(2014)=2014-x-y$. So 3 devides $2014-(x+y)$ but as $x,y\in\{0,1\}$ so $x+y \in\{0,1,2\}$.So $x+y$ must me equal 1.So $f(2014)=(2014-... | [
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1,834,753 | Let $n$ be a natural number and let $P (t) = 1 + t + t^2 + ... + t^{2n}$.
If $x \in R$ such that $P (x)$ and $P (x^2)$ are rational numbers, prove that $x$ is rational number. | We know that $P(x)=1+x+x^2+...x^{2n}$ and $P(x^2)=1+x^2+x^4+....x^{4n}$ are rational. Now, consider the following identity:
$(1+x+x^2+....+x^{2n})(1-x+x^2-......+x^{2n})=1+x^2+x^4+...x^{4n}$. This comes from the simple fact that we can rewrite this identity to be $\frac{(1+x^{2n+1})(1-x^{2n+1})}{(1-x)(1+x)}=\frac{1-x^{... | [
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1,834,764 | Let $p$ be a prime number. The natural numbers $m$ and $n$ are written in the system with the base $p$ as $n = a_0 + a_1p +...+ a_kp^k$ and $m = b_0 + b_1p +..+ b_kp^k$. Prove that
$${n \choose m} \equiv \prod_{i=0}^{k}{a_i \choose b_i} (mod p)$$ | This is [url=https://en.wikipedia.org/wiki/Lucas%27s_theorem]Lucas's theorem[/url]. | [
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1,122,795 | Let $A_{1}A_{2} \dots A_{3n}$ be a closed broken line consisting of $3n$ lines segments in the Euclidean plane. Suppose that no three of its vertices are collinear, and for each index $i=1,2,\dots,3n$, the triangle $A_{i}A_{i+1}A_{i+2}$ has counterclockwise orientation and $\angle A_{i}A_{i+1}A_{i+2} = 60º$, using the ... | Here is a solution which is completely different from the one proposed as an official solution:
[hide=Solution]
For each of the three directions of lines, we have $n$ parallel segments, no two of them on the same line. So we can order them, from outside to inside the figure (note that this is not related to the orderin... | [
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-0.... | 2025-08-11 |
1,279,371 | Determine all pairs $(a, b)$ of real numbers for which there exists a unique symmetric $2\times 2$ matrix $M$ with real entries satisfying $\mathrm{trace}(M)=a$ and $\mathrm{det}(M)=b$.
(Proposed by Stephan Wagner, Stellenbosch University) | We look at the matrix $M=\begin{bmatrix}x&y\\y&z\end{bmatrix}$.
We want $x+z=a$ and $xz-y^2=b$
So $x,z$ are roots of $u^2-au+(y^2+b)=0$
Thus $x=\frac{a\pm\sqrt{a^2-4(y^2+b)}}{2}$ and $y=\frac{a\mp\sqrt{a^2-4(y^2+b)}}{2}$
Solutions can only be unique if $a^2=4(y^2+b)\implies y=\pm\frac{\sqrt{a^2-4b}}{2}$ where we mu... | [
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-0.001017... | [
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-0.004323... | 2025-08-11 |
1,279,373 | Consider the following sequence
$$(a_n)_{n=1}^{\infty}=(1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,1,\dots)$$
Find all pairs $(\alpha, \beta)$ of positive real numbers such that $\lim_{n\to \infty}\frac{\displaystyle\sum_{k=1}^n a_k}{n^{\alpha}}=\beta$.
(Proposed by Tomas Barta, Charles University, Prague) | What I would do is observe that if $T(n)=\frac{n(n+1)}{2}$ then, $\sum_{k=1}^{T(n)} a_k=\sum_{k=1}^{n-1}k(n-k)=\frac{n(n+1)(n-1)}{6}$
Thus $\frac{1}{[T(n)]^\alpha}\sum_{k=1}^{T(n)} a_k=\frac{2^\alpha n^{1-\alpha}(n+1)^{1-\alpha}(n-1)}{6}\sim\frac{2^\alpha n^{3-2\alpha}}{6}$
This converges only if $\alpha\geq\frac{3}{... | [
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-0.... | 2025-08-11 |
1,279,375 | Let $n$ be a positive integer. Show that there are positive real numbers $a_0, a_1, \dots, a_n$ such that for each choice of signs the polynomial
$$\pm a_nx^n\pm a_{n-1}x^{n-1} \pm \dots \pm a_1x \pm a_0$$
has $n$ distinct real roots.
(Proposed by Stephan Neupert, TUM, München) | Sometimes explaining a simple idea behind a problem is better than writing a complete solution.
In this case suppose we have found such positive numbers $a_{n-1},\dots, a_0$. Fix some polynomial $P=\pm a_{n-1}x^{n-1} \pm \dots \pm a_1x \pm a_0$. Draw a picture in your head of the graph of $P$ - how it goes up and dow... | [
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-0.0... | [
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-0.... | 2025-08-11 |
1,279,376 | Let $n>6$ be a perfect number, and let $n=p_1^{e_1}\cdot\cdot\cdot p_k^{e_k}$ be its prime factorisation with $1<p_1<\dots <p_k$. Prove that $e_1$ is an even number.
A number $n$ is [i]perfect[/i] if $s(n)=2n$, where $s(n)$ is the sum of the divisors of $n$.
(Proposed by Javier Rodrigo, Universidad Pontificia Comillas... | [hide=Solution]We know that $\prod_{i=1}^k (1+p_i+p_i^2+\cdot\cdot\cdot+p_i^{e_i}) = 2n = 2p_1^{e_1}\cdot\cdot\cdot p_k^{e_k}$. Suppose for the sake of contradiction, $e_1$ is an odd number. This would imply that $p_1+1|1+p_1+p_1^2+\cdot\cdot\cdot p_1^{e_1}|2n$. So $p_1+1$ would be a factor of $2n$. Since clearly $p_1... | [
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-0.0... | 2025-08-11 |
1,279,385 | Let $A=(a_{ij})_{i, j=1}^n$ be a symmetric $n\times n$ matrix with real entries, and let $\lambda _1, \lambda _2, \dots, \lambda _n$ denote its eigenvalues. Show that
$$\sum_{1\le i<j\le n} a_{ii}a_{jj}\ge \sum_{1\le i < j\le n} \lambda _i \lambda _j$$
and determine all matrices for which equality holds.
(Proposed by ... | We have
\begin{align*}
\left(\operatorname{tr}(A)\right)^2 &= 2\sum_{1\le i<j\le n} a_{ii}a_{jj} +\sum a_{ii}^2=\\
&=2\sum_{1\le i < j\le n} \lambda _i \lambda _j +\sum \lambda_{i}^2
\end{align*}
Thus, to prove the given inequality, it's enough to check:
\[\sum \lambda_{i}^2 \geq \sum a_{ii}^2 \]
It follows from this c... | [
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0.0017... | 2025-08-11 |
1,279,387 | Let $f(x)=\frac{\sin x}{x}$, for $x>0$, and let $n$ be a positive integer. Prove that $|f^{(n)}(x)|<\frac{1}{n+1}$, where $f^{(n)}$ denotes the $n^{\mathrm{th}}$ derivative of $f$.
(Proposed by Alexander Bolbot, State University, Novosibirsk) | This is kind of an old familiar problem to me - I've seen it around, and I know the quickest solution by heart. I'm sure it's appeared here on AoPS.
Write $f(x)=\int_0^1\cos(tx)\,dt.$
Then, since $\cos(tx)$ is a $C^{\infty}$ function of $t$ with bounded partial derivatives, we can justify interchanging limit and inte... | [
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0.0008... | 2025-08-11 |
1,279,395 | For every positive integer $n$, denote by $D_n$ the number of permutations $(x_1, \dots, x_n)$ of $(1,2,\dots, n)$ such that $x_j\neq j$ for every $1\le j\le n$. For $1\le k\le \frac{n}{2}$, denote by $\Delta (n,k)$ the number of permutations $(x_1,\dots, x_n)$ of $(1,2,\dots, n)$ such that $x_i=k+i$ for every $1\le i\... | Let $\Delta(n,k,j)$ be the number of permutations in $S_n$ with $x_i=k+i$ for $1 \le i \le k$ and exactly $j$ fixed points. So $\Delta(n,k)=\Delta(n,k,0)$ and clearly $\sum_j \Delta(n,k,j)=(n-k)!$. On the other hand, we clearly have $\Delta(n,k,j)=\binom{n-2k}{j} \cdot \Delta(n-j,k)$ and hence we obtain the identity
\[... | [
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-0... | 2025-08-11 |
596,927 | Convex quadrilateral $ABCD$ has $\angle ABC = \angle CDA = 90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. Points $S$ and $T$ lie on sides $AB$ and $AD$, respectively, such that $H$ lies inside triangle $SCT$ and \[
\angle CHS - \angle CSB = 90^{\circ}, \quad \angle THC - \angle DTC = 90^{\cir... | First note that $(CHS)$ is orthogonal to $AB$, so after inverting with center $H$ the problem becomes:
Let $ABD$ be a triangle and $H$ be the foot of the perpendicular to $BD$ from $A$. $C$ is a point on $ABD$ such that $(AHC)$ is orthogonal to $(ABD)$ and $S,T$ are intersections of $CM_D, CM_B$ with the circles with ... | [
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... | 2025-08-11 |
596,929 | Let $n \ge 2$ be an integer. Consider an $n \times n$ chessboard consisting of $n^2$ unit squares. A configuration of $n$ rooks on this board is [i]peaceful[/i] if every row and every column contains exactly one rook. Find the greatest positive integer $k$ such that, for each peaceful configuration of $n$ rooks, there ... | The claim is this largest number $k$ is $\boxed{k=\lfloor \sqrt{n-1}\rfloor}$, thus $k^2 < n \leq (k+1)^2$. Label the rows and columns with the numbers from $0$ to $n-1$.
Let $i$ be the label of the row containing a rook on column $n-1$, and let $I$ be any group of $k$ contiguous labels, including $i$. There exist $k$... | [
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596,930 | Let $a_0 < a_1 < a_2 < \dots$ be an infinite sequence of positive integers. Prove that there exists a unique integer $n\geq 1$ such that
\[a_n < \frac{a_0+a_1+a_2+\cdots+a_n}{n} \leq a_{n+1}.\]
[i]Proposed by Gerhard Wöginger, Austria.[/i] | The old $\Delta$ trick, again!
Let $ a_n = a_0 + \Delta_1 + \Delta_2 + ... + \Delta_n $ for each $ n \ge 1$. Notice that $ \Delta_n \ge 1 $
Now, the condition $a_n < \frac{a_0 + ... +a_n}{n} \le a_{n+1}$ can be rewritten as:
\[ a_0 + \Delta_1 + \Delta_2 + ... + \Delta_n < \frac{(n+1)a_0 + n\Delta_1 +(n-1)\Delta_2+ .... | [
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608,489 | Let $A$ be a subset of the irrational numbers such that the sum of any two distinct elements of it be a rational number. Prove that $A$ has two elements at most. | Say $A$ contains distinct irrationals $a,b,c$. Then $a+b=r$, $b+c=s$, $c+a=t$ for some rational $r,s,t$. So $b=\dfrac {r+s-t}{2} \in \mathbb{Q}$, absurd. | [
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608,492 | Let $(X,d)$ be a metric space and $f:X \to X$ be a function such that $\forall x,y\in X : d(f(x),f(y))=d(x,y)$.
$\text{a})$ Prove that for all $x \in X$, $\lim_{n \rightarrow +\infty} \frac{d(x,f^n(x))}{n}$ exists, where $f^n(x)$ is $\underbrace{f(f(\cdots f(x)}_{n \text{times}} \cdots ))$.
$\text{b})$ Prove that the a... | The main use of the given property is this:
$$d(x , f^{m+n}(x)) \leq d(x , f^{m}(x))+d(f^{m}(x) , f^{m+n}(x)) = d(x , f^{m}(x))+d(x , f^{n}(x)) $$
Lemma. For any sequence of positive real numbers satisfying $a_{m+n} \leq a_m+a_n$ , $\lim_{n \rightarrow +\infty} \frac{a_n}{n}$ exists.
Proof.Just divide $n$ by $m$. The... | [
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-0.000... | 2025-08-11 |
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