id int64 41 3.22M | problem stringlengths 38 4.44k | solution stringlengths 8 38.6k |
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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 $\omega_P$ again at $X$. Similarly, suppose the circle with diameter $\overline{AQ}$ intersects $\omega_Q$ again at $N$, and line $AN$ intersects $\omega_Q$ again at $Y$.
Prove that lines $MN$ and $XY$ are parallel.
(Here, the points $P$ and $Q$ are [i]isogonal conjugates[/i] with respect to $\triangle ABC$ if the internal angle bisectors of $\angle BAC$, $\angle CBA$, and $\angle ACB$ also bisect the angles $\angle PAQ$, $\angle PBQ$, and $\angle PCQ$, respectively. For example, the orthocenter is the isogonal conjugate of the circumcenter.)
[i]Proposed by Sammy Luo[/i] | [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 $\measuredangle BPC=\measuredangle BMC.$ Then, \begin{align*}&\measuredangle MCP=\measuredangle QCN=\measuredangle QBN\\&\implies \measuredangle CBA-\measuredangle NBA-\measuredangle CBQ+\measuredangle MAC=\measuredangle ACB-\measuredangle PCB-\measuredangle ACM+\measuredangle BAN\\&\implies \measuredangle CBA+\measuredangle MAC+\measuredangle ACM-\measuredangle CBQ=\measuredangle ACB+\measuredangle BAN+\measuredangle NBA-\measuredangle PCB\\&\implies \measuredangle CBA+180-\measuredangle CMA-\measuredangle CNQ=\measuredangle ACB+180-\measuredangle ANB-\measuredangle PMB\\&\implies \measuredangle QNA=\measuredangle AMP=90.\end{align*} $\blacksquare$
Take the $\sqrt{bc}$ inversion, denoting $Z^*$ as the inverse of $Z.$
[b][color=red]Claim:[/color][/b] $\omega_P^*=\omega_Q.$
[i]Proof.[/i] Let $P'=\overline{AQ}\cap\omega_Q$ and note $$\measuredangle QP'B=\measuredangle QCB=\measuredangle ACP$$ so $\triangle AP'B\sim\triangle ACP$ or $P^*=P'.$ $\blacksquare$
Hence, $P$ and $Q$ lie on $\omega_Q^*$ and $\omega_P^*,$ respectively. Also, $\measuredangle N^*Q^*A=\measuredangle QNA=90$ so $\overline{PN^*}$ is the diameter of $\omega_Q^*.$ Hence, $Y^*$ is the foot from $P$ to $\overline{AN^*}=\overline{AM}$ so $Y^*=M.$ Similarly, $X^*=N.$ Thus, $$AX^*\cdot AN^*=AB\cdot AC=AY^*\cdot AM^*$$ and $\triangle AX^*Y^*\sim\triangle AM^*N^*.$ $\square$ |
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 $CD$.
Define $\omega_2$ analogously as the circle passing through $H$ and the feet of the perpendiculars from $H$ to $BC$ and $DA$.
Show that the midpoint of $GH$ lies on the radical axis of $\omega_1$ and $\omega_2$.
[i]Proposed by Yang Liu[/i] | 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 that line $ CD $ goes to $ \omega_2 $ so $ E $ goes to $ G $. Similarly $ F $ goes to $ H $. Moreover, $ \omega_1 $ and $ \omega_2 $ are the circles with diameters $ EG $ and $ FH $ respectively. Now since $ M $ is the midpoint of $ GH $ and since $ O_1 $ is the midpoint of $ GE $ we have that $ O_{1}M \parallel HE $ and similarly $ O_{2}M \parallel GF $.
But since $ GF $ is the polar of $ E $ with respect to $ \omega $ we have that $ GF \perp OE $ so $ O_{2}M \perp OE $ which implies $ O_{2}M \perp OO_{1} $. Similarly $ O_{1}M \perp OO_{2} $ and so $ M $ is the orthocenter of triangle $ OO_{1}O_{2} $. This means that $ OM \perp O_{1}O_{2} $ so to show that $ M $ is on the radical axis of $ \omega_1 $ and $ \omega_2 $ it suffices to show that $ O $ is on this radical axis.
However, this is clear since both $ \omega_1 $ and $ \omega_2 $ are orthogonal to $ \omega $, so we are done.
Interestingly after doing the inversion step, a complex number solution is somewhat doable as well. |
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 $\omega_A,\omega_B,\omega_C$ have a common point $X$ other than $I$, and that $\angle AXO = \angle OXA'$.
[i]Proposed by Sammy Luo[/i] | 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 Sawayama's Theorem, but I shall provide a more elementary proof. Let $ B_2, C_2 $ be the midpoints of arcs $ CA, AB $ respectively. Then by Archimedes' Lemma we have that $ A', X, C_2 $ and $ A', Y, B_2 $ are collinear. Then by Pascal's Theorem on hexagon $ A'C_{2}CABB_{2} $ we have that $ X, I, Y $ are collinear. But since $ AX = AY $ and $ AI \perp XY $ this immediately implies the desired result.
Lemma 2: If $ A_1 $ denotes the midpoint of arc $ BAC $ then $ A', I, A_1 $ are collinear
[i]Proof:[/i] Consider the homothety centered at $ A' $ that takes the $ A $-mixtilinear incircle to the circumcircle of triangle $ ABC $. This takes $ X $ to $ C_2 $ and $ Y $ to $ B _2 $ so by Lemma 1 it takes $ I $ to $ M $, the midpoint of $ B_{2}C_{2} $. So it suffices to show that $ A_1, I, M $ are collinear. Some quick angle-chasing yields the fact that $ A_{1}C_{2}IB_{2} $ is a parallelogram which implies the desired result.
Now, continuing as in XmL's post, taking the inversion centered at $ I $ with radius $ \sqrt {A'I * A_{1}I} = \sqrt {B'I * B_{1}I} = \sqrt {C'I * C_{1}I} $, we have that $ \omega_A $ is mapped to the line parallel to $ AI $ passing through the reflection of $ A_1 $ over $ I $. We obtain two similar lines, and then reflecting them about $ I $ we obtain the configuration in XmL's post.
So, letting $ l_A $ be the line through $ A_1 $ parallel to $ AI $ and defining $ l_B, l_C $ similarly it suffices to show that these three lines concur at the circumcenter of the excentral triangle of triangle $ ABC $. Let $ I_A, I_B, I_C $ be the $ A, B, C $-excenters of triangle $ ABC $ respectively. It is well-known that $ A_1 $ is the midpoint of $ I_{B}I_{C} $ and that $ AI \perp I_{B}I_{C} $ so $ l_A $ is the perpendicular bisector of $ I_{B}I_{C} $. This immediately implies the desired result, and so we are done. |
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 intersect at a point $X$, and the lines $AR,BS,CT$ intersect at a point $Y$, such that $O,X,Y$ are collinear.
[i]Proposed by Sammy Luo[/i] | 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, CF $ are concurrent ,
so by [b]Terquem theorem[/b] we get $ AR, BS, CT $ are concurrent at $ Y $ .
Since $ \triangle RST $ is the pedal triangle of $ X' $ ($ X' $ is the isogonal conjugate of $ X $ ) ,
so the line passing $ A, B, C $ and perpendicular to $ ST, TR, RS $ are concurrent at $ X $ ,
hence by [b]Sondat theorem[/b] (for $\triangle ABC $ and $\triangle RST $ ) we get $ Y, X', X $ are collinear.
ie. $ X, O, Y $ are collinear
Q.E.D
[b]Remark:[/b]
(1) Easy to see $ X' = AD \cap BE \cap BF $ which is the orthocenter of $\triangle DEF $
(2) Another interesting property in this configuration :
Denote $ O' $ as the circumcenter of $ \triangle ABC $ , then the isogonal conjugate of $ Y $ lie on $ O'H $ .
(see http://www.artofproblemsolving.com/community/c6h366335 ) |
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 $\ell$ where $F_\ell \equiv -1$ (mod $b^k$). Thus, $F_\ell$ will have $k$ digits of value $b-1$ at the end. Implying that
\[s_b(F_\ell) \geq k(b-1) \]
But $k(b-1)>c$ for sufficiently large $k$. |
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]Proposed by Shashwat Kishore[/i] | 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 $a$ is also a non-$d$th power. Thus $a \le b < p/a + 1$ and therefore $a < \sqrt{p} + 1$.
Now $\log r = \epsilon - \frac{1}{\gcd (d,p-1)} > -\frac{1}{2}$ and hence $r > \frac{1}{\sqrt{e}} > \frac{1}{2}$. Therefore there exists a non-$d$th power modulo $p$ less that $p^r$. |
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 $1 + 1 + 1 \le a + b + c$ by AM-GM, we can strengthen the last inequality above into \[a^2 + b^2 + c^2 \ge a + b + c.\] Then since $p - q \ge 0$, we can weight these two inequalities as follows:
\begin{align*}
q\left(a^2 + b^2 + c^2 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) &\ge q\left(2(a + b + c)\right) \\
(p - q)\left(a^2 + b^2 + c^2\right) &\ge (p - q)(a + b + c).
\end{align*}
Summation yields the desired result. $\square$ |
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 this pretty cleanly. The idea is to retain use of symmetric sums.
First off, we can expand the $a(b^2+ca)(c^2+ab)$ on the LHS easily. The $(a^2+bc)(b^2+ca)(c^2+ab)$ on the RHS is also easy to deal with; see what we get when we multiply all squares, 2,1, or 0 squares. So we want to show that \[\left(\sum_{\text{sym}} a^2b\right)\left(ab^2c^2+a^2bc^2+a^2b^2c+a^2b^3+a^2c^3+b^2c^3+b^2a^3+c^2a^3+c^2b^3+a^3bc+ab^3c+abc^3\right)\le \]\[\le (a^2+b^2+c^2+2ab+2bc+2ca)\left(\sum_{\text{cyc}}a^3b^3 +\sum_{\text{cyc}}a^4bc+2a^2b^2c^2\right).\]It's faster to use symmetric sums on the RHS, but I like to avoid fractions. Now on the LHS, we can just multiply $a^2b$ by everything and take the symmetric sum of that; on the RHS, we can multiply our expanded $(a+b+c)^2$ by each of the three terms, then group into symmetric sums. After we do this and clean things up, we get \[\sum_{\text{sym}}a^5b^3+\sum_{\text{sym}}2a^5b^2c+\sum_{\text{sym}}2a^4b^2c^2+\sum_{\text{sym}}3a^4b^3c+\sum_{\text{sym}}a^4b^4+\sum_{\text{sym}}3a^3b^3c^2\le\]\[\le \tfrac{1}{2}\sum_{\text{sym}}a^6bc+\sum_{\text{sym}}a^5b^3+\sum_{\text{sym}}2a^5b^2c+\sum_{\text{sym}}2a^4b^2c^2+\sum_{\text{sym}}3a^4b^3c+\sum_{\text{sym}}a^4b^4+\tfrac{5}{2}\sum_{\text{sym}}a^3b^3c^3.\]Miraculously, nearly everything cancels! Leaving \[\tfrac{1}{2}\sum_{\text{sym}}a^6bc\ge \tfrac{1}{2}\sum_{\text{sym}}a^3b^3c^2,\]which is quite clear by Muirhead![/hide]
[hide="Equality Cases"]By doing the AM-GM in our Muirhead, it's clear that equality can only be achieved when $a=b=c$ or when $a=0$ (and permutations). Thankfully we don't have to try to analyze equality in the Holder; we can just plug back into the original and see that $a=b=c$ works and $a=0$ reduces to $b^3+c^3\ge b^2c+bc^2$, which has equality at $b=c$. So the equality cases are $(a,a,a),(0,a,a)$ and permutations of the second.[/hide]
Out of a 144 term mess we end up with 6. We win! :) |
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 of $RST$ with respect to the triangle $ABC$ be $Q$. Show that $OP = OQ$.
[i]Proposed by Yang Liu[/i] | [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 \} $ and $ Q $ be the Miquel point of $ \{ R, S, T \} $ .
Then $ KP=KQ $ .
[b]Proof:[/b]
Let $ P_x, P_y, P_z $ be the reflection of $ P $ in $ KX, KY, KZ $ , respectively .
Let $ Q_x=P_xR \cap \odot (P_xP_yP_z), Q_y=P_yS \cap \odot (P_xP_yP_z), Q_z=P_zT \cap \odot (P_xP_yP_z) $ .
From $ \angle Q_xP_xP=\angle RDP=\angle SEP=\angle Q_yP_yP \Longrightarrow Q_x \equiv Q_y $ .
Similarly, we can prove $ Q_x \equiv Q_y \Longrightarrow P_xR, P_yS, P_zT $ are concurrent at $ Q' \in \odot (K, KP) $ .
From $ \angle TAS=\angle TQ'S \Longrightarrow A, T, S, Q' $ are concyclic .
Similarly, we can prove $ Q' \in \odot (BRT) $ and $ Q' \in \odot (CRS) $ ,
so $ Q' \equiv Q $ which is the Miquel point of $ \{ R, S, T \} $ and lie on $ \odot (K, KP) $ .
i.e. $ KP=KQ $
Q.E.D |
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 $k = 0,1, ..., m$.
[i]Proposed by Michael Kural[/i] | 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 estimate, suppose some $m>n$ works. Applying the finite differences operator, we obtain $$\Delta^{(n)} P(x)=\sum^n_{j=0} (-1)^{n-j}\binom{n}{j}P(x+j)$$ is the constant function. Note that $t \mid \Delta^{(n)} P(1)$ but $t \nmid \Delta^{(n)} P(0)$ giving us the desired contradiction. $\, \square$
P.S. I am writing this at a late hour; hopefully there are no errors. |
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 $$\sum_{k=0}^{p-1} \frac{\binom{p-1}{k}}{(1-kp)^2}=\sum_{k=0}^{p-1} \binom{p-1}{k} \left(\sum_{i \geq 0} (i+1)(kp)^i \right) \equiv \sum_{k=0}^{p-1} \binom{p-1}{k}(1+2kp+3k^2p^2) \pmod{p^3}$$
Now, from the identity $r\binom{n}{r}=n\binom{n-1}{r-1}$, we get $$\sum_{k=0}^{p-1} \binom{p-1}{k}=2^{p-1}$$
$$\sum_{k=0}^{p-1} k\binom{p-1}{k}=\sum_{k=0}^{p-1} (p-1)\binom{p-2}{k-1}=(p-1)2^{p-2}$$
$$\sum_{k=0}^{p-1} k^2\binom{p-1}{k}=\sum_{k=0}^{p-1} k(k-1)\binom{p-1}{k}+\sum_{k=0}^{p-1} k\binom{p-1}{k}=\sum_{k=0}^{p-1} (p-1)(p-2) \binom{p-3}{k-2}+\sum_{k=0}^{p-1} (p-1)\binom{p-2}{k-1}$$
$$\Rightarrow \sum_{k=0}^{p-1} k^2\binom{p-1}{k}=(p-1)(p-2)2^{p-3}+(p-1)2^{p-2}=p(p-1)2^{p-3}$$
Thus, we get $$\sum_{k=0}^{p-1} \frac{\binom{p-1}{k}}{(1-kp)^2} \equiv 2^{p-1}+2p(p-1)2^{p-2}+3p^3(p-1)2^{p-3} \equiv 2^{p-1}(p^2-p+1) \pmod{p^3}$$
But, then we easily get the result by using the fact that $$(p+1)(p^2-p+1)=p^3+1 \equiv 1 \pmod{p^3} \Rightarrow 2^{p-1}(p^2-p+1) \equiv \frac{2^{p-1}}{p+1} \pmod{p^3} \text{ } \blacksquare$$ |
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 $ |
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 we let $x^2+1 = t \cdot 13^l$ then it becomes $13 | t + 2xz$. It is obvious that we can find such an integer $z$, as $13$ does not divide $2x$.
(this is basically Hensel Lemma :P) |
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 $p \neq 3$ and $p|b-1$. Note that $b^{2n}+b^n(8-b)+1 \equiv 9 \ncong 0 \pmod{p}$. Suppose $p \neq 2$. Then, equating the powers of $p$ on either side using LTE, we get $3v_p(x_n) = v_p(n)$ for sufficiently large $n$. Choosing large $n$ with $v_p(n)=1$ does the job. Otherwise, if $p=2$, again equate powers of $2$ on either side using LTE. We get $v_2(b+1)+v_2(n)-1=3v_3(x_n)$, for all sufficiently large $n$. But, we can easily choose large $n$ such that the LHS is not a multiple of $3$, contradiction.
Therefore, $b-1=3^a$. Since $b \ge 9$, we have $3^a \ge 8 \implies a \ge 2$. Suppose that $a\ge 3$. Then, note that \[v_3(b^{2n}+b^n(8-b)+1) = v_3((b^{2n}-b^{n+1})+8(b^n-1)+9) = 2\]because $27|b-1 \implies 27|b^n-1$ and $27|b^{n+1}(b^{n-1}-1)=b^{2n}-b^{n+1}$. Again, looking at the powers of $3$ on either side, we have $3v_3(x_n)=2+v_3(n)$. Choosing large $n$ not divisible by $3$ gives a contradiction. Therefore, $a=2$, which means $b=10$. And in this case we do have \[ x_n^3 = \frac{(10^n-1)(10^{2n}-2.10^n+1)}{27}= \left( \frac{10^n-1}{3} \right)^3 \] |
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 Sammy Luo[/i] | 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)!} $ |
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 the polar of $X$ then $J \equiv J'$ and we are done.If $AC$ is the polar of $X$ then certainly $XA,XC$ are tangents to $\omega$.Thus $G \equiv A,C \equiv H$ and thus $AC$ and $GH$ are same.We are done in both the cases. |
1,908,062 | Determine the value of the expression $x^2 + y^2 + z^2$,
if $x + y + z = 13$ , $xyz= 72$ and $\frac1x + \frac1y + \frac1z = \frac34$. | $$\sum_{cyc}\frac{1}{x}=\frac{\sum_{cyc}xy}{xyz}=\frac{\sum_{cyc}xy}{72}=\frac{3}{4}\text{ therefore }\sum_{cyc}xy=54$$
$$\text{ Furthermore }\sum_{cyc}x^2=\left(\sum_{cyc}x\right)^2-2\sum_{cyc}xy=13^2-2\cdot 54=61$$
$$\blacksquare.$$ |
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 $(a,b)=1$
$\implies q=2 \implies q^n=2^n=2ab \implies a=2^{n-1},b=1$
$\implies 3|p^m=2^{2(n-1)}-1 \implies p=3$
$\implies 3^m=2^{2(n-1)}-1$
$\implies m=1,n=2$
$\implies r=5$ |
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 $i=2013k$ and $k$ be the smallest integer such that the assumption is valid. Then we have
\[ 2013=\frac{2013k}{a_i} \Leftrightarrow a_i=k. \]
Because of the monotonicity of $(a_n)$ we can conclude
\[ a_1,a_2,\dots,a_i=a_{2013k} \leq k. \]
Let's now look at some cases.
$a_{20}$: $\tfrac{20}{a_{20}}=20$ if and only if $a_{20}=1$. By our assumption ($20$ isn't [i]good[/i]) that implies $a_{20}\neq 1$ and hence $a_{20} \geq 2$.
$a_{40}$: By the same argument $a_{40} \geq 3$.
.
.
.
$a_{20(k-1)} \geq k$ and therefore equality must hold and $a_{20(k-1)},a_{20(k-1)+1},\dots,a_{2013k}=k$ because if the monotonicity.
But then
\[ \frac{20k}{a_{20k}}=\frac{20k}{k}=20. \]
Contradition!
$20$ must be [i]good[/i], if $2013$ is [i]good[/i] and obviously $2013$ can be good.
Obviously, the same thing holds for any $2$ numbers with one being greater than the other one. So as a generalisation:
Let $a_1\leq a_2 \leq \dots$ be a monotonically increasing sequence with positive integers. A positive integer $n$ is called [i]good[/i] if there is an index such that $n=\tfrac{i}{a_i}$.
If a positive integer $k$ is good, then all positive integers $<k$ are good aswell.[/hide] |
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 while their leader claimed that it actually isn't possible.
Prove that they didn't get any coin with the value $12$ Kulotnik. | 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 $(0,45),(12,34),(24,23),(36,12),(44,1)$.
Now consider $11A+12B=5040$. Now $B=45a+34b+23c+12d+e$, and $a+b+c+d+e=11$. This gives $B=11(4a+3b+2c+d+1)$.
Again $A=12b+24c+36d+44e$. Clearly, $e=3,6$. Then, if $e=3$,
$132(b+2c+3d+1)+132(4a+3b+2c+d+1)=5040\iff 4a+4b+4c+4d+2=45$ (no solution).
If $e=6$ then $132(b+2c+3d+2)+132(4a+3b+2c+d+1)=5040\iff 4a+4b+4c+4d+3=45$ (no solution).
That means, since we started with the assumption that both types of coins existed, and did not make any further guess in any other part of problem, it must follow that one of the coins did not exist.
If $11$ did not exist then each gets $540$ kulotnik, i.e. $45$ coins of denomination $12$ which is possible.
If $12$ did not exist then each gets $540$ kulotnik, which can't be provided with $11$ kulotnik coins only. |
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 $x-y =1$
So $(x,y) = (3,2),(-2,-3)$ and since we assumed Wlog hence the other 2 solutions are $(2,3) , (-3,-2)$
|
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 $A=\{8,4\},B=\{7,5\},C=\{6,1,3,2\}$ hence we have a solution for $n\equiv 8(9)$ which includes 2015.
For $n=11$ we have $A=\{11,10,1\},B=\{9,8,5\},C=\{7,6,4,3,2\}$ hence soluble for $n\equiv 11(9)$ which includes 2018.
|
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)$.
For $x=0$ we thus have $Q(0) = 0$. Therefore $Q(x) = xR(x)$. Plugging in and cancelling equal factors, we are left with $ R(x)=R(x-2)$. This implies $R$ is a constant polynomial. Thus the solutins are $\boxed{P(X) = Rx^2(x^2-4)}$, for any $R\in \mathbb{R}$ |
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
\[\begin{aligned}
(2x_{n+1}-3x_n)^2-5x_n^2=-4\implies x_{n+1}^2-3x_{n+1}x_n+x_n^2=-1.
\end{aligned}\]
Thus, we have $ x_{n+2}^2-3x_{n+2}x_{n+1}+x_{n+1}^2=-1$. It follows that $x_{n+2}^2-3x_{n+2}x_{n+1}+x_{n+1}^2= x_{n+1}^2-3x_{n+1}x_n+x_n^2$, so we get $(x_{n+2}-x_n)(x_{n+2}+x_n-3x_{n+1})=0$. As $x_{n+2}>x_{n+1}>x_n$, we get $x_{n+2}=3x_{n+1}-x_n$ for all $n\geq 1$. As $x_1, x_2$ are natural numbers, by induction we deduce that $x_n$ is a natural number for all $n\geq 1$.
[/hide] |
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 possible order for the steps of each "movement").The spider completes $l$ "movements" and afterwards it moves without limitation (it still moves rightwards and upwards only).If $n=m\cdot l$,find the number of the possible paths the spider can follow to reach $C$.Note that $n,m,k,l\in \mathbb{N^{*}}$ with $k<m$. | 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 which $a(m-k)$ must be up. Thus, the answer is $\sum_a \binom{\ell}{a}\binom{\ell(m-k)}{a(m-k)}$. This sum is easy to evaluate when $d:=m-k=0,1$ but already at 2 I don't think there is a nice closed form. (It is also the coefficient of $x^{d\ell}$ in $(1+x^d)^\ell(1+x)^{d\ell}$, and probably can be rewritten lots of other ways, too.) |
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 the line segment $BC$.
The line $AX$, produced beyond $X$, meets the line through $B$ which is perpendicular to $BC$ at $U$.
Also the line $AY$, produced beyond $Y$, meets the line through $C$ which is perpendicular to $BC$ at $V$.
Given that $BY = BA$ and $CX = CA$, determine the angle $\angle VAU$. | 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, $AC=CP$. |
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 is the probability that Alex and Zhu end up getting into the same section for the class? | [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 $3 * (1/5) * (19/99) = \boxed{\dfrac{19}{165}}$.
[/hide]
EDIT: whoops fixed mistake. |
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 $A's$ in the string. | [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 is
\begin{align*}
\frac{1024-243-405-270}{1024-243-405} &= \frac{106}{376} \\
&= \boxed{\frac{53}{188}}.
\end{align*}
$\square$
[/hide] |
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 of arbitrary button presses? (It is permitted for the number displayed to exceed 2014 during the sequence. Here, $\lfloor y \rfloor$ denotes the greatest integer less than or equal to the real number $y$). | [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 \[1+{11\choose 1}+{10\choose 2}+{9\choose 3}+{8\choose 4}+{7\choose 5}+{6\choose 6}=1+11+45+84+70+21+1=\boxed{233}\][/hide] |
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}{12} \\
&\implies 12y+\frac{12y}{10y-1} = 5 \\
&\implies 120y^2=50y-5 \\
&\implies 24y^2=10y-1 \\
&\implies 24y^2-10y+1=0 \\
&\implies (6y-1)(4y-1) = 0 \\
&\implies y = \frac{1}{6}, \frac{1}{4}.
\end{align*}
Finally, substituting these two values into the first equation and then solving for $x$ yields $x=\boxed{4,6}.$ $\square$
[/hide]
|
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^2-12a-8+a^3+9a^2+27a+27) \\
&= \frac{1}{6}(12a+18) \\
&= 2a+3.
\end{align*}
Therefore, the answer is
\begin{align*}
2^A &= 2^{2a+3} \\
&= 8 \cdot \left(2^a\right)^2 \\
&= 8 \cdot \left(2^{\log_2 3}\right)^2 \\
&= 8 \cdot 3^2 \\
&= 8 \cdot 9 \\
&= \boxed{72}.
\end{align*}
$\square$
[/hide] |
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 \[ \left(-\frac{1}{4}+\frac{c}{2}\right) \cdot \left( \frac{3}{2} \right) = c \implies\frac{4c-2}{8} \cdot \frac {3}{2} = c \implies c = \boxed {-\frac{3}{2}}. \]
[/hide] |
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 greater than $\tfrac12$. Hence two roots total. |
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. | $|w^3+z^3|=|w+z|\cdot |w^2-wz+z^2|=\frac{1}{2}|3(w^2+z^2)-(w+z)^2|$
$\ge \frac{1}{2}|3|w^2+z^2|-|w+z|^2|=\frac{41}{2}.$ |
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{align*} Similarly, $MD=\sqrt{21}$. Therefore the segment in total has length $\boxed{2\sqrt{21}}$.[/hide] |
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 to solve: PT/AT = PB/PT, so 13/(2r) = 12/13, so r=169/24. [/hide]
|
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 bisects the other angle. | [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] |
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 $\angle{ACD}=\angle{BDC}=180^{\circ}-70^{\circ}-82^{\circ}=28^{\circ}$. |
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 through distance formula, getting 4rt(5) and 10.
r=A/s=3-sqrt(5).[/hide] |
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 is just $\sqrt{3}$. |
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 $AX = \frac{2}{3}AO = \frac{2}{3}R = \boxed{\frac{65}{12}}$. |
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
\[
\frac{GY}{GM}=\frac{100}{3\sqrt{73}}/\frac{\sqrt{73}}{3}=\frac{100}{73}
\]
Now note that $[NAY]=\frac{NA}{GA}[GAY]=\frac{3}{2}[GAY]$, and $[GAY]=\frac{GY}{GM}[GAM]$. But $[GAM]=\frac{1}{6}[ABC]=\frac{1}{6}\cdot 24=4$, so
\[
[NAY]=\frac{NA}{GA}\cdot \frac{GY}{GM}\cdot 4 = \frac{3}{2}\cdot \frac{100}{73}\cdot 4 = \frac{600}{73}.
\] |
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 through $P$ and is orthogonal to both $\omega_1$ and $\omega_2$. A fourth circle $\omega_4$, orthogonal to $\omega_3$, is externally tangent to $\omega_1$ and $\omega_2$. Compute the radius of $\omega_4$. | 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 $2\sqrt{2}$, therefore
\begin{align*}
2\sqrt{2} = \dfrac{[C_1C_2C_4]}{\tfrac{1}{2} \cdot \left(C_1C_2 + C_2C_4 + C_4C_1\right)} = \dfrac{\sqrt{(c+23) \cdot c \cdot 10 \cdot 13}}{c+23},
\end{align*}
by Heron's Formula. Squaring yields $8 \cdot (c+23)^2 = 130 \cdot (c+23) \cdot c$ or $8(c+23) = 130c$ which is equivalent to $c = \boxed{\tfrac{92}{61}}$. |
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 $\omega_1$ at a point $T$. Ray $AT$ meets segment $BC$ at point $S$, such that $BS - CS = \dfrac4{15}$. Find the radius of $\omega_2$ | 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 insimilicenter of the two circles, so the radius in question is $\frac{65}{8}\cdot \frac{ON}{MN}$.
Let $X$ be the midpoint of $BC$, and let $D$ be the foot of the perpendicular from $A$. Note that $BD=5$, $BX=7$, so $DX=2$, and also $AD=12$. Note that $XB=\frac{2}{15}$, and $DSA\sim XSN$, so $XN=\frac{2}{15}/\frac{32}{15}\cdot 12=\frac{3}{4}$. We can easily compute $R=\frac{65}{8}$, $XM=4$, so $MN=\frac{38}{8}$ and $NO=\frac{27}{8}$. Thus the radius of $\omega_2$ is
\[
\frac{65}{8}\cdot \frac{27}{38}=\frac{1755}{304}
\]
[/hide] |
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 $P(\text{lowest number is exactly 3})=\frac{16}{81}-\frac{1}{16}=\frac{175}{1296}$.[/hide] |
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, the difference of two perfect cubes factors into $(a - b)(a^2 + ab + b^2)$.
So, in this case the numerator can be written as $(n + 2)(n^2 - 2n + 4)$.
The denominator is clearly the difference of two squares, which is very easy to factor: $(n + 2)(n - 2)$.
The term $(n + 2)$ is common in both the numerator and denominator. We know $n$ cannot equal $-2$ because the denominator would then be $0$. For all $n$ besides $-2$, the like terms cancel out.
We are now left with $\frac{n^2 - 2n + 4}{n - 2}$.
Note that $\frac{n^2 - 2n}{n - 2}$ reduces to $n$. On the other hand, the remaining component $\frac{4}{n - 2}$ cannot be reduced.
So, the complete reduction of the original expression is $n + \frac{4}{n - 2}$.
Given that we are asked for all values of $n$ that make the above expression an integer, the problem is basically suggesting that you list all the values of $n$ such that $n - 2$ is a factor of $4$ or $-4$. This will make $\frac{4}{n - 2}$ an integer.
The factors of $4$ and $-4$ are: ${4, 2, 1, -1, -2, -4}$.
The possible $n$-values are: ${6, 4, 3, 1, 0, -2}$.
But, as it was mentioned earlier, $n = -2$ does not work because we got the completely reduced expression from canceling out the $(n + 2)$ in the numerator and denominator, making the assumption that $n$ cannot equal $-2$.
So, the final list of $n$-values is $\boxed{{6, 4, 3, 1, 0}}$ |
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] |
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 get:
$\frac{\frac{s^2\pi}{4}}{s^2\sqrt{3}} = \frac{\pi}{4\sqrt{3}} = \boxed{\frac{\pi\sqrt{3}}{12}}$ |
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 distance they can travel to visit both pipes and then return to their headquarters? | 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 $(5, 1)$. Similarly, reflect $(5, 1)$ across the same line, and then reflect the new point over $y = 7$. The reason for reflecting this point twice is so that the line segment connecting $(5, 1)$ and the final image point intersects the two lines of reflection, which are $y = x$, and $y = 7$, the reflection of $x = 7$. Then, our final image is located at $(1, 9)$.
Now, the answer to the problem is just the distance between $(5, 1)$ and $(1, 9)$.
That line segment is the hypotenuse of a right triangle with legs of $4$ and $8$.
$\sqrt{4^2 + 8^2} = \sqrt{16 + 64} = \sqrt{80} = \boxed{4\sqrt{5}}$ |
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 $\frac{1}{3}(100+20+4)=\frac{124}{3}$. But as the $2$'s $5$'s are independent, the probability is $\frac{\frac{175}{3}\cdot\frac{124}{3}}{100^2}=\frac{7\cdot 31}{900}=\boxed{\frac{217}{900}}$[/hide] |
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 $x=\boxed{\frac{10}3}$ |
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, $\left\lfloor\frac{2^{100}}{2^{50}+2^k}\right\rfloor+\left\lfloor\frac{2^{100}}{2^{50}+2^{100-k}}\right\rfloor$ will be lower than $2^{50}$ and is also an integer. Clearly, this value is either $1$ less or $2$ less than $2^{50}$. Clearly, $\left\lfloor\frac{2^{100}}{2^{50}+2^k}\right\rfloor+\left\lfloor\frac{2^{100}}{2^{50}+2^{100-k}}\right\rfloor = 2^{50}-1$ since $$\frac{2^{100}}{2^{50}+2^k}-1+\frac{2^{100}}{2^{50}+2^{100-k}}-1=2^{50}-2<\left\lfloor\frac{2^{100}}{2^{50}+2^k}\right\rfloor+\left\lfloor\frac{2^{100}}{2^{50}+2^{100-k}}\right\rfloor<\frac{2^{100}}{2^{50}+2^k}+\frac{2^{100}}{2^{50}+2^{100-k}}=2^{50}$$ which is true because $\frac{2^{100}}{2^{50}+2^k}$ is not an integer for $k\in \{0,1,2,...,100\}\setminus \{50\}$. So it follows that our answer is just $50(2^{50}-1)+\frac{2^{100}}{2^{50}+2^{50}}=100\cdot 2^{49}-50+2^{49}=\boxed{101\cdot 2^{49}-50}$.[/hide] |
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 \[\left(\dfrac{c^{n/L}-1}{c^{1/L}-1}\right)^L,\] where $\phi=\tfrac{1+\sqrt 5}2$ and $L=1+\log_\phi(c)$. | 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 $\left(ca+c^2b\right)^{1/L}\ge a^{1/L}+b^{1/L}$.
By H\"older's inequality, \[ \left[\left(c^{1/L}a^{1/L}\right)^L+\left(c^{2/L}b^{1/L}\right)^L\right]^{1/L}\left[\left(c^{-1/L}\right)^{L/(L-1)}\left(c^{-2/L}\right)^{L/(L-1)}\right]^{(L-1)/L}\ge a^{1/L}+b^{1/L}.\]
However $c=\phi^{L-1}$ and $1=\phi^{-1}+\phi^{-2}$, so \[\left(c^{-1/L}\right)^{L/(L-1)}+\left(c^{-2/L}\right)^{L/(L-1)}=c^{-1/(L-1)}+c^{-2/(L-1)}=1,\]
and we are done. |
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] |
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
\[\frac{ED}{AD} = \frac{BH}{CH} \Longrightarrow \frac{a/2}{x} = \frac{x}{2013a} \Longrightarrow a^2 = x^2 \cdot \frac{2}{2013}\] Then by Pythagorean,
\[BC = \sqrt{(2013a)^2 + x^2} = \sqrt{2013 \cdot x^2 \cdot 2 + x^2} = x\sqrt{4027}\] so the answer is $\boxed{\sqrt{4027}}$.
[/hide] |
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. |
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. |
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 coefficients), so the only possible real solutions to this equation are $1$ and $-1$. (since $1/1=1, 1/-1=-1$) Substituting in, we find $k=9/4$ and $-9/4$. |
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 comparing leading coefficients, we get that \[2\left(2^{2014}a\right) = 2\cdot \dfrac{3^{2014}}{2014!}\Longrightarrow a = \boxed{\dfrac{3^{2014}}{2^{2014}2014!}.}\]
[/hide] |
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 $x>\tfrac12$ she colors the intervals $[x,1]$ and $[0,x-\tfrac12]$ with her marker.
What is the expected value of the number of steps Natalie will need to color the entire interval black? | 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 number of steps from a position that has $x$ of it's area colored with black. Note that our final answer will be $1+f\left(\frac12\right)$.
We now set up the recurrence relation which is true for $x\in [\frac12, 1)$.
\[f(x) = 1+ \left(\left(x-\frac12\right)\cdot f(x) + \int_{0}^{1-x} f(x+t) \mathrm dt + (x-\frac12)\cdot f(1) + \int_0^{1-x} f(x+t)\mathrm dt\right) = 1+\left(x-\frac12\right)\cdot f(x) + 2 \int_0^{1-x} f(x+t) \mathrm dt\]
Taking the derivative of both sides gives
\[f'(x) = f(x) + \left(x-\frac12\right) \cdot f'(x) - 2f(x)\]
Rearranging,
\[f(x) = \left(x-\frac32\right) \cdot f'(x)\Longrightarrow \frac{1}{f(x)}\cdot f'(x) = \frac{2}{2x-3}\]
Integrating both sides
\[\ln \left(f(x)\right) = \ln(2x-3) +C\]
Thus, if $C_1=e^C$, we have $f(x)= C\cdot (2x-3)$
Now, in order to solve this differential equation, note that $\lim_{x\to 1}f(x) = 2$.This is true because as $x\to 1$, we're left with a single white point, which is dealt with with probability $\frac12$, from which it is clear that the expected number of turns from here is $\frac{1}{1-\frac12}=2$. Thus, we should have
\[2=\lim_{x\to 1} f(x) = \lim_{x\to 1} C\cdot (2x-3)=-C\]
Thus, $C=-2$, so $f(x) = -4x+6$. We may now finish by noting that the answer we want is
\[1+f\left(\frac12\right) = 1 + -2+6 = 5\]
$\blacksquare$. |
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 \[\cos A+\cos B + \cos C = 1+\dfrac rR.\] |
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) $ |
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 $S_1+S_2=\frac{1}{3}-\frac{(1-\frac{t}{2})(2t-t^2)}{2}=\frac{1}{3}-\frac{t^3-4t^2+4t}{4}$.
Let $f(t)=t^3-4t^2+4t$, $t \in (0,1)$, we want to maximize $f(t)$.
$f'(t)=3t^2-8t+4$, and set it to $0$ we obtain $t=2$ or $t=\frac{2}{3}$.
Simple calculation gives $f(0)=0$, $f(1)=1$ and $f(\frac{2}{3})=\frac{32}{27}$.
Thus the minimum area $S_1+S_2=\frac{1}{3}-\frac{8}{27}=\frac{1}{27}$. |
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, we have $a = b+1$, $b$ and $b-10 = b-1$. Therefore at least one of $a$, $b$ and $b-10$ is divisible by 3. It follows that $a$, $b$, $b-10$ are all primes only if $b-10 = 3$. Thus $b = 13$ and $a = b+10 = 23$. The only triplets of primes is $(a, b, c) = (23, 13, 2)$.
(ii) $c$ is odd and > 3.
Since $b$ and $c$ are odd, $b-c-8 = 2$ or $b = c+10$. So $c$, $b = c+10$ and $a = b+10 = c+20$ are all odd primes. Under modulo 3, we have $a = c-1$, $b = c+1$ and $c$. Therefore at least one of $a$, $b$, $c$ is divisible by 3. It follows that $a$, $b$, $c$ are all odd primes only if $c = 3$. Thus $b = c+10 = 13$ and $a= c+20 = 23$. The only triplets of primes is $(a, b, c) = (23, 13, 3)$.
In all cases, the triplets of primes such that $a-b-8$ and $b-c-8$ are primes are $(a, b, c) = (23, 13, 2)$ and $(23, 13, 3)$. |
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 $r^2-1=\frac{1}{r^2-1}\ (r>1)\Longleftrightarrow r=\sqrt{2}.$
$\therefore S_{min}=S(\sqrt{2})=8\pi.$ |
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 $T$ the operation such that the point $P'(s',\ t')$ is obtained from the point $P(s,\ t)$ on $C$.
(1) Express $s',\ t'$ as the polynomials of $s$ and $t$ respectively.
(2) Let $P_n$ be the point obtained by $n$ operations of $T$ for $P$.
For $P\left(\frac{\sqrt{3}}{2},\ \frac{1}{2}\right)$, plot the points $P_1,\ P_2$ and $P_3$.
(3) For a positive integer $n$, find the number of $P$ such that $P_n=P$. | 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 $P_1\left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right)$, $P_2\left( -\frac{1}{2}, \frac{\sqrt{3}}{2} \right)$, $P_3 =P_2$.
(3) We may consider the problem in a different point of view. If we corresponds each point $P$ to a unit sized complex number with argument $\theta$, $z=\exp (i \theta)$. (This point of view may apply to the previous questions) Here $\theta = \angle PON$. Then $P_n$ corresponds $z^{2^n}=\exp (i 2^n \theta)$. Therefore, to find the point $P$ corresponds to solve the equation $z^{2^n} =z$, or equivalently $z^{2^n-1} =1$. There exists exactly $2^n -1$ such point, therefore. |
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 digits (say $k=a_3 a_2 a_1 a_0$), we claim that $k=9999$. We know that $s(1001k)=s(k)$. Writing $1001k$ it as:
$a_3- a_2- a_1- a_0 -0 -0-0$
$+0-0-0-a_3 -a_2 -a_1 -a_0$
we see that we must have 1 carrier after the sum $a_3+a_0$ (we are summing right to left), because if that's not the case, the sum of the digits of $1001k$ will be $2(a_3+a_2+a_1+a_0)>s(k)$, absurd. Also, if $a_1<9$, the sum of the digits of $1001k$ is at least $a_3+a_2+(a_1+1)+0+a_2+a_1+a_0>s(k)$, contradiction. So, $a_1=9$ and if $a_2<9$, the sum of digits of $1001k$ is at least $a_3+(a_2+1)+0+0+a_2+a_1+a_0>s(k)$, contradiction. So, $a_1=a_2=9$ and if $a_3<9$, the sum of digits of $1001k$ is at least $(a_3+1)+0+0+0+a_2+a_1+a_0>s(k)$, contradiction. Thus, $k=999a_0$, and notice that $a_0 \ne 0$, because if so, $1001k=9990999$, implying $s(1001k)=54=2s(k)$, contradiction. Also, $k$ is divisble by 9 (because $k \equiv s(k)=s(2k) \equiv 2k (mod. 9)$. So, we have $k=9999$, showing us that the minimum is at least 9999.
Now we show that $k=9999$ solves the problem, by showing that $s(ak)=36$, for all $k=1,2,...,2014$ (actually, that works or every $a$ with at most 4 digits). Let $a=a_3a_2a_1a_0$ (we can have $a_3=0$), and first suppose that $a_0 \ne 0$. Then, we can write $9999a$ as $a_3 a_2 a_1 a_0 0000-0000 a_3 a_2 a_1 a_0 = a_3 a_2 a_1 (a_0-1)(9-a_3)(9-a_2)(9-a_1)(10-a_0)$ , and the sum of the digits is precisely 36. Now, if $a_0=0$, $s(ak)=s((a/10)k)=36$ if $a_1 \ne 0$. If $a_0=a_1=0$, $s(ak)=s((a/100)k)=36$, if $a_2=0$. If $a_0=a_1=a_2=0$, $s(ak)=s((a/1000)k)=36$, since $a_3 \ne 0$. Thus, $s(ak)=36$, for all $k=1,2,...,2014$, showing that $k=9999$ is solution, and its the minimum because of the preceding paragraph. |
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 use roza2010's idea.
Suppose that $c$ is not a divisor of $2014$.
For $x=2014$ we get $P(2014-c)=0$
For $x=2014-c$ we get $P(2014-2c)=0$
For $x=2014-2c$ we get $P(2014-3c)=0$
Inductively:
For $x=2014-nc$ we get $P(2014-(n+1)c)=0$ (we can only get this because $2014-nc \ne 0, \forall n \in \mathbb{N}$).
Polynomial $P$ has infinite roots, so $P(x) \equiv 0$, a contradiction, since $P(2014)=1$.
With this, we get that $c$ is a divisor of $2014$. So, we can write $2014=ac$, with $a, c \in \mathbb{Z}$.
For $x=2014+c=c(a+1)$ we get $P(c(a+1))=a+1$
For $x=2014+2c=c(a+2)$ we get $P(c(a+2))=\frac{(a+2)(a+1)}{2}$
Inductively, we get $P(c(a+n))=\frac{(a+n)(a+n-1) \dots (a+1)}{n!}={ a+n \choose n}={a+n \choose a}$.
From here, we can see that $a<0$ is impossible, since $P(c(a+n)) = { a+n \choose n} = 0$, for $n \ge -a$. This implies that $P$ has infinitely many roots, hence $P \equiv 0$, absurd, because $P(2014)=1$. Thus, $a>0$, which means that $c$ is a positive divisor of $2014$.
Now, observe that from $P(c(a+n))={a+n \choose a}$, we have that $P(x)={\frac{x}{c} \choose \frac{2014}{c}}$ for infinitely many $x$, so $P(x)={\frac{x}{c} \choose \frac{2014}{c}}$ for all $x$. This is a souction indeed, because:
$(x-2014)P(x)=(x-c(\frac{2014}{c})).{\frac{x}{c} \choose \frac{2014}{c}}=$
$=c.\frac{(\frac{x}{c}-\frac{2014}{c})(\frac{x}{c})(\frac{x}{c}-1) \dots (\frac{x}{c}-\frac{2014}{c}+1)}{(\frac{2014}{c})!}=$
$=c(\frac{x}{c}).((\frac{x}{c}-1) \dots (\frac{x}{c}-\frac{2014}{c}+1)\frac{(\frac{x}{c}-\frac{2014}{c})}{(\frac{2014}{c})!}=$
$=c(\frac{x}{c}).{\frac{x}{c}-1 \choose \frac{2014}{c}}=x{\frac{x-c}{c} \choose \frac{2014}{c}}=xP(x-c)$.
Finally, for $c$ being a positive divisor of $2014$, we have $P(x) \equiv {\frac{x}{c} \choose \frac{2014}{c}}$, and for any other $c$ there is no solutions. |
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 value for the sum of all the written numbers. | 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 $n/2$-segment. For $k=1,2,...,n/2$, let $A_k$ be the sum of the numbers in all $k$-segments. We'll proof the following:
Lemma 1: If $k_1, \dots, k_l$ are positive integers with $l \ge 3$, $k_1, \dots, k_l < n/2$ and $k_1 + k_2 + \dots +k_l =n$, then $A_{k_1} + \dots + A_{k_l} \le n$.
Proof: Consider all the $n$ convex polygons that have $l$ sides $a_1, \dots, a_l$, in clockwise orientation, and such that $a_1$ is a $k_1$-segment, $a_2$ is a $k_2$ segment and so on. Every $k_1$ side appears exactly once on those polygons, Every $k_2$ side appears exactly once on those polygons, and so on. Thus, the sum of the numbers on those $n$ polygons is exactly $A_{k_1} + \dots + A_{k_l}$. But this sum is at most $n$, so $A_{k_1} + \dots + A_{k_l} \le n$.
Lemma 2: If $k_1, \dots, k_l$ are positive integers with $l \ge 3$, $k_1, \dots, k_{l-1} < n/2$, $k_l=n/2$ and $k_1 + k_2 + \dots +k_l =n$, then $A_{k_1} + \dots + A_{k_{l-1}}+2A_{k_l} \le n$
Proof: Similar to Lemma 1, but you must note that each $n/2$-segment appears twice on the $n$ polygons, instead of once.
With that in mind, we have that, from Lemma 1, $A_t+A_{n/2-t}+A_t+A_{n/2-t} \le n$, for all $t=2,3,...,n/2-1$, hence $A_t+A_{n/2-t} \le n/2$, and summing this from $t=2$ to $t=n/2-1$, we have, calling $S=A_1+A_2+...+A_{n/2}$
$(S-A_1-A_{n/2})+(S-A_{n/2-1}-A_{n/2}) \le (n/2-2).n/2 \Rightarrow$
$\Rightarrow 2S \le n^2/4 +(A_1 + A_{n/2-1} +A_{n/2}-n)$
But from Lemma 2, $A_1 + A_{n/2-1} +A_{n/2} \le n$, where we finnaly get $2S \le n^2/4 \Rightarrow S \le n^2/8$.
The maximum is really $n^2/8$, because if each $k$-segment have $k/n$ written on it, every polygon'll have the sum of the numbers on their sides at most 1, because if the polygon contains the center of the circunference, it's easy to see that the sum is exactly 1, and if not, the sum will be twice the number of points of the minor arc determined by the longest side (includind vértices), and that's at most 1. Now, notice that $A_1=1; A_2=2; ...;A_{n/2-1}=n/2-1$ and $A_{n/2}=n/4$. Therefore, the sum of all numbers is $(n/2-1)(n/2)/2+n/4=n^2/8$, finishing the problem. |
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, they will be inmediately discarded and may no longer be used in subsequent weighings. Determine all $N$ for which the goal is achievable. (There are no limits regarding how many times one may use the balance).
Note: the only difference between genuine and fake coins is their weight; otherwise, they are identical. | 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, because you can't say which coin is false, for $N=2$, and when you can say which coin is false for $N=3$, you find out that the two other coins are genuine, and then you can't use then to determine if the fake coin is lighter or heavier.
Now, we'll prove that for $N=2n+1$, with $n$ a positive integer, it's impossible to find the fake coin and its wheight compared with the other coins. We'll prove this by induction on $n$. The initial case $n=1$ is already solved. Now, suppose that we have $2n+1$ coins, and the weights are distributed in such a way that when we put $a$ coins in a fan of the balance and $b$ coins in the another, with $a>b$, then the fan with $a$ coins will go down. So, with this configuration of wheights, the only useful move is to put a equal number of coins in each fan.
Now, suppose that we put $c<n$ coins in each fan, and suppose that we got equilibrium. Then, all of the $2c$ coins on the balance are genuine, and now we must operate with $2(n-c)+1$ coins in order to discover what we want about the fake coin, but it is impossible by hypothesis. So, we must put first $n$ coins on each fan. If we get the equilibrium, then the only coin outside the balance is the fake coin. However, we can't determine if it is heavier or lighter, because all the $2n$ genuine coins are gone. Thus, we proved that for all $N$ odd it may be impossible to achieve the goal.
Next, we'll prove that for every $N=2n+2$, $n$ positive integer, is solution, by induction on $n$. The first case ($N=4$) is proved as follows: Let $a,b,c,d$ be the four coins. Put $a,b$ in the first fan, and $c,d$ on the second. The balance'll be cleary unbalanced. Suppose that the fan with $a,b$ went up. Now, we remove all the coins and we put $a$ on a fan and $b$ on another. If the balance is unbalanced, then the false coin is $a$ or $b$, which impies that the fake coin is ligther, since the fan with $a,b$ went up on the fist wheight. On the second wheight, if $a$ went up, the $a$ is the fake coin, and it is lighter than the others. Now, suppose that the wheight of $a$ equals to $b$. Then, $a$ and $b$ are genuine coins (so we can't use then anymore), and the fake coin is $c$ or $d$, and it is heavier, because the fan with $c,d$ went down on the first wheight. Now, put $c$ on a fan and $d$ on another, and suppose that $c$ went down. Then, $c$ is the fake coin, and it is heavier than the others. It solves the case $n=1$.
Now, suppose that we have $2n+2$ coins $a_1, a_2, \dots a_{2n+2}$. Put $a_1, \dots a_{n+1}$ on a fan and $a_{n+2}, \dots a_{2n+2}$ on another. Clearly, the balance'll be unbalanced, and we'll suppose that the fan with $a_{n+2}, \dots a_{2n+2}$ went up. Now, compare $a_{2n+1}$ and $a_{2n+2}$. If their weights are equal, they're genuine, and then we can operate only with $a_1, a_2, \dots a_{2n}$, and we can find the fake coin by hypothesis. If, however, $a_{2n+1}$ and $a_{2n+2}$ have different wheights (let us suppose without loss of generality a_{2n+1} lighter), then we conclude that the fake coin is on $a_{n+2}, \dots a_{2n+2}$, and so it is lighter, since the corresponding fan went up. Then, a_{2n+1} is the fake coin and it is lighter the the other. So, by induction, we proved that for all $N$ even, $N>2$, one can always achieve the goal.
Finnaly, the answer is $N=2n+2$, $n \in \mathbb{N}$. |
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 positive integer $n$, we say that $f: \{1,2, \dots, n\} \rightarrow \{1,2, \dots, n\}$ is [i]catracha[/i] if $f^{f(k)}(k)=k$, for every $k=1, 2, \dots n$. Prove that:
(a) If $f$ is catracha, $f$ has at least $\pi (n) -\pi (\sqrt{n}) +1$ fixed points.
(b) If $n \ge 36$, there exists a catracha function $f$ with exactly $ \pi (n) -\pi (\sqrt{n}) + 1$ fixed points. | (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 $f^{f(f^{a}(k))}(f^{a}(k))=f^{a}(k) \Rightarrow f^{f^{a+1}(k)+a}(k)=f^{a}(k)$, so $a_k|(f^{a+1}(k)+a)-a$, or $a_k|f^{a+1}(k)$, for all $a \in \mathbb{Z}, a \ge 0$. Now, if $a=a_k-1$, we have $f^{a+1}(k)=k$, so $a_k|k$, for all $k \in [n]$. In particular, $a_1=1$, so $1$ is a fixed point of $f$.
Let $p> \sqrt{n}$ be a prime number. Because $a_p|p$, we have $a_p = 1$ or $a_p = p$. If $a_p=p$, then the $p$ distinct numbers $p, f(p), f^2(p), \dots, f^{p-1}(p)$ are divisible by $p$, and they're on $[n]$, which is a contradiction, since $p^2>n$. So, $a_p=1$, which implies that $p$ is a fixed point of $f$, for all prime $p> \sqrt{n}$,. Because there are $ \pi (n) - \pi (\sqrt{n})$ such primes, and because $1$ is a fixed point of $f$, we conclude that $f$ has at least $\pi (n) - \pi (\sqrt{n}) +1 $ fixed points.
(b) For each $n \in \mathbb{N}$, we say that a set $A \subset [n]$ is a $\sigma$-set if each element of $A$ is divisible by $|A|$. If we show that $[n]$ can be partitioned in some $\sigma$-sets, with exactly $\pi (n) - \pi (\sqrt{n}) +1$ of them having only one element, then such $f$ does exists, because given a $\sigma$-subset $A= \{ a_1 ,a_2 , \dots, a_{|A|} \} $, we can construct $f$ as $f(a_1)=a_2, f(a_2)=a_3, \dots, f(a_n)=a_1$. So, it remains to prove that such a partition always exists for $n \ge 36$.
Let $S_0=\{ 1 \} \cup \{ q_1 \} \cup \dots \cup \{ q_b \}$, where $q_1, \dots, q_b$ are all the primes between $\sqrt{n}$ and $n$, greater than $\sqrt{n}$. The set $[n]-S_0$ is equal to $L_1 \cup L_2 \cup \dots \cup L_a$, where $a=\pi (\sqrt{n})$, $p_1 < p_2 < \dots < p_a$ are all the primes from 1 to $\sqrt{n}$ and $L_k$ is the set of all numbers in $[n]$ such that the least prime divisor of it is equal to $p_k$, for each $k=1,2, \dots , a$.
Now, we'll run the algoritim below:
Step 1: For $t=a, a-1, \dots, 1$, let $d_t=k_t . p_t + r_t$ be the number of multiples of $p_t$ in $[n]-S_{a-t}$. We'll construct the $\sigma$-sets $ \{ a_1, a_2 ,..., a_{p_t} \}, \dots, \{ a_{p_t(k_t-1)+1}, a_{p_t(k_t-1)+2} ,..., a_{p_t.k_t} \}$ as the following:
$\{ a_1, \dots , a_{|L_t \cap ([n]-S_{a-t})|} \} = L_t \cap ([n]-S_{a-t})$, and $a_1 > \dots > a_{|L_t \cap ([n]-S_{a-t})|}$;
$\{ a_{|L_t \cap ([n]-S_{a-t})|+1}, \dots , a_{k_t . p_t} \} \rightarrow (([n]-S_{a-t})\cap \{p_t, 2p_t, \dots \}$ and $a_{|L_t \cap ([n]-S_{a-t})|+1}> \dots > a_{k_a . p_a}$, where the $\rightarrow$ means that we take all we can from $(([n]-S_{a-t})\cap \{p_t, 2p_t, \dots \}$ in a decreasing order.
Now, let $S_{a-t+1}= S_{a-t}\cup \{ a_1, a_2 ,..., a_{p_a} \} \cup \dots \cup \{ a_{p_a(k_a-1)+1}, a_{p_a(k_a-1)+2} ,..., a_{p_a.k_a} \}$.
Step 2: If you ran the preceding step to $t$, run it to $t-1$ (remember to start on $t=a$).
When we do the step 1 for some $t$ on the algorithm, we take all the numbers on $[n]$ in which the least prime factor is $p_t$ to create $S_{a-t+1}$ (and some other numbers divisible by $p_t$), since we take all numbers in $L_t$, because the numbers in $S_{a-t}$ that belong to $L_t$ were taken before. Also, note that $\{ a_1,a_2, \dots, a_{|L_t \cap ([x]-S_{a-t}| \}}$ is NOT a $\sigma$-subset (we take all we can from $L_t \cap ([x]-S_{a-t})$), so, at first, we can always take all the elements from $L_t \cap ([x]-S_{a-t})$ (if $L_t \cap ([x]-S_{a-t})= \emptyset$, we take nothing, for example), unless the number of elements in $([x]-S_{a-t})$ is grater than the number of elements we want to take, that is, $p_t.k_t$.
It only remains to prove that there's at most $p_t.k_t$ elements in $L_t \cap ([x]-S_{a-t})$, in order to take all the numbers in $L_t$. We can do it by induction (making $t=a,a-1,\dots, 2$). First, observe that $S_{a-t} -S_0$ have only numbers divisible by $p_a, \dots, p_{t+1}$, so if the the numbers $2p_t , 3p_t , \dots , (p_t-1)p_t, (p_t+1)p_t$ are in [n], they aren't in $S_{a-t} -S_0$, and clearly they aren't on $S_0$, so they aren't on $S_{a-t}$. Also, for $p_t$ odd, they aren't in $L_t$. So, because $ L_t \subset \{ p_t, 2p_t, \dots \}$ and $\{ p_t, 2p_t, \dots \} - L_t$ has at least $p_t-1 \ge r_t$ elements, $L_t$ has at most $p_t.k_t$ elements, as we whised. Observe that, for $p_1=2$, this can't be truth. Because of that, we have that $S_a$ can be $[n]$ or $[n]- \{ 2 \}$, since we took the numbers in a decreasing order.
Edited: It may be possible that $2p_t , 3p_t , \dots , (p_t-1)p_t$ are in $[n]$, but $(p_t+1)p_t$ isn't in $[n]$. If this is the case, we have that, by writing $n=q^2+r$, with $q,r \ge 0 $ integers such that $0 \le r \le 2q$, $p_t \le q$ (by choice) and $p_t > q - 1$ (because $p_t(p_t+1)>n>q(q-1)$. Thus, $p_t=q$, which implies that $n<p_t(p_t+1)=q^2+q$, or $r<q$. So, the only numbers in $[n]$ that are divisible by $p_t$ are $p_t, 2p_t, \dots, p_t^2$, and that means that the only numbers in $L_t \cap ([x]-S_{a-t})$ are $p_t$ and $p_t^2$, so $|L_t \cap ([x]-S_{a-t})|=2$ and that's at most $p_tk_t$ elements anyway.
At the end, we obtain a set $S_a$ that can be written as a partiton of $\sigma$-subsets, and it can be equal to $[n]$ or $[n]- \{ 2 \}$, since in the last step of our algorithm, the $\sigma$-subsets to be attached on $S_{a-1}$ has 2 elements each, so at the end we can attach all the remaining numbers ou we can't only attach 2, since 2 is the least number divisible by $p_1=2$ and we construct the $\sigma$-subsets in decreasing order. If $S_a=[n]$, then we reached our goal. else, let's see more...
Let $\{ a, b, c \}$ be the last $\sigma$-set created on the penultimate step of the algorithm. Because $n \ge 36$, the 5 least numbers to enter in the $\sigma$-subsets in the penultimete step are $36, 24, 18, 12, 6$, since those five numbers are the least ones that are divisible by 3 and that belong to $(([n]-S_{a-2})\cap \{3, 6, \dots \}$. Hence, we have that $\{ a, b, c \} = \{36, 24, 18 \}$, if $r_2=2$; $\{ a, b, c \} = \{24, 18, 12 \}$, if $r_2=1$; $\{ a, b, c \} = \{18, 12, 6 \}$, if $r_2=0$. In each case, if we replace $\{ a, b, c \}$ and $ \{ 2 \}$ by $ \{ a, 2 \}$ and $\{ b, c \}$, we have now a partition of $[n]$ in $\sigma$-subsets. Finnaly, the problem is solved! |
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 even,implying d(6)=2d(4),so d(3) = 13 ,d(4) = 19,d(5)= 26 and d(6) = 38.The minimum n for which these are satisfied is 494 |
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 construct an increasing sequence $(x_k)_{k=1}^\infty$ of positive integers for which ${\rm gcd}(f(x_k),f(x_\ell))=1$ for any $k\neq \ell$. Set $x_1=a$ and $x_2=b$. Now, let $\mathcal{P}_2=\{p\text{ prime }:\exists i \in\{1,2\}, p\mid f(x_i)\}$. For any prime $p\in \mathcal{P}_2$, let $n_p$ be an integer such that $f(n_p)\not\equiv 0\pmod{p}$. Now, select $x_3\equiv n_p\pmod{p}$ for every $p\in\mathcal{P}_2$ (the existence of such an $x_3$ is well-justified due to Chinese remainder theorem). In particular, assuming $x_1,\dots,x_{n-1}$ is constructed, set $\mathcal{P}_{n-1}$ set of all primes dividing one of $f(x_i)$ with $1\leqslant I\leqslant n-1$. Then, for any $p\in\mathcal{P}_{n-1}$, let $n_p$ be such that $f(n_p)\not\equiv 0\pmod{p}$. Then setting $x_n\equiv n_p\pmod{p}$ for every $p\in\mathcal{P}_{n-1}$ ensures $f(x_n)$ is coprime to all $f(x_1),\dots,f(x_{n-1})$. This completes the argument. |
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 $\angle BAC = \angle ABC = 30^{\circ}$. |
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 $(x+5y+2z)^2=(x^2+25y^2+4z^2)+(10xy+20yz+4xz)>4(xy+yz+zx)$. |
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 $k$. Also notice that if $f(n) \leq 2014$, we get a direct contradiction, as by statement, $f(f(n)+m)-f(m+2014)=n$, and for large $m$, the order is ascending, thus $n<0$.
Step 5- For large $k$, minimum difference between $f(k)$ and $f(k+2014-f(n))$ would be $f(n)-2014$, as our dey is ascending with integer steps. Thus $n \geq f(n)-2014=>n +2014 \geq f(n)$. But we know $f(n) \geq n+2014$ by ordering for $n>2014$, thus $f(n)=n+2014$ for $n>2014$.
Step 6- $P(m,n)$, for large $m$ and $n<2014$:
As for large inputs, we know the value of $f(m)$, thus $n=f(n)+m-m-2014 => f(n)=n+2014$ for all $n$. |
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 \( m_0 \ge n_0 \).
Let \( k = 2^{k_0} k_1 \) be the order of 2 modulo \( n \) (here \( k_0 \) is a non-negative integer, and \( k_1 \) is an odd positive integer). From \( k \mid \varphi(n) \), it follows that \( k_0 \le n_0 \). From \( 2^{\varphi(m)} \equiv -1 \pmod{n} \), it follows that \( 2^{2\varphi(m)} \equiv 1 \pmod{n} \), and hence \( k \mid 2\varphi(m) \), i.e., \( 2^{k_0} k_1 \mid 2^{m_0 + 1} m_1 \). Moreover, \( n \nmid 2^{\varphi(m)} - 1 \), and therefore \( k \nmid \varphi(m) \), i.e., \( 2^{k_0} k_1 \nmid 2^{m_0} m_1 \). This can happen only if \( k_0 = m_0 + 1 \).
We deduce \( m_0 \ge n_0 \ge k_0 = m_0 + 1 \), a contradiction. Therefore, at least one of the numbers \( m \) and \( n \) is equal to 1, and now it directly follows that solutions are \( (1,1), (1,3) \), \( (3,1) \). |
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. |
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 another lattice point (this happens because adding a number between $ 0 $ and $ \sqrt 2, $ the circle passes the vertex of an unit square which has diagonal $ \sqrt 2 $). Having in mind that no two points are equidistant from $ P, $ we´ve obtained a circle containing exactly $ n+1 $ points.
By induction, we're done. |
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 $6x^2+2x+1996$ with finite applications of the upper operations?
| Discriminant is invariant, so no. |
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 not have a mutual friend, contradiction.\\
2) If $A_1$ has two friends (let these be $A_2$ and $A_3$), then the common friend of $A_1$ and $A_2$ can only be $A_3$ and therefore $A_2$ and $A_3$ are friends. Then each of $A_2$ and $A_3$ have at most two friends out of $A_4,\dots, A_{12}$. This means that someone among $A_4,\dots, A_{12}$ (let it be $A_4$) is not known by $A_2$ and $A_3$. Now $A_1$ and $A_4$ do not have a common friend, contradiction.\\
3) Suppose $A_1$ has three friends, let them be $A_2, A_3$ and $A_4$. If among $A_2, A_3$ and $A_4$ there are only two friends (let it be $A_2$ and $A_3$), then $A_1$ and $A_4$ do not have a common friend, contradiction. Therefore, there are at least two pairs of friends between $A_2, A_3$ and $A_4$. Each of $A_2, A_3$ and $A_4$ has at most 4 friends, a total of at most 12 friends. That means the friends of $A_2, A_3$ or $A_4$ among the other 8 competitors $A_5,\dots, A_{12}$ are at most $12 -3 -2\cdot 2 = 5$ (12 friends without the three friendships with $A_1$ and without the two pairs of friends between $A_2, A_3$ and $A_4$, each of which counted twice). Hence someone among $A_5,\dots,A_{12}$ (let it be $A_5$) is not a friend with any of $A_2, A_3$ and $A_4$ and then $A_1$ and $A_5$ do not have a common friend, contradiction.
The argument above implies everyone has exactly $4$ friends.
[b]Lemma[/b]. Each competitor has exactly four friends, and among them there exist two non-intersecting pairs of friendship.\\
[b]Proof[/b]. It suffices to prove the lemma for $A_1$. Let the friends of $A_1$ be $A_2, A_3, A_4$ and $A_5$. Suppose there exist three pairs of friends between $A_2, A_3, A_4$ and $A_5$. Then the number of students among $A_6,\dots, A_{12}$ which are friends with at least one of $A_2, A_3, A_4,A_5$ are at most $4\cdot 4- 4 -3\cdot 2 = 6$. This means that there is a competitor (let it be $A_6$) among $A_6,A_7,\dots,A_{12}$ who is not friends with $A_2, A_3, A_4$ and $A_5$ and then $A_1$ and $A_6$ have no common friend, contradiction. Therefore there are at most two pairs of friends among $A_2, A_3, A_4, A_5$. Since each of $A_2, A_3, A_4,A_5$ has a friend among the rest tree of them, WLOG the pairs of friends are $A_2,A_3$ and $A_4,A_5$. $\blacksquare$
Further, each of $A_2, A_3, A_4,A_5$ has two friends among $A_6,\dots,A_{12}$ and each of $A_6,\dots,A_{12}$ must have a friend among $A_2, A_3, A_4,A_5$. Then, exactly one of $A_6,\dots,A_{12}$ (let it be $A_6$) has two friends among $A_2, A_3, A_4,A_5$. If $A_6$ is a friend with $A_2$ and $A_3$ (respectively with $A_4$ and $A_5$), we get a contradiction with the Lemma applied to $A_2$ (respectively to $A_4$). Thus, WLOG we can consider that the friendships among $A_2, A_3, A_4$ and $A_5$ are: $A_2$ is a friend with $A_6$ and $A_7$; $A_3$ is friends with $A_8$ and $A_9$; $A_4$ is friends with $A_9$ and $A_{10}$; $A_5$ is friends with $A_{11}$ and $A_{12}$. The common friend of $A_4$ and $A_6$, as well as that of $A_4$ and $A_7$ can only be $A_{10}$. Therefore, $A_{10}$ is friend with $A_6$ and $A_7$, which contradicts the Lemma applied to $A_6$. |
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. |
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 $AD, BE$, and $CQ$ are concurrent. | 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. |
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)
$(2) \;\; y_0^3 \equiv 2^{x_0} = 2^{3q+r} = 8^q \cdot 2^r \equiv 1^q \cdot 2^r = 2^r \equiv 1,2,4 \pmod{7}$.
We know that $n^3 \equiv 0,1,6 \pmod{7}$ for all integers $n$, which combined with (2) implies $2^r \equiv 1 \pmod{7}$, i.e. $r=0$. Hence $x_0=3q$ with $q>0$, which inserted in (1) result in
$(3) \;\; (2^q + 21^q)(2^{2q} - 42^q + 21^{2q}) = y_0^3$.
Let $d = GCD(2^q + 21^q, 2^{2q} - 42^q + 21^{2q})$. Then $d | 3 \cdot 2^q$ and $d | 3 \cdot 21^q$, yielding $d | 3 \cdot GCD(2^q,21^q) = 3 \cdot 1 = 3$, i.e. $d \in \{1,3\}$. If $d=3$, then $3 | 2^q + 21^q$, which means $3 | 2^q$. Hence $d \neq 3$ by contradiction. Therefore $d=1$, which according to (3) give us
$(4) \;\; (2^q + 21^q, 2^{2q} - 42^q + 21^{2q}) = (a^3,b^3)$,
where $a$ and $b$ are two coprime natural numbers s.t. $ab=y_0$. From (4) we have
$(5) \;\; 2^q + 21^q = a^3$.
The fact that $x_0 = 3q + r \geq 3q$ means (since $x_0>0$) ${\textstyle q \leq \frac{x_0}{3} < x_0}$. In other words, $(x,y) = (q,a)$ is a solution of equation (1) where $q<x_0$, contradicting the minimality of $x_0$. Consequently equation (1) has no solutions in natural numbers. [b]q.e.d[/b] |
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 $f(2014)=(2014-1)/3=671$. |
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 $\frac{(1+x^{2n+1})(1-x^{2n+1})}{(1-x)(1+x)}=\frac{1-x^{4n+2}}{1-x^2}$ after simplifying the terms and the identity is now trivial. This identity implies that the term $1-x+x^2-.....+x^{2n}$ is rational. From the fact that $P(x)$ is rational, we can get that $x+x^3+...x^{2n-1}$ is rational by subtracting the two terms. Now $P(x)-1=x+x^2+x^3+...x^{2n}=(1+x)(x+x^3+....x^{2n-1})$ is rational. We can see that this immediately implies that $1+x$ and hence $x$ is rational. Proved. |
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]. |
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 notation $A_{3n+1} = A_{1}$ and $A_{3n+2} = A_{2}$. Prove that the number of self-intersections of the broken line is at most $\frac{3}{2}n^{2} - 2n + 1$
| 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 ordering of the indices of their vertices). Let $A_1,\dots,A_n$ and $B_1,\dots,B_n$ be the resulting lines from two different directions. Then each $A_i$ is linked to some $B_{\pi(i)}$ i.e. we can describe the linking by a permutation $\pi$. It is then easily seen that a line $A_i$ intersects a line $B_{\pi(j)}$ if and only if $i<j$ and $\pi(i)>\pi(j)$. So the number of intersections is just $\text{inv}(\pi)$, the number of inversions of $\pi$. If we denote by $\pi_1,\pi_2,\pi_3$ the three appearing permutations, the total number of intersections is just $\text{inv}(\pi_1)+\text{inv}(\pi_2)+\text{inv}(\pi_3)$. A priori, this could be as large as $3 \cdot \frac{n(n-1)}{2}$ since of course each permutation has $\text{inv}(\pi) \le \frac{n(n-1)}{2}$. This is already pretty close to what we want to show but still too large.
The following picture shows an example for $n=4$ where equality is achieved. What's the problem here? This is not a connected line! And looking at our argument we haven't yet used that the line should be connected. This is easily seen to amount to the condition that $\pi_1 \circ \pi_2 \circ \pi_3$ is a permutation with just one cycle of length $n$.
So we really need to show that whenever $\pi_1 \circ \pi_2 \circ \pi_3$ is a permutation with just one cycle of length $n$ we have $\text{inv}(\pi_1)+\text{inv}(\pi_2)+\text{inv}(\pi_3) \le \frac{3}{2} n^2-2n+1$. Of course at this point it is not at all clear that this is a good reformulation of the problem.
To get an intution for the problem, let us consider the toy example where two of the permutations have the maximal number of inversions i.e. $\pi_1=\pi_2=\tau$ where $\tau(k)=n+1-k$ is the permutation with $\text{inv}(\tau)=\frac{n(n-1)}{2}$. Since $\tau^2=1$, the condition is just that $\pi_3=\pi$ has one cycle and we need to prove that $\text{inv}(\pi) \le \frac{n(n-1)}{2}-\frac{n}{2}+1$.
How could we possibly prove such a bound? Well, a permutation with a lot of inversions should be close to $\tau$. Indeed, if $\text{inv}(\pi)=\frac{n(n-1)}{2}-k$ then $\pi \circ\tau$ has exactly $k$ inversions and hence can be written as a product of $k$ transpositions. So $\pi$ can be written as $\tau \alpha_1\dots \alpha_k$. On the other hand, composing with a transposition alters the number of cycles by exactly $1$: If the two transposed elements are in the same cycle, then this one is broken up into two cycles and if they are in different cycles, the two are merged together.
In particular, the number of cycles can only drop by one when composing with some transposition. Since $\tau$ has at least $\frac{n}{2}$ cycles, we see that $\pi$ has at least $\frac{n}{2}-k$ cycles. On the other hand, it has one cycle by assumption so that $\frac{n}{2}-k \le 1$ and the claim follows.
So we have solved our toy problem, but can we use this in the general case? Writing $\text{inv}(\pi_1)=\frac{n(n-1)}{2}-k, \text{inv}(\pi_2)=\frac{n(n-1)}{2}-\ell$ and $\text{inv}(\pi_3)=\frac{n(n-1)}{2}-m$ we need to prove that $k+m+n \ge \frac{n}{2}-1$. As before, we can certainly write $\pi_1=\tau \alpha_1\dots \alpha_k, \pi_2=\tau \beta_1\dots \beta_{\ell}$ and $\pi_3=\tau \gamma_1\dots \gamma_m$ for transpositions $\alpha_i,\beta_i,\gamma_i$. So the condition is that
\[\tau \alpha_1\dots \alpha_k\tau \beta_1\dots \beta_{\ell}\tau \gamma_1\dots \gamma_m\]
has only one cycle. As before, this implies that
\[\tau \alpha_1\dots \alpha_k\tau \beta_1\dots \beta_{\ell}\tau\]
has at least $m+1$ cycles. However, here it seems that we are stuck since there is a $\tau$ at both sides and we cannot cut off any more transpositions. However, we have already noted that $\tau=\tau^{-1}$ and of course conjugation by any permutation does not change the cycle type. So actually we can conclude that
\[\alpha_1\dots \alpha_k\tau \beta_1\dots \beta_{\ell}\]
has at most $m+1$ cycles and then we can proceed as before to see that $\tau$ has at most $k+\ell+m+1$ cycles from where the claim follows as before.
[/hide] |
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 must have $x=z=\frac{a}{2}$.
However, $y$ will only be unique if $a^2=4b$, so the solution set is $(a,b)=(2t,t^2)$ where $t\in\mathbb{R}$. |
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 $\alpha\geq\frac{3}{2}$.
Thus since the limit is $0$ when $\alpha>\frac{3}{2}$. The only pair with $\beta>0$ is $(\alpha,\beta)=\bigg(\frac{3}{2},\frac{\sqrt{2}}{3}\bigg)$ |
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 down and intersect the abscisse axe in the points $x_1<x_2<\dots<x_{n-1}$. There are some cases that have to be considered. For example let $n-1$ is even. Consider first the case $P$ is positive at the left side of $x_1$. In this case it must be also positive at the right side of $x_{n-1}$.
Now, take $a_{n}>0$ small enough so that $P' := a_nx^n+P(x)$ don't disturb too much the sign of $P$ in the interval $[x_1-\epsilon, x_{n-1}+\epsilon]$. That's, $P'$ will have $n-1$ different roots in that interval. Also $P(x)$ would be positive slightly to the left of $x_1$. But since $n$ is odd and $a_n x^n$ goes to $-\infty$ as $x\to -\infty$ with a higher rate than any polynomial of degree $<n$, it follows $\lim_{x\to-\infty}P(x)=-\infty$. Hence, $P(x)$ have to have an additional root $x', x'<x_1$.
The case $P$ is negative at the left side of $x_1$ goes in the similar way.
Considering all possible polynomials $P$ (there are 2^n of them), we can choose small enough positive $a_n$ appropriate for all of them.
The case $n-1$ is odd is similar. That's all. |
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 $p_1>1$, we know that one of the other prime factors needs to divide $p_1+1$. Clearly, $p_1+1<p_3,p_4,\cdot\cdot\cdot,p_k$. The only way for $p_2$ to divide $p_1+1$ is if $p_2=p_1+1$ and only one solution could exist for this since they are different parities, $(p_1,p_2)=(2,3)$. We have $p_1=2$ and $p_2=3$. Since both $2$ and $3$ are prime factors, we know that $6|n$. Now consider the following inequality, $$s(n)=2n>n+\frac{n}{2}+\frac{n}{3}+\frac{n}{6}+1=2n+1$$, a contradiction.[/hide] |
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 Matrin Niepel, Comenius University, Bratislava) | 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 chain:
\[\operatorname{tr}(A^2)=\sum \lambda_{i}^2 = \sum a_{ii}^2 +2\sum_{i<j} a_{ij}^2\]
The equality holds only in case $A$ is a diagonal matrix. |
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 integral. That gives us
\[f^{[n]}(x)=\int_0^1t^ng(tx)\,dt\] where $g(u)=\pm \sin u$ or $\pm \cos u.$ In particular, $|g(tx)|\le 1.$ and, as long as $x>0,$ $|g(tx)|$ is a continuous function that is not identically equal to $1.$ Hence we may conclude that
\[|f^{[n]}(x)|<\int_9^1t^n\cdot 1\,dt=\frac1{n+1}.\]
We can extend this to the whole real line, including $x=0,$ by defining $f(0)=1.$ The integral representation remains correct, but, once we include $x=0,$ we must modify the conclusion to $\[|f^{[n]}(x)|\le\frac1{n+1}.$
(Key words for the proof: Fourier transform.) |
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 $1\le i\le k$ and $x_j\neq j$ for every $1\le j\le n$. Prove that
$$\Delta (n,k)=\sum_{i=0}^{k=1} \binom{k-1}{i} \frac{D_{(n+1)-(k+i)}}{n-(k+i)}$$
(Proposed by Combinatorics; Ferdowsi University of Mashhad, Iran; Mirzavaziri) | 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
\[\sum_j \binom{n-2k}{j}\Delta(n-j,k)=(n-k)! \quad (1)\]
Note that by plugging in $n=2k$, then $n=2k+1$ etc. we can compute the $\Delta(n,k)$ recursively only from this equation, so this equation (holding for all choices of $n,k$) determines the $\Delta(n,k)$ uniquely. In particular, it suffices to prove that the RHS of the claimed identity also satisfies that, i.e. we are left to prove that
\[\sum_j \binom{n-2k}{j} \sum_i \binom{k-1}{i} \frac{D_{n-k-i-j-1}}{n-k-i-j}=(n-k)!\]
But using Vandermonde, the LHS can be easily rearranged to
\[\sum_s \frac{D_{n-k-s-1}}{n-k-s} \sum_{i+j=s} \binom{n-2k}{j} \binom{k-1}{i}=\sum_s \frac{D_{n-k-s-1}}{n-k-s} \binom{n-k-1}{s}=\sum_j \frac{D_{j+2}}{j+1} \binom{n-k-1}{j}\]
and hence we are left to prove
\[\sum_j \frac{D_{j+2}}{j+1} \binom{m}{j}=(m+1)!\]
or equivalently
\[\sum_j D_{j+1} \cdot \binom{m}{j}=m \cdot m!\]
for all $m$.
But we have $\binom{m}{j}=\binom{m+1}{j+1}-\binom{m}{j+1}$ and $m \cdot m!=(m+1)!-m!$, so it suffices to prove
\[\sum_j D_j \binom{m}{j}=m!\]
which is true by a similar counting argument as above ($\binom{m}{j}D_{n-j}$ is the number of permutations in $S_n$ with exactly $j$ fixed points) or by just plugging in $k=0$ in (1) and noting that $\Delta(n,0)=D_n$. Done.
[i]Wow, this was surprisingly straightforward for a Problem 10 in IMC...[/i] |
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 \[
\angle CHS - \angle CSB = 90^{\circ}, \quad \angle THC - \angle DTC = 90^{\circ}. \] Prove that line $BD$ is tangent to the circumcircle of triangle $TSH$. | 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 diameter $AB, AD$ respectively, where $M_D, M_B$ are the midpoints of $AB, AD$. Then $ST$ is parallel to $BD$.
Indeed, this is true for any $C$ on the circle $\omega$ through $A,H$ orthogonal to $(ABD)$; the center $O_A$ of $\omega$ lies on $M_B M_D$ and $AO_A^2= O_A M_B \cdot O_A M_D$, so that $\omega$ is an $M_D, M_B$ -Apollonius circle of ratio $AB/AD$, which is then the same as $CM_D/CM_B= CS/CT$, as desired. |
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 is a $k \times k$ square which does not contain a rook on any of its $k^2$ unit squares. | 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 $k$ disjoint $k\times k$ squares, made by the rows in $I$ and the columns in $J=\{0,1,\ldots, k^2-1\}$, and only at most $k-1$ rooks that may belong to them, hence one of these squares contains no rook.
A counter-model for $n=m^2$ is given by rooks on positions $(mi+j,mj+i)$ for $0\leq i,j\leq m-1$. An immediate check shows there is no $m\times m$ square empty of rooks. And for any $n'\times n'$ sub-table with $m\leq n'< n$, a fortiori there exists no $m\times m$ square empty of rooks (since the $n'\times n'$ table may always be completed, need be, in order to become peaceful). |
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:
\[ a_0 + \Delta_1 + \Delta_2 + ... + \Delta_n < \frac{(n+1)a_0 + n\Delta_1 +(n-1)\Delta_2+ ... + \Delta_n}{n} \le \]
\[ \le a_0+\Delta_1 + ... +\Delta_n + \Delta_{n+1} \]
After a little calculation, one can have that $\Delta_2 +2\Delta_3 +...+(n-1)\Delta_n < a_0 \le \Delta_2 +2\Delta_3 +...+(n-1)\Delta_n +n\Delta_{n+1}$
Now, defining $f: \mathbb{N} \rightarrow \mathbb{Z}$ by $f(1) = 0$ and $f(n)=\Delta_2 + 2\Delta_3 +.. +(n-1)\Delta_n$, for all $n \ge 2$, we see that $f$ is increasing and unbounded ($f(n+1)-f(n)=n\Delta_{n+1} \ge n$).
Now, it remains to prove that exists a unique $n\in \mathbb{N}$ such that $f(n)<a_0<f(n+1)$ This is clear, since $f(1)=0<a_0$ and $f$ is estrictely increasing and unbounded. Problem Solved! |
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. |
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 amount of the limit does [b][u]not[/u][/b] depend on choosing $x$. | 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 $m$. Then $\frac{a_n}{n}$ is less than $c.\frac{a_m}{m} + \frac{a_r}{n}$ where $r \leq m-1$ and $ c \leq 1$. Just fix $m$ and take $n$ to infinity
So we're done by the lemma since by the first line ,the given sequence satisfies the condition of lemma.
For part b, just write triangle inequality like this:
$$d(x,y)+d(x,y) = d(x,y) + d(f^n(x),f^n(y)) \geq d(x,f^n(x))-d(y,f^n(y))$$
But since this is symmetric on $(x,y)$ , we in fact have $$2d(x,y) \geq max\{ d(x,f^n(x))-d(y,f^n(y) , d(y,f^n(y))-d(x,f^n(x)) \}$$Finally by dividing by $n$ and taking $n$ to infinity, we're done since in the recent inequality, after taking $n$ to infinity $0 = LHS \geq RHS \geq 0$ |
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