New answers tagged

10

Due to symmetry, assume wlog that $a\geqslant b\geqslant c\geqslant d\geqslant 0$. Since I will use this fact later, I am proving it before the cases: Lemma: We have that $2\geqslant a+d$. In fact, this follows from Cauchy-Schwarz: $$2= 2\sqrt{\frac{a^2+b^2+c^2+d^2}{3}}\geqslant 2\sqrt{\frac{a^2+3d^2}{3}}=\sqrt{1+\frac13}\cdot \sqrt{a^2+3d^2}\geqslant a+d$...


3

Since $a^2+b^2+c^2+d^2 = 3$ defines a compact set in $\mathbb{R}^4$, $f(a,b,c,d)=abcd+3-a-b-c-d$ will have a global minimum and maximum over that set, that will occur (can be easily justified) in critical points of the Lagrangian $$ L(a,b,c,d,\lambda) = abcd+3-a-b-c-d -\lambda(a^2+b^2+c^2+d^2-3) $$ So, just compute the critical points, compute the value of $...


1

For any $a, b > 0$, $x=a, y=-b$ gives $\max(f(a+b), f(a-b)) | \min(af(-b)+bf(a), -ab)$. Since $f$ takes positive values, $\max(f(a + b), f(a-b)) | ab$. For $a=b=1$ this shows that $f(0)=f(2)=1$. For $a=b$ this shows that $f(2a) | a^2$, and for $b=1$ this shows that $\max(f(a+1), f(a-1))|a$. Combined together these imply that $f(x) = 1$ for all even $x \...


2

Elementary solution: Suppose $n=t-1$, putting in we get: $A=n^3-3n+1=t^3-3t^2+3$ For $t=3m$ we get: $A=27m^3-27+3=3[9(m^3-m^2)+1]$ We can assume $m^3-m^2=k$ so $9k+1|A$ Then $n=3m-1$ For example: $m=2$, $\rightarrow:$, $p=37$, $n=5$ $m=3$, $\rightarrow:$, $p=163$, $n=8$ $m=4$, $\rightarrow:$, $p=433$, $n=11$ Hence values of n make an arithmetic progression .


5

Hint: $ x^3 - 3x + 1 $ is the minimal polynomial of $ \zeta_9 + \zeta_9^{-1} $ over $ \mathbf Q $, and if $ p \equiv 1 \pmod{9} $ then there is an element of order $ 9 $ in $ (\mathbf Z/p \mathbf Z)^{\times} $.


0

A direct application of C-S and AM-GM works here. We have: $$\left(\sum_{\text{cyc}} \dfrac{a^3}{b+c+d} \right) \left(\sum_{\text{cyc}} a(b+c+d)\right) \geq \left(\sum_{\text{cyc} }a^2 \right)^2.$$ It thus suffices to prove that $$\dfrac{\left(\displaystyle \sum_{\text{cyc} }a^2 \right)^2}{\left(\displaystyle \sum_{\text{cyc}} a(b+c+d)\right) } \geq \dfrac{1}...


1

$(a-d)(b-c)>0$ implies $ab+cd>ac+bd$ implies $ab+cd+bc+ad>ac+bd+bc+ad$ implies $(a+c)(b+d)>(a+b)(c+d)$ implies $(a+c)(c+b)(b+d)>(a+b)(b+c)(c+d)$ This means when we perform a transpose action, the consecutive product of pairwise sums strictly increases. Let the numbers be $x_1,x_2,x_3,...,x_n$. We can deduce that the product $(x_1+x_2)(x_2+x_3)(...


0

WLOG, assume $abcd = 1$. The desired inequality is written as $$\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)} + 2 \ge ab + bc + cd + da + ac + bd.$$ Since $ab + bc + cd + da + ac + bd \ge 6$, it suffices to prove that $$(a^2+1)(b^2+1)(c^2+1)(d^2+1) - (ab + bc + cd + da + ac + bd - 2)^2 \ge 0.$$ Using $d = \frac{1}{abc}$, we have $$\mathrm{LHS} = \frac{(ab - 1)^2(bc-1)^...


1

So first, let me say that it is quite easy to see geometrically/show using Lagrange multipliers, that the maximum of $\sum a_i^2$ on the simplex $\{\sum a_i = 1 : a_i \ge 0\}$ occurs when all but one of the variables are $0$. The Lagrange condition $\lambda \nabla f = \nabla g$ is $2\lambda a_i = 1$ so either $a_i = 0$ or $a_i = \frac{1}{2\lambda}$ (all the ...


3

This is a special case of Lagrange's identity: $$ \Big(\sum_{k=1}^n a_k^2\Big)\Big(\sum_{k=1}^n b_k^2\Big) - \Big(\sum_{k=1}^n a_k b_k\Big)^2 = \sum_{k=1}^{n-1}\sum_{j=i+1}^n (a_ib_j-a_jb_i)^2 \tag{1} $$ where all of the $\,b_k=1\,$ which results in $$ n \Big(\sum_{k=1}^n a_k^2\Big) - \Big(\sum_{k=1}^n a_k\Big)^2 = \sum_{1\le i<j\le n} (a_i-a_j)^...


0

I'm posting this as a community wiki so that this question doesn't appear in the unanswered queue. As verified by @HansEngler in the comments, the last paragraph of my proof can be summarized by noting that: All monotonic sequences in $\Bbb N$ are either non-decreasing and unbounded or eventually constant.


2

If you sum all equations, you will get $\sum_{k=1}^n x_k^2+(a-1)x_k+(\frac{a-1}{2})^2=\sum_{k=1}^n (x_k+\frac{a-1}{2})^2=0$, from where we get $x_k=\frac{1-a}{2}$ for $k=1,\dots n$.


1

My second solution: Remark: Using the Buffalo Way as well, but it is simpler. Use @chenbai's substitutions $x=\frac{\sqrt{3}}{a}, y=\frac{\sqrt{3}}{b}, z=\frac{\sqrt{3}}{c}$. We have $xy+yz+zx=3$. We need to prove that $$\frac{xy}{x + 7y}+\frac{yz}{y + 7z}+\frac{zx}{z+7x}\leq\frac{3}{8}.$$ After homogenization, it suffices to prove that $$\Big(\frac{xy}{x + ...


2

Observe that, since $6.25$ is the length of the longest chord(diameter) of the circle, $a$, $b$ and $c$ will belong to $\{1,2,3,4,5,6\}$. Also, notice that, $25$ will divide $abc$. Then, at least $2$ of these terms will be equal to $5$($\Delta $ will be rational ).Let $a=5$ and $b=5$ without loss of generality. Now, just plug in those values into the ...


1

Bashing is usually a way of using brute force to answer a question. In most cases, it doesn't require any mathematical knowledge of theorems and techniques. The way to go in basing is usually just trial and error but for higher efficiency, you can use theorems to set upper and lower bounds for your answer. This will increase efficiency while bashing. The ...


6

Your idea is very good and we will use that :) As $a+b+c=0$ we can find a pair such that they have the same sign hence WLOG $ab\ge 0$. Now from your attempt we have to prove $$\Leftrightarrow \dfrac{(a+2)^2}{a^2+8}+\dfrac{(b+2)^2}{b^2+8}+\dfrac{(c+2)^2}{c^2+8} \geq \dfrac{3}{2}$$ by C-S $$\dfrac{(a+2)^2}{a^2+8}+\dfrac{(b+2)^2}{b^2+8}\ge \frac{{(a+b+4)}^2}{a^...


1

No : $501$ cannot be removed before $500$ so there is a losing position when faced with $500,501,999,1000$ remaining


2

We know that for each $x\in\mathbb R$, either $f(x)=x^2$ or $f(x)=0$. Let $A=\{x\neq0\mid f(x)=0\}$ and $B=\{x\neq0\mid f(x)=x^2\}$. Then $A$, $B$ partition $\mathbb R\setminus\{0\}$. Suppose both are non-empty. Take $a\in A$ and $b\in B$. Sub $x=a,y=-b$ to get $f(a^2-b)=b^2$. Since $b\neq0$, we must have $b^2=(a^2-b)^2\implies a^2=2b$ (as $a\neq0$). This ...


-1

Hint Take \begin{equation} y = \frac{f(x)-x^2}{2} \end{equation} This leads to $\forall x, f(x)\in \{0, x^2\}$, hence we can write $f(x) = x^21_x$ where $\forall x\in{\mathbb R}, 1_x\in\{0,1\}$. Let us rewrite the functional equation as \begin{equation} y^2(1_{x^2+y}-1_{x^2 1_x - y})+2x^2 y(1_{x^2+y}+1_{x^2 1_x-y}-2. 1_x) + x^4(1_{x^2+y}-1_{x^2 1_x-y}) = 0 \...


0

The naive Titu also works WTS $$ \sum_{cyc} \frac{a^3}{b+c+d} = \sum_{cyc} \frac{ a^4 } { ab+ac+ad } \geq \frac{ (\sum a^2 )^2 } { 2 ( ab+ac+ad+bc+bd+da) } \geq \frac{1}{3}.$$ This follows because


2

A stronger lower bound Define the functions $a_n: [0, \infty) \to \Bbb R$ recursively as $$ \begin{align} a_0(x) &= 1 \, ,\\ a_{n+1}(x) &= \sqrt{1+xa_n(x+1)} \, . \end{align} $$ In Nested Radical of Ramanujan it is shown that for all $x \ge 0$ $$ \lim_{n\to \infty} a_n(x) = x+1 \, . $$ On the other hand, $f(x) \ge a_n(x)$ follows from the given ...


1

Observe that, points $I$ and $H$ will always stay on one side of point $O$$[$ Drop perpendiculars from point $A$, $O$ and $I$ on side $BC$ and then calculate the distances from point $B$ to the feet of these perpendiculars.$]$ and thereafter stay on the arc $BO$. Hence, to maximize the area, we need to evenly distribute points $I$ and $H$ over the arc $BO$.


1

Note that $O$ is fixed as midpoint of arc $BC$. Since the perpendicular bisector of chord $BC$ passes through $O$.


0

Begin with a circle with center $O$, and the three triangle points $A$, $B$, and $C$ have to lie on the circle. How are you going to get the incenter point $I$ on the arc $BC$? The incenter is formed by the bisectors of the angles of the triangle, and thus you would have to have a degenerate null triangle to have the bisector ray moving along one of the ...


0

Let's say your number $n$ is between $2^{k-1}$ and $2^k-1$. On one hand, you can make the partition of size $k$. This can be proven by induction on $k$. Namely, $k=1$ means $n=1$, which is its own partition of size $1$. The inductive step (sketch): "halve" $n$, i.e. take $m=\lceil n/2\rceil$, which gives $n-m=\lfloor n/2\rfloor$; prove that $2^{k-2}...


0

Suppose this holds for a particular value of $n$ (not all $n$). Let's try $f(x) = x^a$. Then $f^{[2]}(x) =f(f(x)) =(x^a)^a = x^{a^2} $, $f^{[3]}(x) =f^{[2]}(f(x)) =(x^a)^{a^2} = x^{a^3} $, and, in general, $f^{[n]}(x) = x^{a^n} $. Therefore $n = a^n$ so $a = n^{1/n} $. If we compose the other way so that $f^{[n+1]}(x) =f(f^{[n]}(x)) $ then, assuming that $f(...


0

Fix $n\ge 2$. $f^{[2n]}(x)=x^{2n}$ $f^{[2n]}(x)=f^{[2n-1]}(f(x))=f(x)^{2n-1}$. So $f(x)=x^{\frac{2n}{2n-1}}$. Since the expression of $f(x)$ depends on $n$, such $f$ cannot exists.


0

By taste, I generally prefer solutions based on structural properties given implicitly in the hypotheses. Here the natural structure is that of the field $\mathbf F_7=\mathbf Z/7\mathbf Z$, more precisely of its multiplicative group $\mathbf F_7^*$. From the properties of finite fields, we know that $\mathbf F_7^*$ is cyclic of order 6 , hence $\cong C_2 \...


0

We will prove the homogeneous inequality (where we used $abc=1$) $$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{a}+\sqrt{b}+\sqrt{c}+3\sqrt[6]{abc}$$ Since equality is attainable for $a=b=c$, this would show that the desired $k$ is $3$. Observe that, in virtue of AM-GM $$\frac{\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}}+\frac{b}{\sqrt{c}...


3

When we saw this problem years ago, the idea was to involve Beatty sequences. IIRC, We couldn't solve it without using Beatty sequences (or its ideas in some form). Here's how the wishful thinking went: Let the sequence be $a_n$. Define $ b_n = a_n + n$, which is the complement. IF (and that's a big wishful if) $a_n = \lfloor \alpha \times n \rfloor$, then $...


2

Comments only: Since you have already known the answer, the following is just hindsight. First of all we know that $\{ a_n \}$ and $\{ b_n = a_n+n \}$ is a partition of $\mathbb N$. On the other hand if a formula of $\{ a_n \}$ ensures that $\{a_n\}$ and $\{a_n+n\}$ partition $\mathbb N$ then we are done (Edit: not quite, we still need $a_{n-1}+n-1>a_n$, ...


3

An alternative way: By the rearrangement inequality we have $$\frac{b}{\sqrt{a}}+\frac{c}{\sqrt{a}}+\frac{c}{\sqrt{b}}+\frac{a}{\sqrt{b}}+\frac{a}{\sqrt{c}}+\frac{b}{\sqrt{c}}\geq \frac{a}{\sqrt{a}}+\frac{a}{\sqrt{a}}+\frac{b}{\sqrt{b}}+\frac{b}{\sqrt{b}}+\frac{c}{\sqrt{c}}+\frac{c}{\sqrt{c}}=2(\sqrt{a}+\sqrt{b}+\sqrt{c})$$ Now proceed by AM-GM as in the ...


1

By the CRT it’s enough to figure out $F_{2020}$ mod $8$ and $125$. It’s easy to check that six applications of the linear recurrence relation turn $(0,1)=(F_0,F_1)$ into $(F_6,F_7)=(8,13)=5 \cdot (0,1)$ mod $8$. As the multiplicative order of $5$ mod $8$ is two, it follows that $F_n$ mod $8$ has a period of $12$. Now note that $2020$ is congruent to $4$ mod $...


6

By AM-GM $$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge 2(\sqrt{\frac{bc}{a}} + \sqrt{\frac{ac}{b}} + \sqrt{\frac{ab}{c}} )=2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$$ Now using $x^2+y^2+z^2\ge xy+yz+zx$ $$\begin{align*}2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \ge 2(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ac}})=2(\sqrt{a}+\...


6

This is specific to chess. Hint: At the start, there is a chess piece that you can move twice in the beginning to return to the starting board position. Effectively passing your turn. We show white (the first player) has a non-losing strategy. Suppose not, then there is a winning strategy for black (the second player). But as white, at the very beginning ...


6

Hint: $$(a + b + c)(ab + ac + bc) - abc=(a + b)(b + c)(c + a)$$


5

If you set the expression equal to $x$ and square the equation, you get $$x^2=25+2\sqrt{n}.$$ This is equivalent to $$n=\left(\frac{x^2-5^2}{2}\right)^2.$$ So just pick $x$ equal to an odd integer and then this above expression is $n.$ As Crostul points out, in order for the original expression to be well-defined, we need $n$ to be less than or equal to $\...


6

If it's a more optimized proof you're looking for, these exist. Again, we notice that $$a^3=a^{(7-1)/2}\equiv\left(\frac a7\right)\in\{-1,0,1\}\pmod7,$$ which we can also see by direct casework. We note that $a^3\equiv0$ if and only if $a\equiv0.$ The advantage of saying $\{-1,0,1\}$ is that we can say the sum $x^3+y^3-z^3$ lives in $[-3,3],$ and further, if ...


1

$\newcommand\abs[1]{\lvert#1\rvert}$Fix $a_i \in \mathbb C$. You can see the following fact by drawing a picture: if $\abs{a_1} = \abs{a_2}$, then the circles of radius $\abs{a_i}$ centred at $a_i$ intersect at $0$ and $a_1 + a_2$, and the intersection of the corresponding discs is contained in the circle of radius $\abs{a_1 + a_2}$ centred at $0$. Here's ...


1

If $b=0,$ the equation holds. Else, let $c=a/b$ and divide by $b^n$ so that we have to show $\prod\limits_{k=0}^{n-1} (c+\epsilon_k^2) = c^n + 1.$ The LHS is simply $\prod\limits_{k=0}^{n-1} (c+\epsilon_k)$ because the $\epsilon_k^2$ form a permutation of the roots of unity as you observed. Now do you see how to finish off with Vieta's Theorem? Extra hint: ...


2

As you correctly identify, since the product of the divisors of $n$ is $n^{1994}$, $n$ must have $1994$ factor pairs, i.e. $3988$ divisors. We also know that $n$ must be divisible by prime $997$, as must the number of divisors. So to get this we need a factor of $p^{\large{\color{red}{996}}}$ which you narrowly miss, and obviously this should use $p=2$ to ...


1

Your idea is really good, but at the end you want the smallest $k$ such that $d(997\cdot 2\cdot k)=997\cdot 2^{2}$. You said that $k=2^{997}$ does the job, but this doen't, because $d(997\cdot 2^{998})=(1+1)(998+1)\neq 997\cdot 2^{2}$. To fix this, note that if you write $k=2^{m}\cdot p$, where $p$ doen't have factors of $2$, then $$d(\color{blue}{997}\cdot ...


2

I will use the shortcuts $\angle A=\alpha, \angle B=\beta, \angle C=\gamma$. Let also $\angle BAD=\varphi$. It follows easily from angle-chasing that $$\angle DBA=\alpha-\varphi,\; \angle CBD=\beta-(\alpha-\varphi),\; \angle DCB=\alpha-\varphi,\; \angle ACD=\gamma-(\alpha-\varphi),\; \angle DAC=\alpha-\varphi$$ This will be helpful due to the Sine Law, since ...


1

Let $a_i=1$ if $i$ is a girl, $0$ otherwise. Let $f(n,k)=\sum_{i=n}^{n+k-1}a_i$, so we want to show that $f(n,k)=f(n+k,k)$ for some $1\leq n\leq1000$ and $100\leq k\leq 300$. Define $g(n,k)=f(n+k,k)-f(n,k)$, and assume for contradiction that $g(n,k)\neq0$. Then as $\lvert f(n+1,k)-f(n,k)\rvert\leq1$, it follows that $\lvert g(n+1,k)-g(n,k)\rvert\leq 2$. Now $...


7

First every such integer is divisible by $7, 11, 13$. Highest power of $2$ of $13!$ is even, so you need to find an integer product such that it's odd. This can be easily done. Then the highest power of $3$ of $13!$ is odd, so you need to find an integer product such that it's divisible by $3$ but not by $9$. This is easy too. The highest power of $5$ of $13!...


2

You can split the $24$ bordering cells into 4 groups of $6\times 1$ rectangles and then represent each rectangle as a set difference between an union of two $3\times 3$ squares and an union of three $2\times 2$ squares. Sometimes a picture worth a thousand words....


4

On the one hand, observe that there can be no arrangement of 4 or more consecutive coins that do not have a value of 3. Equivalently, let $a_j$ be the value of the $j$-th coin in the sequence; $j=1,2,\ldots,2001$. Then for each $i=0,1,\ldots,1997$ at least one of $a_{i+1},a_{i+2},a_{i+3},a_{i+4}$ must be 3. So the number $X$ of $j \in \{1,2,\ldots, 2001\}$ ...


1

Consider a $3 \times 3$ square such as this one. You can see that we can put a $2\times 2$ square in it (in white here). The sum of the numbers in the "boomerang" shape colored in red is the sum of the numbers in the $3 \times 3$ square minus the sum of the numbers in the $2 \times 2$ square which is zero. As such the sum of the numbers in every ...


5

Hints: Notice that any $4$ consecutive coins can have at most one $3$-kopeyka. Show that if you choose any $4$ consecutive coins, say, $a_k, a_{k+1}, a_{k+2}, a_{k+3}$, then exactly one of them is $3$-kopeyka.


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