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231,038
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Let \(a_1\) be a natural number not divisible by \(5\). The sequence \(a_1,a_2,a_3,\dots\) is defined by
\[
a_{n+1}=a_n+b_n,
\]
where \(b_n\) is the last digit of \(a_n\). Prove that the sequence contains infinitely many powers of two. [quote="outback"]Let $ a_1$ be a natural number not divisible by $ 5$. The sequence $ a_1, a_2, a_3, . . .$ is defined by $ a_{n \plus{} 1} \equal{} a_n \plus{} b_n$, where $ b_n$ is the last digit of $ a_n$. Prove that the sequence contains infinitely many powers of two.[/quote]
[hide="Solution"]
The second number of the sequence must be even since it is congruent to $ 2a_1\bmod 10$. Notice that if $ a_2$ ends with a $ 2$, then we add $ 2$, so $ a_3$ ends with a $ 4$, so we add $ 4$, so $ a_4$ ends with an $ 8$, so we add $ 8$, and then $ a_5$ ends with a $ 6$, so we add $ 6$, and then $ a_6$ ends with a two again. Thus, $ a_6 \minus{} a_2 \equal{} 20$. In fact, we can do the same if $ a_2$ ends with a $ 4$, $ 6$, or $ 8$. Analogously, $ a_{k \plus{} 1} \equal{} a_{k \plus{} 5} \plus{} 20$ for a positive integer $ k$. Now, if $ a_{k \plus{} 1}\equiv 2^n\bmod 20$ for some $ n$, then we can continously add $ 20$ and get a number in the sequence that is a power of $ 2$. In fact, we can do this infinitely times to get infinitely many powers of $ 2$ since $ 2^n\equiv 2^{n \plus{} 4}\bmod 20$ when $ n\ge 2$. Thus, if $ a_{k \plus{} 1}$ is congruent to $ 4$, $ 8$, $ 12$, or $ 16\bmod 20$, then we can continuously add twenty. This means that we are done if $ a_{k \plus{} 1}\equiv 0\bmod 4$. If not, then if $ a_{k \plus{} 1}$ must be congruent to $ 2$, $ 6$, $ 14$, or $ 18\bmod 20$. If it is congruent to $ 2\bmod 20$, then it ends with $ 2$, so we add $ 2$, so $ a_{k \plus{} 2}\equiv 4\bmod 20$, meaning we can repeat the process described above with $ a_{k \plus{} 2}$ rather than $ a_{k \plus{} 1}$. Similarly, if $ a_{k \plus{} 1}\equiv 6\bmod 20$, then $ a_{k \plus{} 2}\equiv 12\bmod 20$, so we can do the process with $ a_{k \plus{} 2}$. Now, if $ a_{k \plus{} 1}\equiv 14\bmod 20$, then $ a_{k \plus{} 2}\equiv 18\bmod 20$, so $ a_{k \plus{} 3}\equiv 6\bmod 20$, so $ a_{k \plus{} 4}\equiv 12\bmod 20$ and we can do the process with $ a_{k \plus{} 4}$. Finally, if $ a_{k \plus{} 1}\equiv 18\bmod 20$, then $ a_{k \plus{} 3}\equiv 12\bmod 20$. All the cases are finished, and for each case, there is a way to get infinitely many powers of two, so we are done. [/hide]
|
2,310,426
|
Find the number of complex solutions to
\[
\frac{z^3 - 1}{z^2 + z - 2} = 0.
\] [quote=GameBot]Find the number of complex solutions to
\[\frac{z^3 - 1}{z^2 + z - 2} = 0.\][/quote]
Firstly, let us factor the top with difference of cubes. We rewrite as $(z-1)(z^2+z+1)$. The bottom is just $(z+2)(z-1)$, so the fraction rewrites as $\frac{z^2+z+1}{z+2}=0$. This means that $z^2+z+1=0$. The Quadratic Formula gives $z = \frac{-1 \pm \sqrt{-3}}{2}$. Therefore there are $\boxed{2}$ complex solutions, since $\sqrt{-3}$ is complex.
|
2,310,435
|
Let \(ABC\) be a triangle with circumcircle \((O)\). Let \(P\) be an arbitrary interior point of \((O)\) with \(P\neq O\). Let \(X,Y,Z\) be the symmedian points of triangles \(BPC,\; CPA,\; APB\), respectively. Let \(D,E,F\) be the intersections of the tangents to \((O)\) at \(B\) and \(C\), at \(C\) and \(A\), and at \(A\) and \(B\), respectively. Prove that the lines \(DX,\; EY,\; FZ\) concur on the line \(OP\). I have labelled $X,Y,Z$ as $K_a,K_b,K_c$ in my solution.
[quote=tutubixu9198]Let $ABC$ be a triangle with circumcircle $(O)$. $P$ be an arbitrary interior point of $(O)$ $(P\neq O)$. Let $X,Y,Z$ be the symmedian point of triangle $BPC,CPA,APB$ respectively. Let $D,E,F$ be the intersection of the tangents of $(O)$ at $B$ and $C$, at $C$ and $A$, at $A$ and $B$ respectively. Prove that $DX,EY,FZ$ concur on $OP$.[/quote]
$\textbf{LEMMA:}$ $ABC$ be a triangle and $P$ be an arbitary point on the plane. Let the $A-$ symmedians of $\Delta PAB$ and $\Delta PCA$ meet $\odot(ABC)$ at $U,V$ respectively. Then $\overline{CU},\overline{BV},\overline{AP}$ concur.
Fix a line $\ell$ passing through $A$ and animate $P$ on $\ell$. Let $M,N$ be the midpoints of $BP,CP$. Then $P\mapsto M\mapsto AM\mapsto AU\mapsto U$ is a homography between $\overline{BP}$ and $\odot(ABC)$. Similarly $P\mapsto V$ is a homography. Thus, $\overline{BV}\mapsto\overline{CU}$ is a homography between pencil of lines through $B$ and $C$ and $\overline{BV}\equiv\overline{CU}$ when $P=\infty_{\ell}$. So by degenerate steiner conic theorem we have that $\overline{BV}\cap\overline{CU}$ traces out a line. Let $\overline{AP}\cap\odot(ABC)=T$. Let $\overline{CU}\cap\overline{AP}=J$ and $\overline{BV}\cap\overline{AP}=J^*$. If $P=\ell\cap\odot(ABC)$, then $-1=(A,U;B,T)\overset{C}{=}(A,J;\overline{BC}\cap\overline{AP},T)=(A,V;C,T)\overset{B}{=}(A,J^*;\overline{BC}\cap\overline{AP},T)\implies J^*\equiv J\implies \overline{CU},\overline{BV},\overline{AT}$ are concurrent and when $P\equiv A$, $U,V\equiv T$. Thus, $\overline{CU}\cap\overline{BV}$ trace out the line $\overline{AP}$. $\square$
__________________________________________________________________________________________
Let $\overline{PK_a}\cap\overline{BC}=X$ ; $\overline{PK_b}\cap\overline{AC}=Y$ ; $\overline{PK_c}\cap\overline{AB}=Z$. Now notice that $$\frac{BX}{XC}\cdot\frac{CY}{YA}\cdot\frac{AZ}{ZB}=\left(\frac{BP}{CP}\right)^2\cdot\left(\frac{CP}{AP}\right)^2\cdot\left(\frac{AP}{BP}\right)^2=1$$ Hence, $\overline{AX},\overline{BY},\overline{CZ}$ are concurrent with Perspector $Q$. So, by Cevian Nest Theorem we get $\Delta DEF,\Delta XYZ$ are Perspective triangles. Let the Perspectix $\mathcal{T}$ of $\Delta XYZ,\Delta K_aK_bK_c$ cut $\overline{YZ},\overline{ZX},\overline{XY}$ at $X^*,Y^*,Z^*$ respectively. Now let $\{\overline{AK_b},\overline{AK_c}\}\cap\odot(ABC)=\{V,U\}$ . So by our $\textbf{LEMMA}$ we have that $\overline{CU},\overline{BV},\overline{AP}$ are concurrent. Now by Dual of Desargues Involution Theorem on the Complete Quadrilateral $\mathbf{Q}\equiv\{\overline{K_bK_c},\overline{K_cZ},\overline{ZY},\overline{YK_b}\}$ we get that $\{(\overline{AZ},\overline{AK_b}),(\overline{AY},\overline{AK_c}),(\overline{AP},\overline{AX^*})\}$ are Pairs of an Involution $\Psi$. From $\textbf{LEMMA}$ we get $\{(B,V),(C,U),(A,T)\}$ are pairs of an Involution $\Phi$ on $\odot(ABC)$. Projecting from $A$ we get $\{(\overline{AZ},\overline{AK_b}),(\overline{AY},\overline{AK_c}),(\overline{AP},\overline{EF})\}$ are Pairs of the Involution $\Psi$. Hence, $\overline{AX^*}\equiv\overline{EF}\implies X^*\in\overline{EF}\implies \overline{XY},\overline{EF},\overline{K_bK_c}$ are concurrent at $X^*$. Let the tangents at $P$ to $\odot(BPC),\odot(CPA),\odot(APB)$ intersect $\overline{BC},\overline{CA},\overline{AB}$ at $G_a,G_b,G_c$ respectively. Notice that as $(B,C;X,G_a)=-1$ which implies $\overline{DX}$ is the Polar of $G_a$. Similarly $\overline{EY}$ is the Polar of $G_b$ and $\overline{FZ}$ is the Polar of $G_c$. So by La Hire we get that $\overline{G_aG_bG_c}$ is the Polar of $Q\implies\overline{OQ}\perp\overline{G_aG_bG_c}$. Also $\overline{G_aG_bG_c}$ is the Radical Axis of $\odot(ABC)$ and the Degenerate Circle $\odot(P)\implies\overline{OP}\perp\overline{G_aG_bG_c}$. Hence, $O,P,Q$ are collinear $(\bigstar\bigstar)$. Now Notice that $\Delta DEF,\Delta XYZ,\Delta K_aK_bK_c$ are Pairwise Perspective triangles with a Common Perspectix $\mathcal T\equiv\overline{X^*Y^*Z^*}$, so their Pairwise Perspectors must be collinear. So from $(\bigstar\bigstar)$ the Perspector of $\Delta DEF,\Delta K_aK_bK_c\in\overline{OP}$. $\blacksquare$
|
2,310,441
|
Prove that
\[
\operatorname{Area}(\triangle ABC)=(s-a)r_a,
\]
where \(s=\dfrac{a+b+c}{2}\), \(a=BC\), \(b=CA\), \(c=AB\), and \(r_a\) is the exradius of the A-excircle. [quote=franzliszt]You have $[ABC]=rs$ where $r$ is the inradius. Consider the homothety sending the incircle to the excircle with scale factor $\frac{s}{s-a}$. Then the exradius is $\frac{rs}{s-a}$. Multiply this by $(s-a)$ gives the desired result.[/quote]
Can you please solve it in any other way as I am just a beginner in geometry and don't know homothety?
|
231,045
|
A rectangular cow pasture is enclosed on three sides by a fence and the fourth side is part of the side of a barn that is \(400\) feet long. The fence costs \(\$5\) per foot, and \(\$1{,}200\) altogether. To the nearest foot, find the length of the side parallel to the barn that will maximize the area of the pasture. The maximum amount of fence the farmer can have is 1200/5=240 feet. Since a square maximizes area, we want to split the 240 feet equally into 3 parts, so each section is 80 feet. Thus, the area is 80^2=6400 sq ft.
|
23,105
|
Problem: Find all positive integer $n$ and prime number $p$ such that :
Any $a_1,a_2,...,a_n\in\{1,2,...,p-1\}$ we have $\sum_{k=1}^na_k^2\not\equiv0\pmod{p}$
Repaired ....! I think that $n<3$ and $p=4k+3$.
Let's first show that for $n\ge 3$, all residues $\pmod p$ can be written as $\sum_{i=1}^k a_i^2$. Let $A$ be the set of quadratic residues $\pmod p$. The Cauchy-Davenport Theorem states that (we're working in $\mathbb Z_p$ here) $|A+A|\ge \min(p,2|A|-1)=2|A|-1=p-2$. From here we can show by induction on $n$ that $|A+A+A+\ldots+A|\ge p$ where there are at least $3$ $A$'s in that sum of sets. In particular, $0\in A+A+A+\ldots+A$.
On the other hand, it's well-known that $-1$ is a quadratic residue of $p$ iff $p=4k+1$ (this deals with the case $n=2$). Of course, $n=1$ always works.
Is there something wrong with this?
P.S. Could you please elaborate on that "p-adic analysis needs it thing"? What's the connection with p-adic analysis?
Edit: I only considered $p>2$. For $p=2$ $n$ has to be odd in order to satisfy the requirement $\sum_{i=1}^n a_i^2\ne 0,\ \forall a_i\in\{1\}$ :D.
|
231,050
|
A rectangular piece of paper \(ADEF\) is folded so that corner \(D\) meets the opposite edge \(EF\) at \(D'\), forming a crease \(BC\) where \(B\) lies on edge \(AD\) and \(C\) lies on edge \(DE\). If \(AD=25\) cm, \(DE=16\) cm, and \(AB=5\) cm, find \(BC^2\). mathwizarddude, i believe your wrong.
[u]Here is my solution[/u]
Here is my diagram
http://img72.imageshack.us/my.php?image=paperdiagramks0.jpg
We construct the diagram as the question dictates, as we do, we note that
$ BD \equal{} BD' \equal{} 20$ and
$ DC \equal{} D'C$
this is because its the same side just moved
We also note that
$ BD \equal{} 20$ as $ AD \equal{} 25$ and $ AB \equal{} 5$
We now construct a line perpendicular to $ AD$ at point $ B$.
Let the point on $ FE$ made by the perpend. of $ AD$ at $ B$ be $ G$ (i didnt put this on my diagram)
We have a right angle triangle with sides of $ 16$ and hypotenuse $ 20$
We find $ GD'$ using Pythagoras theorem to be $ 12$.
Now,
$ FG \plus{} FD' \plus{} D'E \equal{} AD \equal{} 25$
$ FG \equal{} AB \equal{} 5$ and
$ GD' \equal{} 12$
$ \therefore D'E \equal{} 8$
Now,
Let $ DC \equal{} D'C \equal{} n$
$ \therefore DE \equal{} 16 \minus{} DC \equal{} 16 \minus{} n$
Now we have the equation
$ 8^2 \plus{} (16 \minus{} n)^2 \equal{} n^2$
solving gives,
$ 8^2 \plus{} 16^2 \plus{} n^2 \minus{} 32n \equal{} n^2$ subtracting $ n^2$ and rearranging gives
$ 32n \equal{} 8^2 \plus{} 16^2$
$ 32n \equal{} 320$
$ \therefore n \equal{} 10$
Now we have another equation for which we can get $ BC^2$
$ BC^2 \equal{} 20^2 \plus{} 10^2$
$ \therefore BC^2 \equal{}\boxed{500}$
|
2,310,536
|
For each positive integer \(n\), the mean of the first \(n\) terms of a sequence is \(n\). What is the 2008th term of the sequence? [hide=Better Solution]We notice that: $$a_1+a_2+\cdots+a_n=n^2.$$ Therefore: $$a_1+a_2+\cdots+a_{2008}=2008^2.$$ Similarly, $$a_1+a_2+\cdots+a_{2007}=2007^2.$$ Hence, $$a_{2008}=2008^2-2007^2=(2008-2007)(2008+2007)=1\cdot4015=\boxed{4015}.$$ [/hide] Lol, mine's like identical to the alcumus solution. No, I didn't copy.
|
231,054
|
Two regular polygons with the same number of sides have side lengths 48 m and 55 m, respectively. A third regular polygon with the same number of sides has area equal to the sum of the areas of the first two. What is the side length of the third polygon? [hide="most likely incorrect"]The area of the regular $ n$-gon is directly proportional to the square of its side length. Thus (where $ c$ is some constant), the area of two polygons are $ 48^2c$ and $ 55^2c$, respectively. The sum of their areas is $ (48^2 \plus{} 55^2)c$, and the $ n$-gon with this area has a side length of $ \sqrt{48^2 \plus{} 55^2} \approx
\boxed{71.13}$.[/hide]
|
2,310,555
|
Let \(f:\mathbb{R}\to\mathbb{R}\) be a differentiable function such that
\[
\lim_{x\to 2}\frac{x^2+x-6}{\sqrt{1+f(x)}-3}\ge 0
\]
and \(f(x)\ge -1\) for all \(x\in\mathbb{R}\). If the line \(6x-y=4\) intersects \(y=f(x)\) at \(x=2\), find \(f'(2)\). If $y=f(x)$ intersects the line $6x-y=4$, then $f(2)=8$ (i). Upon inspection, we have:
$\lim_{x \rightarrow 2} \frac{x^2+x-6}{\sqrt{1+f(x)}-3} = \lim_{x \rightarrow 2} \frac{(x+3)(x-2)}{\sqrt{1+f(x)}-3} = \frac{0}{0}$ (ii)
which is indeterminate. By L'Hopital's Rule, we next obtain:
$\lim_{x \rightarrow 2} \frac{2x+1}{f'(x)/2\sqrt{1+f(x)}} = \lim_{x \rightarrow 2} \frac{2(2x+1)\sqrt{1+f(x)}}{f'(x)} = \frac{30}{f'(2)} \ge 0$ (iii);
of which we require $f'(2) > 0$ to satisfy (iii). If $y=6x-4$ is the tangent line to $f(x)$ at $P(2,8)$, then:
$y -8 = 6(x-2) = f'(2)(x-2) \Rightarrow \textcolor{red}{f'(2)=6}$.
|
2,310,567
|
Solve the equation
\[
- x^2 = \frac{3x+1}{x+3}.
\]
Enter all solutions, separated by commas. Cross multiplying gives $-x^2(x+3)=3x+1$. Expanding the LHS results in $-x^3-3x^2=3x+1$. Now, we add $x^3+3x^2$ to both sides. This gives $x^3+3x^2+3x+1=0$. We can factor $x^3+3x^2+3x+1$ as $(x+1)^3$, so the answer is just $\boxed{-1}$.
|
2,310,595
|
If \(a\) and \(b\) are complex numbers such that \(|a| = 6\) and \(|b| = 4\), find \(\left| \frac{a}{b} \right|\). [quote=OlympusHero][quote=GameBot]If $a$ and $b$ are complex numbers such that $|a| = 6$ and $|b| = 4,$ then find $\left| \frac{a}{b} \right|.$[/quote]
@above, nice cheese.
Notice that $\frac{|a|}{|b|}=\left|\frac{a}{b}\right|$, so the answer is $\boxed{\frac{3}{2}}$.[/quote]
FTFY
|
231,062
|
Determine whether \(100,\!895,\!598,\!169\) is prime. Provide a solution. [quote="stevenmeow"]uhh testing either 1) all odd #s
or 2) all primes in a database
yeah thats probably how
[/quote]
I would bet that's not how it was done. There are several (advanced) techniques for determining [url=http://en.wikipedia.org/wiki/Primality_testing]if a number is prime[/url]. They can tell you if a number is prime without explicitly determing any factors of the number. Then there are also advanced techniques for [url=http://en.wikipedia.org/wiki/Integer_factorization]actually factoring numbers[/url] (aside from trial division), but they are comparably slower, by computer standards, than the techniques for just testing for primality.
|
2,310,658
|
Find all real values of \(r\) that satisfy
\[
\frac{1}{r}>\frac{1}{r-1}+\frac{1}{r-4}.
\]
Give your answer in interval notation. ans is [hide]$(-\infty,-2)\cup(0,1)\cup(2,4)$[/hide], you multiply both sides of the inequality by [hide]$r(r-1)(r-4)$[/hide], being careful of the sign and doing casework on whether you flip the inequality or not
might post full sol later idk
|
2,310,660
|
Let \(a,b,c\) be nonnegative real numbers such that \(a+b+c=1\). Find the maximum value of
\[
\frac{ab}{a+b}+\frac{ac}{a+c}+\frac{bc}{b+c}.
\] [quote=GameBot]Let $a,$ $b,$ $c$ be nonnegative real numbers such that $a + b + c = 1.$ Find the maximum value of
\[\frac{ab}{a + b} + \frac{ac}{a + c} + \frac{bc}{b + c}.\][/quote]
very intuitive that maximal is reached when a=b=c
|
2,310,680
|
For constants \(x\) and \(a\), the third, fourth, and fifth terms in the expansion of \((x+a)^n\) are \(84\), \(280\), and \(560\), respectively. Find \(n\). We can see that the third, fourth, and fifth terms are $\binom n2x^{n-2}a^2=84,\binom n3x^{n-3}a^3=280,$ and $\binom n4x^{n-4}a^4=560,$ which is a system of equations with solutions $x=1,a=2,$ and $n=\boxed7.$
|
2,310,688
|
Solve
\[
\frac{1}{x-5}>0.
\]
Enter your answer using interval notation. [quote=bestzack66][quote name="HIA2020" url="/community/p18362019"]
Not too good at this, but $(5,infinity]$?
Am I right?
[/quote]
when we write interval notation, we write ) or ( before/after infinity.[/quote]
thanks :)
|
2,310,694
|
Given triangle \(ABC\) inscribed in \((O)\). Let \(M\) be the midpoint of \(BC\), and let \(H\) be the foot of the perpendicular from \(A\) to \(BC\). Let \(OH\) meet \(AM\) at \(P\). Prove that \(P\) lies on the radical axis of \((BOC)\) and the nine-point circle of triangle \(ABC\). Let $\Omega$ be the nine-point circle. For any point $\bullet$ define a function $$f(\bullet) = \text{Pow}(\bullet, (BOC)) - \text{Pow}(\bullet, \Omega).$$ We calculate \begin{align*} \left\lvert \frac{f(A)}{f(M)} \right\rvert &= \left\lvert \frac{\text{Pow}(A, (BOC)) - \text{Pow}(A, \Omega)}{\text{Pow}(M, (BOC)) - \text{Pow}(M, \Omega)} \right\rvert \\ &= \frac{AB\cdot \tfrac{AC}{2 \cos A} - \tfrac{AB}{2} \cdot AC \cos A}{\tfrac{1}{4} BC^2} \\ &= \frac{2AB \cdot AC \cdot (\sec A - \cos A)}{BC^2} \\ &= \frac{2AB \cdot AC \cdot \tfrac{\sin^2 A}{\cos A}}{BC^2} \\ &= \frac{4[ABC] \tan A}{BC^2}\end{align*} and $$\frac{AP}{MP} = \frac{AH}{MO} = \frac{2[ABC]/BC}{BC/(2 \tan A)} = \frac{4[ABC] \tan A}{BC^2}.$$ These are equal, so we may conclude by linearity of PoP.
|
2,310,715
|
If \(\log_2 x + \log_2 x^2 = 6\), find the value of \(x\). [quote=Dragonslayer118]The answer is 4[/quote]
Dragonslayer118, the community appreciates that you want to participate, but remember, in the FAQ, it says that if you don't have anything to add to the discussion, you don't need to post anything.
|
2,310,753
|
Find constants \(A,B,C,\) and \(D\) so that
\[
\frac{x^3 + 3x^2 - 12x + 36}{x^4 - 16} = \frac{A}{x - 2} + \frac{B}{x + 2} + \frac{Cx + D}{x^2 + 4}.
\]
Enter the ordered quadruple \((A,B,C,D)\). [quote=dajeff][hide=solution]
As in the AoPS solution, we multiply both sides by $(x-2)(x+2)(x^2+4)$ and substitute $x=2,-2$ to get $A=1,B=-2$.
We can substitute in $x=2i$, which satisfies $x^2+4=0$, to get
\begin{align*}
(2iC+D)(2i-2)(2i+2)&=(2i)^3+3(2i)^2-12(2i)+36
\\(2iC+D)((2i)^2-4)&=-8i-12-24i+36
\\(2iC+D)(-8)&=24-32i
\\D+2iC&=-3+4i.
\end{align*}
Thus, $D=-3,C=2$, and we get the quadruple $\boxed{(1,-2,2,-3)}$.
We can confirm that $C,D$ are real by plugging in $x=-2i$ and solving a system of equations in $C,D$.
[/hide][/quote]
You do not need to confirm by plugging in $x=2i.$ If whatever on the right side is $f(a),$ then the first result is $f(x)=D+2iC.$ Plugging in $x=2i$ yields $f(\overline{x})=D-2iC=\overline{D+2iC}.$ Obviously, $f(\overline{x})=\overline{f(x)}$ for all polynomials $f(x).$
|
2,310,754
|
Two real numbers \(x\) and \(y\) satisfy \(x-y=4\) and \(x^3-y^3=28\). Compute \(xy\). Cubing gives $x^3-3x^2y+3xy^2-y^3=64$. So $-3x^2y+3xy^2=36$. Dividing each side by $-3$ gives $x^2y-xy^2=-12$, so $xy(x-y)=-12$. Therefore, $xy=-3$.
lol i forgot a negative sign :P
|
2,310,758
|
The fourth-degree polynomial equation
\[
x^4 - 7x^3 + 4x^2 + 7x - 4 = 0
\]
has four real roots \(a,b,c,d\). What is the value of
\[
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}?
\]
Express your answer as a common fraction. [quote=eduD_looC]We have $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{abc+acd+abd+bcd}{abcd}=\boxed{\frac{7}{4}}$ by Vieta's.[/quote]
Yes, this is correct. Very good! :thumbup:
[quote=hamster9]hmmmmmmmm[/quote]
Do you need help? Is it confusing?
|
2,310,764
|
Let \(E\) and \(F\) be points on side \(BC\) of triangle \(\triangle ABC\). Points \(K\) and \(L\) are chosen on segments \(AB\) and \(AC\), respectively, so that \(EK\parallel AC\) and \(FL\parallel AB\). The incircles of \(\triangle BEK\) and \(\triangle CFL\) touch segments \(AB\) and \(AC\) at \(X\) and \(Y\), respectively. Lines \(AC\) and \(EX\) intersect at \(M\), and lines \(AB\) and \(FY\) intersect at \(N\). Given that \(AX=AY\), prove that \(MN\parallel BC\). [quote=2018 Thailand TST 6.1]Let $E$ and $F$ be points on side $BC$ of a triangle $\vartriangle ABC$. Points $K$ and $L$ are chosen on segments $AB$ and $AC$, respectively, so that $EK \parallel AC$ and $FL \parallel AB$. The incircles of $\vartriangle BEK$ and $\vartriangle CFL$ touches segments $AB$ and $AC$ at $X$ and $Y$ , respectively. Lines $AC$ and $EX$ intersect at $M$, and lines $AB$ and $FY$ intersect at $N$. Given that $AX = AY$, prove that $MN \parallel BC$.[/quote]
*typo edited
Easy, nice, but okay for a p1.
Notice some pairs of similar triangles $\{\Delta XKE, \Delta XAM\}$ and $\{\Delta YLF, \Delta YAN\}$ and $\{\Delta KBE, \Delta LFC, \Delta ABC\}$, also if the incircle of $\Delta BKE$ touches $KE$ at $U$ , $\{\Delta KBU, \Delta LFY\}$ , thus \[\frac{AM}{AN}=\frac{KE}{KX}\cdot \frac{LY}{LF}=\frac{KE}{KB}=\frac{AC}{AB}\] as desired. $\textbf{Q.E.D.}$
|
2,310,792
|
Given vectors \(\mathbf{a}\) and \(\mathbf{b}\) such that \(\|\mathbf{a}\| = 6\), \(\|\mathbf{b}\| = 8\), and \(\|\mathbf{a} + \mathbf{b}\| = 11\). Find \(\cos\theta\), where \(\theta\) is the angle between \(\mathbf{a}\) and \(\mathbf{b}\). [quote=OlympusHero]Given vectors $\mathbf{a}$ and $\mathbf{b}$ such that $\|\mathbf{a}\| = 6,$ $\|\mathbf{b}\| = 8,$ and $\|\mathbf{a} + \mathbf{b}\| = 11.$ Find $\cos \theta,$ where $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}.$
I need help please! Hint only. Thanks![/quote]
Um did you know $\frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} = \cos \theta$?
|
2,310,805
|
The roots of the polynomial
\[
x^3 - 52x^2 + 581x - k
\]
are distinct prime numbers. Find \(k\). [quote=jasperE3]How did you find what the roots were?[/quote]
[hide=spoiler]
one prime must be 2 by common sense
then the other two primes add to 50
lets call thm $p$ and $q$
we have that $p + q = 50$
and that $pq + 2p + 2q + 4 = 585$
or $(p+2)(q+2) = 585$
from there we see that $585 = 39 * 15$
therefore the two primes are 37 and 13
|
231,084
|
Show that if \(f:\mathbb{R}\to\mathbb{R}\) is a monotone increasing function, then the set of points where \(f\) is discontinuous has Lebesgue measure zero. [quote="Kalle"]I was just loosely saying that an "uncountable sum" of positive numbers is infinite. Index all the discontinuities in $ (a,b)$ as $ x_j$ where $ j$ runs through some index, and let the corresponding jump size at $ x_j$ be $ K_j$. Then the set of $ x_j$ such that $ K_j \geq 1/n$ is finite. The set of all $ x_j$ is the countable union of the former sets as $ n$ runs through the positive integers. The countable union of finite sets is countable, so the number of $ x_j$ is countable.[/quote]
why is the set of $ x_j$ s.t. $ K_j \geq \frac{1}{n}$ finite?
|
231,086
|
How many combinations of pennies, nickels, and dimes are there with a total value of 25 cents? [quote="ernie"]The formula which mewto talks about is this:
$ 3 \plus{} \frac {(\lfloor\frac {n}{5}\rfloor)(\lfloor\frac {n}{5}\rfloor \minus{} 1)}{2}$
This works for $ 10$ through $ 50$ cents, I believe.
$ n \equal{} \text{number of cents}$[/quote]
21:
3+(4*3)/2=3+6=9
Using my way:
0 dimes: 0-4 nickels, so 5
1: 0-2, so 3
2: 0, so 1
equals 9...
o_O
|
2,310,868
|
Find the sum of the real solutions for \(x\) to the equation
\[
\frac{1}{x-1} + \frac{1}{x+1} = 17.
\] [hide=Sol]$\frac{1}{x-1} + \frac{1}{x+1} = 17$
$(x+1)+(x-1)=17(x-1)(x+1)$
$2x=17(x^2-1)$
$2x=17x^2-17$
$17x^2-2x-17=0$
$x=\frac{2\pm\sqrt{2^2+4\cdot17\cdot17}}{34}=\frac{2\pm\sqrt{1160}}{34}=\frac{1\pm\sqrt{290}}{17}$[/hide]
|
23,109
|
Find the minimal constant \(k\) such that for all positive real numbers \(a,b,c\) the following inequality holds:
\[
k\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{8abc}{(a+b)(b+c)(c+a)} \ge 1+k.
\] The minimal k such that the inequality
$k\frac{a^2+b^2+c^2}{bc+ca+ab}+\frac{8abc}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}\geq 1+k$
holds for any positive numbers a, b, c is k = 1. In fact, proving that every k smaller than 1 doesn't work is almost trivial (take a = 1, b = 1, c = 0), and the fact that every $k\geq 1$ will work if k = 1 works follows from $\frac{a^2+b^2+c^2}{bc+ca+ab}\geq 1$ (since $a^2+b^2+c^2\geq bc+ca+ab$). So it is enough to prove the inequality for k = 1. In other words, it is enough to prove the inequality
$\frac{a^2+b^2+c^2}{bc+ca+ab}+\frac{8abc}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}\geq 2$
for any positive numbers a, b, c. Here is how I prove this inequality:
WLOG assume that $a\geq b\geq c$. Then, $a-b\geq 0$, and thus $a^2\left(b+c\right)-b^2\left(c+a\right)=\left(a-b\right)\left(bc+ca+ab\right)\geq 0$, so that $a^2\left(b+c\right)\geq b^2\left(c+a\right)$. Similarly, $b^2\left(c+a\right)\geq c^2\left(a+b\right)$. Hence, we have $a^2\left(b+c\right)\geq b^2\left(c+a\right)\geq c^2\left(a+b\right)$. Combining this with $a\geq b\geq c$, we see that we have the right conditions to apply the Vornicu-Schur inequality (Lemma 1 in http://www.mathlinks.ro/Forum/viewtopic.php?t=24555 post #9) with $x=a^2\left(b+c\right)$, $y=b^2\left(c+a\right)$, $z=c^2\left(a+b\right)$; thus, we get
$a^2\left(b+c\right)\left(a-b\right)\left(a-c\right)+b^2\left(c+a\right)\left(b-c\right)\left(b-a\right)+c^2\left(a+b\right)\left(c-a\right)\left(c-b\right)\geq 0$.
Now, what does this have to do with the inequality in question? Well, some algebra shows
$\frac{a^2+b^2+c^2}{bc+ca+ab}+\frac{8abc}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}-2$
$=\frac{a^2\left(b+c\right)\left(a-b\right)\left(a-c\right)+b^2\left(c+a\right)\left(b-c\right)\left(b-a\right)+c^2\left(a+b\right)\left(c-a\right)\left(c-b\right)}{\left(b+c\right)\left(c+a\right)\left(a+b\right)\left(bc+ca+ab\right)}$;
thus,
$a^2\left(b+c\right)\left(a-b\right)\left(a-c\right)+b^2\left(c+a\right)\left(b-c\right)\left(b-a\right)+c^2\left(a+b\right)\left(c-a\right)\left(c-b\right)\geq 0$
implies
$\frac{a^2+b^2+c^2}{bc+ca+ab}+\frac{8abc}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}-2\geq 0$,
and thus $\frac{a^2+b^2+c^2}{bc+ca+ab}+\frac{8abc}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}\geq 2$, and the proof is complete.
[[b]EDIT:[/b] Now I see that the inequality
$\frac{a^2+b^2+c^2}{bc+ca+ab}+\frac{8abc}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}\geq 2$
is known - look at http://www.mathlinks.ro/Forum/viewtopic.php?t=6403 !]
Darij
|
2,310,918
|
The four positive integers \(a,b,c,d\) satisfy
\[
a\times b\times c\times d = 10!.
\]
Find the smallest possible value of \(a+b+c+d\). Wrong.
[hide=Solution.]Note that $\sqrt[4]{10!}\approx43.6,$ so we try to find factors of $10!$ around that number, and $a=40,b=42,c=45,d=48$ works, giving $a+b+c+d=\boxed{175}.$
Then, note that, by AM-GM, $a+b+c+d\ge4\sqrt[4]{10!}\approx174.58.$ Since $a,b,c,d$ are integers, $a+b+c+d\ge175.$ So $175$ is the minimum.[/hide]
|
2,310,935
|
Solve the nonlinear system
\[
\begin{cases}
\dfrac{x^2}{16}+\dfrac{y^2}{9}=1,\\[6pt]
3x+4y=12.
\end{cases}
\]
Solution:
From \(3x+4y=12\) we have \(y=\dfrac{12-3x}{4}=3-\dfrac{3x}{4}\). Substitute into the ellipse equation:
\[
\frac{x^2}{16}+\frac{\bigl(3-\tfrac{3x}{4}\bigr)^2}{9}=1.
\]
Compute the second term:
\[
\bigl(3-\tfrac{3x}{4}\bigr)^2=9-\tfrac{18x}{4}+\tfrac{9x^2}{16}=9-\tfrac{9x}{2}+\tfrac{9x^2}{16}.
\]
Thus
\[
\frac{x^2}{16}+\frac{1}{9}\Bigl(9-\frac{9x}{2}+\frac{9x^2}{16}\Bigr)=1
\]
\[
\frac{x^2}{16}+1-\frac{x}{2}+\frac{x^2}{16}=1.
\]
Combine terms:
\[
\frac{x^2}{8}-\frac{x}{2}=0.
\]
Factor:
\[
\frac{x}{8}(x-4)=0,
\]
so \(x=0\) or \(x=4\).
For \(x=0\): \(y=3-\tfrac{3\cdot 0}{4}=3\), giving \((0,3)\).
For \(x=4\): \(y=3-\tfrac{3\cdot 4}{4}=0\), giving \((4,0)\).
Thus the solutions are \((x,y)=(0,3)\) and \((4,0)\). I believe that there is a geometric way to solve this but whatever I'm going to algebra bash it.
There are two ways to do it:
[hide=Solution 1]Guess and check is not that hard in this problem. We can easily see that the solutions are $(0,3)$ and $(4,0)$. [/hide]
[hide=Solution 2]Do substitution. Let's find $x$ in terms of $y$.
From the second equation, we can find that $x=(12-4y)/3$
Let's substitute that value into the first equation.
$((12-4y)/3)^2/16+y^2/9=1.$
After simplification, we get a fairly simple quadratic which yields the solutions $(0,3)$ and $(4,0)$.[/hide]
|
2,310,955
|
Find the smallest solution to the equation
\[
\frac{2x}{x-2}+\frac{2x^2-24}{x}=11.
\] Why hasn't anyone posted a solution yet? Are we all scared of the cubic? [img]https://artofproblemsolving.com/assets/images/smilies/tongue.gif[/img]
[hide=Solution.]The Integer Root Theorem says that for an integer $r$ to be a root of $2x^3-13x^2-2x+48=0,$ we need $r\mid48.$ Hence, we have to test $r=-48,-24,-16,-12,-8,-6,-4,-3,-2,-1,1,2,3,4,6,8,12,16,24,48.$ Testing each reveals that $r=6.$ Now, we divide $2x^3-13x^2-2x+48$ by $x-6$ to get $2x^2-x-8=0.$ Using the Quadratic Formula gives $x=\frac{1\pm\sqrt{65}}{4}.$ Note that since $\sqrt{65}>1,$ we have $\frac{1-\sqrt{65}}{4}<0$ and thus $\frac{1-\sqrt{65}}{4}<6$ and $\frac{1-\sqrt{65}}{4}<\frac{1+\sqrt{65}}{4}.$ Hence, our smallest root is $x=\boxed{\frac{1-\sqrt{65}}{4}}.$
|
231,097
|
Let \(a,b,c\ge 0\). Prove that
\[
\sum_{\text{cyc}}\frac{1}{a^2+ab+b^2}\ge\frac{9}{(a+b+c)^2}.
\] [quote="conan_naruto236"]
Suppose $ a \plus{} b \plus{} c \equal{} 1$. We have :
$ \sum\frac {1}{1 \minus{} (ab \plus{} bc \plus{} ca) \minus{} a} \geq 9$
$ \leftrightarrow 1 \minus{} 4(ab \plus{} bc \plus{} ca) \plus{} 9abc \geq 0$
[/quote]
Why? :huh:
[quote="karis"]Let$ a;b;c\geq 0$.Prove that:
$ \sum\frac {1}{a^2 \plus{} ab \plus{} b^2}\geq\frac {9}{(\sum a)^2}$[/quote]
Try $ \sum_{cyc}\frac {1}{a^2 \plus{} ab \plus{} b^2} \equal{} \sum_{cyc}\frac {(3c \plus{} a \plus{} b)^2}{(3c \plus{} a \plus{} b)^2(a^2 \plus{} ab \plus{} b^2)}\geq$
$ \geq\frac {25(a \plus{} b \plus{} c)^2}{\sum(3c \plus{} a \plus{} b)^2(a^2 \plus{} ab \plus{} b^2)}.$ :wink:
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6364
|
2,310,975
|
In a geometric sequence \(a_1,a_2,a_3,\dots\), where all the terms are positive, \(a_5-a_4=576\) and \(a_2-a_1=9\). Find \(a_1+a_2+a_3+a_4+a_5\). Because $a_1, a_2, a_3, \cdots, a_n$ form a geometric series, we can write the sequence as $a_1, a_1 \times r, a_1 \times r^2, \cdots.$ Thus, $a_1 \times r^4 - a_1 \times r^3=576,$ and $a_1 \times r - a_1 =9.$ Factoring the first and second expressions gives $a_1 \times r^3 (r-1)=576,$ and $a_1 (r-1) =9.$ Dividing these two equations gives $r^3 =8,$ so $r=2.$ Plugging $r=2$ into the second equation gives $2a_1 -a_1 =9,$ so $a_1 =9.$ Now it is merely a matter of computation. We have $9+18+36+72+144=279.$
|
2,310,982
|
Let \(a_1,a_2,a_3\) be the first three terms of a geometric sequence. If \(a_1=1\), find the smallest possible value of \(4a_2+5a_3\). Because $a_1,a_2, a_3$ form a geometry sequence, and $a_1 =1,$ we can write the sequence as $1,r,r^2.$ Thus, we seek to find the minimum value of $4r+5r^2.$ The minimum value of this occurs at $r= -\frac{2}{5},$ so the smallest value of $4a_2 + 5a_3$ is $4*-\frac{2}{5} +5* \frac{4}{25}=\boxed{-\frac{4}{5}}.$
|
2,310,983
|
Find all values of \(z\) such that
\[
z^4 - 4z^2 + 3 = 0.
\]
Enter all the solutions, separated by commas. [hide = Solution]
Two numbers that multiply to $3$ and add to $-4$ are $-3$ and $-1$ so this factors to $(z^2 - 3)(z^2 - 1).$ The solutions are then $\pm \sqrt{3}$ and $\pm 1.$ [/hide]
|
23,110
|
Find the best constant \(k\) such that for all positive real numbers \(a,b,c\) the following holds:
\[
\frac{a^3}{ka^2+b^2}+\frac{b^3}{kb^2+c^2}+\frac{c^3}{kc^2+a^2}\ge\frac{a+b+c}{1+k}.
\] [quote="manlio"]Find the best costant $ k$ such that for all positive reals $ x,y,z$ we have
$ \frac {x^3}{kx^2 \plus{} y^2} \plus{} \frac {y^3}{ky^2 \plus{} z^2} \plus{} \frac {z^3}{kz^2 \plus{} x^2} \ge \frac {x \plus{} y \plus{} z}{k \plus{} 1}.$[/quote]The magnificent inequality holds if and only if $ 0\leq k\leq k_0 \equal{} 2.6032\cdots$,
where $ k_0$ is the a root of the following irreducible polynomial
$ 64k^{15} \minus{} 1504k^{14} \plus{} 13076k^{13} \minus{} 38833k^{12} \minus{} 49126k^{11} \plus{} 146301k^{10} \plus{} 316414k^9$
$ \minus{} 466451k^8 \minus{} 61696k^7 \plus{} 1051234k^6 \plus{} 162556k^5 \plus{} 290137k^4 \plus{} 768982k^3 \plus{} 157577k^2$
$ \plus{} 262722k \plus{} 258147;$
with equality if $ k \equal{} k_0,x \equal{} 1,y \equal{} 0.41933\cdots$ is the a root of the following irreducible polynomial
$ 139968y^{45} \minus{} 1029280y^{44} \minus{} 755376y^{43} \plus{} 20810424y^{42} \minus{} 26548844y^{41} \minus{} 87189854y^{40}$
$ \plus{} 155203309y^{39} \minus{} 331344676y^{38} \plus{} 67222529y^{37} \plus{} 2055632070y^{36} \minus{} 2312464392y^{35}$
$ \minus{} 535304560y^{34} \plus{} 2644721111y^{33} \minus{} 5103932592y^{32} \plus{} 12480795729y^{31}$
$ \minus{} 11808250280y^{30} \minus{} 18018288682y^{29} \plus{} 34234003174y^{28} \minus{} 21540086197y^{27}$
$ \plus{} 19245195548y^{26} \minus{} 11931880935y^{25} \minus{} 51416384304y^{24} \plus{} 104600257767y^{23}$
$ \minus{} 41600120400y^{22} \minus{} 35545315746y^{21} \plus{} 71738916790y^{20} \minus{} 55559387145y^{19}$
$ \plus{} 12194200988y^{18} \plus{} 35983391529y^{17} \minus{} 38854216216y^{16} \plus{} 13076286123y^{15}$
$ \plus{} 11185657736y^{14} \minus{} 17746315010y^{13} \plus{} 5098178378y^{12} \plus{} 5537942177y^{11}$
$ \minus{} 3559408644y^{10} \plus{} 867431795y^9 \plus{} 1485730262y^8 \minus{} 443412688y^7 \minus{} 91331564y^6$
$ \plus{} 184690029y^5 \minus{} 14412236y^4 \minus{} 23098411y^3 \plus{} 3498884y^2 \minus{} 1834368y \minus{} 139968,$
$ z \equal{} 0.66956\cdots$ is the a root of the following irreducible polynomial
$ 139968z^{45} \plus{} 1834368z^{44} \minus{} 3498884z^{43} \plus{} 23098411z^{42} \plus{} 14412236z^{41}$
$ \minus{} 184690029z^{40} \plus{} 91331564z^{39} \plus{} 443412688z^{38} \minus{} 1485730262z^{37} \minus{} 867431795z^{36}$
$ \plus{} 3559408644z^{35} \minus{} 5537942177z^{34} \minus{} 5098178378z^{33} \plus{} 17746315010z^{32}$
$ \minus{} 11185657736z^{31} \minus{} 13076286123z^{30} \plus{} 38854216216z^{29} \minus{} 35983391529z^{28}$
$ \minus{} 12194200988z^{27} \plus{} 55559387145z^{26} \minus{} 71738916790z^{25} \plus{} 35545315746z^{24}$
$ \plus{} 41600120400z^{23} \minus{} 104600257767z^{22} \plus{} 51416384304z^{21} \plus{} 11931880935z^{20}$
$ \minus{} 19245195548z^{19} \plus{} 21540086197z^{18} \minus{} 34234003174z^{17} \plus{} 18018288682z^{16}$
$ \plus{} 11808250280z^{15} \minus{} 12480795729z^{14} \plus{} 5103932592z^{13} \minus{} 2644721111z^{12}$
$ \plus{} 535304560z^{11} \plus{} 2312464392z^{10} \minus{} 2055632070z^9 \minus{} 67222529z^8 \plus{} 331344676z^7$
$ \minus{} 155203309z^6 \plus{} 87189854z^5 \plus{} 26548844z^4 \minus{} 20810424z^3 \plus{} 755376z^2 \plus{} 1029280z$
$ \minus{} 139968.$
|
2,311,001
|
The function \(f(x)\) satisfies
\[
f(x) + 2f(1 - x) = 3x^2
\]
for all real numbers \(x\). Find \(f(3)\). Very good, @above! :thumbup: This is called a cyclic function. When you see questions like this, you want to first get $f(3)$ in the first $f$ and then $f(3)$ in the second $f$, it will give you a system of equations.
|
231,102
|
1. Construct a circle tangent to two given lines that passes through a given point.
2. Construct a circle tangent to a given line and passing through two given points. [quote]Get the perpendiclar lines on B,C repectively [/quote]
the two lines through B,C perpendicular to which other two lines??? To WY (see my picture) an YZ, or to PA and PB?
Also, I had thought for sure that one of the steps would be to construct the angle bisector of $ \angle WYZ$ since the center O of the circle we want to construct clearly must be on this angle bisector.
Ok, I made a not so great geogebra picture, so just move the points around yourself if you want the circle to look like its tangent to WY and YZ and goes through point P. (also, there is some extra stuff you can ignore off to the right that I used to simply construct the angle bisector - I don't know how to hide objects)
[geogebra]94980c9ce3a647fb9135b9d28b5847ca3bb63796[/geogebra]
Now, hopefully it's easier to talk about this problem
|
2,311,025
|
Find the ordered pair \((a,b)\) of positive integers, with \(a<b\), for which
\[
\sqrt{1+\sqrt{21+12\sqrt{3}}}=\sqrt{a}+\sqrt{b}.
\] [quote=pog][quote=OlympusHero]$\sqrt{21+12\sqrt{3}}=3+2\sqrt{3}$. Now we need $\sqrt{4+2\sqrt{3}}$ which is $\sqrt{3}+1$. So it is $\boxed{(1,3)}$.[/quote]
Could you elaborate on how you found them?[/quote]
Just do $\sqrt{21+12\sqrt3} = a+\sqrt{b},$ square it, and solve.
|
2,311,075
|
Two real numbers \(x\) and \(y\) satisfy
\[
x - y = 4
\]
and
\[
x^3 - y^3 = 28.
\]
Find \(xy\). I confused.
1) [url=https://artofproblemsolving.com/community/c63h2310754]Same Alcumus question?[/url]
2) My solution wrong somehow :what?:
Cubing gives $x^3-3x^2y+3xy^2-y^3=64$. So $-3x^2y+3xy^2=36$. Dividing each side by $-3$ gives $x^2y-xy^2=-12$, so $xy(x-y)=-12$. Therefore, $xy=-3$.
Oh a found a problem in your solution @above. $x=2$ and $y=1$ doesn't satisfy $x-y=4$.
|
2,311,081
|
The equation \(x^3 - 4x^2 + 5x - \frac{19}{10} = 0\) has real roots \(r,\ s,\) and \(t.\) Find the length of the long diagonal of a box with side lengths \(r,\ s,\) and \(t.\) [hide=Solution]Lets first write what we know using Vieta's. We have $$r+s+t = 4$$, $$rs+rt+st = 5$$, and $rst = \dfrac{19}{10}$. We want to find the space diagonal which can be expressed as $\sqrt{r^s+s^2+t^2}$. We can find this by doing: $$(r+s+t)^2 = r^2+s^2+t^2 + 2(rs+rt+st) \Longrightarrow r^2+s^2+t^2 = 16-10 = 6$$. Therefore, we have an answer of $\boxed{\sqrt{6}}$[/hide]
|
231,109
|
Let \(ABC\) be a triangle. Construct outside it two similar isosceles triangles \(ABD\) and \(ACE\), with \(AB=AD\) and \(AC=AE\). Let \(I=BE\cap CD\), and let \(O\) be the circumcenter of triangle \(IDE\). Prove that \(AO\perp BC\). [hide="Solution"]
Let $ T$ be the midpoint of arc $ DE$ on the circumcircle of $ \triangle EID$. Notice that $ \angle DOE \equal{} 2\angle BID \equal{} 2\angle BAD$, so $ \angle DOT \equal{} \angle BAD$. This, combined with the fact that $ DO \equal{} OT$ gives us the spiral similarity: $ \triangle BAD\sim \triangle TOD$, which yields that $ (AO, BT) \equal{} \angle ADB$. We then observe that $ \triangle ABD\sim \triangle AEC$ with $ AB \equal{} AD$, which gives the rotation: $ \triangle ABE\cong \triangle ADC$, meaning that $ DC \equal{} BE$. We have that $ \angle TDC \equal{} \angle TEB$ and $ TD \equal{} TE$, yielding the rotation: $ \triangle TEB\cong \triangle TDC$. This gives that $ \angle TBI \equal{} \angle TCI$, so $ TIBC$ is cyclic, giving that\[ 90 \minus{} \angle ADB \equal{} \frac {\angle BAD}{2} \equal{} \frac {\angle DOT}{2} \equal{} \angle TED \equal{} \angle TIC \equal{} \angle TBC\]so we have that $ (BT, AO) \plus{} (BT, BC) \equal{} 90$, from which we conclude that $ AO\perp BC$. [/hide]
EDIT: Look here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=841255#841255
I don't believe that my exact solution was posted in the above link, but the idea was the same.
|
2,311,102
|
Find the sum of all solutions to \(2^{|x|} + 3|x| = 18.\) See. Let's say $a$ is a solution. We know
$$2^{|a|} + 3|a| = 18$$.
Now since $|a|=|-a|$, we can substitute to get $2^{|-a|} + 3|-a| = 18$, which implies that $-a$ is also a solution.
EDIT: @wamofan :P Get it?
|
231,117
|
The line \(y - x\sqrt{3} + 3 = 0\) cuts the parabola \(y^{2} = x + 2\) at \(A\) and \(B\). If \(P=(\sqrt{3},0)\), find the value of \(PA\cdot PB\). hello,inserting $ y\equal{}x\sqrt{3}\minus{}3$ in $ y^2\equal{}x\plus{}2$ we get the quadratic equation
$ 3x^2\minus{}6x\sqrt{3}\minus{}x\plus{}7\equal{}0$, with the solutions
$ x_1\equal{}\sqrt{3}\plus{}\frac{1}{6}\plus{}\frac{1}{6}\sqrt{25\plus{}12\sqrt{3}}$
$ x_2\equal{}\sqrt{3}\plus{}\frac{1}{6}\minus{}\frac{1}{6}\sqrt{25\plus{}12\sqrt{3}}$
Inserting this in the equation above we get
$ y_1\equal{}\frac{\sqrt{3}}{6}\plus{}\frac{\sqrt{3}}{6}\sqrt{25\plus{}12\sqrt{3}}$
$ y_2\equal{}\frac{\sqrt{3}}{6}\minus{}\frac{\sqrt{3}}{6}\sqrt{25\plus{}12\sqrt{3}}$
Now we have
$ |PA|\equal{}\sqrt{\left(\sqrt{3}\minus{}\sqrt{3}\minus{}\frac{1}{6}\minus{}\frac{1}{6}\sqrt{25\plus{}12\sqrt{3}}\right)^2\plus{}\left(\frac{\sqrt{3}}{6}\plus{}\frac{\sqrt{3}}{6}\sqrt{25\plus{}12\sqrt{3}}\right)^2}$
$ |PB|\equal{}\sqrt{\left(\sqrt{3}\minus{}\sqrt{3}\minus{}\frac{1}{6}\plus{}\frac{1}{6}\sqrt{25\plus{}12\sqrt{3}}\right)^2\plus{}\left(\frac{\sqrt{3}}{6}\minus{}\frac{\sqrt{3}}{6}\sqrt{25\plus{}12\sqrt{3}}\right)^2}$
and the product is $ |PA|\cdot|PB|\equal{}\frac{8}{3}\plus{}\frac{4}{3}\sqrt{3}$.
Sonnhard.
|
2,311,181
|
The ellipse
\[
\frac{(x-6)^2}{25} + \frac{(y-3)^2}{9} = 1
\]
has two foci. Find the one with the larger \(x\)-coordinate. Enter your answer as an ordered pair, like \((2,1)\). [hide=Sol]Standard form:
$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$
$h=6$, $k=3$, $a=5$, $b=3$
Distance from center to foci is $\sqrt{a^2-b^2}=4$.
Center is $(h,k)=(6,3)$. Foci are $(6\pm4,3)$. The answer is $(10,3)$.[/hide]
|
2,311,213
|
Compute the exact value of the expression
\[
\left|\pi - \left|\pi - 7\right|\right|.
\]
Write your answer using only integers and \(\pi\), without any absolute value signs. Starting from the inner absolute value signs, we know $|\pi-7|$ can we written as $7-\pi$. Now we have $|\pi-(7-\pi)|.$ Simplifying this expression, we have [hide=sol]$|2\pi-7|$ which is equivalent to $7-2\pi$ and we are done.[/hide].
|
2,311,254
|
Let \(t\) be a real number such that
\[
\begin{aligned}
\lfloor t \rfloor &= 4,\\
\lfloor t+\{t\}\rfloor &= 4,\\
\lfloor t+2\{t\}\rfloor &= 5,
\end{aligned}
\]
where \(\{t\}=t-\lfloor t\rfloor\). Find all possible values of \(t\). (Enter your answer in interval notation.) [hide=alternate way, hint]you could start by bringing $\lfloor t \rfloor $ out of the last 2 equations (it's already an integer) and solving for a bound on $\{t\}$, then adding $\lfloor t \rfloor $ back.[/hide]
|
2,311,260
|
Find the smallest solution to the equation
\[
\frac{1}{x-2} + \frac{1}{x-4} = \frac{3}{x-3}.
\] [hide=Solution]After some manipulation and factoring, you'll reach a quadratic solution $$x = \frac{6 \pm \sqrt{6^2 - 4 \cdot 6}}{2} = 3 \pm \sqrt{3},$$ whose least is $\boxed{3-\sqrt{3}}.$ [I don't why the topic is FE and ineq where I found this.][/hide]
|
23,113
|
For \(a,b,c\) positive real numbers, prove that
\[
\frac{a^4}{a^3+b^3}+\frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3}\ge\frac{a+b+c}{2}.
\] it's the same as $\sum_{cyc} a\frac{a^3-b^3}{a^3+b^3}\geq 0$.
1. case: $\frac{a}{b},\frac{b}{c},\frac{c}{a}\leq \frac 12 (\sqrt{13}+1)$. in that case we can use $\frac{1-x^3}{1+x^3}\geq \frac 32 (1-x)$ for $x\geq \frac 16 (\sqrt{13}-1)$; taking $x=\frac {b}{a}$, ... we get the desired inequality.
2. case: $\frac{a}{b}\geq \frac 12 (\sqrt{13}+1)$ (or something cyclic).
2.1. : $a\geq b\geq c$.
2.1.1. : $\frac{a}{b}\geq \frac 12 (\sqrt{13}+1)\geq 2$.
$\frac{a^4}{a^3+b^3}+\frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3}\geq \frac 89 a+\frac 12 b\geq \frac{a+b+c}{2}$
2.1.2. : $\frac{b}{c}\geq \frac 12 (\sqrt{13}+1)\geq 2$.
$\frac{a^4}{a^3+b^3}+\frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3}\geq \frac 12 a+\frac 89 b\geq \frac{a+b+c}{2}$
2.2. : $a\leq b\leq c$.
it follows that $\frac{c}{a}\geq \frac 12 (\sqrt{13}+1)$.
$\frac{a^4}{a^3+b^3}+\frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3}-\frac{a+b+c}{2}\geq \frac{a^4}{a^3+b^3}+\frac{b^4}{b^3+c^3}+\frac c5 -\frac b2\geq \frac{b^4}{b^3+c^3}+\frac c5 -\frac b2=b(\frac{1}{1+x^3}+\frac x5 -\frac 12)$
where $x=\frac cb$. it's left to show that $\frac{1}{1+x^3}+\frac x5 -\frac 12\geq 0$.
i just plotted this using mathematica, and wow, it is true, but you don't guess how sharp it is :D ... about $0.01$ from being false :lol:
this is not really interesting, however let's do this:
let $f(x)=10+(1+x^3)(2x-5)=2x^4-5x^3+2x+5$. the inequality is trivial for $x\leq 1$ and $x\geq \frac 52$.
therefore we only consider the $x$ between.
we have $f'(x)=8x^3-15x^2+2$. since $f'(1)<0$ and $f'(\frac 52)>0$, the minimum occurs somewhere where $f'(x)=0$, therefore $8x^3-15x^2+2=0$. but then
$32f(x)=8x(15x^2-2)-160x^3+64x+160=-40x^3+48x+160=-5(15x^2-2)+48x+160=-75x^2+48x+170\geq 0$
for the zero $x_0$ of $8x^3-15x^2+2$ as one easily checks: $x_0\leq 1.8$, but $-75(1.8)^2+48(1.8)+170\geq 0$.
omg, what an ugly proof :D
Peter
|
2,311,321
|
The function \(f\) takes nonnegative integers to real numbers, such that \(f(1)=1\), and
\[
f(m+n)+f(m-n)=\frac{f(2m)+f(2n)}{2}
\]
for all nonnegative integers \(m\ge n\). Find the sum of all possible values of \(f(10)\). [hide=different sol]We notice that $n=0$ gives $f(2m)=4f(m).$ The fakesolve way would be to notice $x^2,$ but here’s the rigorous way. We plug in $5,3$ to get a system with $f(2),f(8),f(6),$ and $f(10).$ We plug in $6,4$ to get a system with $f(10),f(2),f(8),$ and $f(12).$ We use the relation $f(2m)=4f(m)$ to calculate $f$ for powers of $2,$ and also to write $f(12)$ in terms of $f(6).$ Now we have a 2 var 2 eq linear system in $f(6),f(10)$ which is easily solvable to confirm $f(10)=100.$[/hide]
|
2,311,349
|
Let \(a\) and \(b\) be real numbers. One of the roots of \(x^3 + a x + b = 0\) is \(1 + i\sqrt{3}\). Find \(a + b\). The [hide=Alcumus solution]Since the coefficients are real, another root is $1 - i \sqrt{3}.$ By Vieta's formulas, the sum of the roots is 0, so the third root is $-2.$ Hence, the cubic polynomial is
\begin{align*}
(x - 1 - i \sqrt{3})(x - 1 + i \sqrt{3})(x + 2) &= ((x - 1)^2 - (i \sqrt{3})^2)(x + 2) \\
&= ((x - 1)^2 + 3)(x + 2) \\
&= x^3 + 8.
\end{align*}Thus, $a = 0$ and $b = 8,$ so $a + b = \boxed{8}.$[/hide] isn’t really complete, so I will present [hide=my solution.]Since the coefficients are real, then the conjugate of $1+i \sqrt3$, which is $1-i
\sqrt3$ must also be a root. Because the sum of the roots of our cubic is $0$, thus another root must be $-2$. Solving, we find that our cubic is $x^3+8.\quad\blacksquare$[/hide]
|
231,136
|
Let \(G\) be a non-commutative group. Consider all bijections \(a\mapsto a'\) of \(G\) onto itself such that \((ab)'=b'a'\) for all \(a,b\in G\) (i.e. the anti-automorphisms of \(G\)). Prove that these mappings together with the automorphisms of \(G\) constitute a group which contains the group of automorphisms of \(G\) as a direct factor. this is also true for commutative groups, but then it's, of course, boring ;-).
so let $ G$ be non-commutative. then there is no antiautomorphism, which is also an automorphism. there is a canonical antiautomorphism, namely the inversion $ i : x \mapsto x^{ \minus{} 1}$ which satisfies $ i^2 \equal{} 1,i \neq 1$ and commutes with all automorphisms. furthermore, $ f \mapsto fi$ defines a bijection between the automorphisms and the antiautomorphisms, the inverse is also given by $ f \to fi$. from this we see that
$ Aut(G) \times \langle i \rangle \to Sym(G) , (f,g) \mapsto fg$
is an injective group homomorphism, whose image consists exactly of the automorphisms and the antiautomorphisms. in particular, these form a group isomorphic to $ Aut(G) \times \mathbb{Z}/2$.
|
2,311,366
|
In the coordinate plane, the graph of
\[
\lvert x+y-1\rvert + \bigl\lvert\,\lvert x\rvert - x\bigr\rvert + \bigl\lvert\,\lvert x-1\rvert + x-1\bigr\rvert = 0
\]
is a certain curve. Find the length of this curve. [hide=Solution]
Each of the absolute values must equal $0$, so we have $x+y-1=0$, $|x|-x=0$, and $|x-1|+x-1=0$. Solving these equations, we find $0\leq x\leq 1$, and $y=-x+1$. So the curve is the line $y=-x+1$ and the domain is $x\in [0,1]$, thus we find the length of the line to be $\boxed{\sqrt2}$.
[/hide]
|
231,137
|
Solve the equation
\[
\sin x+\sin 2x+\sin 3x=1.
\] I've tried but i couldn't get the solution...however here it is
$ \sin x \plus{} \sin 2x \plus{} \sin 3x \equal{} 1$
$ \sin x \plus{} 2\sin x \cos x \plus{} \sin x(3\minus{}4\sin^2 x) \equal{} 1$
$ \sin x (1 \plus{} 2\cos x \plus{} 3 \minus{} 4 \sin^2 x) \equal{} 1$
$ \sin x (4 \plus{} 2\cos x\minus{} 4 (1\minus{}\cos^2 x)) \equal{} 1$
$ \sin x (4 \plus{} 2\cos x\minus{} 4 \plus{} 4\cos^2 x)) \equal{} 1$
$ \sin x (2\cos x \plus{} 4\cos^2 x)) \equal{} 1$
$ 2\sin x\cos x (1 \plus{} 2\cos x) \equal{} 1$
$ \sin 2x (1 \plus{} 2\cos x) \equal{} 1$
and then?
or
$ \sin x (4 \plus{} 2\cos x\minus{} 4 \sin^2 x) \equal{} 1$
$ 2\cos x \equal{} \frac{1}{\sin x} \minus{} 4 \plus{} 4 \sin^2 x$
$ (2\cos x)^2 \equal{} (\frac{1}{\sin x} \minus{} 4 \plus{} 4 \sin^2)^2$
$ 4(1 \minus{} \sin^2 x) \equal{} \frac{1}{\sin^2 x} \minus{}16 \plus{} 16\sin^4 x \minus{} \frac{8}{\sin x} \plus{} 8\sin x \minus{} 32 \sin^2 x$
$ 1 \minus{} 16\sin^2 x \plus{} 16 \sin^6 x \minus{} 8 \sin x \plus{} 8 \sin^3 x \minus{}32 \sin^4 x \minus{} 4 \sin^2 x \plus{} 4 sin^4 x\equal{}0$
$ 16\sin^6 x \minus{} 28 \sin^4 x \plus{} 8 \sin^3 x \minus{} 20\sin^2 x \minus{} 8\sin x \plus{} 1 \equal{} 0$
But i can't solve this 6th degree equation...
|
2,311,372
|
Let real numbers \(a,b,c\) satisfy
\[
a+b+c=ab+bc+ca>0.
\]
Prove that
\[
\sqrt{a^{2}-a+1}+\sqrt{b^{2}-b+1}+\sqrt{c^{2}-c+1}\ge a+b+c.
\] [quote=anhduy98]Let three real numbers $ a, b, c $ satisfy : $ a + b + c = ab + bc + ca> 0 $ . Prove that:$$\sqrt{a^2-a+1}+\sqrt{b^2-b+1}+\sqrt{c^2-c+1}\ge a+b+c$$[/quote]
In fact inequaity is true for all real numbers $a,b,c$ not just for $a+b+c>0$ since for $a+b+c\leq 0$ clearly hold. Now we can consider $a+b+c>0$. We can easily prove $a^2-a+1=(c+a-1)(a+b-1)$ same for the others. Also we can prove that $a+b-1=\frac{a^2+ab+b^2}{a+b+c}$ same for the others. So, inequality it is equivalent to prove,
$$\sum_{cyc}\sqrt{(a^2+ab+b^2)(b^2+bc+c^2)}\geq (a+b+c)^2.$$
Since $(a^2+ab+b^2)(b^2+bc+c^2)-\left(b^2+ca+\frac{ab+bc}{2}\right)^2=\frac{3}{4}b^2(c-a)^2\geq 0$ hence we have,
$$\sum_{cyc}\sqrt{(a^2+ab+b^2)(b^2+bc+c^2)}\geq \sum_{cyc}\left|b^2+ca+\frac{ab+bc}{2}\right|\geq \sum_{cyc}\left(b^2+ca+\frac{ab+bc}{2}\right)=(a+b+c)^2.$$
|
2,311,374
|
Evaluate the infinite geometric series:
\[
\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\dots
\] [quote=Stormbreaker7984][quote=selinapan][hide]Do you need a solution or hint.... since it's an alcumus problem I'm assuming hint
Use the geometric series sum formula, $\frac{a}{1-r}$, where a is your first term and r is the common ratio. [/hide][/quote]
would that be 2/3?[/quote]Um definitely not[color=#ccc][size=50]jkjk lol[/size]
[/color]
|
231,138
|
Let $2n$ distinct points on a circle be given. Arrange them into disjoint pairs in an arbitrary way and join each pair by a chord. Determine the probability that no two of these $n$ chords intersect. (All possible arrangements into pairs are equally likely.) The total number $ a_n$ of "good" pairings of $ 2n$ vertices is given by the $ n$th Catalan number $ \frac{(2n)!}{n!(n\plus{}1)!}$: it's quite clear that $ a_0 \equal{} a_ 1 \equal{} 1$, and then choose a fixed vertex and connect it to in all possible ways to see that $ a_{n \plus{} 1} \equal{} \sum_{i\equal{} 0}^n a_i a_{n \minus{} i}$. The total number of pairings is just $ (2n \minus{} 1)!! \equal{} \frac{(2n)!}{2^n n!}$, so the answer is the ratio of these two, $ \frac{2^n}{(n \plus{} 1)!}$.
|
231,142
|
Let \(n\) be a positive integer. Prove that, for \(0<x<\dfrac{\pi}{n+1}\),
\[
\sin x - \frac{\sin 2x}{2} + \frac{\sin 3x}{3} - \cdots + (-1)^{n+1}\frac{\sin n x}{n} - \frac{x}{2}
\]
is positive if \(n\) is odd and negative if \(n\) is even. $ f_n(x) \equal{} \sin{x} \minus{} \frac {\sin{2x}}{2} \plus{} \cdots \plus{} ( \minus{} 1)^{n \plus{} 1}\frac {\sin{nx}}{n} \minus{} \frac {x}{2}$.
$ f_n'(x) \equal{} \minus{} \mbox{Re}(\sum_{n \equal{} 1}^{n}z^n) \minus{} \frac12$ with $ z \equal{} e^{i(\pi \plus{} x)}$.
After some simplifications we get
$ f_n'(x) \equal{} \frac {( \minus{} 1)^{n \plus{} 1}}{2}((1 \minus{} \cos(x))\frac {\sin((n \plus{} 1)x)}{\sin(x)} \plus{} \cos((n \plus{} 1)x))$
and $ f_n''(x) \equal{} \frac {( \minus{} 1)^{n}}{2}\frac {(n \plus{} 1)\sin(nx) \plus{} n\sin((n \plus{} 1)x)}{1 \plus{} \cos(x)}$
The formula for $ f_n''$ shows that $ ( \minus{} 1)^n f$ is convex for $ 0 < x < \frac {\pi}{n \plus{} 1}$
$ f_n(0) \equal{} 0$ and $ f_n'(0) \equal{} \frac {( \minus{} 1)^{n \plus{} 1}}{2}$.
We are ready when we can show that $ ( \minus{} 1)^{n \plus{} 1}f_n(\frac {\pi}{n \plus{} 1}) > 0$.
We have to distinct between two different, but very similar cases, namely $ n$ is odd, and $ n$ is even.
Let's restrict to the case $ n$ is even.
We prove $ f_{2n}(\frac {\pi}{2n \plus{} 1}) < 0$.
$ f_{2n}(\frac {\pi}{2n \plus{} 1}) \equal{} \sum_{k \equal{} 1}^{2n} ( \minus{} 1)^{k \plus{} 1}\frac {\sin(\frac {k\pi}{2n \plus{} 1})}{k} \minus{} \frac {\pi}{2(2n \plus{} 1)} \equal{}$
$ \frac {\pi}{2n \plus{} 1}(\sum_{k \equal{} 1}^{n} \frac {\sin(\frac {(2k \minus{} 1)\pi}{2n \plus{} 1})}{\frac {(2k \minus{} 1)\pi}{2n \plus{} 1}} \minus{} \sum_{k \equal{} 1}^{n} \frac {\sin(\frac {2k\pi}{2n \plus{} 1})}{\frac {2k\pi}{2n \plus{} 1}}) \minus{} \frac {\pi}{2(2n \plus{} 1)}$.
The function $ x \mapsto \frac {\sin(x)}{x}$ is descending on $ [0,\pi]$, thus
both sums lay between $ a$ and $ a \plus{} \frac {2\pi}{2n \plus{} 1}$, where $ a \equal{} \int_0^{\pi}\frac {\sin(x)}{x}\,dx$.
Thus $ f_{2n}(\frac {\pi}{2n \plus{} 1}) < \frac {\pi}{2n \plus{} 1}\cdot\frac {2\pi}{2n \plus{} 1} \minus{} \frac {\pi}{2(2n \plus{} 1)} < 0$.
|
2,311,429
|
If \(x\) is a real number and \(\lfloor x \rfloor = -9\), how many possible values are there for \(\lfloor 5x \rfloor\)? Let's say our test values were $-9.1,-9.2,-9.3,...,-9.9$ (note that these numbers satisfy $\lfloor x \rfloor = -9$).
If we plug in $-9.1$ and $-9.2$ into $\lfloor 5x \rfloor,$ we get $-46.$
If we plug in $-9.3$ and $-9.4$ into $\lfloor 5x \rfloor,$ we get $-47.$
If we plug in $-9.5$ and $-9.6$ into $\lfloor 5x \rfloor,$ we get $-48.$
If we plug in $-9.7$ and $-9.8$ into $\lfloor 5x \rfloor,$ we get $-49.$
If we plug in $-9.9$ into $\lfloor 5x \rfloor,$ we get $-50.$
[hide=answer]Therefore there are only $5$ possible values for $\lfloor 5x \rfloor.$[/hide]
|
2,311,431
|
Solve
\[
\frac{1}{x+9}+\frac{1}{x+7}=\frac{1}{x+10}+\frac{1}{x+6}.
\] Why should this be in Intermediate Algebra? This is so simple!
[hide=My Steps]
1. Multiply all terms by all 4 denominators
2. Simplify
3. Subtract $2x^3$ from both sides
4. Subtract $48x^2$ from both sides
5. Subtract $382x$ from both sides
6. Subtract $960$ from both sides
7. After you do all those steps, you [b][u][I]should[/i][/u][/b] get $-6x=48$
8. $x=-8$
[/hide]
Try it out!
|
2,311,445
|
Let \(x\diamondsuit y=xy+x+y+1\) for positive real \(x,y\). Given \(x+y=4\), compute the maximum possible value of \(x\diamondsuit y\). [hide=Sol]$x\diamondsuit y=xy+x+y+1=xy+5$, so we want to maximize $xy$, and this is done when $x=y=2$. $x\diamondsuit y=\boxed9$[/hide]
|
2,311,447
|
The function \(f(x)=x+1\) generates the sequence
\[
1,2,3,4,\dots
\]
in the sense that plugging any number in the sequence into \(f(x)\) gives the next number in the sequence.
What rational function \(g(x)\) generates the sequence
\[
\frac{1}{2},\ \frac{2}{3},\ \frac{3}{4},\ \frac{4}{5},\ \dots
\]
in this manner? Let $a_1=\frac{1}{2}, a_2=\frac{2}{3}, a_3=\frac{3}{4}, \dots, a_n = \frac{n}{n+1}$.
So, $a_n = \frac{n}{n+1} \implies na_n+a_n=n \implies n-na_n = a_n \implies n(1-a_n) = a_n \implies n = \frac{a_n}{1-a_n}$.
We seek a function such that $f(a_n)=a_{n+1}=\frac{n+1}{n+2}$. Plugging in $n = \frac{a_n}{1-a_n}$ gives:
$f(a_n) = \frac{\frac{a_n}{1-a_n}+1}{\frac{a_n}{1-a_n}+2} = \frac{a_n+1-a_n}{a_n+2-2a_n} = \frac{1}{2-a_n}$.
Hence, the function we seek is $g(x) = \frac{1}{2-x}$, and we are done.
|
231,147
|
Let \(a,b,c\in[1,2]\). Find the maximum value of
\[
(a-b)^4+(b-c)^4+(c-a)^4+(a-b)(b-c)(c-a).
\] [quote="Dr Sonnhard Graubner"]hello, i have found with calculus $ P_{Max} \equal{} 2$ for $ a \equal{} 1,b \equal{} 1,c \equal{} 2$.
Sonnhard.[/quote]
Yes, :lol: ,and the proof ?(note the equality holds also when $ a\equal{}b\equal{}2.c\equal{}1$
|
231,148
|
Find a closed-form expression for the sum
\[
1^k + 2^k + \cdots + n^k,
\]
where \(k\) and \(n\) are positive integers. Thank you very much hsiljak.
I have another question about this.
Is it true that $ 1^k\plus{}2^k\plus{}...\plus{}n^k$ is a polynomial of degree $ k\plus{}1$, there is no constant term, the least degree term in the polynomial is always $ n$, and the leading coefficient is $ 1/k$?
|
2,311,489
|
Let \(a_n\) be the first (leading) digit of \(2^n\).
Prove that the sequence \((a_n)_{n\ge1}\) is not periodic. Let $a_n=k$ be the first digit of $2^n$, then $k\cdot 10^s \le 2^n <(k+1)10^s \implies s+\log k\le \log(2^n)\implies (k+1)10^s
\implies \alpha_n:={\rm fr}(n\log 2)\in [\log k,\log(k+1))$.
Calculating modulo 1, fixing $n$ and $p\ge 1$ integer we have $\alpha_{n+tp}=\alpha_n+t (p\log 2)$,
Since $p\log 2\not\in {\bf Q}$, these numbers lie dense on the unit circle when $t=1,2,\cdots$
Hence $p$ cannot be a period.
|
231,149
|
For any sets \(C\) and \(D\), prove that
\[
\text{i)}\quad C = (C\cup D)\cap D,
\qquad
\text{ii)}\quad C = (C\cap D)\cup C.
\] Well, what can I use in the proof? If you can use the axioms of Boolean algebra, then simple absorption (or the distribution) rule should do the trick-they're both applicable here (BTW, check your i) ;) ).
On the other hand, if you want to use truth tables instead of axioms (although the tables are generally proved using the axioms of Boolean algebra), then you can do the same as in logic (just take the cap as conjunction, and cup as disjunction), or you can simply do the following: draw a Venn diagram. Then you have four possibilities where to put an element (just in C, just in D, in the common part of C and D, or outside both of them), and show that if the elemnt is on the LHS of your equality, then and only then is on the RHS. That's basically same as truth table.
|
2,311,493
|
Solve \(\tan x = \sin x\) for \(0 \le x \le 2\pi.\) Enter all the solutions, separated by commas. [hide=solution]Expressing $\tan x$ as $\dfrac{\sin x}{\cos x}$ gives $$\dfrac{\sin x}{\cos x} = \sin x.$$
If $\sin x = 0$, then the equality holds, giving $x = 0, \pi, 2\pi$.
If $\sin x \neq 0$, then simplifying the equality gives $\cos x = 1$, which gives the same solutions as the first case.
Therefore, our answer is $\boxed{0, \pi, 2\pi}$.[/hide]
|
2,311,494
|
Let \(\mathbf{a}=\begin{pmatrix}2\\1\\5\end{pmatrix}.\) Find the vector \(\mathbf{b}\) such that
\[
\mathbf{a}\cdot\mathbf{b}=11
\]
and
\[
\mathbf{a}\times\mathbf{b}=\begin{pmatrix}-13\\-9\\7\end{pmatrix}.
\] [quote=ilovepizza2020][quote=mop][quote=cryptographer][hide=AoPS Solution #1]Let $\mathbf{b} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}.$ Then the equation $\mathbf{a} \cdot \mathbf{b} = 11$ gives us $2x + y + 5z = 11.$ Also,
\[\mathbf{a} \times \mathbf{b} = \begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix} \times \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -5y + z \\ 5x - 2z \\ -x + 2y \end{pmatrix}.\]Comparing entries, we obtain
\begin{align*}
-5y + z &= -13, \\
5x - 2z &= -9, \\
-x + 2y &= 7.
\end{align*}Solving this system, along with the equation $2x + y + z = 5z = 11,$ we find $x = -1,$ $y = 3,$ and $z = 2.$ Hence, $\mathbf{b} = \boxed{\begin{pmatrix} -1 \\ 3 \\ 2 \end{pmatrix}}.$[/hide]
[hide=AoPS Solution #2]By the vector triple product, for any vectors $\mathbf{p},$ $\mathbf{q},$ and $\mathbf{r},$
\[\mathbf{p} \times (\mathbf{q} \times \mathbf{r}) = (\mathbf{p} \cdot \mathbf{r}) \mathbf{q} - (\mathbf{p} \cdot \mathbf{q}) \mathbf{r}.\]From the equation $\mathbf{a} \times \mathbf{b} = \begin{pmatrix} -13 \\ -9 \\ 7 \end{pmatrix},$
\[\mathbf{a} \times (\mathbf{a} \times \mathbf{b}) = \begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix} \times \begin{pmatrix} -13 \\ -9 \\ 7 \end{pmatrix} = \begin{pmatrix} 52 \\ -79 \\ -5 \end{pmatrix}.\]Hence,
\[(\mathbf{a} \cdot \mathbf{b}) \mathbf{a} - (\mathbf{a} \cdot \mathbf{a}) \mathbf{b} = \begin{pmatrix} 52 \\ -79 \\ -5 \end{pmatrix}.\]Then
\[11 \mathbf{a} - 30 \mathbf{b} = \begin{pmatrix} 52 \\ -79 \\ -5 \end{pmatrix}.\]Solving, we find $\mathbf{b} = \boxed{\begin{pmatrix} -1 \\ 3 \\ 2 \end{pmatrix}}.$[/hide][/quote]
here we go again...
don post aops solutions lol, this happened in another thread so go there for better context :rotfl:[/quote]
Don't worry, I already reported all of the posts he made with the solutions in them.[/quote]
he? You sure bout that?
|
2,311,498
|
The points \((0,0)\), \((a,11)\), and \((b,37)\) are the vertices of an equilateral triangle. Find the value of \(ab\). Compute the matrix for rotation by $60$ degrees about the origin (which is a linear transformation). Then, solve for $a$ (the $y$ coordinate of the rotation of $(a, 11)$ should be $37$). Then solve for $b$.
|
2,311,499
|
Find the curve defined by the equation
\[
r = 2.
\]
(A) Line
(B) Circle
(C) Parabola
(D) Ellipse
(E) Hyperbola
Enter the letter of the correct option. [quote=GameBot]Find the curve defined by the equation
\[r = 2.\]
(A) Line
(B) Circle
(C) Parabola
(D) Ellipse
(E) Hyperbola
Enter the letter of the correct option.[/quote]
Assuming it is in polar coordinates, answer is $B$
|
2,311,504
|
Find the vector \(\mathbf{v}\) such that
\[
\begin{pmatrix}
2 & 3 & -1\\[4pt]
0 & 4 & 5\\[4pt]
4 & 0 & -2
\end{pmatrix}
\mathbf{v}
=
\begin{pmatrix}
2\\[4pt]
27\\[4pt]
-14
\end{pmatrix}.
\] Oh, no! An empty GameBot thread?
Let $\mathbf{\text v} = \begin{pmatrix} a \\ b \\ c \end{pmatrix}.$
By the matrix dot product rule, we get
$2x+3y-z=2,$
$4y+5z=27,$
$4x-2z=-14.$
Solving, we get $a=-2,~b=3,~$ and $c=3,$ so v = $ \begin{pmatrix} -2 \\ 3 \\ 3 \end{pmatrix}.$
|
2,311,506
|
For real numbers \(t\) where \(\tan t\) and \(\sec t\) are defined, the point
\[
(x,y)=(\tan t,\sec t)
\]
is plotted. All the plotted points lie on what kind of curve?
(A) Line
(B) Circle
(C) Parabola
(D) Ellipse
(E) Hyperbola
Enter the letter of the correct option. [hide=Solution]Note that $$\sec^2{t}-\tan^2{t}=\frac{1}{\cos^2{t}}-\frac{\sin^2{t}}{\cos^2{t}}=\frac{1-\sin^2{t}}{\cos^2{t}}=\frac{\cos^2{t}}{\cos^2{t}}=1.$$ This means that $x^2-y^2=1.$ Thus, the shape of all the plotted points is a $\text{Hyperbola}\Longrightarrow\boxed{(\text{E})}.$[/hide]
|
2,311,507
|
When \(\begin{pmatrix} a \\ b \end{pmatrix}\) is projected onto \(\begin{pmatrix} \sqrt{3} \\ 1 \end{pmatrix}\), the resulting vector has magnitude \(\sqrt{3}\). Also, \(a = 2 + b\sqrt{3}\). Enter all possible values of \(a\), separated by commas. [hide=Why does Alcumus solution always overkill?]
You just draw the line and a circle centered at the origin with radius sqrt(3), and draw the line perpendicular to that line and see where it is tangent to the circle.
[/hide]
|
231,151
|
Να βρεθεί ο φυσικός αριθμός \(n\) αν ο αριθμός
\[
A=2^{17}+17\cdot 2^{12}+2^{n}
\]
είναι τέλειο τετράγωνο ακεραίου αριθμού. [b][color=red]Καλησπέρα μετά από καιρό! Καταρχήν εύχομαι σε όλους καλή σχολική και καλή ΟΛΥΜΠΙΑΚΗ χρονιά! Καλή επιτυχία στον ερχόμενο διαγωνισμό της ΕΜΕ και καλή σταδιοδρομία στους (πλέον) φοιτητές μας![/color] [/b]
Για την άσκηση είμαι σχεδόν σίγουρος ότι θα υπάρχει πιο σύντομος δρόμος με κάποια άλλη διαδικασία αλλά σας παραθέτω τις δικές μου σκέψεις:
Η παράσταση Α είναι ίση με $ 49\cdot 2^{12} + 2^n$. Θέλουμε λοιπόν να λύσουμε τη διοφαντική εξίσωση $ 49\cdot 2^{12} + 2^n = x^2$ για κάποιο ακέραιο $ x$ (ο οποίος προφανώς είναι μεγαλύτερος από $ 7\cdot 2^6$).
Ισοδύναμα $ (x - 7\cdot 2^6)(x + 7\cdot 2^6) = 2^n$. To $ n$ αποκλείεται να είναι ίσο με 0, άρα πρέπει να λύσουμε το σύστημα
$ \left\{\begin{array}{ll} x - 7\cdot 2^6 = 2^{n_1} , & \hbox{} \\
x + 7\cdot 2^6 = 2^{n_2} , & \hbox{.} \end{array} \right.$ με $ n_1,n_2 \geq 1$ και $ n_1 + n_2 = n$
Με αφαίρεση κατά μέλη των παραπάνω παίρνουμε $ 7\cdot 2^7 = 2^{n_2} - 2^{n_1}$. Εύκολα φαίνεται από την τελευταία, ότι $ n_1,n_2 \geq 7$ διότι διαφορετικά καταλήγουμε σε άτοπο [Τί γίνεται εάν τουλάχιστον ένας εκ των $ n_1, n_2$ (ή ακόμη χειρότερα [b]και οι δύο[/b]) είναι μικρότερος από 7 ?] Άρα καταλήγουμε στην εξίσωση
$ 7 = 2^{n_2 - 7} - 2^{n_1 - 7}$ η οποία είναι αδύνατη αν οι $ n_1,n_2$ είναι (και οι δύο) μεγαλύτεροι από 7. Άρα τουλάχιστον ένας είναι ίσος με 7 (αφού ούτε μικρότεροι από 7 μπορεί να είναι).
Αν $ n_2 = 7$ τότε παίρνουμε αδύνατη εξίσωση, ενώ αν $ n_1 = 7$, τότε πρέπει $ 8 = 2^{n_2 - 7}$ άρα $ n_2 = 10$, συνεπως
$ n = n_1 + n_2 \Rightarrow \boxed{n = 17}$
Ελπίζω να μη ξέχασα κάποια περίπτωση
Αλέξανδρος
|
2,311,522
|
For real numbers \(t\), the point
\[
(x,y) = (5\cos 2t,\, 3\sin 2t)
\]
is plotted. All the plotted points lie on what kind of curve?
(A) Line
(B) Circle
(C) Parabola
(D) Ellipse
(E) Hyperbola
Enter the letter of the correct option. [hide=Solution]We have $x=5\cos2t$ and $y=3\sin2t.$ Then, $x^2=25\cos^22t$ and $y^2=9\sin^22t,$ so $\frac{x^2}{25}=\cos^22t$ and $\frac{y^2}{9}=\sin^22t.$ Adding these two equations gives $\frac{x^2}{25}+\frac{y^2}{9}=\sin^22t+\cos^22t=1,$ which is clearly the equation of an ellipse, which is option $\boxed{\textbf{(D)}}.$
|
2,311,547
|
If \(a\ge b\ge c\ge 0\) and \(a^2+b^2+c^2=3\), prove that
\[
abc-1+\sqrt{\tfrac{2}{3}}\,(a-c)\ge 0.
\] [quote=csav10]If $a\ge b\ge c\ge 0$ and $a^2+b^2+c^2=3$, then
$abc-1+\sqrt\frac 2{3}\ (a-c)\ge 0$.[/quote]
$$LHS \ge abc-1+\sqrt{\frac{2}{3}}\sqrt{(a-c)^2+(a-b)(c-b)}=r-1+\sqrt{\frac{2(p^2-3q)}{3}}$$
$\bullet Let : x=\sqrt{\frac{2(p^2-3q)}{3}} \rightarrow q=\frac{6-3x^2}{2},p=\sqrt{9-3x^2}, \left(0\le x \le \sqrt{2}\right),p=a+b+c;q=ab+bc+ca;r=abc$
[b][u][i]+Case 1:[/i][/u][/b] $1 \le x \le \sqrt{2} $
$$\rightarrow LHS \ge r-1+x \ge -1+x \ge 0$$
[b][u][i]+Case 2:[/i][/u][/b] $0 \le x \le 1 , (Schur) \rightarrow 9r \ge 4pq-p^3=(3-3x^2)\sqrt{9-3x^2} $
$$\rightarrow LHS \ge (1-x^2)\sqrt{1-\frac{x^2}{3}}-1+x =\frac{x(1-x)\left[x^2(1-x)+3x(1-x)+(1-x)+5 \right]}{3+3(x+1)\sqrt{1-\dfrac{x^2}{3}}} \ge 0$$
|
2,311,557
|
The matrix
\[
\mathbf{M} = \begin{pmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{pmatrix}
\]
satisfies \(\mathbf{M}^T \mathbf{M} = \mathbf{I}\). Find \(x^2 + y^2 + z^2\).
For a matrix \(\mathbf{A}\), \(\mathbf{A}^T\) is the transpose of \(\mathbf{A}\). Here,
\[
\mathbf{M}^T = \begin{pmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{pmatrix}.
\] [hide=shorter solution]Since $\mathbf{M}^T$ is the inverse of $\mathbf{M}$, it's also true that $\mathbf{MM}^T=\mathbf{I}$. Therefore, the product-sum of the second row of $\mathbf{M}$ and the second column of $\mathbf{M}^T$ is $x^2+y^2+z^2=\mathbf{I}_{22}=\boxed{1}$.[/hide]
|
231,156
|
\[
\int \log_{3} x \, dx
\]
\[
\int_{0}^{\pi} \sin^{2} x \, dx
\]
\[
\int \frac{x^{2}}{x^{2}-4x+5} \, dx
\] While it's true that these don't belong here (for future reference, try http://www.artofproblemsolving.com/Forum/index.php?f=296 ) I'll provide hints for the other two that JRav hasn't done.
1) I'll assume that's a $ \log_3 x$. What do you know about using $ \log$ in an integral? Isn't it usually much easier to have $ \ln$ to work with? Perhaps there's some way to convert bases.
3) As far as I can see from looking at this one, it's probably going to entail a partial-fractions approach combined with a trig substitution.
|
2,311,563
|
Simplify \(\sin(x-y)\cos y + \cos(x-y)\sin y.\) if you write it out it becomes ${cos(y)*sin(x)*cos(y) - cos(x)*sin(y)*cos(y)}$ +
${sin(y)*cos(x)*cos(y)+sin(x)*sin(y)*sin(y)}$
which simplified to $sin(x)cos^2(y)+sin(x)sin^2(y).$
factor out the $sin(x)$ to get
$sin(x)*{cos^2(y)+sin^2(y)} = sin(x)*1$ :D
|
2,311,567
|
Does there exist any function \(f:\mathbb{C}\to\mathbb{C}\) such that
\[
f(f(z)) = z^{2}\quad\text{for all }z\in\mathbb{C}?
\] How would you define $i^{\sqrt{2}}$ ? ;)
In fact, the answer is no. Note first that
$$f(z^2) = f(f(f(z))) = f(z)^2 \ \forall z \in \mathbb{C}.$$
This implies that $f(0)^2 = f(0)$ and $f(1)^2 = f(1)$, so $f(0), f(1) \in \{0,1 \}$. If $f(0)=1$ then $f(1)=f(f(0))=0$, which implies that $f(-1)^2 = f(1)= 0 \Longrightarrow f(-1)=0$. This in turn yields $f(i)^2 = f(-1) = 0 \Longrightarrow f(i)=0$. But then $-1=f(f(i))=f(0)=1$, which is absurd.
Thus we must have $f(0)=0$ and $f(1)=1$, whence $f(-1)^2 = 1$. If $f(-1)=-1$ then $f(f(-1))=-1$, a contradiction. Hence $f(-1)=1$, but then $f(i)^2 = 1 \Longrightarrow f(i) \in \{-1, 1 \} \Longrightarrow f(f(i))=1$, which can't be true either. Therefore, such a function does not exist.
If $\mathbb{C}$ is replaced by $\mathbb{R}$ then we can take $f(x)=|x|^{\sqrt{2}}$, which works.
|
2,311,570
|
In triangle \(ABC\), \(\sin A = \frac{3}{5}\) and \(\cos B = \frac{5}{13}\). Find \(\cos C\). We recognise the sines and cosines as 3-4-5 triangle and 5-12-13 right triangle. Now, WLOG, we choose AC to be 5. Then we use the cosine law to find that the answer is [hide]$\boxed{\frac{16}{65}}$[/hide]. idk how to hide attachment, let me know if you know how
|
2,311,577
|
Compute \(\arcsin\!\left(\frac{1}{\sqrt{2}}\right)\). Express your answer in radians. [hide=Solution]
$\sin = \dfrac{\text{opposite}}{\text{hypotenuse}}$, so the opposite is $1$ and the hypotenuse is $\sqrt{2}$. This should remind you of a $45-45-90$ triangle, and $45$ degrees is $\boxed{\dfrac{\pi}{4}}$ radians.
[/hide]
|
2,311,578
|
Find the area of the parallelogram generated by
\[
\begin{pmatrix}3\\1\\-2\end{pmatrix}
\quad\text{and}\quad
\begin{pmatrix}1\\-3\\4\end{pmatrix}.
\] [hide]
The area of a parallelogram is equal to twice the area of one of the triangles formed by drawing its diagonal. Thus, the area is $\frac{ab\sin\theta}{2}\cdot2=ab\sin\theta$, where $a$ and $b$ are the distinct side lengths of the parallelogram and $\theta$ is the angle between them. Note that the formula for the magnitude of the cross product of two three-dimensional vectors is $\|\mathbf{v}\|\|\mathbf{w}\|\sin\theta$, which is equal to our answer for $\mathbf{v}=\begin{pmatrix}3\\1\\-2\end{pmatrix}$ and $\mathbf{w}=\begin{pmatrix}1\\-3\\4\end{pmatrix}$. We have $\left\|\begin{pmatrix}3\\1\\-2\end{pmatrix}\times\begin{pmatrix}1\\-3\\4\end{pmatrix}\right\|=\boxed{10\sqrt{3}}$ as our final answer.
|
2,311,581
|
Find the matrix \(\mathbf{R}\) such that for any vector \(\mathbf{v}\), \(\mathbf{R}\mathbf{v}\) is the reflection of \(\mathbf{v}\) through the \(xy\)-plane. [hide=solution sketch] Find the images for $i, j, k$, then write it out as a matrix. (You could make a general point but this is easier since no variables). It gives you $\mathbf{R} = \boxed{\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}}.$
[/hide]
|
2,311,587
|
Solve over \(\mathbb{C}\) the system of equations
\[
\begin{cases}
x\bigl(y^{2}+1\bigr)=2y,\\[4pt]
y\bigl(z^{2}+1\bigr)=2z,\\[4pt]
z\bigl(x^{2}+1\bigr)=2x.
\end{cases}
\] [quote=vralex]Solve over $\mathbb{C}$ the system of equations $x(y^2+1)=2y, y(z^2+1)=2z, z(x^2+1)=2x$.[/quote]
Let $f(x)=\frac {2x}{x^2+1}$ so that solutions are $(f(f(x)),f(x),x)$ where $x$ is any real or complex root of $f(f(f(x)))=x$
Equation $f(f(f(x)))=x$ is $x^9+20x^7+14x^5-28x^3-7x=0$
With three easy real roots $-1,0,1$
Plus the six complex roots of $x^6+21x^4+35x^2+7=0$ which are easy to get (course method for solving cubics) :
$\pm i\sqrt{7-8\sqrt{\frac 73}\cos\left(\frac{\arccos\left(-\frac 32\sqrt{\frac 37}\right)+2k\pi}3\right)}$ where $k\in\{0,1,2\}$
|
2,311,616
|
Let \(S\) be the set of complex numbers \(z\) such that \(\operatorname{Re}\!\left(\frac{1}{z}\right)=\frac{1}{6}\). This set forms a curve. Find the area of the region inside the curve. [hide]Let z = a + bi, then Re(1/z) = Re(1/(a+bi)) = a / (a^2+b^2). Given this is 1/6, we arrange to get a^2 - 6a + b^2 = 0. This is a circle centered at (3,0) with a radius of 3, for an area of 9pi. [/hide]
|
231,162
|
Let p be a prime and let A = Z / p^r Z, where r ∈ N. Let G be the group of units of A (the residue classes modulo p^r represented by integers prime to p). Show that G is cyclic except when p = 2 and r ≥ 3, in which case G ≅ C2 × C_{2^{r−2}}. Show $ x\in (1 \plus{} p^k\mathbb{Z})\setminus (1 \plus{} p^{k \plus{} 1}\mathbb{Z})\Rightarrow x^p \in (1 \plus{} p^{k \plus{} 1}\mathbb{Z})\setminus (1 \plus{} p^{k \plus{} 2}\mathbb{Z})$
for all primes $ p$ and $ k\in\mathbb{N}$ except for $ (p,k) \equal{} (2,1)$.
For that calculate $ (1 \plus{} tp^k)^p\pmod{p^{k \plus{} 2}}$ for any $ t\in\mathbb{Z}\setminus p\mathbb{Z}$ and use $ p|\binom{p}{2}$ for $ p > 2$.
Using this we get $ 5 \plus{} 2^r\mathbb{Z} \equal{} 1 \plus{} 2^2 \plus{} 2^r\mathbb{Z}$ of order $ 2^{r \minus{} 2}$ in $ (\mathbb{Z}/2^r\mathbb{Z})^\times$ for $ r\ge 2$.
Then $ \langle \minus{} 1, 5\rangle \equal{} \langle \minus{} 1\rangle \times\langle 5\rangle$ has order $ 2^{r \minus{} 1}$ in $ (\mathbb{Z}/2^r\mathbb{Z})^\times$ with $ |(\mathbb{Z}/2^r\mathbb{Z})^\times| \equal{} \varphi(2^r) \equal{} 2^{r \minus{} 1}$,
so $ (\mathbb{Z}/2^r\mathbb{Z})^\times \equal{} \langle \minus{} 1\rangle \times\langle 5\rangle$ of type $ (2,2^{r \minus{} 2})$.
Also for $ p > 2$ we get $ 1 \plus{} p \plus{} p^{r}\mathbb{Z}$ of order $ p^{r \minus{} 1}$ with $ |(\mathbb{Z}/p^r\mathbb{Z})^\times| \equal{} \varphi(p^r) \equal{} (p \minus{} 1)p^{r \minus{} 1}$,
so $ \langle 1 \plus{} p\rangle$ is the unique Sylow $ p$-subgroup of the abelian group $ G \equal{} (\mathbb{Z}/p^r\mathbb{Z})^\times$,
and $ G$ has a normal Hall $ p'$-subgroup $ H$ with $ G \equal{} \langle 1 \plus{} p\rangle\times H$.
But the canonical epi $ G\to (\mathbb{Z}/p\mathbb{Z})^\times,x \plus{} p^r\mathbb{Z}\mapsto x \plus{} p\mathbb{Z}$ has kernel $ N \equal{} \{1 \plus{} pt \plus{} p^r\mathbb{Z}: t\in\mathbb{Z}\}$ of order $ p^{r \minus{} 1}$,
so we must have $ N \equal{} \langle 1 \plus{} p\rangle$ and $ H\cong G/N\cong (\mathbb{Z}/p\mathbb{Z})^\times \equal{} \mathbb{F}_p^\times$ cyclic of order $ p \minus{} 1$.
So $ G \equal{} \langle 1 \plus{} p\rangle\times H$ is cyclic of order $ (p \minus{} 1)p^{r \minus{} 1}$.
|
2,311,623
|
In polar coordinates, the point \(\left(-2,\frac{3\pi}{8}\right)\) is equivalent to what other point, in the standard polar coordinate representation? Enter your answer in the form \((r,\theta)\), where \(r>0\) and \(0\le\theta<2\pi.\) Or you could simply add $\pi$ to the angle.
|
2,311,633
|
Solve the functional equation: find all functions \(f:\mathbb{R}\to\mathbb{R}\) satisfying
\[
f(xy)=f(x)+f(y)\quad\text{for all }x,y\in\mathbb{R}.
\] I will let $P(x,\, y)$ be the assertion of this functional equation.
Eh, this question is ill-defined and up to interpretation, because the domain and range are not stated. Picking the domain and range to be $\mathbb{R}$ yields the boring solution $f(x)=0$. It can be much more fun! I'll change the domain and range to be:
(1) Domain is $\mathbb{R} / \{ 0 \}$ and range is $\mathbb{R}$. This eliminates the possibility of requiring $f(x)=0$ only. To get the non-trivial solutions, let $f(x)=g(\log|x|)$. Rewriting our functional equation to be: $$g(\log(|xy|)=g(\log|x|)+g(\log|y|)$$ $$g(\log|x|+\log|y|)=g(\log|x|)+g(\log|y|)$$ Note that since the range of the logarithm is $\mathbb{R}$, we have that $g(x+y)=g(x)+g(y)$ for all real $x,\, y$. This already yields uncountably infinitely many solutions, many of which are not nice. However, if we restrict ourselves for continuous $g$, we have that $$g(x)=mx,\, \forall m\in\mathbb{R}$$ Replacing $x$ with $\log|x|$, we get $$f(x)=m\log|x|,\, \forall m\in\mathbb{R}$$ And we are done. $\blacksquare$
(2) Domain is $\mathbb{Q} / \{0 \}$ and range is $\mathbb{Q}$. This problem becomes more and more interesting. Let $r=\frac{p}{q}$. Obviously $f(1)=0$ and $f(-1)=0$. First, by $P\left(p, \frac{1}{p}\right)$, we have $$f(1)=f(p)+f\left(\frac{1}{p}\right)$$ $$f\left(\frac{1}{p}\right)=-f(p)$$ Next, by $P\left(p, \frac{1}{q}\right)$, we have $$f\left(\frac{p}{q}\right)=f(p)-f(q)$$ Thus, if we really need a rational function that satisfies the functional equation, we need to define the function over the integers. Luckily, since $f(-1)=0$, we have that $f(x)=f(-x)$ (i.e. $f$ is even), so we have to worry about the natural numbers only. This construction will be delightful, as this will connect the problem with number theory!
Let $f$ be a function that such $f(1)=0$ and such that $f(p)$ can be any arbitrary value for prime $p$. (e.g. we can set $f(2)=69$, $f(3)=42$, $f(5)=420$, and etc. if we wanted to.) Then calculate the values of $f(c)$ for composite $c$ based on the functional equation. (Proofs that it satisfies the equation and is general solution are trivial. I will write them when I have the time.)
Note 1: By the fundamental theorem of arithmetic, we may write $c=\prod_{i=1}^{k} p_i^{c_i}$ for prime $p_i$ and natural $a_i$ and $k$. Hence we can write $$f(c)=f\left(\prod p_i^{c_i}\right)=f(p_1^{c_1})+f(p_2^{c_2})+\ldots = \sum_{i=1}^{k} c_i f(p_i)$$
Note 2: There is a very nice function that follows this form. It is when $f(p)=1$ for all primes $p$. This function counts the number of (including repeated) prime factors. For example, $f(4)=2$ because $4=2\times 2$, $f(6)=2$, $f(8)=3$, $f(9)=2$, etc. It is nice that it solves this functional equation.
This is as far as I went with this question. Things could become more crazy, but I'll stop it at that.
|
2,311,634
|
Evaluate
\[
\begin{vmatrix}
\cos\alpha\cos\beta & \cos\alpha\sin\beta & -\sin\alpha\\[4pt]
-\sin\beta & \cos\beta & 0\\[4pt]
\sin\alpha\cos\beta & \sin\alpha\sin\beta & \cos\alpha
\end{vmatrix}.
\] [hide=Solution.]Expanding the determinant with focus towards the third column, we have
\begin{align*}
\begin{vmatrix} \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \end{vmatrix}&= -\sin\alpha\begin{vmatrix}-\sin\beta & \cos\beta \\ \sin\alpha\cos\beta & \sin\alpha\sin\beta \\\end{vmatrix} - 0\begin{vmatrix}\cos\alpha\cos\beta & \cos\alpha\sin\beta \\ \sin\alpha\cos\beta & \sin\alpha\sin\beta \\\end{vmatrix} + \cos\alpha\begin{vmatrix}\cos\alpha\cos\beta & \cos\alpha\sin\beta \\ -\sin\beta & \cos\beta \\\end{vmatrix} \\
&=-\sin\alpha\left(-\sin\alpha\sin^2\beta-\sin\alpha\cos^2\beta\right)+\cos\alpha\left(\cos\alpha\cos^2\beta+\cos\alpha\sin^2\beta\right) \\
&= -\sin\alpha\left(-\sin\alpha\right)+\cos\alpha\left(\cos\alpha\right) \\
&=\sin^2\alpha+\cos^2\alpha \\
&=\boxed{1}. \\
\end{align*}[/hide]
|
2,311,636
|
The matrices
\[
\mathbf{A}=\begin{pmatrix}1 & x\\[4pt] y & -\frac{9}{5}\end{pmatrix}
\qquad\text{and}\qquad
\mathbf{B}=\begin{pmatrix}\frac{12}{5} & \frac{1}{10}\\[4pt] 5 & z\end{pmatrix}
\]
satisfy \(\mathbf{A}+\mathbf{B}=\mathbf{A}\mathbf{B}.\) Find \(x+y+z.\) [hide=who decided to invent matrices, i will find them]
using addition and multiplication properties of matrices, we have that
$ \renewcommand{\arraystretch}{1.5} \begin{pmatrix} 3.2 & x+0.1 \\ y+5 & z-1.4 \end{pmatrix} \renewcommand{\arraystretch}{1} \quad = \renewcommand{\arraystretch}{1.5} \begin{pmatrix} 2.4+5x & xz+0.1 \\ 2.4y-9 & 0.1y-1.4z \end{pmatrix} \renewcommand{\arraystretch}{1} \quad$.
then this whole thing becomes a system of equations.
we can easily see that $x=0.2$. with that information, we then can find $z=1$, and then after more work, we find that $y=10$.
so, we have $x+y+z=0.2+10+1=\boxed{11.2}$.
|
2,311,641
|
For a constant \(c\), in spherical coordinates \((\rho,\theta,\phi)\), find the shape described by the equation
\[
\theta = c.
\]
(A) Line
(B) Circle
(C) Plane
(D) Sphere
(E) Cylinder
(F) Cone
Enter the letter of the correct option. [quote=ThriftyPiano][hide=Answer]Since the angle must stay constant and everything else can move around, everything on the plane with that angle is accessible, so it is a plane.[/hide][/quote]
it says answer the letter of the correct option. Thus, the answer is C.
|
2,311,651
|
The point \((4 + 7\sqrt{3},\, 7 - 4\sqrt{3})\) is rotated \(60^\circ\) counterclockwise about the origin. Find the resulting point. [hide=sol]
Let $\mathbf{M}$ be the matrix that corresponds to the transformation. Since $\mathbf{M}$ is the rotation, we can use the formula for a rotation matrix. Therefore $\mathbf{M}=\begin{pmatrix}
\cos60 & -\sin60 \\
\sin60 & \cos60
\end{pmatrix}$. Multiplying this by the vector $\begin{pmatrix}
4+7\sqrt3 \\
7-4\sqrt3
\end{pmatrix}$, we find that the transformed point is $\boxed{(8, 14)}$.
[/hide]
|
2,311,652
|
Express \(\sin(a+b)-\sin(a-b)\) as a product of trigonometric functions. [hide]
I can't believe I put $2\sin a\cos b$ as my first answer
Anyways, $\sin(a+b)-\sin(a-b)=2\sin b\cos a$ by sum to product (or trig addition formulas then cancellation of a bunch of stuff)
[/hide]
|
2,311,654
|
Find the values of the following expressions:
\[
{}^{\pi}C_{r},\qquad {}^{i}C_{r},\qquad {}^{n}C_{\pi},\qquad {}^{n}C_{i},
\]
\[
{}^{\pi}P_{r},\qquad {}^{i}P_{r},\qquad {}^{n}P_{\pi},\qquad {}^{n}P_{i},
\]
where \(i=\sqrt{-1}\). [quote name="HumanCalculator9" url="/community/p18375315"]
I think this involves the gamma function
[/quote]
Yep; for complex numbers $a,b$, we have $\binom{a}{b}=\frac{\Gamma(1+a)}{\Gamma(1+b)\Gamma(1+a-b)}$
|
2,311,661
|
The volume of the parallelepiped generated by \(\begin{pmatrix}2\\3\\4\end{pmatrix}, \begin{pmatrix}1\\k\\2\end{pmatrix},\) and \(\begin{pmatrix}1\\2\\k\end{pmatrix}\) is \(15\). Find \(k\), where \(k>0\). [hide=Solution]Note that the volume of the parallelpiped is $$\left|\begin{matrix}2&3&4\\1&k&2\\1&2&k\end{matrix}\right|=(2)(k)(k)+(3)(2)(1)+(4)(1)(2))-((4)(k)(1)+(2)(2)(2)+(k)(3)(1)=(2k^2+6+8)-(4k+8+3k)=2k^2-7k+6=15\Leftrightarrow2k^2-7k-9=0\Leftrightarrow(k+1)(2k-9)=0\Leftrightarrow k=-1,\frac{9}{2}.$$ Since $k>0,$ we conclude that $\boxed{\frac{9}{2}}$ is the answer.[/hide]
|
2,311,672
|
When every vector on the line \(y=\tfrac{5}{2}x+4\) is projected onto a certain vector \(\mathbf{w}\), the result is always the vector \(\mathbf{p}\). Find the vector \(\mathbf{p}\). [hide=sol]The same vector p will be the projection of w onto the line.
The head of p will be the closest point on $y=\frac{5}{2}x+4$ to the origin, so we can let $(x,y)=(2t,5t+4)$. Then we want to minimize $x^2+y^2=29t^2+40t+16$ which is minimimized when $t=-20/29$.
Therefore the head of p is going to be $(-40/29,16/29)$. Therefore \begin{align*}
\mathbf{p} &= \boxed{ \begin{pmatrix} -40/29 \\ 16/29 \end{pmatrix} }
\end{align*}.[/hide]
|
2,311,678
|
Let \(a,b,c,d\) be nonzero integers such that
\[
\begin{pmatrix} a & b \\ c & d \end{pmatrix}^2
=
\begin{pmatrix} 7 & 0 \\ 0 & 7 \end{pmatrix}.
\]
Find the smallest possible value of \(|a|+|b|+|c|+|d|\). first post
so you just expand out the left hand side by multiplying the matrices, and you get that $a+d=0$. The only other restriction you have is that $a^2+bc=7$, and you can do some casework to find values that minimize the sum in the problem.
|
2,311,685
|
Find the point of intersection of the line
\[
\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}
\]
and the plane
\[
x-y+z=5.
\] [hide=Solution]Multiplying by $12$ gives $4(x-2)=3(y+1)=z-2$, or $4x-8=3y+3=z-2$. From $4x-8=z-2$, we have $4x=z+6$, so $x=\frac{z+6}{4}$. From $3y+3=z-2$, we have $3y=z-5$, so $y=\frac{z-5}{3}$. Then $\frac{z+6}{4}-\frac{z-5}{3}+z=5$, meaning $z=2$. Then $x=2$ and $y=-1$, so the answer is $\boxed{(2,-1,2)}$.[/hide]
|
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