aops_id int64 3 3.62M | problem stringlengths 17 4.71k | best_solution stringlengths 7 38.7k |
|---|---|---|
231,038 | 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... |
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 F... |
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 ... | 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... |
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 begin... |
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... |
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 sam... |
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 le... |
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{2... |
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... |
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)) - \t... |
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)&=2... |
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 \(Y\), resp... | [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 $X$... |
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{\mat... |
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 prime... |
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 ... |
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,... |
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 ev... |
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.... |
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:
\[
\b... | 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 te... |
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... |
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^... |
2,310,975 | 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 thes... |
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 ... |
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 circ... |
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 yo... |
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$... |
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 ... |
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}... |
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{... |
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... |
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 +... |
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\... | 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... |
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{... |
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 c... |
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=5... |
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... |
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'(... |
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... |
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]... |
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:
$... |
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 $p\log 2\not\in {\bf Q}$,... |
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 us... |
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 ... |
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{pmat... |
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 αλλ... |
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}\... |
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 ... | [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 ha... |
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... |
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 ... |
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 root... |
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... |
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 ... |
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}... |
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 & x... |
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\s... |
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 $k>0,$ w... |
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... |
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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