Prove that
.
and 
Prove that
and 
Prove that
,
,
is not greater than 
.
.
.
and 
be points on a plane such that 
,
is a positive integer. Let 
be the set of all points 
such that 
,
is a real number. The path that 
to the cross product 
reveals that a negative determinant [of 
] corresponds to a matrix which reverses the direction of the cross product of two vectors.[/quote]
and 
,
and 
although its supposed to be a quick comparison.
Y by CXNJ.AoPS, tofubear, Yrock, Sedro
As far as I know, it seems that no one on HSM has organized an equation marathon before. So why not give it a try? Let's start one!
Some basic rules need to be clarified:
If a problem has not been solved within 
days, then others are eligible to post a new probkem.
Not only simple one-variable equations, but also systems of equations are allowed.
The difficulty of these equations should be no less than that of typical quadratic one-variable equations. If the problem involves higher degrees or more variables, please ensure that the problem is solvable (i.e., has a definite solution, rather than an approximate one).
Please indicate the domain of the solution to the equation (e.g., solve in 
, solve in 
).
Here's an simple yet fun problem, hope you enjoy it
:
Some basic rules need to be clarified:
If a problem has not been solved within 
days, then others are eligible to post a new probkem. Not only simple one-variable equations, but also systems of equations are allowed. The difficulty of these equations should be no less than that of typical quadratic one-variable equations. If the problem involves higher degrees or more variables, please ensure that the problem is solvable (i.e., has a definite solution, rather than an approximate one). Please indicate the domain of the solution to the equation (e.g., solve in 
, solve in 
).Here's an simple yet fun problem, hope you enjoy it
:This post has been edited 2 times. Last edited by steven_zhang123, Jan 21, 2025, 12:37 AM
.
,
.
,![$$\sqrt[3]{15-x^3+3x^2-3x}=2\sqrt{x^2-4x+2}+3-x$$](http://latex.artofproblemsolving.com/e/f/4/ef457a31ebab89cbe07898a85d61030748a74f75.png)
![$\sqrt[3]{a}-\sqrt[3]{b}=\frac{a-b}{\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2}}$](http://latex.artofproblemsolving.com/c/2/e/c2eb20e1113244696a46430f9199291b17151324.png)
and 
of real numbers that satisfy the system of equations![\[\begin{cases}x^3 = 3x - 12y + 50, \\ y^3 = 12y + 3z - 2, \\ z^3 = 27z + 27x. \end{cases}\]](http://latex.artofproblemsolving.com/0/7/9/079062f832872bdf8b57377d8388254ffa379522.png)
;
is a set of solutions to the original equation.
,
contradiction 
.
,
,
gives:
![$x\in [0,\pi]$](http://latex.artofproblemsolving.com/3/e/c/3ecbeef3f695ea04d27c5cafc03c70ec34b52c0f.png)
:
,
in ![$[0,\pi]$](http://latex.artofproblemsolving.com/e/c/e/ece27ba559bd93cdf2416aa0f88fc8d7d85e5791.png)
.
thus 
and using the identity 
we have the solutions 
.
:
for positive values.
or 
.
,
,
is nonnegative. This obviously has no solution since on ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
and 
.
,
.
.
.![$x=\sqrt[3]{a}+\sqrt[3]{a^2}$](http://latex.artofproblemsolving.com/5/5/c/55c5e858342befc8d9deee8c809d5923857452c6.png)
when 
and 
,
,
is the set of natural numbers, 
is the positive reals
is obviously too large, 
is obviously too small, so we guess 
and are correct.
to get 
,
.
satisfying the equations 
and 
.
,
and 
are monotonically increasing,
,
.
,
!=(x+2)(x+3)(x+4), when x=7, LHS>RHS, when x≥7, (x+1)!/x!=x>6>(x+5)/(x+2), so LHS>RHS,
the system 
.
and 
.
,
.
.
and 
so that 
for 
.
is the solution set for 
,
![$S=-1+e^{2\pi ki/3}\sqrt[3]{1+\sqrt{2}}-e^{-2\pi ki/3}\sqrt[3]{-1+\sqrt{2}}$](http://latex.artofproblemsolving.com/2/5/9/25930639fec336b09360889394210265cf83f815.png)
![\begin{align*}
S^2-4P
&=S^2-\frac{4(S^3-1)}{3S}\\
&=\frac{4}{3S}-\frac{S^2}{3}\\
&=\frac{2(-S^2-3S-6)}{3}-\frac{S^2}{3}\\
&=-S^2-2S-4\\
&=-(S+1)^2-3\\
&=-\left(e^{2\pi ki/3}\sqrt[3]{1+\sqrt{2}}-e^{-2\pi ki/3}\sqrt[3]{-1+\sqrt{2}}\right)^2-3\\
&=-1-e^{-2\pi ki/3}\sqrt[3]{3+2\sqrt{2}}-e^{2\pi ki/3}\sqrt[3]{3-2\sqrt{2}}
\end{align*}](http://latex.artofproblemsolving.com/3/f/2/3f2fea41e600054b5fc44a6a12e0d63453565974.png)
is the solution set to 
,
.
gives![$x,y=-\frac{1}{2}+\frac{e^{2\pi ki/3}}{2}\sqrt[3]{1+\sqrt{2}}-\frac{e^{-2\pi ki/3}}{2}\sqrt[3]{-1+\sqrt{2}}\pm\frac{i}{2}\sqrt{1+e^{-2\pi ki/3}\sqrt[3]{3+2\sqrt{2}}+e^{2\pi ki/3}\sqrt[3]{3-2\sqrt{2}}}$](http://latex.artofproblemsolving.com/c/0/3/c03ebea13fdfe509f85ac40aa13fb43a176fe69b.png)
.
for ![$z=1+e^{-2\pi i/3}\sqrt[3]{3+2\sqrt{2}}+e^{2\pi i/3}\sqrt[3]{3-2\sqrt{2}}$](http://latex.artofproblemsolving.com/0/c/1/0c13f0ce34afe12f73cbd1cef5f6b21b65fa34b0.png)
.![$z=1-\frac{1}{2}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)-\frac{\sqrt{3}}{2}\left(\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}\right)i$](http://latex.artofproblemsolving.com/c/a/a/caaab9bacb1cebca0a3130f2c0fcfd7601701c31.png)
![\begin{align*}
|z|^2&=\left(1-\frac{1}{2}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\right)^2+\frac{3}{4}\left(\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}\right)^2\\
&=1-\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)+\frac{1}{4}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)^2+\frac{3}{4}\left(\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}\right)^2\\
&=-\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}+\sqrt[3]{\left(3+2\sqrt{2}\right)^2}+\sqrt[3]{\left(3-2\sqrt{2}\right)^2}\\
\end{align*}](http://latex.artofproblemsolving.com/b/3/5/b3596d78d2341c3c6ca2494e03908b6f83e2f5a1.png)
where ![$w=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}$](http://latex.artofproblemsolving.com/1/b/3/1b345f994f118c97599db9cc919013495e401dee.png)
obeying 
.
,
.
obeys 
.
and 
gives 
as roots of 
.
and ![$|z|^2=2\sqrt[3]{1+\sqrt{2}}-2\sqrt[3]{-1+\sqrt{2}}$](http://latex.artofproblemsolving.com/8/8/2/8820e7845852d894620109a826004d42577ae18c.png)
![$\sqrt{z}=\frac{1}{2}\sqrt{2-\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}+2\sqrt{2\sqrt[3]{1+\sqrt{2}}-2\sqrt[3]{-1+\sqrt{2}}}}-\frac{i}{2}\sqrt{-2+\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}+2\sqrt{2\sqrt[3]{1+\sqrt{2}}-2\sqrt[3]{-1+\sqrt{2}}}}$](http://latex.artofproblemsolving.com/4/9/3/4933a20ddc6c5613476066225b32b8bf31f401be.png)
![$z=1+e^{2\pi i/3}\sqrt[3]{3+2\sqrt{2}}+e^{-2\pi i/3}\sqrt[3]{3-2\sqrt{2}}$](http://latex.artofproblemsolving.com/9/b/e/9be712ca8c573731089c4303fb981e79e02fe328.png)
,![$\sqrt{z}=\frac{1}{2}\sqrt{2-\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}+2\sqrt{2\sqrt[3]{1+\sqrt{2}}-2\sqrt[3]{-1+\sqrt{2}}}}+\frac{i}{2}\sqrt{-2+\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}+2\sqrt{2\sqrt[3]{1+\sqrt{2}}-2\sqrt[3]{-1+\sqrt{2}}}}$](http://latex.artofproblemsolving.com/1/1/5/115111b4951ee3bb5f2bc6f8564cb79d76ca9cd8.png)
,
pairs as 
,
,
,
,
where:![\begin{align*}
A_\pm&=-\frac{1}{2}+\frac{1}{2}\sqrt[3]{1+\sqrt{2}}-\frac{1}{2}\sqrt[3]{-1+\sqrt{2}}\pm\frac{i}{2}\sqrt{1+\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}}\\
B_\pm&=-\frac{1}{2}-\frac{1}{4}\sqrt[3]{1+\sqrt{2}}+\frac{1}{4}\sqrt[3]{-1+\sqrt{2}}+\frac{1}{4}\sqrt{-2+\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}+2\sqrt{2\sqrt[3]{1+\sqrt{2}}-2\sqrt[3]{-1+\sqrt{2}}}}\\
&\pm\left(\frac{\sqrt{3}}{4}\sqrt[3]{1+\sqrt{2}}+\frac{\sqrt{3}}{4}\sqrt[3]{-1+\sqrt{2}}+\frac{1}{4}\sqrt{2-\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}+2\sqrt{2\sqrt[3]{1+\sqrt{2}}-2\sqrt[3]{-1+\sqrt{2}}}}\right)i\\
C_\pm&=-\frac{1}{2}-\frac{1}{4}\sqrt[3]{1+\sqrt{2}}+\frac{1}{4}\sqrt[3]{-1+\sqrt{2}}-\frac{1}{4}\sqrt{-2+\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}+2\sqrt{2\sqrt[3]{1+\sqrt{2}}-2\sqrt[3]{-1+\sqrt{2}}}}\\
&\pm\left(\frac{\sqrt{3}}{4}\sqrt[3]{1+\sqrt{2}}+\frac{\sqrt{3}}{4}\sqrt[3]{-1+\sqrt{2}}-\frac{1}{4}\sqrt{2-\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}+2\sqrt{2\sqrt[3]{1+\sqrt{2}}-2\sqrt[3]{-1+\sqrt{2}}}}\right)i\\
\end{align*}](http://latex.artofproblemsolving.com/c/d/9/cd91da70d0eefc538da1bdd6b233619787771c33.png)
