Probabiliy II: Who has the winning strategy?
Consider a polynomial
$$P(x)=a_0+a_1x+\cdots+a_{2011}x^{2011}+x^{2012}$$
Nigel and Jessica are playing the following game. In turn, they choose one of the coefficients $a_0,\,\cdots,\,a_{2011}$ and assign a real value to it. Nigel has the first move. Once a value is assigned to a coefficient, it cannot be changed any more. The game ends after all the coefficients have been assigned values.
$$P(x)=a_0+a_1x+\cdots+a_{2011}x^{2011}+x^{2012}$$
Nigel and Jessica are playing the following game. In turn, they choose one of the coefficients $a_0,\,\cdots,\,a_{2011}$ and assign a real value to it. Nigel has the first move. Once a value is assigned to a coefficient, it cannot be changed any more. The game ends after all the coefficients have been assigned values.
Find all pairs of integer solutions $x(y^2 + 9) + y(x^2 − 9) + x^2(x − 6) = 0$
My solution:
Before revealing my method of solving, I wish to tell you how I encountered students kept asking me why should they study quadratic function. How can they be useful.. They said quadratic functions have nothing special, and it's really easy peasy to find for its discriminant, and to completing the square to look for its optimal point and factoring it to investigate its roots.
Yes, that's all that to it for quadratic functions, but when you progress into higher grade, you would encounter problem like solving the equation for integer solutions.
That is when the concept of quadratic function creeps in to assist us in finding all possible integer solutions. How? Continue reading to figure out the answer.
Analysis: Quiz 15
Question 1: Which of the following represents the graph of $y=\sqrt{2}x$?
A.
B.
C.
B.
C.
Is there a real number $x$, that the expressions $\tan x + \sqrt{3}$ and $\cot x+ \sqrt{3}$ are both integers?
Is there a real number $x$, that the expressions $\tan x + \sqrt{3}$ and $\cot x+ \sqrt{3}$ are both integers?
My solution:
First, let's assume $\tan x + \sqrt{3}=a$ and $\cot x+ \sqrt{3}=b$ where $a,\,b$ are both integers.
My solution:
First, let's assume $\tan x + \sqrt{3}=a$ and $\cot x+ \sqrt{3}=b$ where $a,\,b$ are both integers.
Probabiliy: Who has the winning strategy?
Consider a polynomial
$$P(x)=a_0+a_1x+\cdots+a_{2011}x^{2011}+x^{2012}$$
Nigel and Jessica are playing the following game. In turn, they choose one of the coefficients $a_0,\,\cdots,\,a_{2011}$ and assign a real value to it. Nigel has the first move. Once a value is assigned to a coefficient, it cannot be changed any more. The game ends after all the coefficients have been assigned values.
$$P(x)=a_0+a_1x+\cdots+a_{2011}x^{2011}+x^{2012}$$
Nigel and Jessica are playing the following game. In turn, they choose one of the coefficients $a_0,\,\cdots,\,a_{2011}$ and assign a real value to it. Nigel has the first move. Once a value is assigned to a coefficient, it cannot be changed any more. The game ends after all the coefficients have been assigned values.
Third Solution: Find $k$ if $k\sin 6x=\sin 2x$ given $\dfrac{\cos 6x}{\cos 2x}=\dfrac{1}{6}$.
Find $k$ if $k\sin 6x=\sin 2x$ given $\dfrac{\cos 6x}{\cos 2x}=\dfrac{1}{6}$.
Third method:
Second Method: Find $k$ if $k\sin 6x=\sin 2x$ given $6\cos 6x=\cos 2x$.
Find $k$ if $k\sin 6x=\sin 2x$ given $\dfrac{\cos 6x}{\cos 2x}=\dfrac{1}{6}$.
Second method:
Note that we can rewrite the given equality $6\cos 6x=\cos 2x$ as $\dfrac{\cos 6x}{\cos 2x}=\dfrac{1}{6}$, also, our target expression as $\dfrac{\sin 6x}{\sin 2x}=\dfrac{1}{k}$.
Second method:
Note that we can rewrite the given equality $6\cos 6x=\cos 2x$ as $\dfrac{\cos 6x}{\cos 2x}=\dfrac{1}{6}$, also, our target expression as $\dfrac{\sin 6x}{\sin 2x}=\dfrac{1}{k}$.
Find $k$ if $k\sin 6x=\sin 2x$ given $6\cos 6x=\cos 2x$.
Find $k$ if $k\sin 6x=\sin 2x$ given $\dfrac{\cos 6x}{\cos 2x}=\dfrac{1}{6}$.
There are at least three different methods of solving for this particular problem. One is rather straightforward, and the other two methods are quite special.
There are at least three different methods of solving for this particular problem. One is rather straightforward, and the other two methods are quite special.
Floor function system
Solve the following equation in the real number system:
$\left\lfloor{\log_2 x}\right\rfloor+\left\lfloor{\log_4 x}\right\rfloor=3$
$\left\lfloor{\log_2 x}\right\rfloor+\left\lfloor{\log_4 x}\right\rfloor=3$
Olympiad Math Problem: Find the maximum and minimum of P (Heuristic Solution)
Find the minimum and maximum of $P=\dfrac{y−x}{x+8y}$ for all real $x$ and $y$ that satisfy the equation $y^2(6-x^2)-xy-1=0$.
My solution:
Note that the given equality $y^2(6-x^2)-xy-1=0$ has the terms $y^2,\,x^2y^2$ and $xy$ while the target expression, $P$ is a rational function with the terms $x$ and $y$.
My solution:
Note that the given equality $y^2(6-x^2)-xy-1=0$ has the terms $y^2,\,x^2y^2$ and $xy$ while the target expression, $P$ is a rational function with the terms $x$ and $y$.
Prove the equality : $\sqrt{33 − 16\sqrt{3}\sin80^\circ}= 1+8\sin10^\circ$
Prove the equality :
$\sqrt{33 − 16\sqrt{3}\sin80^\circ}= 1+8\sin10^\circ$
First, note that
$\begin{align*}\cos^2 20^\circ&=\cos20^\circ(2\cdot\dfrac{1}{2}\cos20^\circ)\\&=\cos20^\circ(2\cos60^\circ\cos20^\circ)\\&=\cos20^\circ((\cos(60^\circ+20^\circ)+\cos(60^\circ-20^\circ))\\&=\cos20^\circ(\cos80^\circ+\cos40^\circ)\\&=\cos20^\circ\cos80^\circ+\cos20^\circ\cos40^\circ\\&=\dfrac{1}{2}(\cos(20^\circ+80^\circ)+\cos(80^\circ-20^\circ))+\dfrac{1}{2}(\cos(20^\circ+40^\circ)+\cos(40^\circ-20^\circ))\\&=\dfrac{1}{2}(\cos100^\circ+\cos60^\circ)+\dfrac{1}{2}(\cos60^\circ+\cos20^\circ)\\&=\dfrac{1}{2}\left(\cos(90^\circ+10^\circ)+\dfrac{1}{2}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}+\cos 20^\circ\right)\\&=\dfrac{1}{2}(-\sin 10^\circ)+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{2}\cos 20^\circ\\&=\dfrac{1}{2}(\cos20^\circ-\sin10^\circ+1)\end{align*}$
$\sqrt{33 − 16\sqrt{3}\sin80^\circ}= 1+8\sin10^\circ$
First, note that
$\begin{align*}\cos^2 20^\circ&=\cos20^\circ(2\cdot\dfrac{1}{2}\cos20^\circ)\\&=\cos20^\circ(2\cos60^\circ\cos20^\circ)\\&=\cos20^\circ((\cos(60^\circ+20^\circ)+\cos(60^\circ-20^\circ))\\&=\cos20^\circ(\cos80^\circ+\cos40^\circ)\\&=\cos20^\circ\cos80^\circ+\cos20^\circ\cos40^\circ\\&=\dfrac{1}{2}(\cos(20^\circ+80^\circ)+\cos(80^\circ-20^\circ))+\dfrac{1}{2}(\cos(20^\circ+40^\circ)+\cos(40^\circ-20^\circ))\\&=\dfrac{1}{2}(\cos100^\circ+\cos60^\circ)+\dfrac{1}{2}(\cos60^\circ+\cos20^\circ)\\&=\dfrac{1}{2}\left(\cos(90^\circ+10^\circ)+\dfrac{1}{2}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}+\cos 20^\circ\right)\\&=\dfrac{1}{2}(-\sin 10^\circ)+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{2}\cos 20^\circ\\&=\dfrac{1}{2}(\cos20^\circ-\sin10^\circ+1)\end{align*}$
Olympiad Math Problem: Find the maximum and minimum of P (First Attempt)
Find the minimum and maximum of $P=\dfrac{y−x}{x+8y}$ for all real $x$ and $y$ that satisfy the equation $y^2(6-x^2)-xy-1=0$.
On one hand, this is not a unmanageable Olympiad Mathematics optimization problem, on the other hand, this problem allows us to show students how powerful algebraic manipulation is when we use it diligently and how effective the accurate solution we could have arrived compared to all the alternatives.
On one hand, this is not a unmanageable Olympiad Mathematics optimization problem, on the other hand, this problem allows us to show students how powerful algebraic manipulation is when we use it diligently and how effective the accurate solution we could have arrived compared to all the alternatives.
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