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 #1
avatar+19651 
+1

Suppose that a,b, and c are positive integers satisfying (a+b+c)^3 - a^3 - b^3 - c^3 = 150. 

Find a+b+c.

 

\(\small{ \begin{array}{|rcll|} \hline \boxed{(a+b+c)^3 - a^3 - b^3 - c^3 = 150} \\ \hline (a+b+c)^3 - a^3 - b^3 - c^3 &=& \Big( (a+b)+c \Big)^3 -a^3-b^3-c^3 \\ && \boxed{ ( (a+b)+c )^3 = (a+b)^3 + 3(a+b)^2c+3(a+b)c^2+c^3 \\= (a+b)^3+c^3+3c(a+b)(a+b+c) } \\ &=& (a+b)^3+c^3+3c(a+b)(a+b+c) -a^3-b^3-c^3 \\ && \boxed{ (a+b)^3 = a^3+3a^2b+3ab^2+b^3 \\ = a^3+b^3+3ab(a+b) } \\ &=& a^3+b^3+3ab(a+b)+c^3+3c(a+b)(a+b+c) -a^3-b^3-c^3 \\ &=& a^3+b^3+c^3+ 3ab(a+b)+3c(a+b)(a+b+c) -a^3-b^3-c^3 \\ &=& 3ab(a+b)+3c(a+b)(a+b+c) \\ &=& 3(a+b) \Big( ab+c(a+b+c) \Big) \\ &=& 3(a+b)(ab+ac+bc+c^2) \\ &=& 3(a+b)\Big( a(b+c)+c(b+c) \Big) \\ &=& 3(a+b)(b+c)(c+a) \\ \hline \boxed{3(a+b)(b+c)(c+a)=150} \\ \hline \end{array} } \)

 

\(\begin{array}{|lrcll|} \hline & 3(a+b)(b+c)(c+a) &=& 150 \quad & | \quad :3 \\ & (a+b)(b+c)(c+a)&=& 50 \quad & | \quad 50 = 2\cdot 5^2 \\ & (a+b)(b+c)(c+a)&=& 2\cdot 5^2 \\\\ (1)& 2\cdot 5 \cdot 5 &=& 2\cdot 5^2 \\ (2)& 5\cdot 2 \cdot 5 &=& 2\cdot 5^2 \\ (3)& 5\cdot 5 \cdot 2 &=& 2\cdot 5^2 \\\\ (1) & (a+b) &=& 2 \\ & (b+c) &=& 5 \\ & (c+a) &=& 5 \\ & (a+b) + (b+c) +(c+a) &=& 2 + 5 + 5 \\ & 2(a+b+c) &=& 12 \quad & | \quad :2 \\ & \mathbf{ (a+b+c) }& \mathbf{=} & \mathbf{ 6 } \\\\ (2) & (a+b) &=& 5 \\ & (b+c) &=& 2 \\ & (c+a) &=& 5 \\ & (a+b) + (b+c) +(c+a) &=& 5 + 2 + 5 \\ & 2(a+b+c) &=& 12 \quad & | \quad :2 \\ & \mathbf{ (a+b+c) }& \mathbf{=} & \mathbf{ 6 } \\\\ (3) & (a+b) &=& 5 \\ & (b+c) &=& 5 \\ & (c+a) &=& 2 \\ & (a+b) + (b+c) +(c+a) &=& 5 + 5 + 2 \\ & 2(a+b+c) &=& 12 \quad & | \quad :2 \\ & \mathbf{ (a+b+c) }& \mathbf{=} & \mathbf{ 6 } \\\\ \hline \Rightarrow &&& \{a=1, b=1, c=4\} \text{ or }\\ &&& \{a=4, b=1, c=1\} \text{ or } \\ &&& \{a=1, b=4, c=1\} \\ \hline \end{array}\)

 

laugh

heureka 5 hours ago
 #3
avatar+19651 
+1
heureka Jul 18, 2018
 #3
avatar+19651 
0

1. Let $r(\theta) = \frac{1}{1-\theta}$. What is $r(r(r(r(r(r(30))))))$ (where $r$ is applied $6$ times)?

 

\(\begin{array}{|rcll|} \hline r(\theta) &=& \dfrac{1}{1-\theta} \\ \hline r(r(\theta)) &=& \dfrac{1}{1-\dfrac{1}{1-\theta}} \\\\ &=& \dfrac{1-\theta}{1-\theta-1 } \\\\ &=& \dfrac{1-\theta}{ -\theta } \\\\ &=& \dfrac{\theta-1}{ \theta } = 1-\dfrac{1}{\theta} \\\\ \hline r(r(r(\theta))) &=& \dfrac{1}{1- (1-\dfrac{1}{\theta}) } \\\\ &=& \dfrac{1}{1- 1 + \dfrac{1}{\theta} }\\\\ &=& \dfrac{1}{ \dfrac{1}{\theta} } \\\\ &=& \theta \\ \hline r(r(r(r(\theta)))) &=& \dfrac{1}{1-\theta} \\ \hline \end{array}\)

 

This is a cycle:

\(\begin{array}{|r|r|c|} \hline \text{cycle} & & \ldots r(r(\theta) \\ \hline 1 & \text{once} & \color{red}\dfrac{1}{1-\theta} \\ \hline 1 & \text{twice} & \color{green}\dfrac{\theta-1}{ \theta } \\ \hline 1 & 3\text{ times} & \color{blue}\theta \\ \hline\hline 2 & 4\text{ times} & \color{red}\dfrac{1}{1-\theta} \\ \hline 2 & 5\text{ times} & \color{green}\dfrac{\theta-1}{ \theta } \\ \hline 2 & 6\text{ times} & \color{blue}\theta \\ \hline \ldots & \ldots& \ldots \\ \hline \end{array} \)

 

\(\text{So $r(r(r(r(r(r(\theta)))))) = \theta $ and $ r(r(r(r(r(r(30)))))) = \mathbf{30} $ } \)

 

2. If $f(a) = \frac{1}{1-a}$, find the product $f^{-1}(a) \times a \times f(a)$. (Assume $a \neq 0$ and $a \neq 1$.)

 

\(\begin{array}{|rcll|} \hline f{\color{red}(}f^{-1}(a){\color{red})} &=& a \\\\ f {\color{red}\Big(}\dfrac{a-1}{a}{\color{red}\Big)} &=& a \quad & | \quad \text{see table above } r\left(\dfrac{\theta-1}{\theta} \right) = \theta \\\\ \text{so } \\\\ f^{-1}(a) &=& \dfrac{a-1}{a} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline && f^{-1}(a) \times a \times f(a) \\\\ &=& \dfrac{a-1}{a}\times a \times \dfrac{1}{1-a} \\\\ &=& \dfrac{a-1}{1-a} \\\\ &=& -\dfrac{1-a}{1-a} \\\\ &=& -1 \\ \hline \end{array}\)

 

 

laugh

heureka Jul 18, 2018
 #1
avatar+19651 
+2

Circle gamma intersects the hyperbola y=1/x  at (1,1), (3, 1/3) and two other points.

What is the product of the y coordinates of the other two points?

 

\(\text{Let $P_1 = (x_1 = 1,\ y_1 = 1)$} \\ \text{Let $P_2 = (x_2 = 3,\ y_2 = \frac13 )$} \\ \text{Center of the circle $= (x_0,\ y_0)$} \\ \text{Radius of the circle $= r $} \)

 

Formula of the circle:

\(\begin{array}{|lrcll|} \hline & (x-x_0)^2+(y-y_0)^2 &=& r^2 \\ \hline P_1(x_1,y_1): & (x_1 - x_0)^2 + (y_1 - y_0)^2 &=& r^2 \\ P_2(x_2,y_2): & (x_2 - x_0)^2 + (y_2 - y_0)^2 &=& r^2 \\ \hline (1)=(2): & (x_1 - x_0)^2 + (y_1 - y_0)^2 &=& (x_2 - x_0)^2 + (y_2 - y_0)^2 \\ & \Rightarrow \\ & ax_0+by_0 &=& c \\ & y_0 &=&\dfrac{c-ax_0}{b} \\ & &=& 3x_0-\dfrac{16}{3} \\ & \boxed{a=2(x_2-x_1) = 4 } \\ & \boxed{b=2(y_2-y_1) = -\dfrac{4}{3} } \\ & \boxed{c=x_2^2 +y_2^2-x_1^2-y_1^2=\dfrac{64}{9} } \\ \hline \end{array} \)

 

Circle intersects the hyperbola \(y=\dfrac{1}{x} \text{ or } x=\dfrac{1}{y}\)

\(\begin{array}{rcll} (x-x_0)^2+(y-y_0)^2 &=& r^2 \quad & | \quad r^2 = (x_1-x_0)^2+(y_1-y_0)^2 \\ (x-x_0)^2+(y-y_0)^2 &=& (x_1-x_0)^2+(y_1-y_0)^2 \quad & | \quad x=\dfrac{1}{y} \\ \left(\dfrac{1}{y}-x_0 \right)^2+(y-y_0)^2 &=& (x_1-x_0)^2+(y_1-y_0)^2 \\ \Rightarrow \\ \end{array} \\ \begin{array}{rcll} \boxed{ y^4-y^3\left(6x_0-\frac{32}{3}\right)+y^2\left(8x_0-\frac{38}{3}\right)-y\cdot2x_0+1= 0} \\ \end{array} \)

 

Examples calculated by WolframAlpha:

\(\begin{array}{|l|l|lr|l|} \hline x_0 & y_0 & y_3 & y_4 & y_3\cdot y_4 \\ \hline 1.4226 & -\dfrac{3.1966}{3} & -1.75493 & -1.70947 & 3.00000\ldots \\ 1.2 & -\dfrac{5.2}{3} & -4.06132 & -0.738675 & 2.99999\ldots \\ 1 & -\dfrac{7}{3} & -3-\sqrt{6} & \sqrt{6}-3 & 3 \\ 0.5 & -\dfrac{11.5}{3} & -8.65331 & -0.346688 & 2.99999\ldots \\ \hline \end{array} \)
 

 

I assume theproduct of the y coordinates of the other two points is 3

 

laugh

heureka Jul 18, 2018