heureka

2)
Let a, b, and c be the roots of \(x^3 - 5x + 7 = 0\).
Find the monic polynomial, in \(x\), whose roots are \(a - 2\), \(b - 2\), and \(c - 2\).

\(\begin{array}{|rcll|} \hline && x'^3 - 5x' + 7 \quad | \quad x'=x+2 \quad \text{moving the polynomial by two units} \\ && (x+2)^3 - 5(x+2) + 7 \\ &=& x^3+3x^2\cdot 2+3x\cdot 2^2+2^3-5x-10+7 \\ &=& x^3+ 6x^2+7x+5 \\ \hline \end{array} \)

2)
Let a, b, and c be the roots of \(x^3 - 5x + 7 = 0\).
Find the monic polynomial, in \(x\), whose roots are \(a - 2\), \(b - 2\), and \(c - 2\).

\(\begin{array}{|lrcll|} \hline \text{Let $a$, $b$, and $c$ be the roots} \\ & x^3 - 5x + 7 &=& (x-a)(x-b)(x-c) \\\\ \text{vieta:} \\ \text{coefficient}\ x^2: & -(a+b+c) &=& 0 \\ \text{coefficient}\ x^1: & ab+bc+ac &=& -5 \\ \text{coefficient}\ x^0: & -abc &=& 7 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \text{Let $a-2$, $b-2$, and $c-2$ be the roots} \\ x^3+Ax^2+Bx+C &=& \Big(x-(a-2)\Big)\Big(x-(b-2)\Big)\Big(x-(c-2)\Big) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \Big(x-(a-2)\Big)\Big(x-(b-2)\Big)\Big(x-(c-2)\Big) \\ &=& -a b c + a b x + 2 a b + a c x + 2 a c - a x^2 - 4 a x - 4 a + b c x \\ && + 2 b c - b x^2 - 4 b x - 4 b - c x^2 - 4 c x - 4 c + x^3 + 6 x^2 + 12 x + 8 \quad | \quad \text{Expanded form} \\ &=& x^3+x^2\Big(6-(a+b+c)\Big)+x\Big(ab+bc+ac-4(a+b+c)+12\Big)-abc+2(ab+bc+ac)-4(a+b+c)+8 \\ && \boxed{ a+b+c=0\\ ab+bc+ac=-5 \\ -abc=7 } \\ &=& x^3+x^2(6-0)+x(-5-0+12)+7+2(-5)-0+8 \\ &=& \mathbf{x^3+6x^2+7x+5} \\ \hline \end{array}\)

The monic polynomial, in \(x\), whose roots are \(a - 2\), \(b - 2\), and \(c - 2\) is \(\mathbf{x^3+6x^2+7x+5}\)

Ms. Forsythe gave the same algebra test to her three classes.
The first class averaged 80%, the second class averaged 85%, and the third 89%.
Together, the first two classes averaged 83%, and the second and third classes together averaged 87%.
What was the average for all three classes combined?
Express your answer to the nearest hundredth.

\(\text{The students in the first class $ = s_1$ } \\ \text{The students in the second class $ = s_2$ } \\ \text{The students in the third class $ = s_3$ } \\ \text{The sum of the points in the first class $ = p_1$ } \\ \text{The sum of the points in the second class $ = p_2$ } \\ \text{The sum of the points in the third class $ = p_3$ } \\ \text{The maximal points of the test $ = p$ } \)

\(\begin{array}{|lrcl|lrcl||lrcl|} \hline & \dfrac{\frac{p_1}{s_1}} {p} &=& 80\% & & \dfrac{\frac{p_2}{s_2}} {p} &=& 85\% & & \dfrac{\frac{p_3}{s_3}} {p} &=& 89\% \\ (1) & p_1 &=& 80\%ps_1 & (2) & p_2 &=& 85\%ps_2 &(3) & p_3 &=& 89\%ps_3 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \dfrac{\frac{p_1+p_2}{s_1+s_2}} {p} &=& 83\% \quad | \quad p_1 = 80\%ps_1,\ p_2 = 85\%ps_2 \\ \dfrac{\frac{80\%ps_1+85\%ps_2}{s_1+s_2}} {p} &=& 83\% \\ \dfrac{80\%ps_1+85\%ps_2}{(s_1+s_2)p} &=& 83\% \\ \dfrac{80\%s_1+85\%s_2}{(s_1+s_2)} &=& 83\% \\ 80\%s_1+85\%s_2 &=& 83\% (s_1+s_2) \\ 80\%s_1+85\%s_2 &=& 83\%s_1+83\%s_2 \\ 85\%s_2-83\%s_2 &=& 83\%s_1-80\%s_1 \\ 2\%s_2&=& 3\%s_1 \\ 2s_2&=& 3s_1 \\ \mathbf{s_2} &=& \mathbf{1.5s_1} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \dfrac{\frac{p_2+p_3}{s_2+s_3}} {p} &=& 87\% \quad | \quad p_2 = 85\%ps_2,\ p_3 = 89\%ps_3 \\ \dfrac{\frac{85\%ps_2+89\%ps_3}{s_2+s_3}} {p} &=& 87\% \\ \dfrac{85\%ps_2+89\%ps_3}{(s_2+s_3)p} &=& 87\% \\ \dfrac{85\%s_2+89\%s_3}{(s_2+s_3)} &=& 87\% \\ 85\%s_2+89\%s_3 &=& 87\%(s_2+s_3) \\ 85\%s_2+89\%s_3 &=& 87\%s_2+87\%s_3 \\ 89\%s_3-87\%s_3 &=& 87\%s_2-85\%s_2 \\ 2\%s_3 &=& 2\%s_2 \\ \mathbf{s_3} &=& \mathbf{s_2} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{\frac{p_1+p_2+p_3}{s_1+s_2+s_3}} {p} &=& x \quad | \quad p_1 = 80\%ps_1,\ p_2 = 85\%ps_2,\ p_3 = 89\%ps_3 \\ \dfrac{\frac{80\%ps_1+85\%ps_2+89\%ps_3}{s_1+s_2+s_3}} {p} &=& x \\ \dfrac{80\%ps_1+85\%ps_2+89\%ps_3}{(s_1+s_2+s_3)p} &=& x \\ \dfrac{80\%s_1+85\%s_2+89\%s_3}{(s_1+s_2+s_3)} &=& x \\ 80\%s_1+85\%s_2+89\%s_3 &=& x(s_1+s_2+s_3) \quad | \quad s_2 =1.5s_1,\ s_3=s_2=1.5s_1 \\ 80\%s_1+85\%1.5s_1+89\%1.5s_1 &=& x(s_1+1.5s_1+1.5s_1) \\ 80\%s_1+85\%1.5s_1+89\%1.5s_1 &=& x(4s_1) \\ 80\% +85\%1.5 +89\%1.5 &=& 4x \\ 80\% +127.5\% +133.5\% &=& 4x \\ 341\% &=& 4x \quad | \quad : 4 \\ 85.25\% &=& x \\ \mathbf{x} &=& \mathbf{85.25\%} \\ \hline \end{array}\)

The average for all three classes combined is \(\mathbf{85.25\%}\)

1)
The sequence \(a_1, a_2, a_3,\ \ldots\) satisfies \(a_1 = 19,\ a_9 = 99\), and for all \(n \ge 3\),

\(a_n\) is the arithmetic mean of the first \(n - 1\) terms.
Find a_2.

I assume:

\(\begin{array}{|rcll|} \hline a_3 &=& \dfrac{a_1+a_2}{2} &\text{or }\qquad 2a_3 = a_1+a_2 \\ \hline a_4 &=& \dfrac{a_1+a_2}{3} + \dfrac{a_3}{3} &\text{or }\qquad 3a_4 = a_1+a_2+a_3 \\\\ &=& \dfrac{2a_3}{3} + \dfrac{a_3}{3} \\\\ &=& \dfrac{3a_3}{3} \\\\ \mathbf{a_4} &=& \mathbf{a_3} \\ \hline a_5 &=& \dfrac{a_1+a_2+a_3}{4} + \dfrac{a_4}{4} &\text{or }\qquad 4a_5 = a_1+a_2+a_3+a_4 \\\\ &=& \dfrac{3a_4}{4} + \dfrac{a_4}{4} \\\\ &=& \dfrac{4a_4}{4} \\\\ \mathbf{a_5} &=& \mathbf{a_4} \\ \hline a_6 &=& \dfrac{a_1+a_2+a_3+a_4}{5} + \dfrac{a_5}{5} &\text{or }\qquad 5a_6 = a_1+a_2+a_3+a_4+a_5 \\\\ &=& \dfrac{4a_5}{5} + \dfrac{a_5}{5} \\\\ &=& \dfrac{5a_5}{5} \\\\ \mathbf{a_6} &=& \mathbf{a_5} \\ \hline \ldots \\ a_3&=&a_4=a_5=a_6=a_7=a_8=a_9 \\ a_3 &=& a_9 \quad | \quad a_3 = \dfrac{a_1+a_2}{2},\ a_9 = 99 \\\\ \dfrac{a_1+a_2}{2} &=& 99 \quad | \quad a_1 = 19 \\\\ \dfrac{19+a_2}{2} &=& 99 \quad | \quad \cdot 2 \\\\ 19+a_2 &=& 198 \\\\ a_2 &=& 198-19 \\ \mathbf{ a_2 } &=& \mathbf{179} \\ \hline \end{array}\)

The sequence: \(19,\ 179,\ 99,\ 99,\ 99,\ 99,\ 99,\ 99,\ 99,\ \ldots\)

3)

Let a,b,c be positive real numbers such that \(\log_a b + \log_b c + \log_c a = 0\).
Find \((\log_a b)^3 + (\log_b c)^3 + (\log_c a)^3\).

\(\text{Let $\log_a b = \mathbf{x} $} \\ \text{Let $\log_b c = \mathbf{y} $} \\ \text{Let $\log_c a = \mathbf{z} $} \)

\(\begin{array}{|rcll|} \hline \log_a b + \log_b c + \log_c a &=& 0 \\ \mathbf{x+y+z} &=& \mathbf{0} \qquad (1) \\ \hline \end{array}\)

\(\begin{array}{|rclrcl|} \hline \log_a b &=&\dfrac{\log_c b}{\log_c a} \quad&| \quad \log_c b &=&\dfrac{\log_b b}{\log_b c} \\\\ \log_a b &=&\dfrac{\log_b b}{\log_b c\log_c a} \quad&| \quad \log_b b = 1 \\\\ \log_a b &=&\dfrac{1}{\log_b c\log_c a} \\\\ \log_a b\log_b c\log_c a &=& 1 \\\\ \mathbf{xyz} &=& \mathbf{1} \qquad (2) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (x+y+z)^3 &=& (x+y+z)^2(x+y+z) \\ &=& \left(x^2+y^2+z^2+2(xy+yz+xz)\right)(x+y+z) \\ &=& (x^2+y^2+z^2)(x+y+z)+ 2(x+y+z)(xy+yz+xz) \\\\ &=& x^3+y^3+z^3 \\ && +x^2y+x^2z+y^2x+y^2z+z^2x+z^2y + 2(x+y+z)(xy+yz+xz) \\\\ &=& x^3+y^3+z^3 \\ && +(x+y+z)(xy+yz+xz) -3xyz + 2(x+y+z)(xy+yz+xz) \\\\ (x+y+z)^3&=& x^3+y^3+z^3 -3xyz + 3(x+y+z)(xy+yz+xz) \\ \hline \mathbf{x^3+y^3+z^3} &=& \mathbf{(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3xyz} \qquad (3) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{x^3+y^3+z^3} &=& \mathbf{(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3xyz} \\ && \boxed{x+y+z = 0} \\ x^3+y^3+z^3 &=& 0^3-3\cdot 0\cdot (xy+yz+xz)+3xyz \\ x^3+y^3+z^3 &=& 3xyz \\ && \boxed{xyz = 1} \\ x^3+y^3+z^3 &=& 3\cdot 1 \\ x^3+y^3+z^3 &=& 3 \\ \mathbf{(\log_a b)^3 + (\log_b c)^3 + (\log_c a)^3} &=& \mathbf{3} \\ \hline \end{array}\)

Twelve 1 by 1 squares form a rectangle, as shown. What is the total area of the shaded region?

\(\text{Let the area of the lower triangle is $\dfrac{2\cdot 4}{2} = 4 $ }\\ \text{Let the area of the upper triangle is $\dfrac{3\cdot 4}{2} = 6 $ }\\ \text{The area of the shaded region is $4+6=\mathbf{10} $ }\)

A square and a triangle have the same area.

If the square has a side length of 6 units and the triangle has a base of 8 units, what is the length, in units, of the altitude to that base of the triangle?

\(\begin{array}{|rcll|} \hline \dfrac{8h}{2} &=& 6^2 \\ 4h &=& 36 \\ h &=& \dfrac{36}{4} \\ \mathbf{h} &=& \mathbf{9} \\ \hline \end{array} \)

The altitude  of the triangle is 9 units

A square is divided, as shown. What fraction of the area of the square is shaded? Express your answer as a fraction.

\(\text{Let $s$ is the side of the square }\\ \text{Let $s^2$ is the area of the square }\\ \text{Let $s\sqrt{2}$ is the diagonal of the square } \\ \text{Let $s\dfrac{\sqrt{2}}{2}$ is the parallel diagonal of the square } \)

\(\begin{array}{|rcll|} \hline \mathbf{A_{\text{trapezoid}}} &=& \left(\dfrac{s\sqrt{2}+s\dfrac{\sqrt{2}}{2}}{2}\right)\cdot s\dfrac{\sqrt{2}}{4} \quad &| \quad s\dfrac{\sqrt{2}}{4} \text{ is the height of the trapezoid } \\ &=& \left(\dfrac{1+ \dfrac{1}{2} }{2}\right)\cdot s\sqrt{2} s\dfrac{\sqrt{2}}{4} \\ &=& \left(\dfrac{\dfrac{3}{2} }{2}\right)\cdot s^2 \dfrac{2}{4} \\ &=& \left(\dfrac{3}{4}\right)\cdot s^2 \dfrac{1}{2} \\ &=& \mathbf{ \dfrac{3}{8} s^2} \\\\ \mathbf{A_{\text{shaded}}} &=& \dfrac{A_{\text{trapezoid}}}{2} \\ &=& \dfrac{\dfrac{3}{8} s^2}{2} \\ \mathbf{A_{\text{shaded}}} &=& \mathbf{ \dfrac{3}{16} s^2} \\ \hline \end{array}\)

The fraction of the area of the square shaded is \(\dfrac{\dfrac{3}{16} s^2}{s^2} = \dfrac{3}{16}\)

Thank You, CPhill !

Thank you, GingerAle !