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In triangle ABC, AB=c, BC=a, CA=b,
\(\angle A = \alpha\),
\(\angle B = \beta\), and \(\angle C = \gamma\)
where all angles are measured in degrees.
Suppose that \(b^2-a^2=ac\) and \(\beta^2 = \alpha \gamma\).
Find \(\lfloor 10\beta \rfloor\).

1.
\(\begin{array}{|rcll|} \hline b^2-a^2 &=&ac \\ (b-a)(b+a) &=& ac \\ \mathbf{(a-b)(a+b)} &=& \mathbf{-ac} \\ \hline \end{array}\)

Mollweide's formula

\(\begin{array}{|rcll|} \hline (a-b)\cos\left(\dfrac{\gamma}{2}\right) &=& c\cdot \sin\left(\dfrac{\alpha-\beta}{2} \right) \qquad (1) \\ (a+b)\sin\left(\dfrac{\gamma}{2}\right) &=& c\cdot \cos\left(\dfrac{\alpha-\beta}{2} \right) \qquad (2) \\ \hline \end{array}\)

(1) x (2)

\(\begin{array}{|rcll|} \hline (a-b)(a+b)\sin\left(\dfrac{\gamma}{2}\right)\cos\left(\dfrac{\gamma}{2}\right) &=& c^2\sin\left(\dfrac{\alpha-\beta}{2} \right)\cos\left(\dfrac{\alpha-\beta}{2} \right) \\ -ac\dfrac{\sin(\gamma)}{2} &=& c^2\dfrac{\sin(\alpha-\beta)}{2} \\ -a\sin(\gamma) &=& c\sin(\alpha-\beta) \\ a\sin(\gamma) &=& c\left(-\sin(\alpha-\beta)\right) \\ a\sin(\gamma) &=& c\left(\sin(-(\alpha-\beta))\right) \\ a\sin(\gamma) &=& c\sin(\beta-\alpha) \\ \dfrac{a}{c} &=& \dfrac{\sin(\beta-\alpha)}{\sin(\gamma)} \\ && \boxed{\dfrac{a}{c}=\dfrac{\sin(\alpha)} {\sin(\gamma)} } \\ \dfrac{\sin(\beta-\alpha)}{\sin(\gamma)} &=& \dfrac{\sin(\alpha)} {\sin(\gamma)} \\ \sin(\beta-\alpha) &=& \sin(\alpha) \\ \beta-\alpha &=& \alpha \\ \beta &=& 2\alpha \quad \text{or} \quad \mathbf{\alpha = \dfrac{\beta}{2}} \\ \hline \end{array}\)

2.

\(\begin{array}{|rcll|} \hline \beta^2 &=& \alpha \gamma \quad | \quad \mathbf{\alpha = \dfrac{\beta}{2}} \\ \beta^2 &=& \dfrac{\beta}{2} \gamma\\ \beta&=& \dfrac{\gamma}{2} \\ \mathbf{\gamma} &=& \mathbf{2\beta} \\ \hline \end{array}\)

3.

\(\begin{array}{|rcll|} \hline \mathbf{ \alpha + \beta + \gamma } &=& \mathbf{ 180^\circ } \quad | \quad \alpha = \dfrac{\beta}{2} \\ \dfrac{\beta}{2} + \beta + \gamma &=& 180^\circ \\ \dfrac{3}{2}\cdot \beta + \gamma &=& 180^\circ \quad | \quad \gamma = 2\beta \\ \dfrac{3}{2}\cdot\beta + 2\beta &=& 180^\circ \quad | \quad :2 \\ \dfrac{3}{4}\cdot\beta + \beta &=& 90^\circ \\ \dfrac{7}{4}\cdot\beta &=& 90^\circ \\ \beta &=& \dfrac{4}{7}\cdot 90^\circ \\ \beta &=& 51.4285714286^\circ \\ 10\cdot \beta &=& 514.285714286^\circ \\ \mathbf{ \lfloor 10\beta \rfloor } &=& \mathbf{ 514 } \\ \hline \end{array}\)

function

\(f_3=-f_2+3f_1=-3+3*(-1)=-6\\ f_4=-f_3+3f_2=6+3*3=15\\f_5=-f_4+3f_3=-15+3*(-6)=-33\)

Let \((x_1,y_1),\ (x_2,y_2),\ \dots \ ,\ (x_n,y_n)\)
be the real solutions to the system
\(\begin{align*} x + 8y &= 7, \\ x^3 + 8y^3 &= 7. \end{align*}\)

Enter \(x_1 + y_1 + x_2 + y_2 + \dots + x_n + y_n\).

Use Vieta:

\(\begin{array}{|rcll|} \hline \mathbf{x^3 + 8y^3} &=& \mathbf{7} \quad | \quad x = 7-8y \\ \left( 7-8y \right)^3 + 8y^3 &=& 7 \\ 7^3- 3*7^2*8y+ 3*7*(8y)^2 - (8y)^3 + 8y^3 &=& 7 \\ \ldots \\ y^3*(8-8^3) + y^2*(3*7*8^2) - y*(3*7^2*8) + 7^3 - 7 &=& 0 \\ y^3*8*(1-8^2) + y^2*(3*7*8^2) - y*(3*7^2*8) + 7^3 - 7 &=& 0 \\ y^3*(-8*63) + y^2*(3*7*8^2) - y*(3*7^2*8) + 7^3 - 7 &=& 0 \quad | \quad : (-8*63) \\\\ y^3 - y^2*\dfrac{3*7*8^2} {8*63} + y*\dfrac{3*7^2*8} {8*63} -\dfrac{7^3 - 7} {8*63} &=& 0 \quad | \quad : (-8*63) \\\\ y^3 - y^2*\dfrac{3*7*8} {3*21} + y*\dfrac{3*7^2*8} {8*63} -\dfrac{7^3 - 7} {8*63} &=& 0 \quad | \quad : (-8*63) \\\\ y^3 - y^2*\dfrac{7*8} {21} + y*\dfrac{3*7^2*8} {8*63} -\dfrac{7^3 - 7} {8*63} &=& 0 \quad | \quad : (-8*63) \\\\ y^3 - y^2*\dfrac{56} {21} + y*\dfrac{3*7^2*8} {8*63} -\dfrac{7^3 - 7} {8*63} &=& 0 \quad | \quad : (-8*63) \\\\ \hline \mathbf{y_1+y_2+y_3} &=& \mathbf{\dfrac{56} {21}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{x^3 + 8y^3} &=& \mathbf{7} \quad | \quad y = \dfrac{7-x}{8} \\\\ x^3 + 8*\left( \dfrac{7-x}{8} \right)^3 &=& 7 \\\\ x^3 + \dfrac{(7-x)^3}{8^2} &=& 7 \quad | \quad *8^2 \\\\ 8^2x^3 + (7-x)^3 &=& 7*8^2 \\ 8^2x^3 + (7-x)^3 - 7*8^2 &=& 0 \\ \ldots \\ x^3*(8^2-1) + x^2*(3*7) - x*3*7^2 +7^2 - 7*8^2 &=& 0 \\ x^3*63 + x^2*(3*7) - x*3*7^2 +7^2 - 7*8^2 &=& 0 \quad | \quad : 63 \\\\ x^3 + x^2*\dfrac{3*7}{63} - \dfrac{x*3*7^2}{63} + \dfrac{7^2 - 7*8^2}{63} &=& 0 \\\\ x^3 + x^2*\dfrac{3*7}{3*21} - \dfrac{x*3*7^2}{63} + \dfrac{7^2 - 7*8^2}{63} &=& 0 \\\\ x^3 + x^2*\dfrac{7}{21} - \dfrac{x*3*7^2}{63} + \dfrac{7^2 - 7*8^2}{63} &=& 0 \\\\ \hline \mathbf{x_1+x_2+x_3} &=& -\mathbf{\dfrac{7} {21}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{x_1 + y_1 + x_2 + y_2 + x_3 + y_3} &=& \dfrac{56} {21} - \dfrac{7} {21} \\\\ &=& \dfrac{56-7} {21} \\\\ &=& \dfrac{49} {21} \\\\ \mathbf{x_1 + y_1 + x_2 + y_2 + x_3 + y_3} &=& \mathbf{\dfrac{7} {3}} \\ \hline \end{array}\)

If \(z^2 + z + 1 = 0\), find
\(z^{49} + z^{50} + z^{51} + z^{52} + z^{53}\).

1)

\(\begin{array}{|rcll|} \hline && \mathbf{z^{49} + z^{50} + z^{51} + z^{52} + z^{53}} \\ &=& \left(z^{49} + z^{50} + z^{51} \right) + z^{52} + z^{53} \\ &=& z^{49}\left(1+ z + z^2 \right) + z^{52} + z^{53} \quad | \quad \mathbf{z^2 + z + 1 = 0} \\ &=& z^{49}\cdot 0 + z^{52} + z^{53} \\ &=& \mathbf{z^{52} + z^{53}} \\ \hline \end{array}\)

2)

\(\begin{array}{|rcll|} \hline && \mathbf{z^{49} + z^{50} + z^{51} + z^{52} + z^{53}} \\ &=& z^{49} + \left(z^{50} + z^{51} + z^{52}\right) + z^{53} \\ &=& z^{49}+ z^{50}\left(1+ z + z^2 \right) + z^{53} \quad | \quad \mathbf{z^2 + z + 1 = 0} \\ &=& z^{49}+ z^{50}\cdot 0 + z^{53} \\ &=& \mathbf{z^{49} + z^{53}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline z^{52} + z^{53} &=& z^{49} + z^{53} \\ z^{52} &=& z^{49} \quad | \quad :z^{49}\\ \dfrac{z^{52}}{z^{49}} &=& 1 \\ z^{52-49} &=& 1 \\ \mathbf{z^{3}} &=& \mathbf{1} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{z^{49} + z^{50} + z^{51} + z^{52} + z^{53}} &=& \mathbf{z^{49} + z^{53}} \\ &=& z^{3\cdot16+1} + z^{3\cdot17+2} \\ &=& z^{3\cdot16}z + z^{3\cdot17}z^2 \\ &=& \left(z^3\right)^{16}z + \left(z^3\right)^{17}z^2 \quad | \quad \mathbf{z^{3} = 1 } \\ &=& 1^{16}z + 1^{17}z^2 \\ &=& z + z^2 \\ && \boxed{z^2 + z + 1 = 0 \\ z^2 + z = -1 } \\ \mathbf{z^{49} + z^{50} + z^{51} + z^{52} + z^{53}} &=& \mathbf{-1} \\ \hline \end{array}\)

Summation notation of a sequence
The sequence of \(7 + 14x + 42x^2 + 168x^3 + 840x^4 + \ldots\)
\(n=0\), and goes to infinity.

\(\begin{array}{|rcll|} \hline && \mathbf{ 7 + 14x + 42x^2 + 168x^3 + 840x^4 + \ldots} \\ && \boxed{7 = 7\cdot 1!\\ 14 = 7\cdot 2!\\ 42 = 7\cdot 3!\\ 168 = 7\cdot 4!\\ 840 = 7\cdot 5!\\ \ldots } \\ &=& 7\cdot 1! + 7\cdot 2!x + 7\cdot 3!x^2 + 7\cdot 4!x^3 + 7\cdot 5!x^4 + \ldots \\ &=& \mathbf{ 7\sum \limits_{n=0}^{\infty} (n+1)!x^n} \\ \hline \end{array} \)

What is the largest integer n such that \(7^n\) divides 1000!

Alternate form
One may also reformulate Legendre's formula in terms of the base-p expansion of m.
Let s_{p}(m) denote the sum of the digits in the base-p expansion of m.

Given
\(\begin{array}{|rcll|} \hline px+qy+rz&=&1 \\ p+qx+ry&=&z \\ pz+q+rx&=&y \\ py+qz+r&=&x \\ p+q+r &=&-3 \\ \hline \end{array}\)

Find x+y+z.

\(\small{ \begin{array}{|lrcll|} \hline &px+qy+rz&=&1 \\ &p+qx+ry&=&z \\ &pz+q+rx&=&y \\ &py+qz+r&=&x \\ \hline \text{sum} & (px+py+pz) +(qx+qy+qz) \\ & +(rx+ry+rz) + (p+q+r) \\ & &=& 1+x+y+z \\\ & p(x+y+z) +q(x+y+z) \\ & +r(x+y+z) + (p+q+r) \\ & &=& 1+x+y+z \\ & (x+y+z)(p+q+r)+ (p+q+r) &=& 1+x+y+z \\ & (p+q+r)+(x+y+z)(p+q+r) &=& 1+x+y+z \\ & (p+q+r)(1+x+y+z) &=& 1+x+y+z \quad | \quad p+q+r=-3 \\ & -3(1+x+y+z) &=& (1+x+y+z) \quad | \quad -(1+x+y+z) \\ & -4(1+x+y+z) &=& 0 \quad | \quad : (-4) \\ & 1+x+y+z &=& 0 \\ & \mathbf{x+y+z} &=& \mathbf{-1} \\ \hline \end{array} }\)

what is the sum of \(i^{2017}\) , \(i ^{2018}\) , and \(i ^{2019}\) where i = square root of -1?

\(Formula: \boxed{\mathbf{i^n+i^{2+n}=0}}\)

\(\begin{array}{|rcll|} \hline && i^{2017}+i^{2018}+i^{2019} \\ &=& i^{2017}+i^{2019}+i^{2018} \quad | \quad \mathbf{i^{2017}+i^{2019} = 0} \\ &=& 0 + i^{2018} \\ &=& i^{2*1009} \\ &=& \left(i^{2}\right)^{1009} \quad | \quad \mathbf{i^2 = -1} \\ &=& \left(-1\right)^{1009} \\ &=& \mathbf{-1} \\ \hline \end{array}\)

If (a_1, a_2..... a_17) satisfy

1 + a_2 + a_3 = 1,
a_2 + a_3 + a_4 = 3,
a_3 + a_4 + a_5 = 5,

...

a_{15} + a_{16} + a_{17} = 29,
a_{16} + a_{17} + a_{1} = 31,
a_{17} + a_{1} + a_{2} = 33,

what is the value of a_17?

\(\begin{array}{|lrcll|} \hline (1) & 1 + a_2 + a_3 &=& 1 \\ (2) & a_2 + a_3 + a_4 &=& 3 \\ (3) & a_3 + a_4 + a_5 &=& 5 \\ (4) & a_4 + a_5 + a_6 &=& 7 \\ (5) & a_5 + a_6 + a_7 &=& 9 \\ \ldots \\ (14) &a_{14} + a_{15} + a_{16} &=& 27 \\ (15) &a_{15} + a_{16} + a_{17} &=& 29 \\ (16) &a_{16} + a_{17} + a_{1} &=& 31 \\ (17) &a_{17} + a_{1} + a_{2} &=& 33 \\ \hline \end{array} \begin{array}{|rcll|} \hline (1) & 1 + a_2 + a_3 &=& 1 \\ & \mathbf{a_2 + a_3} &=& \mathbf{0} \ \text{or}\ \mathbf{a_3 =-a_2} \\ \hline (2) & a_2 + a_3 + a_4 &=& 3 \\ & 0 + a_4 &=& 3 \\ & \mathbf{a_4} &=& \mathbf{3} \\ \hline \end{array}\)

\(\begin{array}{|l|rcll|} \hline (3)-(2) & a_5 &=& 2 + a_2 \\ (4)-(3) & a_6 &=& 2 + a_3 \\ (5)-(4) & a_7 &=& 2 + a_4 \\ (6)-(5) & a_8 &=& 2 + a_5 \\ (7)-(6) & a_9 &=& 2 + a_6 \\ (8)-(7) & a_{10} &=& 2 + a_7 \\ (9)-(8) & a_{11} &=& 2 + a_8 \\ (10)-(9) & a_{12} &=& 2 + a_9 \\ (11)-(10) & a_{13} &=& 2 + a_{10} \\ (12)-(11) & a_{14} &=& 2 + a_{11} \\ (13)-(12) & a_{15} &=& 2 + a_{12} \\ (14)-(13) & a_{16} &=& 2 + a_{13} \\ (15)-(14) & a_{17} &=& 2 + a_{14} \\ (16)-(15) & a_1 &=& 2 + a_{15} \\ (17)-(16) & a_2 &=& 2 + a_{16} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a_3 &=& -a_2 \\ a_4 &=& 3 \\ \hline a_5 &=& 2 + a_2 \\ a_6 = 2 + a_3 &=& 2 - a_2 \\ a_7 = 2 + a_4 &=& 2 + 3 = 5 \\ \hline a_8 = 2 + a_5 &=& 2 + 2 + a_2 = 4+a_2\\ a_9 = 2 + a_6 &=& 2+2 - a_2 = 4 - a_2 \\ a_{10} = 2 + a_7 &=& 2 + 5 = 7 \\ \hline a_{11} = 2 + a_8 &=& 2 + 4 + a_2 = 6 + a_2\\ a_{12} = 2 + a_9 &=& 2 + 4 - a_2 = 6 - a_2 \\ a_{13} = 2 + a_{10} &=& 2 + 7 = 9 \\ \hline a_{14} = 2 + a_{11} &=& 2 + 6 + a_2 = 8 + a_2\\ a_{15} = 2 + a_{12} &=& 2 + 6 - a_2 = 8 - a_2 \\ a_{16} = 2 + a_{13} &=& 2 + 9 = 11 \\ \hline a_{17} = 2 + a_{14} &=& 2 + 8 + a_2 = 10 + a_2\\ a_{1} = 2 + a_{15} &=& 2 + 8 - a_2 = 10 - a_2 \\ \mathbf{a_2} = 2 + a_{16} &=& 2 + 11 \mathbf{= 13} \\ \hline a_{17} &=& 10 + a_2 \quad | \quad a_2 = 13 \\ a_{17} &=& 10 + 13 \\ \mathbf{a_{17}} &=& \mathbf{23} \\ \hline \end{array}\)

... so ...

\(\begin{array}{|rcll|} \hline a_1 &=& -3 \\ \hline a_2 &=& 13 \\ a_3 &=& - 13 \\ a_4 &=& 3 \\ a_5 &=& 15 \\ a_6 &=&-11 \\ a_7 &=& 5 \\ a_8 &=& 17 \\ a_9 &=& -9 \\ a_{10} &=& 7 \\ a_{11} &=& 19 \\ a_{12} &=& -7 \\ a_{13} &=& 9 \\ a_{14} &=& 21 \\ a_{15} &=& -5 \\ a_{16} &=& 11 \\ a_{17} &=& 23 \\ a_1 &=& -3 \\ a_2 &=& 13 \\ \hline \text{check} && a_{17} + a_1 + a_2 = 23-3+13 = 33\ \checkmark \\ \hline \end{array}\)

Suppose a, b, c, and d are real numbers which satisfy the system of equations
a + 2b + 3c + 4d = 10
4a + b + 2c + 3d = 4
3a + 4b + c + 2d = 10
2a + 3b + 4c + d = −4.
Find a + b + c + d.

\(\begin{array}{|lrcll|} \hline (1) & a + 2b + 3c + 4d &=& 10 \\ (2) & 4a + b + 2c + 3d &=& 4 \\ (3) & 3a + 4b + c + 2d &=& 10 \\ (4) &2a + 3b + 4c + d &=& −4 \\ \hline \hline \text{sum} & 10a+10b+10c+10d &=& 10 + 4 + 10 - 4 \\ & 10(a+b+c+d) &=& 20 \quad | \quad : 10 \\ & a+b+c+d &=& \dfrac{20}{10} \\ & \mathbf{a+b+c+d} &=& \mathbf{2} \\ \hline \end{array} \)