heureka

Two 3x4 rectangles overlap, as shown below. Find the area of the overlapping region. (Which is shaded)

\(\begin{array}{|rcll|} \hline 3^2+x^2 &=& (4-x)^2 \\ 3^2+x^2 &=& 16 - 2*4x + x^2 \\ 3^2+x^2 &=& 16 {\color{red}{-8}}x + x^2 \\ 9 &=& 16 -8x \\ 8x &=& 16-9 \\ 8x &=& 7\\\\ x &=& \mathbf{\dfrac{7}{8}} \\ \hline \end{array}\)

area

\(\begin{array}{|rcll|} \hline A &=& 3*(4-x) \\ &=& 12-3x \\\\ &=& 12 -\dfrac{3*7}{8} \\\\ &=& 12 -\dfrac{21}{8} \\\\ &=& \dfrac{8*12-21}{8} \\\\ \mathbf{A }&=& \mathbf{\dfrac{75}{8}} \qquad \text{or} \quad \dfrac{150}{16} \\ \hline \end{array}\)

What is the remainder when the sum

\(1^2+2^2+3^2+...+2016^2 + 2017^2 + 2018^2 +2019^2 +2020^2\) 

is divided by 17?

\(\begin{array}{|rcll|} \hline 1^2 \pmod {17} &=& 1 \\ 2^2 \pmod {17} &=& 4 \\ 3^2 \pmod {17} &=& 9 \\ 4^2 \pmod {17} &=& 16 \\ 5^2 \pmod {17} &=& 8 \\ 6^2 \pmod {17} &=& 2 \\ 7^2 \pmod {17} &=& 15 \\ 8^2 \pmod {17} &=& 13 \\ 9^2 \pmod {17} &=& 13 \\ 10^2 \pmod {17} &=& 15 \\ 11^2 \pmod {17} &=& 2 \\ 12^2 \pmod {17} &=& 8 \\ 13^2 \pmod {17} &=& 16 \\ 14^2 \pmod {17} &=& 9 \\ 15^2 \pmod {17} &=& 4 \\ 16^2 \pmod {17} &=& 1 \\ 17^2 \pmod {17} &=& 0 \\ \hline \end{array}\)

cycle mod 17: \((1, 4, 9, 16, 8, 2, 15, 13, 13, 15, 2, 8, 16, 9, 4, 1, 0)\)

circle mod 17

\(\begin{array}{|rcll|} \hline && 1+ 4+ 9+ 16+ 8+ 2+ 15+ 13+ 13+ 15+ 2+ 8+ 16+ 9+ 4+ 1+ 0\pmod {17} \\ &=& 136 \pmod {17} \\ &=& \mathbf{0 \pmod {17}} \\ \hline \end{array}\)

2020/17 = 118 cycles remainder 14

remainder 14, this is the sum in the cycle from 1 to 14  mod 17

\(\begin{array}{|rcll|} \hline && 1+ 4+ 9+ 16+ 8+ 2+ 15+ 13+ 13+ 15+ 2+ 8+ 16+ 9\pmod {17} \\ &=& 131 \pmod {17} \\ &=&\mathbf{ 12 \pmod {17}} \\ \hline \end{array}\)

\( \begin{array}{|rcll|} \hline && 1^2+2^2+3^2+...+2016^2 + 2017^2 + 2018^2 +2019^2 +2020^2\pmod {17} \\ &=& \underbrace{0*118 \pmod {17}}_{118 \text{ cycles}} + \underbrace{\mathbf{ 12 \pmod {17}}}_{\text{remainder 14}} \\ &=& \mathbf{ 12 \pmod {17}} \\ \hline \end{array}\)

The remainder when the sum \(1^2+2^2+3^2+...+2016^2 + 2017^2 + 2018^2 +2019^2 +2020^2 \) is divided by 17 is \(\mathbf{12}\)

Find all complex solutions such that z^4 = -4
Note: All solutions should be expressed in the form a+bi,
where a and b are real numbers help!

\(\huge{z^2 = \pm 2i}\)

\(\begin{array}{|rcll|} \hline z^2 &=& \pm 2i \quad | \quad z = a+bi \\ \left(a+bi\right)^2 &=& \pm 2i \\ a^2+2abi+b^2i^2 &=& \pm 2i \quad | \quad i^2 = -1 \\ a^2+2abi-b^2 &=& \pm 2i \\ \mathbf{a^2-b^2+2abi} &=& \mathbf{\pm 2i} \\ \begin{array}{|rcll|} \hline a^2-b^2+2abi &=& \mathbf{+2i}\\ a^2-b^2+2abi &=& 0+2i \quad | \quad \text{compare}\\ \begin{array}{|rcll|} \hline 2ab&=& 2 \\ \mathbf{b} &=& \dfrac1a \\ \hline \end{array} &\text{and}& \begin{array}{|rcll|} \hline a^2-b^2 &=& 0 \\ a^2 &=& b^2 \\ a^2 &=& \left(\dfrac1a\right)^2 \\ a^4 &=& 1 \\ \mathbf{a} &=& \pm \mathbf{1} \\ a_1 &=& 1 \\ a_2 &=& -1 \\ b_1 &=& \dfrac1a_1 \\ b_1 &=& 1 \\ b_2 &=& \dfrac1a_2 \\ b_2 &=& -1 \\ \hline \end{array}\\ \hline \end{array} &\text{or}& \begin{array}{|rcll|} \hline a^2-b^2+2abi &=& \mathbf{-2i}\\ a^2-b^2+2abi &=& 0-2i\quad | \quad \text{compare}\\ \begin{array}{|rcll|} \hline 2ab&=& -2 \\ \mathbf{b} &=& -\dfrac1a \\ \hline \end{array} &\text{and}& \begin{array}{|rcll|} \hline a^2-b^2 &=& 0 \\ a^2 &=& b^2 \\ a^2 &=& \left(-\dfrac1a\right)^2 \\ a^2 &=& \left(\dfrac1a\right)^2 \\ a^4 &=& 1 \\ \mathbf{a} &=& \pm \mathbf{1} \\ a_3 &=& 1 \\ a_4 &=& -1 \\ b_3 &=& -\dfrac1a_3 \\ b_3 &=& -1 \\ b_4 &=& -\dfrac1a_4 \\ b_4 &=& 1 \\ \hline \end{array}\\ \hline \end{array} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline z&=& a_1+b_1i = 1+i\\ z&=& a_2+b_2i = -1-i\\ z&=& a_3+b_3i = 1 -i\\ z&=& a_4+b_4i = -1+i\\ \hline \end{array}\)

Let a and b be the solutions to 5x^2 + 11x + 14 = 0.
Find 1/a + 1/b

\(\begin{array}{|rcll|} \hline \mathbf{5x^2 + 11x + 14} &=& \mathbf{0} \\ \boxed{x\rightarrow \frac1x} \\ 5\left(\dfrac{1}{x}\right)^2 + 11\left(\dfrac{1}{x}\right) + 14 &=& 0 \\\\ \dfrac{5}{x^2} + \dfrac{11}{x} + 14 &=& 0 \quad | \quad \times x^2 \\\\ 5 + 11x + 14x^2 &=& 0 \quad | \quad :14 \\\\ \dfrac{5}{14} + \dfrac{11}{14}x + x^2 &=& 0 \\\\ x^2 + \dfrac{11}{14}x + \dfrac{5}{14} &=& 0 \\\\ \boxed{\text{By Vieta:}} \\\\ x^2 + \underbrace{\dfrac{11}{14}}_{=-\left(\dfrac1a+\dfrac1b \right)}x + \underbrace{\dfrac{5}{14}}_{=\dfrac1a\times\dfrac1b} &=& 0 \\\\ \hline -\left(\dfrac1a+\dfrac1b \right) &=& \dfrac{11}{14} \\\\ \mathbf{\dfrac1a+\dfrac1b} &=& \mathbf{-\dfrac{11}{14}} \\ \hline \end{array}\)

For positive integer values of N, let boxed(N) be equal to:
 
boxed(N) = 2+4+6+8+...+N if N is even,
and
boxed(N) = 1+3+5+7+...+N if N is odd.

What is the value of boxed(2009) minus boxed(2008)?

\(\begin{array}{|rcll|} \hline s &=& boxed(2009) \quad minus \quad boxed(2008) \\ s &=& (1 + 3 + 5 + ... + 2009) - (2 + 4 + ... + 2008) \\ s &=& 1 - 2 + 3 - 4 + \dots + 2007 - 2008 + 2009 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline s &=& 1 - 2 + 3 - 4 + \dots + 2007 - 2008 + 2009 \\ s &=& 2009 - 2008 +2007-2006 + \dots +3-2+1 \\ \hline 2s &=& 2010-2010+2010-2010 + \dots + 2010-2010+2010 \\ 2s &=& 0+0+ \dots +0+2010 \\ 2s &=& 2010 \quad | \quad : 2\\ s &=& 1005 \\ \hline \end{array}\)

boxed(2009) minus boxed(2008)  = \( \mathbf{1005}\)

If z is a complex number satisfying z + 1/z =1, calculate z^12+ 1/z^12.

Compute

\(\mathbf{\dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{10^3 - 10}}\)

\(\begin{array}{|rcll|} \hline s&=& \dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{10^3 - 10} \\\\ &=& \dfrac{1}{2(2^2-1)} + \dfrac{1}{3(3^2-1)} + \dfrac{1}{4(4^2-1)} + \dots + \dfrac{1}{10(10^2-19} \\\\ &=& \dfrac{1}{(2-1)2(2+1)} + \dfrac{1}{(3-1)3(3+1)} + \dfrac{1}{(4-1)4(4+1)} + \dots + \dfrac{1}{(10-1)10(10+1)} + \dots + \dfrac{1}{(n-1)n(n+1)} \\\\ \hline s &=&\sum \limits_{n=2}^{10}\dfrac{1}{(n-1)n(n+1)} \\ \\ &&\boxed{ \frac{1}{(n-1)n} =\frac{1}{n-1}-\frac{1}{n} \\ \frac{1}{(n+1)n} =\frac{1}{n}-\frac{1}{n+1}\\\ldots\\ \frac{1}{(n-1)n(n+1)} = \frac12\left( \frac{1}{n-1}-\frac{2}{n} +\frac{1}{n+1} \right) } \\\\ s &=&\dfrac12\sum \limits_{n=2}^{10}\left( \dfrac{1}{n-1}-\dfrac{2}{n} +\dfrac{1}{n+1} \right) \\ \\ &=&\dfrac12\left( \sum \limits_{n=2}^{10}\dfrac{1}{n-1}-\sum \limits_{n=2}^{10}\dfrac{2}{n} +\sum \limits_{n=2}^{10}\dfrac{1}{n+1} \right) \\ \\ &=&\dfrac12\left( \sum \limits_{n=1}^{9}\dfrac{1}{n} -\sum \limits_{n=2}^{10}\dfrac{2}{n} +\sum \limits_{n=3}^{11}\dfrac{1}{n} \right) \\ \\ &&\boxed{ \sum \limits_{n=1}^{9}\frac{1}{n} = \frac11 + \frac12 + \sum \limits_{n=3}^{9}\frac{1}{n} \\ -\sum \limits_{n=2}^{10}\frac{2}{n} = -\frac{2}{2}-\frac{2}{10}-2\sum \limits_{n=3}^{9}\frac{1}{n} \\ \sum \limits_{n=3}^{11}\frac{1}{n} = \frac{1}{10}+\frac{1}{11}+\sum \limits_{n=3}^{9}\frac{1}{n} } \\\\ s &=&\dfrac12\left( \dfrac11 + \dfrac12 + \sum \limits_{n=3}^{9}\dfrac{1}{n} -\dfrac{2}{2}-\dfrac{2}{10}-2\sum \limits_{n=3}^{9}\dfrac{1}{n} +\dfrac{1}{10}+\dfrac{1}{11}+\sum \limits_{n=3}^{9}\dfrac{1}{n}\right) \\\\ &=&\dfrac12\left( \dfrac11 + \dfrac12 -\dfrac{2}{2}-\dfrac{2}{10}+\dfrac{1}{10}+\dfrac{1}{11}\right) \\ \\ &=&\dfrac12\left( \dfrac12 -\dfrac{2}{10}+\dfrac{1}{10}+\dfrac{1}{11}\right) \\ \\ &=&\dfrac12\left( \dfrac12 -\dfrac{1}{10}+\dfrac{1}{11}\right) \\ \\ &=&\dfrac14 -\dfrac{1}{20}+\dfrac{1}{22} \\ \\ &=&\dfrac{5}{20} -\dfrac{1}{20}+\dfrac{1}{22} \\ \\ &=&\dfrac{4}{20}+\dfrac{1}{22} \\ \\ &=&\dfrac{1}{5}+\dfrac{1}{22} \\ \\ &=&\dfrac{22+5}{5*22}\\ \\ \mathbf{s}&=&\mathbf{\dfrac{27}{110}}\\ \hline \end{array}\)

\(\dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{10^3 - 10} = \mathbf{\dfrac{27}{110}}\)

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

\(\begin{array}{|lrcll|} \hline & z^2 + z + 1 &=& 0 \\ \Rightarrow & z + 1 &=& -z^2 \\ \Rightarrow & z^2 + z &=& -1 \\ \hline \end{array} \begin{array}{|rcll|} \hline z^2 + z + 1 &=& (1+z)^2-z \quad | \quad z + 1 = -z^2 \\ z^2 + z + 1 &=& (-z^2)^2-z \\ z^2 + z + 1 &=& z^4-z \quad | \quad z^2 + z + 1 = 0 \\ 0 &=& z^4-z \\ \mathbf{z^4} &=& \mathbf{z} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline z^{49} &=& z^{48}z \\ z^{49} &=& \left(z^{4}\right)^{12}z \quad | \quad z^4 = z \\ z^{49} &=& \left(z\right)^{12}z \\ z^{49} &=& \left(z^{4}\right)^{3}z \quad | \quad z^4 = z \\ z^{49} &=& \left(z\right)^{3}z \\ z^{49} &=& z^{4} \\ \mathbf{z^{49}} &=& \mathbf{z} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && z^{49} + z^{50} + z^{51} + z^{52} + z^{53} \\ &=& z^{49}\left(1 + z + z^2+ z^3 + z^4\right) \\ && \boxed{ z^{49}=z \\ 1 + z + z^2 = 0 \\z^4 = z } \\ &=& z\left(0+ z^3 + z\right) \\ &=& z^4+z^2 \quad | \quad z^4 = z \\ &=& z+z^2 \quad | \quad z^2 + z = -1 \\ &=& -1 \\ \hline \end{array}\)

\(\boxed{z^{49} + z^{50} + z^{51} + z^{52} + z^{53} = -1}\)

Help pls

\(\text{In triangle $ABC$, points $D$ and $F$ are on $\overline{AB}$,$\\$ and $E$ is on $\overline{AC}$ such that $\overline{DE}\parallel \overline{BC}$ and $\overline{EF}\parallel \overline{CD}$.$\\$ If $AF = 9$ and $DF = 3$, then what is $BD$?} \)

\(\begin{array}{|lrcll|} \hline (1) & \dfrac{\overline{AC}}{\overline{AE}} &=& \dfrac{9+3}{9} \\\\ (2) & \dfrac{\overline{AC}}{\overline{AE}} &=& \dfrac{9+3+BD}{9+3} \\ \hline \\ (1)=(2): &\dfrac{\overline{AC}}{\overline{AE}}= \dfrac{9+3}{9} &=& \dfrac{9+3+BD}{9+3} \\\\ & \dfrac{9+3}{9} &=& \dfrac{9+3+BD}{9+3} \\\\ & \dfrac{12}{9} &=& \dfrac{12+BD}{12} \\\\ & \dfrac{4}{3} &=& 1+\dfrac{BD}{12} \\\\ & \dfrac{BD}{12}&=& \dfrac{4}{3}-1 \\\\ & \dfrac{BD}{12}&=& \dfrac{1}{3}\\\\ & BD &=& \dfrac{12}{3} \\\\ & \mathbf{BD} &=& \mathbf{4} \\ \hline \end{array}\)

Find the value of:

\(x = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}\) ?

\(\begin{array}{|rcll|} \hline x &=& 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}} \\ x-1 &=& \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}} \\ x-1 &=& \cfrac{1}{2 + \cfrac{1}{2 + \ddots}} \\ && \boxed{x-1= \cfrac{1}{2 + \ddots} } \\ x-1 &=& \cfrac{1}{2 + (x-1) } \\ x-1 &=& \cfrac{1}{x+1} \\ (x-1)(x+1) &=& 1 \\ x^2-1 &=& 1 \\ x^2 &=& 2 \\ \mathbf{x} &=& \mathbf{\sqrt{2}} \\ \hline \end{array}\)