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As shown in the figure, circle of diameter 4 touches the other two circles and the bottom of the rectangle.

Circle of diameter 3 touches the top at A and left side of the rectangle.

Circle of diameter 6 touches the top at B and two other sides of the rectangle.

Find the length of AB.

\(AB = x+y\)

\(\begin{array}{|rcll|} \hline (1.5+2)^2 &=& x^2+(6-1.5-2)^2 \\ 3.5^2 &=& x^2+2.5^2 \\ x^2 &=& 3.5^2-2.5^2 \\ x^2 &=& 6 \\ x &=& \sqrt{6} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline (2+3)^2 &=& y^2 + (3-2)^2 \\ 5^2 &=& y^2 +1 \\ y^2 &=& 25-1\\ y^2 &=& 24 \\ y^2 &=& 4*6 \\ y &=& 2\sqrt{6} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline AB &=& x+y \\ AB &=& \sqrt{6}+2\sqrt{6} \\ \mathbf{AB} &=& \mathbf{3\sqrt{6}} \\ \hline \end{array}\)

Find the sum 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ... + 301 + 302.  
(The pattern of signs is + + - - + + - - ...)

\(\begin{array}{|rll|} \hline && 1 \\ && \underbrace{+2 - 3}_{-1} \underbrace{ - 4 + 5}_{+1}\\ && \underbrace{ + 6 - 7}_{-1} \underbrace{ - 8 + 9}_{+1} \\ && + \ldots \\ && \underbrace{+298 - 299}_{-1} \underbrace{-300 + 301}_{+1} \\ && + 302 \\ &=& 1+0+302 \\ &=& \mathbf{303} \\ \hline \end{array} \)

For a certain arithmetic progression, the sum of the first 1729 terms is equal to the sum of the first 29 terms. 

Find, with proof, the sum of the first 1758 terms. 

Show all your steps!

Formula arithmetic progeression: \(a_n = a_1 + (n-1)d\)

\(\begin{array}{|rcll|} \hline s_{1729} &=& \left(\dfrac{a_1+a_{1729}}{2}\right)*1729 \quad & | \quad a_{1729} = a_1 + 1728d \\ s_{1729} &=& \left(\dfrac{a_1+a_1 + 1728d}{2}\right)*1729 \\ s_{1729} &=& \left(\dfrac{2a_1+ 1728d}{2}\right)*1729 \quad & | \quad \text{the sum of the first 1729 terms} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline s_{29} &=& \left(\dfrac{a_1+a_{29}}{2}\right)*29 \quad & | \quad a_{29} = a_1 + 28d \\ s_{29} &=& \left(\dfrac{a_1+a_1 + 28d}{2}\right)*29 \\ s_{29} &=& \left(\dfrac{2a_1+ 28d}{2}\right)*29 \quad & | \quad \text{the sum of the first 29 terms} \\ \hline \end{array} \)

\(\mathbf{s_{1729}=s_{29}}\)

\(\begin{array}{|rcll|} \hline \mathbf{s_{1729} } &=& \mathbf{s_{29}} \\\\ \left(\dfrac{2a_1+ 1728d}{2}\right)*1729 &=& \left(\dfrac{2a_1+ 28d}{2}\right)*29 \\ \left( 2a_1+ 1728d \right)*1729 &=& \left( 2a_1+ 28d \right)*29 \\ 2*1729a_1+1728*1729d &=& 2*29a_1+28*29d \\ 2a_1(1729-29) &=& (28*29-1728*1729)d \\ 2a_1*1700 &=& -2986900d \\ 2a_1*17 &=& -29869d \\ \mathbf{ 2a_1 } &=& \mathbf{-1757d} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline s_{1758} &=& \left(\dfrac{a_1+a_{1758}}{2}\right)*1758 \quad & | \quad a_{1758} = a_1 + 1757 \\ s_{1758} &=& \left(\dfrac{a_1+a_1 + 1757d}{2}\right)*1758 \\ s_{1758} &=& \left(\dfrac{2a_1+ 1757d}{2}\right)*1758 \quad & | \quad \text{the sum of the first 1758 terms} \\ && \boxed{\mathbf{ 2a_1 = -1757d}} \\ s_{1758} &=& \left(\dfrac{-1757d+ 1757d}{2}\right)*1758 \\ s_{1758} &=& \left(\dfrac{0}{2}\right)*1758 \\ \mathbf{s_{1758}} &=& \mathbf{0} \\ \hline \end{array}\)

The sum of the first 1758 terms is 0.

The sides of a triangle are 37, 684, and 685.  Find its area.

\(\begin{array}{|rcll|} \hline \sqrt{37^2+684^2} &=& 685 \\ \hline \end{array}\)

This is a right angular triangle!

The area is: \( \dfrac{37*684}{2} = 12654 \)

Compute \(2*\arctan\left(\dfrac{1}{3}\right) + \arctan\left(\dfrac{1}{7}\right)\).

\(\begin{array}{|rcll|} \hline && 2*\arctan\left(\dfrac{1}{3}\right) + \arctan\left(\dfrac{1}{7}\right) \\ &=& \arctan\left(\dfrac{1}{3}\right)+\arctan\left(\dfrac{1}{3}\right) + \arctan\left(\dfrac{1}{7}\right) \\ &=& \arctan\left( \dfrac{1}{3} + \dfrac{1}{3} \above 1pt 1-\dfrac{1}{3}\cdot \dfrac{1}{3} \right)+ \arctan\left(\dfrac{1}{7}\right) \\ &=& \arctan\left( \dfrac{3}{4} \right)+ \arctan\left(\dfrac{1}{7}\right) \\ &=& \arctan\left( \dfrac{3}{4} + \dfrac{1}{7} \above 1pt 1-\dfrac{3}{4}\cdot \dfrac{1}{7} \right)+ \arctan\left(\dfrac{1}{7}\right) \\ &=& \arctan(1) \\ &=& 45^\circ \quad & | \quad \left(\dfrac{\pi}{4}\right) \\ \hline \end{array}\)

If x^2 + 1/x^2 = 171, then what are the possible values of x - 1/x?

\(\begin{array}{|rcll|} \hline \left( x - \dfrac{1}{x} \right)^2 &=& x^2-2x\dfrac{1}{x} + \dfrac{1}{x^2} \\ \left( x - \dfrac{1}{x} \right)^2 &=& x^2+ \dfrac{1}{x^2} -2 \quad &|\quad x^2 + \dfrac{1}{x^2} = 171 \\ \left( x - \dfrac{1}{x} \right)^2 &=&171 -2 \\ \left( x - \dfrac{1}{x} \right)^2 &=&169 \\ x - \dfrac{1}{x} &=& \pm \sqrt{169} \\ \mathbf{ x - \dfrac{1}{x} } &=& \mathbf{\pm 13} \\ \hline \end{array} \)

If a + b + c = 8 and ab + ac + bc = -2, what is a^2 + b^2 + c^2?

\(\begin{array}{|rcll|} \hline (a + b + c)^2 &=& a^2+b^2+c^2+2(ab + ac + bc) \\ 8^2 &=& a^2+b^2+c^2+2(-2) \\ 64 &=& a^2+b^2+c^2-4 \\ \mathbf{68} &=& \mathbf{a^2+b^2+c^2} \\ \hline \end{array} \)

please help
\(a_1, a_2, a_3, \cdots , a_{99}\) is an arithmetic progeression.  
If \(a_2 + a_5 + a_8 + \cdots+ a_{98} = 205\),
then find \(a_1 + a_2 + a_3 + \cdots + a_{99}\).

Formula arithmetic progeression: \(a_n = a_1 + (n-1)d \)

\(\begin{array}{|lrcll|} \hline a_2 =& a_1 + (2-1)*d &=& a_1 + d \\ a_5 =& a_1 + (5-1)*d &=& a_1 + 4d \\ a_8 =& a_1 + (8-1)*d &=& a_1 + 7d \\ a_{11} =& a_1 + (11-1)*d &=& a_1 + 10d \\ \cdots \\ a_{98} =& a_1 + (98-1)*d &=& a_1 + 97d \\ \hline \text{terms:} & 2+3(i-1) = 98 \\ & 3(i-1) = 96 \\ & i-1 = 32 \\ & i = 33 \\ \hline \text{sum}& 205&=& 33*a_1+d(1+4+7+10+\cdots + 97 ) \\ & 205&=& 33*a_1+d\left(\dfrac{1+97}{2}\right)\times 33 \\ &\mathbf{ 205 }&=& \mathbf{ 33*a_1+49\cdot 33d } \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline a_1 =& a_1 &=& a_1 \\ a_2 =& a_1 + (2-1)*d &=& a_1 + d \\ a_3 =& a_1 + (3-1)*d &=& a_1 + 2d \\ a_4 =& a_1 + (4-1)*d &=& a_1 + 3d \\ a_5 =& a_1 + (5-1)*d &=& a_1 + 4d \\ \cdots \\ a_{99} =& a_1 + (99-1)*d &=& a_1 + 98d \\ \hline \text{terms:} & 99 \\ \hline & \text{sum} &=& 99*a_1+d(1+2+3+4+\cdots + 98 ) \\ & \text{sum}&=& 99*a_1+d\left(\dfrac{1+98}{2}\right)\times 98 \\ &\mathbf{ \text{sum} }&=& \mathbf{ 99*a_1+49\cdot 99d } \\ & \text{sum} &=& 3\times \left( \underbrace{33*a_1+49\cdot 33d}_{=205} \right) \\ & \text{sum} &=& 3\times 205 \\ & \mathbf{ \text{sum} } &=& \mathbf{615} \\ \hline \end{array}\)

\(a_1 + a_2 + a_3 + \cdots + a_{99} = 615 \)

Completely factor x^8 - y^8.

\(\begin{array}{|rcll|} \hline \mathbf{x^8 - y^8} &=& \left( x^\frac{8}{2} - y^\frac{8}{2} \right)\left( x^\frac{8}{2} + y^\frac{8}{2} \right) \\ &=& \left( x^4 - y^4 \right)\left( x^4 + y^4 \right) \\ &=& \left( x^\frac{4}{2} - y^\frac{4}{2} \right)\left( x^\frac{4}{2} + y^\frac{4}{2} \right)\left( x^4 + y^4 \right) \\ &=& \left( x^2 - y^2 \right)\left( x^2 + y^2 \right) \left( x^4 + y^4 \right)\\ &=& \left( x^\frac{2}{2} - y^\frac{2}{2} \right)\left( x^\frac{2}{2} + y^\frac{2}{2} \right) \left( x^2 + y^2 \right) \left( x^4 + y^4 \right) \\ \mathbf{x^8 - y^8} &=&\mathbf{ (x - y)(x + y)\left( x^2 + y^2 \right) \left( x^4 + y^4 \right) } \\ \hline \end{array} \)

Solve the system \(x^2 = y + 2,\ y^2 = x + 2\).

\(\begin{array}{|lrcll|} \hline (1) & x^2 &=& y+2 \quad \text{or} \quad \boxed{y = x^2-2} \\\\ (2) & y^2 &=& x+2 \\ & \left(x^2-2\right)^2 &=& x+2 \\ & x^4-4x^2+4 &=& x+2 \\ & x^4-4x^2-x+2 &=& 0 \\\\ \boxed{x_1= 2}: & x^4-4x^2-x+2 &\overset{?}=& 0 \\ & 2^4-4*2^2-2+2 &=& 0 \ \checkmark \\ & \text{Polynomial long division}: & \\ & (x^4-4x^2-x+2) &=& (x-2)(x^3+2x^2-1) \\\\ \boxed{x_2= -1}: & x^3+2x^2-1 &\overset{?}=& 0 \\ & (-1)^3+2(-1)^2-1 &=& 0 \ \checkmark \\ & \text{Polynomial long division}: & \\ & (x^3+2x^2-1) &=& (x+1)(x^2+x-1) \\\\ & x^2+x-1 &=& 0 \\\\ & x &=& \dfrac{-1\pm \sqrt{1-4(-1)}} {2} \\ & x &=& \dfrac{-1\pm \sqrt{5}} {2} \\ \boxed{x_3 = \dfrac{-1+\sqrt{5}} {2}} \\ \boxed{x_4 = \dfrac{-1-\sqrt{5}} {2}} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline && y = x^2-2 \\\\ x_1=1 && y_1 = -1 \\ x_2 =2 && y_2 = 2 \\ x_3 = \dfrac{-1+\sqrt{5}} {2} && y_3 = \dfrac{-1-\sqrt{5}} {2} \\ x_4 = \dfrac{-1-\sqrt{5}} {2} && y_4 = \dfrac{-1+\sqrt{5}} {2} \\ \hline \end{array} \)