heureka

1: 

The line 

\(y = (3x + 7)/4 \)

y = (3x + 7)/4

intersects the circle

\(x^2 + y^2 = 25 \)

x^2 + y^2 = 25

at A and B.

Find the length of chord .

\(\begin{array}{|rcll|} \hline x^2+y^2 &=& 25 \quad & | \quad y = \dfrac{3x+7}{4} \\ x^2+ \left(\dfrac{3x+7}{4}\right)^2 &=& 25 \\\\ x^2+ \dfrac{(3x+7)^2}{16} &=& 25 \quad & | \quad \cdot 16 \\\\ 16x^2+ (3x+7)^2 &=& 25\cdot 16 \\ 16x^2+ (3x+7)^2 &=& 400 \\ 16x^2+ 9x^2+42x + 49 &=& 400 \\ 25x^2+42x + 49-400 &=& 0 \\ 25x^2+42x - 351 &=& 0 \\ x &=& \dfrac{-42\pm \sqrt{42^2-4\cdot 25\cdot(-351) } } {2\cdot 25} \\\\ x &=& \dfrac{-42\pm \sqrt{1764+100\cdot(351) } } {50} \\\\ x &=& \dfrac{-42\pm \sqrt{36864} } {50} \\\\ x &=& \dfrac{-42\pm 192} {50} \\\\ x_a & =& \dfrac{-42 + 192} {50} \\ \mathbf{x_a} &\mathbf{=}& \mathbf{3} \\ y_a &=& \dfrac{3\cdot 3+7}{4} \\ \mathbf{y_a} & \mathbf{=}& \mathbf{ 4 } \\\\ \mathbf{x_b} & \mathbf{=}& \mathbf{\dfrac{-42 - 192} {50} } \\ x_b &=& -4.68 \\ y_b &=& \dfrac{3\cdot (-4.68)+7}{4} \\ \mathbf{y_b} & \mathbf{=}& \mathbf{-1.76} \\ \hline \end{array}\)

The length of chord AB:

\(\begin{array}{|rcll|} \hline && \sqrt{(x_a-x_b)^2+(y_a-y_b)^2} \\ &=& \sqrt{[3-(-4.68)]^2+[4-(-1.76)]^2} \\ &=& \sqrt{(3+4.68)^2+(4+1.76)^2} \\ &=& \sqrt{7.68^2+5.76^2} \\ &=& \sqrt{92.16} \\ &=& 9.6 \\ \hline \end{array}\)

The length of chord AB is 9.6

2:  Points A(0,0), B(9,6) and C(6,12) are vertices of triangle ABC.
Point D is on segment AB such that 2(AD) = DB,
point E is on segment BC such that 2(BE) = EC and
point F is on segment CA such that 2(CF) = FA.
What is the ratio of the area of triangle DEF to the area of triangle ABC?
Express your answer as a common fraction.

\(\begin{array}{|rcll|} \hline D &=& \dbinom{9}{6}\cdot \dfrac13 \\ &=& \dbinom{3}{2} \\\\ E &=& \dbinom{9}{6} + \left( \dbinom{6}{12} - \dbinom{9}{6} \right)\cdot \dfrac13 \\ &=& \dbinom{9}{6} + \dbinom{-3}{6}\cdot \dfrac13 \\ &=& \dbinom{9}{6} + \dbinom{-1}{2} \\ &=& \dbinom{8}{8} \\\\ F &=& \dbinom{6}{12}\cdot \dfrac23 \\ &=& \dbinom{12}{24}\cdot \dfrac13 \\ &=& \dbinom{4}{8} \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \text{Area}_\text{ABC} &=& \frac12 \left( 0\cdot 6 - 9\cdot 0 + 9 \cdot 12 - 6 \cdot 6 + 6\cdot 0 - 0 \cdot 12 \right) \\ &=& \frac12 \left( 9 \cdot 12 - 6 \cdot 6 \right) \\ &=& \frac12 \left( 72 \right) \\ &=& 36 \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \text{Area}_\text{DEF} &=& \frac12 \left( 3\cdot 8 - 8 \cdot 2 + 8 \cdot 8 - 4 \cdot 8 + 4\cdot 2 - 3 \cdot 8 \right) \\ &=& \frac12 \left( 24-16+64-32+8-24 \right) \\ &=& \frac12 \left( 24 \right) \\ &=& 12 \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \dfrac { \text{Area}_\text{DEF} } {\text{Area}_\text{ABC}} &=& \dfrac{12}{36} = \dfrac13 \\ \hline \end{array}\)

1:  A line with slope 6 bisects the area of a unit square with vertices (0,0), (1,0), (1,1), and (0,1).
What is the y-intercept of this line?

\(\begin{array}{|lrcll|} \hline & y &=& mx + b \quad & | \quad m = 6 \\ & y &=& 6x + b \\\\ P_1 = (a,0): & 0 &=& 6a+b \\ & -6a &=& b \\\\ P_2 = (1-a,1): & 1 &=& 6(1-a)+b \\ & 1 &=& 6-6a + b \quad & | \quad -6a = b \\ & 1 &=& 6 +b + b \\ & 1 &=& 6 +2b \\ & -5 &=& 2b \\ & 2b &=& -5 \\\\ & b &=& -\dfrac{5}{2} \\ & b &=& -2.5 \\ \hline \end{array}\)

The y-intercept of this line is -2.5

4:  The lines

\(y = \frac{5}{12} x \)

and 

\(y = \frac{4}{3} x \)

are drawn in the coordinate plane.

Find the slope of the line that bisects the angle between these lines.

\(\text{Let $\tan(\varphi_1) = \frac{5}{12}$ } \\ \text{Let $\tan(\varphi_2) = \frac{4}{3}$ }\)

\(\begin{array}{|rcll|} \hline \text{slope}_{bisect} &=&\tan \left( \dfrac{\arctan(\varphi_1)+\arctan(\varphi_2) } {2} \right) \\\\ &=&\tan \left( \dfrac{22.6198649480^{\circ}+53.1301023542^{\circ} } {2} \right) \\\\ &=&\tan \left( \dfrac{75.7499673022^{\circ} } {2} \right) \\\\ &=&\tan (37.8749836511^{\circ} ) \\\\ &=& 0.77777777778 \\\\ &=& \dfrac{7}{9} \\ \hline \end{array}\)

3:  Find the largest real number x for which there exists a real number y such that x^2 + y^2 = 2x + 2y.
 

\(\begin{array}{|rcll|} \hline x^2+y^2 &=& 2x+2y \\ x^2-2x + y^2-2y &=& 0 \\ (x-1)^2 -1 + (y-1)^2 -1 &=& 0 \\ (x-1)^2 + (y-1)^2 &=& 2 \\\\ \text{circle}_{\text{center}} &=& (1,1) \\ \text{circle}_{\text{radius}} &=& \sqrt{2} \\ x_{\text{max}} &=& x_{\text{center}} + \text{ radius} \\ x_{\text{max}} &=& 1 + \sqrt{2} \\ x_{\text{max}} &=& 2.41421356237 \\ \hline \end{array}\)

4)
Find AC.
Picture: https://latex.artofproblemsolving.com/9/1/d/91d520533f0fb8cb325932013b2aebf0db9bde54.png

cos-rule

\(\begin{array}{|rcll|} \hline \mathbf{(7\sqrt{3})^2 }& \mathbf{=}& \mathbf{ 9^2 + AC^2 -2\cdot 9 \cdot AC \cdot \cos(150^{\circ}) } \\ 49\cdot 3 &=& 81 + AC^2 - 18 \cdot AC \cdot \cos(150^{\circ}) \quad & | \quad \small{\cos(150^{\circ}) = -\dfrac{\sqrt{3}}{2}} \\ 49\cdot 3 &=& 81 + AC^2 + 18 \cdot AC \cdot \dfrac{\sqrt{3}}{2} \\ 147 &=& 81 + AC^2 + 9 \cdot \sqrt{3} \cdot AC \quad & | \quad - 81 \\ 66 &=& AC^2 + 9 \sqrt{3} \cdot AC \\ AC^2 + 9 \sqrt{3} \cdot AC - 66 &=& 0 \\\\ AC &=& \dfrac{-9 \sqrt{3} \pm \sqrt{ (9\sqrt{3})^2 - 4\cdot (-66) } } {2} \\\\ AC &=& \dfrac{-9 \sqrt{3} \pm \sqrt{ (81\cdot 3 + 264 } } {2} \\\\ AC &=& \dfrac{-9 \sqrt{3} \pm \sqrt{507} } {2} \\\\ AC &=& \dfrac{-9 \sqrt{3} \pm \sqrt{13^2\cdot 3} } {2} \\\\ AC &=& \dfrac{-9 \sqrt{3} \pm 13\sqrt{3} } {2} \\\\ AC &=& \dfrac{-9 \sqrt{3} + 13\sqrt{3} } {2} \quad & | \quad \text{AC}> 0\ !\\\\ AC &=& \dfrac{4\sqrt{3} } {2} \\\\ \mathbf{AC} & \mathbf{=} & \mathbf{ 2\sqrt{3} } \\ \hline \end{array}\)

8)
Find DB in the diagram below.
Picture: https://latex.artofproblemsolving.com/f/6/f/f6f917a5c86b363a5cdf8871300d061dcd54fc0b.png

\(\begin{array}{|rcll|} \hline \angle C &=& 180^{\circ} - ( 30^{\circ} + 45^{\circ}) \\ \angle C &=& 105^{\circ} \\\\ \angle ACD &=& \angle C - 60^{\circ} \\ \angle ACD &=& 105^{\circ} - 60^{\circ} \\ \angle ACD &=& 45^{\circ} \\\\ \dfrac{DC} {\sin(30^{\circ})} &=& \dfrac{2}{\sin(ACD)} \\\\ \dfrac{DC} {\sin(30^{\circ})} &=& \dfrac{2}{\sin(45^{\circ}) } \\\\ DC &=& \dfrac{2\cdot \sin(30^{\circ}) }{\sin(45^{\circ}) } \quad & | \quad \sin(30^{\circ}) = \dfrac{1}{2} \\\\ DC &=& \dfrac{2\cdot \dfrac{1}{2} }{\sin(45^{\circ}) } \\\\ DC &=& \dfrac{1}{\sin(45^{\circ}) } \quad & | \quad \sin(45^{\circ}) = \dfrac{\sqrt{2}}{2} \\\\ DC &=& \dfrac{1}{\dfrac{\sqrt{2}}{2} } \\\\ DC &=& \dfrac{2}{ \sqrt{2} }\cdot \dfrac{\sqrt{2}}{\sqrt{2}} \\\\ DC &=& \dfrac{2\sqrt{2}}{2 } \\\\ \mathbf{DC} & \mathbf{=} & \mathbf{ \sqrt{2} } \\\\ \dfrac{DB} {\sin(60^{\circ})} &=& \dfrac{\sqrt{2}}{\sin(45^{\circ})} \\\\ DB &=& \dfrac{\sqrt{2}\cdot \sin(60^{\circ})}{\sin(45^{\circ})} \quad & | \quad \sin(45^{\circ}) = \dfrac{\sqrt{2}}{2} \\\\ DB &=& \dfrac{\sqrt{2}\cdot \sin(60^{\circ})}{\dfrac{\sqrt{2}}{2}} \\\\ DB &=& 2\cdot \sin(60^{\circ}) \quad & | \quad \sin(60^{\circ}) = \dfrac{\sqrt{3}}{2} \\\\ DB &=& 2\cdot \dfrac{\sqrt{3}}{2} \\\\ \mathbf{DB} & \mathbf{=} & \mathbf{ \sqrt{3} } \\ \hline \end{array}\)