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2.

In triangle ABC, S is a point on side BC such that BS:SC = 1:2,

and T is a point on side AC such that AT:TC = 4:3.
Let U be the intersection of AS and BT.

Find \(\dfrac{AU}{US}\)

\(\begin{array}{|rcll|} \hline \vec{u} &=& \vec{A}-\vec{C} \\ \vec{v} &=& \vec{B}-\vec{C} \\ \hline \vec{U}-\vec{C} &=& \dfrac{3}{7}\vec{u}+ \lambda\left( \vec{v}-\dfrac{3}{7}\vec{u} \right) \quad | \quad \lambda \text{ is a real value} \\ \vec{U}-\vec{C} &=& \dfrac{2}{3}\vec{v}+ \dfrac{US}{AS}\left( \vec{u}-\dfrac{2}{3}\vec{v} \right) \quad | \quad \dfrac{US}{AS} \text{ is a real value} \\ \hline \vec{U}-\vec{C} = \dfrac{3}{7}\vec{u}+ \lambda\left( \vec{v}-\dfrac{3}{7}\vec{u} \right) &=& \dfrac{2}{3}\vec{v}+ \dfrac{US}{AS}\left( \vec{u}-\dfrac{2}{3}\vec{v} \right) \\ \dfrac{3}{7}\vec{u}+ \lambda\left( \vec{v}-\dfrac{3}{7}\vec{u} \right) &=& \dfrac{2}{3}\vec{v}+ \dfrac{US}{AS}\left( \vec{u}-\dfrac{2}{3}\vec{v} \right) \\ \dfrac{3}{7}\vec{u}+ \lambda \vec{v}-\dfrac{3}{7}\lambda\vec{u} &=& \dfrac{2}{3}\vec{v}+ \dfrac{US}{AS}\vec{u}-\dfrac{2}{3}\dfrac{US}{AS}\vec{v} \\ \vec{u}\left(\underbrace{\dfrac{3}{7}-\dfrac{3}{7}\lambda-\dfrac{US}{AS} }_{=0}\right) &=& \vec{v}\left(\underbrace{\dfrac{2}{3}-\dfrac{2}{3}\dfrac{US}{AS}-\lambda }_{=0} \right) \quad | \quad \vec{u} \text{ and } \vec{v}\text{ are independent} \\ \hline \dfrac{3}{7}-\dfrac{3}{7}\lambda-\dfrac{US}{AS} &=& 0 \\ \quad |\quad \cdot \dfrac{7}{3} \\ 1- \lambda-\dfrac{7}{3}\dfrac{US}{AS} &=& 0 \qquad (1) \\\\ \dfrac{2}{3}-\dfrac{2}{3}\dfrac{US}{AS}-\lambda &=& 0 \qquad (2) \\ \hline (1)-(2): \qquad 1- \lambda-\dfrac{7}{3}\dfrac{US}{AS} - \left( \dfrac{2}{3}-\dfrac{2}{3}\dfrac{US}{AS}-\lambda \right) &=& 0 \\ 1- \lambda-\dfrac{7}{3}\dfrac{US}{AS} - \dfrac{2}{3}+\dfrac{2}{3}\dfrac{US}{AS}+\lambda &=& 0 \\ \dfrac{1}{3} -\dfrac{5}{3}\dfrac{US}{AS} &=& 0 \\ \dfrac{5}{3}\dfrac{US}{AS} &=& \dfrac{1}{3} \quad | \quad \cdot \dfrac{3}{5} \\ \mathbf{ \dfrac{US}{AS} } &=& \mathbf{\dfrac{1}{5}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{AU}{AS} &=& \left( 1-\dfrac{US}{AS} \right) \\ \dfrac{AU}{AS} &=& \left( 1-\dfrac{1}{5} \right) \\ \mathbf{ \dfrac{AU}{AS} } &=& \mathbf{\dfrac{4}{5}} \\ \hline \dfrac{AU}{US} &=& \dfrac{ \dfrac{AU}{AS} }{ \dfrac{US}{AS} } \\ \dfrac{AU}{US} &=& \dfrac{ \dfrac{4}{5} }{ \dfrac{1}{5} } \\ \mathbf{ \dfrac{AU}{US} } &=& \mathbf{4} \\ \hline \end{array}\)

1.

In triangle ABC, S is a point on side BC such that BS:SC = 1:2,

and T is a point on side AC such that AT:TC = 4:3.
Let U be the intersection of AS and BT.

We can write \( \vec{T} = w \vec{A} + x \vec{C}\), \(\vec{S} = y \vec{B} + z \vec{C}\) for some real values of w, x, y, and z.
Find w, x, y, and z.

\(\begin{array}{|rcll|} \hline \vec{T} &=& \vec{A}+ \dfrac{4}{7}\left(\vec{C}-\vec{A} \right) \\ \vec{T} &=& \vec{A}+ \dfrac{4}{7}\vec{C} -\dfrac{4}{7}\vec{A} \\ \vec{T} &=& \dfrac{3}{7}\vec{A}+ \dfrac{4}{7}\vec{C} \quad &|\quad \vec{T} = w \vec{A} + x \vec{C}\\ \hline \mathbf{w}&=& \mathbf{\dfrac{3}{7}} \\ \mathbf{x}&=& \mathbf{\dfrac{4}{7}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \vec{S} &=& \vec{C}+ \dfrac{2}{3}\left(\vec{B}-\vec{C} \right) \\ \vec{S} &=& \vec{C}+ \dfrac{2}{3}\vec{B} -\dfrac{2}{3}\vec{C} \\ \vec{S} &=& \dfrac{2}{3}\vec{B}+ \dfrac{1}{3}\vec{C} \quad &|\quad \vec{S} = y \vec{B} + z \vec{C}\\ \hline \mathbf{y}&=& \mathbf{\dfrac{2}{3}} \\ \mathbf{z}&=& \mathbf{\dfrac{1}{3}} \\ \hline \end{array}\)

Find the value of \(k\) for which the binomial \((2x – 1)\) is a factor of \(2x^3 + kx^2 – 2x + 1\)

\(\begin{array}{|rcll|} \hline (2x – 1)&=& 2\left(x – \mathbf{\dfrac12} \right) \\ \hline \end{array}\)

The root is at \(\mathbf{x=\dfrac12}\)

\(\begin{array}{|rcll|} \hline 2x^3 + kx^2 – 2x + 1 \quad &| \quad \mathbf{x=\dfrac12} \\ 2\left(\dfrac12\right)^3 + k\left(\dfrac12\right)^2 – 2\left(\dfrac12\right) + 1 &=& 0 \\ \dfrac{2}{8} + \dfrac{k}{4} – \dfrac{2}{2} + 1 &=& 0 \\ \dfrac{1}{4} + \dfrac{k}{4} – 1 + 1 &=& 0 \\ \dfrac{1}{4} + \dfrac{k}{4} &=& 0 \quad &| \quad \cdot 4\\ 1+k &=& 0 \quad &| \quad -1 \\ \mathbf{k} &=& \mathbf{-1} \\ \hline \end{array}\)

For a certain value of k, the system
\(\begin{align*} x + y + 3z &= 10, \\ -4x + 2y + 5z &= 7, \\ kx + z &= 3 \end{align*}\)
has no solutions. What is this value of \(k\)?

\(\begin{array}{|rcll|} \hline \begin{vmatrix} 1&1&3\\ -4 & 2 & 5\\ k & 0 &1 \\ \end{vmatrix} &=& 0 \\\\ k\begin{vmatrix} 1&3\\ 2 & 5\\ \end{vmatrix} + \begin{vmatrix} 1&1\\ -4 & 2\\ \end{vmatrix} &=& 0 \\\\ k*(5-6)+2+4 &=& 0\\ -k &=& -6 \\ \mathbf{k} &=& \mathbf{6} \\ \hline \end{array}\)

(a)

Compute the sums of the squares of Rows 1-4 of Pascal's Triangle.

Central binomial coefficients: \(\dbinom{2n}{n}\)

Regular octagon \(ABCDEFGH\) has a side length of 2 cm.

Find the area of square \(ACEG\).

\(\text{ Let the side length of the square $=s$ } \\ \text{ Let the area of the square $=s^2$ } \\ \text{ Let $\angle ABC =135^\circ \qquad 135^\circ = 180^\circ-\dfrac{360^\circ}{8}$ } \)

cosine formula:

\(\begin{array}{|rcll|} \hline s^2 &=& 2^2+2^2-2\cdot 2\cdot 2\cdot \cos(135^\circ) \\ s^2 &=& 8-8\cdot \cos(135^\circ) \\ && \boxed{ \cos(135^\circ) \\ =\cos(180^\circ-45^\circ)\\ = -\cos(45^\circ)\\=-\dfrac{\sqrt{2}}{2} } \\ s^2 &=& 8-8\cdot \left( -\dfrac{\sqrt{2}}{2}\right) \\ \mathbf{s^2} &=& \mathbf{8+4\cdot\sqrt{2}} \\ \mathbf{s^2} &=& \mathbf{13.6568542495\ cm^2} \\ \hline \end{array}\)

The area of square \(ACEG\) is \(\mathbf{\approx 13.7\ cm^2}\)

There are values \(A\) and \(B\) such that \(\dfrac{Bx-11}{x^2-7x+10}=\dfrac{A}{x-2}+\dfrac{3}{x-5}\).
Find \(A+B\).

\(x^2-7x+10 = (x-2)(x-5)\)

\(\begin{array}{|lrcll|} \hline & \dfrac{Bx-11}{x^2-7x+10} &=& \dfrac{A}{x-2}+\dfrac{3}{x-5} \\\\ & \dfrac{Bx-11}{(x-2)(x-5)} &=& \dfrac{A}{x-2}+\dfrac{3}{x-5} \quad & | \quad \times (x-2)(x-5) \\\\ & Bx-11 &=& A(x-5) + 3(x-2) \\\\ \hline x=5: & 5B-11 &=& A(5-5) + 3(5-2) \\ & 5B-11 &=& 0 + 3*3 \\ & 5B-11 &=& 9 \\ & 5B &=& 9+11 \\ & 5B &=& 20 \\ & B &=& \dfrac{20}{5} \\ & \mathbf{ B} &=& \mathbf{4} \\ \hline x=2: & 2B-11 &=& A(2-5) + 3(2-2) \\ & 2B-11 &=& -3A + 0 \\ & 3A &=& 11-2B \quad &|\quad B=4 \\ & 3A &=& 11-2\cdot4 \\ & 3A &=& 11-8 \\ & 3A &=& 3 \quad &|\quad :3 \\ & \mathbf{A} &=& \mathbf{1} \\ \hline \end{array}\)

\(\mathbf{A+B} = 1+4 \mathbf{= 5}\)

If \(\sqrt{5 + x} + \sqrt{20 - x} = 7\), what is the value of \((5 + x)(20 - x)\)?

\(\begin{array}{|rcll|} \hline \sqrt{5 + x} + \sqrt{20 - x} &=& 7 \quad &|\quad \text{square both sides} \\ \left(\sqrt{5 + x} + \sqrt{20 - x}\right)^2 &=& 7^2 \\ \left(\sqrt{5 + x}\right)^2 +2\sqrt{5 + x} \sqrt{20 - x} + \left(\sqrt{20 - x}\right)^2 &=& 49 \\ 25 +2\sqrt{(5 + x)(20 - x)} &=& 49 \quad &|\quad -25 \\ 2\sqrt{(5 + x)(20 - x)} &=& 49-25 \\ 2\sqrt{(5 + x)(20 - x)} &=& 24 \quad &|\quad : 2 \\ \sqrt{(5 + x)(20 - x)} &=& 12 \quad &|\quad \text{square both sides} \\ \left( \sqrt{(5 + x)(20 - x)}\right)^2 &=& 12^2 \\ \mathbf{(5 + x)(20 - x)} &=& \mathbf{144} \\ \hline \end{array}\)

Find the remainder when the polynomial \(\large{x^{1000}}\) is divided by the polynomial \((x^2+1)(x+1)\). 

\(\begin{array}{|lrcll|} \hline &\mathbf{ x^{1000} }&=& \mathbf{(x^2+1)(x+1)q(x) +\underbrace{ax^2+bx+c}_{=r(x)}} \\ \hline x=-1:& (-1)^{1000} &=& \left((-1)^2+1\right)(-1+1)q(x) + a(-1)^2+b(-1)+c \\ & 1 &=& 0 + a-b+c \\ &\mathbf{ a-b+c} &=& \mathbf{1} \qquad (1) \\ \hline x=i:& (i)^{1000} &=& (i^2+1)(i+1)q(x) + ai^2+bi+c \quad | \quad \boxed{i^2=-1\\i^{1000}=\left(i^2\right)^{500}=\left(-1\right)^{500}=1} \\ & 1 &=& 0 -a+bi+c \\ &\mathbf{-a+bi+c} &=& \mathbf{1} \qquad (2) \\ \hline x=-i:& (-i)^{1000} &=& \left((-i)^2+1\right)(-i+1)q(x) + a(-i)^2+b(-i)+c \quad | \quad \boxed{(-i)^2=-1\\ (-i)^{1000}=1} \\ & 1 &=& 0 -a-bi+c \\ &\mathbf{-a-bi+c} &=& \mathbf{1} \qquad (3) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline &\mathbf{ a-b+c} &=& \mathbf{1} \qquad (1) \\ &\mathbf{-a+bi+c} &=& \mathbf{1} \qquad (2) \\ &\mathbf{-a-bi+c} &=& \mathbf{1} \qquad (3) \\ \hline (2)+(3): & -a+bi+c+-a-bi+c &=& 1+1 \\ &-2a+2c&=& 2 \quad &| \quad :2 \\ &- a+ c&=& 1 \\ & \mathbf{c} &=& \mathbf{1+a} \qquad (4) \\ \hline (1) : & a-b+c &=& 1 \quad &| \quad \boxed{c = 1+a} \\ & a-b+1+a &=& 1 \\ & 2a-b &=& 0 \\ & \mathbf{b} &=& \mathbf{2a} \qquad (5) \\ \hline (2) : & -a+bi+c &=& 1 \quad &| \quad \boxed{b = 2a,\ c = 1+a}\\ & -a+2ai+1+a &=& 1 \\ & 2ai &=& 0 \quad &| \quad :2i \\ &\mathbf{a} &=& \mathbf{0} \\ & b &=& 2a \\ & b &=& 2*0 \\ &\mathbf{b} &=& \mathbf{0} \\ & c&=& 1+a \\ & c &=& 1+0 \\ &\mathbf{c} &=& \mathbf{1} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline r(x) &=& ax^2+bx+c \\ r(x) &=& 0*x^2+0*x +1 \\ \mathbf{r(x)} &=& \mathbf{1} \\ \hline \end{array}\)

The remainder is 1

In the diagram, \(D\) and \(E\) are the midpoints of \(\overline{AB}\) and \(\overline{BC}\) respectively.

Determine the sum of the \(x\) and \(y\) coordinates of \(F\), the point of intersection of \(\overline{AE}\) and \(\overline{CD}\).

\(\text{Let $\vec{A} =\dbinom06 $} \\ \text{Let $\vec{D} =\dfrac{\vec{A}}{2} = \dbinom03 $} \\ \text{Let $\vec{C} =\dbinom80 $} \\ \text{Let $\vec{E} =\dfrac{\vec{C}}{2} = \dbinom40 $} \\ \text{Let $\vec{F} = \dbinom x y $} \)

\(\begin{array}{|lrcll|} \hline (1) & \vec{F} &=& \vec{E}+\lambda(\vec{A}-\vec{E}) \quad \lambda \text{ is a scalar } \\ (2) & \vec{F} &=& \vec{D}+\mu(\vec{C}-\vec{D}) \quad \mu \text{ is a scalar } \\ \hline &\vec{F} = \vec{E}+\lambda(\vec{A}-\vec{E}) &=& \vec{D}+\mu(\vec{C}-\vec{D}) \\ & \mathbf{ \vec{E}+\lambda(\vec{A}-\vec{E}) } &=& \mathbf{\vec{D}+\mu(\vec{C}-\vec{D})} \\ & \lambda(\vec{A}-\vec{E})-\mu(\vec{C}-\vec{D}) &=& \vec{D}-\vec{E} \\\\ &&& \boxed{ \vec{A}-\vec{E} =\dbinom06-\dbinom40=\dbinom{-4}{6} \\ \vec{C}-\vec{D} =\dbinom80-\dbinom03=\dbinom{8}{-3} \\ \vec{D}-\vec{E} =\dbinom03-\dbinom40=\dbinom{-4}{3} }\\\\ & \lambda\dbinom{-4}{6}-\mu\dbinom{8}{-3} &=& \dbinom{-4}{3} \quad | \quad \times \dbinom{3}{8} \\\\ & \lambda\dbinom{-4}{6}\dbinom{3}{8}-\mu\underbrace{\dbinom{8}{-3}\dbinom{3}{8}}_{=0} &=& \dbinom{-4}{3}\dbinom{3}{8} \\ & \lambda\dbinom{-4}{6}\dbinom{3}{8} &=& \dbinom{-4}{3}\dbinom{3}{8} \\ & \lambda(-12+48) &=& -12+24 \\ & \lambda(36) &=& 12 \quad | \quad : 12 \\ & 3\lambda &=& 1 \\ &\mathbf{ \lambda } &=& \mathbf{ \dfrac{1}{3} } \\ \hline & \vec{F} &=& \vec{E}+\lambda(\vec{A}-\vec{E}) \\ & \vec{F} &=& \dbinom40+\dfrac{1}{3}\dbinom{-4}{6} \\ & \vec{F} &=& \begin{pmatrix} 4-\dfrac{4}{3} \\ \dfrac{6}{3} \end{pmatrix} \\ & \mathbf{\vec{F}} &=& \mathbf{ \begin{pmatrix} \dfrac{8}{3} \\ 2 \end{pmatrix} } \\\\ & x+y &=& \dfrac{8}{3} + 2 \\ &\mathbf{ x+y } &=& \mathbf{\dfrac{14}{3}} \\ \hline \end{array}\)