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OSL4#23

\(\large{ \begin{array}{r|r|r|r|r|r|r|r|r|r|} \hline ^{1^\nearrow} & ^{6^\nearrow} & ^{11^\nearrow} & \color{red}- & {\color{red}-}^{20^\nearrow}& ^{42^\nearrow}& ^{92^\nearrow}& ^{192^\nearrow}&^{B=372^\nearrow} \\ \hline ^{1^\nearrow} & ^{5^\nearrow} & {\color{red}-}^{5^\nearrow} & \color{red}- & {\color{red}-}^{9^\nearrow}& ^{22^\nearrow}& ^{50^\nearrow}& ^{100^\nearrow}&^{180^\nearrow} \\ \hline ^{1^\nearrow} & ^{4^\nearrow} & \color{red}- & \color{red}-& {\color{red}-}^{4^\nearrow}& ^{13^\nearrow}& ^{28^\nearrow}& ^{50^\nearrow}&^{80^\nearrow} \\ \hline ^{1^\nearrow} & ^{3^\nearrow} & \color{red}- & \color{red}-& {\color{red}-}^{4^\nearrow}& ^{9\nearrow} & ^{15^\nearrow}& ^{22^\nearrow}&^{30^\nearrow} \\ \hline ^{1^\nearrow} & ^{2^\nearrow} & ^{3^\nearrow} & \color{red}-& {\color{red}-}^{4^\nearrow}& ^{5^\nearrow} & ^{6^\nearrow} & ^{7^\nearrow}& ^{8^\nearrow} \\ \hline ^{A^\nearrow} & ^{1^\nearrow} & ^{1^\nearrow} & & ^{1^\nearrow}& ^{1^\nearrow} & ^{1^\nearrow} & ^{1^\nearrow}& ^{1^\nearrow}\\ \end{array} }\)

We have triangle \(\triangle ABC\) where \(AB = AC\) and \(AD\) is an altitude.
Meanwhile, \(E\) is a point on \(AC\) such that \(AB \parallel DE\).
If \(BC = 12\) and the area of  \(\triangle ABC\) is \(180\), what is the area of \(ABDE\)?

\(\text{Let $AB = AC $ } \\ \text{Let $DB = CD =\dfrac{BC}{2} = 6$ } \)

\(\begin{array}{|rcll|} \hline \text{area of } \triangle ABC =180 &=& \dfrac{BC*AD}{2} \\ 180 &=& \dfrac{12*AD}{2} \\ 180 &=& 6AD \\ \mathbf{ AD } &=& \mathbf{30} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \dfrac{CD}{ED} &=& \dfrac{BC}{AB} \\\\ \dfrac{6}{ED} &=& \dfrac{12}{AB} \\\\ \dfrac{ED}{6} &=& \dfrac{AB}{12} \\\\ \mathbf{ED }&=& \mathbf{\dfrac{AB}{2}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{area of } ABDE &=& \dfrac{AB+ED}{2}\times H \quad| \quad H=\dfrac{DB*AD}{AB}=\dfrac{6*30}{AB} \\\\ \text{area of } ABDE &=& \left(\dfrac{AB+ED}{2}\right)\times \dfrac{6*30}{AB} \\\\ \text{area of } ABDE &=& \left(\dfrac{AB+\dfrac{AB}{2}}{2}\right)\times \dfrac{6*30}{AB} \\\\ \text{area of } ABDE &=& \dfrac{3}{4}AB\times \dfrac{6*30}{AB} \\\\ \text{area of } ABDE &=& 3* 3*15 \\\\ \mathbf{\text{area of } ABDE} &=& \mathbf{135} \\ \hline \end{array}\)

Let \(F_1=(10,2)\) and  \(F_2 = (-16,2)\). Then the set of points \(P\) such that \(|PF_1 - PF_2| = 24\) form a hyperbola.
The equation of this hyperbola can be written as \(\dfrac{(x - h)^2}{a^2} - \dfrac{(y - k)^2}{b^2} = 1\).
Find \(h + k + a + b\).

\(\begin{array}{|rcll|} \hline \mathbf{|PF_1 - PF_2|} &=& \mathbf{2a} \\ |PF_1 - PF_2| &=& 24 \\ 2a&=& 24 \\ \mathbf{a} &=& \mathbf{12} \\ \hline \end{array}\)

\(\begin{array}{|lrc|} \hline \mathbf{\text{Focus}_1:}& \mathbf{F_1(h+ae,k)} \\ & F_1(10,2): & h+ae = 10 \quad (1) \\ && \mathbf{ k =2} \\ \hline \mathbf{\text{Focus}_2:}& \mathbf{F_2(h-ae,k)} \\ & F_2(-16,2): & h-ae = -16\quad (2) \\ && \mathbf{ k =2} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline (1)+(2): & h+ae + h-ae &=& 10-16 \\ & 2h &=& -6 \\ & \mathbf{ h }&=& \mathbf{-3} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline (1) & h+ae &=& 10 \quad |\quad h=-3 \\ & -3+ae &=& 10 \\ & \mathbf{ ae }&=& \mathbf{13} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a^2 + b^2 &=& (ae)^2 \\ 12^2+b^2 &=& 13^2 \\ b^2 &=& 13^2-12^2 \\ b^2 &=& 169-144 \\ b^2 &=& 25 \\ \mathbf{ b }&=& \mathbf{5} \\ \hline \end{array}\)

\(\begin{array}{rcll} && \mathbf{h + k + a + b} \\ &=& -3 + 2 +12+5 \\ &=& \mathbf{16} \\ \end{array}\)

The asymptotes of a hyperbola are \(y = 2x - 3\) and \(y = 17 - 2x\).
Also, the hyperbola passes through the point \((4,7)\).
Find the distance between the foci of the hyperbola.

Formula hyperbola:

\(\begin{array}{|l|rcll|} \hline \mathbf{1. \text{ asymptote}} & 2x - 3 &=& y \\ & 2x - 3-y &=& 0 \\\\ \hline \mathbf{2. \text{ asymptote}} & 17 - 2x &=& y \\ & 17 - 2x-y &=& 0 \\\\ \hline \text{hyperbola: } & (2x - 3-y)(17 - 2x-y) + c &=& 0 \\\\ P(x=4,y=7) & (8-3-7)(17-8-7)+c &=& 0 \\ & (-2)(2)+c &=& 0 \\ & c &=& 4 \\ \mathbf{\text{hyperbola: }} &\mathbf{ (2x - 3-y)(17 - 2x-y) + 4} &\mathbf{=}& \mathbf{0} \\\\ & (2x - 3-y)(17 - 2x-y) + 4 &=& 0 \\ & 34x-4x^2-2xy-51+6x+3y-17y+2xy+y^2+4 &=& 0 \\ & -4x^2+40x+y^2-14y &=& 47 \\ &\ldots \\ & \mathbf{ \dfrac{(x-5)^2}{1} - \dfrac{(y-7)^2}{4}} &=& \mathbf{ 1 } \\ & \boxed{\mathbf{\text{Formula hyperbola:} } \dfrac{ (x-h)^2 }{a^2}- \dfrac{ (x-k)^2 }{b^2} = 1 } \\\\ & a^2 = 1 \\ & \mathbf{a= 1} \\\\ & b^2 = 4 \\ & \mathbf{b= 2} \\ \hline \end{array}\)

\(\begin{array}{|l|rcll|} \hline \text{Focus}_1: & F_1(h+ae,k) \\ \text{Focus}_2: & F_2(h-ae,k) \\ \hline \text{Foci distance}: & && h+ae-(h-ae) \\ & &=&h+ae-h+ae \\ & &=& 2ae \quad | \quad e=\dfrac{\sqrt{a^2+b^2}}{a} \\ & &=& 2a\dfrac{\sqrt{a^2+b^2}}{a} \\ & &=&\mathbf{ 2 \sqrt{a^2+b^2}} \quad | \quad a^2=1, \quad b^2= 4 \\ & &=& 2 \sqrt{1+4} \\ & &\mathbf{=}& \mathbf{2\sqrt{5} } \\ \hline \end{array} \)

The distance between the foci of the hyperbola is \(\mathbf{2\sqrt{5} }\)

Let P be the intersection point of the line through points D = (1, 1, 2) and E = (2, 3, 4) 

with the plane through A = (0,1,1), B = (1,1,0) and C = (1,0,3).

What is P?

\(\begin{array}{|rcl|rclrcl|} \hline \text{plane} &&& \text{line} \\ \hline && A = \begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix} \qquad B = \begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix} \qquad C = \begin{pmatrix} 1\\ 0\\ 3 \end{pmatrix} \qquad & && D = \begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix} \qquad E = \begin{pmatrix} 2\\ 3\\ 4 \end{pmatrix} \\ \vec{x} &=& \vec{B}+s(\vec{A}-\vec{B})+t(\vec{C}-\vec{B}) & \vec{x} &=& \vec{D}+r(\vec{E}-\vec{D}) \\\\ \vec{x} &=&\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix} +s\left(\begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix}-\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}\right) +t\left(\begin{pmatrix} 1\\ 0\\ 3 \end{pmatrix}-\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}\right) & \vec{x} &=& \begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix} +r\left(\begin{pmatrix} 2\\ 3\\ 4 \end{pmatrix}-\begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}\right) \\\\ \vec{x} &=&\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}+s\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}+t\begin{pmatrix} 0\\ -1\\ 3 \end{pmatrix} & \vec{x} &=& \begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}+r\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix}\\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \vec{x} =\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}+s\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}+t\begin{pmatrix} 0\\ -1\\ 3 \end{pmatrix} &=& \begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}+r\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix} \\\\ \begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}+s\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}+t\begin{pmatrix} 0\\ -1\\ 3 \end{pmatrix} &=&\begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}+r\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix} \\\\ s\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}+t\begin{pmatrix} 0\\ -1\\ 3 \end{pmatrix} -r\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix} &=& \begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}-\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix} \\\\ \mathbf{ s\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}+t\begin{pmatrix} 0\\ -1\\ 3 \end{pmatrix} -r\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix} } &=& \mathbf{ \begin{pmatrix} 0\\ 0\\ 2 \end{pmatrix} } \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1) & -s-r &=& 0 \\ & \mathbf{r} &=& \mathbf{ -s} \\ \hline (2) & -t-2r &=& 0 \\ & t &=& -2r \\ & t &=& -2(-s) \\ & \mathbf{t} &=& \mathbf{2s} \\ \hline (3) & s+3t-2r &=& 2 \\ & s +3(2s)-2(-s) &=& 2 \\ & s+6s+2s &=& 2 \\ & 9s &=& 2 \\ & \mathbf{s} &=& \mathbf{\dfrac{2}{9}} \\\\ & t &=& 2s \\ & t &=& 2\left(\dfrac{2}{9}\right) \\ &\mathbf{t} &=& \mathbf{\dfrac{4}{9}} \\\\ & r &=& -s \\ & r &=& -\left(\dfrac{2}{9}\right)\\ &\mathbf{r} &=& \mathbf{- \dfrac{2}{9} } \\ \hline \end{array}\)

\(\mathbf{\vec{P}=\ ?}\)

\(\begin{array}{|rcll|} \hline \vec{x} &=& \begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}+r\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix} \\\\ \vec{P} &=& \begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}- \dfrac{2}{9}\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix} \\\\ \vec{P} &=& \begin{pmatrix} 1-\dfrac{2}{9}\\ 1-\dfrac{4}{9}\\ 2-\dfrac{4}{9} \end{pmatrix} \\\\ \mathbf{\vec{P}} &=& \mathbf{\begin{pmatrix} \dfrac{7}{9}\\ \dfrac{5}{9}\\ \dfrac{14}{9} \end{pmatrix} } \\ \hline \end{array} \)

Let A=(10,-10) and O=(0,0).

Determine the sum of all x and y-coordinates of all points Q on the line y=-x+6

such that angle OQA = 90.

\(\begin{array}{|rcll|} \hline -\dbinom{x_Q}{y_Q}\dbinom{x_A-x_Q}{y_A-y_Q} &=& 0 \\\\ \dbinom{x_Q}{y_Q}\dbinom{x_A-x_Q}{y_A-y_Q} &=& 0 \quad | \quad y_Q = 6-x_Q \\\\ \dbinom{x_Q}{6-x_Q}\dbinom{x_A-x_Q}{y_A-(6-x_Q)} &=& 0 \quad | \quad x_A = 10,\ y_A=-10 \\\\ \dbinom{x_Q}{6-x_Q}\dbinom{10-x_Q}{-10-(6-x_Q)} &=& 0 \\\\ \dbinom{x_Q}{6-x_Q}\dbinom{10-x_Q}{-10-6+x_Q} &=& 0 \\\\ \dbinom{x_Q}{6-x_Q}\dbinom{10-x_Q}{-16+x_Q} &=& 0 \\\\ x_Q(10-x_Q) + (6-x_Q)(-16+x_Q) &=& 0 \\ 10x_Q-x_Q^2 -96+6x_Q+16x_Q-x_Q^2 &=& 0 \\ -2x_Q^2 + 32x_Q-96 &=& 0 \quad | \quad : (-2) \\ \mathbf{x_Q^2 - 16x_Q + 48} &=& \mathbf{0} \\ x_Q &=& \dfrac{16\pm \sqrt{16^2-4\cdot 48} }{2} \\ x_Q &=& \dfrac{16\pm \sqrt{64} }{2} \\ x_Q &=& \dfrac{16\pm 8 }{2} \\ x_Q &=& 8 \pm 4 \\\\ x_Q &=& 8+4 \\ \mathbf{x_Q} &=& \mathbf{12} \\ y_Q &=& 6-x_Q \\ y_Q &=& 6-12 \\ \mathbf{y_Q} &=& \mathbf{-6} \\\\ x_Q &=& 8-4 \\ \mathbf{x_Q} &=& \mathbf{4} \\ y_Q &=& 6-x_Q \\ y_Q &=& 6-4 \\ \mathbf{y_Q} &=& \mathbf{2} \\ \hline \end{array}\)

The sum of all x and y-coordinates of all points Q
\(12-6+4+2 = \mathbf{12}\)

math help

Consider parallelogram ABCD with points S and T chosen such that CS:SD = BT:TC = 2:1, as in the picture.
Let\( \vec{AB} = v\) and \(\vec{AD} = w\).
Then there exist constants r, s, t, u such that
\(\vec{AT} = r v + s w\),
\(\vec{BS} = t v + u w\).

2.

Also, what is\( \mathbf{\dfrac{AT^2+BS^2}{AC^2+BD^2}}\) equal to?

\(\begin{array}{|rcll|} \hline \left(\vec{AT}\right)^2 &=& \left(v +\dfrac{2}{3}w \right)^2 \\ \mathbf{\left(\vec{AT}\right)^2} &=& \mathbf{v^2+2\cdot \dfrac{2}{3}vw +\dfrac{4}{9}w^2 } \\\\ \left(\vec{BS}\right)^2 &=& \left(w -\dfrac{2}{3}v \right)^2 \\ \mathbf{\left(\vec{BS}\right)^2} &=& \mathbf{w^2 - 2\cdot \dfrac{2}{3}vw +\dfrac{4}{9}v^2 } \\\\ \left(\vec{AC}\right)^2 &=& \left(v+w \right)^2 \quad | \quad \vec{AC}=v+w \\ \mathbf{\left(\vec{AC}\right)^2} &=& \mathbf{v^2 + 2vw + w^2 } \\\\ \left(\vec{BD}\right)^2 &=& \left(w-v \right)^2 \quad | \quad \vec{BD}=w-v \\ \mathbf{\left(\vec{BD}\right)^2} &=& \mathbf{w^2 - 2vw + v^2 } \\ \hline \dfrac{AT^2+BS^2}{AC^2+BD^2} &=& \dfrac{v^2+2\cdot \dfrac{2}{3}vw +\dfrac{4}{9}w^2+w^2 - 2\cdot \dfrac{2}{3}vw +\dfrac{4}{9}v^2} {v^2 + 2vw + w^2+w^2 - 2vw + v^2} \\\\ \dfrac{AT^2+BS^2}{AC^2+BD^2} &=& \dfrac{v^2+\dfrac{4}{3}vw +\dfrac{4}{9}w^2+w^2 - \dfrac{4}{3}vw +\dfrac{4}{9}v^2} {v^2 + w^2+w^2 + v^2} \\\\ \dfrac{AT^2+BS^2}{AC^2+BD^2} &=& \dfrac{v^2 +\dfrac{4}{9}w^2+w^2 +\dfrac{4}{9}v^2} {v^2 + w^2+w^2 + v^2} \\\\ \dfrac{AT^2+BS^2}{AC^2+BD^2} &=& \dfrac{v^2 +\dfrac{4}{9}v^2+w^2 +\dfrac{4}{9}w^2} {2(v^2 + w^2)} \\\\ \dfrac{AT^2+BS^2}{AC^2+BD^2} &=& \dfrac{\dfrac{13}{9}(v^2+w^2)} {2(v^2 + w^2)} \\\\ \mathbf{\dfrac{AT^2+BS^2}{AC^2+BD^2}} &=& \mathbf{\dfrac{13}{18}} \\ \hline \end{array}\)

math help

Consider parallelogram ABCD with points S and T chosen such that \(CS:SD = BT:TC = 2:1\), as in the picture.
Let \( \vec{AB} = v\) and \(\vec{AD} = w\).

1.
Then there exist constants r, s, t, u such that
\(\vec{AT} = r v + s w\),
\(\vec{BS} = t v + u w\).

\(\text{Let $\vec{AB} = \vec{DC}$ } \\ \text{Let $\vec{AD} = \vec{BC}$ }\)

\(\begin{array}{|rcll|} \hline \vec{AT} &=& v +\dfrac{2}{3}w \quad | \quad \vec{AT} = r v + s w \\ \mathbf{r} &=& \mathbf{1} \\ \mathbf{s} &=& \mathbf{\dfrac{2}{3}} \\\\ \vec{BS} &=& w -\dfrac{2}{3}v \quad | \quad \vec{BS} = t v + u w \\ \mathbf{t} &=& \mathbf{-\dfrac{2}{3}} \\ \mathbf{u} &=& \mathbf{1} \\ \hline \end{array}\)

3.

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 also write \(\vec{U} = x \vec{A} + y \vec{B} + z \vec{C}\)
for some real values of x, y, z. Find x, y, z.

\(\begin{array}{|rcll|} \hline \vec{U}-\vec{C} &=& \dfrac{2}{3}\vec{v}+ \dfrac{US}{AS}\left( \vec{u}-\dfrac{2}{3}\vec{v} \right) \quad | \quad \vec{u} = \vec{A}-\vec{C},\ \vec{v} = \vec{B}-\vec{C} \\ \vec{U}-\vec{C} &=& \dfrac{2}{3}\left(\vec{B}-\vec{C}\right)+ \dfrac{US}{AS}\left( \vec{A}-\vec{C}-\dfrac{2}{3}\left(\vec{B}-\vec{C}\right)\right) \\ \vec{U}-\vec{C} &=& \dfrac{2}{3}\vec{B}-\dfrac{2}{3}\vec{C} + \dfrac{US}{AS}\left( \vec{A}-\vec{C}-\dfrac{2}{3}\vec{B}+\dfrac{2}{3}\vec{C}\right) \\ \vec{U}-\vec{C} &=& \dfrac{2}{3}\vec{B}-\dfrac{2}{3}\vec{C} + \dfrac{US}{AS}\left( \vec{A}-\dfrac{2}{3}\vec{B}-\dfrac{1}{3}\vec{C}\right) \quad | \quad \dfrac{US}{AS}=\dfrac{1}{5} \\ \vec{U}-\vec{C} &=& \dfrac{2}{3}\vec{B}-\dfrac{2}{3}\vec{C} + \dfrac{1}{5}\left( \vec{A}-\dfrac{2}{3}\vec{B}-\dfrac{1}{3}\vec{C}\right) \\ \vec{U}-\vec{C} &=& \dfrac{2}{3}\vec{B}-\dfrac{2}{3}\vec{C} + \dfrac{1}{5}\vec{A}-\dfrac{2}{15}\vec{B}-\dfrac{1}{15}\vec{C} \\ \vec{U} &=& \vec{C}+\dfrac{2}{3}\vec{B}-\dfrac{2}{3}\vec{C} + \dfrac{1}{5}\vec{A}-\dfrac{2}{15}\vec{B}-\dfrac{1}{15}\vec{C} \\ \vec{U} &=& \dfrac{1}{5}\vec{A} +\dfrac{8}{15}\vec{B}+\dfrac{4}{15}\vec{C} \quad |\quad \vec{U} = x \vec{A} + y \vec{B} + z \vec{C} \\ \hline \mathbf{x}&=& \mathbf{\dfrac{1}{5}} \\ \mathbf{y}&=& \mathbf{\dfrac{8}{15}} \\ \mathbf{z}&=& \mathbf{\dfrac{4}{15}} \\ \hline \end{array}\)