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Given triangle ABC,

sinC=(sinA+sinB)/(cosA+cosB),

What is the shape of triangle ABC? 

\(\begin{array}{|rcll|} \hline \sin(C) &=& \dfrac{\sin(A)+\sin(B)}{\cos(A)+\cos(B)} \\ && \boxed{\sin(A)+\sin(B) =2\cdot \sin\left(\dfrac{A+B}{2}\right)\cdot \cos\left(\dfrac{A-B}{2}\right)} \\ && \boxed{\cos(A)+\cos(B) =2\cdot \cos\left(\dfrac{A+B}{2}\right)\cdot \cos\left(\dfrac{A-B}{2}\right)} \\ \sin(C) &=& \dfrac{2\cdot \sin\left(\dfrac{A+B}{2}\right)\cdot \cos\left(\dfrac{A-B}{2}\right)} {2\cdot \cos\left(\dfrac{A+B}{2}\right)\cdot \cos\left(\dfrac{A-B}{2}\right)} \\ \sin(C) &=& \dfrac{ \sin\left(\dfrac{A+B}{2}\right) } { \cos\left(\dfrac{A+B}{2}\right) } \\ && \boxed{\sin(C)=\sin\Big(180^\circ-(A+B)\Big)=\sin(A+B)} \\ \sin(A+B) &=& \dfrac{ \sin\left(\dfrac{A+B}{2}\right) } { \cos\left(\dfrac{A+B}{2}\right) } \\ && \boxed{\sin(A+B) =2\cdot \sin\left(\dfrac{A+B}{2}\right)\cdot \cos\left(\dfrac{A+B}{2}\right) } \\ 2\cdot \sin\left(\dfrac{A+B}{2}\right)\cdot \cos\left(\dfrac{A+B}{2}\right) &=& \dfrac{ \sin\left(\dfrac{A+B}{2}\right) } { \cos\left(\dfrac{A+B}{2}\right) } \\ 2\cdot \cos\left(\dfrac{A+B}{2}\right) &=& \dfrac{ 1 } { \cos\left(\dfrac{A+B}{2}\right) } \\ \cos^2\left(\dfrac{A+B}{2}\right) &=& \dfrac{ 1 }{2} \\ \cos\left(\dfrac{A+B}{2}\right) &=& \dfrac{ 1 }{\sqrt{2}} \\ \cos\left(\dfrac{A+B}{2}\right) &=& \dfrac{ \sqrt{2} }{2} \\ \dfrac{A+B}{2} &=& \arccos\left( \dfrac{ \sqrt{2} }{2} \right) \\ \dfrac{A+B}{2} &=& 45^\circ \\ \mathbf{ A+B } & \mathbf{=} & \mathbf{ 90^\circ} \quad \text{ so } C=90^\circ \\ \hline \end{array} \)

The shape is a right-angled triangle

Triangle ABC has a right angle at C. sinA sinB and sinC forms a geometric sequence.

Please determine the value of sinA. 

\(\text{Let $r=$ ratio } \)

\(\begin{array}{lcll} C=90^\circ \\ \sin(C) = \sin(90^\circ) = 1 \\ \end{array} \)

\(\begin{array}{lcll} B=90^\circ-A \\ \sin(B) = \sin(90^\circ-A) = \cos(A) \\ \end{array} \)

Geometric sequence:

\(\begin{array}{|rcll|} \hline a_1 &=& a \\ &=& \sin(A) \\\\ a_2 &=&a\cdot r \quad &| \quad a=\sin(A) \\ &=& \sin(A)\cdot r \quad &| \quad a_2 = \sin(B)=\cos(A)\\ \mathbf{\cos(A)} & \mathbf{=}& \mathbf{\sin(A)\cdot r} \qquad (1) \\\\ a_3 &=& a\cdot r^2 \quad &| \quad a=\sin(A) \\ &=& \sin(A)\cdot r^2 \quad &| \quad a_3 =\sin(C)=1 \\ \mathbf{1} &\mathbf{=}& \mathbf{\sin(A)\cdot r^2} \qquad (2) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \cos(A) &=&\sin(A)\cdot r & (1) \\ r &=& \dfrac{\cos(A)}{\sin(A)} \\ \hline 1 &=& \sin(A)\cdot r^2 & (2) \quad | \quad r = \dfrac{\cos(A)}{\sin(A)} \\ 1 &=& \sin(A)\cdot \left(\dfrac{\cos(A)}{\sin(A)}\right)^2 \\ 1 &=& \sin(A)\cdot \dfrac{\cos^2(A)}{\sin^2(A)} \\ 1 &=& \dfrac{\cos^2(A)}{\sin(A)} \\ \sin(A) &=& \cos^2(A) & \quad | \quad \cos^2(A)=1-\sin^2(A) \\ \sin(A) &=& 1-\sin^2(A) \\ \sin^2(A)+\sin(A) -1 &=& 0 \\\\ \sin(A) &=& \dfrac{-1\pm \sqrt{1-4\cdot(-1)} } {2} \\ \sin(A) &=& \dfrac{-1\pm \sqrt{5} } {2} \\ \sin(A) &=& \dfrac{-1{\color{red}+}\sqrt{5} } {2} \\ \mathbf{\sin(A) } &\mathbf{=}& \mathbf{\dfrac{-1 + \sqrt{5} } {2} } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline r^2 &=& \dfrac{1}{\sin(A)} \\ r &=& \dfrac{1}{\sqrt{\sin(A)}} \\ \mathbf{r} &\mathbf{=}& \mathbf{ \dfrac{1}{\sqrt{\dfrac{-1 + \sqrt{5} } {2}}} }\\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline a_1 = \sin(A) &=& \dfrac{-1 + \sqrt{5} } {2} \\ a_2 = \sin(B) &=& \sqrt{\dfrac{-1 + \sqrt{5} } {2}} \\ a_3 = \sin(C) &=& 1 \\ \hline \end{array} \)

The value of \(\sin(A)\) is  \(\mathbf{\dfrac{-1 + \sqrt{5} } {2} } \) 

1.
Let z be a complex number such that \(|z - 12| + |z - 5i| = 13\).
Find the smallest possible value of \(|z|\).

\(\text{Let $z=a+bi$ } \\ \text{Let $|z|=\sqrt{a^2+b^2}$ } \)

\(\begin{array}{|rcll|} \hline z - 12 &=& a+bi - 12 \\ &=&(a-12) +bi \\ |z-12| &=& \sqrt{(a-12)^2+b^2 } \\ \hline z - 5i &=& a+bi - 5i \\ &=& a + (b-5)i \\ |z - 5i| &=& \sqrt{a^2+(b-5)^2} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{|z - 12| + |z - 5i|} & \mathbf{=}& \mathbf{13} \\\\ \sqrt{(a-12)^2+b^2 } + \sqrt{a^2+(b-5)^2} &=& 13 \\ \sqrt{(a-12)^2+b^2 } &=& 13- \sqrt{a^2+(b-5)^2} \quad | \quad \text{square both sides} \\ (a-12)^2+b^2 &=& \left( 13- \sqrt{a^2+(b-5)^2} \right)^2 \\ (a-12)^2+b^2 &=& 13^2-26\sqrt{a^2+(b-5)^2}+ a^2+(b-5)^2 \\ 26\sqrt{a^2+(b-5)^2} &=& 13^2+ a^2+(b-5)^2-(a-12)^2-b^2 \\ 26\sqrt{a^2+(b-5)^2} &=& 13^2+ a^2+b^2-10b+25-b^2-a^2+24a-12^2 \\ 26\sqrt{a^2+(b-5)^2} &=& 13^2+ -10b+25+24a-12^2 \quad | \quad 13^2-12^2 = 5^2=25 \\ 26\sqrt{a^2+(b-5)^2} &=& 50 -10b +24a \quad | \quad : 2 \\ 13\sqrt{a^2+(b-5)^2} &=& 25 -5b +12a \quad | \quad \text{square both sides} \\ 13^2\left(a^2+(b-5)^2\right) &=& (25-5b+12a)^2 \\ 13^2\left(a^2+(b-5)^2\right) &=& 25^2+25b^2+12^2a^2-250b+24\cdot 25a-120ab \\ 13^2a^2+13^2(b-5)^2 &=& 25^2+25b^2+12^2a^2-250b+24\cdot 25a-120ab \\ 13^2a^2+13^2(b^2-10b+25) &=& 25^2+25b^2+12^2a^2-250b+24\cdot 25a-120ab \\ 13^2a^2+13^2b^2-13^2\cdot 10b+13^2\cdot 25 &=& 25^2+25b^2+12^2a^2-250b+24\cdot 25a-120ab \\ (13^2-12^2)a^2+(13^2-5^2)b^2+13^2\cdot 25 &=& (13^2\cdot 10-250)b+600a-120ab+25^2-13^2\cdot 25 \\ 5^2a^2+12^2b^2&=& 1440b+600a-120ab-3600 \\ (5^2a^2+120ab+12^2b^2) &=& 120(5a+12b)-3600 \\ (5a+12b)^2 &=& 120(5a+12b)-3600 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline (5a+12b)^2 &=& 120(5a+12b)-3600 \\ (5a+12b)^2 -120(5a+12b)+3600 &=& 0 \\\\ 5a+12b &=& \dfrac{120\pm \sqrt{120^2-4\cdot 3600}}{2} \\ 5a+12b &=& \dfrac{120\pm 0}{2} \\ 5a+12b &=& \dfrac{120}{2} \\ \mathbf{ 5a+12b } &\mathbf{=}& \mathbf{60} \quad \text{this is a line!!! }\\ \text{or}\quad b&=&5-\dfrac{5}{12}a \\ \hline \end{array} \)

The normal vector of this line \(5a+12b-60=0\), perpendicular to the line is \(\vec{n}=\dbinom{5}{12}\)

\(\begin{array}{|rcll|} \hline \vec{n_0} &=& \dfrac{1}{\sqrt{5^2+12^2}} \dbinom{5}{12} \\\\ &=& \dfrac{1}{\sqrt{13^2}} \dbinom{5}{12} \\\\ &=& \dfrac{1}{13} \dbinom{5}{12} \\\\ &=& \begin{pmatrix} \dfrac{5}{13}\\ \dfrac{12}{13} \end{pmatrix} \\\\ |z|_{\text{min}} &=& \dbinom{0}{5}\cdot \begin{pmatrix} \dfrac{5}{13}\\ \dfrac{12}{13} \end{pmatrix} \\ &=& 0\cdot\dfrac{5}{13}+5\cdot \dfrac{12}{13} \\\\ & = & \dfrac{5\cdot 12}{13} \\\\ & = & \dfrac{60}{13} \\\\ \mathbf{ |z|_{\text{min}} } & \mathbf{=} & \mathbf{4.61538461538} \\ \hline \end{array}\)

The smallest possible value of |z| is 4.61538461538

Solve:

\(|2x - 1| = |3x + 5|\).

Write your answers as a list of numbers, separated by commas,
e.g. '23, 19' (but without the quotes).

\(\begin{array}{|rcll|} \hline |2x - 1| &=& |3x + 5| \\ (2x - 1)^2 &=& (3x + 5)^2 \\ 4x^2-4x+1&=& 9x^2+30x+25 \\ \ldots \\ 5x^2 +34x+24 &=& 0 \\\\ x &=& \dfrac{-34\pm \sqrt{34^2-4\cdot 5 \cdot 24} }{2\cdot 5} \\ &=& \dfrac{-34\pm \sqrt{34^2-480} }{10} \\ &=& \dfrac{-34\pm \sqrt{676} }{10} \\ &=& \dfrac{-34\pm 26 }{10} \\\\ x &=& \dfrac{-34+ 26 }{10} \\ \mathbf{x} & \mathbf{=} & \mathbf{-0.8}\\\\ x &=& \dfrac{-34- 26 }{10} \\ \mathbf{x} & \mathbf{=} & \mathbf{-6}\\ \hline \end{array} \)

\(\mathbf{"-0.8, -6"}\)

For a certain value of k , the system

x + y + 3z = 10
-4x + 2y + 5z = 7

kx + z = 3
has no solutions. What is this value of k ?

For a certain value of k , the system
\(x + y + 3z = 10 \\ -4x + 2y + 5z = 7 \\ kx + z = 3 \)
has no solutions.

What is this value of k ?

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

A polynomial P(x) is divided by \(x^2+x-2\).

\(P(-2)=5,\ P(1)=20\).

Find the remainder of the division said above.

\(\text{Let $x^2+x-2 = (x-1)(x+2)$} \)

\(P(x)\) is divided by \((x-1)(x+2)\) it's is a polynomial of degree two.
Reminder should be a polynomial of degree less then 2. Say... \(r(x)=ax+b\)

\(\text{Let $r(x)=ax+b$ is the remainder of the division } \)

Now... \(P(x) = (x-1)(x-2) Q(x) + r(x)\)

\(\begin{array}{|lrcll|} \hline & \mathbf{P(x)} & \mathbf{=} & \mathbf{(x-1)(x+2) Q(x) + r(x)} \\ \hline x=-2: & P(-2)=5 & =& (-2-1)(-2+2) Q(-2) + r(-2) \\ & 5 & =& 0 + r(-2) \\ & \mathbf{r(-2)} &\mathbf{=}& \mathbf{5} \\ \hline x=1: & P(1)=20 & =& (1-1)(1+2) Q(1) + r(1) \\ & 20 & =& 0 + r(1) \\ & \mathbf{r(1)} &\mathbf{=}& \mathbf{20} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{r(x)} &\mathbf{=}& \mathbf{ax+b} \\ \hline r(-2)=5 &=& a\cdot (-2)+b \\ 5 &=& -2a+b \\ \hline r(1)=20 &=& a\cdot 1 +b \\ 20 &=& a+b \\ \hline \end{array}\)

Solve two equations and get the value of \(a\) and \(b\)

\(\begin{array}{|lrcll|} \hline & 5 &=& -2a+b \\ (1) & b&=& 5+2a \\\\ (2) & 20&=& a+b \quad | \quad b= 5+2a \\ & 20&=& a+5+2a \\ & 15&=& 3a \\ & \mathbf{a} &\mathbf{=}& \mathbf{5} \\\\ & b&=& 5+2a \\ & b &=& 5+2\cdot 5 \\ & \mathbf{b} &\mathbf{=}& \mathbf{15} \\ \\ & r(x) &=& ax+b \\ & \mathbf{ r(x) } &\mathbf{=}& \mathbf{5x+ 15} \\ \hline \end{array}\)

The remainder of the division is \(\mathbf{5x+ 15}\)

Let F1 and F2  be the foci of the ellipse kx^2+y^2=1 where k is greatr than 1 is a constant.
Suppose that there is a circle which passes through F1  and F2  and
which lies tangent to the ellipse at two points on the x-axis.
Compute k.

We rearrange \(kx^2+y^2=1\):

\(\begin{array}{|rcll|} \hline \mathbf{kx^2+y^2} & \mathbf{=} & \mathbf{1} \\ \\ \dfrac{x^2}{\dfrac{1}{k}}+\dfrac{y^2}{1} &=& 1 \\\\ \dfrac{x^2}{ \left(\color{red}{\dfrac{1}{\sqrt{k}}} \right)^2 }+\dfrac{y^2}{{\color{red}1}^2} &=& 1 \quad | \quad \boxed{\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 } \\\\ \mathbf{a} &\mathbf{=} & \mathbf{\dfrac{1}{\sqrt{k}}} \\ \mathbf{b} &\mathbf{=} & \mathbf{1} \\ \hline \end{array} \)

The formula with the focus of this ellipse is \(c^2= b^2-a^2\)
where \(c\) is the distance from the focus to origin.

\(\begin{array}{|rcll|} \hline c^2 &=& b^2-a^2 \quad & | \quad b=1,\ a=\dfrac{1}{\sqrt{k}} \\\\ &=& 1-\dfrac{1}{k} \\\\ \mathbf{c} &\mathbf{=}& \mathbf{\sqrt{1-\dfrac{1}{k}} } \\ \hline \end{array} \)

Circle: \(r = c=a\)

\(\begin{array}{|rcll|} \hline \sqrt{1-\dfrac{1}{k}} &=& \dfrac{1}{\sqrt{k}} \\\\ 1-\dfrac{1}{k} &=& \dfrac{1}{k} \\\\ 1 &=& \dfrac{2}{k} \\\\ \mathbf{k} &\mathbf{=}& \mathbf{2} \\ \hline \end{array}\)

Let ABCD be a convex quadrilateral, and let M and N be the midpoints of sides AD and BC, respectively. Prove that MN<=(AB+CD)/2.

Let $ABCD$ be a convex quadrilateral, and let $M$ and $N$ be the midpoints of sides $\overline{AD}$ and $\overline{BC}$, respectively.
Prove that $MN \le (AB + CD)/2$.

When does equality occur?

Rotate the convex quadrilateral:

In the triangle  C'AB we must have \(\overline{AB}+\overline{C'D'} < 2\overline{NM}\)

Since \(\overline{C'D'} = \overline{CD}\)

\(\begin{array}{|rcll|} \hline \overline{AB}+\overline{CD} < 2\overline{NM} \\\\ \dfrac{\overline{AB}+\overline{CD}}{2} < \overline{NM} \\ \hline \end{array}\)

Equality occur if the convex quadrilateral is a trapezoid:

\(\begin{array}{|rcll|} \hline \overline{AB}+\overline{CD} &=& 2 \overline{NM} \\\\ \dfrac{\overline{AB}+\overline{CD}}{2}& =& \overline{NM} \\ \hline \end{array}\)