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 #2
avatar+21848 
+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

 

laugh

Mar 22, 2019
 #1
avatar+21848 
+3

For all real numbers $r$ and $s$, define the mathematical operation $\#$ such that the following conditions apply:

$r\ \#\ 0 = r, r\ \#\ s = s\ \#\ r$, and $(r + 1)\ \#\ s = (r\ \#\ s) + s + 1$.

What is the value of $11\ \#\ 5$

 

 

\(\begin{array}{|rcll|} \hline \mathbf{(r + 1)\ \#\ s }& \mathbf{=} & \mathbf{(r\ \#\ s) + s + 1} \quad \text{ or } \quad \mathbf{r\ \#\ s = \Big((r-1)\ \#\ s\Big) + s + 1} \\ \hline \\ r\ \#\ s &=& \Big((r-1)\ \#\ s\Big) + s + 1 \quad | \quad (r-1)\ \#\ s = \Big((r-2)\ \#\ s\Big) + s + 1 \\ &=& \Big((r-2)\ \#\ s\Big) + s + 1 + s + 1 \\ &=& \Big((r-2)\ \#\ s\Big) + 2s + 2 \quad | \quad (r-2)\ \#\ s = \Big((r-3)\ \#\ s\Big) + s + 1 \\ &=& \Big((r-3)\ \#\ s\Big) + s + 1 + 2s + 2 \\ &=& \Big((r-3)\ \#\ s\Big) + 3s + 3 \quad | \quad (r-3)\ \#\ s = \Big((r-4)\ \#\ s\Big) + s + 1 \\ &=& \Big((r-4)\ \#\ s\Big) + s + 1 + 3s + 3 \\ &=& \Big((r-4)\ \#\ s\Big) + 4s + 4 \\ \ldots \\ \mathbf{r\ \#\ s} &\mathbf{=}& \mathbf{ \Big((r-n)\ \#\ s\Big) + ns + n} \qquad n\in \mathbb{Z} \\ \hline \end{array} \)

 

\(\mathbf{11\ \#\ 5 = \ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{r\ \#\ s} &\mathbf{=}& \mathbf{ \Big((r-n)\ \#\ s\Big) + ns + n} \quad &| \quad r=n=11,\ s=5 \\ 11\ \#\ 5 & = & \Big((11-11)\ \#\ 5\Big) + 11\cdot 5 + 11 \\ 11\ \#\ 5 & = & \Big(0\ \#\ 5\Big) + 11\cdot 5 + 1 \quad &| \quad 0\ \#\ 5 = 5\ \#\ 0 \\ 11\ \#\ 5 & = & \Big(5\ \#\ 0\Big) + 11\cdot 5 + 11 \quad &| \quad 5\ \#\ 0 = 5 \\ 11\ \#\ 5 & = & 5 + 11\cdot 5 + 11 \\ \mathbf{11\ \#\ 5} & \mathbf{=} & \mathbf{71} \\ \hline \end{array}\)

 

laugh

Mar 21, 2019
 #1
avatar+21848 
+4

2013 NS 29

 

1.

cos-theorem:

\(\begin{array}{|rcll|} \hline \mathbf{ b^2 } &\mathbf{=}& \mathbf{d^2+(a-c)^2-2d(a-c)\cos(L)} \\\\ 65^2&=& 60^2+58^2-120\cdot 58\cos(L) \\ \cos(L)&=& \dfrac{60^2+58^2-65^2 }{120\cdot 58} \\ \cos(L)&=& \dfrac{2739}{6960} \\ \cos(L)&=& 0.39353448276 \\ L &=& \arccos(0.39353448276) \\ L &=& 66.8253951955^\circ \\ \mathbf{\sin( L)} &\mathbf{=}& \mathbf{0.91930985575} \\ \hline \end{array}\)

 

2.

cos-theorem:

\(\begin{array}{|rcll|} \hline \mathbf{ d^2 } &\mathbf{=}& \mathbf{b^2+(a-c)^2-2b(a-c)\cos(K)} \\\\ 60^2&=& 65^2+58^2-130\cdot 58\cos(K) \\ \cos(K)&=& \dfrac{65^2+58^2-60^2 }{130\cdot 58} \\ \cos(K)&=& \dfrac{3989}{7540} \\ \cos(K)&=& 0.52904509284 \\ K &=& \arccos(0.52904509284 ) \\ K &=& 58.0590417221^\circ \\ \mathbf{\sin(K)} &\mathbf{=}& \mathbf{0.84859371300} \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline 1. & \sin(K) &=& \dfrac{r}{x} \qquad \text{ or } \qquad r = x\sin(K) \\\\ 2. & \sin(L) &=& \dfrac{r}{80-x} \quad | \quad r = x\sin(K) \\ & \sin(L) &=& \dfrac{x\sin(K)}{80-x} \\ & (80-x)\sin(L) &=& x\sin(K) \\ & 80\sin(L)-x\sin(L) &=& x\sin(K) \\ & x\sin(L)+x\sin(K) &=& 80\sin(L) \\ & x\Big(\sin(L)+\sin(K)\Big) &=& 80\sin(L) \\ & \mathbf{x} &\mathbf{=}& \mathbf{80\cdot \dfrac{ \sin(L)} {\sin(L)+\sin(K)} }\\\\ & x & = & 80\cdot \dfrac{0.91930985575} {0.91930985575+0.84859371300} \\ & x & = & 80 \cdot 0.52 \\ & x & = & 41.6 \\ & \mathbf{x} &\mathbf{=}& \mathbf{41 \ \dfrac{3}{5}} \\ \hline \end{array}\)

 

The length of segment KA is \(\mathbf{41 \ \dfrac{3}{5}}\)

 

laugh

Mar 20, 2019
 #1
avatar+21848 
+4

2015 SS 28

 

1.

cos-theorem:

\(\begin{array}{|rcll|} \hline BM^2&=&x^2+\left(\dfrac{x}{2}\right)^2-2\cdot x \cdot \dfrac{x}{2}\cos(108^\circ) \\ BM^2 &=& x^2\left(1+\dfrac{1}{4}-\cos(108^\circ)\right) \quad |\quad \cos(108^\circ)=\cos(180^\circ-72^\circ)=-\cos(72^\circ) \\ BM^2 &=& x^2\left( \dfrac{5}{4}+\cos(72^\circ)\right) \\ \mathbf{BM} &\mathbf{=}& \mathbf{x \sqrt{\dfrac{5}{4}+\cos(72^\circ)}} \\ \hline \end{array} \)

 

2.
sin-theorem:

\(\begin{array}{|rcll|} \hline \dfrac{\sin(z)}{x} &=& \dfrac{\sin(36^\circ)}{ZB} \\\\ \mathbf{ZB} &\mathbf{=}& \mathbf{\dfrac{\sin(36^\circ)}{\sin(z)}x }\\ \hline \end{array} \)

 

3.

sin-theorem:

\(\begin{array}{|rcll|} \hline \dfrac{\sin(180^\circ-z)}{\frac{x}{2}} &=& \dfrac{\sin(72^\circ)}{ZM} \\\\ \dfrac{2\sin(z)}{x} &=& \dfrac{\sin(72^\circ)}{ZM} \\\\ \mathbf{ZM} &\mathbf{=}& \mathbf{\dfrac{\sin(72^\circ)}{2\sin(z)}x }\\ \hline \end{array} \)

 

\(\mathbf{z=\ ?}\)

\(\begin{array}{|rcll|} \hline BM &=& ZB+ZM \\ BM&=& \dfrac{\sin(36^\circ)}{\sin(z)}x+\dfrac{\sin(72^\circ)}{2\sin(z)}x \\\\ BM&=& \dfrac{x}{\sin(z)} \left( \sin(36^\circ) + \dfrac{\sin(72^\circ)}{2}\right) \quad | \quad BM=x \sqrt{\dfrac{5}{4}+\cos(72^\circ)} \\\\ x \sqrt{\dfrac{5}{4}+\cos(72^\circ)}&=& \dfrac{x}{\sin(z)} \left( \sin(36^\circ) + \dfrac{\sin(72^\circ)}{2}\right) \\\\ \sqrt{\dfrac{5}{4}+\cos(72^\circ)}&=& \dfrac{1}{\sin(z)} \left( \sin(36^\circ) + \dfrac{\sin(72^\circ)}{2}\right) \\\\ \mathbf{\sin(z)} & \mathbf{=} & \mathbf{ \dfrac{ \sin(36^\circ) + \dfrac{\sin(72^\circ)}{2} }{\sqrt{\dfrac{5}{4}+\cos(72^\circ)}} } \\\\ z &=& 180^\circ- \arcsin\left( \dfrac{ \sin(36^\circ) + \dfrac{\sin(72^\circ)}{2} }{\sqrt{\dfrac{5}{4}+\cos(72^\circ)}}\right) \\ &=& 180^\circ- \arcsin\left( \dfrac{ 1.06331351044 }{1.24860602048} \right) \\ &=& 180^\circ- \arcsin\left( 0.85160049928 \right) \\ &=& 180^\circ- 58.3861775592^\circ \\ \mathbf{z} &\mathbf{=}& \mathbf{121.613822441^\circ} \\ \hline \end{array}\)

 

\(\text{Let $\angle ABM =180^\circ-(36^\circ+z) $} \\ \text{Let $ 36^\circ+z = 36^\circ+ 121.613822441^\circ=157.613822441^\circ$} \)

 

4.
sin-theorem:

\(\begin{array}{|rcll|} \hline \dfrac{\sin(z)}{x} &=& \dfrac{\sin(ABM)}{3} \quad | \quad \angle ABM =180^\circ-(36^\circ+z) \\\\ \dfrac{\sin(z)}{x} &=& \dfrac{\sin\Big(180^\circ-(36^\circ+z)\Big)}{3} \\\\ \dfrac{\sin(z)}{x} &=& \dfrac{\sin( 36^\circ+z )}{3} \\\\ \mathbf{x} &\mathbf{=}& \mathbf{3\cdot \dfrac{\sin(z)}{\sin( 36^\circ+z )} } \\\\ x & = & 3\cdot \dfrac{0.85160049928}{\sin( 157.613822441^\circ )} \\\\ x & = & 3\cdot \dfrac{0.85160049928}{0.38084732121} \\\\ x & = & 3\cdot 2.23606797751 \quad | \quad 2.23606797751 = \sqrt{5} \\ \mathbf{x} &\mathbf{=}& \mathbf{3\cdot \sqrt{5}} \\ \hline \end{array}\)

 

laugh

Mar 20, 2019
 #3
avatar+21848 
+2

2.

In cyclic quadrilaterla ABCD, AB=2, BC=3, CD=10, and DA=6.

Let P be the intersection of lines AB and CD.

Find the length BP.

\(\text{Let $BP =x$} \\ \text{Let $PA =2+x$} \\ \text{Let $PC =y$} \\ \text{Let $PD =10+y$} \)


\(\text{Let $\angle BPC = P $} \\ \text{Let $\angle CBP = B $} \\ \text{Let $\angle ABC = 180^\circ -B $} \\ \text{Let $\angle ADC = 180^\circ - \angle ABC =180^\circ-(180^\circ -B)=B $} \)

 

1.

Intersecting secants theorem:

see: https://en.wikipedia.org/wiki/Intersecting_secants_theorem

\(\begin{array}{|rcll|} \hline \mathbf{BP\cdot PA} &\mathbf{=}& \mathbf{PC\cdot PD} \quad | \quad \text{Intersecting secants theorem} \\\\ x\cdot (2+x) &=& y\cdot (10+y) \\ \hline \end{array} \)

 

2.
sin-theorem:

\(\begin{array}{|lrcll|} \hline 1. & \dfrac{\sin(P)}{3} &=& \dfrac{\sin(B)}{y} \\ & \dfrac{\sin(P)}{\sin(B)} &=& \dfrac{3}{y} \\\\ 2. & \dfrac{\sin(P)}{6} &=& \dfrac{\sin(B)}{2+x} \\ & \dfrac{\sin(P)}{\sin(B)} &=& \dfrac{6}{2+x} \\ \hline & \dfrac{\sin(P)}{\sin(B)} = \dfrac{3}{y} &=& \dfrac{6}{2+x} \\\\ & \dfrac{3}{y} &=& \dfrac{6}{2+x} \\\\ & \dfrac{y}{3} &=& \dfrac{2+x}{6} \\\\ & y &=& \dfrac{3}{6}\cdot(2+x) \\\\ & \mathbf{y} & \mathbf{=}& \mathbf{\dfrac{1}{2}\cdot(2+x)} \\\\ & x\cdot (2+x) &=& y\cdot (10+y) \quad | \quad y = \dfrac{1}{2}\cdot(2+x) \\\\ & x\cdot (2+x) &=& \dfrac{1}{2}\cdot(2+x)\cdot \left(10+\dfrac{1}{2}\cdot(2+x)\right) \\\\ & x &=& \dfrac{1}{2} \cdot \left(10+ \dfrac{1}{2}\cdot(2+x) \right) \quad | \quad \cdot 2 \\\\ & 2x &=& 10+ \dfrac{1}{2}\cdot(2+x) \quad | \quad \cdot 2 \\\\ & 4x &=& 20+ 2+x \\\\ & 3x &=& 22 \\\\ & x &=& \dfrac{22}{3} \\\\ & \mathbf{ x } & \mathbf{=} & \mathbf{7.\bar{3}} \\ \hline \end{array}\)

 

laugh

Mar 19, 2019
 #3
avatar+21848 
+2
Mar 15, 2019
 #1
avatar+21848 
+3

There are many ways to circumscribe a rectangle R about a 5 x 10 rectangle so
that each vertex of the 5 x 10 rectangle is on a different side of R.
Rectangle R's area is 110. what is the maximum area of a rectangle R
that can be circumscribed about a 5 x 10 rectangle?

Let:

\(\begin{array}{|rcll|} \hline a &=& W\sin(\theta) \\ c &=& W\cos(\theta) \\ \hline \end{array} \begin{array}{|rcll|} \hline d &=& L\sin(\theta) \\ b &=& L\cos(\theta) \\ \hline \end{array} \)

 

The area A of the circumscribing rectangle:

\(\begin{array}{|rcll|} \hline \mathbf{A(\theta)} &\mathbf{=}& \mathbf{(a+b)(c+d)} \\\\ A(\theta) &=& \Big( W\sin(\theta)+L\cos(\theta) \Big) \Big( W\cos(\theta)+L\sin(\theta) \Big) \\ &=& W^2\sin(\theta)\cos(\theta) +L\cdot W\sin^2(\theta) + LW\cos^2(\theta) + L^2\sin(\theta)\cos(\theta) \\ &=& (W^2+L^2)\sin(\theta)\cos(\theta) +L\cdot W\Big((\sin^2(\theta) + \cos^2(\theta)\Big) \quad | \quad \sin^2(\theta) + \cos^2(\theta) = 1 \\ &=& (W^2+L^2)\sin(\theta)\cos(\theta) +L\cdot W \quad | \quad \sin(\theta)\cos(\theta) = \dfrac{\sin(2\theta)}{2} \\ &=& (W^2+L^2)\dfrac{\sin(2\theta)}{2} +L\cdot W \\ && \color{red}\underline{\text{maximize}}:\\ && \qquad \color{red}\sin(2\theta) = 1 \\ && \qquad \color{red}2\theta = \arcsin(1) \\ && \qquad \color{red} 2\theta = 90^\circ \\ && \qquad \color{red} \theta = 45^\circ \\ A_{\text{max}}(45^\circ) &=& (W^2+L^2)\dfrac{\sin(2\cdot 45^\circ)}{2} +L\cdot W \\ &=& (W^2+L^2)\dfrac{\sin(90^\circ)}{2} +L\cdot W \quad | \quad \sin(90^\circ) = 1 \\ &=& \dfrac{1}{2} (W^2+L^2) +L\cdot W \quad | \quad \sin(90^\circ) = 1 \\ &=& \dfrac{1}{2} (W^2+L^2+2LW) \\ &=& \dfrac{1}{2} (W+L)^2 \\ \hline \end{array}\)

 

The maximum area:

\(\begin{array}{|rcll|} \hline A_{\text{max}} &=& \dfrac{1}{2} (W+L)^2 \\\\ &=& \dfrac{1}{2} (5+10)^2 \\\\ &=& \dfrac{15^2}{2} \\\\ &=& \dfrac{225}{2} \\\\ &=& 112.5 \\ \hline \end{array}\)

 

The maximum area is 112.5


source:
https://www.youtube.com/watch?v=q3ZnOvhEJWo

 

laugh

Mar 15, 2019