heureka

Suppose functions g and f have the properties that

$g(x)=3f^{-1}(x)$

and

$f(x)=\dfrac{24}{x+3}$.
For what value of x does g(x)=15?

$\begin{array}{|rclcrcl|} \hline f\Big(f^-1(x)\Big) &=& x & | & \quad f^-1(x) &=& \dfrac{g(x)}{3} \\ && & | & \quad &=& \dfrac{15}{3} \\ && & | & \quad &=& 5 \\ f(5) &=& x & | & \quad f(5) &=& \dfrac{24}{5+3} \\ \dfrac{24}{5+3} &=& x \\ \dfrac{24}{8} &=& x \\ \mathbf{x} & \mathbf{=} & \mathbf{3} \\ \hline \end{array}$

How many whole numbers less than 18,632 are congruent to 23(mod 37)?

$\begin{array}{|lrcll|} \hline & 18632 &\equiv& 23 \pmod {37} \\ & 18632 -23 &=& n\cdot 37 \\ & n &=& \dfrac{18632 -23}{37} \\ & n &=& 502.945945946 \quad | \quad 0 \le n \le 502 \\ \Rightarrow \text{There are } \mathbf{503} \text{ numbers}. \\ \Rightarrow \text{The lowest number is } 37\cdot 0 + 23 = 23 \\ \Rightarrow \text{The highest number is } 37\cdot 502 + 23 = 18597 \\ \hline \end{array}$

If f(x)=5x-12, find a value for x so that f^-1(x)=f(x+1).

$\begin{array}{|rclcrcl|} \hline f\Big(f^-1(x)\Big) &=& x & | & \quad f^-1(x)&=& f(x+1) \\ f\Big(f(x+1)\Big) &=& x & | & \quad f(x+1) &=& 5(x+1) - 12\\ && & | & \quad &=& 5x - 7 \\ f\Big(5x - 7)\Big) &=& x \\ 5(5x-7)-12 &=& x \\ 25x-35-12 &=& x \\ 24x &=& 47 \\ \mathbf{x} & \mathbf{=} & \mathbf{\dfrac{47}{24}} \\ \hline \end{array}$

The Lucas sequence is the sequence 1, 3, 4, 7, 11, where the first term is 1,

the second term is 3 and each term after that is the sum of the previous two terms.

What is the remainder when the 100th term of the sequence is divided by 8?

$\begin{array}{|r|r|r|} \hline n & L_n & L_n \pmod{8} \\ \hline 1 & 1 & \color{green}1 \\ 2 & 3 & \color{green}3 \\ 3 & 4 & \color{green}4 \\ 4 & 7 & \color{red}-1 \\ 5 & 11 & \color{red}3 \\ 6 & 18 & \color{red}2 \\ 7 & 29 & \color{red}-3 \\ 8 & 47 & \color{red}-1 \\ 9 & 76 & \color{red}-4 \\ 10 & 123 & \color{red}3 \\ 11 & 199 & \color{red}-1 \\ 12 & 322 & \color{red} 2 \\ \hline 13 & 521 & \color{green}1 \\ 14 & 843 & \color{green}3 \\ 15 & 1364 & \color{green}4 \\ 16 & 2207 & \color{red}-1 \\ \ldots \\ 100 & 792070839848372253127 & -1 \text{ or } 7 \\ \hline \end{array}$

$\text{The cycle of $L_n \pmod{8}$ is $\mathbf{12}$ } : \color{green}1,\ \color{green}3,\ \color{green}4,\ \color{red}-1,\ \color{red}3,\ \color{red}2,\ \color{red}-3,\ \color{red}-1,\ \color{red}-4,\ \color{red}3,\ \color{red}-1,\ \color{red} 2$

$\text{To $n = 100$ we have $\left\lfloor\dfrac{100}{8}\right\rfloor = 12$ full cycles and a remainder of $4$.} \\ \text{And the $4th$ value in the cycle is $ \mathbf{-1}$}$

$\text{The remainder when the $100th$ term of the sequence $L_{100}=792070839848372253127$ is divided by $8$ }\\ \text{is $\mathbf{-1}$ or what's the same is $ \mathbf{7}$ }$

Find the value of $\mathbf{ {\color{red}x}}$ if

 
$\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}}}}=\dfrac{67}{96}$

$\begin{array}{|rcll|} \hline \dfrac{67}{96} &=& \dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}}}} \\ \left( 1+\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}}} \right)67 &=& 96 \\\\ 67+\dfrac{67}{2+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}}} &=& 96 \\\\ \dfrac{67}{2+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}}} &=& 96-67 \\\\ \dfrac{67}{2+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}}} &=& 29 \\\\ \left( 2+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}} \right)29 &=& 67 \\\\ 58+\dfrac{29}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}} &=& 67 \\\\ \dfrac{29}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}} &=& 67-58 \\\\ \dfrac{29}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}} &=& 9 \\\\ \left( 3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}} \right)9&=& 29 \\\\ 27+\dfrac{9}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}&=& 29 \\\\ \dfrac{9}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}&=& 29-27 \\\\ \dfrac{9}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}&=& 2 \\\\ \left( 4+\dfrac{1}{\mathbf{ {\color{red}x}}} \right)2&=& 9 \\\\ 8+\dfrac{2}{\mathbf{ {\color{red}x}}}&=& 9 \\\\ \dfrac{2}{\mathbf{ {\color{red}x}}}&=& 9-8 \\\\ \dfrac{2}{\mathbf{ {\color{red}x}}}&=& 1 \\\\ \mathbf{2} & \mathbf{=} & \mathbf{ {\color{red}x} } \\ \hline \end{array}$

Consider the vectors

$\mathbf{\vec{v}} = \dbinom{1}{3}$ and  $\mathbf{\vec{w}} = \dbinom{3}{2}$
Can you write $\mathbf{\vec{u}} = \dbinom{1}{2}$
as a linear combination of  $\vec{v}$ and $\vec{w}$ ?
If no, answer with 0.
If yes, find coefficients a and b such $a \dbinom{1}{3} + b \dbinom{3}{2} = \dbinom{1}{2}$
and answer with $\dfrac{a}{b}$.

$\mathbf{\vec{v}_{\perp} = \ ?}$

$\begin{array}{|rcll|} \hline \vec{v} \cdot \vec{v}_{\perp} &=& 0 \\ \dbinom{1}{3} \cdot \underbrace{\dbinom{-3}{1}}_{=\vec{v}_{\perp}} &=& -3+3 = 0 \\ \hline \end{array}$

$\mathbf{\vec{w}_{\perp} = \ ?}$

$\begin{array}{|rcll|} \hline \vec{w} \cdot \vec{w}_{\perp} &=& 0 \\ \dbinom{3}{2} \cdot \underbrace{\dbinom{-2}{3}}_{=\vec{w}_{\perp}} &=& -6+6 = 0 \\ \hline \end{array}$

$\mathbf{a,b = \ ?}$

$\begin{array}{|lrclccccc|} \hline & & & & & I. & & II. \\ & a\vec{v} + b\vec{w} &=& \vec{u} & | & \cdot \vec{v}_{\perp}& | & \cdot \vec{w}_{\perp}& | \\ \hline I. & \underbrace{a\vec{v}\vec{v}_{\perp}}_{=0} + b\vec{w}\vec{v}_{\perp} &=& \vec{u}\vec{v}_{\perp} \\ & b\vec{w}\vec{v}_{\perp} &=& \vec{u}\vec{v}_{\perp} \\ & \mathbf{b} &\mathbf{=}& \mathbf{\dfrac{ \vec{u}\vec{v}_{\perp} } { \vec{w}\vec{v}_{\perp} } } \\\\ & b &=& \dfrac{ \dbinom{1}{2}\dbinom{-3}{1} } { \dbinom{3}{2}\dbinom{-3}{1} } \\\\ & b &=& \dfrac{ -3+2 } { -9+2 } \\\\ & b &=& \dfrac{ -1 } { -7} \\\\ & \mathbf{b} &\mathbf{=}& \mathbf{\dfrac{ 1 } { 7}} \\ \hline II. & a\vec{v}\vec{w}_{\perp} + \underbrace{b\vec{w}\vec{w}_{\perp}}_{=0} &=& \vec{u}\vec{w}_{\perp} \\ & a\vec{v}\vec{w}_{\perp} &=& \vec{u}\vec{w}_{\perp} \\ & \mathbf{a} &\mathbf{=}& \mathbf{\dfrac{ \vec{u}\vec{w}_{\perp} } { \vec{v}\vec{w}_{\perp} } } \\\\ & a &=& \dfrac{ \dbinom{1}{2}\dbinom{-2}{3} } { \dbinom{1}{3}\dbinom{-2}{3} } \\\\ & a &=& \dfrac{ -2+6 } { -2+9 } \\\\ & \mathbf{a} &\mathbf{=}& \mathbf{\dfrac{ 4 } { 7 }} \\ \\ \hline & \dfrac{a}{b} &=& \dfrac{\dfrac{ 4 } { 7 }}{\dfrac{ 1 } { 7}} \\\\ & &=& \dfrac{ 4 } { 7 } \cdot \dfrac{ 7 } { 1} \\\\ & \mathbf{\dfrac{a}{b}} &\mathbf{=}& \mathbf{4} \\ \hline \end{array}$

Let

$\mathbf{v} \text{ and } \mathbf{w} \text{ be vectors such that } \operatorname{proj}_{\mathbf{w}}( \mathbf{v} )= \begin{pmatrix} 4 \\ -7 \end{pmatrix}.$

Find

$\operatorname{proj}_{-2 {\mathbf{w}}} (3 {\mathbf{v}}).$

$\begin{array}{|rcll|} \hline \vec{p} &=& \operatorname{proj}_{\mathbf{w}}( \mathbf{v} ) \\\\ &=& \dbinom{4}{-7} \\\\ &=& \dfrac{\vec{w}\vec{v}}{w^2}\vec{w} \\ \hline \vec{P} &=& \operatorname{proj}_{\mathbf{-2w}}( \mathbf{3v} ) \\\\ &=& \dfrac{(-2\vec{w})(3\vec{v})}{|-2w|^2}(-2\vec{w}) \\\\ &=& 6\dfrac{\vec{w}\vec{v}}{2w^2}\vec{w} \\\\ &=& 3\dfrac{\vec{w}\vec{v}}{w^2}\vec{w} \\\\ &=& 3\vec{p} \\\\ &=& 3\dbinom{4}{-7} \\\\ &=& \dbinom{12}{-21} \\ \hline \end{array}$

$\operatorname{proj}_{-2 {\mathbf{w}}} (3 {\mathbf{v}}) = \dbinom{12}{-21}.$

A square $DEFG$ varies inside equilateral triangle $ABC,$
so that $E$ always lies on side $\overline{AB},$ $F$ always lies on side $\overline{BC},$
and $G$ always lies on side $\overline{AC}.$
The point $D$ starts on side $\overline{AB},$ and ends on side $\overline{AC}.$
The diagram below shows the initial position of square $DEFG,$ an intermediate position, and the final position.

Show that as square DEFG varies, the height of point D above line BC remains constant.

$\text{Let $\overline{DH}=h $ the height of point $D$ above line $\overline{BC}$ } \\ \text{Let ${\color{red}s} $ the side of the square $DEFG$ } \\ \text{Let $\overline{DF}=\sqrt{2}\color{red}s$ } \\ \text{Let $\angle EFD = 45^{\circ}$ } \\ \text{Let $\angle BFE = \alpha$ } \\ \text{Let $\angle CFG = 90^{\circ}-\alpha$ } \\ \text{The equilateral triangle $ABC$ has the angles $60^{\circ}$, $60^{\circ}$, $60^{\circ}$ and the sides $a$, $a$, $a$. }$

In the rectangular triangle DHF the following applies:

$\begin{array}{|rcll|} \hline \sin(\alpha + 45^{\circ}) &=& \dfrac{h}{\sqrt{2}\color{red}s} \\\\ \sqrt{2}\sin(\alpha + 45^{\circ}) &=& \dfrac{h}{\color{red}s} \quad | \quad \sqrt{2}\sin(\alpha + 45^{\circ}) = \sin(\alpha) + \cos(\alpha) \\\\ \sin(\alpha) + \cos(\alpha) & = & \dfrac{h}{\color{red}s} \\\\ \mathbf{h}& \mathbf{=} & \mathbf{{\color{red}s}\sin(\alpha) + {\color{red}s}\cos(\alpha)} \\ \hline \end{array}$

$\text{Let $\overline{BC}=a $ is the side of the equilateral triangle $ABC$. }$

$\begin{array}{|rcll|} \hline a &=& \overline{BM} + \overline{MF} +\overline{FN} + \overline{NC} \\ && \overline{BM} = \dfrac{s\sin(\alpha)}{\tan(60^{\circ})} \\ && \overline{MF} = s\cos(\alpha) \\ && \overline{FN} = s\sin(\alpha) \\ && \overline{NC} = \dfrac{s\cos(\alpha)}{\tan(60^{\circ})} \\ a &=& \dfrac{ {\color{red}s} \sin(\alpha)}{\tan(60^{\circ})} + {\color{red}s}\cos(\alpha) +{\color{red}s}\sin(\alpha) + \dfrac{{\color{red}s}\cos(\alpha)}{\tan(60^{\circ})} \\ a &=& {\color{red}s}\sin(\alpha) +{\color{red}s}\cos(\alpha) + \Big({\color{red}s}\sin(\alpha) +{\color{red}s}\cos(\alpha)\Big) \dfrac{1}{ \tan(60^{\circ}) } \\ a &=& \Big({\color{red}s}\sin(\alpha) +{\color{red}s}\cos(\alpha)\Big) \left( 1+\dfrac{1}{ \tan(60^{\circ}) } \right) \quad | \quad \mathbf{h={\color{red}s}\sin(\alpha) + {\color{red}s}\cos(\alpha)} \\ a &=& h \left( 1+\dfrac{1}{ \tan(60^{\circ}) } \right) \quad | \quad \tan(60^{\circ}) = \sqrt{3} \\ a &=& h \left( 1+\dfrac{1}{ \sqrt{3} } \right) \\ a &=& h \left( \dfrac{ \sqrt{3}+1}{ \sqrt{3} } \right) \\ \mathbf{h}& \mathbf{=} & \mathbf{a \left( \dfrac{ \sqrt{3} }{ 1+\sqrt{3}} \right)} \\ \hline \end{array}$

So the height of point D above line BC remains constant.

Mrs. Roland distributed a bag of skittles to her students. She gave each student five skittles,

then ate the 8 that remained in the bag.

If there were originally143 skittles in the bag. How many students does she have?

Let s = students

$\begin{array}{|rcll|} \hline 5s+8 &=& 143\\ 5s &=& 143 - 8\\ 5s &=& 135\\ s&=& 27 \\ \hline \end{array}$

She does have 27 students.

Maximum Function Help!
If $x$ and $y$ are real and $x^2 + y^2 = 1$, compute the maximum value of $(x+y)^2$.

$x^2 + y^2 = 1$  is a circle with a radius of 1.

A circle can be defined as the locus of all points that satisfy the equations
$x = r \cos(\alpha) \qquad y = r \sin(\alpha)$
where x,y are the coordinates of any point on the circle, r is the radius of the circle and
$\alpha$ is the parameter - the angle subtended by the point at the circle's center.

We substitute:

$\begin{array}{|rcll|} \hline x^2 + y^2 &=& 1 \quad | \quad x = \cos(\alpha) \qquad y = \sin(\alpha) \\ \mathbf{\sin^2(\alpha) + \cos^2(\alpha)} & \mathbf{=}& \mathbf{ 1 } \\ \hline \end{array}$

Maximul value:

$\begin{array}{|rcll|} \hline (x+y)^2_\text{max} && \qquad | \quad x = \cos(\alpha) \qquad y = \sin(\alpha) \\ &=& \Big(\sin(\alpha) + \cos(\alpha)\Big)^2 \\ &=& \sin^2(\alpha) + \cos^2(\alpha) + 2\sin(\alpha)\cos(\alpha) \quad | \quad \sin^2(\alpha) + \cos^2(\alpha) = 1 \\ &=& 1+2\sin(\alpha)\cos(\alpha) \quad | \quad 2\sin(\alpha)\cos(\alpha) = \sin(2\alpha) \\ &=& 1+ \sin(2\alpha) \quad | \quad \text{The sin-function is max at 1 } \\ &=& 1+ 1 \\ (x+y)^2_\text{max}&\mathbf{=}& \mathbf{2} \\ \hline \end{array}$

The maximum value of $(x+y)^2$ is 2