heureka

Hallo ilorty,

i use:

\(\begin{array}{|rcll|} \hline Example~line~1 \ldots \\ Example~line~2 \ldots \\ \hline \end{array}\)

or with \boxed{}:

\(\boxed{Box~Example~line~ 1 \\Box~Example~line~ 2 }\)

Next line is "\\"

Six squares are inscribed in an 11 by 13 rectangle.  Find the shaded area.

\(\text{Let $P_2=\dbinom{x_2}{y_2} $ } \\ \text{Let $P_4=\dbinom{x_4}{y_4} $ }\)

\(\begin{array}{|rcll|} \hline P_1 &=& a\dbinom{\sin(\varphi)}{\cos(\varphi)} \\ \hline \end{array} \begin{array}{|rcll|} \hline P_2 &=& P_1 + 3a\dbinom{\cos(\varphi)}{-\sin(\varphi)} \\\\ P_2 &=& a\dbinom{\sin(\varphi)}{\cos(\varphi)} + 3a\dbinom{\cos(\varphi)}{-\sin(\varphi)} \\\\ P_2 &=& a\dbinom{\sin(\varphi)+3\cos(\varphi)} {\cos(\varphi)-3\sin(\varphi)} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \mathbf{x_2} &=& \mathbf{a\Big( \sin(\varphi)+3\cos(\varphi) \Big)} \quad | \quad x_2 = 11 \\ a\Big( \sin(\varphi)+3\cos(\varphi) \Big) &=& 11 \\ a &=& \dfrac{11}{ \Big( \sin(\varphi)+3\cos(\varphi) \Big) } \qquad (1) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline P_3 &=& 2a\dbinom{\cos(\varphi)}{-\sin(\varphi)} \\ \hline \end{array} \begin{array}{|rcll|} \hline P_4 &=& P_3 - 3a\dbinom{\sin(\varphi)}{\cos(\varphi)} \\\\ P_4 &=& 2a\dbinom{\cos(\varphi)}{-\sin(\varphi)} - 3a\dbinom{\sin(\varphi)}{\cos(\varphi)} \\\\ P_4 &=& a\dbinom{2\cos(\varphi)-3\sin(\varphi)} {-2\sin(\varphi)-3\cos(\varphi)} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \mathbf{|y_4|} &=& \mathbf{a\Big( 2\sin(\varphi)+3\cos(\varphi)\Big)} \quad | \quad y_4 = 13 \\ a\Big( 2\sin(\varphi)+3\cos(\varphi) \Big) &=& 13 \\ a &=& \dfrac{13} { \Big( 2\sin(\varphi)+3\cos(\varphi) \Big) } \qquad (2) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1): & a &=& \dfrac{11}{ \Big( \sin(\varphi)+3\cos(\varphi) \Big) } \\ (2): & a &=& \dfrac{13} { \Big( 2\sin(\varphi)+3\cos(\varphi) \Big) } \\ \hline & a = \dfrac{11}{ \Big( \sin(\varphi)+3\cos(\varphi) \Big) } &=& \dfrac{13} { \Big( 2\sin(\varphi)+3\cos(\varphi) \Big) } \\\\ & \dfrac{11}{ \Big( \sin(\varphi)+3\cos(\varphi) \Big) } &=& \dfrac{13} { \Big( 2\sin(\varphi)+3\cos(\varphi) \Big) } \\\\ & 11 \Big( 2\sin(\varphi)+3\cos(\varphi) \Big) &=& 13 \Big( \sin(\varphi)+3\cos(\varphi) \Big) \\\\ & 22\sin(\varphi)+33\cos(\varphi) &=& 13 \sin(\varphi)+39\cos(\varphi) \\\\ & 22\sin(\varphi)- 13 \sin(\varphi) &=& 39\cos(\varphi)-33\cos(\varphi) \\\\ & 9\sin(\varphi) &=& 6\cos(\varphi) \\\\ & \mathbf{\tan(\varphi)} &=& \mathbf{\dfrac{2}{3}} \\ \hline \end{array}\)

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

\(\begin{array}{|rclrcl|} \hline a &=& \dfrac{11}{ \Big( \sin(\varphi)+3\cos(\varphi) \Big) } \\ &&& \sin(\varphi) &=& \dfrac{\tan(\varphi) }{\sqrt{1+\tan^2(\varphi)}} \quad | \quad \mathbf{\tan(\varphi)=\dfrac{2}{3}} \\ &&& \sin(\varphi) &=& \dfrac{\dfrac{2}{3}}{\sqrt{1+\dfrac{4}{9}}} \\ &&& \mathbf{\sin(\varphi)} &=& \mathbf{\dfrac{2\sqrt{13}}{13}} \\\\ &&& \cos(\varphi) &=& \dfrac{1}{\sqrt{1+\tan^2(\varphi)}} \quad | \quad \mathbf{\tan(\varphi)=\dfrac{2}{3}} \\ &&& \cos(\varphi) &=& \dfrac{1}{\sqrt{1+\dfrac{4}{9}}} \\ &&& \mathbf{\cos(\varphi)} &=& \mathbf{\dfrac{3\sqrt{13}}{13}} \\\\ a &=& \dfrac{11}{ \dfrac{2\sqrt{13}}{13}+3*\dfrac{3\sqrt{13}}{13} } \\\\ a &=& \dfrac{11}{ \dfrac{2\sqrt{13}}{13}+\dfrac{9\sqrt{13}}{13} } \\\\ a &=& \dfrac{11}{ \dfrac{11\sqrt{13}}{13} } \\\\ a &=& \dfrac{13}{\sqrt{13}} \\\\ \mathbf{a} &=& \mathbf{\sqrt{13}} \\ \hline \end{array}\)

The shaded area:

\(\begin{array}{|rcll|} \hline \text{The shaded area} &=& 11*13 - 6a^2 \quad | \quad \mathbf{a=\sqrt{13}} \\ \text{The shaded area} &=& 11*13 - 6*13 \\ \text{The shaded area} &=& 5*13 \\ \mathbf{\text{The shaded area}} &=& \mathbf{65} \\ \hline \end{array}\)

Prove that there do not exist integers m and n such that

\(5m^2 − 6mn + 7n^2 = 2011\).

\(\begin{array}{|rclrcl|} \hline \mathbf{5m^2 - 6mn + 7n^2} &=& \mathbf{2011} \quad | \quad *5 \\\\ 25m^2 - 30mn + 35n^2 &=& 2011*5 \\ 25m^2 - 30mn + 9n^2+ 26n^2 &=& 2011*5 \\ (5m-3n)^2+ 26n^2 &=& 2011*5 \quad | \quad \mathbf{2011*5} \equiv \mathbf{6 \pmod{13}} \\ \hline \\ (5m-3n)^2+ 26n^2 &\equiv& 6 \pmod{13} \quad | \quad \mathbf{26n^2\pmod{13}} \equiv \mathbf{0} \\ (5m-3n)^2 +0 &\equiv& 6 \pmod{13} \\ (5m-3n)^2 &\equiv& 6 \pmod{13} \quad | \quad x= 5m-2n \\ \mathbf{x^2} &\equiv& \mathbf{6 \pmod{13}} \\ \hline \end{array}\)

The Quadratic Reciprocity Law with Legendre symbol:

Let \(\mathbf{p}\) be an odd prime. The integer \(\mathbf{a}\), prime to \(\mathbf{p}\), is said to
be a \(\mathbf{\text{quadratic residue}}\) or \(\mathbf{\text{nonresidue}}\) of \(\mathbf{p}\) according as the congruence

\(x^2 \equiv a\pmod{p}\)

is of is not solvable. The Legendre symbol \(\left(\dfrac{a}{p}\right)\) is defined to be \(+1\) or \(-1\)
according as \(\mathbf{a}\) is a quadratic residue or nonresidure of \(\mathbf{p}\).

\(\begin{array}{|rcll|} \hline \mathbf{x^2} &\equiv& \mathbf{6 \pmod{13}} \quad | \quad a = 6,\ p = 13 \\\\ \left(\dfrac{a}{p}\right) &=& \left(\dfrac{6}{13}\right) \\ \hline \\ \mathbf{\left(\dfrac{6}{13}\right)} &=& \left(\dfrac{2*3}{13}\right) \\\\ &=& \left(\dfrac{2}{13}\right)\left(\dfrac{3}{13}\right) \quad &| \quad \left(\dfrac{3}{13}\right)=\left(\dfrac{13}{3}\right) \\\\ &=& \left(\dfrac{2}{13}\right)\left(\dfrac{13}{3}\right) \\\\ &=& \left(\dfrac{2}{13}\right)\left(\dfrac{13\pmod{3}}{3}\right) \\\\ &=& \left(\dfrac{2}{13}\right)\left(\dfrac{1}{3}\right) \quad &| \quad \left(\dfrac{1}{3}\right) = 1 \\\\ &=& \left(\dfrac{2}{13}\right)*1 \\\\ &=& \left(\dfrac{2}{13}\right) \\\\ &=& \left(-1\right)^{ \frac{13^2-1}{8} } \\\\ &=& \left(-1\right)^{21} \\\\ &=& \mathbf{-1} \\ \hline \end{array}\)

\(\mathbf{\left(\dfrac{6}{13}\right)}=-1\), there do not exist integers m and n such that
\(5m^2 - 6mn + 7n^2 = 2011\).

In the month of June, the temperature in Johannesburg, South Africa, varies over the day in a periodic way that can be modeled approximately by a trigonometric function.

The lowest temperature is usually around 3C, and the highest temperature is around 18C.

The temperature is typically halfway between the daily high and daily low at both 10 a.m., and 10 p.m.

and the highest temperatures are in the afternoon.

Find the formula of the trigonometric function that models the temperature T in Johannesburg t hours after midnight.

Define the function using radians.

\(\begin{array}{|rcll|} \hline \dfrac{18^\circ C + 3^\circ C}{2} &=& 10.5^\circ \\ \\ \dfrac{18^\circ C - 3^\circ C}{2} &=& 7.5^\circ \\ \\ \hline \mathbf{T} &=& \mathbf{10.5^\circ C+7.5^\circ C\cdot\sin\Bigg(\left(\frac{t+2}{12}-1\ \right)\cdot\pi \Bigg)} \\ \hline \end{array}\)

Image below with two squares, ABCD and BFGE, sharing a vertex.
Given that AE = 5, what is the length of DG?

My attempt:

\(\begin{array}{|rcll|} \hline DG^2 &=& 5^2+5^2 \\ DG^2 &=&2*5^2 \\ \mathbf{DG} &=& \mathbf{5\sqrt{2}} \\ \hline \end{array}\)

Suppose that \(f(x)=4x+5\).

What is \(f^{-1}\left(f^{-1}(9)\right)\)?

\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{4x+5} \quad | \quad \text{change } f(x) \text{ and } x \\\\ x &=& 4f^{-1}(x)+5 \quad | \quad x=9 \\ 9 &=& 4f^{-1}(9)+5 \\ 4 &=& 4f^{-1}(9) \\ \mathbf{f^{-1}(9)} &=& \mathbf{1} \\\\ x &=& 4f^{-1}(x)+5 \quad | \quad x=1 \\ 1 &=& 4f^{-1}(1)+5 \\ -4 &=& 4f^{-1}(1) \\ \mathbf{f^{-1}(1)} &=& \mathbf{-1} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{f^{-1}\left(f^{-1}(9)\right)} \quad &| \quad \mathbf{f^{-1}(9)=1} \\ &=& f^{-1} (1) \quad &| \quad \mathbf{f^{-1}(1)=-1} \\ &=& \mathbf{-1}\\ \hline \end{array}\)

Find the area of the shaded region if r = 1/4 and each side of the square is 1.

\(\begin{array}{|rcll|} \hline \mathbf{AC^2} &=& \mathbf{(1-r)^2+(1-r)^2} \\ AC^2 &=& 2(1-r)^2 \\ AC &=& (1-r)\sqrt{2} \\ \mathbf{AC} &=& \mathbf{\sqrt{1.125}} \\ \\ \mathbf{AB^2+r^2}&=& \mathbf{AC^2} \\ AB^2+0.0625&=& 1.125 \\ AB^2&=& 1.0625 \\ \mathbf{AB} &=& \mathbf{\sqrt{1.0625}} \\ \hline \end{array} \)

cos-rule:

\(\begin{array}{|rcll|} \hline \mathbf{r^2} &=& \mathbf{AC^2+AB^2-2*AC*AB*\cos(\varphi)} \\\\ \cos(\varphi) &=& \dfrac{AC^2+AB^2-r^2}{2*AC*AB} \\\\ \cos(\varphi) &=& \dfrac{1.125+1.0625-0.0625}{2*\sqrt{1.125}*\sqrt{1.0625}} \\\\ \cos(\varphi) &=& \dfrac{2.125}{2*\sqrt{1.1953125}} \\\\ \cos(\varphi) &=& 0.97182531581 \\\\ \varphi &=& 13.6330222254^\circ \\ \hline 45^\circ - \varphi &=& 31.3669777746^\circ \\ \mathbf{\tan(45^\circ - \varphi)} &=& \mathbf{\dfrac{x}{1} } \\ \tan(45^\circ - \varphi) &=& x \\ x &=& \tan(45^\circ - \varphi) \\ x &=& \tan(31.3669777746^\circ) \\ \mathbf{x} &=& \mathbf{0.60961179680} \\ \hline \text{Area of the shaded region} &=& \dfrac{x*1}{2} \\ \text{Area of the shaded region} &=& \dfrac{0.60961179680}{2} \\ \mathbf{\text{Area of the shaded region}} &=& \mathbf{0.30480589840} \\ \hline \end{array}\)

Let \(x\), \(y\), and \(z\) be positive real numbers such that  \(xyz=1\)
Find the minimum value of \((x+2y)(y+2z)(xz+1)\).

\(\mathbf{\huge{AM \geq GM }}\)

\(\begin{array}{|rcll|} \hline \dfrac{x+2y}{2} & \ge & \sqrt{x*2y} \\\\ \dfrac{y+2z}{2} & \ge & \sqrt{y*2z} \\\\ \dfrac{xz+1}{2} & \ge & \sqrt{xz*1} \\\\ \hline \dfrac{(x+2y)(y+2z)(xz+1)}{8} & \ge & \sqrt{x*2y}\sqrt{y*2z} \sqrt{xz*1} \\\\ \dfrac{(x+2y)(y+2z)(xz+1)}{8} & \ge & \sqrt{x*2y*y*2z*xz*1} \\\\ \dfrac{(x+2y)(y+2z)(xz+1)}{8} & \ge & \sqrt{4x^2y^2z^2} \\\\ \dfrac{(x+2y)(y+2z)(xz+1)}{8} & \ge & 2xyz \\\\ (x+2y)(y+2z)(xz+1) & \ge & 16*xyz \quad | \quad xyz=1 \\\\ (x+2y)(y+2z)(xz+1) & \ge & 16*1 \\\\ \mathbf{(x+2y)(y+2z)(xz+1)} & \ge & \mathbf{16} \\ \hline \end{array}\)

The minimum value of \((x+2y)(y+2z)(xz+1)\) is \(\mathbf{16}\)

If

\(f(c)=\dfrac{3}{2c-3}\), find \(\dfrac{kn^2}{lm} \)

when \(f^{-1}(c)\times c \times f(c)\) equals the simplified fraction \(\dfrac{kc+l}{mc+n}\),
where \(k\), \(l\), \(m\), and \(n\) are integers.

My attempt:

\(\begin{array}{|rcll|} \hline f(c) &=& \dfrac{3}{2c-3} \quad | \quad c=f^{-1}(c) \\\\ f\Big(f^{-1}(c)\Big) &=& \dfrac{3}{2f^{-1}(c)-3} \quad | \quad f\Big(f^{-1}(c)\Big) = c \\\\ c &=& \dfrac{3}{2f^{-1}(c)-3} \\\\ c\Big(2f^{-1}(c)-3\Big) &=& 3 \\\\ 2cf^{-1}(c)-3c &=& 3 \\\\ 2cf^{-1}(c) &=& 3c+3 \\\\ cf^{-1}(c) &=& \dfrac{3c+3}{2} \\\\ \mathbf{f^{-1}(c)\times c} &=& \mathbf{\dfrac{3c+3}{2}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline f^{-1}(c)\times c \times f(c) &=& \dfrac{(3c+3)}{2}\times \dfrac{3}{(2c-3)} \\\\ f^{-1}(c)\times c \times f(c) &=& \dfrac{3(3c+3)}{2(2c-3)} \\\\ \mathbf{f^{-1}(c)\times c \times f(c)} &=& \mathbf{\dfrac{9c+9}{4c-6}} \\ \hline && \boxed{k=9,\ l=9,\ m=4,\ n=-6} \\\\ \dfrac{kn^2}{lm} &=& \dfrac{9(-6)^2}{9*4} \\\\ \dfrac{kn^2}{lm} &=& \dfrac{36}{4} \\\\ \mathbf{\dfrac{kn^2}{lm}} &=& \mathbf{9} \\ \hline \end{array}\)