heureka

Thank you, CPhill

Hello CPhill,

"how was that method derived":

1.)

This is the Legendre Theorem from 1808:

Der Grenzwert ist -0,25, allerdings kann ich das irgendwie nicht zeigen :O
\(\lim \limits_{x\rightarrow 0} \dfrac{1-\sqrt{1+2x^2}}{4x^2}\)

\(\begin{array}{|rcll|} \hline && \mathbf{ \lim \limits_{x\rightarrow 0} \dfrac{1-\sqrt{1+2x^2}}{4x^2} } \\\\ &=& \dfrac14 \cdot \lim \limits_{x\rightarrow 0} \dfrac{1-\sqrt{1+2x^2}}{x^2} \\\\ &=& \dfrac14 \cdot \lim \limits_{x\rightarrow 0} \dfrac{1-\sqrt{1+2x^2}}{x^2}\cdot \left( \dfrac{1+\sqrt{1+2x^2}}{1+\sqrt{1+2x^2}} \right) \\\\ &=& \dfrac14 \cdot \lim \limits_{x\rightarrow 0} \dfrac{1-(1+2x^2)}{x^2(1+\sqrt{1+2x^2})} \\\\ &=& \dfrac14 \cdot \lim \limits_{x\rightarrow 0} \dfrac{1-1-2x^2}{x^2(1+\sqrt{1+2x^2})} \\\\ &=& \dfrac14 \cdot \lim \limits_{x\rightarrow 0} \dfrac{-2x^2}{x^2(1+\sqrt{1+2x^2})} \\\\ &=& \dfrac14 \cdot \lim \limits_{x\rightarrow 0} \dfrac{-2}{(1+\sqrt{1+2x^2})} \\\\ &=& -\dfrac14 \cdot \lim \limits_{x\rightarrow 0} \dfrac{2}{(1+\sqrt{1+2x^2})} \quad | \quad x=0 \\\\ &=& -\dfrac14 \cdot \dfrac{2}{(1+\sqrt{1+0})} \\\\ &=& -\dfrac14 \cdot \dfrac{2}{2} \\\\ &\mathbf{=}& \mathbf{-\dfrac14} \\ \hline \end{array}\)

int the prime factorization of 109! what is the exponent of 3?

\(\begin{array}{|rcl|} \hline \text{The exponent of $3$ is } \\ && \lfloor \frac{109}{3} \rfloor +\lfloor \frac{109}{9} \rfloor +\lfloor \frac{109}{27} \rfloor +\lfloor \frac{109}{81} \rfloor \\ &=& 36 + 12 +4 +1\\ &=& 53\\ \hline \end{array}\)

7. 

As shown in the diagram, points B and D are on different sides of line AC. 

We know that Angle B=2*Angle D=60 degrees and that AC=4sqrt(3). 

What is the distance between the circumcenters of Triangle ABC and Triangle ADC?

line segments:

\(\text{Let $AC =4\sqrt{3}$ } \\ \text{Let $AG=GC =2\sqrt{3}$ } \\ \text{Let $AE=EC =r$ } \\ \text{Let $AF=FC =R$ } \\ \text{Let $FE = EG+GF $ } \)

The distance between the circumcenters of Triangle ABC and Triangle ADC \(= FE\)

angle:

\(\text{Let $\angle ABC = 60^{\circ} $ } \\ \text{Let $\angle ADC = \frac{\angle ABC}{2} = 30^{\circ} $ } \\ \text{Let $\angle AEC = 2*\angle ABC = 120^{\circ} $ } \\ \text{Let $\angle AFC = 2*\angle ADC = 60^{\circ} $ } \)

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

\(\begin{array}{|rcll|} \hline AC^2 &=& 2r^2\Big(1-\cos(120^{\circ})\Big) \quad \text{$\cos$-rule} \quad | \quad \cos(120)^{\circ} = -0.5 \\ (4\sqrt{3})^2 &=& 2r^2(1+0.5) \\ 48 &=& 2r^2(1.5) \\ 24 &=& r^2(1.5) \\ r^2 &=& 16 \\\\ AG^2 + EG^2 &=& r^2 \\ (2\sqrt{3})^2 + EG^2 &=& 16 \\ 12 + EG^2 &=& 16 \\ EG^2 &=& 4 \\ \mathbf{EG} & \mathbf{=}& \mathbf{2} \\ \hline \end{array}\)

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

\(\begin{array}{|rcll|} \hline AC^2 &=& 2R^2\Big(1-\cos(60^{\circ})\Big) \quad \text{$\cos$-rule} \quad | \quad \cos(60)^{\circ} = 0.5 \\ (4\sqrt{3})^2 &=& 2R^2(1-0.5) \\ 48 &=& 2R^2(0.5) \\ 24 &=& R^2(0.5) \\ R^2 &=& 48 \\\\ AG^2 + GF^2 &=& R^2 \\ (2\sqrt{3})^2 + GF^2 &=& 48 \\ 12 + GF^2 &=& 48 \\ GF^2 &=& 36 \\ \mathbf{GF} & \mathbf{=}& \mathbf{6} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline FE &=& EG+GF \\ &=& 2+6 \\ \mathbf{FE} & \mathbf{=} & \mathbf{8} \\ \hline \end{array}\)

The distance between the circumcenters of Triangle ABC and Triangle ADC is 8

George is about to get a certain amount of change less than one dollar from the cash register.

If he gets the most quarters possible and the rest in pennies, he would need to receive 3 pennies to meet the amount.

If he gets the most dimes possible and the rest in pennies, he would need to receive 8 pennies to meet the amount.

What is the sum, in cents, of the possible amounts of change that he is trying to get?

\(\text{Let $n =$ quarters}\\ \text{Let $m =$ dimes}\)

Formula:

\(\begin{array}{|rclrcl|} \hline \mathbf{25n+3} &=& \mathbf{10m+8} \\ 10m &=& 25n-5 \\\\ m &=& \dfrac{25n-5}{10} \\\\ &=& \dfrac{20n+5n-5}{10} \\\\ &=& \dfrac{20n}{10}+\dfrac{5n-5}{10} ~|~ \text{cut off whole parts} \\\\ m &=& 2n+ \underbrace{\dfrac{5n-5}{10}}_{=a} \qquad m = 2n+a \qquad (1) \\\\ & a &= \dfrac{5n-5}{10} \\\\ &10a &= 5n-5 \\\\ &5n &= 10a+5 \quad | \quad :5 \\\\ &\mathbf{n} & \mathbf{=} \mathbf{2a+1} \qquad (2) \\\\ m &=& 2n +a \\ &=& 2(2a+1) +a \\\\ \mathbf{m} & \mathbf{=}& \mathbf{5a+2} \qquad (3) \\ \hline \end{array}\)

conclusion:

\(\begin{array}{|rcll|} \hline \mathbf{n} & \mathbf{=}& \mathbf{2a+1} \qquad a \in \mathbb{Z} \\ \mathbf{m} & \mathbf{=}& \mathbf{5a+2} \qquad a \in \mathbb{Z} \\ \hline \end{array}\)

solution:

\(\begin{array}{|c|r|r|r|} \hline a & \text{quarters} & \text{dimes} & \text{pennies} \\ \hline 0 & 1 & 2& 28 \\ 1 & 3 & 7& 78 \\ 2 & 5 & 12 & 128 > $1~ ! \\ \hline \end{array} \)

G is the centroid of equilateral Triangle ABC. D, E, and F are midpoints of the sides as shown.
P, Q, and R are the midpoints of Line AG, Line BG, and Line CG, respectively.
If AB=sqrt(3) what is the perimeter of hexagon DREPFQ?

\(\text{Let $AB=\sqrt{3}$} \\ \text{Let $CE=\frac12\sqrt{3}$} \\ \text{Let Point $C=(0,0)$} \\ \text{Let Point $A=(\sqrt{3},0)$} \\ \text{Let Point $B=(\frac12\sqrt{3},\frac32)$} \\\)

\(\begin{array}{|rcll|} \hline BE^2 + \left(\frac12 \sqrt{3} \right)^2 &=& (\sqrt{3})^2 \\ BE^2 &=& 3 - \frac34 \\ BE^2 &=& \frac94 \\ \mathbf{BE} & \mathbf{=} & \mathbf{\frac32} \\ \hline \end{array} \)

\(\text{Let Point $B=(\frac12\sqrt{3},\frac32)$}\)

\(\begin{array}{|rcll|} \hline G &=& \left( \frac{0+\sqrt{3}+\frac12\sqrt{3}}{3}, \frac{0+\sqrt{0}+\frac32}{3} \right) \\ G &=& \left( \frac12\sqrt{3}, \frac12\right) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline s &=& \frac12 \\ \text{perimeter}_{\text{of hexagon }DREPFQ} &=& 6s \\ &=& 6\cdot \frac12 \\ &\mathbf{=}& \mathbf{3} \\ \hline \end{array}\)

What is the residue modulo 13 of the sum of the modulo 13 inverses of the first 12 positive integers?
Express your answer as an integer from 0 to 12 , inclusive.

\(\begin{array}{|rcll|} \hline 1\cdot { \color{red}1} &\equiv& 1 \pmod{13} \\ 2\cdot 2^{-1} \equiv \pmod{13} \equiv 2\cdot { \color{red}2^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 3\cdot 3^{-1} \equiv \pmod{13} \equiv 3\cdot { \color{red}3^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 4\cdot 4^{-1} \equiv \pmod{13} \equiv 4\cdot { \color{red}4^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 5\cdot 5^{-1} \equiv \pmod{13} \equiv 5\cdot { \color{red}5^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 6\cdot 6^{-1} \equiv \pmod{13} \equiv 6\cdot { \color{red}6^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 7\cdot 7^{-1} \equiv \pmod{13} \equiv 7\cdot { \color{red}7^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 8\cdot 8^{-1} \equiv \pmod{13} \equiv 8\cdot { \color{red}8^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 9\cdot 9^{-1} \equiv \pmod{13} \equiv 9\cdot { \color{red}9^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 10\cdot 10^{-1} \equiv \pmod{13} \equiv 10\cdot { \color{red}10^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 11\cdot 11^{-1} \equiv \pmod{13} \equiv 11\cdot { \color{red}11^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 12\cdot 12^{-1} \equiv \pmod{13} \equiv 12\cdot { \color{red}12^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ \hline \end{array}\)

\(\begin{array}{|ll|} \hline & \text{sum}_{\text{of the modulo 13 inverses of the first 12 positive integers}} \\ =& 1^{11}+2^{11}+3^{11}+4^{11}+5^{11}+6^{11}+7^{11}+8^{11}+9^{11}+10^{11}+11^{11}+12^{11} \pmod{13} \\ =& 1+7+9+10+8+11+2+5+3+4+6+12 \pmod{13} \\ =& 1+2+3+4+5+6+7+8+9+10+11+12 \pmod{13} \\ =& \dfrac{1+12}{2}\cdot 12 \pmod{13} \\ =& 13 \cdot 6 \pmod{13} \\ \mathbf{=}& \mathbf{0 \pmod{13}} \\ \hline \end{array}\)

Let \(\large{p(x) = 2x^2 + 7x + c}\),
where \(c\) is not equal to \(0\).
If \(\large{r_1}\) and \(\large{r_2}\) are the roots of \(\large{p(x)}\),

then find the value of \(\large{\dfrac{1}{r_1}} + \large{\dfrac{1}{r_2}}\)in terms of \(\large{c}\).

\(\begin{array}{|rcll|} \hline p\left(\dfrac{1}{x}\right) = 2\left(\dfrac{1}{x}\right)^2 + 7\dfrac{1}{x} + c &=& 0\\\\ \dfrac{2}{x^2} + \dfrac{7}{x} + c &=& 0 \quad | \quad \cdot x^2 \\ 2 + 7x + cx^2 &=& 0 \\ cx^2+7x+2 &=& 0 \\\\ x &=& \dfrac{-7\pm\sqrt{7^2-4\cdot c\cdot 2}}{2c} \\ x &=& \dfrac{-7\pm\sqrt{49-4c}}{2c} \\ \dfrac{1}{r_1} + \dfrac{1}{r_2} &=& \dfrac{-7+\sqrt{49-4c}}{2c} + \dfrac{-7-\sqrt{49-4c}}{2c} \\\\ &=& \dfrac{-7+\sqrt{49-4c}-7-\sqrt{49-4c}}{2c} \\\\ &=& \dfrac{-14}{2c} \\\\ \mathbf{\dfrac{1}{r_1} + \dfrac{1}{r_2} } &\mathbf{=}& \mathbf{ -\dfrac{7}{c} } \\ \hline \end{array}\)

If

\(\large{f(x)=\dfrac{a}{x+2}}\),

solve for the value of a so that

\(\large{f(0)=f^{-1}(3a)}\).

\(\begin{array}{|rcll|} \hline f\left(~ f^{-1}(x) ~\right) &=& x \quad & | \quad x = 3a \\ f\left(~ f^{-1}(3a) ~\right) &=& 3a \quad & | \quad f^{-1}(3a) = f(0) \\ f\left(~ f(0) ~\right) &=& 3a \quad & | \quad f(0) = \dfrac{a}{0+2}=\dfrac{a}{2} \\ f\left(~ \dfrac{a}{2} ~\right) &=& 3a \quad & | \quad f\left(\dfrac{a}{2} \right) = \dfrac{a}{\dfrac{a}{2}+2}=\dfrac{2a}{a+4} \\ \dfrac{2a}{a+4} &=& 3a \\ \dfrac{2 }{a+4} &=& 3 \\ 2 &=& 3 (a+4) \\ 2 &=& 3a+12 \\ 3a &=& -10 \\ \mathbf{a} &\mathbf{=}& \mathbf{-\dfrac{10}{3}} \\ \hline \end{array}\)