heureka

How many weeks are there in \(8!\) minutes?

\(\begin{array}{|rcll|} \hline 8!\ \text{min.} &=& 1*2*3*4*5*6*7*8\ \text{min.} \\ &=& 40320\ \text{min.} \\ \hline && 40320\ \text{min.} \times \dfrac{1\ \text{h}}{60\ \text{min.}} \times \dfrac{1\ \text{d}}{24\ \text{h}} \times \dfrac{1\ \text{week}}{7\ \text{d}} \\ &=& \dfrac{ 40320} {60\cdot 24\cdot 7} \ \text{weeks} \\ &=& \mathbf{4 \ \text{weeks} } \\ \hline \end{array} \)

8! minutes are 4 weeks.

Thank you, CPhill !

In triangle ABC, side AB = 20, AC = 11, and BC = 13.  

Find the diameter of the semicircle inscribed in triangle ABC, whose diameter lies on AB, and that is tangent to AC and BC. 

1. sin-rule:

\(\begin{array}{|lrcll|} \hline (1) & \dfrac{\sin(A)}{13} &=& \dfrac{\sin(B)}{11} \\ & \mathbf{\sin(A)} &=& \mathbf{\dfrac{13}{11} \sin(B)} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (2) & 20 &=& x+y+2r \\ & \mathbf{x} &=& \mathbf{20-y-2r} \\ \hline \end{array}\)

2. 

\(\begin{array}{|lrcll|} \hline (3) & \mathbf{\sin(B)} &=& \mathbf{\dfrac{r}{y+r}} \\\\ (4) & \sin(A) &=& \dfrac{r}{x+r} \quad | \quad \mathbf{ \sin(A)= \dfrac{13}{11} \sin(B) } \\\\ & \dfrac{13}{11} \sin(B) &=& \dfrac{r}{x+r} \quad | \quad \mathbf{x=20-y-2r} \\\\ & \dfrac{13}{11} \sin(B) &=& \dfrac{r}{20-y-2r+r} \\\\ & \mathbf{\dfrac{13}{11} \sin(B)} &=& \mathbf{\dfrac{r}{20-y-r}} \\ \hline \end{array}\)

3. cos-rule:

\(\begin{array}{|lrcll|} \hline (5) & 11^2 &=& 13^2+20^2-2*13*30*\cos(B) \\\\ & \cos(B) &=& \dfrac{13^2+20^2-11^2}{2*13*30} \\ & \cos(B) &=& \dfrac{56}{65} \\\\ & \sin(B) &=& \sqrt{1-\cos^2(B)} \\ & \sin(B) &=& \sqrt{1- \left(\dfrac{56}{65}\right)^2 } \\ & \mathbf{\sin(B)} &=& \mathbf{ \dfrac{33}{65}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (3) & \mathbf{\sin(B)} &=& \mathbf{\dfrac{r}{y+r}} \quad | \quad \mathbf{\sin(B)=\dfrac{33}{65}} \\ & \dfrac{33}{65} &=& \dfrac{r}{y+r} \\ & 33(y+r) &= 65r \\ & 33y &=& 65r - 33r \\ & 33y &=& 32r \\ (6) & \mathbf{y} &=& \mathbf{ \dfrac{32}{33}r } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline (4) & \mathbf{\dfrac{13}{11} \sin(B)} &=& \mathbf{\dfrac{r}{20-y-r}} \quad | \quad \mathbf{\sin(B)=\dfrac{33}{65}},\ \mathbf{y=\dfrac{32}{33}r } \\\\ & \dfrac{13}{11}*\dfrac{33}{65} &=& \dfrac{r}{20-\dfrac{32}{33}r-r} \\\\ & \dfrac{13*3}{65} &=& \dfrac{r}{20-\dfrac{65}{33}r } \\\\ & \dfrac{39}{65} &=& \dfrac{r}{20-\dfrac{65}{33}r } \\\\ & \left( 20-\dfrac{65}{33}r\right)*39 &=& 65r \\\\ & 20*39 - \dfrac{65*39}{33}r &=& 65r \\\\ & 65r + \dfrac{65*39}{33}r &=& 20*39 \\ \\ & 65r*\left(1 + \dfrac{ 39}{33} \right) &=& 20*39 \\ \\ & \dfrac{65*72}{33}r &=& 20*39 \\ \\ & r &=& \dfrac{33*20*39 } {65*72}\\ \\ & \mathbf{r} &=& \mathbf{5.5} \\ \hline \end{array}\)

The diameter of the semicircle \(= 2r = 2*5.5 = \mathbf{11}\)

Vollständige Induktion \(\sum \limits_{k=1}^{2^n} \dfrac{1}{k} \geq 1 + \dfrac{n}{2}\)

Induktionsanfang:

\(\begin{array}{|lrcll|} \hline n=1: & \sum \limits_{k=1}^{2^1} \dfrac{1}{k} = \dfrac{1}{1} + \dfrac{1}{2} = \dfrac{3}{2} \geq \dfrac{3}{2} = \dfrac{1}{1} + \dfrac{1}{2} \ \checkmark \\ \hline \end{array}\)

Induktionsannahme: \(\sum \limits_{k=1}^{2^n} \dfrac{1}{k} \geq 1 + \dfrac{n}{2}\)

Induktionsschritt:

\(\sum \limits_{k=1}^{2^{n+1}} \dfrac{1}{k} \geq 1 + \dfrac{n+1}{2} \\ \)

\(\begin{array}{|rcll|} \hline \mathbf{\sum \limits_{k=1}^{2^{n+1}} \dfrac{1}{k} } &\geq& \mathbf{ 1 + \dfrac{n+1}{2} =1 + \dfrac{n}{2} +\dfrac{1}{2} } \\\\ \hline \\ \sum \limits_{k=1}^{2^{n+1}} \dfrac{1}{k} &=& \sum \limits_{k=1}^{2^{n}} \dfrac{1}{k} + \sum \limits_{k=2^{n}+1}^{2^{n+1}} \dfrac{1}{k} \\ &\geq& 1 + \dfrac{n}{2} + \sum \limits_{k=2^{n}+1}^{2^{n+1}} \underbrace{\dfrac{1}{k}}_{\geq \dfrac{1}{2^{n+1}} ^{1)} } \quad \text{nach Induktionsannahme} \\ &\geq& 1 + \dfrac{n}{2} + \sum \limits_{k=2^{n}+1}^{2^{n+1}} \dfrac{1}{2^{n+1} } \quad 2^n \text{Summanden} ^{2)} \\ &=& 1 + \dfrac{n}{2} + 2^n\left( \dfrac{1}{2^{n+1} } \right) \\ &=& 1 + \dfrac{n}{2} + 2^{n-n-1} \\ &=& 1 + \dfrac{n}{2} +\dfrac{1}{2} \ \checkmark\\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline ^{1)} \text{Beispiel n=2: } & \sum \limits_{k=2^{2}+1}^{2^{2+1}} \dfrac{1}{k} = \sum \limits_{k=5}^{8} \dfrac{1}{k} = \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8} \geq 4\cdot \dfrac{1}{8} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline ^{2)} & (2^n+1)-2^n &\ldots& 2^{n+1} - (2^n) \\ & 1 &\ldots& 2\cdot 2^{n} - (2^n) \\ & 1 &\ldots& 2^n \\ \hline \end{array}\)

1. How many paths of minimum length are there from a to b  in the grid below?

Find the polynomial \(p(x)\), with real coefficients, such that  \(p(x^3) - p(x^3 - 2) = [p(x)]^2 + 12\)
for all real numbers \(x\).

\(p(x) = 6x^3-6\)

John has an ample supply of 10-cent stamps and 15-cent stamps. 

He will buy an ample supply of stamps of one more denomination, which is a whole number. 

He needs to be able to combine the stamps to make any value of 30 cents or more. 

In cents, what is the greatest denonmination of stamp he can buy which will enable him to do this?

My attempt:

\(\begin{array}{|rcll|} \hline \text{cent} & \text{any value of } x \text{ cents or more} \\ \hline 2 & x = 14 \\ 3 & x = 18 \\ 4 & x = 22 \\ \color{red}6 & \color{red}x = 30 \\ 7 & x = 34 \\ 8 & x = 38 \\ 9 & x = 42 \\ 11 & x = 90 \\ \hline \end{array} \)

Find the product of all real numbers \(n\) which satisfy \(|n^2 - 9n + 20| = |16 - n^2|\).

\(\begin{array}{|lrcll|} \hline & \mathbf{|n^2 - 9n + 20|} &=& \mathbf{|16 - n^2|} \quad | \quad \text{square both sides} \\\\ & (n^2 - 9n + 20)^2 &=& (16 - n^2)^2 \quad | \quad n^2 - 9n + 20 =(n-5)(n-4) \\ & (n-5)^2(n-4)^2 &=& (16 - n^2)^2 \quad | \quad (16 - n^2) =(4-n)(4+n) \\ & (n-5)^2(n-4)^2 &=& (4-n)^2(4+n)^2 \\ & (n-5)^2(n-4)^2 - (4-n)^2(4+n)^2 &=& 0 \\ & \mathbf{(n-4)^2\Big( (n-5)^2 - (4+n)^2 \Big)} &=& \mathbf{0} \\ \hline 1. & \mathbf{(n-4)^2} &=& \mathbf{0} \\ & n-4 &=& 0 \\ & \mathbf{n_1} &=& \mathbf{4} \\ \hline 2. & \mathbf{(n-5)^2 - (4+n)^2} &=& \mathbf{0} \\ & (n-5)^2 &=& (4+n)^2 \\ & n-5 &=& \pm \sqrt{(4+n)^2 } \\ & \mathbf{n-5} &=& \mathbf{\pm (4+n)} \\\\ 3. & n-5 &=& 4+n \\ & -5 &\neq& 4 \quad | \quad \text{no solution} \\\\ 4. & n-5 &=& -(4+n) \\ & n-5 &=& -4 - n \\ & 2n &=& 1 \\ & \mathbf{n_2} &=& \mathbf{\dfrac{1}{2}} \\ \hline \end{array}\)

The product of all real numbers \(n\) is \(n_1\times n_2 = 4\times \dfrac{1}{2} = \mathbf{2}\)

There are 8 ways that 1 x 1 and 2 x 2 squares can be arranged in a 2 x 5 grid:

Find the number of ways that 1 x 1 and 2 x 2 squares can be arranged in a 2 x 50 grid.

\(\begin{array}{|c|c|c|c|} \hline & \text{Starts with} & \text{Starts with} \\ \text{grid:} & 2\times 1 & 2\times 2 & \text{sum} \\ \hline 2 \times 1 & 1 & 0 & 1 = F_{2} \\ \hline 2 \times 2 & 1 & 1 & 2 = F_{3} \\ \hline 2 \times 3 & 2 & 1 & 3 = F_{4} \\ \hline 2 \times 4 & 3 & 2 & 5 = F_{5} \\ \hline 2 \times 5 & 5 & 3 & 8 = F_{6} \\ \hline 2 \times 6 & 8 & 5 & 13 = F_{7} \\ \hline 2 \times 7 & 13 & 8 & 21 = F_{8} \\ \hline \ldots & \ldots \\ 2 \times 10 & 55 & 34 & 89 = F_{11} \\ \hline \ldots & \ldots \\ \hline 2 \times k & F_k & F_{k-1} & F_{k+1} \\ \hline 2\times 50 & F_{50} & F_{49} & 20365011074= F_{51} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \mathbf{ F_k \text{ is the } k^{th} \text{ Fibonacci Number:}} \\ F_1 = 1,\ F_2 = 1,\ F_3 = 2,\ F_4 = 3,\ F_5 = 5, \\ F_6 = 8,\ F_7 = 13,\ F_8 = 21,\ F_9 = 34,\ F_{10} = 55,\ldots , \\ F_{49} = 7778742049,\ F_{50} =12586269025,\ F_{51} = 20365011074 \\ \hline \end{array}\)

hint:

Fibonacci Number as Sum of Binomial Coefficients:

\(\begin{array}{rcll} \displaystyle F_n &=& \displaystyle \sum_{k \mathop = 0}^{\Big\lfloor {\dfrac {n - 1} 2}\Big\rfloor } \dbinom {n - k - 1} k \\ &=&\displaystyle \binom {n - 1} 0 + \binom {n - 2} 1 + \binom {n - 3} 2 + \dotsb + \binom {n - j} {j - 1} + \binom {n - j - 1} j \quad \text{ where } \quad j = \Big\lfloor {\dfrac {n - 1} 2} \Big\rfloor \\ \end{array}\)

Thank you, Melody !