heureka

If the midpoints of the sides AB and AC of triangle ABC are (3,5) and (-3,-3), respectively, then find the length of BC.

\(\text{Let $P_1 =(x_1,\ y_1)=(3,\ 5)$ } \\ \text{Let $P_2 =(x_2,\ y_2)=(-3,\ -3)$ }\)

\(\begin{array}{|rcll|} \hline \dfrac{x_A+x_B}{2} &=& x_1 \\ x_A &=& 2x_1-x_B \\ &=& 2*3-x_B \\ \mathbf{x_A} &=& \mathbf{6-x_B} \\\\ \dfrac{y_A+y_B}{2} &=& y_1 \\ y_A &=& 2y_1-y_B \\ &=& 2*5-x_B \\ \mathbf{y_A} &=& \mathbf{10-y_B} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \dfrac{x_A+x_C}{2} &=& x_2 \\ x_A &=& 2x_2-x_C \\ &=& 2*(-3)-x_C \\ \mathbf{x_A} &=& \mathbf{-6-x_C} \\\\ \dfrac{y_A+y_C}{2} &=& y_2 \\ y_A &=& 2y_2-y_C \\ &=& 2*(-3)-y_C \\ \mathbf{y_A} &=& \mathbf{-6-y_C} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline x_A = 6-x_B &=& -6-x_C \\ \mathbf{x_B -x_C} &=& \mathbf{12} \\\\ y_A = 10-y_B &=& -6-y_C \\ \mathbf{y_B -y_C} &=& \mathbf{16} \\\\ BC &=& \sqrt{12^2+16^2} \\ \mathbf{BC} &=& \mathbf{20} \\ \hline \end{array}\)

\(P(x)\) is a polynomial with real coefficients such that \(P(2+i)=4-3i\). Find \(P(2-i)\).

\(\begin{array}{|rcll|} \hline \mathbf{P(x)} &=& \mathbf{ax+b} \\\\ P(2+i) &=& a(2+i)+b \\ P(2+i) &=& 2a+ai+b \\ P(2+i) &=& 2a+b+ai \quad | \quad P(2+i) = 4-3i \\ 4-3i &=& \underbrace{2a+b}_{=4}+\underbrace{a}_{=-3}i \qquad \text{compare} \\\\ \mathbf{a} &=& -3 \\\\ 2a+b &=& 4 \\ 2(-3)+b &=& 4 \\ -6+b &=& 4 \\ \mathbf{b} &=& 10 \\\\ \mathbf{P(x)} &=& \mathbf{-3x+10} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{P(x)} &=& \mathbf{-3x+10} \\ P(2-i) &=& -3(2-i)+10 \\ P(2-i) &=& -6+3i +10 \\ \mathbf{P(2-i)} &=& \mathbf{4+3i} \\ \hline \end{array}\)

1.)

\(8\mid 5^{2n} - 3^{2n} \)

\(\text{$5^{2n} – 3^{2n}$ ist durch 8 teilbar für alle $n \geq 0$} \)

Induktionsanfang:

\(\text{$n = 0: 5^0 – 3^0 = 1 – 1 = 0$ ist durch $8$ ohne Rest teilbar.}\)

Induktionsschluss:

\(\begin{array}{|rcll|} \hline 5^{2(n+1)} – 3^{2(n+1)} &=& 5^{2n+2} – 3^{2n+2} \\ &=& 25·5^{2n} – 9·3^{2n} \\ &=& 24·5^{2n} + 1·5^{2n} – 8·3^{2n} – 1·3^{2n} \\ &=& (5^{2n} – 3^{2n}) + 8(3·5^{2n} – 3^{2n}) \\ \hline \end{array}\)

ist durch 8 teilbar, da der erste Summand durch 8 teilbar ist nach Induktionsvoraussetzung
und der zweite Summand ein ganzzahliges Vielfaches von 8 ist.

Two trains started moving one from point A and the other from point B against each other at the same time
for a trip length of 240 km.
The first train speed is 80 km/h and the second one is 100 km/h.
What is the distance between the two trains 30 minutes before they meet?

\(\begin{array}{|rcll|} \hline y_B &=& -100t+240 \\ y_A &=& 80t \\ \hline \end{array}\)

meeting time

\(\begin{array}{|rcll|} \hline \mathbf{y_A} &=& \mathbf{y_B} \\\\ 80t_{meet} &=& -100t_{meet}+240 \\ 180t_{meet} &=& 240 \\ t_{meet} &=& \dfrac{240}{180} \\ \mathbf{t_{meet}} &=& \mathbf{ \dfrac{4}{3}\ h } \qquad ( t_{meet}= 80\ \text{minutes} )\\ \hline \end{array}\)

\(t_{meet}- 30\ \text{minutes} = 50\ \text{minutes}\)

\(\begin{array}{|rcll|} \hline y_B &=& -100 \left(\dfrac{50}{60}\right)+240 \\ y_A &=& 80\left(\dfrac{50}{60}\right) \\ \hline y_B - y_A &=& -100 \left(\dfrac{5}{6}\right)+240 - 80\left(\dfrac{5}{6}\right) \\ &=& -180 \left(\dfrac{5}{6}\right)+240 \\ &=& -150+240 \\ &=& \mathbf{90\ \text{km}} \\ \hline \end{array}\)

The distance between the two trains 30 minutes before they meet is \(\mathbf{90\ \text{km}}\)

How many even natural-number factors does \(n = (2^2)(3^1)(7^2)\) have?

\(\begin{array}{|llcll|} \hline 2^{\color{red}2} * & 3^{\color{blue}1} * 7^{\color{green}2} \\\\ ({\color{red}2}+1 ) *& ({\color{blue}1}+1) * ({\color{green}2}+1) = 3*2*3 &=& 18~ \text{divider} \\ & ({\color{blue}1}+1) * ({\color{green}2}+1) = 2*3 &=& 6~ \text{odd divider} \qquad (\text{odd*odd=odd})\\ \hline &&& 18-6 = \mathbf{12}~ \text{even divider} \\ \hline \end{array} \)

Find all solutions of the equation \(|x^2-20x-1|=20\).

a)
\(x^2-20x-1 = 20\)

\(\begin{array}{|rcll|} \hline x^2-20x-1 &=& 20 \\ x^2-20x-21 &=& 0 \\\\ x &=& \dfrac{20\pm\sqrt{400-4(-21)} }{2} \\ x &=& \dfrac{20\pm\sqrt{484} }{2} \\ x &=& \dfrac{20\pm 22 }{2} \\ \\ x_1 &=& \dfrac{42}{2} \\ \mathbf{x_1} &=& \mathbf{21} \\\\ x_2 &=& \dfrac{-2}{2} \\ \mathbf{x_2} &=& \mathbf{-1} \\ \hline \end{array}\)

b)
\(x^2-20x-1 = -20\)

\(\begin{array}{|rcll|} \hline x^2-20x-1 &=& -20 \\ x^2-20x+19 &=& 0 \\\\ x &=& \dfrac{20\pm\sqrt{400-4(19)} }{2} \\ x &=& \dfrac{20\pm\sqrt{324} }{2} \\ x &=& \dfrac{20\pm 18 }{2} \\ \\ x_3 &=& \dfrac{38}{2} \\ \mathbf{x_3} &=& \mathbf{19} \\\\ x_4 &=& \dfrac{2}{2} \\ \mathbf{x_4} &=& \mathbf{1} \\ \hline \end{array}\)

Let f(x) be a polynomial with integer coefficients such that \(f(5) = f(7) = 20\).
What is the smallest possible value of \(|f(0)|\)?

a) 

\(f(x) = 20 + (5-x)(7-x) \qquad \Big(~ f(x)=x^2-12x+55 ~\Big)\)

\(\begin{array}{|rcll|} \hline f(0) &=& 20 + (5-0)(7-0) \\ f(0) &=& 20 + 35 \\ f(0) &=& 55 \\ \mathbf{|f(0)|} &=& \mathbf{55} \\ \hline \end{array}\)

b)
\(f(x) = 20 - (5-x)(7-x) \qquad \Big(~ f(x)=-x^2+12x-15 ~\Big)\)

\(\begin{array}{|rcll|} \hline f(0) &=& 20 - (5-0)(7-0) \\ f(0) &=& 20 - 35 \\ f(0) &=& -15 \\ \mathbf{|f(0)|} &=& \mathbf{15} \\ \hline \end{array}\)

The smallest possible value of \(|f(0)|\) is 15

Thank you, CPhill !

Find the equation of the line that passes through the points \((-2,1)\) and \((3,3)\).
Write your answer in the form \(y=mx+b\) .

\(\begin{array}{|rcll|} \hline \dfrac{y-1}{x-(-2)} &=& \dfrac{3-1}{3-(-2)} \\\\ \dfrac{y-1}{x+2} &=& \dfrac{2}{5} \\\\ y-1 &=& \dfrac{2}{5}(x+2) \\\\ y &=& \dfrac{2}{5}(x+2) + 1 \\\\ y &=& \dfrac{2}{5}x + \dfrac{4}{5} + 1 \\\\ \mathbf{y} &=& \mathbf{\dfrac{2}{5}x + \dfrac{9}{5}} \\ \hline \end{array}\)

Find the x-intercept of the line 3x-5y=7.

\(\begin{array}{|rcll|} \hline 3x-5y &=& 7 \quad | \quad x\text{-intercept}~(y=0) \\\\ 3x-5*0 &=& 7 \\ 3x &=& 7 \\ \mathbf{x} &=& \mathbf{\dfrac{7}{3}} \qquad (x=2.\bar{3}) \\ \hline \end{array} \)

\((2.\bar{3},~0)\)