heureka

A clock each time it increased 20 minutes , it stopped for 10min.
How long will it take the minutes arrow to do 2 full rounds ?

\(\begin{array}{|rcll|} \hline \text{Time} &=& {\color{red}20}\ \text{min.} + 10\ \text{min.} +{\color{red}20}\ \text{min.} + 10\ \text{min.} + {\color{red}20}\ \text{min.} \quad | \quad \text{minutes arrow first round } 60\ \text{min.}\\ && + 10\ \text{min.}+{\color{red}20}\ \text{min.} + 10\ \text{min.} + {\color{red}20}\ \text{min.} + 10\ \text{min.}+ {\color{red}20}\ \text{min.} \quad | \quad \text{minutes arrow second round } 120\ \text{min.}\\ \text{Time} &=& 170\ \text{min.} \\ \hline \end{array}\)

Help with absolute value

1)

Find all real numbers x that satisfy the equation \(\large{\mathbf{|x+4| + |x-7| = |2x-1|}}\).

There are three discontinuities at \(x=-4,\ x=7,\ x=\dfrac{1}{2}\).
These digits divide the number range from \(-\infty\) to \(+\infty\) into the four ranges \(x<-4\),   \(-4\leq x < \dfrac{1}{2}\),   \(\dfrac{1}{2}\leq x <7\)  and  \(x\geq 7\).

\(\begin{array}{lrcll} x<-4 :& -(x+4)-(x-7)+(2x-1) &=& 0 \\ & -x-4-x+7+2x -1 &=& 0 \\ &6 &=& 0 \quad | \quad \text{no solution!} \\ -4\leq x < \dfrac{1}{2} : & (x+4)-(x-7)+(2x-1) &=& 0 \\ & x+4-x+7+2x-1 &=& 0 \\ & 2x+10 &=& 0 \\ & 2x &=& -10 \\ & x &=& -5 \quad | \quad \text{no solution! Because }-4\leq x < \dfrac{1}{2} \\ \\ \dfrac{1}{2}\leq x <7 : & (x+4)-(x-7)-(2x-1) &=& 0 \\ & x+4-x+7-2x+1 &=& 0 \\ &-2x+12 &=& 0 \\ & 2x &=& 12 \\ & \mathbf{x} &=& \mathbf{6} \\ \\ x\geq 7 : & (x+4)+(x-7)-(2x-1) &=& 0 \\ & x+4+x-7-2x+1 &=& 0 \\ & -2 &=& 0 \quad | \quad \text{no solution!} \\ \end{array}\)

\(\mathbf{x = 6}\)  satisfy the equation \(\large{\mathbf{|x+4| + |x-7| = |2x-1|}}\).

how to solve this

\(\sin^4 ( A ) + \sin^2 ( A ) \cos^2 ( A ) + \cos^2 ( A ) = 1\)

\(\begin{array}{|rcll|} \hline \sin^4 ( A ) + \sin^2 ( A ) \cos^2 ( A ) + \cos^2 ( A ) &=& 1 \\ \sin^2 ( A )\sin^2 ( A )+ \sin^2 ( A ) \cos^2 ( A ) + \cos^2 ( A ) &=& 1 \\ \sin^2 ( A ) \left(\sin^2 ( A )+ \cos^2 ( A )\right) + \cos^2 ( A ) &=& 1 \\ \sin^2 ( A ) \left(\underbrace{\sin^2 ( A )+ \cos^2 ( A )}_{=1} \right) + \cos^2 ( A ) &=& 1 \\ \sin^2 ( A )*1 + \cos^2 ( A ) &=& 1 \\ \underbrace{\sin^2 ( A ) + \cos^2 ( A )}_{=1} &=& 1 \\ \hline \end{array} \)

If \(a, b, c\) are positive integers less than \(13\) such that

\(\begin{align*} 2ab+bc+ca&\equiv 0\pmod{13}\\ ab+2bc+ca&\equiv 6abc\pmod{13}\\ ab+bc+2ca&\equiv 8abc\pmod {13} \end{align*}\)

then determine the remainder when \(a+b+c\) is divided by \(13\).

Let a, b, c be vectors such that
\(\mathbf{a} \times \mathbf{b} = \begin{pmatrix} -1 \\- 1 \\ -1 \end{pmatrix},\ \mathbf{a} \times \mathbf{c} = \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix},\ \text{ and } \mathbf{b} \times \mathbf{c} = \begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix}\).

2.
Then evaluate
\((\mathbf{b} + \mathbf{c})\times \mathbf{b},\ \mathbf{a}\times(\mathbf{b} + 4 \mathbf{a}),\ (\mathbf{a} + \mathbf{b} + \mathbf{c})\times \mathbf{a}\)

\(\begin{array}{|l|rcll|} \hline \mathbf{1.} & (\mathbf{b} + \mathbf{c})\times \mathbf{b} \\ & &=& \mathbf{b}\times \mathbf{b} + \mathbf{c}\times \mathbf{b} \\ & &=& 0 + \mathbf{c}\times \mathbf{b} \\ & &=& - (\mathbf{b}\times \mathbf{c}) \\ & &=& - \begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix} \\ & &=& \begin{pmatrix} 0 \\ -2 \\ -3 \end{pmatrix} \\ \hline \mathbf{2.} & \mathbf{a}\times(\mathbf{b} + 4 \mathbf{a}) \\ & &=& \mathbf{a}\times \mathbf{b} + \mathbf{a}\times (4 \mathbf{a}) \\ & &=& \mathbf{a}\times \mathbf{b} + 4(\mathbf{a}\times \mathbf{a}) \\ & &=& \mathbf{a}\times \mathbf{b} + 4 * 0 \\ & &=& \mathbf{a}\times \mathbf{b} \\ & &=& \begin{pmatrix} -1 \\- 1 \\ -1 \end{pmatrix} \\ \hline \mathbf{3.} & (\mathbf{a} + \mathbf{b} + \mathbf{c})\times \mathbf{a} \\ & &=& \mathbf{a}\times \mathbf{a} + \mathbf{b}\times \mathbf{a} + \mathbf{c}\times \mathbf{a} \\ & &=& \mathbf{a}\times \mathbf{a} - \mathbf{a}\times \mathbf{b} - \mathbf{a}\times \mathbf{c} \\ & &=& 0 - (\mathbf{a}\times \mathbf{b}) - (\mathbf{a}\times \mathbf{c}) \\ & &=& - (\mathbf{a}\times \mathbf{b}) - (\mathbf{a}\times \mathbf{c}) \\ & &=& - \begin{pmatrix} -1 \\- 1 \\ -1 \end{pmatrix} - \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} \\ & &=& \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} - \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} \\ & &=& \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix} \\ \hline \end{array}\)

Let a, b, c be vectors such that

\(\mathbf{a} \times \mathbf{b} = \begin{pmatrix} -1 \\- 1 \\ -1 \end{pmatrix},\ \mathbf{a} \times \mathbf{c} = \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix},\ \text{ and } \mathbf{b} \times \mathbf{c} = \begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix} \).
1.
Then evaluate
\((\mathbf{a} + \mathbf{b})\times \mathbf{c},\ \mathbf{a}\times(2\mathbf{b} -3 \mathbf{c}),\ (\mathbf{a} + 2 \mathbf{c})\times \mathbf{b}\)

\(\begin{array}{|l|rcll|} \hline \mathbf{1.} & (\mathbf{a} + \mathbf{b})\times \mathbf{c} \\ & &=& \mathbf{a}\times \mathbf{c} + \mathbf{b} \times \mathbf{c} \\ &&=& \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix}+\begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix} \\ &&=& \begin{pmatrix} -1 \\ 4 \\ 4 \end{pmatrix} \\ \hline \mathbf{2.} & \mathbf{a}\times(2\mathbf{b} -3 \mathbf{c}) \\ & &=& \mathbf{a}\times(2\mathbf{b}) - \mathbf{a}\times (3 \mathbf{c}) \\ & &=& 2(\mathbf{a}\times\mathbf{b}) - 3(\mathbf{a}\times \mathbf{c}) \\ & &=& 2 \begin{pmatrix} -1 \\- 1 \\ -1 \end{pmatrix} - 3 \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} \\ & &=& \begin{pmatrix} -2 \\- 2 \\ -2 \end{pmatrix} - \begin{pmatrix} -3 \\ 6 \\ 3 \end{pmatrix} \\ & &=& \begin{pmatrix} 1 \\- 8 \\ -5 \end{pmatrix} \\ \hline \mathbf{3.} & (\mathbf{a} + 2 \mathbf{c})\times \mathbf{b} \\ & &=& \mathbf{a}\times \mathbf{b} + (2 \mathbf{c})\times \mathbf{b} \\ & &=& \mathbf{a}\times \mathbf{b} + 2 (\mathbf{c}\times \mathbf{b}) \\ & &=& \mathbf{a}\times \mathbf{b} - 2 (\mathbf{b}\times \mathbf{c}) \\ & &=& \begin{pmatrix} -1 \\- 1 \\ -1 \end{pmatrix} - 2 \begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix} \\ & &=& \begin{pmatrix} -1 \\- 5 \\ -7 \end{pmatrix} \\ \hline \end{array}\)

Thank you, Melody !

Let O and H be the circumcenter and orthocenter of triangle ABC, respectively.
Let a, b, and c denote the side lengths as in the picture below:

2.
Also, Find \(AH^2 + BH^2 + CH^2\) if the circumradius is equal to 7 and \(a^2 + b^2 + c^2 = 432\).

\(\begin{array}{|lrcll|} \hline (1) & \vec{OA} + \vec{AH} &=& \vec{OH} \\ (2) & \vec{OB} + \vec{BH} &=& \vec{OH} \\ (3) & \vec{OC} + \vec{CH} &=& \vec{OH} \\ \hline (1)+(2)+(3): & \underbrace{\vec{OA}+\vec{OB}+\vec{OC}}_{=\vec{OH}} +\vec{AH}+\vec{BH}+\vec{CH} &=& 3\vec{OH} \\ & \vec{OH} +\vec{AH}+\vec{BH}+\vec{CH} &=& 3\vec{OH} \\ & \mathbf{\vec{AH}+\vec{BH}+\vec{CH}} &=& \mathbf{2\vec{OH}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \left(\vec{AH}+\vec{BH}+\vec{CH}\right)^2 &=& 4\left(\vec{OH}\right)^2 \\ AH^2+BH^2+CH^2\\ +2\vec{AH}\vec{BH}\\ +2\vec{AH}\vec{CH}\\ +2\vec{BH}\vec{CH}&=& 4\left(\vec{OH}\right)^2 \\\\ && 2\vec{AH}\vec{BH} = 2AH*BH\ \cos(\angle AHB) & c^2 = AH^2+BH^2-2AH*BH\ \cos(\angle AHB) \\ &&& 2AH*BH\ \cos(\angle AHB)=AH^2+BH^2-c^2 \\ && \mathbf{2\vec{AH}\vec{BH} = AH^2+BH^2-c^2} \\\\ \\ && 2\vec{AH}\vec{CH} = 2AH*CH\ \cos(\angle AHC) & b^2 = AH^2+CH^2-2AH*CH\ \cos(\angle AHC) \\ &&& 2AH*CH\ \cos(\angle AHC)=AH^2+CH^2-b^2 \\ && \mathbf{2\vec{AH}\vec{CH} = AH^2+CH^2-b^2} \\\\ \\ && 2\vec{BH}\vec{CH} = 2BH*CH\ \cos(\angle BOC) & a^2 = BH^2+CH^2-2BH*CH\cos(\angle BOC) \\ &&& 2BH*CH\cos(\angle BOC)=BH^2+CH^2-a^2 \\ && \mathbf{2\vec{BH}\vec{CH} = BH^2+CH^2-a^2} \\\\ AH^2+BH^2+CH^2\\ +AH^2+BH^2-c^2\\ +AH^2+CH^2-b^2\\ +BH^2+CH^2-a^2 &=& 4\left(\vec{OH}\right)^2 \\ 3\left( AH^2+BH^2+CH^2\right)-\left(a^2+b^2+c^2\right) &=& 4\left(\vec{OH}\right)^2 \\ 3\left( AH^2+BH^2+CH^2\right) &=& 4\left(\vec{OH}\right)^2+\left(a^2+b^2+c^2\right) \\ 3\left( AH^2+BH^2+CH^2\right) &=& 4*9+432 \\ 3\left( AH^2+BH^2+CH^2\right) &=& 468 \quad | \quad :3 \\ \mathbf{AH^2+BH^2+CH^2} &=& \mathbf{156} \\ \hline \end{array}\)

Let O and H be the circumcenter and orthocenter of triangle ABC, respectively.

Let a, b, and c denote the side lengths as in the picture below:

1.
Find OH if the circumradius is equal to 7 and a^2 + b^2 + c^2 = 432.

\(\text{Let circumradius $R=7$} \\ \text{Let $OA=OB=OC=R$}\)

\(\vec{OH}=\vec{OA}+\vec{OB}+\vec{OC} \quad \text{This is the Sylvester’s relation}\)

\(\begin{array}{|rcll|} \hline \left(\vec{OH}\right)^2 &=& \left(\vec{OA}+\vec{OB}+\vec{OC}\right)^2 \\ OH^2 &=& \underbrace{OA^2+OB^2+OC^2}_{=3R^2} +2\vec{OA}\vec{OB}+2\vec{OA}\vec{OC}+2\vec{OB}\vec{OC} \\ OH^2 &=& 3R^2 +2\vec{OA}\vec{OB}+2\vec{OA}\vec{OC}+2\vec{OB}\vec{OC} \\\\ && 2\vec{OA}\vec{OB} = 2R^2\ \cos(\angle AOB) & c^2 = 2R^2-2R^2\cos(\angle AOB) \\ &&& 2R^2\cos(\angle AOB)=2R^2-c^2 \\ && \mathbf{2\vec{OA}\vec{OB} = 2R^2-c^2} \\\\ && 2\vec{OA}\vec{OC} = 2R^2\ \cos(\angle AOC) & b^2 = 2R^2-2R^2\cos(\angle AOC) \\ &&& 2R^2\cos(\angle AOC)=2R^2-b^2 \\ && \mathbf{2\vec{OA}\vec{OC} = 2R^2-b^2} \\\\ && 2\vec{OB}\vec{OC} = 2R^2\ \cos(\angle BOC) & a^2 = 2R^2-2R^2\cos(\angle BOC) \\ &&& 2R^2\cos(\angle BOC)=2R^2-a^2 \\ && \mathbf{2\vec{OB}\vec{OC} = 2R^2-a^2} \\\\ OH^2 &=& 3R^2 +2R^2-c^2+2R^2-b^2+2R^2-a^2 \\\\ \mathbf{OH^2} &=& \mathbf{9R^2 -\left(a^2+b^2+c^2\right)} \\ \hline OH^2 &=& 9R^2 -\left(a^2+b^2+c^2\right) \quad | \quad R=7,\ a^2 + b^2 + c^2 = 432 \\ OH^2 &=& 9*7^2 -432 \\ OH^2 &=& 9 \\ \mathbf{OH} &=& \mathbf{3} \\ \hline \end{array}\)

Thank you dgfgrafgdfge111