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 #4
avatar+20585 
+2

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

 

laugh

heureka Dec 11, 2018 7:47:13 AM
 #3
avatar+20585 
+2

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.

 

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


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

laugh

heureka Dec 10, 2018
 #1
avatar+20585 
+3

Calculate:

\(\large{\frac{1}{2*4}+\frac{1}{6*4}+\frac{1}{6*8}+\frac{1}{10*8}+\cdots+\frac{1}{96*98}+\frac{1}{100*98}}\)

 

\(\begin{array}{|rcll|} \hline && \frac{1}{2*4}+\frac{1}{4*6}+\frac{1}{6*8}+\frac{1}{8*10}+\frac{1}{10*12}+\frac{1}{12*14}+\cdots+\frac{1}{94*96}+\frac{1}{96*98}+\frac{1}{98*100} \\\\ &=& \frac12\left( \frac{1}{2}-\frac{1}{4}\right) +\frac12\left( \frac{1}{4}-\frac{1}{6}\right) +\frac12\left( \frac{1}{6}-\frac{1}{8}\right) +\frac12\left( \frac{1}{8}-\frac{1}{10}\right) \\ &&+\frac12\left( \frac{1}{10}-\frac{1}{12}\right) +\frac12\left( \frac{1}{12}-\frac{1}{14}\right)+\cdots\\ && +\frac12\left( \frac{1}{94}-\frac{1}{96}\right) +\frac12\left( \frac{1}{96}-\frac{1}{98}\right) +\frac12\left( \frac{1}{98} -\frac{1}{100} \right)\\\\ &=& \frac14 \\ &&-\frac12\left(\frac{1}{4}\right)+ \frac12\left(\frac{1}{4}\right)\\ &&-\frac12\left(\frac{1}{6}\right)+\frac12\left(\frac{1}{6}\right) \\ &&-\frac12\left(\frac{1}{8}\right)+\frac12\left(\frac{1}{8}\right) \\ &&-\frac12\left( \frac{1}{10}\right)+\frac12\left( \frac{1}{10}\right) \\ &&-\frac12\left( \frac{1}{12}\right)+\frac12\left( \frac{1}{12}\right) \\ &&-\frac12\left( \frac{1}{14}\right)+\frac12\left( \frac{1}{14}\right)+\cdots \\ &&-\frac12\left( \frac{1}{94}\right)+\frac12\left( \frac{1}{94}\right)\\ &&-\frac12\left( \frac{1}{96}\right)+\frac12\left( \frac{1}{96}\right) \\ &&-\frac12\left( \frac{1}{98}\right)+\frac12\left( \frac{1}{98}\right) \\ &&-\frac12\left( \frac{1}{100} \right)\\\\ &=& \frac14-\frac12\left( \frac{1}{100} \right) \\\\ &=& \frac14- \frac{1}{200} \\\\ &\mathbf{=}& \mathbf{\frac{49}{200}} \\\\ &\mathbf{=}& \mathbf{0.245} \\ \hline \end{array}\)

 

laugh

heureka Dec 10, 2018
 #2
avatar+20585 
+4

All the positive integers greater than 1 are arranged in five columns (A, B, C, D, E) as shown.

Continuing the pattern, in what column will the integer 800 be written?

 

We rearrange:

\(\small{ \begin{array}{|cccc|cccc|cccc|cccc|cccc|c|} \hline B&C&D&E& D&C&B&A& B&C&D&E& D&C&B&A& B&C&D&E& \ldots\\ \hline 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 &10 & 11 & 12 & 13 &14 & 15 & 16 & 17 &18 & 19 & 20 & 21 & \ldots \\ \hline \end{array} }\)

 

\(\begin{array}{|r|r!c|c|c|c|l|} \hline \text{Position} ~&(& 1 & 2 & 3 & 4\text{ or }0& ) \\ \hline \text{Row odd} ~&(& B & C & D & E &) \\ \text{Row even} ~&(& D & C & B & A &) \\ \hline \end{array}\)

 

Formula:

\(\begin{array}{|rcll|} \hline \text{Row} &=& 1 + \left\lfloor \dfrac{n-2}{4} \right\rfloor \quad & \quad \left\lfloor\ldots \right\rfloor \text{Function floor or Integerpart} \\\\ \text{Position} &=& (n-1) \pmod{4} \quad & \quad \text{If the position is zero, so the position is 4}\\ \hline \end{array}\)

 

\(\text{Row} =\ ?, ~n = 800:\)

\(\begin{array}{|rcll|} \hline \text{Row} &=& 1 + \left\lfloor \dfrac{800-2}{4} \right\rfloor \\ &=& 1 + \left\lfloor \dfrac{798}{4} \right\rfloor \\ &=& 1 + \left\lfloor 199.5 \right\rfloor \\ &=& 1 + 199 \\ &=& 200 \quad & \quad \text{the row is even!} \\ \hline \end{array} \)

 

The number 800 is in row = 200. And the row is an even number.

 

\(\text{Position} =\ ?, ~n = 800:\)

\(\begin{array}{|rcll|} \hline \text{Position} &=& (800-1) \pmod{4} \\ &=& 799 \pmod{4} \\ &=& 3 \\ \hline \end{array}\)

 

\(\text{Column} =\ ?, ~\text{Row is even}:\)

\(\begin{array}{|rcll|} \hline \hline \text{Position} ~&(& 1 & 2 & \color{red}3 & 4\text{ or }0& ) \\ \hline \text{Row even} ~&(& D & C & \color{red}B & A &) \\ \hline \end{array}\)

 

The integer 800 will be written in column B

 

laugh

heureka Dec 7, 2018
 #7
avatar+20585 
+5

The polynomial f(x) has degree 3. If f(-1) = 15, f(0) = 0, f(1) = -5, and f(2) = 12,

 then what are the x-intercepts of the graph of f?

 

\(\boxed{f(x) =ax^3+bx^2+cx+d}\) degree 3

 

\(\begin{array}{|lrcll|} \hline f(0) = 0: & f(0) = 0 &=& a\cdot 0^3+ b\cdot 0^2+ c\cdot 0 +d \\ & 0 &=& d \\ & \mathbf {d }& \mathbf{=}& \mathbf{0} \\ \hline \end{array} \)

 

So  \(\boxed{f(x) =ax^3+bx^2+cx}\)

 

\(\begin{array}{|lrcll|} \hline f(1) = -5: & f(1) = -5 &=& a\cdot 1^3+ b\cdot 1^2+ c\cdot 1 \\ & -5 &=& a + b + c \\ & \mathbf {a + b + c }& \mathbf{=}& \mathbf{-5} \qquad (1) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline f(-1) = 15: & f(-1) = 15 &=& a\cdot (-1)^3+ b\cdot (-1)^2+ c\cdot (-1) \\ & 15 &=& -a + b - c \\ & \mathbf {-a + b - c }& \mathbf{=}& \mathbf{15}\qquad (2) \\ \hline \end{array}\)

 

\(\mathbf{(1)+(2):}\)

\(\begin{array}{|lrcll|} \hline (1) & \mathbf {a + b + c }& \mathbf{=}& \mathbf{-5} \\ (2) & \mathbf {-a + b - c }& \mathbf{=}& \mathbf{15} \\ \hline (1)+(2): & 2b &=& -5+15 \\ & 2b &=& 10 \\ & \mathbf {b }& \mathbf{=}& \mathbf{5} \\ \hline (1) : & a+b+c &=& -5 \quad | \quad b=5 \\ & a+5+c &=& -5 \\ & \mathbf {a+c} &\mathbf {=}& \mathbf {-10} \qquad (3) \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline f(2) = 12: & f(2) = 12&=& a\cdot (2)^3+ b\cdot (2)^2+ c\cdot (2) \\ & 12 &=& 8a + 4b +2c \quad | \quad b=5 \\ & 12 &=& 8a + 4\cdot 5 +2c \\ & 12 &=& 8a + 20 +2c \\ & 8a + 2c &=& 12-20 \\ & 8a + 2c &=& -8 \quad & | \quad :2 \\ & \mathbf {4a + c}& \mathbf{=}& \mathbf{-4}\qquad (4) \\ \hline \end{array}\)

 

\(\mathbf{(4)-(3):}\)

\(\begin{array}{|lrcll|} \hline (4) & \mathbf {4a + c}& \mathbf{=}& \mathbf{-4} \\ (3) & \mathbf {a+c} &\mathbf {=}& \mathbf {-10}\\ \hline (4)-(3): & 3a &=& -4+-(-10)\\ & 3a &=& 6 \\ & \mathbf {a }& \mathbf{=}& \mathbf{2} \\ \hline (3) : & a+c &=& -10 \quad | \quad a=2 \\ & 2+c &=& -10 \\ & \mathbf {c} &\mathbf {=}& \mathbf {-12} \\ \hline \end{array}\)

 

 

\(\mathbf{\text{The $x$-intercepts of the graph of $~\boxed{f(x)=2x^3+5x^2-12x}$:}}\)

 

\(\begin{array}{|rcll|} \hline 2x^3+5x^2-12x &=& 0 \\ x\cdot (2x^2+5x-12) &=& 0 \\ \hline \mathbf{x_1} & \mathbf{=}& \mathbf{ 0 } \\ \hline 2x^2+5x-12 &=& 0 \\ x &=& \dfrac{-5\pm \sqrt{25-4\cdot 2 \cdot (-12)} } {2\cdot 2} \\\\ x &=& \dfrac{-5\pm \sqrt{121} } {4} \\\\ x &=& \dfrac{-5\pm 11 } {4} \\\\ x_2 &=& \dfrac{-5+ 11 } {4} \\ \mathbf{x_2} & \mathbf{=}& \mathbf{ \dfrac32 } \\\\ x_3 &=& \dfrac{-5- 11 } {4} \\ \mathbf{x_3} & \mathbf{=}& \mathbf{ -4 } \\ \hline \end{array}\)

 

The x-intercepts of the graph of f(x) are \(0,\dfrac32, -4\)

 

 

laugh

heureka Dec 7, 2018