heureka

3. 

As shown in the diagram Line AD, Line BE  and Line CF are concurrent. We know that CD=CE=BF=4 and BD=AF=6. What is AE?

As shown in the diagram Line AD, Line BE  and Line CF are concurrent.

We know that CD=CE=BF=4 and BD=AF=6. 

What is AE?

Ceva's Theorem:

\(\begin{array}{|rcll|} \hline \dfrac{AF}{BF}\cdot\dfrac{BD}{CD}\cdot\dfrac{CE}{AE} &=& 1 \\ \\ \dfrac{6}{4}\cdot\dfrac{6}{4}\cdot\dfrac{4}{AE} &=& 1 \\ \\ \dfrac{36}{4AE} &=& 1 \\ \\ AE&=& \dfrac{36}{4} \\ \\ \mathbf{AE}&\mathbf{=}& \mathbf{9} \\ \hline \end{array}\)

Calculate:

\(\large{\frac{1}{11*15}+\frac{1}{19*15}+\frac{1}{19*23}+\frac{1}{23*27}+\cdots+\frac{1}{1103*1107}+\frac{1}{1111*1107}}\)

\(\begin{array}{|rcll|} \hline && \frac{1}{11*15}+\frac{1}{15*19}+\frac{1}{19*23}+\frac{1}{23*27}+\cdots+\frac{1}{1103*1107}+\frac{1}{1107*1111} \\\\ &=& \frac14\left( \frac{1}{11}-\frac{1}{15}\right) +\frac14\left( \frac{1}{15}-\frac{1}{19}\right) +\frac14\left( \frac{1}{19}-\frac{1}{23}\right) +\frac14\left( \frac{1}{23}-\frac{1}{27}\right) +\cdots\\ && +\frac14\left( \frac{1}{1103}-\frac{1}{1107}\right) +\frac14\left( \frac{1}{1107} -\frac{1}{1111} \right)\\\\ &=& \frac{1}{4*11} \\ &&-\frac14\left(\frac{1}{15}\right)+ \frac14\left(\frac{1}{15}\right)\\ &&-\frac14\left(\frac{1}{19}\right)+\frac14\left(\frac{1}{19}\right) \\ &&-\frac14\left(\frac{1}{23}\right)+\frac14\left(\frac{1}{23}\right) \\ &&-\frac14\left( \frac{1}{27}\right)+\frac14\left( \frac{1}{27}\right)+\cdots \\ &&-\frac14\left( \frac{1}{1103}\right)+\frac14\left( \frac{1}{1103}\right)\\ &&-\frac14\left( \frac{1}{1107}\right)+\frac14\left( \frac{1}{1107}\right) \\ &&-\frac14\left( \frac{1}{1111} \right)\\\\ &=& \frac{1}{44}-\frac14\left( \frac{1}{1111} \right) \\\\ &=& \frac{1}{44}- \frac{1}{4444} \\\\ &\mathbf{=}& \mathbf{\frac{275}{12221}} \\\\ &\mathbf{=}& \mathbf{\frac{25}{1111}} \\\\ &\mathbf{=}& \mathbf{0.\overline{0225}} \\ \hline \end{array}\)

Calculate:

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

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

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

th 5 prob 5

The diagram shows a calculator screen on which the parabolas y=1/4(x-3)(x-8) and y=1/2(x+1)(x-3) have been graphed. The window setting consists of two inequalities, A is less than or equal to X is less than or equal to B and C is less than or equal to Y is less than or equal to D. What are the values of a, b, c, and d? Explain your reasoning, please!!

\(\mathbf{y=\dfrac14(x-3)(x-8)}\\ \begin{array}{|lrcll|} \hline x =8 ~\Rightarrow~ y = 0 ~\Rightarrow~ b=8 \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline x =0 ~\Rightarrow~ &y&=&\dfrac14(-3)(-8) \\ & y&=&\dfrac14\cdot 3 \cdot 8 \\ & y &=& 6 ~\Rightarrow~ d=6 \\ \hline \end{array}\)

\(\mathbf{y=\dfrac12(x+1)(x-3)}\\ \begin{array}{|lrcll|} \hline x_{\text{vertex}} &=& \dfrac{-1+3}{2} \\ &=& 1 \\\\ y_{\text{vertex}} &=&\dfrac12(1+1)(1-3) \\ &=&-2 ~\Rightarrow~ c=-2 \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline y = 6 ~\Rightarrow~ &6& =&\dfrac12(x+1)(x-3) \\ &12&=& (x+1)(x-3) \\ & x^2-2x-3 &=& 12 \\ & x^2-2x-15 &=& 0 \\ & x &=& \dfrac{2\pm \sqrt{4-4\cdot (-15)}}{2} \\ & x &=& \dfrac{2\pm \sqrt{64}}{2} \\ & x &=& \dfrac{2\pm 8}{2} \\\\ & x &=& \dfrac{2+ 8}{2} \\ & x &=& 5 \\\\ & x &=& \dfrac{2- 8}{2} \\ & x &=& -3 ~\Rightarrow~ a=-3 \\ \hline \end{array}\)

\(-3 \le x \le 8 \\ -2 \le y \le 6\)

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 \(0,\dfrac32, -4\)

The fifth term of an arithmetic sequence is 9 and the 32nd term is -84.
What is the 23rd term?

Formula of an arithmetic sequence:

\(\begin{array}{|rcll|} \hline \begin{vmatrix} a_i & a_j & a_k \\ i & j & k \\ 1 & 1 & 1 & \\ \end{vmatrix} = 0 \\\\ a_i(j-k)+a_j(k-i)+a_k(i-j) &=& 0 \\ \hline \end{array} \)

\(\text{Set $i=5$, $\ j=32$, $\ k=23$ } \\ \text{Set $a_i=a_5=9$, $\ a_j=a_{32}=-84$, $\ a_k=a_{23}$ } \)

\(\begin{array}{|rcll|} \hline a_i(j-k)+a_j(k-i)+a_k(i-j) &=& 0 \\\\ 9(32-23)+(-84)(23-5)+a_{23}(5-32) &=& 0 \\\\ 9(9)+(-84)(18)-a_{23}(27) &=& 0 \\\\ 81-1512-27a_{23} &=& 0 \\\\ 27a_{23} &=& 81-1512 \\\\ 27a_{23} &=& -1431 \\\\ a_{23} &=& -\dfrac{1431}{27} \\\\ \mathbf{a_{23}} &\mathbf{=}& \mathbf{-53} \\ \hline \end{array}\)

The lengths of the perpendiculars drawn to the sides of a regular hexagon from an interior point are 4, 5, 6, 8, 9, and 10 centimeters.
What is the number of centimeters in the length of a side of this hexagon?
Express your answer as a common fraction in simplest radical form.

With Aera A:

\(\begin{array}{|rcll|} \hline A &=& \dfrac{s^2\cdot \sin(60^{\circ})}{2} \cdot 6 \quad & | \quad \sin(60^{\circ}) = \dfrac{\sqrt{3}}{2} \\\\ &=& \dfrac{s^2\cdot \dfrac{\sqrt{3}}{2}}{2} \cdot 6 \\\\ \mathbf{A} & \mathbf{=} & \mathbf{\dfrac{3}{2} \sqrt{3}s^2 } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline A &=& \dfrac{s\cdot 4}{2}+ \dfrac{s\cdot 5}{2}+ \dfrac{s\cdot 6}{2}+ \dfrac{s\cdot 8}{2}+ \dfrac{s\cdot 9}{2}+ \dfrac{s\cdot 10}{2} \\\\ &=& \dfrac{1}{2}(4+5+6+8+9+10)s \\\\ &=& \dfrac{1}{2}\cdot 42s \\\\ \mathbf{A} & \mathbf{=} & \mathbf{21s} \\ \hline \end{array}\)

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

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{3}{2} \sqrt{3}s^2 } &=& \mathbf{21s} \\\\ \dfrac{3}{2} \sqrt{3}s &=& 21 \\\\ s &=& 21 \dfrac{2}{3\sqrt{3}} \\\\ s &=& \dfrac{14}{\sqrt{3}} \\\\ s &=& \dfrac{14}{\sqrt{3}}\cdot \dfrac{\sqrt{3}}{\sqrt{3}} \\\\ \mathbf{s} & \mathbf{=} & \mathbf{\dfrac{14}{3}\cdot \sqrt{3} } \\ \hline \end{array} \)

The number of centimeters in the length of a side of this hexagon is \(\mathbf{\dfrac{14}{3}\cdot \sqrt{3} } \ \text{cm}\)

The lengths of the perpendiculars drawn to the sides of a regular hexagon from an interior point are 4, 5, 6, 8, 9, and 10 centimeters.

What is the number of centimeters in the length of a side of this hexagon?

Express your answer as a common fraction in simplest radical form.

\(\begin{array}{|rcll|} \hline 7 = \dfrac{4+10}{2} = \dfrac{5+9}{2}= \dfrac{6+8}{2} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline {\color{red} s}^2 &=& \left({\color{red}\dfrac{s}{2}}\right)^2 +7^2 \\\\ s^2 &=&\dfrac{s^2}{4} + 7^2 \\\\ s^2-\dfrac{s^2}{4} &=& 7^2 \\\\ \dfrac{3}{4} s^2 &=& 7^2 \\\\ s^2 &=& \dfrac{4}{3} \cdot 7^2 \quad & | \quad \text{sqrt both sides}\\\\ s &=& \dfrac{2}{\sqrt{3}} \cdot 7 \\\\ s &=& \dfrac{14}{\sqrt{3}} \\\\ s &=& \dfrac{14}{\sqrt{3}}\cdot \dfrac{\sqrt{3}}{\sqrt{3}} \\\\ \mathbf{s} & \mathbf{=} & \mathbf{\dfrac{14}{3}\cdot \sqrt{3} } \\ \hline \end{array}\)

The number of centimeters in the length of a side of this hexagon is \( \mathbf{\dfrac{14}{3}\cdot \sqrt{3} } \ \text{cm}\)