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We have a right triangle, triangle ABC where the legs AB and AC have lengths 6 and 3sqrt3 respectively.

Medians AM and CN meet at point P.

What is the length of CP?

Let \(\vec{A} = \binom{0}{0}\)
Let \(\vec{B} = \binom{6}{0}\)
Let \(\vec{C} = \binom{0}{3\sqrt{3}}\)

\(\mathbf{\vec{P} = \ ?}\)

\(\begin{array}{|rcll|} \hline \vec{P} &=& \frac13 ( \vec{A}+\vec{B}+\vec{C} ) \\ \vec{P} &=& \frac13 \left( \binom{0}{0}+\binom{6}{0}+\binom{0}{3\sqrt{3}} \right) \\ \vec{P} &=& \frac13 \cdot \binom{0+6+0}{0+0+3\sqrt{3} } \\ \vec{P} &=& \frac13 \cdot \binom{6}{3\sqrt{3} } \\ \vec{P} &=& \dbinom{2}{ \sqrt{3} } \\ \hline \end{array}\)

CP = ?

\(\begin{array}{|rcll|} \hline CP &=& |~\vec{C}-\vec{P}~| \\ CP &=& |~\binom{0}{3\sqrt{3}}-\binom{2}{ \sqrt{3} }~| \\ CP &=& |~\binom{0-2}{3\sqrt{3}-\sqrt{3} } ~| \\ CP &=& |~\binom{-2}{2\sqrt{3} } ~| \\ CP &=& \sqrt{(-2)^2+(2\sqrt{3})^2 } \\ CP &=& \sqrt{4+ 4\cdot3 } \\ CP &=& \sqrt{4+ 12 } \\ CP &=& \sqrt{16} \\ \mathbf{ CP } & \mathbf{=} & \mathbf{4} \\ \hline \end{array} \)

Point X is on overline AC such that AX = 3 CX = 12. If angle ABC = angle BXA = 90 degrees, then what is BX?

\(\begin{array}{|rcll|} \hline \overline{BX}^2 &=& \overline{AX}\cdot \overline{CX} \\ \overline{BX}^2 &=& 3\cdot 12 \\ \overline{BX}^2 &=& 36 \\ \mathbf{\overline{BX}} &\mathbf{=}& \mathbf{ 6 } \\ \hline \end{array}\)

Right \(\triangle{ABC}\) has AB = 3, BC = 4, and AC = 5.
Square XYZW is inscribed in triangle ABC with X and Y on \(\overline{AC}\),
W on \(\overline{AB}\), and
Z on \(\overline{BC}\).
What is the side length of the square?

Let s is the side length oft the square \( = \overline{XY} = \overline{YZ} = \overline{ZW} = \overline{WX}\)

Let h = \(\overline{BT}\)

Let A the area of \(\triangle{ABC}\)

h = ?

\(\begin{array}{|rcll|} \hline A &=& \frac{\overline{AB} \cdot \overline{BC} }{2} \\ A &=& \frac{3\cdot 4}{2} \\ \mathbf{A} &\mathbf{=}& \mathbf{6} \\\\ A &=& \frac{\overline{AC}\cdot h}{2} \\ A &=& \frac{5\cdot h}{2} \quad & | \quad \mathbf{A=6} \\ 6 &=& \frac{5\cdot h}{2} \\ \mathbf{h} &\mathbf{=}& \mathbf{ \frac{12}{5} } \\ \hline \end{array}\)

\(\mathbf{\overline{BW} =\ ?}\)

\(\begin{array}{|rcll|} \hline \frac{ \overline{BW} } {s} &=& \frac{ \overline{AB} } { \overline{AC} } \\ \frac{ \overline{BW} } {s} &=& \frac{ 3 } { 5 } \\ \mathbf{ \overline{BW} } & \mathbf{=} & \mathbf{ \frac{3}{5}s } \\ \hline \end{array}\)

\(\mathbf{\overline{BZ} =\ ?} \)

\(\begin{array}{|rcll|} \hline \frac{ \overline{BZ} } {s} &=& \frac{ \overline{BC} } { \overline{AC} } \\ \frac{ \overline{BZ} } {s} &=& \frac{ 4 } { 5 } \\ \mathbf{\overline{BZ}} &\mathbf{=}& \mathbf{\frac{4}{5}s } \\ \hline \end{array}\)

s = ?

\(\begin{array}{|rcll|} \hline A_{\triangle{ZBW}} = \frac{ \overline{BW}\cdot \overline{BZ} } {2} &=& \frac{s\cdot(h-s)} {2} \\ \overline{BW}\cdot \overline{BZ} &=& s\cdot(h-s) \quad & | \quad \mathbf{ \overline{BW} =\frac{3}{5}s } \quad \mathbf{ \overline{BZ} =\frac{4}{5}s } \quad \mathbf{h=\frac{12}{5}} \\ \frac{3}{5}s \cdot \frac{4}{5}s &=& s\cdot(\frac{12}{5}-s) \\ \frac{12}{25}s &=& \frac{12}{5}-s \\ s+\frac{12}{25}s &=& \frac{12}{5} \\ s \cdot \left(1+\frac{12}{25} \right) &=& \frac{12}{5} \\ s \cdot \left(\frac{25+12}{25} \right) &=& \frac{12}{5} \\ s \cdot \left(\frac{37}{25} \right) &=& \frac{12}{5} \\ s &=& \frac{25}{37} \cdot \frac{12}{5} \\ s &=& \frac{5}{37} \cdot 12 \\ s &=& \frac{60}{37} \\ \mathbf{s} &\mathbf{=}& \mathbf{1.\overline{621}} \\ \hline \end{array}\)

The side length oft the square is \(\mathbf{1.\overline{621}}\)

The three points (3,-5), (-a + 2, 3), and (2a+3,2) lie on the same line.

What is a?

Intercept theorem:

\(\begin{array}{|rcll|} \hline \dfrac{(2a-3)-(3)}{ 2-(-5) } &=& \dfrac{ (-a+2) - 3 } { 3-(-5) } \\\\ \dfrac{2a}{ 7 } &=& \dfrac{ -a -1 } { 8 } \\\\ 16a &=& -7a-7 \\ 23a &=& -7 \\ \mathbf{a} &\mathbf{=}& \mathbf{ -\frac{7}{23} } \\ \hline \end{array} \)

Let triangle ABC be an isosceles triangle such that BC = 30 and AB = AC.
We have that I is the incenter of triangle ABC, and IC = 18.
What is the length of the inradius of the triangle?

\(\begin{array}{|rcll|} \hline \left( \frac{30}{2} \right)^2 + r^2 &=& 18^2 \\ 15^2 + r^2 &=& 18^2 \\ r^2 &=& 18^2-15^2 \\ r^2 &=& 324-225 \\ r^2 &=& 99 \\ r^2 &=& 9\cdot 11 \\ r &=& 3\cdot \sqrt{11} \\ \hline \end{array}\)

r = 9.95

Please give 3 more terms of the following sequence and the formula or method used, if possible:

1, 2/3, 2/3, 4/5, 16/15, 32/21, 16/7.......etc.

Recursive Sequence:

\(\begin{array}{|rcll|} \hline a_n &=& 2^n \cdot \frac{1}{n(n+1)} \\ a_{n+1} &=& 2^{n+1} \cdot \frac{1}{(n+1)(n+2)} \\\\ \frac{a_{n+1}} {a_n} &=& \frac{ 2^{n+1} \cdot \frac{1}{(n+1)(n+2)} } { 2^n \cdot \frac{1}{n(n+1)} } \\ \frac{a_{n+1}} {a_n} &=& 2^{n+1-n} \cdot \frac{n(n+1)} {(n+1)(n+2)} \\ \frac{a_{n+1}} {a_n} &=& 2 \cdot \frac{n} {(n+2)} \\ \frac{a_{n+1}} {a_n} &=& \frac{2n} {(n+2)} \\\\ \mathbf{ a_{n+1} }& \mathbf{=} & \mathbf{ a_n\cdot \frac{2n} {(n+2)} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a_7 &=& \frac{16}{7} \\\\ a_8 &=& \frac{16}{7} \cdot \frac{2\cdot 7} {(7+2)} \\ &=& \frac{16}{7} \cdot \frac{2\cdot 7} {9} \\ &=& \frac{16\cdot 2}{9} \\ & \mathbf{=} & \mathbf{ \frac{32}{9} } \\\\ a_9 &=& \frac{32}{9} \cdot \frac{2\cdot 8} {(8+2)} \\ &=& \frac{32}{9} \cdot \frac{16} {10} \\ &=& \frac{16\cdot 16}{9\cdot 5} \\ & \mathbf{=} & \mathbf{ \frac{256}{45} } \\\\ a_{10} &=& \frac{256}{45} \cdot \frac{2\cdot 9} {(9+2)} \\ &=& \frac{256}{5\cdot 9} \cdot \frac{2\cdot 9} {11} \\ &=& \frac{256\cdot 2}{5\cdot 11} \\ & \mathbf{=} & \mathbf{ \frac{512}{55} } \\ \hline \end{array}\)

Let triangle ABC be a triangle such that AB=13, BC=14, and AC=15. 

Meanwhile, D is a point on BC such that AD bisects angle A.

Find the area of triangle ADC

Let area of triangle \(\text{ADC} = A_\text{ADC} \)

Let area of triangle \( \text{ADB} = A_\text{ADB}\)
Let area of triangle \(\text{ABC} = A\)

Let \(A_\text{ADB} = A - A_\text{ADC}\)

\(\begin{array}{|rcll|} \hline \frac { A_\text{ADB} } { A_\text{ADC} } &=& \frac{ \overline{AD}\cdot 13 \cdot \sin(\frac{A}{2}) \cdot \frac12 } { \overline{AD}\cdot 15 \cdot \sin(\frac{A}{2}) \cdot \frac12 } \\ \frac { A_\text{ADB} } { A_\text{ADC} } &=& \frac{ 13 } { 15 } \quad & | \quad A_\text{ADB} = A - A_\text{ADC} \\ \frac { A - A_\text{ADC} } { A_\text{ADC} } &=& \frac{ 13 } { 15 } \\ \frac { A } { A_\text{ADC} } - 1 &=& \frac{ 13 } { 15 } \\ \frac { A } { A_\text{ADC} } &=& 1+ \frac{ 13 } { 15 } \\ \frac { A } { A_\text{ADC} } &=& \frac{ 28 } { 15 } \\ \frac { A_\text{ADC} } { A } &=& \frac{ 15 } { 28 } \\\\ \mathbf{ A_\text{ADC } } & \mathbf{=} & \mathbf{ \frac{ 15 } { 28 } A } \\ \hline \end{array} \)

Heron's formula states that the area of a triangle whose sides have lengths a, b, and c is

\( A = \sqrt{s(s-a)(s-b)(s-c)}\)

where s is the semiperimeter of the triangle; that is,
\(s=\frac{a+b+c}{2}\).

\(\begin{array}{|rcll|} \hline s &=& \frac{a+b+c}{2} \quad & | \quad a= 14, \ b= 15, \ c= 13 \\ s &=& \frac{15+15+13}{2} \\ s &=& \frac{42}{2} \\ \mathbf{ s }& \mathbf{=} & \mathbf{21} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline A &=& \sqrt{s(s-a)(s-b)(s-c)} \quad & | \quad s= 21, \ a= 14, \ b= 15, \ c= 13 \\ A &=& \sqrt{21(21-14)(21-15)(21-13)} \\ A &=& \sqrt{21\cdot 7 \cdot 6 \cdot 8 } \\ A &=& \sqrt{3\cdot 7 \cdot 7 \cdot 2\cdot 3 \cdot 2\cdot 4 } \\ A &=& \sqrt{4^2\cdot 7^2 \cdot 3^2 } \\ A &=& 4\cdot 7 \cdot 3 \\ \mathbf{ A }& \mathbf{=} & \mathbf{84} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{ A_\text{ADC } } & \mathbf{=} & \mathbf{ \frac{ 15 } { 28 } A } \quad & | \quad \mathbf{ A } \mathbf{=} \mathbf{84} \\ A_\text{ADC } & = & \frac{ 15 } { 28 } \cdot 84 \\ A_\text{ADC } & = & 15 \cdot 3 \\\\ \mathbf{ A_\text{ADC } }& \mathbf{=} & \mathbf{45} \\ \hline \end{array}\)

Let A,B  be the points on the coordinate plane with coordinates \(\tbinom{t-4}{-1}\)  and \(\tbinom{-2}{t+3}\), respectively.
The square of the distance between the midpoint of  \( \overline{AB}\) and an endpoint of   \(\overline{AB}\) is equal to  \(\tfrac{t^2}{2}\).
What is the value of \(t\) ?

\(\begin{array}{|rcll|} \hline \left ( \frac{ \overline{AB} } {2} \right)^2 &=& \frac{t^2}{2} \\ \frac{ \overline{AB}^2 } {4} &=& \frac{t^2}{2} \\ \mathbf{ \overline{AB}^2 } & \mathbf{=} & \mathbf{2 t^2 } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \overline{AB}^2 &=& [~ (t-4)-(-2) ~ ]^2 + [~ (-1) - (t+3) ~ ]^2 \\ \overline{AB}^2 &=& (~ t-4+2 ~)^2 + (~ -1 - t -3 ~ )^2 \\ \overline{AB}^2 &=& (~ t-2 ~)^2 + (~ - t -4 ~ )^2 \\ \overline{AB}^2 &=& (~ t-2 ~)^2 + [~ - (t +4) ~ ]^2 \\ \overline{AB}^2 &=& (~ t-2 ~)^2 + (~ t +4 ~)^2 \quad & | \quad \overline{AB}^2 = 2t^2\\ 2t^2 &=& (~ t-2 ~)^2 + (~ t +4 ~)^2 \\ 2t^2 &=& t^2 -4t + 4 + t^2 + 8t + 16 \\ 0 &=& -4t + 4 + 8t + 16 \\ 0 &=& 4t + 20 \\ 4t &= -20 \\ t &=& -\frac{20}{4} \\ \mathbf{ t } & \mathbf{=} & \mathbf{-5} \\ \hline \end{array}\)

Calculate the volume of an atom of diameter 0.000000001metres?

\(\text{Let diameter } d = 0.000000001\ m = 1\cdot 10^{-9}\ m = 10^{-9}\ m \)

\(\begin{array}{|rcll|} \hline V_{\text{Atom}} &=& \frac16\pi d^3 \quad & | \quad d = 10^{-9}\ m \\ &=& \frac16\pi \cdot (10^{-9}\ m)^3 \\ &=& \frac16\pi \cdot 10^{-9\cdot 3}\ m^3 \\ &=& 0.5235987756 \cdot 10^{-27}\ m^3 \\ &=& 5.235987756 \cdot 10^{-1} \cdot 10^{-27}\ m^3 \\\\ \mathbf{ V_{\text{Atom}} } & \mathbf{=} & \mathbf{ 5.235987756 \cdot 10^{-28}\ m^3 } \\ \hline \end{array} \)

Using triangle inequity prove |x+y|>=||x|-|y||

\(\begin{array}{|rcll|} \hline |x+y| & \stackrel{?} \ge & ||x|-|y|| \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline &|y| = |(x+y)-x| & \le & |x+y| + |x| \\ & |y| & \le & |x+y| + |x| \\ (1) & |y|-|x| & \le & |x+y| \\\\ & |x| = |(x+y)-y| & \le & |x+y| + |y| \\ & |x| & \le & |x+y| + |y| \\ (2) & |x|-|y| & \le & |x+y| \\\\ \hline (1) \text{ and } (2) & ||x|-|y| | & \le & |x+y| \\ & |x+y| & \ge & ||x|-|y| | \\ \hline \end{array}\)