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Points M, N, and O are the midpoints of sides \(\overline{KL}\), \(\overline{LJ}\), and \(\overline{JK}\), respectively, of triangle JKL.
Points P, Q, and R are the midpoints of \(\overline{NO}\), \(\overline{OM}\), and \(\overline{MN}\) respectively.
If the area of triangle PQR is 21, then what is the area of triangle LPQ?

Points M, N, and O are the midpoints of sides \(\overline{KL}\), \(\overline{LJ}\), and \(\overline{JK}\), respectively, of triangle JKL.
Points P, Q, and R are the midpoints of \(\overline{NO}\), \(\overline{OM}\), and \(\overline{MN}\) respectively.
If the area of triangle PQR is 21, then what is the area of triangle LPQ?

\(\begin{array}{|rclcrcl|} \hline \text{area}\ [JKL] &=& A \\ \text{area}\ [MNO] &=& \frac14 A \\ \text{area}\ [PQR] &=& \frac14 *[MNO] \\ &=& \frac14 \cdot \frac14 A \\ [PQR] &=& \frac{1}{16} A \quad | \quad [PQR]= 21 \\ 21 &=& \frac{1}{16} A \\ 21*16 &=& A \\ \mathbf{A} &=& \mathbf{16\cdot 21} \\\\ \text{area}\ [LPQ] &=& \frac{JK}{4} \cdot \frac{h}{2} \quad & | & \sin(J) &=& \frac{h}{ \frac{JL}{2} +\frac{JL}{4} } \\ & & & | & \sin(J) &=& \frac{h}{ \frac34 * JL } \\ & & & | & h &=& \frac34 *JL \sin(J) \\ \triangle LPQ &=& \frac{JK}{4} \cdot \frac{ \frac34 JL \sin(J) }{2} \\ &=& \frac{3}{32} \cdot JK\cdot JL\cdot \sin(J) \quad & | & JK\cdot JL\cdot \sin(J) &=& 2A \\ &=& \frac{3}{32} \cdot 2A \\ &=& \frac{6}{32} A \quad & | & A &=& 16\cdot 21 \\ &=& \frac{6}{32} \cdot 16\cdot 21 \\ &=& \frac{6}{2} \cdot 21 \\ &=& 3 \cdot 21 \\ \mathbf{\text{area}\ [LPQ]} &=& \mathbf{63} \\ \hline \end{array}\)

Point D is the midpoint of median \(\overline{AM}\) of triangle ABC.
Point E is the midpoint of \(\overline{AB}\), and point T is the intersection of \(\overline{BD}\) and \(\overline{ME}\).
Find the area of triangle DMT if [ABC] =150. [ABC] is the area of triangle ABC

Point D is the midpoint of median \(\overline{AM}\) of triangle ABC.
Point E is the midpoint of \(\overline{AB}\), and point T is the intersection of \(\overline{BD}\) and \(\overline{ME}\).
Find the area of triangle DMT if [ABC] =150. [ABC] is the area of triangle ABC

My attempt:

\(\begin{array}{|rcll|} \hline 2[ABC] &=& 2u*2w*\sin(M) \\ [ABC] &=& 2uw\sin(M)\quad \text{ or } \quad \mathbf{\sin(M) = \dfrac{[ABC]}{2uw}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{In $\triangle BDM$ } \\ \hline \dfrac{\sin(M)}{\overline{BD}} &=& \dfrac{\sin(D)}{u} \quad | \quad \mathbf{\sin(M) = \dfrac{[ABC]}{2uw}} \\ \dfrac{[ABC]}{2uw\overline{BD}} &=& \dfrac{\sin(D)}{u} \\\\ \dfrac{[ABC]}{2w\overline{BD}} &=& \sin(D) \\\\ \mathbf{\dfrac{[ABC]}{2\overline{BD}}} &=& \mathbf{w\sin(D)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 2[DMT] &=& w*\overline{DT}*\sin(D) \\ 2[DMT] &=& w\sin(D)*\overline{DT} \quad | \quad \mathbf{w\sin(D) = \dfrac{[ABC]}{2\overline{BD}}} \\ 2[DMT] &=& \dfrac{[ABC]}{2\overline{BD}}*\overline{DT} \\\\ \mathbf{[DMT]} &=& \mathbf{\dfrac{[ABC]}{4} * \dfrac{\overline{DT}} {\overline{BD}}} \\ \hline \end{array} \)

The Centroid of a Triangle ABM Divides Each Median in the Ratio \(1:2\)

\(\begin{array}{|rcll|} \hline \overline{DT} &=& \dfrac{1}{3}*\overline{BD} \\\\ \overline{BT} &=& \dfrac{2}{3}*\overline{BD} \\\\ \dfrac{\overline{DT}} {\overline{BD}} &=& \dfrac{\overline{DT}} { \overline{DT}+\overline{BT}} \\\\ \dfrac{\overline{DT}} {\overline{BD}} &=& \dfrac{\dfrac{1}{3}*\overline{BD}} { \dfrac{1}{3}*\overline{BD}+\dfrac{2}{3}*\overline{BD}} \\\\ \dfrac{\overline{DT}} {\overline{BD}} &=& \dfrac{\dfrac{1}{3}*\overline{BD}} { \dfrac{3}{3}*\overline{BD}} \\\\ \dfrac{\overline{DT}} {\overline{BD}} &=& \dfrac{\dfrac{1}{3} } { \dfrac{3}{3} } \\\\ \dfrac{\overline{DT}} {\overline{BD}} &=& \dfrac{\dfrac{1}{3} } { 1 } \\\\ \mathbf{\dfrac{\overline{DT}} {\overline{BD}} } &=& \mathbf{ \dfrac{1}{3}} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{[DMT]} &=& \mathbf{\dfrac{[ABC]}{4} * \dfrac{\overline{DT}} {\overline{BD}}} \quad | \quad \mathbf{\dfrac{\overline{DT}} {\overline{BD}} = \dfrac{1}{3}} \\\\ [DMT] &=& \dfrac{[ABC]}{4}* \dfrac{1}{3} \\\\ [DMT] &=& \dfrac{[ABC]}{12} \quad | \quad [ABC] =150 \\\\ [DMT] &=& \dfrac{150}{12} \\\\ \mathbf{[DMT]} &=& \mathbf{12.5} \\ \hline \end{array}\)

Use an Addition or Subtraction Formula to write the expression as a trigonometric function of one number.
\(\sin(12^\circ) \cos(18^\circ) + \cos(12^\circ) \sin(18^\circ) \)

\(\begin{array}{|rcll|} \hline && \sin(12^\circ) \cos(18^\circ) + \cos(12^\circ) \sin(18^\circ) \\ &=& \sin(12^\circ+18^\circ) \\ &=& \sin(30^\circ) \\ &=& \mathbf{0.5} \\ \hline \end{array}\)

Let \(A = (3, \theta_1)\) and \(B = (9, \theta_2)\) in polar coordinates. If  \(\theta_1 - \theta_2 = \dfrac{\pi}{2}\), then  find the distance \(AB\)

Formula:
We have two points \((r_1,\ \theta_1)\) and \(( r_2,\ \theta_2)\).
The distance between two polar coordinates \(\mathbf{d =\sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta_2 - \theta_1})}\)

\(\begin{array}{|rcll|} \hline d &=& \sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta_2 - \theta_1) } \\ d &=& \sqrt{r_1^2+r_2^2-2r_1r_2\cos\Big(-(\theta_1 - \theta_2)\Big) } \\ d &=& \sqrt{r_1^2+r_2^2+2r_1r_2\cos(\theta_1 - \theta_2) } \\ d &=& \sqrt{3^2+9^2+2*3*9\cos\left( \dfrac{\pi}{2} \right) } \\ d &=& \sqrt{9+81+54\cos\left( \dfrac{\pi}{2} \right) } \quad | \quad \cos\left( \dfrac{\pi}{2} \right) = 0 \\ d &=& \sqrt{9+81+0 } \\ d &=& \sqrt{90} \\ d &=& \sqrt{9*10} \\ d &=& \sqrt{3^2*10} \\ \mathbf{d} &=& \mathbf{3\sqrt{10}} \\ \hline \end{array} \)

Let \(a,b,c,d\) be positive real numbers.

Find the minimum value of \(\dfrac{(a + b)(a + c)(b + c)}{abc}\)

\(\mathbf{\huge{AM \geq GM }}\)

\(\begin{array}{|rcll|} \hline \dfrac{a+b}{2} & \ge & \sqrt{ab} \\\\ \dfrac{a+c}{2} & \ge & \sqrt{ac} \\\\ \dfrac{b+c}{2} & \ge & \sqrt{bc} \\\\ \hline \left( \dfrac{a+b}{2} \right) \left( \dfrac{a+c}{2} \right) \left( \dfrac{b+c}{2} \right) & \ge & \sqrt{ab}\sqrt{ac} \sqrt{bc} \\\\ \dfrac{(a + b)(a + c)(b + c)}{8} & \ge & \sqrt{abacbc} \\\\ \dfrac{(a + b)(a + c)(b + c)}{8} & \ge & \sqrt{a^2b^2c^2} \\\\ \dfrac{(a + b)(a + c)(b + c)}{8} & \ge & abc \\\\ \dfrac{(a + b)(a + c)(b + c)}{abc} & \ge & 8 \\\\ \mathbf{ 8 } & \le & \mathbf{\dfrac{(a + b)(a + c)(b + c)}{abc}} \\ \hline \end{array}\)

The minimum value is 8

Find the number of ordered pairs \((a,b)\) of integers such that \(\dfrac{a + 2}{a + 5} = \dfrac{b}{3}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{a + 2}{a + 5}} &=& \mathbf{\dfrac{b}{3}} \\\\ 3(a+2) &=& b(a+5) \\ 3a+6 &=& ab+5b \\ -3a+5b+ab &=& 6 \\ -3a+5b+ab+(-3*5) &=& (-3*5)+6 \\ -3a+5b+ab-15 &=& -15+6 \\ \mathbf{(a+5)(b-3)} &=& \mathbf{-9} \\ \mathbf{m*n} &=& \mathbf{-9} \\ && \text{find all integer products }\ m*n = -9, \\ && \text{where}\ m=a+5,\ \text{ and }\ n=b-3 \\ \hline \end{array} \)

all integer products

\(\begin{array}{|ccl|r|r|r|r|} \hline && m*n & m & n & a=m-5 & b=n+3 \\ \hline -9 &=& 1\times -9 & 1 & -9 & -4 & -6 \\ -9 &=& -1\times 9 & -1 & 9 & -6 & 12 \\ -9 &=& 3\times -3 & 3 & -3 & -2 & 0 \\ -9 &=& -3\times 3 & -3 & 3 & -8 & 6 \\ -9 &=& 9\times -1 & 9 & -1 & 4 & 2 \\ -9 &=& -9\times 1 & -9 & 1 & -14 & 4 \\ \hline \end{array} \)

There are 6 integer products.

The number of ordered pairs (a,b) are 6.

\((-4,-6),\ (-6,12),\ (-2,0),\ (-8,6),\ (4,2),\ (-14,4) \)

If

\(\dfrac{p}{q}=\dfrac{q}{r}\) ,

then prove the given problem.

\(\begin{array}{|rcll|} \hline \dfrac{p}{q} &=& \dfrac{q}{r} \\ pr &=& q^2 \\ \mathbf{2pr} &=& \mathbf{2q^2} \qquad (1) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && (p+q+r)(p-q+r) \\ &=& \Big((p+r)+q\Big) \Big((p+r)-q\Big) \\ &=& (p+r)^2-q^2 \\ &=& p^2+2pr+r^2-q^2 \quad &| \quad \mathbf{2pr=2q^2} \\ &=& p^2+2q^2+r^2-q^2 \\ \mathbf{(p+q+r)(p-q+r)} &=& \mathbf{p^2+q^2+r^2}\ \checkmark \\ \hline \end{array}\)

find f 1/(2x+3)^2   d(x)

I assume: Derivative: \(\dfrac{d\ \left( \frac{1}{(2 x + 3)^2} \right) }{dx}\)

\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{ \dfrac{1}{\left(2 x + 3\right)^2} } \\\\ f'(x) &=& \dfrac{1}{\left(2 x + 3\right)^2} \times \left( \dfrac{0}{1} - \dfrac{2*(2x+3)*2}{(2 x + 3)^2} \right) \\\\ f'(x) &=& \dfrac{1}{\left(2 x + 3\right)^2} \times \left( 0 - \dfrac{4*(2x+3)}{(2 x + 3)^2} \right) \\\\ f'(x) &=& \dfrac{1}{\left(2 x + 3\right)^2} \times \left( - \dfrac{4}{(2 x + 3)} \right) \\\\ \mathbf{f'(x)} &=& \mathbf{-\dfrac{4}{\left(2 x + 3\right)^3}} \\ \hline \end{array}\)