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O is the centre and R the radius of the circle circumscribing the triangle ABC, AO, BO, and CO meet the opposite sides in L, M, N respectively.

Show that

\((i)\quad AL = \dfrac{ b\sin C} { \cos(B-C) }\)

\(\text{Let $\angle BAO= \angle ABO = w$ } \\ \text{Let $\angle CAO= \angle ACO = v$ } \\ \text{Let $\angle CBO= \angle BCO = u$ } \\ \text{Let $\angle A= v+w$ } \\ \text{Let $\angle B= w+u$ } \\ \text{Let $\angle C= u+v$ } \\ \text{Let $\angle CLA = \varphi$} \)

\(\begin{array}{|rcll|} \hline A+B-C &=& (v+w)+(w+u)-(u+v) \\ A+B-C &=& 2w \quad | \quad A+B = 180^\circ - C \\ 180^\circ - C-C &=& 2w \\ 180^\circ -2C &=& 2w \quad | \quad : 2 \\ 90^\circ -C &=& w \\ \mathbf{w} &=& \mathbf{90^\circ -C} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline A-B+C &=& (v+w)-(w+u)+(u+v) \\ A-B+C &=& 2v \quad | \quad A+C = 180^\circ - B \\ 180^\circ - B-B &=& 2v \\ 180^\circ -2B &=& 2v \quad | \quad : 2 \\ 90^\circ -B &=& v \\ \mathbf{v} &=& \mathbf{90^\circ -B} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline -A+B+C &=& -(v+w)+(w+u)+(u+v) \\ -A+B+C &=& 2u \quad | \quad B+C = 180^\circ - A \\ 180^\circ - A-A &=& 2u \\ 180^\circ -2A &=& 2u \quad | \quad : 2 \\ 90^\circ -A &=& u \\ \mathbf{u} &=& \mathbf{90^\circ -A} \\ \hline \end{array}\)

In triangle ALC:

\(\begin{array}{|rcll|} \hline \varphi &=& 180^\circ-(v+C) \quad | \quad \mathbf{v=90^\circ -B} \\ \varphi &=& 180^\circ-(90^\circ -B+C)\\ \varphi &=& 180^\circ-90^\circ -(-B+C)\\ \varphi &=& 180^\circ-90^\circ -(C-B)\\ \varphi &=& 90^\circ -(C-B)\\ \mathbf{\varphi} &=& \mathbf{90^\circ +(B-C)} \\ \hline \end{array}\)

In triangle ALC - sin rule:

\(\begin{array}{|rcll|} \hline \dfrac{ \sin{C} } {AL} &=& \dfrac{\sin(\varphi)}{b} \\\\ \dfrac{ \sin{C} } {AL} &=& \dfrac{\sin\Big(90^\circ +(B-C)\Big)}{b} \\\\ \dfrac{ \sin{C} } {AL} &=& \dfrac{\cos\Big(B-C\Big)}{b} \quad | \quad \updownarrow \\\\ \dfrac{AL}{ \sin{C} } &=& \dfrac{b} {\cos\Big(B-C\Big)}\\\\ \mathbf{AL} &=& \mathbf{\dfrac{b\sin{C}} {\cos\Big(B-C\Big)}} \\ \hline \end{array}\)

The foci of the ellipse\( \dfrac{x^2}{49} + \dfrac{y^2}{33} = 1\) are \(F_1\) and \(F_2\) as shown below.
Let P be a point on the circle \(x^2 + (y - 3)^2 = 4\).
Line \(F_2P\) intersects the ellipse again at Q,
where the y-coordinate of Q is positive.
Find the maximum value of \(PQ + F_1 Q\).

\(\text{1. Parameter of the ellipse:}\\ \begin{array}{rcll} a^2 &=& 49 \\ \mathbf{a} &=& \mathbf{7} \\\\ \mathbf{b^2} &=& \mathbf{33} \\\\ c^2 &=& a^2-b^2 \\ c^2 &=& 49-33\\ c^2 &=& 16 \\ \mathbf{c} &=& \pm4 \\\\ F_2 &=& (-c,0) = (-4,0) \\ F_1 &=& (c,0) = (4,0) \\ \end{array}\)

\(\text{2. Find the maximum value:}\\ \begin{array}{|rcll|} \hline (PQ + F_1 Q)_{max} &=& F_2Q+F_1Q-F_2P \quad | \quad F_1Q+F_2P = 2a \\ (PQ + F_1 Q)_{max} &=& 2a-F_2P \quad | \quad a=7 \\ (PQ + F_1 Q)_{max} &=& 14-F_2P \\ && F_2P= \sqrt{(x_p-(-4))^2+(y_p-0)^2} \\ && \mathbf{F_2P= \sqrt{(x_p+4)^2+y_p^2}} \\ && \boxed{x_p^2+(y_p-3)^2=4\\ (y_p-3)^2=4-x_p^2 \\ \mathbf{y_p=3+\sqrt{4-x_p^2} } \\ \ldots \\ y_p^2 = 13-x_p^2 +6\sqrt{4-x_p^2} } \\ && F_2P= \sqrt{(x_p+4)^2+13-x_p^2 +6\sqrt{4-x_p^2}} \\ && F_2P= \sqrt{x_p^2+8x_p+16+13-x_p^2 +6\sqrt{4-x_p^2}} \\ && F_2P= \sqrt{29+8x_p+6\sqrt{4-x_p^2}} \\ (PQ + F_1 Q)_{max} &=& 14-\sqrt{29+8x_p+6\sqrt{4-x_p^2}} \\\\ (PQ + F_1 Q)_{max} &=& 14-( 29+8x_p+6(4-x_p^2)^{\frac12} )^{\frac12} \\\\ \text{derivation:} \\ && \boxed{-\dfrac{1}{2}\left(29+8x_p+6(4-x_p^2)^{\frac12} \right)^{\frac12-1} \\ \times \left(8+\dfrac{6}{2}(4-x_p^2)^{\frac12-1}(-2x_p) \right)} \\\\ \text{derivation $= 0$:} \\ && -\dfrac{8-\dfrac{6x_p}{\sqrt{4-x_p^2}} } {2\sqrt{29+8x_p+6\sqrt{4-x_p^2} } } = 0 \\\\ && 8-\dfrac{6x_p}{\sqrt{4-x_p^2}} = 0 \quad | \quad : 2 \\\\ && 4-\dfrac{3x_p}{\sqrt{4-x_p^2}} = 0 \\\\ && \dfrac{3x_p}{\sqrt{4-x_p^2}} = 4 \\\\ && 3x_p = 4\sqrt{4-x_p^2} \quad | \quad \text{square both sides} \\ && 9x_p^2 = 16(4-x_p^2) \\ && 9x_p^2 = 64-16x_p^2 \\ && \ldots \\ && \mathbf{x_p} = \pm\dfrac{8}{5} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline y_p &=& 3\pm \sqrt{4-x_p^2} \quad | \quad x_p = \pm\dfrac{8}{5} \\ \ldots \\ y_p &=& 3\pm\dfrac{6}{5} \\\\ \mathbf{y_p= \dfrac{21}{5} } &\text{or}&\mathbf{y_p= \dfrac{9}{5} } \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline x_p=\dfrac{8}{5},~y_p=\dfrac{9}{5}: & F_2P &=& \sqrt{ \left(\dfrac{8}{5}+4\right)^2 + \left(\dfrac{9}{5}\right)^2 } \\ & F_2P &=& \dfrac{ \sqrt{28^2+9^2 } } {5} \\ & \mathbf{F_2P} &=& \mathbf{5.8821764679} \\ \hline x_p=\dfrac{8}{5},~y_p=\dfrac{21}{5}: & F_2P &=& \sqrt{ \left(\dfrac{8}{5}+4\right)^2 + \left(\dfrac{21}{5}\right)^2 } \\ & F_2P &=& \dfrac{ \sqrt{28^2+21^2 } } {5} \\ & \mathbf{F_2P} &=& \mathbf{7} \\ \hline x_p=-\dfrac{8}{5},~y_p=\dfrac{9}{5}: & F_2P &=& \sqrt{ \left(-\dfrac{8}{5}+4\right)^2 + \left(\dfrac{9}{5}\right)^2 } \\ & F_2P &=& \dfrac{ \sqrt{12^2+9^2 } } {5} \\ & \mathbf{F_2P} &=& \mathbf{3} \\ \hline x_p=-\dfrac{8}{5},~y_p=\dfrac{21}{5}: & F_2P &=& \sqrt{ \left(-\dfrac{8}{5}+4\right)^2 + \left(\dfrac{21}{5}\right)^2 } \\ & F_2P &=& \dfrac{ \sqrt{12^2+21^2 } } {5} \\ & \mathbf{F_2P} &=& \mathbf{4.8373546490} \\ \hline \end{array}\)

\(\text{For the maximum we need the smallest value of $F_2P$: $\quad F_2P = 3$ }\\ \begin{array}{|rcll|} \hline (PQ + F_1 Q)_{max} &=& 14-F_2P \\ (PQ + F_1 Q)_{max} &=& 14-3 \\ \mathbf{(PQ + F_1 Q)_{max}} &=& \mathbf{11} \\ \hline \end{array}\)

Thank you, DragonLord !

Thank you Guest,

Ellipses help

The figure below shows the ellipse \(\dfrac{(x-20)^2}{20}+\dfrac{(y-16)^2}{16}=2016\).

Let  \(R_1,~R_2,~R_3~\)and \(R_4\) denote those areas within the ellipse that are in the first, second, third, and fourth quadrants, respectively.
Determine the value of \(R_1-R_2+R_3-R_4\).

The center of the ellipse is \((20,~16)\)

\(\text{Let the area of ellipse $= A$ } \\ \text{Let the area of D $= 20*16=320$ }\)

\(\begin{array}{|rcll|} \hline R_1 &=& \dfrac{1}{4}A + B + D + E \\ R_2 &=& \dfrac{1}{4}A - B + C \\ R_3 &=& \dfrac{1}{4}A - C - D - F \\ R_4 &=& \dfrac{1}{4}A - E + F \\\\ R_1-R_2+R_3-R_4 &=& \dfrac{1}{4}A + B + D + E \\ && -\left(\dfrac{1}{4}A - B + C \right) \\ && + \dfrac{1}{4}A - C - D - F \\ && -\left(\dfrac{1}{4}A - E + F \right) \\\\ R_1-R_2+R_3-R_4 &=& \dfrac{1}{4}A + B + D + E \\ && -\dfrac{1}{4}A + B - C \\ && + \dfrac{1}{4}A - C - D - F \\ && -\dfrac{1}{4}A + E - F \\\\ R_1-R_2+R_3-R_4 &=& B + D + E \\ && B - C \\ && - C - D - F \\ && E - F \\\\ R_1-R_2+R_3-R_4 &=& 2(B - C + E - F ) \quad | \quad B = D+F,~ E=C+D \\ R_1-R_2+R_3-R_4 &=& 2(D+F - C + C+D - F ) \\ R_1-R_2+R_3-R_4 &=& 2(D+D) \\ R_1-R_2+R_3-R_4 &=& 2(2D) \\ R_1-R_2+R_3-R_4 &=& 4D \quad | \quad D =320 \\ R_1-R_2+R_3-R_4 &=& 4*320 \\ \mathbf{R_1-R_2+R_3-R_4} &=& \mathbf{1280} \\ \hline \end{array}\)

Tyace needs at least 40 hot dogs and 40 buns for a cookout.
Packages of hotdogs cost $3.50 and contains 10 hotdogs.
Packages of buns cost $2.50 and contains 8 buns.
The maximum amount of money Tyace can spend on hot dogs and buns is $70.
What are possible combinations of packages that Tyace can buy?

I assume:

\(\text{Let packages of hotdogs $=x$ and $x_{min} = 4$ packages } \\ \text{Let packages of buns $=y$ and $y_{min} = 5$ packages }\)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{$3.50}{10~hotdogs} * 10x + \dfrac{$2.50}{8~buns} * 8y} &=& \mathbf{$70} \\\\ \dfrac{$3.50}{ hotdogs} * x + \dfrac{$2.50}{ buns} * y &=& $ 70 \\\\ 3.50x + 2.50y &=& 70 \quad | \quad * 2 \\\\ \mathbf{7x + 5y} &=& \mathbf{140} \\ \hline \end{array}\)

Euler :

\(\begin{array}{|rclrclrcl|} \hline \mathbf{7x + 5y} &=& \mathbf{140} \\ 5y &=& 140 - 7x \\ y &=& \dfrac{140 - 7x}{5} \\ y &=& \dfrac{140 - 5x-2x}{5} \\ y &=& 28 - x - \underbrace{\dfrac{2x}{5}}_{=a} \\ y &=& 28 - x - a & a&=&\dfrac{2x}{5} \\ & & & 5a&=&2x \\ & & & 2x&=&5a \\ & & & x&=&\dfrac{5a}{2} \\ & & & x&=&\dfrac{4a+a}{2} \\ & & & x&=&2a+\underbrace{\dfrac{a}{2}}_{=b} \\ & & & x&=&2a+b & b&=& \dfrac{a}{2} \\ & & & & & & 2b&=& a \\ & & & & & & \mathbf{a}&=& \mathbf{2b} \\ & & & x&=&2(2b)+b \\ & & & \mathbf{x}&=&\mathbf{5b} \\ y &=& 28 - 5b - 2b \\ \mathbf{y} &=& \mathbf{28 - 7b} \\ \hline \end{array}\)

The possible combinations of packages:

\(\begin{array}{|c|r|r|} \hline b & x=5b & y = 28 - 7b \\ \hline 1 & 5 & 21 \\ 2 & 10 & 14 \\ 3 & 15 & 7 \\ \hline \end{array}\)

1.   5 packages of hotdogs and 21 packages of buns

2. 10 packages of hotdogs and 14 packages of buns

3. 15 packages of hotdogs and 7 packages of buns

One focus of the ellipse \(\dfrac{x^2}{2} + y^2 = 1 \)is at \(F = (1,0)\)
There exists a point \(P = (p,0)\), where p > 0 such that for any chord \(\overline{AB}\) that passes through \(F\)
angles \(\angle APF\) and \(\angle BPF\) are equal.
Find \(p\).

\(\text{Let $ \angle APF = \angle BPF = \varphi$ }\)

\(\begin{array}{|rcll|} \hline \tan{\varphi} = \dfrac{y_A}{x_P-x_A} &=& \dfrac{-y_B}{x_P-x_B} \\\\ \dfrac{y_A}{x_P-x_A} &=& \dfrac{-y_B}{x_P-x_B} \\\\ \mathbf{\dfrac{y_A}{y_B}} &=&\mathbf{ -\dfrac{(x_P-x_A)}{(x_P-x_B)}} \\ \hline \\ \tan{F} = \dfrac{y_A}{x_F-x_A} &=& \dfrac{-y_B}{x_B-x_F} \\\\ \dfrac{y_A}{x_F-x_A} &=& \dfrac{-y_B}{x_B-x_F} \\\\ \dfrac{y_A}{y_B} &=& -\dfrac{(x_F-x_A)}{(x_B-x_F)} \quad | \quad x_F = 1 \\ \\ \mathbf{\dfrac{y_A}{y_B}} &=&\mathbf{ -\dfrac{(1-x_A)}{(x_B-1)}} \\ \hline \dfrac{y_A}{y_B} = -\dfrac{(x_P-x_A)}{(x_P-x_B)} &=& -\dfrac{(1-x_A)}{(x_B-1)} \\\\ -\dfrac{(x_P-x_A)}{(x_P-x_B)} &=& -\dfrac{(1-x_A)}{(x_B-1)} \\\\ \dfrac{x_P-x_A}{x_P-x_B} &=& \dfrac{1-x_A}{x_B-1} \\\\ (x_B-1)(x_P-x_A) &=& (1-x_A)(x_P-x_B) \\ \cdots \\ \mathbf{x_P} &=& \mathbf{ \dfrac{2x_Ax_B-(x_A+x_B)} {x_A+x_B-2} } \\ \hline \end{array}\)

line \(y=mx+b\):

\(\begin{array}{|rcll|} \hline y &=& mx+b \quad | \quad F = (1,0) \text{ is on the line } \\ y_F &=& mx_F +b \quad | \quad y_F =0 \quad x_F = 1 \\ 0 &=& m*1 +b \\ b &=& -m \\\\ y &=& mx+b \quad | \quad b = -m \\ y &=& mx-m \\ y &=& m(x-1) \quad | \quad \text{equation of the line through F } \\ \mathbf{y^2} &=& \mathbf{m^2(x-1)^2} \\ \hline \end{array}\)

ellipse \(\dfrac{x^2}{2} + y^2 = 1\):

\(\begin{array}{|rcll|} \hline \dfrac{x^2}{2} + y^2 &=& 1 \\ \mathbf{ y^2} &=& \mathbf{1-\dfrac{x^2}{2}} \\ \hline \end{array}\)

intersection line - ellipse:

\(\begin{array}{|rcll|} \hline y^2 = m^2(x-1)^2 &=& 1-\dfrac{x^2}{2} \\\\ m^2(x-1)^2 &=& 1-\dfrac{x^2}{2} \\\\ m^2(x^2-2x+1) &=& 1-\dfrac{x^2}{2} \\\\ m^2 x^2-2m^2x+m^2 &=& 1-\dfrac{x^2}{2} \\\\ m^2 x^2-2m^2x+m^2 -1+\dfrac{x^2}{2} &=& 0 \\\\ x^2\left(m^2+\dfrac{1}{2}\right)-2m^2x+ (m^2 -1) &=& 0 \\\\ x &=& \dfrac{2m^2\pm \sqrt{4m^4 - 4 \left(m^2+\dfrac{1}{2}\right)(m^2 -1)} }{2\left(m^2+\dfrac{1}{2}\right)} \\\\ x &=& \dfrac{2m^2\pm 2\sqrt{m^4 - \left(m^2+\dfrac{1}{2}\right)(m^2 -1)} }{\left(2m^2+\dfrac{1}{2}\right)} \\\\ x &=& \dfrac{m^2\pm \sqrt{m^4 - \left(m^2+\dfrac{1}{2}\right)(m^2 -1)} }{\left(m^2+\dfrac{1}{2}\right)} \\\\ x &=& \dfrac{m^2\pm \sqrt{ \dfrac{1+m^2}{2} } }{\left(m^2+\dfrac{1}{2}\right)} \\\\ \mathbf{ x_A} &=& \mathbf{\dfrac{m^2+ \sqrt{ \dfrac{1+m^2}{2} } }{\left(m^2+\dfrac{1}{2}\right)}} \\\\ \mathbf{ x_B} &=& \mathbf{\dfrac{m^2- \sqrt{ \dfrac{1+m^2}{2} } }{\left(m^2+\dfrac{1}{2}\right)} }\\ \hline \end{array}\)

\(\mathbf{x_A =\ ?}\quad \mathbf{x_B =\ ?}\)
For example we set  \(m = 1 \text{$(~$for any chord $\overline{AB}$ that passes through $F$ angles $\angle APF$ and $\angle BPF$ are equal$~)$}\)

\(\begin{array}{|rcll|} \hline x_A &=& \dfrac{1^2+ \sqrt{ \dfrac{1+1^2}{2} } }{\left(1^2+\dfrac{1}{2}\right)} \\\\ x_A &=& \dfrac{1+ \sqrt{ \dfrac{2}{2} } }{\left(1+\dfrac{1}{2}\right)} \\\\ x_A &=& \dfrac{1+ 1}{\left(1+\dfrac{1}{2}\right)} \\\\ x_A &=& \dfrac{2}{ \dfrac{3}{2} } \\\\ \mathbf{x_A} &=& \mathbf{\dfrac{4}{3}} \\ \hline \\ x_B &=& \dfrac{1^2- \sqrt{ \dfrac{1+1^2}{2} } }{\left(1^2+\dfrac{1}{2}\right)} \\\\ x_B &=& \dfrac{1- \sqrt{ \dfrac{2}{2} } }{\left(1^2+\dfrac{1}{2}\right)} \\\\ x_B &=& \dfrac{1- 1}{\left(1^2+\dfrac{1}{2}\right)} \\\\ x_B &=& \dfrac{0}{\left(1^2+\dfrac{1}{2}\right)} \\\\ \mathbf{x_B} &=& \mathbf{0} \\ \hline \end{array}\)

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

\(\begin{array}{|rcll|} \hline \mathbf{x_P} &=& \mathbf{ \dfrac{2x_Ax_B-(x_A+x_B)} {x_A+x_B-2} } \\\\ x_P &=& \dfrac{2*\dfrac{4}{3}*0-\left(\dfrac{4}{3}+0\right)} {\dfrac{4}{3}+0-2} \\\\ x_P &=& \dfrac{ -\dfrac{4}{3}} {\dfrac{4}{3}-2} \\\\ x_P &=& \dfrac{ -\dfrac{4}{3}} {\dfrac{4-6}{3}} \\\\ x_P &=& \dfrac{ -4} {-2} \\\\ \mathbf{ x_P} &=& \mathbf{2} \\ \hline \end{array}\)

\(P = (p,~0)\\ p=x_P=2 \\ P = (2,~0)\)

In general:

\(\begin{array}{|rcll|} \hline x^2\left(m^2+\dfrac{1}{2}\right)-2m^2x+ (m^2 -1) &=& 0 \quad | \quad :\left(m^2+\dfrac{1}{2}\right) \\\\ x^2-\dfrac{2m^2}{\left(m^2+\dfrac{1}{2}\right)}x+ \dfrac{(m^2 -1)}{\left(m^2+\dfrac{1}{2}\right)} &=& 0 \\\\ x^2\underbrace{-\dfrac{2m^2}{\left(m^2+\dfrac{1}{2}\right)}}_{=-(x_A+x_B)}x+ \underbrace{\dfrac{(m^2 -1)}{\left(m^2+\dfrac{1}{2}\right)}}_{=x_A*x_B} &=& 0 \\ \boxed{\text{Vieta's formulas:}\\ \text{The roots } r_1 ,\ r_2\text{ of the quadratic polynomial } \\ P(x)=a x^2 + b x + c \text{ satisfy}\\ r_1 + r_2 = -\dfrac{b}{a},\ r_1 r_2 = \dfrac{c}{a}. } \\ \hline \mathbf{x_A+x_B} &=& \mathbf{\dfrac{2m^2}{\left(m^2+\dfrac{1}{2}\right)}} \\\\ \mathbf{x_A*x_B} &=& \mathbf{\dfrac{(m^2 -1)}{\left(m^2+\dfrac{1}{2}\right)}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{x_P} &=& \mathbf{ \dfrac{2x_Ax_B-(x_A+x_B)} {x_A+x_B-2} } \\\\ && \mathbf{x_A+x_B} = \mathbf{\dfrac{2m^2}{\left(m^2+\dfrac{1}{2}\right)}} \\\\ && \mathbf{x_A*x_B} = \mathbf{\dfrac{(m^2 -1)}{\left(m^2+\dfrac{1}{2}\right)}} \\\\ x_P &=& \dfrac{2\dfrac{(m^2 -1)}{\left(m^2+\dfrac{1}{2}\right) } -\dfrac{2m^2}{\left(m^2+\dfrac{1}{2}\right)}} {\dfrac{2m^2}{\left(m^2+\dfrac{1}{2}\right)}-2} \\\\ x_P &=& \dfrac{\dfrac{2(m^2 -1)-2m^2}{\left(m^2+\dfrac{1}{2}\right) } } {\dfrac{2m^2}{\left(m^2+\dfrac{1}{2}\right)} -2\dfrac{\left(m^2+\dfrac{1}{2}\right)}{\left(m^2+\dfrac{1}{2}\right)}} \\\\ x_P &=& \dfrac{\dfrac{2(m^2 -1)-2m^2}{\left(m^2+\dfrac{1}{2}\right) } } {\dfrac{2m^2-2\left(m^2+\dfrac{1}{2}\right)}{\left(m^2+\dfrac{1}{2}\right)} } \\\\ x_P &=& \dfrac{2(m^2 -1)-2m^2} {2m^2-2\left(m^2+\dfrac{1}{2}\right) } \\\\ x_P &=& \dfrac{2m^2 -2-2m^2}{2m^2-2m^2-1} \\\\ x_P &=& \dfrac{-2}{-1} \\\\ \mathbf{x_P} &=& \mathbf{2} \quad (p=2) \\ \hline \end{array}\)

homework questions

Using only the paths and directions shown, how many different routes are there from M to N?