Triangle Inside The Ellipse

Relevant wiki: 2D Coordinate Geometry - Problem Solving

For any given slope $t\in\mathbb{R}$ , the equation of the line passing through $A$ is given by $\ell : y=t(x-2)$ . Let $B(x_1,y_1)$ and $C(x_2,y_2)$ be the intersection points of the line $\ell$ and ellipse $E$ . Then the coordinates of $B$ and $C$ satisfy the system of equations $\begin{cases} x^2+9y^2=9\\ y = t(x-2). \end{cases}$ Thus $x_1$ and $x_2$ are the roots of the equation $(1+9t^2)x^2-36t^2x+36t^2-9=0$ with $x_1+x_2 = \dfrac{36t^2}{1+9t^2}\text{ and }x_1x_2 = \dfrac{36t^2-9}{1+9t^2}.\qquad(\star)$ Now, $\overrightarrow{PB}=\langle x_1, y_1-1\rangle=\langle x_1, tx_1-(2t+1)\rangle$ and $\overrightarrow{PC}=\langle x_2, y_2-1\rangle=\langle x_2, tx_2-(2t+1)\rangle$ . Since $\angle{BPC}=90^\circ$ , we have the dot product $\overrightarrow{PB}\bullet\overrightarrow{PC}=0$ . Using the equations of (\star), this dot product can be simplified to $\begin{aligned} 0 = \overrightarrow{PB}\bullet\overrightarrow{PC} &= x_1x_2+(tx_1-(2t+1))(tx_2-(2t+1))\\ &=(1+t^2)x_1x_2-t(2t+1)(x_1+x_2)+(2t+1)^2\\ &= (1+t^2)\cdot\dfrac{36t^2-9}{1+9t^2}-t(2t+1)\cdot\dfrac{36t^2}{1+9t^2}+(2t+1)^2\\ &= \dfrac{4(5t-2)(2t+1)}{1+9t^2}, \end{aligned}$ which gives $t=\frac{2}{5}$ or $t=-\frac{1}{2}$ . If $t=-\frac{1}{2}$ , then $P$ will lie on the line $y=-(x-2)/2$ , which means one of the points $B$ and $C$ will coincide $P$ , a contradiction. Hence, we obtain $t=\frac{2}{5}$ . It follows that (using $\star$ ) $\begin{aligned} BC^2 &= (x_1-x_2)^2+(t(x_1-2)-t(x_2-2))^2\\ &= (1+t^2)(x_1-x_2)^2 = (1+t^2)((x_1+x_2)^2-4x_1x_2)\\ &= (1+t^2)\left[\left(\dfrac{36t^2}{1+9t^2}\right)^2-4\cdot\dfrac{36t^2-9}{1+9t^2}\right]\\ &= \dfrac{36(1+t^2)(1+5t^2)}{(1+9t^2)^2}\\ BC &= \dfrac{6\sqrt{(1+t^2)(1+5t^2)}}{1+9t^2}=\dfrac{18\sqrt{145}}{61} \end{aligned}$ and the distance $d_P$ from $P$ to line $BC$ is given by $d_P=\dfrac{|t(0)-1-2t|}{\sqrt{1+t^2}}=\dfrac{9}{\sqrt{29}}.$ Therefore, $[BPC]=\dfrac{d_P\cdot BC}{2} = \dfrac{1}{2}\cdot\dfrac{9}{\sqrt{29}}\cdot\dfrac{18\sqrt{145}}{61}=\dfrac{81\sqrt{5}}{61}$ and $m+n+p=81+5+61=\boxed{147}$ .

Relevant wiki: 2D Coordinate Geometry - Problem Solving

The line with gradient $m$ passing through the point $(2,0)$ has equation $y \,=\, m(x-2)$ . Since the line cannot pass through $P$ , we must have $m \neq -\tfrac12$ . This line meets the ellipse $x^2 + 9y^2 = 9$ at the points $(x_\pm,y_\pm)$ , where $x_\pm \; =\; \frac{18m^2 \pm 3\sqrt{5m^2+1}}{9m^2+1} \qquad y_\pm \; = \; \frac{-2m \pm 3m\sqrt{5m^2+1}}{9m^2+1} \;.$ The gradients of the lines joining these two points to $(0,1)$ are $m_\pm \; = \; \frac{y_\pm - 1}{x_\pm} \; = \; \frac{-(9m^2 + 2m + 1) \pm 3m\sqrt{5m^2 + 1}}{18m^2 \pm 3\sqrt{5m^2+1}} \;.$ These lines must be perpendicular, so that $m_+ m_- = -1$ , and hence $\frac{(9m^2 + 2m + 1)^2 - 9m^2(5m^2 + 1)}{324m^4 - 9(5m^2 + 1)} \; = \; -1 \;;$ This equation simplifies to $360m^4 + 36m^3 - 32m^2 + 4m - 8 \; =\; 4(9m^2 + 1)(2m+1)(5m-2) \; = \; 0 \;,$ and hence $m=\tfrac25$ , since $m \neq -\tfrac12$ . Thus the points $B,C$ have coordinates $\left( \tfrac{9}{61}\big(8 \pm 5\sqrt{5}\big)\,,\,\tfrac{2}{61}\big(-10 \pm 9\sqrt{5}\big)\right)$ and so the distances $BP,CP$ are $\tfrac{9}{61}\sqrt{290 \pm 44\sqrt{5}} \;.$ The area of the right-angled triangle $BPC$ is $\tfrac12 \times \tfrac{9}{61}\sqrt{290 + 44\sqrt{5}} \times \tfrac{9}{61}\sqrt{290 - 44\sqrt{5}} \; = \; \tfrac{81}{61}\sqrt{5} \;,$ so the answer is $81 + 61 + 5 = 147$ .

Shift the origin to point P.Let the equation of BC in shifted frame be lX+mY=1.Point A becomes (2,-1) in the new frame.Homogenize equation of BC with new equation of ellipse to get joint equation of BP and PC in the new frame.Since BP and PC are perpendicular coefficients of x^2 and y^2 sum to zero.From this we can get equation of BC in new frame.We have the formula for area of triangle between ax^2+2hxy+by^2=0 and lx+my+n=0 as (n^2*sqrt(h^2-ab))/(bl^2-2hlm+am^2).Using this we can find the required area of triangle.

With gradient of line B C = m,

$(1+9m^2) x^2 - 36 m^2 x + 36 m^2 - 9 = 0$

$d$ = $\frac{|m x - y - 2 m|}{\sqrt{m^2 +1}}$ = $\frac{|1 + 2 m|}{\sqrt{m^2 + 1}}$

$D = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$m_1 m_2 = -1$

Excel simulation can handle these figures numerically by ease:

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m = 0.4 
a = 2.44    
b = -5.76   
c = -3.24   
C:  2.82988621290968,   0.331954485163872
B:  -0.469230475204763, -0.987692190081905
d = 1.67125804359347    
B C =   3.55325741669281    
Area1 = 2.969205019303; TRUE
A B =   2.90766930511517;   2.969205019303
A C =   2.04232648745823;   147
Area2 = 2.969205019303; 2.23606797749979
m_AC =  -0.23606797749979;  1.32786885245902
m_AB =  4.23606797749979;   1  20/61 
mp =    -1; 1.32786885245902

Area = $\frac{81 \sqrt5}{61}$

81 + 5 + 61 = 147

Answer: $\boxed{147}$