A Question of Area

We note that $\triangle BDE$ and $\triangle CDE$ are congruence. Hence $BE = CE$ . Let $\angle EFA = \angle CFE = \theta$ and $t = \tan \theta$ . Then $AE = 3 \tan \theta = 3t$ and $AC = 3 \tan 2\theta = \dfrac {6t}{1-t^2}$ . By Pythagorean theorem ,

$\begin{aligned} BE^2 & = AE^2 + AB^2 = (3t)^2 + 7^2 = 9t^2 + 49 \\ \implies BE & = \sqrt{9t^2+49} & \small \blue{\text{Since }BE=CE=AC-AE} \\ AC - AE & = \sqrt{9t^2+49} \\ \frac {6t}{1-t^2} - 3t & = \sqrt{9t^2+49} \\ \frac {3t(1+t^2)}{1-t^2} & = \sqrt{9t^2+49} \\ 3t(1+t^2) & = (1-t^2)\sqrt{9t^2+49} & \small \blue{\text{Squaring both sides}} \\ 9t^6 + 18t^4 + 9t^2 & = 9t^6+31t^4-89t^2+49 \\ 13t^4 - 98t^2 + 49 & = 0 \\ (t^2-7)(13t^2-7) & = 0 \\ t & = \pm \sqrt 7, \pm \sqrt{\frac 7{13}} & \small \blue{\text{Since }\theta < 45^\circ} \\ \implies t & = \sqrt{\frac 7{13}} \end{aligned}$

The area of $\triangle ABC$ , $[ABC] = \dfrac 12 AB \cdot AC = \dfrac 12 \cdot 7 \cdot \dfrac {6t}{1-t^2} = \dfrac {21\sqrt{\frac 7{13}}}{1-\frac 7{13}} = \dfrac {7\sqrt{91}}2$ . Therefore $a+b+c = 7+91+2 = \boxed{100}$ .

Plot this figure on the Cartesian plane with coordinates $A(0,0), F(3,0), B(7,0), E(0,k), C(0,h),$ where $h > k > 0.$

Since $ED$ is the perpendicular bisector of $CB$ , then $D( \tfrac72, \tfrac h2) .$

The equation of the straight line $CB$ is $\dfrac x7 + \dfrac yh = 1 \Leftrightarrow y = -\dfrac h7 x + h$ . So its gradient $m_{CB} = -\dfrac h7$ .

Also the gradient of the straight line $ED$ is $m_{ED} = \dfrac{k- \frac h2}{0 - \frac 72} = \dfrac{h - 2k}7$ .

Since $ED$ and $CB$ are perpendicular to each other, $m_{ED} \cdot m_{CB} = -1$ , or $\dfrac{h - 2k}7 = \dfrac 7h \tag1$

And because $EF$ bisects $\measuredangle AFC$ , $\begin{array} { r c l} \measuredangle CFB &=& \measuredangle EFA \\ \\ \tan^{-1} \dfrac h3 &=& 2 \cdot \tan^{-1} \dfrac k3 \\ \\ \dfrac h3 &=& \tan \left [ 2 \cdot \tan^{-1} \dfrac k3 \right ] = \dfrac{6k}{9-k^2} \\ \\ h &=& \dfrac{18k}{9-k^2} \end{array}$

Substitute $h$ into $(1)$ and solve the equation yields $(h,k) = (\pm \sqrt7, \mp 3\sqrt7 ) , \left (\pm \sqrt{91}, \mp \sqrt{\dfrac{63}{13}} \right).$

But since $h>k > 0$ , then the only solution is $(h,k) = \left (\sqrt{91}, \sqrt{\dfrac{63}{13}} \right)$ .

Therefore, the area of the triangle $ABC$ is $\dfrac12 \cdot 7 \cdot h = \dfrac{7 \sqrt{91}}2$ . The answer is $7 + 91 + 2 = \boxed{100}.$

Let's find an algebraic expression for the area. If $AF = x$ and $FB = y$ , then the following program suggests that

$\triangle ABC = \dfrac{(x+y)\sqrt {3x^2 + 4xy + y^2}}{2} = \dfrac{7\sqrt 91}{2} \implies 7+91+2 = \boxed{100}$

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"""
diagram: https://www.geogebra.org/m/atn4hkzy

let AF = x, FB = y, and C = (0, ht) where 'ht' is the triangle height.
The strategy is to allow E1 and E2 to be different expressions for E, then equate them and solve for the area.
"""

from sympy import *
x, y, ht = symbols(('x', 'y', 'ht'))

A = Point(0, 0)
B = Point(x + y, 0)
C = Point(0, ht)

F = Point(x, 0)
D = Segment(B, C).midpoint

t = Triangle(A, F, C)
bisector = t.bisectors()[F]
normal = Line(B, C).perpendicular_line(D)
E1 = normal.intersection(Line(A, C))[0]
E2 = bisector.points[1]

ht = solve(E1.y - E2.y, ht)[1]    # set E1 = E2 = E
area = ht*(x + y)/2

print(area)
print(area.subs(x, 3).subs(y, 4))

1
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(x + y)*sqrt(3*x**2 + 4*x*y + y**2)/2
7*sqrt(91)/2