All Three Concurrency

By Ceva's theorem , $AB||DE$ . $\Delta DEF=\Delta BDE$ and $\angle DAE=\angle BAD=\angle ADE$ -we get $AE=DE$

again Ceva gives, $\displaystyle \frac{AE}{CE}.\frac{CD}{BD}=1$ .......... (1)

$\displaystyle \frac{AE}{CE}=\frac{BD}{CD}= \frac{AB}{AC}$

$\displaystyle \therefore \frac{CE}{AC}=\frac{AE}{AB}=sinC=\frac{AB}{AC}$ $\Rightarrow CE=AB$

from (1) , $\displaystyle \frac{AE}{CE}=\frac{BD}{CD}= \frac{AB}{AC}$ $\therefore \frac{AE}{CE}=\frac{BD}{CD}=\frac{AB}{AC}$

$\therefore \frac{AC}{CE}=\frac{AB+AC}{AC}$ $\Rightarrow AC^2=CE(AB+AC)$ $\Rightarrow AB^2+BC^2=AB(AB+AC)$ $\Rightarrow BC^2=AB.AC=AB$

$\therefore AB^2+AB=1^2 \Rightarrow AB = \frac{\sqrt{5}-1}{2} , BC=\sqrt{ \frac{\sqrt{5}-1}{2}}$

, $\displaystyle \therefore DE=AE=AC-CE=1-AB=1-\frac{\sqrt{5}-1}{2}=\frac{3-\sqrt{5}}{4}$

By chasing angle bisecting theorem , $\displaystyle BD=\frac{AB.BC}{AB+AC}$

Finally , $\Delta DEF=\Delta BDE$ $=\frac{1}{2}.BD.DE$

$\displaystyle =\frac{1}{2}.\frac{AB.BC}{AB+AC}.\frac{3-\sqrt{5}}{4}$

$\displaystyle =\frac{1}{2}.\frac{\frac{\sqrt{5}-1}{2}.\sqrt{ \frac{\sqrt{5}-1}{2}} }{\frac{\sqrt{5}-1}{2}+1}.\frac{3-\sqrt{5}}{4}$

and it gets $\displaystyle \sqrt{\frac{1}{68\sqrt{5}+152}}$ , $a+b+c=225$

Solution outline:

Find that $AB\parallel DE$ by Ceva's Theorem. Now we can effectively ignore $C$ since we are guaranteed to have $CF$ be concurrent to $AD$ and $BE$ as long as $CF$ passes through the midpoint of $DE$ .

Note that $AE=DE$ because $\angle ADE = \angle BAD = \angle DAE$ .

Let $\angle BAD = \theta$ and $BD=x$ . We represent every line segment in terms of $x$ and $\theta$ , and find that we get the equation $1+\dfrac{2}{\sin^2 2\theta}=\dfrac{1}{\tan^2\theta}$ which resolves into $\cos\theta=\dfrac{\sqrt{\sqrt{5}+1}}{2}$ after some algebra.

Thus, $\cos 2\theta = \dfrac{\sqrt{5}-1}{2}$ and so $AB=\dfrac{\sqrt{5}-1}{2}$ . Compute $BC=\sqrt{\dfrac{\sqrt{5}-1}{2}}$ .

We compute that $DE=\dfrac{3-\sqrt{5}}{2}$ by some similar triangle bashing. Draw a altitude from $F$ to $DE$ with foot $H$ and compute that $FH=\sqrt{\dfrac{5\sqrt{5}-11}{2}}$ with some more similar triangle bashing.

Thus, $[DEF]=\dfrac{1}{2}\cdot \left(\dfrac{3-\sqrt{5}}{2}\right)\left(\sqrt{\dfrac{5\sqrt{5}-11}{2}}\right)=\dfrac{1}{\sqrt{152+68\sqrt{5}}}$ and our answer is $152+68+5=\boxed{225}$ .

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I would love to see an elegant solution, since my solution is the first thing I thought of and is the straightforward, expected solution from bashing everything at first glance.

The ratio of AE/EC is in fact the golden ratio. The length of AB has a value of golden ratio as well. You can find the lengths of all segments resulting from intersection of cevians with original triangle sides quite simply. Then you get the area of the triangle DEF = Area ( ABC) - Area (AFE) - Area (FBD) - Area ( EDC). The result has a value of ${\frac{-2 \; \sqrt{\varphi} \; \varphi^{3} + \sqrt{\varphi} \; \varphi}{2}}$

My approach is not the same. Although it is not as preferred, I find that it is special for a numerical analysis.

The critical fact is the area wanted can be found by two different expressions that coincide for fixed dimensions. One is obtainable from the whole triangle minus 3 triangles at each corner, while another one is obtainable from knowing 3 sides, all with one unknown and looking for a known.

Let AF = FB = h. To save from complicated workings, let me just elaborate the results.

$\Delta = h^2 \sqrt{1 - 4 h^2}(1 - \frac {4 h^2}{1+ 2 h})$ ,

Sides of it are $h, h \sqrt {\frac{5 - 6 h}{1 + 2 h}}$ and $2 h (1 - 4 h^2)$ ; $\Delta = \sqrt{s (s - a)(s - b)(s - c)}$ . {Correctness to all terms are guaranteed for comfortable readings.}

Knowing $\displaystyle \frac{1}{\sqrt {a + b \sqrt{c}}}$ , analysis is done for (a, b, c) of:

(152, 68, 5), (285, 11, 3), (1055, -531, 2), (513, -63, 11), (342, -12, 10), (2672, -895, 7) and (3677, -1377, 6).

Actual Excel's proximity:

h = 0.309016994374948

a = 0.309016994374948

b = 0.430884493294898

c = 0.381966011250105

s = 0.560933749459975

Area1 = 0.0573489701224549

Area2 = 0.0573489701224549

Only $\displaystyle \frac{1}{\sqrt {152 + 68 \sqrt{5}}}$ with parameters above shows that "All Three Concurrency" is true!

Area = 0.05734897012245476879526277200116+

It is interesting to meet with this question because all the 7 listed (a, b, c)'s can satisfy Area1 = Area2. Concurrency becomes the only fact to determine the sole answer such that (152, 68, 5) as all others cannot be concurrent.

$\Delta = h^2 \sqrt{1 - 4 h^2}(1 - \frac {4 h^2}{1+ 2 h})= \sqrt{s (s - a)(s - b)(s - c)}$ is not easy to solve if not by iteration from substitutions, while other factors to determine h were not found. f (h) formed seems to be a very high ordered function. With origin at B:

F(0, 0.309016994374948-), C(0.786151377757422+, 0)

A(0, 0.618033988749896-), D(0.300283106000778-, 0)

B(0, 0), E(0.300283106000778-, 0.381966011250105+)

Lengths and angles are all determined as a result. We can draw an exact triangle to see!

152 + 68 + 5 = $\boxed{225}$