Christmas Streak 23/88: Try Angle Chase

$From~the~left~bottom~fig. ~we~have,\\ Area~\Delta~ABC=1/4*\sqrt{(5+7+8)*(5+7-8)*(5-7+8)*(-5+7+8)}=1/4*\sqrt{20*4*6*10}=10\sqrt3.\\ Area~\Delta~PQR=Area~\Delta~DCR*RQ/RC*RP/RD=Area~\Delta~DCR*3Z/6Z*3Z/4Z=3/8.\\ Area~\Delta~DCR=Area~\Delta~DCA*RD/AD=Area~2/3*\Delta~ABC*4Z/7Z\\ \therefore~Area~\Delta~PQR=Area~3/8*2/3*10\sqrt3*4/7=10\sqrt3/7=0+10\sqrt3/7=a+b\sqrt {c}/d.\\ a+b+c+d=0+10+3+7=\Large~~\color{#D61F06}{20}\\ ~~~~~\\$
Correction:- In the fig to the right, point G and so line EG is not required.
On second line in the note under this fig , please replace
$\color{#3D99F6}{'~~ by~~ \Delta}$ .
That is the start of line two:- " $So~\Delta~EBC ~SD ||.....$ "

THE APPROACH:-
All three trisectors are near to a side in a similar way. AD near AB, BE near BC, CF near CA. So each trisector is divided
by other two in the same ratio, with constant of proportionality z for AD, x for BC and y for CF.
We found first BP:PE::3:4. Then DP:DR::1:4
Thus for any one trisector say BE we have, distances between,
( vertex to nearest intersection point ):(two intersecting points):(second intersecting point and intersection with opposite side):BP:PQ:QE::3:3:1.
Similarly for CF and AD.
There after calculated the area of triangle ABC and linked it to triangle PQR. This was done knowing Ratio of areas of any two triangles with one common vertex and then one common height between two triangles, using ratios found before. We did by using trisector BE and AD.

When all sides are not divided in equal ratios we will have to use different method for getting ratio we obtained as 1:4. We have to stick to trisector BE for this.
Drawing line EG||AB as shown in the fig. to the right plus another line QN||AB, N on BC. Point K as intersection of EG and QC. First using Triangle QCN and KN||QN. Then using triangle BEG and QN||EG. Using same method for other two cevians. In fact at first I had used this method but switched to above since it is shorter.

This explanation to make it easy for those not familiar with such problems.

Let the three trisecting points be D, E, and F. Further let the vector $\vec{u} = \vec{AB}$ and $\vec{v} = \vec{BC}$ , and let's take point $A$ as the origin of the reference frame.

Then, it follows that points on the line $AD$ are given by:

$\vec{p} = x ( u + \dfrac{1}{3} v )$

and points on the line BE are given by

$\vec{q} = u + y (v + \dfrac{1}{3} (- u - v) ) = u + y (-\dfrac{1}{3} u+ \dfrac{2}{3} v)$

and points on the line CF are given by

$\vec{r} = u + v + z (-u-v + \dfrac{1}{3} u ) = u + v + z (-\dfrac{2}{3} u - v)$

Intersecting the first two, we obtain the following system of equations,

$x = 1 - \dfrac{1}{3} y , and \dfrac{1}{3} x = \dfrac{2}{3} y$

The second equation implies $y = \dfrac{1}{2} x$ . Substituting into the first equation,

$x = 1 - \dfrac{1}{6} x \Rightarrow x = \dfrac{6}{7}$

Hence, the point $P_1$ is given by,

$P_1 = \dfrac{6}{7} ( u + \dfrac{1}{3} v ) = \dfrac{6}{7} u + \dfrac{2}{7} v$

From symmetry we can deduce the expressions for $P_2$ and $P_3$ ,

$P_2 = u + \dfrac{6}{7} v + \dfrac{2}{7} (-u-v) = \dfrac{5}{7} u + \dfrac{4}{7} v$

and

$P_3 = u + v + \dfrac{6}{7} (-u-v) + \dfrac{2}{7} u = \dfrac{3}{7} u + \dfrac{1}{7} v$

To compute the area, we compute the following vectors,

$P_1 - P_3 = \dfrac{3}{7} u + \dfrac{1}{7} u$

and

$P_2 - P_3 = \dfrac{2}{7} u + \dfrac{3}{7} v$

And the area of the small triangle is given by

$\text{Area} = \dfrac{1}{2} | (P_1 - P_3) \times ( P_2 - P_3) |$

Note that $u \times u = 0$ and $v \times v = 0$ , hence

$\text{Area} =\dfrac{1}{2} | \dfrac{9}{49} (u \times v) + \dfrac{2}{49} (v \times u) |$

$\text{Area} =\dfrac{1}{2} | (\dfrac{9}{49}-\dfrac{2}{49}) (u \times v) |$

Simplifying, and noting that $\dfrac{1}{2} | u \times v | = \text{Area of triangle ABC}$

$\large \boxed {[P_1 P_2 P_3] = \dfrac{1}{7} [ABC] }$

From Heron's formula, we can easily compute the area of ABC, the semi-perimeter is given by,

$s = \dfrac{a + b + c}{2} = 10$ , therefore,

$[ABC] = \sqrt{s(s-a)(s-b)(s-c)} =\sqrt{ 10 \cdot 2 \cdot 3 \cdot 5 } = 10 \sqrt{3}$

And finally,

$[P_1 P_2 P_3 ] = \dfrac{ 10 \sqrt{3} }{7}$

making the answer $0 + 10 + 3 + 7 = 20$ .

Assume that the triangle is like this:

This problem is application of Routh's Theorem.

So, the area of green triangle in the problem ( or orange in the figure ) = $Area_{\Delta ABC} \left( \dfrac{(xyz - 1)^2}{(xy+y+1)(yz + z + 1)(zx + x + 1)} \right)$

In this problem, we will find out that $x = y = z = 2$ . So,

$\begin{aligned} \text{Area of green triangle} \ & =\sqrt{\dfrac{7+5+8}{2} (10 - 7)(10 - 5)(10 - 8)} \left( \dfrac{((2)(2)(2) - 1)^2}{((2*2) + 2 + 1)((2*2) + 2 + 1)((2*2) + 2 + 1)} \right) \\ & =\sqrt{10(3)(5)(2)} \left( \dfrac{7^2}{7^3} \right) \\ & = \dfrac{10\sqrt{3}}{7} \\ & \implies \boxed{a + b + c + d = 0 + 10 + 3 + 7 = 20}. \end{aligned}$

• Suggestion only: I guess, you may use weights here? Is it applicable?