As shown in the diagram below, a star sign consists of five straight lines. It produces five triangles and a pentagon. If areas of five triangles are 3, 10, 7, 15, and 8 square unit respectively. Find the area of the pentagon.
Solution: Draw line AB, BC, CD, DE, EA, and label the area of $5$ newly formed triangles as $a$, $b$, $c$, $d$, and $e$ respectively. Additionally, label the area of the pentagon $FGHIJ$ as $f$, we have the following diagram:
Based on the common ratios of areas for triangles sharing the same bases, we have the following:
$$\dfrac{Area\ \triangle{CDG}}{Area\ \triangle{DEG}}=\dfrac{Area\ \triangle{CBG}}{Area\ \triangle{BEG}}$$
i.e. $$\dfrac{a}{b+15}=\dfrac{e+7}{f+18}\tag{1}$$
Similarly, we have the following additional equations:
$$\dfrac{b}{a+15}=\dfrac{c+8}{f+10}\tag{2}$$
$$\dfrac{c}{b+8}=\dfrac{d+3}{f+25}\tag{3}$$
$$\dfrac{d}{c+3}=\dfrac{e+10}{f+15}\tag{4}$$
$$\dfrac{e}{d+10}=\dfrac{a+7}{f+18}\tag{5}$$
Since there are $6$ variables in the above 5 equations, we can add co-linear constraint $C$, $F$, and $G$.
By using Barycentric Coordinate System for triangles, we have the following non-normalized coordinates for $C$, $F$, and $G$, relative to $\triangle{ABD}$:
$$C=(d+e+10,-(a+e+7),a+f+25)\ \ \ \ \ \ F=(c+8,b,0)\ \ \ \ \ \ G=(e+7,0,a)$$
Because $C$, $F$, and $G$ are co-linear, with Barycentric Coordinate System, we have:
$$\begin{vmatrix} d+e+10 & -(a+e+7) & a+f+25\\ c+8 & b & 0 \\ e+7 & 0 & a \end{vmatrix}=0\tag{6}$$
Expanding the above 6 equations, we have:
$$\begin{cases} \ \ a(f+18)=(e+7)(b+15)\\ \ \ b(f+10)=(a+15)(c+8)\\ \ \ c(f+25)=(b+8)(d+3)\\ \ \ d(f+15)=(c+3)(e+10)\\ \ \ e(f+18)=(d+10)(a+7)\\ \ \ ab(d+e+10)+a(c+8)(a+e+7)-b(e+7)(a+f+25)=0 \end{cases}$$
By using a symbolic Computer Algebra System, such as Wolfram Alpha or Wolfree Alpha, with an input such as:
a>0,b>0,c>0,d>0,e>0,f>0,a(f+18)=(b+15)(e+7), b(f+10)=(c+8)(a+15),
c(f+25)=(d+3)(b+8), d(f+15)=(e+10)(c+3), e(f+18)=(a+7)(d+10),
ab(d+e+10)+a(c+8)(a+e+7)-b(e+7)(a+f+25)=0We have the following unique solution:
$$a=21,\ \ \ \ \ b=20,\ \ \ \ \ c=7,\ \ \ \ \ d=\dfrac{15}{2},\ \ \ \ \ e=14,\ \ \ \ \ f=17$$
Therefore the area of the pentagon is $\boxed{17}$ square unit.
Note: Please refer to Barycentric Coordinates in Olympiad Geometry or Barycentric Coordinates for the Impatient for additional information related to Barycentric Coordinates.