triangles , pentagon area

Question

A star sign consists of five straight lines. It produces five big and five small total ten triangles and a pentagon. If areas of five small triangles are 30, 25, 24, 25 and 10 square unit respectively then what is the area of pentagon?

Answer 1

(I am very indetebted to @Oscar Lanzi who has pinpointed an error of mine on barycentric coordinates signs, yielding a false conclusion).

1) First method

Fig.1.

Let us use the natural notations one can find on Fig. 1: $f$ for the central pentagon area, $a,b,c,d,e$ for triangular areas between spikes, etc.

We can write the following equations expressing common ratios of areas for triangles sharing the same bases. For example, for the first one, we have the equality of areas' ratios: $\dfrac{[ACB_1]}{[CDB_1]}=\dfrac{[EAB_1]}{[DEB_1]}$, i.e., $\dfrac{f+54}{10+a}=\dfrac{25+c}{b}$:

$$\begin{cases}b(f+54)&=&(25+c)(10+a)\\ c(f+35)&=&(24+d)(25+b)\\ d(f+55)&=&(25+e)(24+c)\\ e(f+34)&=&(30+a)(25+d)\\ a(f+50)&=&(10+b)(30+e)\end{cases}\tag{1}$$

To these equations, we can add the following alignment constraint of points $B,A_1,E_1$

$$\left|\underbrace{\begin{matrix} \ \ c+d+24\\ -(d+e+25)\\ \ \ e+f+80\end{matrix}}_{B} \ \ \underbrace{\begin{matrix}b+10\\a\\ 0\end{matrix}}_{A_1} \underbrace{\begin{matrix}d+25\\0\\ e\end{matrix}}_{E_1}\right|=0\tag{2}$$

using their (non-normalized) barycentrical coordinates with respect to triangle $CEA$.

Submitting system (1)+(2) (6 equations with 6 unknowns) to a symbolic Computer Algebra System gives many $6$-uples of "solutions", some with real values, some with complex values, only one of them being acceptable:

$$f=48.1691,$$

with:

$$a=26.1468, \ \ b=22.9426, \ \ c=39.8474, \ \ d=45.1259, \ \ e=47.9176.$$

2) 2nd method (that has my preference)

Let us take another direction giving the solution and, at the same time, an effective placement of the vertices.

It is possible WLOG to assume that $AC \perp AD$ (by using a certain transvection which doesn't change areas ; see remark 1 below). Taking $A$ as the origin and $AC$ and $AD$ as axes of a coordinate system, we have $12$ unknowns $a,b,c,...$ (see fig. 2) with $11$ equations:

$$\begin{cases}\ \ \ ad&=&48\\ \ \ \ q(c-b)&=&60\\-s(b-a)&=&50\\-t(e-d)&=&50\\ \ \ \ p(f-e)&=&20 \end{cases} \ \ \ \ \ \ \begin{cases}\dfrac{p}{f-q}&=&\dfrac{b}{f}&=&\dfrac{r}{f-s}\\ \dfrac{q}{c-p}&=&\dfrac{e}{c}&=&\dfrac{u}{c-t}\\ \dfrac{-s}{r-a}&=&\dfrac{d}{a}&=&\dfrac{u}{a-t}\end{cases}$$

which are resp. the area constraints and the alignments constraints (or, more intuitively, equalities between slopes):

Fig. 2: A possible realization of the star. The points on the $x$ axis are: $(a=4;0), \ (b=6.216;0), \ (c=10.463;0)$, and on the $y$ axis $(0;d=12), \ (0;e=19.529), \ (0;f=26.438)$ with intersection points $p;q)=(2.894;14.126)$, $(r;s)=(11.521;-22.563)$ and $(t,u)=(-6.641;31.924).$

As there are $11$ equations and $12$ unknowns, we have to fix one of them; this is what we have done here by taking arbitrarily $a=4$. The solution given in Fig. 2 has been obtained by a CAS (as in the first method). It is worth mentionning the equations we have used expressing the values of the areas

Remarks:

1. Given fixed axes, there are two different kinds of transvections (also called shears) are: $\begin{pmatrix}1&\alpha\\0&1\end{pmatrix}$ or $\begin{pmatrix}1&0\\ \beta&1\end{pmatrix}$ for any $\alpha$ or any $\beta$. They preserve areas (and alignments, of course). They belong to the special linear group $SL_2$ of matrices with determinant $1$.

2. See as well: https://mathoverflow.net/q/143664

3. Slightly connected, the beautiful figures and properties in this article of Forum Geometricorum (Volume 16 (2016) 347–354) "Two Six-Circle Theorems for Cyclic Pentagons" authored by Grégoire Nicollet (http://forumgeom.fau.edu/FG2016volume16/FG201644.pdf)

4. Slightly connected too : Does the inner pentagon inside a Robbins pentagon $also$ have a rational area?

5. Here is a recent answer of mine where, as in the first solution, barycentrical coordinates are used in a pentagon.

6. Fig. 2 has a lot in common with the figure given by Christion Blatter here for a cousin issue I discovered today (2022-03-15).