find a certain ratio of areas in a triangle where other areas are given

Question

Given : [BDE]=8, [BDC]=12, [CDF]=9

where [.] is the area of the respective triangles.

Find ratio [AFD]/[AED] from the figure.

NOTE:-Question may have errors please just edit that or comment the error.

Answer 1

Let us give the names $x:=[ADF]$ and $y:=[ADE]$.

Fig. 1.

We are going to prove that

$$x=21 \ \text{and} \ y= 20 \ \ \implies \ \ [ADF]/[ADE]=x/y=21/20.$$

In fact, we have :

$$\dfrac{x+y+9}{20}=\dfrac{9+x}{12} \ (a) \ \ \ \ \text{and} \ \ \ \ \dfrac{x+y+8}{21}=\dfrac{8+y}{12} \ (b)\tag{1}$$

Let us explain the first equality (a)

(1)(a) is equivalent to

$$\dfrac{[ACE]}{[BCE]}=\dfrac{[ACD]}{[BCD]} \tag{2}$$

(2) can be transformed in the following equivalent expression :

$$\dfrac{\tfrac12 CE \times h_1}{\tfrac12 CE \times h_2}=\dfrac{\tfrac12 CD \times h_1}{\tfrac12 CD \times h_2}\tag{3}$$

Explnation for the equivlence between (2) and (3) : We have considered triangles $AEC$ and $BEC$ in particular as having a common base $EC$ ; we have defined $h_1$ (resp. $h_2$) as the height, i.e., the length of the altitude dropped from $A$ (resp $B$) on this common base $EC$ (we have used the classical formula for the area of a triangle : $\tfrac12$ base $\times$ height).

(3) is evidently true ! Thus (1)(a) is established.

The second identity (1)(b) can be explained in exactly the same way.

Now, convert (1) (a) and (b) into a linear system,

$$\begin{cases}-2x&+&3y&=&18 \ \ \ (a)\\ \ \ \ 4x&-&3y&=&24 \ \ \ (b)\end{cases}$$

and solve it to obtain $x$ and $y$.

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Edit : 2 important issues

1) An existence issue could have occured. Indeed, we have been here in the logic of a "necessary condition" : if such a configuration of points exists with these area values $8, 12, 9$ then $x=...$, $y=...$. But, the set of conditions could have been self-contradictory (for example with areas $10,1,10$ instead of $8,12,9$). Happily, this is not the case. There are solutions ; there are even many of them (see Fig. 2) ; this aspect is developed in the next section.

2) As always, it is interesting to ask, in particular for these geometrical issues, what are the invariant transformations.

a) It is clear that horizontal transvections (horizontal shears) :

$$\text{for any} \ u, \ \ S_u=\begin{pmatrix}1&u\\0&1\end{pmatrix}$$

do not modify areas because their determinant is one.

b) But there is another family of transformations, with matrices with determinant 1 as well :

$$\text{for any} \ v \neq 0, \ \ R_v=\begin{pmatrix}v&0\\0&1/v\end{pmatrix}$$

that have the same effect : indeed, starting from a solution like the blue triangle on Fig. 2, one can enlarge its base $BC$ by a factor $v$ while compressing the vertical direction (i.e. the ordinates of all points) using a factor $1/v$.

Combining operations $S_u$ and $R_v$, we generate all triangles that are solutions to the problem. One could add a general rotation $T_{\theta}$ to account for a supplementary degree of freedom in the group $SL(2,\mathbb{R})$ of matrices with determinant 1, which is a necessary and sufficient condition for area preservation. Doing that, combination $T_{\theta}R_vS_u$ is called the Iwasawa decomposition. See http://www.math.polytechnique.fr/xups/xups07-02.pdf.

Fig. 2 : Different solutions obtained from an initial one (blue triangle) by submitting it either to horizontal shears (red triangles, using $S_u$) or to extensions-compressions (magenta triangles, using $R_v$).

A key observation for obtaining explicit general coordinates of points $A,B,C,D,E,F$ (this constitute a second method I will not expand here) is to observe that, once the base $BC$ is fixed, $D$ has to be chosen on a certain horizontal line, the same for $E$ and $F$. One finds in this way the following general form (attesting the dependency on two parameters $d$ and $w$ that can take any $>0$ value) :

$$\begin{cases} A&=&(a;70/w) \ \text{with} \ a=35d/6+w/6\\ B&=&(-w;0)\\ C&=&(w;0)\\ D&=&(d;12/w)\\ E&=&(e;20/w) \ \text{with} \ e=5d/3 - 2w/3\\ F&=&(f;21/w) \ \text{with} \ f=7d/4 + 3w/4 \end{cases}$$

Please note that parameters $d$ and $w$ have a clear geometrical interpretation : $2w$ is the width $BC$ of triangle $ABC$, and $d$ is the abscissa of $D$. Coordinates of $D$ account for the fact that the area of triangle $BCD$ is $12$ ; the ordinates of $E$ and $F$ can be understood in the same way. Abscissas of $A,E$ and $F$ look more complicated : they have been obtained by alignment constraints ($C,D,E$ must be aligned ; the same for $BDF$).