《求生之路2/生存之旅2》免安装简体中文绿色版

Question

The equilateral triangle $ABC$ is subdivided into five smaller triangles as shown in the figure. The areas of three of these triangles are indicated inside them: $△EBF=18?cm^2$, $△FBC=9?cm^2$ and $△DCF=20?cm^2$.
Determine the area of the central equilateral triangle $△DEF$.
Here is my attempt:
I droped three perpendiculars $h_1$, $h_2$ and $h_3$ from the point $F$ to all three sides of the triangle $ABC$ and found these heights in terms of the side of triangle $ABC$.
Further more I knew the sum of these three heights is equal to $H$, the main height of the triangle $ABC$.
Then I equated the area of the triangle $ABC$ both in terms of $H$ and those three heights, but this led me to nothing and I couldn't go any further.
Any help will be highly appreciated.

Answer 1

Assume first that $E\in AB$ is fixed and $D\in AC$ is variable: the vertex $F$ of the equilateral triangle $DEF$ (positively oriented) lies on a line parallel to $AB$. Similarly, if $D$ is fixed and $E$ is variable $F$ travels on a line parallel to $AC$.

This allows us to solve this preliminary problem: given $F$ in the interior of $ABC$, how to find $D\in AC$ and $E\in AB$ such that $DEF$ is equilateral?

Well, fairly simple: if $D'$ is the intersection between $AC$ and the parallel to $AB$ through $F$, and $E'$ is the intersection between $AB$ and the parallel to $AC$ through $F$, the circumcircle of $FD'E'$ meets the sides $AB,AC$ at the wanted $D,E$ points. In particular $D,D'$ and $E,E'$ are inverses with respect to the circle centered at $A$ which is orthogonal to the circumcircle of $FDE$, whose radius equals $E'D'/\sqrt{3}$.

By angle chasing $DE'$ and $ED'$ are both parallel to $BC$.

So, given the trilinear coordinates of $F$ we are able to find the trilinear coordinates of $D$ and $E$, and also the ratios $[FEB]/[FBC]$ and $[FDC]/[FBC]$. If we impose that these ratios are the ones provided by the problem we are able to find the trilinear coordinates of $F$, then the relative positions of $D$ and $E$ on the $AB$ and $AC$ sides, immediately leading to the solution.

In a explicit way, let us assume that the exact trilinear coordinates (the triple of the distances from the sides) of $F$ are $[x,y,z]$. When we move this point along the parallel to $AB$ the third coordinate does not change, and the sum of the coordinates does not change as well by Viviani's theorem. So $D'=[x+y,0,z]$ and, for the exact same reason, $E'=[x+z,y,0]$. By the previous lemma $D=[x+z,0,y]$ and $E=[x+y,z,0]$.

This gives $DA:DC=y:(x+z)$ and $EA:EB=z:(x+y)$, so

$$\frac{[FEB]}{[FBC]}=\frac{z}{x}\cdot \frac{x+y}{x+y+z},\qquad \frac{[FDC]}{[FBC]}=\frac{y}{x}\cdot\frac{x+z}{x+y+z}.$$

By imposing that the first ratio equals $2$ and the second ratio equals $\frac{20}{9}$ we get $\color{blue}{[x,y,z]\propto\left[1,5,3\right]}$.

Since $\frac{BA}{BE}=\frac{x+y+z}{x+y}=\frac{3}{2}$ we have $[BFA]=\frac{3}{2}18=27$. Similarly $\frac{CA}{CD}=\frac{x+y+z}{x+z}=\frac{9}{4}$ leads to $[CFA]=\frac{9}{4}20=45$. By adding $[FAB],[FBC],[FCA]$ we get that $[ABC]=27+45+9=81$ and also $[AED]=\frac{yz}{(x+y+z)^2}[ABC]=15$. Now

$$\begin{eqnarray*}\color{green}{[FED]}&=&[ABC]-[FEB]-[FDC]-[FBC]-[ADE]\\&=&81-18-20-9-15={\large\color{green}{19}}.\end{eqnarray*}$$

Numerical check:

passed.

Answer 2

Note: figure is not to correct scale.

From $F$ draw a line parallel to $BC$ and let it intersect $AB$ and $AC$ at $G$ and $H$, respectively. Note that $\triangle{AGH}$ is equilateral. Also note that $|AE| = |GF| = |HD|$ and $|EG| = |FH| = |DA|$.

For ease of writing, let us say $|AE| = a$, $|EG| = b$, and $|GB| = c$.

Let us also refer to the lengths of perpendiculars from $F$ to $BC$, $CA$, and $AB$ as $h_1$, $h_2$, and $h_3$, respectively.

As shown in the figure, in $\triangle{AGH}$ there are $\triangle{DEF}$, whose area is desired, and three other triangles that are congruent with one another. We have: $$h_1 (a+b+c) = 18$$ $$h_2 (a+c) = 40$$ $$h_3 (b+c) = 36$$ $$h_1 + h_2 + h_3 = \frac{\sqrt{3}}{2} (a+b+c)$$ $$\frac{h_1}{h_2} = \frac{c}{b}$$ $$\frac{h_1}{h_3} = \frac{c}{a}$$ (The last two equations come from area calculations for $\triangle{FHC}$ and $\triangle{FGB}$, respectively)

The above six independent equations form a system with six unknowns. The interested reader can solve the system as an exercise to obtain: $$a = 2 \times 3^{\frac34} \quad b = \frac{10}{3^{\frac14}} \quad c =\frac{2}{3^{\frac14}} $$ $$h_1 = 3^{\frac14} \quad h_2 = 5 \times 3^{\frac14} \quad h_3 = 3 \times 3^{\frac14} $$ Now we can calculate the desired area: $$S_{\triangle{DEF}} = S_{\triangle{AGH}} - 3S_{\triangle{FDH}}$$ $$= \frac{\sqrt3}{4}(a+b)^2 - \frac32 a \times h_2$$ $$=64 - 3 \times 15 = 19 $$

Answer 3

Here is a complex geometry solution.

Let us shrink the figure by an homothety with ratio $1/k$ and rotate it in such a way that

$$A=1, \ \ B=w, \ \ C=w^2 \ \ \text{where} \ \ w:=e^{i \tfrac{2 \pi}{3}} \ \ \text{(3rd roots of unity)}$$

Let

$$D:=(1-d)+dw^2, \ \ E:=(1-e)+ew$$

from which we can deduce the expression of $F$ by using the fact that this point is situated is situated on the perpendicular bissector of line segment $[DE]$ :

$$F=\frac{D+E}{2}+i\sqrt{3}\frac{E-D}{2}$$

Writing :

$$\begin{cases}\text{area}(F,E,w)&=&18 k\\ \text{area}(F,w^2,D)&=&20 k\\ \text{area}(w,w^2,F)&=&9k\end{cases} \ \iff \ \begin{cases} -3/4\sqrt{3}e^2 + 3/4\sqrt{3}e &=& 18k\\ -3/4\sqrt{3}d^2 + 3/4\sqrt{3}d&=&20k\\ -3/4\sqrt{3}d - 3/4\sqrt{3}e + 3/4\sqrt{3}&=&9k\end{cases},\tag{1}$$

we get a system of 3 equations in the 3 unknowns $d,e,k$.

Remark : the (signed) area of a triangle given by the complex coordinates of its vertices can be obtained by the following formula :

$$\text{area}(A,B,C)=\tfrac12\Im((B-A)(C-A)^*)$$

where $\Im$ stands for imaginary part and $*$ for complex conjugation.

The solutions of system (1) are :

$$d=\frac59, \ \ e=\frac13, \ \ k=\frac{\sqrt{3}}{108}.$$

We now have all the ingredients for computing the area of triangle $DEF$, taking into account the shrinking factor $k$, with a final result $S=19$, as awaited.

I have carried out these computations using SAGE, a now widespread (free) Computer Algebra System: :

var('d e k')
w=exp(I*2*pi/3)
ar(w1,w2,w3)=(1/2)*((w3-w2)*((w3-w1).conjugate())).imag()
assume(d,'real')
assume(e,'real')
D=(1-d)+d*w^2
E=(1-e)+e*w
F=(D+E)/2+i*sqrt(3)*(E-D)/2
s=solve([ar(F,E,w)==18*k,ar(F,w^2,D)==20*k,ar(w,w^2,F)==9*k],d,e,k)
v=s[1]
show(v)
k=k.subs(v)
D=D.subs(v)
E=E.subs(v)
F=F.subs(v)
show(ar(D,E,F)/k)