In the figure $\Delta ABC$ and $\Delta DEF$ are both equilateral and $AB = 8$ cm, $DE = 3$ cm.

Question

In the figure $\Delta ABC$ and $\Delta DEF$ are both equilateral and $AB = 8$ cm, $DE = 3$ cm. Find the possible value of $AD + BE + CF$ from the options.
$(1)$ $6.9$ cm
$(2)$ $7.1$ cm
$(3)$ $5.2$ cm
$(4)$ $8.3$ cm

What I Tried: Here is the picture :-

I think applying Triangle Inequality can just solve the problem.
We notice that :- $$BE + AE > AB$$ $$\rightarrow BE + AD + 3 > 8$$ $$\rightarrow BE + AD > 5$$ Similarly we get that $AD + FC > 5$ , and $FC + EB > 5$ . Adding all these up, we get that $$2(AD + BE + CF) > 15$$ $$\rightarrow AD + BE + CF > 7.5$$ So according to me, the correct option should be $(4)$ .

Am I right? Because I doubt that a question like this shouldn't be that easy. I just need to confirm this.

Answer 1

What you have done is correct, you have ruled out answers $1, 2$ and $3$.
However, the question is broken, the $4^{th}$ answer $8.3\,$cm is also wrong.

Let $a = AD$, $b = BE$ and $c = CF$, we have $AE = a + 3$, $BF = b+3$ and $CD = c+3$.

Consider $\triangle ABE$, Since $\angle BEA$ is an exterior angle for $\triangle DEF$, it equals to $120^\circ$.
This implies

$$\begin{align} & 8^2 = AB^2 = BE^2 + EA^2 - 2\cos(120^\circ)(BE)(EA)\\ \iff & 8^2 = b^2 + (a+3)^2 + b(a+3)\\ \iff & a^2 + ab + b^2 + 6a + 3b = 8^2 - 3^2 = 55\\ \iff & (a^2 + ab + b^2) + \frac92(a+b) + \frac32(a-b) = 55\end{align}$$ Multiply both sides by $a-b$, we have $$(a^3 - b^3) + \frac92(a^2-b^2) + \frac32(a-b)^2 = 55(a-b)$$ We have two similar equalities from $\triangle BCF$ and $\triangle CAD$. $$(b^3 - c^3) + \frac92(b^2-c^2) + \frac32(b-c)^2 = 55(b-c)$$ $$(c^3 - b^3) + \frac92(c^2-b^2) + \frac32(c-a)^2 = 55(c-a)$$ "Summing" these $3$ equalities leads to $$\frac32((a-b)^2 + (b-c)^2 + (c-a)^2) = 0\quad\iff\quad a = b = c$$

So $AD + BE + CF = a + b + c = 3a$ and $a$ is a root of the quadratic equation

$$8^2 = a^2 + (a+3)^2 + a(a+3) \iff 3a^2 + 9a - 55 = 0$$ This quadratic equation has only one positive root $\frac{\sqrt{741}-9}{6}$. So the correct answer is $$\frac{\sqrt{741}-9}{2} \approx 9.1106575888162{\,\rm cm}$$

Answer 2

Let $AD=x$, $BE=y$ and $CF=z$.

Thus, by law of cosines we obtain the following system: $$x^2+(z+3)^2+x(z+3)=64,$$$$y^2+(x+3)^2+y(x+3)=64$$ and $$z^2+(y+3)^2+z(y+3)=64,$$ which gives $$x=y=z=\frac{\sqrt{741}-9}{6}$$ and $$x+y+z=\frac{\sqrt{741}-9}{2}\approx9.11...$$

The system we can solve by the following way.

Let $x-y=u$, $y-z=v$ and $z-x=w$.

Thus, $u+v+w=0$.

Now, the first and the second equations give: $$v(x+y+z)=3u+6w$$ and from the first and the third equations we obtain: $$u(x+y+z)=3w+6v.$$ Now, if $u=0$, so $u=v=w=0$, which gives $x=y=z$ and we obtain the previous solution.

Let $uvw\neq0$.

Thus, $$\frac{u}{v}=\frac{w+2v}{u+2w}$$ or $$\frac{u}{v}=\frac{u-v}{u+2v}$$ or $$u^2+uv+v^2=0$$ or $$\left(u+\frac{v}{2}\right)^2+\frac{3}{4}v^2=0,$$ which is impossible.

Thanks to Christian Blatter, we can prove that $x=y=z$ also by the following way. $$\measuredangle BAE=60^{\circ}-\measuredangle DAE=60^{\circ}-(\measuredangle EDF-\measuredangle DCA)=\measuredangle DCA,$$ which says $$\Delta ABE\cong\Delta CAD,$$ which gives $y=z$.

By the same way we obtain $x=y$.