Jtdylktuy

QUESTION:

In a triangle $ABC$, $AD, BE$ and $CF$ are three cevian lines such that $BD:DC = CE:AC = AF:FB = 3:1$. The area of $triangle ABC$ is $100$ unit$^2$. Find the area of $triangle HIG$

Attempt:

First, I thought that it can be done through applying Menelaus's Theorem. Despite having few knowledge about that theorem, I started like that:

As a consequence of Menelaus's theorem, if two cevian $BH$ and $GF$ of $ABG$ meet at $I$ then:

$$frac{HI}{IB} = frac{BF}{AB}.frac{AH}{HG}$$

$implies frac{HI}{IB} = frac{AH}{4HG}......(i)$

And

$$frac{FI}{IG} = frac{3HG}{AG}......(ii)$$

Similarly from $triangle ACI$, I got two more equations and that is:

$frac{CG}{GI} = frac{4GH}{AE}.......(iii)$

$frac{EH}{HI} = frac{IG}{4IC}........(iv)$

And likewise, from $triangle BHC$,

$frac{DG}{HG} = frac{3HI}{BH}........(v)$

$frac{IG}{GC} = frac{BI}{4IH}.........(vi)$

But later on, I thought that it wouldn't be so good for relating with so many sides or segments and also be ineffective instead of relating the area. Then, how could I relate the area using Menelaus's theorem or in any other way like pure geometry?

Thanks in advance.

Let $Nin EC$ such that $DN||BE$ and $NC=y$.

Thus, by Thales $EN=3y$ and $AE=frac{1}{3}EC=frac{4}{3}y,$

which says
$$frac{AH}{HD}=frac{AE}{EN}=frac{frac{4}{3}y}{3y}=frac{4}{9}.$$
Also, let $Min DC$ such that $EM||AD$ and $DM=x$.

Thus, by Thales again we obtain: $MC=3x$, $BD=12x$ and
$$frac{BH}{HE}=frac{BD}{DM}=frac{12x}{x}=12.$$

Similarly, $$AG:GD=CI:IF=12:1$$ and
$$BI:IE=CG:GF=4:9,$$ which gives
$$AH:HG:GD=BI:IH:HE=CG:GI:IF=4:8:1.$$
Now, let $S_{Delta HIG}=s$.

Thus,
$$frac{S_{Delta GFA}}{s}=frac{9cdot12}{8cdot8}=frac{27}{16},$$
which gives
$$S_{Delta GFA}=frac{27}{16}s.$$
Also,
$$frac{S_{Delta AFC}}{frac{27}{16}s}=frac{FC}{FG}=frac{13}{9},$$ which gives
$$S_{Delta AFC}=frac{39}{16}s$$ and since
$$frac{S_{Delta ABC}}{frac{39}{16}s}=frac{4}{3},$$ we obtain:
$$S_{Delta ABC}=frac{13}{4}s$$ and $$s=frac{400}{13}.$$

(Adapted from my proof of the side-trisecting version on AoPS)

First, invert the logic of the construction. Start with an arbitrary triangle $GHI$, and extend $GH$ to $A$ so that $frac{HA}{GH}=frac12$, $HI$ to $B$ so that $frac{IB}{HI}=frac12$, and $IG$ to $C$ so that $frac{GC}{IG}=frac12$. Then, extend $AG$ to meet $BC$ at $D$, $BH$ to meet $AC$ at $E$, and $HI$ to meet $AB$ at $F$. We wish to find the ratio of the area of triangle $GHI$ to that of $ABC$.

(Unlike the figure in the question or that in Anirban Niloy's answer, this figure is actually to scale)

From $GH=2cdot HA$ and the common vertex $I$, the red triangle $GHI$ has twice the area of the yellow triangle $HAI$. Similarly, $GHI$ has twice the area of the other two yellow triangles $IBG$ and $GCH$.

From $GH=2cdot HA$ and the common vertex $C$, the yellow triangle $GHC$ has twice the area of the green triangle $HAC$. Similarly, $HIA$ has twice the area of $IBA$ and $IGB$ has twice the area of $GCB$, so the area of each yellow triangle is twice that of each green triangle.

With three green triangles, three yellow triangles, and one red triangle, the total area is $3+3cdot 2+1cdot 4=13$ times the area of one green triangle. The area of the red triangle $GHI$ is $4$ times the area of a green triangle, for a ratio of $frac{4}{13}$ of the large triangle $ABC$.

That's the area ratios. But we're not done; we still need to show that this is the same as the original configuration. The ratio $frac{AE}{EC}$ is equal to the ratio of areas $frac{ABE}{CBE}$, which is equal to the ratio of areas $frac{ABH}{CBH}$ - one green and one yellow triangle over one green, two yellow, and one red triangle. That's $1+2=3$ times a green triangle in the numerator, and $1+2cdot 2+4=9$ times a green triangle in the denominator, a ratio of $frac13$. Similarly, $frac{BF}{FA}=frac13$ and $frac{CD}{DB}=frac13$. The sides are indeed cut in fourths, and it's the same configuration.

Applying this to the given total area of $100$, the central triangle's area is $frac{4}{13}cdot 100=frac{400}{13}$.

I found a solution but actually it's not mine. I found it on a book and noticed that it is quite different from the Routh's theorem and it has been solved by the pure geometic way. And that is:

Point $C$ and $H$ are connected.

$frac{triangle ADC}{triangle ADB} = frac{1}{3}$ and $frac{triangle HDC}{triangle HDB} = frac{1}{3}$

Hence, $$frac{triangle ACH}{triangle ABH} = frac{1}{3}$$.

Similarly, $$frac{triangle CHB}{triangle ABH} = 3$$

Now, $triangle ABC = triangle ABH + triangle ACH +triangle BHC = triangle ABH + frac{1}{3}triangle ABH + triangle ABH = frac{13}{3} triangle ABH$

So, $$triangle ABH = frac{3}{13} triangle ABC$$

Similarly, $$triangle ACG = frac{3}{13} triangle ABC$$

And, $$triangle BIC = frac{3}{13} triangle ABC$$

Therfore, $triangle HIG = triangle ABC - triangle ACG - triangle BIC - triangle ABH = (1-3*frac{3}{13})triangle ABC = frac{4}{13}*100 = frac{400}{13}$

Hence, we get the area of $triangle HIG = frac{400}{13}$ unit$^2$.