Ufdjrw

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:

$$fracHIIB = fracBFAB.fracAHHG$$

$implies fracHIIB = fracAH4HG......(i)$

And

$$fracFIIG = frac3HGAG......(ii)$$

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

$fracCGGI = frac4GHAE.......(iii)$

$fracEHHI = fracIG4IC........(iv)$

And likewise, from $triangle BHC$,

$fracDGHG = frac3HIBH........(v)$

$fracIGGC = fracBI4IH.........(vi)$

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 $AE=frac13EC=frac43y,$

which says
$$fracAHHD=fracAEEN=fracfrac43y3y=frac49.$$
Also, let $Min DC$ such that $EM||AD$ and $DM=x$.

Thus, by Thales again we obtain: $MC=3x$, $BD=12x$ and
$$fracBHHE=fracBDDM=frac12xx=12.$$

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,
$$fracS_Delta GFAs=frac9cdot128cdot8=frac2716,$$
which gives
$$S_Delta GFA=frac2716s.$$
Also,
$$fracS_Delta AFCfrac2716s=fracFCFG=frac139,$$ which gives
$$S_Delta AFC=frac3916s$$ and since
$$fracS_Delta ABCfrac3916s=frac43,$$ we obtain:
$$S_Delta ABC=frac134s$$ and $$s=frac40013.$$

(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 $fracHAGH=frac12$, $HI$ to $B$ so that $fracIBHI=frac12$, and $IG$ to $C$ so that $fracGCIG=frac12$. 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 $frac413$ 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 $fracAEEC$ is equal to the ratio of areas $fracABECBE$, which is equal to the ratio of areas $fracABHCBH$ - 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, $fracBFFA=frac13$ and $fracCDDB=frac13$. 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 $frac413cdot 100=frac40013$.

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.

$fractriangle ADCtriangle ADB = frac13$ and $fractriangle HDCtriangle HDB = frac13$

Hence, $$fractriangle ACHtriangle ABH = frac13$$.

Similarly, $$fractriangle CHBtriangle ABH = 3$$

Now, $triangle ABC = triangle ABH + triangle ACH +triangle BHC = triangle ABH + frac13triangle ABH + triangle ABH = frac133 triangle ABH$

So, $$triangle ABH = frac313 triangle ABC$$

Similarly, $$triangle ACG = frac313 triangle ABC$$

And, $$triangle BIC = frac313 triangle ABC$$

Therfore, $triangle HIG = triangle ABC - triangle ACG - triangle BIC - triangle ABH = (1-3*frac313)triangle ABC = frac413*100 = frac40013$

Hence, we get the area of $triangle HIG = frac40013$ unit$^2$.