# Inequality regarding areas of triangles

BdMO Nationals 2013:

There is a point O inside ∆ABC. After joining A,O; B,O and C,O extend those line and they will intersect BC, AC and AB at points D, E and F respectively. AF:FB=4:3 and area of ∆BOF and ∆BOD is 60 and 70 square unit respectively. Find the triangle with the largest area among ∆AOF, ∆AOE, ∆COE and ∆COD and write down the area of that one.

NOTE: No trigonometry allowed.

Area of $\triangle AOF$ is obviously $80$.But I can't see how to compute the other areas.I am not even sure if I need to.Here are some of my further works.

$\dfrac{\triangle AFC}{\triangle BCF}=\dfrac{AF}{FB}=\dfrac{4}{3}$

But this implies that $3\triangle AOC=280+4\triangle COD$.But this also tells us that

$3\triangle AOC>282\implies \triangle AOC>94$

$3\triangle AOC=280+4\triangle COD$ also implies that

$3\Delta AOC<282+6\Delta COD$ meaning $\Delta AOC<94+2\Delta COD$.Is this enough to derive that $2\Delta COD>\Delta AOC$? A hint will be appreciated.

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By $S(...)$ I mean area. Suppose $S(AOE)=x,S(OEC)=y,S(OCD)=z$. We have $AO=2OD$ and $AF/BF=4/3$ so:

$x+y=2z \\ \frac{x+y+80}{z+130}=\frac{4}{3}$

Now you can find $z$ and after that you can find $\frac{x}{y}$.

Good luck!

How to find $x/y$:

$\frac{x}{S(AOB)}=\frac{y}{S(BOC)}=\frac{OE}{OB}$

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Can you explain the $AO=2OD$ part? –  rah4927 Dec 14 '13 at 12:55
$S(AOB)=2S(BOD)$ –  hhsaffar Dec 14 '13 at 12:56
Therefore $z=140$,which implies that $x+y=280$. If we figure out $\dfrac{x}{y}$,we can compute them as well. –  rah4927 Dec 14 '13 at 13:09
Yes, it is correct. –  hhsaffar Dec 14 '13 at 13:10
I updated my answer. Hold you mouse over the gray part. –  hhsaffar Dec 14 '13 at 13:23

Refer to a figure of the question. Call $\delta(PQR)$ as the area of some triangle PQR.

Let $BD/DC=m$. Therefore by Ceva's Theorem, $AE/EC=(4m/3)$. (Given $AF/FB=4/3$).

Since we know $\delta(OBD)$ and the ratio in which D divides BC, $\delta(ODC)=70/m$

You have already figured out that $\delta(AOF)=80$. Let $S=$\delta(OAE)+\delta(OEC)$. Consider triangles$AFC$and$BFC$. Since F divides AB in the ratio 4/3, we have$\delta(AFC)/4=\delta(BFC)/3, or (130+70/m)/3=(S+80)/4$which gives$S=(280/3)+(280/(3m))....(1)$. Let$BC=a$and length of altitude from A onto BC be$H$. Now$\delta(ABD)=(m/(m+1))*\delta(ABC)$which gives$[(maH)/(2(m+1))]=80+60+70=210.....(2)$Now$[aH/2]=210+(70/m)+S$. Putting this in$(2)$and using$(1)$, we get$m=1/2$which gives$S=280=\delta(AOE)+\delta(OEC)$. Also$\delta(AOE)/\delta(OEC)=AE/EC=2/3$, which gives$\delta(AOE)=112$and$\delta(OEC)=168$. Also$\delta(ODC)=70/m=140$. Clearly$\delta(OEC)\$ is the maximum.

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don't mind the writing. I don't know latex properly. –  Apurv Dec 14 '13 at 12:31
have edited the answer. Hope the solution is now ok... –  Apurv Dec 14 '13 at 13:03
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