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Note by Rohit Camfar
9 months, 3 weeks ago

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This can be easily done.
Consider a triangle ABC with centroid O and medians AD; BE and CF.
Now;
Let; $$ar(AOE) = ar(COE) = a$$;
$$ar(COD) = ar(BOD) = b$$;
$$ar(BOF) = ar(AOF) = c$$;

But; since $$ar(ADB) = ar(BEA) = \frac{ar(ABC)}{2}$$
$$\implies ar(AOE) = ar(BOD)$$
OR $$a = b$$

Similarly, we can prove that in fact;
$$a = b = c = \frac{ar(ABC)}{6}$$

Now, using the fact that triangles AOF and AOC share the same height but the area of the latter is twice the former; we can easily deduce that CO:FO = 2:1.

Similarly we can do for the other 2 medians.

Sorry for being unable to upload picture. · 9 months, 2 weeks ago

Eureka , Nice approach . But can be done even more simply · 9 months, 2 weeks ago

· 9 months, 2 weeks ago

Keep it pure keep it geometric . · 9 months, 2 weeks ago

this seems to be your signature dialogue hahahaha · 9 months, 1 week ago

No · 9 months, 1 week ago

TRy trig · 9 months, 1 week ago

It is it's a compliment not anything bad · 9 months, 1 week ago

I know that · 9 months, 1 week ago

Has your exam been postponded · 9 months, 1 week ago

No bye · 9 months, 1 week ago

Exams are being held too late · 9 months, 1 week ago