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Regional Olympiad Problem


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Note by Rohit Camfar
11 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.

Yatin Khanna - 11 months, 3 weeks ago

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Eureka , Nice approach . But can be done even more simply

Vishwash Kumar - 11 months, 3 weeks ago

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Vishwash Kumar - 11 months, 3 weeks ago

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Keep it pure keep it geometric .

Vishwash Kumar - 11 months, 3 weeks ago

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this seems to be your signature dialogue hahahaha

Neel Khare - 11 months, 2 weeks ago

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No

Rohit Camfar - 11 months, 2 weeks ago

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@Rohit Camfar TRy trig

Rohit Camfar - 11 months, 2 weeks ago

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@Rohit Camfar It is it's a compliment not anything bad

Neel Khare - 11 months, 2 weeks ago

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@Neel Khare I know that

Rohit Camfar - 11 months, 2 weeks ago

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@Rohit Camfar Has your exam been postponded

Rohit Camfar - 11 months, 2 weeks ago

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@Rohit Camfar No bye

Neel Khare - 11 months, 2 weeks ago

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@Neel Khare Exams are being held too late

Rohit Camfar - 11 months, 2 weeks ago

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@Rohit Camfar Can;t believe

Rohit Camfar - 11 months, 2 weeks ago

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