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# Geometry doubt....?

Update:

O is a random point on MN .

CO is extended upto C' .

BO is extended upto B' .

Update 2:

Great solution sir (@atsuo ) !

Thank you very very much !

_______________________________

Very smart approach sir (@Pope) !

Thank you very very much !

Update 3:

Amazing solution !

Thank you so so much dear sir ! (@Indica) ### 4 Answers

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Let MO = x and ON = y. And let S(@) be the area of @.

We know BC = 2MN = 2(x+y) = 2x+2y, so

BC' : MC' = 2x+2y : x

BM = BC' - MC', so

BC' : BM = 2x+2y : x+2y

BC' = BM*(2x+2y)/(x+2y)

AM = BM, so BM = (1/2)BA. Therefore

BC' = (1/2)BA*(2x+2y)/(x+2y) = BA*(x+y)/(x+2y)

So S(C'BC) = S(ABC)*[(x+y)/(x+2y)] ---(#1)

And C'A = BA - BC' = BA*y/(x+2y) ---(#2)

Similarly, if we exchange x←→y and B←→C  then we can find

CB' = CA*(x+y)/(2x+y)

S(B'BC) = S(ABC)*[(x+y)/(2x+y)] ---(#1')

B'A = CA - CB' = CA*x/(2x+y) ---(#2')

Next, we can find

S(Red) = (1/2)C'A*B'A*sin(∠BAC)

= (1/2)[BA*y/(x+2y)][CA*x/(2x+y)]*sin(∠BAC) ... by (#2),(#2')

= (1/2)[BA*CA*sin(∠BAC)]*[xy/((x+2y)(2x+y))]

= S(ABC)*[xy/((x+2y)(2x+y))] ---(#3)

And the height of △BOC is 1/2 of the height of ABC, so

S(OBC) = S(ABC)*[1/2] ---(#4)

The area of quadrilateral AC'OB' becomes

S(AC'OB') = S(ABC) - [S(C'BC) + S(B'BC) - S(OBC)]

= S(ABC)*[1 - ((x+y)/(x+2y) + (x+y)/(2x+y) - 1/2)] ... by (#1),(#1'),(#4)

= S(ABC)*[1 - (2(x+y)(2x+y) + 2(x+y)(x+2y) - (x+2y)(2x+y))/(2(x+2y)(2x+y))]

= S(ABC)*[1 - (4x^2+7xy+4y^2)/(2(x+2y)(2x+y))]

= S(ABC)*[3xy/(2(x+2y)(2x+y))]

S(Green) = S(AC'OB') - S(Red)

= S(ABC)*[3xy/(2(x+2y)(2x+y)) - xy/((x+2y)(2x+y))]

= S(ABC)*[xy/(2(x+2y)(2x+y)]

Therefore, S(Red)/S(Green)

= [xy/(x+2y)(2x+y)] / [xy/(2(x+2y)(2x+y))]

= 2 <--- the answer

• This might be of interest  .... • Here is a trick I used to employ in exams when I found myself running out of time. I would take a limiting case or a special case. For this question, let me take a special case, where point O is the midpoint of MN. Surely you see all the parallel lines and similar triangle resulting from this relation.

∆OB'N ~ ∆BB'C

B'N : B'C = ON : BC = 1 : 4

B'N : NC = 1 : 3

B'N : AN = 1 : 3

AB' : B'N = 2 : 1

This ratio is equal to the ratio of heights of ∆AB'C' and ∆OB'C', where they have common base B'C'. That therefore is the ratio of areas.

This would never do as a proof for the general case. However, it does solve this one case. If this ratio is not the same in all cases, then the question itself is incomplete. • Anonymous
2 months ago

The question is incomplete as it doesn't explain how points B', C' and O are chosen.

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