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O is a random point on MN .
CO is extended upto C' .
BO is extended upto B' .
Great solution sir (@atsuo ) !
Thank you very very much !
Very smart approach sir (@Pope) !
Thank you very very much !
Amazing solution !
Thank you so so much dear sir ! (@Indica)
- ?Lv 61 month agoFavourite answer
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')
= 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(Green) = S(AC'OB') - S(Red)
= S(ABC)*[3xy/(2(x+2y)(2x+y)) - xy/((x+2y)(2x+y))]
= [xy/(x+2y)(2x+y)] / [xy/(2(x+2y)(2x+y))]
= 2 <--- the answer
- IndicaLv 71 month ago
This might be of interest ....
- PopeLv 71 month ago
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.
- Anonymous1 month ago
The question is incomplete as it doesn't explain how points B', C' and O are chosen.