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In the figure A G = 2 cm , G B = 1 2 cm , B E = 3 cm , E C = 1 0 cm , C F = 4 cm , F A = 1 1 cm , If the area of shaded region can be expressed as G M Where both are co-prime to each other, then find M + G .
Note :The answer is not as sweet as M and G.
The answer is 18167773.
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2 solutions
Nice solution...but can we solve it by any other method?and can you please elaborate that how by menelaus' theorem AP/PE=13/60?
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Menelaus' theorem,
Given a triangle ABE, and a transversal line GC that crosses AB, BE and EA at points G, C and P respectively,
then, | (AP/PE)(EC/CB)(BG/GA) | = 1
AP/PE * 10/13 * 12/2 = 1
AP/PE = 13/60
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Thanks and I think This is the generalised form for solving these type of problems
Good solution!
I always kept thinking a way to solve it by similarity and stuff like that, seeing no success in that I finally decided to use coordinate geometry. I assumed B = ( 0 , 0 ) and C = ( 1 3 , 0 ) . Through these two we get A = ( 1 3 7 0 , 1 3 1 6 8 ) .
Then I found out the points E , F and G . Then I found out the equations for B F , C G and A E . Then their solutions (or intersections as P , Q and R ). And then the area, of △ P Q R . I had had lost my patience till then but still I tried the answer!
The coordinates are, P = ( 9 4 9 4 7 0 7 , 9 4 9 1 0 0 8 0 ) , Q = ( 6 5 2 0 7 , 2 0 1 5 2 0 1 6 ) and R = ( 6 1 1 6 4 1 7 , 6 1 1 2 0 1 6 ) .
Very very long and tiring method.
@Harshi Singh did you used the generalized formula to solve the question in the first place?
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My problems are usually based on whatever i study on internet or from some similar resources , One day i was studying unknown theorem's of geometry,Thereby i came to know this theorem and thus posted a problem on it. So, yes at first place i use the formula but now Sir ahmad has shown a more delicate way to solve it........