Impossible Test Question | Many Students Failed To Solve This Tricky Geometry Problem 

Centre of gravity always 2:1 ratio so evidently 2/3rds of AB= DE, AB must be in between 6 and 10, As DE must belongs to N then AB=7.5 and AB = 9 which is divisible by (3/2) which will give DE 5 and 6 .

Since the centroid divides the median in ratio 2:1 this leads to CD=2*2/3=4/3 and CE=2*8/3=16/3. Now, DE^2=(4/3)^2+(16/3)^2-2*(4/3)*(16/3)*cos(t) where t is the angle between AC CB. Moreover, | cos t| < 1 , it is clear that 4 < DE < 6.6 with the condition DE a natural number. Therefore, DE=5 and DE=6.

At 3:36 you've shown that CD:DA = 2:1, and therefore CD = 2y and DA = y Using the same reasoning, we have CE:EB = 2:1, and so CE = 2z and EB = z Since CA = 2, we get 3y = 2 → y = 2/3 → CD = 2y = 4/3 Since CB = 8, we get 3z = 8 → z = 8/3 → CE 2z = 16/3 By triangle inequality, we get: 16/3 − 4/3 < CD < 16/3 + 4/3 12/3 < CD < 20/3 4 < CD < 6.66... *CD = 5 or 6*

It can also be 4 with the ‘flat’ triangle, where beta=0 degrees. The solution is easier to reach though. Given the conditions, AB can only be in the range of 6-10. Given similarity between DCE and AB and DE being a centroid line, and hence CD is CA* 2/3, and CE is CB* 2.3, DE is AB times 2/3. With AB in the range of 6-10, DE must be 4-6. However, DE of 4 requires the “triangle to be degenerate to a line, and is therefore arguably excluded. Solutions with beta > 0 (not a stated criteria are 5 and 6. In the real world there are examples where the solution would include the degenerate case: e.g. two linkages connecting other things.

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why does it have 2 answers?

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