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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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TopNewestI don't think this is from IMO (International Mathematical Olympiad).

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This is from an exam conducted here in India by a private company.

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Oh, I see. Indian Mathematical Olympiad?

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If the private organisations use names such as "IMO" and mislead the students, then its a very bad tactic to promote themselves.

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Hey guys,I think she is talking of SOF IMO and not the Great and the one you all are thinking IMO!!!!

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YOU,RE RIGHT!. It is a problem from the work book.

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Consider the sides of the rhombus to be of length $x$, i.e., $AD=x$. So, $AE= \frac{x}{2}$.

Let $\angle A= \theta$. So, $cos \theta = \frac{AE}{AD} =\frac{1}{2}$

$\implies \theta =60^o$ $\implies \angle A=60^o$ and $\angle B=180^o-60^o=120^o$(As they are interior opposite angles)

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This que is not from IMO !

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I think is 180 degrees

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Consider- $DA = x$

$AE = x/2$

$EB = x/2$

$DE = x^{2} - (x/2)^{2}$ $= \sqrt{3}x/2$

$DB = x$ (

Pythagoras Theorem)$DA = x , DB = x , AB = x$

$ADB$ is an Equilateral triangle

Thus, $A = 60°$ $B = 120°$

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THANK YOU!!!

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ANGLE A=60 ANGLE B=120

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If this problem is from IMO, please tell me what year and what question.

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Thats Indian Maths Olympiad

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this problem is not from the main exam, it is from the workbook

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