Find the Angle X | Nice Geometry Question | Olympiad Mathematics

Find the Angle X | Nice Geometry Question | Olympiad Mathematics

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[dmtri]

Nice Solution!! I always admire people who know where to draw the auxialliary lines! In case someone does not want this I would suggest the following : Triangle ODC is isosceles so OD = OC (1)
Furthemore Angles AOD + BOC = 180 . Now apply the sin Rule in triangles BOC and AOD : sinx/OC = sin (180 - AOD)/BC but AD= BC (ABCD is parallelogram) and from (1) we get sinx = sin (60-x) which implies x= 30.

[aminimam]

DOC is an isosceles triangle 180 -(30 +120) =30. so angle BCO =ADO. Since it is a parallelogram opposite angles are equal. 360/4 =90 degrees. AOB also will be an equilateral triangle, with angles of 60 degrees. thus X=30

[PeterMcDaid]

When looking at geometry problems, i always ask myself how i would construct the diagram with straight edge and compass. Here there is no specific parallelogram in relation to triangle DCO, so it is apparent that the given information is true for all parallelograms where AB subtends an angle of 60 degrees at O. Therefore it is true for the rectangle. So X=30 degrees with the minimum amount of calculation.

[santiagoarosam]

How ABCD is a parallelogram → AB has the same length and is parallel to DC and the same happens with AD and BC → In a path that conforms to the proposed premises, AB can only be the height of a right triangle with angles ∠ABO =90º ; ∠OAB=30º and ∠AOB=60º → ∠DOB=∠BOC=∠DOC=120º → OD, OC and OB are bisectors of the equilateral triangle ∆BCD → ∠OBC = X=30º
Kind regards

[mrityunjaykumar]

circle approach is wrong...the circle is not circumscribing the parallelogram.. the angle subtended by the arc CD at centre of a circle (120 degree ) is double the angle on the circle(60). from that figure will be totally different