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Triangle at a square corner (Posted on 2023-09-02) Difficulty: 3 of 5
On sides CB and CD of square ABCD are chosen points M and K so that the perimeter of triangle CMK equals double the side of the square. Find angle MAK.

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Better solution | Comment 2 of 3 |
Better in my opinion, since it's purely geometric and doesn't use sum of angle tangent formula.

As before AB=1, CM=a, CK=b.
WLOG let ABCD be counterclockwise.  Rotate an image of segment DK by 90° clockwise about A so that D'=B and K' is the image of K on line CB and KAK'=90.  AK=AK'.

MB=1-a, DK=BK'=1-b, so MK'=2-a-b

From the problem we have a+b+KM=2
so KM=2-a-b and KM=KM'

AMK  AMK' by SSS.
MAK = 90°/2 = 45°.



  Posted by Jer on 2023-09-02 17:56:46
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