A cute problem by Dr. A. Kraemer.

Solution:

Our favorite variation is 1.Qc2 Bh5 2.Qd1 Bxd1 3.exf7#

Some other possibilities are: 1.Qc2 Bc7 [1...Bxc2 2.exf7#; 1...f5 2.Qa4 Bb6 3.Qxd7#; 1...a5 2.Qd3 fxe6 3.Qxg6#; 1...Bb6 2.Qc8+ Bd8 3.exd7#] 2.Qxc7 fxe6 3.Qc8#

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