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[ Information | Background | The CyberTeaser | Quantum News | Miscellaneous Stuff ] Solution to CyberTeaser B192
CyberTeaser contestant Jonathan Devor of Jerusalem, Israel, sent in a nifty JPEG of his solution (in which he used a different notation, but to the same end). CyberTeaser Archive
Copyright © 1996 National
Science Teachers Association |
Let a be the side
of the square that adjoins the black one on the left (see the figure). Then, moving
clockwise, we find the sides of the other squares touching the black one, one after the
other. We find that the side of the red square is a – (a – 4) = 4.
The side of the blue square can be found in a similar fashion. It’s 15. 
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