Enigma 1313: Triangles
29 June 2014
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From New Scientist #2471, 30th October 2004
Draw a triangle ABC. On the side AB mark the point P such that AP=(2/5)AB, on BC mark Q such that BQ=(2/5)BC and on CA mark R such that CR=(2/5)CA.
Draw the lines AQ, BR and CP. Call the point where AQ and BR cross X, the point where BR and CP cross Y and the point where CP and AQ cross Z.
If you did the appropriate calculations you would find that the area of triangle XYZ is 1/19 of the area of triangle ABC.
I went through the whole procedure above again, but, this time with each occurrence of the number 2/5 replaced by the number k, which is between 0 and 1/2. This time I found that the area of triangle XYZ was 1/37 of the area of triangle ABC.
What was the number k?