Friday, April 2, 2010

Bowling Puzzler XI: Equilaterality

I normally reserve these posts for Thursdays, but yesterday's puzzler was a total bust. Maybe it was too difficult. Maybe folks are losing interest. Anyway, this one is a bit more tractable. At first glance, it seems pretty simple, but once you think you have it solved, spend a bit more time with it to make sure you have it down. So, here it is:

By some strange coincidence, in ten pin bowling, ten pins are used. Those ten pins are arranged in the form of an equilateral triangle, a triangle with three equal angles (60 degrees each) and sides of equal length. Each row of pins is comprised of one more pin than the previous row, beginning with one and ending with four. The 1,7, and 10 pins form the corners of the triangle, but within that largest triangle, other triangles can be formed. The question is:

How many total equilateral triangles occur within the pin deck in ten pin bowling? To qualify, the entire triangle must fit with the ten pins and each corner of the triangle must be marked by a pin.

Click on the icon to the right for the answer

4 comments:

  1. Let me be the first to try (and be wrong):

    9 small equilateral triangles formed by adjacent pins (1-2-3, 2-4-5, 2-5-3, 3-5-6, 4-7-8, 4-5-8, 5-8-9, 5-9-6, 6-9-10)

    5 medium equilateral triangles formed by 6 adjacent pins (1-4-6, 2-7-9, 3-8-10, 3-4-9,, 2-8-6)

    So my answer is 14.

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  2. You said it, not me, but it's very close.

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  3. OK my revised answer is 15 if you include the whole 1-7-10 equilateral triangle (which I thought didn't count). Definitely a case of RTFM.

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  4. Nicely done. I'm going to have to figure out something very tricky for the grand finale.

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