Arranging Rose BushesDate: 9/13/95 at 15:42:28 From: Christine Heffernan Subject: Question to be answered Hi Dr. Math - This is the first time I've tried this. Here's a question. A gardener laying out a rosebed found she could plant 7 rose bushes in such a way that they formed 6 straight lines with 3 rose bushes in each line. How was this possible? Show a diagram. (b) How could she plant 10 rosebushes so that she has 5 lines with 4 rosebushes in each? For the questions, the distance between rosebushes does not have to be equal. That's it. Good Luck and thanks in advance. Gavin Date: 9/13/95 at 17:20:56 From: Doctor Steve Subject: Re: Question to be answered Try laying out the bushes in circular patterns. Write us back if you want more hints. -Doctor Steve, The Geometry Forum Date: 9/13/95 at 21:30:52 From: Christine Heffernan Subject: Re: Question to be answered I've been trying that for a while now and I'm still completely stuck on both parts of the question! I think I need more help! Thanks Date: 9/13/95 at 21:42:26 From: Doctor Steve Subject: Re: Question to be answered Try putting six bushes in a circle and one in the middle. Now look for your six straight lines with 3 bushes in each. - Doctor Steve, The Geometry Forum Date: 01/17/2001 at 08:39:20 From: Peter Bradford Subject: Rose Bushes There is an error in your answer. Unless you consider a straight line as TWO straight lines, depending on which end you start, there are only 3 rows by this method. The true solution is to put 3 bushes at the corners of an equilateral triangle. Three more, each at the midpoint of a side of the triangle. Finally, the seventh bush is placed at the centroid of the triangle. Date: 01/18/2001 at 13:45:49 From: Doctor Greenie Subject: Re: Rose Bushes The writer says "the true" solution is with 3 bushes at the corners of an equilateral triangle... The way I read the original problem, it is solved with the first 3 bushes at ANY points forming a triangle, the next 3 bushes at the medians of that triangle, and the 7th bush at the point of intersection of those medians. -Doctor Greenie |
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