bonanova Posted February 27, 2011 Report Share Posted February 27, 2011 The diameters of three circles form a right triangle. Without recourse to a measuring device, cut the circles into no more than five pieces that can be arranged into four groups of equal total area. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 27, 2011 Report Share Posted February 27, 2011 clarification: should we consider the circles separate when cutting? or should we treat the overlaps as part of the same cut? (i.e. does this require two cuts or one cut?) also are we allowed non neat shapes? that is are we just looking for total area to be equal, or total area and nice complete circles? Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted February 28, 2011 Author Report Share Posted February 28, 2011 You can slice as you please, but not using eg a ruler or scales. And no more than 5 pieces. Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted February 28, 2011 Report Share Posted February 28, 2011 By four groups of equal area which do you mean? Cut the shape into pieces A, B, C, D and E with the areas A = B = C = D + E? or A = B; A = C + D; B = C + D; C = D + E ---- I can get the second scenerio using the edges of the circles as the cut line using circles with the radius of 1, 1, and sqrt(2). The first scenerio...I have no idea yet. circles.bmp C = E; A + B = C; A + B = E and A + D = B *shrug* Quote Link to comment Share on other sites More sharing options...
0 araver Posted February 28, 2011 Report Share Posted February 28, 2011 You can slice as you please, but not using eg a ruler or scales. And no more than 5 pieces. Compass-and-straightedge is allowed? Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 1, 2011 Author Report Share Posted March 1, 2011 By four groups of equal area which do you mean? Cut the shape into pieces A, B, C, D and E with the areas A = B = C = D + E? or A = B; A = C + D; B = C + D; C = D + E I can get the second scenerio using the edges of the circles as the cut line using circles with the radius of 1, 1, and sqrt(2). The first scenerio...I have no idea yet. circles.bmp C = E; A + B = C; A + B = E and A + D = B *shrug* Like case 1. Four disjoint groups of equal aggregate area. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 1, 2011 Author Report Share Posted March 1, 2011 Compass-and-straightedge is allowed? Knife, and straight edge if you want. No rulers or compass. If it's useful to know where the centers are, you may assume they're marked. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 1, 2011 Report Share Posted March 1, 2011 there are a few of nice properties, but i don't really see how to make use of them. Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted March 1, 2011 Report Share Posted March 1, 2011 Is the shape one piece or three actual circles? If one piece, are the circles drawn for us along with the centers? Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 2, 2011 Author Report Share Posted March 2, 2011 Is the shape one piece or three actual circles? If one piece, are the circles drawn for us along with the centers? Three actual circles. I originally heard it as three pancakes to be divided equally among four children for breakfast, by cutting them into five pieces. One child obvious gets two pieces, the other three get one each of the remaining pieces. It's not that difficult - maybe an aha! will strike ... Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted March 2, 2011 Report Share Posted March 2, 2011 That makes it a bit easier! I was trying to work with one shape. Total area = pi( x^2 + y^2 + z^2 ) Goal = pi( x^2 + y^2 + z^2 ) / 4 Since the diameters form a right triangle we know the radii(?) also form a right triangle. so x^2 + y^2 = z^2 Goal => pi( z^2 + z^2 ) / 4 = > 2 * pi( z^2 ) / 4 => pi( z^2 ) / 2 or half of the biggest "pancake". Now we have two of our fourths. Next step is to cut the middle "pancake" so that the piece taken off when added to the small "pancake" will make them equal. Best I can think of is using the smallest circle as a template place so the edges are touching. Cut along circumference until you reach the straight line made by the centers. If six pieces were allowed we would just have to cut the two smaller circles in half. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 2, 2011 Author Report Share Posted March 2, 2011 That makes it a bit easier! I was trying to work with one shape. Total area = pi( x^2 + y^2 + z^2 ) Goal = pi( x^2 + y^2 + z^2 ) / 4 Since the diameters form a right triangle we know the radii(?) also form a right triangle. so x^2 + y^2 = z^2 Goal => pi( z^2 + z^2 ) / 4 = > 2 * pi( z^2 ) / 4 => pi( z^2 ) / 2 or half of the biggest "pancake". Now we have two of our fourths. Next step is to cut the middle "pancake" so that the piece taken off when added to the small "pancake" will make them equal. Best I can think of is using the smallest circle as a template place so the edges are touching. Cut along circumference until you reach the straight line made by the centers. If six pieces were allowed we would just have to cut the two smaller circles in half. That's it. Quote Link to comment Share on other sites More sharing options...
0 plainglazed Posted March 2, 2011 Report Share Posted March 2, 2011 urgg - curr3nt beat me to it but spent so much time on the image, gonna post it anyway. Gotta get internet service at my other location. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 2, 2011 Author Report Share Posted March 2, 2011 urgg - curr3nt beat me to it but spent so much time on the image, gonna post it anyway. Gotta get internet service at my other location. Yup - if the triangle is isosceles, you need only one cut and four pieces. Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted March 2, 2011 Report Share Posted March 2, 2011 Why can't I see the picture? Quote Link to comment Share on other sites More sharing options...
Question
bonanova
The diameters of three circles form a right triangle.
Without recourse to a measuring device, cut the
circles into no more than five pieces that can be
arranged into four groups of equal total area.
Link to comment
Share on other sites
14 answers to this question
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.