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TchrWill

USA
79 Posts

 Posted - 05/26/2013 :  12:11:49 Given any triangle. Draw "n" lines from each vertex to "n" points on the opposite sides such that no three of the lines intersect at the same point. How many regions do the 3n lines divide the interior of the triangle into?

someguy

143 Posts

 Posted - 05/29/2013 :  14:34:08 If your stuck on this one, try it when n=1 and when n=2.Just draw the lines one at a time and pay attention to how you get divide a region into two.Start with the triangle.Draw all n lines starting at the first vertex first, but do it one at a time. How does drawing each line increase the number of regions.Next, draw the lines coming out of the second vertex. As you draw each line, pay attention to how you divide a region. How many lines do you cross as you move from the vertex to the opposite edge? Don't forget to note that you divide one of the regions in two when you reach the final edge.Now move on to the third vertex. Draw the n lines one at a time. Since no three lines are allowed to intersect at the same point, how many lines do you cross? How many more regions do you get each time you draw a line (again, don't forget about the opposite edge).Add things up.

someguy

143 Posts

 Posted - 06/14/2013 :  03:31:02 Looks like I killed the board. :(

TchrWill

USA
79 Posts

 Posted - 06/26/2013 :  14:49:02 Wrong problem, wrong solution. Edited by - TchrWill on 06/26/2013 20:05:47

TchrWill

USA
79 Posts

 Posted - 06/30/2013 :  09:56:57 quote:Originally posted by someguyLooks like I killed the board. :(I doubt it. More likely my misstating the problem.Given any triangle. Draw "n" lines from each vertex to "n" points on the opposite sides such that no three of the lines intersect at the same point. How many regions do the 3n lines divide the interior of the triangle into?Lets try a small value of n = 1 and see what happens when we deal with just two sides of the triangle. Draw one line to one point on the opposite side. We immediatley see that we have divided the triangle into 2 regions. Draw a second line and we now have 4 regions. Add the 3rd line and we have 7 regions. Try n = 2. Draw two lines to the opposite side and we obtain 3 regions. Draw two more lines and we now have 9 regions. Draw the last two lines and we now have 19 regions. Try n = 3. Draw three lines to the opposite side and we obtain 4 regions. Draw three more lines and we now have 16 regions. Draw the last three lines and we now have 37 regions.Lets see what this data tells us so far.Pts..................Lines/Regions.1..............1/2.........2/4.........3/7.2..............2/3.........4/9........6/19.3..............3/4........6/16.......9/37Notice anything about these numbers? Clearly, the first line, or lines, divides the triangle up into (n + 1) regions. Just as obvious, the second line, or lines, divides the triangle up into n^2 regions. If you take careful note of the difference between the 2nd and 3rd region numbsrs, you might recognize that the differences between pairs within each set of points are every other triangular number. For 1 point, 7 - 4 = 3. For 2 points, 19 - 9 = 10. For 3 points, 37 - 16 = 21, or every other triangular number. On this basis, my educated guess tells me that the next few sets of data follows the following pattern: Pts..................Lines/Regions.1..............1/2.........2/4.........3/7.2..............2/3.........4/9........6/19.3..............3/4........6/16.......9/37.4..............4/5........8/25......12/61.5..............5/6.......10/36.....15/91.6..............6/7.......12/49.....18/127 and so on.If we just enumerate the region numbers we getPoints...............1.....2.....3.....4.....5.....6................nRegions.............7....19...37...61...91...127.............N1st difference........12...18...24...30...362nd difference...........6.....6.....6.....6Thus, we have a finite difference series where the second differences are constant, which leads to the expression for the nth term being of the second order, i.e., N = an^2 + bn + c. Using the data points (n1, N1), (n2,N2), (n3,N3), etc., we substitute them into N = an^2 + bn + c as follows:(n1,N1) = (1,7) produces a(1^2) + b(1) + c = 7 or a + b + c = 7(n2,N2) = (2,19) produces a(2^2) + b(2) + c = 19 or 4a + 2b + c = 19(n3,N3) = (3,37) produces a(3^2) + b(3) + c = 37 or 9a + 3b + c = 37Solving, a = 3, b = 3, and c = 1 making the nth term of this regional series N = 3n^2 + 3n + 1.Therefore, the number of regions developed within a triangle resulting from the creation of "n" lines being drawn from each vertex to "n" points on the opposite sides derives from N = 3n^2 + 3n + 1.While only three of you took the time to respond, I will continue to post problems if you desire. If you derive solutions, please post them so that I, and others, might benefit from your efforts.

someguy

143 Posts

 Posted - 07/01/2013 :  16:04:17 Hi TchrWill,I enjoy your problems and would like to see more. I also like to see the techniques used by others solve problems (to others: if you have a different solution to a question, we still want to see it ).My method of solution relies on noting that when you draw a line, you increase the number of regions by 1 every time you cross a pre-existing line or make contact with the opposite edge.No matter how you draw the lines, you can get to that configuration by drawing all lines from the first vertex, then drawing all lines from the second vertex, and finally drawing all lines from the third vertex. Drawing the lines in this order allows us to count the number of regions easily.Initially, we start with 1 region.Drawing the n lines from the first vertex increases the number of regions by n. Each line increases the number of regions by 1 when we draw it.Drawing the n lines from the second vertex increases the number of regions by n(n+1).Each line increases the number of regions by n+1 when we draw it because it must cross n lines and intersect the opposite edge.Drawing the n lines from the third vertex increases the number of regions by n(2n+1).Each line increases the number of regions by 2n+1 when we draw it because it must cross 2n lines and intersect the opposite edge.Total regions in the end is 1 + n + n(n+1) + n(2n+1) = 1 + 3n + 3n^2.

TchrWill

USA
79 Posts

 Posted - 11/10/2013 :  15:02:23 Hi someguy,Your words of last July were appreciated:"Hi TchrWill,I enjoy your problems and would like to see more. I also like to see the techniques used by others solve problems (to others: if you have a different solution to a question, we still want to see it )."They led me to feel that the word problem exchange might grow and all that participated would benefit. But, alas, the train came to a screeching halt and there have been no responses for several weeks now.If you know why this has happened, I would deeply appreciate your enlightening me.Thanks for your, the hill1962's ultraglides's and subhotosh Khan's participation and contributions. I hope you all might find time to continue our exchange.Happy holidays.Tchrwill LISTEN TO THE SONG, "THE TWELVE DAYS OF CHRISTMAS".HOW MANY GIFTS IN ALL DID MY TRUE LOVE GIVE TO ME?IS THERE A PATTERN? Not a toughy but cute however. The kids might enjoy it.Solution below.From the lyrics of the song, we get:Day1--Partridge in a pear tree2--Two turtle doves3--Three French hens4--Four calling birds5--Five golden rings 6--Six geese a-laying7--Seven swans a-swimming 8--Eight maids a-milking 9--Nine ladies dancing 10--Ten lords a-leaping11--Eleven pipers piping12--Twelve drummers drumming The number of items grows by one each day. Therefore, the daily number of items is simply the sequence of the counting numbers from 1 to 12. We can represent the daily accumulation of items pictorially by creating equilateral triangles with O's representing the items as follows:Day -...............1...........2................3......................4..............................5Items - ............1...........2................3......................4..............................5, 6, 7, 8, 9, 10, 11, 12......................O...........O...............O.....................O.............................O.................................O..O...........O..O................O..O.........................O..O.................................................O..O..O............O..O..O....................O..O..O......................................................................O..O..O..O................O..O..O..O...................................................................................................O..O..O..O..OAccumulation...1...........3................6.....................10............................15, 21, 28, 36, 45, 55, 66, 78Continuing this on out to the 12th day, we derive the daily accumulations of the items up to that day as 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, and 78. This sequence of numbers is referred to as the triangular numbers. You can readily see that the triangular numbers are simply the sum of the counting numbers from 1 through n, 12 in our case, which results in a total accumulation on the 12th day of 78 items. This sum can also be derived by looking at the sequence of numbers from 1 to 12 as an arithmetic progression of the general form a, a + d, a + 2d, a + 3d, .......a + (n-1)d where a = the first term, 1, d = the common difference, 1, and n = the number of terms, 12. The sum of such a progression is given by S = n(a + L)/2 where L = the last term, a + (n-1)d, giving us the sum of S = n(a + a + (n-1)d)/2 = n(1 + 1 + (n-1)1)/2 = n(2 + n - 1)/2 = n(n + 1)/2 as the sum of any number of integers from 1 to n. For our 12 days, this results in a total number of items of S = 12(12 + 1)/2 = 156/2 = 78. Edited by - TchrWill on 11/15/2013 10:06:34
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