School: Burnaby South Secondary High School
Address: 5455 Rumble Street, Burnaby, BC, Canada.
Teacher: Mr. Swift
OUTPUT FOR A 11X11 CHECKER BOARD (36 triangles) THE SIDES GIVEN BELOW NEED TO BE SQUARED ROOTED FOR CORRECT ANSWER. THE POINTS GIVEN BELOW ARE ONLY ONE EXAMPLE OF EACH TRIANGLE. TRIANGLE # 1 SIDES : 5 5 10 THE POINTS : (0,0) (1,2) (3,1) TRIANGLE # 2 SIDES : 10 10 20 THE POINTS : (0,0) (1,3) (4,2) TRIANGLE # 3 SIDES : 10 25 25 THE POINTS : (0,0) (0,5) (3,1) TRIANGLE # 4 SIDES : 10 65 65 THE POINTS : (0,0) (1,8) (4,7) TRIANGLE # 5 SIDES : 13 13 26 THE POINTS : (0,0) (1,5) (3,2) TRIANGLE # 6 SIDES : 17 17 34 THE POINTS : (0,0) (1,4) (5,3) TRIANGLE # 7 SIDES : 20 20 40 THE POINTS : (0,0) (2,4) (6,2) TRIANGLE # 8 SIDES : 20 25 25 THE POINTS : (0,0) (0,5) (4,2) TRIANGLE # 9 SIDES : 20 50 50 THE POINTS : (0,0) (1,7) (5,5) TRIANGLE # 10 SIDES : 20 85 85 THE POINTS : (0,0) (2,9) (6,7) TRIANGLE # 11 SIDES : 25 25 50 THE POINTS : (0,0) (1,7) (4,3) TRIANGLE # 12 SIDES : 25 25 80 THE POINTS : (0,0) (0,5) (4,8) TRIANGLE # 13 SIDES : 25 25 90 THE POINTS : (0,0) (0,5) (3,9) TRIANGLE # 14 SIDES : 26 26 52 THE POINTS : (0,0) (1,5) (6,4) TRIANGLE # 15 SIDES : 26 65 65 THE POINTS : (0,0) (1,5) (8,1) TRIANGLE # 16 SIDES : 29 29 58 THE POINTS : (0,0) (2,5) (7,3) TRIANGLE # 17 SIDES : 34 34 68 THE POINTS : (0,0) (2,8) (5,3) TRIANGLE # 18 SIDES : 34 85 85 THE POINTS : (0,0) (2,9) (7,6) TRIANGLE # 19 SIDES : 37 37 74 THE POINTS : (0,0) (1,6) (7,5) TRIANGLE # 20 SIDES : 40 40 80 THE POINTS : (0,0) (2,6) (8,4) TRIANGLE # 21 SIDES : 40 50 50 THE POINTS : (0,0) (1,7) (6,2) TRIANGLE # 22 SIDES : 40 100 100 THE POINTS : (0,0) (0,10) (6,2) TRIANGLE # 23 SIDES : 41 41 82 THE POINTS : (0,0) (1,9) (5,4) TRIANGLE # 24 SIDES : 45 45 90 THE POINTS : (0,0) (3,6) (9,3) TRIANGLE # 25 SIDES : 50 50 100 THE POINTS : (0,0) (1,7) (8,6) TRIANGLE # 26 SIDES : 52 52 104 THE POINTS : (0,0) (2,10) (6,4) TRIANGLE # 27 SIDES : 52 65 65 THE POINTS : (0,0) (1,8) (7,4) TRIANGLE # 28 SIDES : 53 53 106 THE POINTS : (0,0) (2,7) (9,5) TRIANGLE # 29 SIDES : 58 58 116 THE POINTS : (0,0) (3,7) (10,4) TRIANGLE # 30 SIDES : 65 65 80 THE POINTS : (0,0) (1,8) (8,4) TRIANGLE # 31 SIDES : 65 65 130 THE POINTS : (0,0) (1,8) (9,7) TRIANGLE # 32 SIDES : 68 68 136 THE POINTS : (0,0) (2,8) (10,6) TRIANGLE # 33 SIDES : 68 85 85 THE POINTS : (0,0) (2,8) (9,2) TRIANGLE # 34 SIDES : 80 100 100 THE POINTS : (0,0) (0,10) (8,4) TRIANGLE # 35 SIDES : 82 82 164 THE POINTS : (0,0) (1,9) (10,8) TRIANGLE # 36 SIDES : 85 85 90 THE POINTS : (0,0) (2,9) (9,3) PROOF a. Possible Slopes First of all, assume m = slope of the axis of symmetry. That tells us that 0 < m < 1 (since it can't be 0, it can't be 1, and anything beyond one, the triangle can be flipped; and anything negative, the triangle can also be flipped.) The slope of the third side (the non-equal side) must be -1 / m. Suppose we place the first point of the triangle at (0,0) Then in order to let the other two points fall on valid points, (integral coordinate), the slope must be able to be expressed in the form of a/b (where a and b are integers, and a & b can not be factored.) Therefore, we limit the possible slopes of axis of symmetry to: (anything beyond, like 31/32, will definitely have the end points outside the 11x11 board) 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8, 1/9 2/3, 2/5, 2/7, 2/9, 3/4, 3/5, 3/7, 3/8... All the way up to 8/9. b. Possible Points Here, we discuss using only slope=1/3; all other slopes can be discussed in the same procedures. Here, the slope of axis of symmetry is 1/3. So the slope of the third line must be -3. So, in order to let the points fall on the valid points, the mininum possible length of the third line is root of 10. And all the other possible lengths are multiples of that: X=root of 10, the possible lengths are x, 2x, 3x, 4x... So for the slope 1/3, we first get a grid paper, and plotted (0,0) as one of the point. Then we draw a line through (0,0) with a slope of 1/3. First, on the line, we find point (3,1), (6,2), (9,3)... These are one set of possible point of intersection between the axis of symmetry and the third side. But they are not all, we divide the line segment in half to get the rest; that is because for (0,0)-(3,1), the third side is 2 times the minimum length; therefore, at half, it will be the minimum length. The only problem with half point is: will the two end points fall on valid coordinates? The answer is: only for a/b where a and b are both non-factorable with each other, and are both odd numbers. The proof is simple. Since a/b is simplest form, cutting it in half, the coordinate will fall between the grid both vertically and horizontally, if and only if, both a and b are odd numbers. Now, to achieve the minimum length, that means both upward and downward has to have 1/2 the minimum length; so both way, the displacement will have .5 more horizontal and .5 more vertically beyond the grid scale. So 1/2 + 1/2 = 1, so the end point fall exactly on the grid line. Therefore, we prove that for 1/3, simplest fraction, the points are (3,1), (6,2)..., and since both 1 and 3 are odd numbers, we also have (1.5, 0.5), (4.5, 1.5)... For every one of them, draw a line thru them with a slope of -3. And this line will intersect the grid at (1,2)-(2,-1), (0,5)-(3,-4) .., and each of these corresponding set will be the two points other than (0,0) So for slope 1/3, we find a sequence of possible third side, and from that, we find a sequence of possible 2nd and 3rd points. Now, all we have to do is to eliminate those triangles that are too big to fit into 11x11 grid. And we can have every single possible triangle that has an axis of symmetry slope = 1/3. c. Getting All Of Them Now, since we have the list of all possible slopes, and for each, we can get all possible triangles. Then we can get all the possible triangles. To ensure that we get them all, we program the computer (in QBasic from MS-DOS 5.0) to automatically check for all possibilities. In this program, in order to reduce the number of testing from 11^6 to 11^4, an important proof is needed. d. Prove that anything triangle (isosclese or not), can always be flipped and shifted, to have one point (0,0), and the other points also inside the 11X11 board: First of all, for any triangle, it must contain three points. Let's call them A, B, and C. Let's sort A B C so that Ax<=Bx, and Ax<=Cx, and Bx<=Cx. So now, let's find the point with lowest Y. If it's point B, then we look at the highest Y. One of the point must be lowest/highest Y, and most left/right X. Now, we can always flip this triangle horizontally and vertically to get this point on bottom left. So we can shift this triangle to the point (0,0). Since it's the bottom most and left most, the other two points MUST fall within this board. Proof is done. Since I limit one point to (0,0), I only have to check 11^4 times. On my IBM-PC 386 SX, it took me about one hour to test through all results; totally there are 36 triangles (non-congruent). The list matches the total triangles we found by slope elimination. e. Listing of the program (using QBasic from MS-DOS Version 5) ' ' MATH SOLVE PUZZLE #1 ' ' QUESTION COMES FROM: pow@mathforum.org (problem of the week) ' COMPUTER PROGRAMMER: FELIX SHENG-HO CHANG ' ' ON AN 11 BY 11 GEOBOARD, FIND ALL NON-CONGRUENT, ISOSCELES TRIANGLES ' WHOSE AXIS OF SYMMETRY IS NEITHER HORIZONTAL, NOR VERTICAL, NOR AT ' A 45 DEGREE ANGLE FROM HORIZONTAL. STATE YOUR SOLUTION BY GIVING THE ' LENGTH OF THE SIDES OF THE TRIANGLES YOU FIND. WHAT TECHNIQUES DID YOU ' USE TO FIND THE ANSWERS? ' CONST MAXX = 10 'since it starts from zero to 10. CONST MAXY = 10 'since it starts from zero to 10. CONST MAXENTRY = 100 CONST OVERFLOW = 150 SCREEN 8 DEFINT A-Z DIM SHARED ALLDATA(1 TO MAXENTRY) AS STRING DIM SHARED ALLDATAS(1 TO MAXENTRY) AS STRING DIM SHARED ALLDATANOW AS INTEGER CLS LINE (0, 30)-(330, 195), 0, BF FOR P = 0 TO MAXX: LINE (P * 30, 30)-(P * 30, MAXY * 15 + 30): NEXT FOR P = 0 TO MAXY: LINE (0, P * 15 + 30)-(MAXX * 30, P * 15 + 30): NEXT ALLDATANOW = 0 RESMUCH& = 0 ' THREE SIDES ALL THE SAME RESCORE& = 0 ' EXACTLY TWO SAME SIDES. TOTAL COUNT. RESREGU& = 0 ' EXACTLY TWO SAME SIDES. VER/HOR/45 AXIS OF SYMMETRY. RESCORR& = 0 ' EXACTLY TWO SAME SIDES. NON REGULAR AXIS. TOTAL COUNT. RESYEAH& = 0 ' EXACTLY TWO SAME SIDES. NON REGULAR AXIS. NO REPEAT. FOR P1X = 0 TO 0: FOR P1Y = 0 TO 0 FOR P2X = 0 TO MAXX: FOR P2Y = 0 TO MAXY FOR P3X = 0 TO MAXX: FOR P3Y = 0 TO MAXY LOCATE 1, 1: PRINT "("; P1X; ","; P1Y; ")"; " ("; P2X; ","; P2Y; ")"; " ("; P3X; ","; P3Y; ")"; SPACE$(10); LOCATE 2, 1: PRINT "ALL :"; RESCORE&; " REG :"; RESREGU&; " NON-REG :"; RESCORR&; " Non-Repeat :"; RESYEAH&; SPACE$(20); KX1 = P1X - P2X: KY1 = P1Y - P2Y: K1 = KX1 ^ 2 + KY1 ^ 2 KX2 = P2X - P3X: KY2 = P2Y - P3Y: K2 = KX2 ^ 2 + KY2 ^ 2 KX3 = P1X - P3X: KY3 = P1Y - P3Y: K3 = KX3 ^ 2 + KY3 ^ 2 'RULE OUT EQUAL POINTS... IF P1X = P2X AND P1Y = P2Y THEN GOTO TRY IF P2X = P3X AND P2Y = P3Y THEN GOTO TRY IF P1X = P3X AND P1Y = P3Y THEN GOTO TRY 'RULE OUT EQUAL SLOPES (THAT IS, AREA OF TRIANGLE = 0) IF KY1 * KX2 = KX1 * KY2 THEN GOTO TRY IF KY1 * KX3 = KX1 * KY3 THEN GOTO TRY IF KY3 * KX2 = KX3 * KY2 THEN GOTO TRY 'RULE OUT ALL THREE EQUAL SIDES IF K1 = K2 AND K2 = K3 THEN RESMUCH& = RESMUCH& + 1: GOTO TRY 'RULE OUT TRIANGLES THAT HAVE NO EQUAL SIDES IF K1 <> K2 AND K2 <> K3 AND K1 <> K3 THEN GOTO TRY 'DETERMINE WHETHER IT'S VER/HOR/45 (K=4) OR NOT (K=1) IF K1 = K2 THEN Z1X = P2X * 2: Z1Y = P2Y * 2: Z2X = P1X + P3X: Z2Y = P1Y + P3Y IF K2 = K3 THEN Z1X = P3X * 2: Z1Y = P3Y * 2: Z2X = P1X + P2X: Z2Y = P1Y + P2Y IF K1 = K3 THEN Z1X = P1X * 2: Z1Y = P1Y * 2: Z2X = P2X + P3X: Z2Y = P2Y + P3Y IF Z1Y = Z2Y OR Z1X = Z2X OR ABS(Z1X - Z2X) = ABS(Z1Y - Z2Y) THEN K = 4 ELSE K = 1 RESCORE& = RESCORE& + 1 IF K = 4 THEN RESREGU& = RESREGU& + 1: GOTO TRY 'DRAW IT LINE (0, 30)-(330, 195), 0, BF FOR P = 0 TO MAXX: LINE (P * 30, 30)-(P * 30, MAXY * 15 + 30): NEXT FOR P = 0 TO MAXY: LINE (0, P * 15 + 30)-(MAXX * 30, P * 15 + 30): NEXT LINE (P1X * 30, P1Y * 15 + 30)-(P2X * 30, P2Y * 15 + 30), K LINE (P2X * 30, P2Y * 15 + 30)-(P3X * 30, P3Y * 15 + 30), K LINE (P1X * 30, P1Y * 15 + 30)-(P3X * 30, P3Y * 15 + 30), K PLAY "L64D" RESCORR& = RESCORR& + 1 IF K3 >= K2 AND K3 >= K1 THEN PK1 = K3 IF K2 >= K1 THEN PK2 = K2: PK3 = K1 IF K1 >= K2 THEN PK2 = K1: PK3 = K2 ELSEIF K2 >= K1 AND K2 >= K3 THEN PK1 = K2 IF K1 >= K3 THEN PK2 = K1: PK3 = K3 IF K3 >= K1 THEN PK2 = K3: PK3 = K1 ELSEIF K1 >= K2 AND K1 >= K3 THEN PK1 = K1 IF K2 >= K3 THEN PK2 = K2: PK3 = K3 IF K3 >= K2 THEN PK2 = K3: PK3 = K2 END IF Z$ = STR$(PK3) + STR$(PK2) + STR$(PK1) PP = 0: FOR P = 1 TO ALLDATANOW IF ALLDATA(P) = Z$ THEN PP = 1: P = OVERFLOW NEXT IF PP = 0 THEN PLAY "L16CDE": ALLDATANOW = ALLDATANOW + 1: RESYEAH& = RESYEAH& + 1 ALLDATA(ALLDATANOW) = Z$ ALLDATAS(ALLDATANOW) = "(" + STR$(P1X) + "," + STR$(P1Y) + ") (" + STR$(P2X) + "," + STR$(P2Y) + ") (" + STR$(P3X) + "," + STR$(P3Y) + ")" END IF LOCATE 2, 1: PRINT "ALL :"; RESCORE&; " REG :"; RESREGU&; " NON-REG :"; RESCORR&; " Non-Repeat :"; RESYEAH&; SPACE$(20); LOCATE 3, 1: PRINT K1; " "; K2; " "; K3; SPACE$(20); REM IF K = 1 THEN A$ = INPUT$(1) TRY: NEXT: NEXT NEXT: NEXT NEXT: NEXT CLS : SCREEN 0, 0, 0 PRINT "THIS IS DONE!!!!!": PLAY "L4O3CDEFGABO4C" ZZ = 0: FOR Z = 1 TO ALLDATANOW: ZZ = ZZ + 1 PRINT "TRIANGLE #"; : PRINT USING "####"; Z; : PRINT " SIDES : "; ALLDATA(Z); " THE POINTS : "; ALLDATAS(Z) IF ZZ = 20 THEN PRINT "--------------------PRESS----------------------": A$ = INPUT$(1): ZZ = 0: CLS NEXT Z: PRINT : PRINT "THE SOLUTION :"; RESYEAH& A$ = INPUT$(1)
COMMENTS: These guys did a great job. They were very methodical and considered many possibilities. They also managed to define the problem well enough to write a computer program to find the answers. Their explanation goes through all the aspects of the problem they considered and what constraints there were on each option. Realizing that each triangle could just as well include the origin certainly helped narrow things down. In evaluating the solutions, I also found it very useful to have the points included, instead of just the lengths of the sides. Overall, excellent job, and well-written explanation.
[Jacob sent along some pictures that he refers to but I haven't included.]
Solutions: Length of Isosceles sides length of other the square root of 5 the square root of 10 the square root of 10 the square root of 20 the square root of 13 the square root of 26 the square root of 17 the square root of 34 the square root of 20 the square root of 40 the square root of 25 the square root of 50 the square root of 26 the square root of 52 the square root of 29 the square root of 58 the square root of 34 the square root of 68 the square root of 41 the square root of 82 the square root of 37 the square root of 7 the square root of 40 the square root of 80 the square root of 45 the square root of 90 the square root of 52 the square root of 104 the square root of 61 the square root of 122 the square root of 50 the square root of 100 the square root of 53 the square root of 106 the square root of 58 the square root of 116 the square root of 65 the square root of 130 the square root of 68 the square root of 136 the square root of 68 the square root of 136 the square root of 73 the square root of 146 the square root of 82 the square root of 164 the square root of 85 the square root of 170 the square root of 101 the square root of 202The way I got these 25 answers is the following. First I realized that it probably wouldn't work to do a triangle such as the one in exhibit a. With all of the sides exactly on the dots of the grid. The problem was if they weren't on the dots how could they be measured? I realized that the Pythagorean theorem could be used to measure it. In exhibit b I show how a side drawn on a diagonal can be measured. I made a right triangle around the line so that the line would become the hypotenuse. I then use a squared+b squared=c squared to find the length. I fooled around with this type of triangle until I found one that meets all of the requirements. It was one where each of the sides had a slope of 2/1 and the none isosceles sides had a slope of 3/1 making the sides the square root of 5, the square root of 5 and the square root of 10 (exhibit c). I then found that this triangle could be enlarged to make a number of similar but none congruent triangles. The pattern for these triangles is that there slope could not be something like 2/2 or 3/3 but the numbers had to be different. Also, the numbers in the slope could not add up to more then 11 to fit on the grid. The none isosceles side, was always the square root of the number two times as high as the number in the isosceles side. I just made out a list of the possible slopes, and figured out what the triangles lengths would be using these slopes. Here is my list:
I also found five other triangles using different methods. I was just thinking about my solutions and I realized that there might be a few triangles with one side on the dots. Because 3 squared (9)+4 squared(16)=5 squared(25) there is a triangle (exhibit d) with one side on five dots, one side not on the dots, with a slope of 4,3 and the other side. I found two different triangles using this approach. I found similar triangles using 10 as the isosceles sides. Using my two methods I have a grand total of 29 solutions.
Hello. My name is Sean Nichols, and I am a student at Burnaby South Secondary School, in Burnaby, British Columbia, Canada. This is my submission for the problem of the month, October 1994.
SOLUTION:
I have found that in an 11x11 geoboard, 24 non-congruent, isosceles
triangles with axes of symmetry non vertical, horizontal, or on a 45/-45
degree angle can be drawn. The triangles' dimensions are as follows:
TRIANGLES:
Triangle Number: Spaces up & Length of Length of Across: Legs: Base: 1 2x1 2.236 3.162 2 3x1 3.162 4.472 3 3x2 3.606 5.099 4 4x1 4.123 5.831 5 4x2 4.472 6.325 6 4x3 5.000 6.083 7 5x1 5.099 7.211 8 5x2 5.385 7.616 9 5x3 5.831 7.280 10 5x4 6.403 9.055 11 6x1 6.083 8.602 12 6x2 6.325 8.944 13 6x3 6.708 8.544 14 6x4 7.211 10.198 15 7x1 7.071 10.000 16 7x2 7.280 10.296 18 7x3 7.616 9.849 19 7x4 -> 8x1 8.062 8.944 20 8x1 -> 7x4 8.062 3.162 21 8x1 8.062 11.402 22 8x2 8.246 11.662 23 9x1 9.055 12.806METHOD:
Once I had this pattern understood, it was a simple matter to use pythagoras' theorem (and a calculator), to figure out the lengths of the sides, and write them down in the table above.