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Superbowl Sunday - January 29 - February 2, 1996 Next Problem || Previous Problem || Contents || All Problems || Search POWs In 1996, Superbowl Sunday fell on January 28 in the United States. Ah, yes, the Super Bowl. Let's talk about field goals. (Fortunately the game did not come down to a field goal, but we can pretend it did to make the problem more exciting, not to mention the game.) The Steelers are losing 20-17. They have a chance to tie it with a field goal from 50 yards away (which is worth three points). Let's look at this from the blimp and from the sidelines. TOP VIEW | SIDE VIEW | goal post | * X_____________X | * * | | * * | | * * | | * * | | * * | | * * | | * * | | * * | | * | _________|_________ | * | line of scrimmage | O_______|_______________________________| | | ball line of goal | | scrimmage post | | | | O | ball | | Here are some numbers that we'll need. The ball is always kicked from seven yards behind the line of scrimmage. The goal posts are 18'6" apart and the cross bar is 10' off the ground (the goal posts look like this: |_| ) The first question is about how accurate the Steelers' kicker Norm Johnson has to be to put it between the goal posts from 50 yards away. If he is kicking on a direct line to the goal posts, as in the picture on the left, how far off that line could the ball be when it crosses the line of scrimmage and still go through the goal posts - not be too far to the left or the right? The second question has to do with how far Johnson has to kick it to get it over the cross bar. To make the problem a little easier, we're going to pretend the path that the ball travels is a triangle shape, as in the picture above right, instead of a curve. If the ball is 9 feet off the ground when it crosses the line of scrimmage, how far must it go before it starts going down to still be high enough to go over the cross bar? (This part is a little complicated. You'll want to draw a good picture, and be careful!) If the ball goes 35 yards before it starts going down, will the score be tied or will the Cowboys still be ahead? - Annie Fetter Solutions Annie says: This problem proved a fair bit harder than I anticipated, which I should have figured out after I discovered that I had made a mistake when originally playing with the numbers! You'll notice two different sets of answers in the solutions. This is because I didn't explain everything I needed to: when a field goal is measured as being 50 yards, that includes the 7 yards behind the line of scrimmage. (For a 50-yard field goal, the line of scrimmage is actually the 33-yard-line - there are 10 yards added for the end zone and seven yards added for the spot of the ball behind the line.) I counted either case as correct, however. In the first part of this problem, when we are measuring the amount that the ball could vary from the center line, a number of students used the Pythagorean theorem to find the outside edgelength of the triangle. Why? Why not just use the center line as the side of the similar triangles? There was a lot of guessing on the second part, and many folks forgot, or didn't notice, the little triangle at the end after the ball passes over the crossbar. That was tricky. Kaitlin and Dustin came up with a slightly different answer because they figured that to clear a 10-foot crossbar, the height of the ball would actually have to be a bit more to allow for the width of the ball. Fair enough - especially since they explained it clearly. Thomas Kuo came up with a different answer for the second part, including a conclusion that the field goal would not be good, because he read the distance I wanted as the distance the ball itself travelled instead of the horizontal distance. Again, he explained this well, and the problem wasn't worded that clearly, so he gets credit. And now a comment completely unrelated to this week's solution. Yesterday I was watching New Yankee Workshop, a furniture-building show on public television, and the host, Norm Abrams, was making screen doors. He nailed them together and then, with the door lying on his worktable, said, "Now we'll check the door for square." He proceeded to measure the two diagonals and, finding that they were equal, pronounced the door perfect! (This was pretty exciting to me, that a Problem of the Week showed up on PBS, though I would understand if you were less than thrilled. :-) Below are the names and the solutions for all the correct submissions. Correct solutions were submitted by: Anna Margush, Grade 4 - Home Schooled Martin County High School, Stuart, Florida Grade 12 Danielle Cegelis Grades 9 & 10 Julia Schumm and Edna Evans, Mike Schmidt and Phillip Pool, Tim Casale, Kaitlin Coffinbarger and Dustin Hicks, Christine Francescani and Julie Conant Murray Junior High School, Ridgecrest, California Thomas S. Kuo, Grade 9 Julia Schumm and Edna Evans Grade 10 Martin County High School, Stuart Florida The Steelers kicker will only be able to be off from the direct line to the goal posts a maximum of 1.295 feet and still kick a field goal. I obtained this answer by figuring that the ratio of 50 yards to the goal line with a 9.25 feet variance (half the width of the field goal posts) is the same as 7 yards at the line of scrimmage to the variable. Or X/7 = 9.25/50. For Johnson to kick the field goal he will have to kick the ball up for at least 86.665 feet, or 28.89 yards at the halfway point before it starts falling. This was obtained by figuring that the 9 feet of height at the line of scrimmage, 7 yards away (or 21 feet), is the same as 10 feet of height at the goal post to 23.33 feet it will go past the goal post before it hits the ground. The goal post is 150 feet from the start, plus the 23.33 feet past, which means the ball has to go 173.33 feet overall. The halfway point where it starts going down is therefore 86.665 feet (173.33/2). If the balls travels 35 yards before starting to fall, it would be able to have enough height to get the field goal, so in fact yes, the score will be tied. Mike Schmidt and Phillip Pool Grades 9 and 10 Martin County High School, Stuart, Florida The answer to your first question about how far left or right the ball has to be when it crosses the line of scrimmage is between 1.295 feet left and 1.295 feet right but not those exact numbers or else the ball would hit the bar. We reached this answer when we figured that the triangle that is created by the path of the ball, the line of scrimmage, and the distance between the kicker and the line of scrimmage is similar to the triangle created by the distance between the goal post and the kicker, the width of the goal post from the center, and the path of the ball to the outside post. Since the larger triangle is similar to the smaller, it must have similar scale degrees, so we figured out the width of the goal posts, and the length between kicker and goal posts must have the same scale degrees as the smaller triangle and come up with the answer of 1.295 feet. We used the same basic priciples with the second and third questions and came up with the answers of 87 feet and yes, it would go over the crossbar. We made some modifications to our drawing and used the same scale factor rule with the second problem. This would be true because the triangle created by the distance between kicker and line of scrimmage, height of ball when it passes line of scrimmage, and path of ball is similar to the larger triangle. Tim Casale Grade 10 Martin County High School, Stuart, Florida The answer to your first problem about the kicker's accuracy was that he could kick it 1.295 feet to either side and still make it. I came to this answer by using similiar triangles to get the ratio of the width of half the goalpost and the ball to the goalposts and then using that ratio, 9.25' to 150', to get the distance out he could kick it. 9.25/150 = x/21 150x = 194.25 x = 1.295 The answers to the second problem about the distance the the ball has to travel before starting to fall to go over the goalpost were 87 ft and No. I arrived at this conclusion by taking the rising ratio, 9' to 21', and using that to find out how far it went before going through and subtracting that from 150'. /\ / \ /21' \9ft /150'--| 150-63=87' Danielle P. Cegelis Grade 12 Martin County High School, Stuart, Florida 1) The ball could not be more than 1.295 ft. to the right or left of the direct line when it crosses the line of scrimmage. 2) a. The ball must get to at least 37.143 ft. high in order to completely clear the cross bar. b. Thus if the ball travels 35 yds. (105 ft.) before it starts to come down then it will have risen to a height of 45 ft. before it begins to come down, which is greater than the 37.143 ft. minimum to clear the bar. EXPLANATION: 1) I drew a picture of the lines from the ball to the goal posts off in both directions, which formed a triangle. I calculated the lengths of these lines which came out to be 150.285ft. Next I drew a line through the larger triangle where the line of scrimmage was located, which created a smaller triangle within the larger triangle that could be used as a proportional triangle to calculate the measurements of the ball when traveling over the line of scrimmage. 2) I used the triangle diagram provided to create to smaller triangles within the larger triangle. The smaller triangles were used to create proportion calculations to the larger triangle in order to find the minimum height of the ball with it still being able to clear the crossbar triangle to calculate the measurements of the ball when traveling over the line of scrimmage. Kaitlin Coffinbarger and Dustin Hicks Grade 9 Martin County High School, Stuart, Florida To answer the first question, first we found half of the width of the goal post. Then we saw that the triangle formed by the middle of the goal post, end of goal post, and ball was similar to the triangle formed by the line of scrimmage, possible path of the ball, and the ball. One of the legs of the large triangle was 50 yards and the other was 9'3". One of the legs of the smaller triangle was 7 yards and the other was x. We made the proportion of 9'3"/50 = x/7. Then we worked that out and got x = 15.54. Then we divided that by 12 to get feet. So the ball could be 1.295 feet off the line of scrimmage and still go through. For the second problem the path would form 2 similar triangles. If at 7 yards, the ball is 9' or 3 yards high and the goal post is 10' or 3 1/3 yards high, the ball would have to be at 11' or 3 2/3 high over the goal post. Since the triangles are similar, 3yards/7 = 3 2/3yards/x. Then x = 8.5 yards. The ball would have to go out 29.28 yards before going down in order to clear the goal post. For the other part, the ball must go at least 29.28 yards, so since 35 yards is longer, the ball would clear and the game would be tied. Christine Francescani and Julie Conant Grade 10 Martin Country High School, Stuart, Florida 1. We first compared the triangle that has sides 9.25 feet and 171 feet with the triangle that has sides y and 21. We then found that they were similar triangles and therefore 9.25' is to y as 171' is to 21'. We solved for y and got 1.14' so the farthest he could have been off was 1.14'. 2. One vertex of the triangle is the ball's halfway point, so we drew an altitude from the vertex (also the ball's halfway point) that divided the larger triangle into two congruent ones. We compared the triangle with sides 9' and 21' to the similar one with sides 10' and x. 9' is to 10' as 21' is to x. We solved for x, which equalled 23.33'. We found that the length of the bottom of the triangle is 194.33'. 194.33' divided by 3 gives you 64.78 yards. To find the ball's halfway point we divided the 64.78 yards by 2, so the point at which the ball begins to fall must be 32.39 yards. 3. If the ball goes 35 yards before it starts going down, it will clear the goal post and the game will be tied. Anna Margush Grade 4 School: Home Schooled Tim Margush The University of Akron, Department of Mathematical Sciences Akron, OH 1. Distances are converted to inches - the football is then kicked 50 yards or 1800" and the right half of the goal post area is 111" We solved the proportion: 111 ? ---- = ---- -> ? = 15.54" 1800 252 2. Ball hits ground 23 1/3 ft past goal. This is due to similar triangles on the left and right (7 yards = 21 feet) 9 10 --- = ---- -> ? = 23 1/3 ft 21 ? Adding the two distances together (150', and 23 1/3') gave a total distance of 173 1/3 ft. Half of this was going up, half going down. The ball must go 86 4/6 feet! 3. The 35 yards is the same as 105 ft. This is more than required, so the goal is good and the score is tied. Thomas S. Kuo Grade: 7 School: Murray Junior High School, Ridgecrest, California 1. The ball can be off no more than 1'2". B C -------*----------*----- * * * * * * * * D * * E -------*----*------- * * * * * * A * Triangle ADE and ABC are similar. Ratio AD/AB = DE/BC. I know that BC = 18'6"/2 = 9.25' AD = 7 yards = 21' DB = 50 yards = 150' and AB = AD + DB = 171'. Then 21/171 = DE/9.25, and DE = 1.136' or approximately 1'2". 2. The answer is 105'9". C * * * * * * * * * * * * * B * H * * * * * ** D * * * * * * * * * * * * * * * * * * * A * * * * * * * * * * * * * * * * * * * * * * I G F E Since C is the highest point the ball can reach and fall down, angle ACF = angle FCD, then angle CAE = angle CDH, then triangle AGB, triangle AFC, and triangle DHC are similar. I can find ratio to build equations. BG/AG = CF/AF and BG/AG = CH/HD. I also know the following: BG = 9. AG = 7 yards = 21'. GF = y. GE = 50 yards = 150'. FE = HD = 150 - y. CF = h'. DE = 10'. CH = h - 10. AF = AG + GF = 21 = y. I then substituted them into equations. 9/21 = h/(21+y) and 9/21 = (h-10)/(150-y). I then solved equations and got y=76.17' and h=41.64'. I also know that AC^2 = AF^2 + CF^2 = (21+76.17)^2 + 41.64^2, and I got AC = 105.72' = 105'9". 3. Since 35 yards = 105' is less than 10'9", the Steelers do not make the field goal and the Cowboys are still ahead. 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