The high point of my Sunday was when Herschel Walker ran a kickoff back 86 yards, giving the Dallas Cowboys a chance to take an early lead against the Philadelphia Eagles. Despite his advanced age (35), he can still run!When Herschel was at the University of Georgia, he ran track in addition to playing "American football." His best events were the 60 yard dash (very useful in football) and the 400m relay. I'm not sure what other events he ran, but let's see if he ran the equivalent of the 100 yard dash when he returned that kick.
He received the kick on the right hash mark at the 7 yard line and ran it back to the other 7 yard line on the left side of the field for a total of 86 yards. We want to know the absolute minimum distance he had to run to get from one point to the other. How far was it? Was it close to a 100 yard dash? (Remember to explain your answers - answers of "yes" or "no," accompanied by a number, no matter how right, will not be considered correct!)
To figure this out, we need to know a few things. The American football field is 160 feet wide. There are hashmarks 70' 9" from each sideline, so the width of the field looks like this:
|<------------------- 160' ------------------>|
|<--- 70' 9" -->| |<--- 70' 9" -->|
sideline | - - - - - - - + - - - - - - + - - - - - - - | sideline
^ ^
hashmark hashmark
The field is 100 yards long, and is marked off starting at the 50 yard line in the middle and decreasing as you get to the goal lines. The goal line is yard zero. The 7 yard line is 7 yards from one end zone and 93 from the other. (Let me know if you have any questions about the field layout.)
Annie says: Solutions
First off, I must say that those of you who dislike the Dallas Cowboys were incredibly polite this week! Thank you.
A lot of you didn't get the answer I was looking for because I was not clear about where Herschel Walker was when he ended his run. We all know he was on the 7 yard line, but some of you assumed he went out of bounds (he did) and some of you assumed he ended at the other hash mark. Consequently, both answers have been considered correct.
The units did not catch as many people as I expected them to - good work! But a number of people did more work than they had to. This of course isn't considered wrong, but when you start dealing with bigger problems, it's worth being careful about what you do and don't have to figure out. For example, a lot of people took the two 7 yard lines and figured out that there are 86 yards between them. But the problem says that right off the bat! You don't have to figure it out. Also, a number of people took the width of the field, subtracted the two 70'9" distances outside the hashmarks to find out how far it was between then, and then added 70'9" back in to get the distance from one hash mark to the other sideline. Seems a bit silly to me - just subtract one 70'9" the first time around and you're all set. The more work you do, the more chances there are for you to make a simple mistake.
As for whether 91 (or 86) yards is close to 100, that's hard to say. Some people said yes. Someone pointed out that in football or track, it is nowhere near 100 yards. It all depends on your perspective.
I have highlighted three solutions below. Jason Yeung from the Iolani School provided a very nice explanation which shows how he approached the problem. Dan Tobin from Germantown Academy did much the same thing (and also mentioned that being a Cowboys fan living in Philadelphia might not be a good idea, but didn't bother to remind me who eventually won that game - nice guy!). Tomomi Nakajima provided a very nice picture that made his explanation very clear.
A list of all the people who got this problem right and most of the solutions are also available.
Jason Yeung Grade: 10 School: Iolani School, Honolulu, Hawaii Answer: The minimum distance is 91 yd, and Walker is pretty close to a 100 yard dash. In Geometry, we learn that the absolute minimum distance is a straight line. Therefore, the shortest distance Walker would run is the hypotenuse of the right triangle. To find the hypotenuse of the right triangle, we must first find the length of the 2 sides of the right triangle. Knowing that he started at the right hash mark and ran to the other 7 yard line on the left side, the distance from the right hash mark to the left sideline on the same 7 yard line is 160'-70'9", or 89.25'. 3 ft = 1 yd, so 89.25' is also 29.75 yd. The distance from the left side of the 7 yard to the left side of the other 7 yard line is 86 yd. Based on the Pythagorean theorem, a square + b square = c square. Thus, 29.75 square + 86 square = c square, or c = 91.000 yd. Therefore, the minimum distance is 91 yd, and he is pretty close to a 100 yard dash.
Dan Tobin School: Germantown Academy, Fort Washington, Pennsylvania Mrs. Carver's Geometry Class First I would like to say that it was not smart revealing that you are a Dallas Cowboy fan to us Eagle fans. Our Eagles are enemies to your Cowboys. Anyway, as soon as I read the problem, I thought I would have to create a right triangle in order to complete this problem. I first figured the distance between the two hashmarks in the middle of the field. I added the distances from the sideline to the hashmark, so 70.75 feet + 70.75 feet gave me 141.5 feet. Then, to get the distance between the hashmarks I did 160 feet - 141.5 feet and received 18.5 feet. This was one leg of my right triangle. I knew from the 7 yard line to the 7 yard line was 86 yards, but I had to convert that into feet because my other leg was in feet. I know that there are three feet in a yard so 86*3 gave me 258 feet. One leg was 18.5 feet and the other was 258 feet. In order to find the hypotenuse, which is also the minimum amount of distance he had to run, I used the Pythagorean theorem. After plugging it into my calculator, the hypotenuse came out to be 258.7 feet. In order to find if this was close to the 100 yard dash, I had to calculate 258.7 feet back into yards. So, 258.7 divided by 3 = 86.2 yards. The minimum amount of distance that Walker ran and the 100 yard dash had a difference of 13.8 yards, which in the world of football and track is quite a lot of distance.
Tomomi Nakajima Grade: 9 School: Newport High School, Bellevue, Washington Art Mabbott To solve this problem, I first drew my own diagram of the problem. B O __________________________________________________ | | | | | | | | | | | |70'9" |________|________________________________|________| | | | | | | | |18'6" |________|________________________________|________| | A| C| | | | | | | | | | 70'9" P |________|________________________________|________| 7 yds 86 yds 7 yds *PO=160' First of all, I got the distance between the 70'9" marks by: 160'-2(70'9")=18'6" I needed to figure out the distance of BC BC=160'-70'9"=89'3"=89.25' Now I needed to convert 86yds into feet: AC=86yds=86x3=258' I used the Pythagorean theorem to find the distance for AB: (AC)^2+(BC)^2=(AB)^2 (89.25)^2+(258)^2=(AB)^2 7965.5625+66.564=(AB)^2 74529.5625=(AB)^2 (AB)=273.0010302' (AB)=273'/3 (AB)=91.000yds So, Herschel Walker was only able to run 91 yards as the absolute minimum distance.
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