Thursday, February 09, 2006
Math Problems
I like math.
There I said it.
In fact, I came -> <- that close to being a math teacher in the fall of 2004. I get along with math people. I think in math. I count steps every time I go up or down stairs, and I take special note when it's a prime number. I like solving equations and then thinking of the other ways I could have solved them. I tutored a student in geometry once, and it was about the most fulfilling thing I've ever done to help someone out. So here are some of my favorite problems of all time:
Easy - worth a donut
A man selling oranges starts his day with a whole number of oranges. To his first customer, he sells half his oranges, plus half and orange. To his send customer, he sells half of what he has left, plus have an orange. To his third customer, he sells half of the remaining oranges, plus half an orange. At this point, he has no more oranges and never had to cut an orange. How many did he start with?
Medium - worth a Wawa latte
There are 625 students enrolled in a school, and each one has a locker. One day, the students decide to wreak some havoc. They all line up and the first student runs down the hall and opens every locker up. The second student then runs after him and closes every other locker (starting with number 2, then 4 and so on). The third student follows and changes the position of the door to every third locker, i.e. closes number 3 (opened by the first student), opens number 6 (closed by the second student), and so on. This pattern is continued for every student in order. At the end of the ordeal, how many lockers are open?
Difficult - worth a Starbucks latte
A piece of railroad track exactly one mile long is securely fastened at both ends during the cool of the night. In the heat of the day, it expands by exactly one foot, and buckles upward in a symmetrical fashion. Assuming for the purposes of this problem a flat earth, how high off the ground is the middle of the bowed track?
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There I said it.
In fact, I came -> <- that close to being a math teacher in the fall of 2004. I get along with math people. I think in math. I count steps every time I go up or down stairs, and I take special note when it's a prime number. I like solving equations and then thinking of the other ways I could have solved them. I tutored a student in geometry once, and it was about the most fulfilling thing I've ever done to help someone out. So here are some of my favorite problems of all time:
Easy - worth a donut
A man selling oranges starts his day with a whole number of oranges. To his first customer, he sells half his oranges, plus half and orange. To his send customer, he sells half of what he has left, plus have an orange. To his third customer, he sells half of the remaining oranges, plus half an orange. At this point, he has no more oranges and never had to cut an orange. How many did he start with?
Medium - worth a Wawa latte
There are 625 students enrolled in a school, and each one has a locker. One day, the students decide to wreak some havoc. They all line up and the first student runs down the hall and opens every locker up. The second student then runs after him and closes every other locker (starting with number 2, then 4 and so on). The third student follows and changes the position of the door to every third locker, i.e. closes number 3 (opened by the first student), opens number 6 (closed by the second student), and so on. This pattern is continued for every student in order. At the end of the ordeal, how many lockers are open?
Difficult - worth a Starbucks latte
A piece of railroad track exactly one mile long is securely fastened at both ends during the cool of the night. In the heat of the day, it expands by exactly one foot, and buckles upward in a symmetrical fashion. Assuming for the purposes of this problem a flat earth, how high off the ground is the middle of the bowed track?
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Comments:
Jason Estock said...
First I set the 5281 length track to describe a portion of a circle , and the 5280 track to be a straight line (chord) within the circle. The intersection of the tracks at the ends, with the middle of the circle, with the midpoint of the 5280 track can describe a triangle. From the circle and the triangle we can derive the following equations. I am solving for the angle from the middle of the circle to the endpoints of the chord, (x is the angle).
5281 = 2*3.1415*R*(X/360)
Sin (.5x) = 2640 / R
R = 2640 / Sin(.5x)
5281 = 2*3.1415(2640/sin(.5x))(x/360)
I then put each side of that equation into a graphing calculator and find the intersection of the two lines, since I dont know any of the quick and easy rules to solving this puzzle. X = 3.8626246. Therefore .5x = 1.9313123.
Now I know a length and an angle for my right triangle, and can use simple trig to solve for the other unknown sides of the triangle, getting length of 78339.8434 for the hypotenuse, and 78295.34776 for the other vertical unknown side. The difference between these numbers is the vertical rise of the track, which is 44.49564 feet. Please tell me an easier way to do this problem.
-J>
First I set the 5281 length track to describe a portion of a circle , and the 5280 track to be a straight line (chord) within the circle. The intersection of the tracks at the ends, with the middle of the circle, with the midpoint of the 5280 track can describe a triangle. From the circle and the triangle we can derive the following equations. I am solving for the angle from the middle of the circle to the endpoints of the chord, (x is the angle).
5281 = 2*3.1415*R*(X/360)
Sin (.5x) = 2640 / R
R = 2640 / Sin(.5x)
5281 = 2*3.1415(2640/sin(.5x))(x/360)
I then put each side of that equation into a graphing calculator and find the intersection of the two lines, since I dont know any of the quick and easy rules to solving this puzzle. X = 3.8626246. Therefore .5x = 1.9313123.
Now I know a length and an angle for my right triangle, and can use simple trig to solve for the other unknown sides of the triangle, getting length of 78339.8434 for the hypotenuse, and 78295.34776 for the other vertical unknown side. The difference between these numbers is the vertical rise of the track, which is 44.49564 feet. Please tell me an easier way to do this problem.
-J>
Excellent work, once again. I completely concur with your analysis and believe that you solved the problem using sound mathematical principles. It would have warmed my soul to see you try to work out the equation, but from my limited understanding, it is not a closed-form solution. There is a long way to approximate the answer that your fancy calculator gives. I accept your answer and award you one (1) Starbucks latte, to be delivered in proximity to the donut you earned for being the first to recognize the title of my previous post as being the message of the first telegram. Congratulations!>
well, i didnt consider it very soul warming but I had already simplified the equation to
(sin (.5x))/x = .008725 but was unable to solve it, so I put it into the calculator at that point. Is there perhaps some way to simplify it further and solve?
-J>
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(sin (.5x))/x = .008725 but was unable to solve it, so I put it into the calculator at that point. Is there perhaps some way to simplify it further and solve?
-J>
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