This weeks Time magazine has an article on the Putnam math competition and it gives an example of an easy problem.
A right circular cone has a base of radius 1 and a height of 3. A cube is inscribed in the cone so that one face of the cube is contained in the base of the cone. What is the length (s) of an edge of the cube?
Any of the math gurus here know how to solve this?
No answer, but I’ll give you a strategy:
Assume the cube has a side length L.
The top of the cube will cut across the cone at that distance from the base of the cone. Find the value for L where the diagonal of the cube face is equal to the radius of the cross section of the cone at distance L from the base.
Assuming I’ve understood the question correctly, I get 3(3[symbol]Ö[/symbol]2-2)/7, about 0.96, using similar triangles
Well, the way I would do it is first draw a right triangle and lable the bottom “1” and the side “3”
Then, I would draw a tall rectangle in the triangle so that the right angle makes up one corner, and the opposite corner touches the hypotenuse. Now lable the tall side “L” and the short side “1/2 L”. You can lable the remaining segments “3-L” and “1 - L/2” if you like.
Now, solve for L.
My trig is fuzzy, but I think you can do it easily enough by saying the ratio of the “opposite” over “adjacent” will be constant, so
3/1 = (3-L)/(L/2), which works out to be L = 6/5.
Other than Jabba I guess none of you guys get Time because they give the answer, it’s s = (9*sqrt(2) - 6)/ 7.
Sorry Jabba-- no credit for you–you didn’t show your work,
I see my mistake.
I forgot the top face of the cube would need to be totally enclosed in the cone. Therefore, if you draw a square inside a circle, you will see that one side of the square is equal to the circle’s diameter times the ã2. If you then draw a rectangle in a triangle (as in my previous post) then the sides of the rectangle would be L and L/ã2, with the segments being equal to (3-L) and (1-L/ã2).
Using the same trig as before,
L = 3ã2/(3 + ã2), which is the same number as above
Again, without attempting to use the Sqrt symbol…
I see my mistake.
I forgot the top face of the cube would need to be totally enclosed in the cone. Therefore, if you draw a square inside a circle, you will see that one side of the square is equal to the circle’s diameter times the sqrt(2).
If you then draw a rectangle in a triangle (as in my previous post) then the sides of the rectangle would be L and L/sqrt(2), with the segments being equal to (3-L) and (1-L/sqrt(2)).
Using the same trig as before,
L = 3sqrt(2)/(3 + sqrt(2)), which is the same number as above.
One way to do it is with a diagram sorta like this:
|
|
| . \
This is a cross section of half the cone, where the line at left goes from the center of the base of the cone to the top of the cone(length 3), the base is the radius (length 1), and the line across in the middle is half the diagonal of the top face of the cube, so its a distance s above the bottom line and its length is s/sqrt(2). Using using similar triangles, 3/1 = (3-s)/(s/sqrt(2)). You can solve to get s = 3/(1+(3/sqrt(2)), which you can rationalize to get what everyone else has (about .96).
This is way easier anything that actually gets put on the Putnam. Usually, the median score of the intelligent college students who take the test is 0.
[hijack]
This is the Putnam problem that’s always given out to us as a rediculously easy one. It was actually put on the Putnam at one point, but the problems are usually much more difficult.
This year was the third year I’ve taken the test. The first two, I got 0 points. (Things are looking up, though. After talking to some of my math-genius friends, I predict that my score this year will be in [10,20]).
Here’s an example of a harder Putnam problem, from this year: “A palindrome in base b is a positive integer whose base-b digits read the same backwards and forwards; for example, 2002 is a 4-digit palindrome in base 10. Note that 200 is not a palindrome in base 10, but it is the 3-digit palindrome 242 in pase 9, and 404 in base 7. Prove that there is an integer which is a 3-digit palindrome in base b for at least 2002 different values of b”
7 semesters of Math under my belt and I don’t even know how to begin to solve that.
I remember this problem. 1999, was it? And as much as I hate to say it, that is a very, very easy problem for the Putnam. 
Here, let me try to redeem myself from my first post. First, let me say there is some ambiguity in the question. Clearly the number 111 is a palindrome for base 2 - infinity. For base 2, the number is equal to 7, base 3 it is equal to 13, base 4 it is equal to 21, etc. I am going to assume that is not what the question is asking, because that is just silly.
I assume the question asks is there a number that can be converted into 2002 palindromes?
I believe the answer is yes.
I recognize that 0x0 is a convenient palindrome. As an example, take 100, which can be represented as 010 in base 100, 020 in base 50, 040 in base 25, and 050 in base 20. There are other palindromes in different formats (e.g., 121 in base 9 is still 100) but I’m not going to worry about those.
The larger the number gets, the more 0x0 palindromes you have. For example, 1000 is 0 1 0 in b(1000), 0 2 0 in b(500), 0 4 0 in b(250), 0 5 0 in b(200), 0 8 0 in b(125), 0 10 0 in b(100), 0 20 0 in b(50), and 0 25 0 in b(40).
Note that in both cases the base b can not be greater than or equal to the square root of the integer in question. It does not make sense to have a number 0 10 0 represent 100 in base 10.
The general statement can be made that a number N has at least Z number of palindromes, where Z equals the number of unique factors that are less than the sqrt(N).
Therefore, there clearly exists a finite number less than infinity that has 2002 unique factors in it whereby each of those factors are less than the square root of N.
I can’t believe that question was on the Putnam. I got 14 and 22 out of 120 when I wrote it, and my best result was the 14: 90th percentile. The one I got 22 on was just over a year ago, and it was (relatively) super easy. There was a gimme that was about as hard as that cone problem. I missed it! :smack:
Well get to work! The palindrome can’t have a leading 0!
I’m gonna think about this for a bit, cause I just handed in my last term paper (woohoo!) and I have some time to kill before I meet my friend.
I spent two hours on that stupid palindrome question last night. It’s going to drive me crazy until I get it. I tried starting with the number that’s 101 in base N, where N is 2002, 2[sup]2002[/sup], 2002!, 2002![sup]2002![/sup], and 2^2^2^2…2^2 with 2002 terms. Also, 2 raised to each of these N’s, plus or minus 1.
No luck yet.
Says who?
Anyway, the same logic goes for 1x1 as 0x0.
I don’t think that’s the right approach. They probably just chose 2002 cause they always put the year in a problem somewhere. I think you need to prove that there is a palindrome in n bases for any n.
If 0 counts as a leading digit, then I nominate the following integer:
0
which is a 3-digit palindrome (000) in every base!
KarmaComa: everywhere I have a 2002, replace it with an n. That’s what I really did. I was going to let n be 2002 when I was done. Certainly, I don’t know any significant properties of 2002! that don’t apply to 2003!, so it’s not like picking that number specifically would have changed my work at all.
But it is not a “positive integer”
I’d love to see the reasoning behind that one!
0’s not positive. Okay fair enough. But still, I don’t think 040 counts as “a three-digit palindrome”. Here’s another answer that is bogus:
N = 100000 × 100001 × 100002 × … × 102001
Consider each of the bases:
b = N / 100000 : N = 0.100000.0
b = N / 100001 : N = 0.100001.0
b = N / 100002 : N = 0.100002.0
…
b = N / 102001 : N = 0.102001.0
Where the . stands for decimal separation.