| Transcript
for the Math
Jam "USAMTS Round 2 Math Jam"
on Nov 26. |
| Math Jam hosted by DPatrick
(Dave Patrick ). |
DPatrick19:28:19
Hello and welcome to the second 2007-08 USA Math Talent Search Math Jam.
DPatrick19:28:28
Before we get started I would like to take a moment to explain our virtual classroom to those who have not previously participated in a Math Jam or one of our online classes.
DPatrick19:28:44
The classroom is moderated, meaning that students can type into the classroom, but only the moderators can choose a comment to drop into the classroom. This helps keep the discussion organized and on track.
DPatrick19:29:03
This also means that only well-written comments will be dropped into the classroom, so please take time writing responses that are complete and easy to read.
DPatrick19:29:19
Also, only moderators can enter into private chats with other users, although due to the size of this Math Jam it is unlikely that this will happen.
DPatrick19:29:36
As there are a lot of users today, it will not be possible for me to respond to each comment or question individually. Please do not take it personally if I do not post or respond to your comments individually.
DPatrick19:29:45
There will be images in this lecture. The images should appear directly in the classroom window, as in the example below:
DPatrick19:29:52
DPatrick19:30:09
If you click on the link, the image may appear in a separate window. (You may have to hold down the Ctrl key while you click on an image link, and/or you may have to disable your popup blocker.)
DPatrick19:30:34
You can view the second round problems as we discuss them by clicking on the following link:
DPatrick19:30:47
Problem 1
DPatrick19:30:50
CatalystOfNostalgia19:31:23
try to construct one to get an estimate for n
DPatrick19:31:33
Yes, this problem lends itself to an "experimental" approach. We can try coloring integers and see how long a sequence we can make with no arithmetic sequence of the same color.
DPatrick19:31:45
How might this go?
krsattack19:31:51
WLOG, assume it starts with Red
DPatrick19:32:02
OK, we can arbitrarily color 1 red; it doesn't matter.
Brut3Forc319:32:24
8 does not work - RRBBRRBB
Xantos C. Guin19:32:24
RRBBRRBB
apple pi19:32:24
rrbbrrbb has no arithmetic sequences
McDutchy19:32:24
Show 1-8 doesnt work, with pattern RRBBRRBB
DPatrick19:32:35
Of course this works, but surely you didn't come up with this out of this air.
DPatrick19:32:51
The point of the Math Jam is to show how we arrive at the solution, systematically.
krsattack19:33:05
3 Cases... RRB, RBR, RBB
DPatrick19:33:23
Right: we can't then have both #2 and #3 be red, because then the arithmetic sequence {1,2,3} would be all red.
McDutchy19:33:28
Obvious things: No groups of 3 consecutive integers of the same color
DPatrick19:33:40
Indeed, we can never have 3 consecutive integers all the same color.
DPatrick19:33:53
So there are 3 cases (with #1 red):
Case 1: 2 is red, 3 is blue
Case 2: 2 is blue, 3 is red
Case 3: 2 is blue, 3 is blue
DPatrick19:34:11
We can just take them in order.
DPatrick19:34:19
Case 1
DPatrick19:34:25
DPatrick19:34:35
Are any other choices constrained?
VDLmath19:34:49
4 can be red or blue
samath19:34:49
not yet
Brut3Forc319:34:49
So far, no.
DPatrick19:34:59
Not at present. So we have two sub-cases, depending on which color we put for #4.
DPatrick19:35:08
Case 1a
DPatrick19:35:11
kostya19:35:41
7 must be blue
kostya19:35:41
6 must be blue
atennis19:35:41
6 can not be red
Xantos C. Guin19:35:43
6 and 7 must be blue
DPatrick19:35:48
#6 must be blue so that {2,4,6} is not all the same color.
Similarly #7 must be blue so that {1,4,7} is not all the same color.
DPatrick19:35:57
DPatrick19:36:08
kostya19:36:20
then 5 and 8 must be red, but that makes 2,5, and 8 red
krsattack19:36:20
6, 7 = B ==> 5, 8 = R... but 2, 5, 8
DPatrick19:36:31
Right --now #5 and #8 must be red, so that {5,6,7} and {6,7,8} are not all blue:
DPatrick19:36:37
Brut3Forc319:36:48
Which makes 5 and 8 red - a contradiction {2,5,8}
samath19:36:48
2 5 8
lingomaniac8819:36:52
Therefore the longest one we can make for case 1a is 7
DPatrick19:37:00
This doesn't work. {2,5,8} is all red. So this gives us a valid sequence of length 7.
McDutchy19:37:08
So now go to case 1b, RRBB
DPatrick19:37:12
Case 1b
DPatrick19:37:24
SigmaTheta19:37:31
5 has to red
apple pi19:37:31
Then 5 must be R, to avoid a sequence
lingomaniac8819:37:31
5 must be red
DPatrick19:37:36
#5 must be red, so that {3,4,5} is not all blue:
DPatrick19:37:40
kostya19:37:56
8 is blue
atennis19:37:56
8 can't be read
electricj19:38:02
9 is blue
DPatrick19:38:08
Now #8 must be blue, so that {2,5,8} is not all red.
Also #9 must be blue, so that {1,5,9} is not all red.
DPatrick19:38:14
apple pi19:38:30
then 7 and 10 are R
krsattack19:38:30
then 6 is red by {4, 6, 8}
DPatrick19:38:46
This means that #7 must be red (by {7,8,9}), and thus #6 must be red (by {4,6,8}):
DPatrick19:38:56
krsattack19:39:10
5, 6, 7 = R
lingomaniac8819:39:10
But then we have {5,6,7}
p4fn2w19:39:13
{5, 6, 7} contradiction
DPatrick19:39:30
But now {5,6,7} is all red, which is not allowed. So we only have a valid sequence of length 6.
Brut3Forc319:39:46
Eliminating 9 and changing 7 to R fixes the error.
apple pi19:39:46
if we don't have a 9, then we can get a sequence of 8
DPatrick19:40:09
Right, we can create a valid sequence of length 8 by ignoring #9 and changing #7 to blue:
DPatrick19:40:32
DPatrick19:40:46
Let's move to Case 2:
DPatrick19:40:50
alkjash19:41:03
5 is blue
p4fn2w19:41:03
5 is blue
perfect62819:41:03
5 must be blue
DPatrick19:41:10
#5 must be blue, so that {1,3,5} is not all red.
Xantos C. Guin19:41:15
5 is blue, and then 8 is red
kostya19:41:16
then 8 is red
DPatrick19:41:21
Then #8 must be red, so that {2,5,8} is not all blue.
DPatrick19:41:31
samath19:41:57
subcases again ...
DPatrick19:42:03
There are no other immediate constraints, so we again have two subcases, depending on how we color #4.
DPatrick19:42:13
Case 2a
DPatrick19:42:17
krsattack19:42:45
6=B
math33319:42:45
7 is blue
jtw19:42:45
7 = B
VDLmath19:42:45
so then 7 is blue
atennis19:42:45
6 is blue
DPatrick19:42:53
#6 must be blue so that {4,6,8} is not all red.
DPatrick19:43:07
And #7 must be blue so that {1,4,7} is not all red.
DPatrick19:43:16
Xantos C. Guin19:43:24
{5,6,7} is blue, a contradiction
lingomaniac8819:43:24
But then we have {5,6,7}
McDutchy19:43:24
but then 5,6,7 Blue
DPatrick19:43:37
But this breaks since {5,6,7} is all red.
DPatrick19:43:50
Case 2b
DPatrick19:43:54
ifckoh19:44:08
you mean blue
DPatrick19:44:17
oops, yeah, I meant "blue" in my last line of text, thanks.
math33319:44:32
6 is R, 7 is B
DPatrick19:44:38
#6 must be red so that {4,5,6} is not all blue.
Then #7 must be blue so that {6,7,8} is not all red.
DPatrick19:44:45
DPatrick19:45:04
This is a valid sequence of length 8.
atennis19:45:14
9 is blue
alkjash19:45:14
9 must be blue
p4fn2w19:45:14
9 cant be red or blue
DPatrick19:45:35
But we cannot color #9.
If we color it red, then {3,6,9} is all red.
If we color it blue, then {5,7,9} is all blue.
perfect62819:45:52
9 always creates a contradiction
SigmaTheta19:45:52
9 cant be anything
SigmaTheta19:45:55
so I guess that is as far as we can go
DPatrick19:46:03
Right...we have one more case to look at.
DPatrick19:46:07
Case 3
DPatrick19:46:10
math33319:46:32
4 is R
fred36919:46:32
4 is R
VDLmath19:46:32
7 is blue
DPatrick19:46:37
We must color #4 red so that {2,3,4} is not all blue.
We must then color #7 blue so that {1,4,7} is not all red.
DPatrick19:46:42
math33319:47:10
5 ir R, 6 is B
alkjash19:47:10
5 is red
Criticalline185919:47:10
5 is red
DPatrick19:47:15
Now #5 must be red so that {3,5,7} is not all blue.
And then #6 must be blue so that {4,5,6} is not all red.
DPatrick19:47:23
mz9419:47:35
8 is red
jun900819:47:35
then 8 is red also?
DPatrick19:47:41
#8 must be red so that {6,7,8} is not all blue:
DPatrick19:47:48
DPatrick19:47:52
This is also a valid coloring of 1 through 8.
math33319:48:03
9 can't be anything
Brut3Forc319:48:03
9 cannot be colored
lingomaniac8819:48:03
And then 9 can't be red or blue
DPatrick19:48:09
However, we cannot color #9.
If it's red, then {1,5,9} is all red.
If it's blue, then {3,6,9} is all blue.
DPatrick19:48:24
So any attempt to color so that no 3-term arithmetic sequence is the same color will fail at coloring #9 (or earlier).
DPatrick19:48:40
Thus, conversely, any coloring of 1-9 must have a 3-term arithmetic sequence the same color. And we have exhibited a coloring of 1-8 that does not have such a sequence.
stanman19:48:47
thus, the answer is 9!
katrina2k259319:48:47
therefore, we have covered all cases, and n must equal 9
DPatrick19:48:51
Therefore, the answer is n=9.
DPatrick19:49:38
There are some more general results for coloring problems such as this...you can do a bit of research on your own, and we'll probably provide some additional commentary when we post the solutions.
DPatrick19:49:48
Problem 2
DPatrick19:49:50
VDLmath19:50:14
let z=-x-y, and then plug it into the other 2 equations
Laplace's_Demon19:50:14
Substitute -(x+y) for z and plug away
DPatrick19:50:38
There are a number of things we could try, but I think the most straightforward is to write z = -(x+y).
DPatrick19:50:46
Let's plug this into the second equation:
DPatrick19:50:50
Aubrey Yang19:51:09
foil it out
DPatrick19:51:19
We can expand this and cancel the x^5 and y^5 terms:
DPatrick19:51:25
Aubrey Yang19:51:40
factor out xy
samath19:51:40
factor!
mathnerd31415919:51:40
Divide by 5.
DPatrick19:51:46
Let's divide by 5 and factor what's left:
DPatrick19:51:53
Criticalline185919:52:05
we can prove x (and y) are not 0 and divide by xy
VDLmath19:52:07
x and y cannnot equal 0
DPatrick19:52:21
Right. We see that one solution is x=0. This means that y = -z.
But in this case, x^3 + y^3 + z^3 = y^3 - y^3 = 0, not 3. So we have to throw this out.
DPatrick19:52:30
Similarly, we can't have y=0.
DPatrick19:52:39
So we can divide by xy:
DPatrick19:52:44
Altheman19:53:03
This is a polynomial in x. Note that x=y gives zero, so (x-y) is a factor.
mathnerd31415919:53:03
Or z=0.
DPatrick19:53:15
Right. For the same reasoning as above, we can't have z=0.
DPatrick19:53:32
But z=0 corresponds to x=y, so we see that we can divide by x-y as well.
DPatrick19:53:52
oops, I meant x=-y, so we'll be able to divide by x+y.
DPatrick19:53:54
DPatrick19:54:01
DPatrick19:54:09
tenniskidperson319:54:26
this is a quadratic in y
DPatrick19:54:35
Sure: one way to finish is to use the quadratic formula to solve for x in terms of y:
DPatrick19:54:42
(or vice versa)
DPatrick19:54:46
DPatrick19:55:00
Xantos C. Guin19:55:17
Multiply by x-y and get x^3=y^3
DPatrick19:55:33
Or..we could have taken our quadratic and multiplied by x-y, to get 0 = x^3 - y^3.
DPatrick19:55:42
In particular, note that x^3 = y^3!
krsattack19:55:51
x^3 = y^3 (= z^3)
DPatrick19:56:02
Right, we also have x^3 = z^3 (by symmetry).
tenniskidperson319:56:09
they all equal 1
CatalystOfNostalgia19:56:09
so x^3+y^3+z^3=3=> x^3=y^3=z^3=1
DPatrick19:56:15
Therefore, x^3 = y^3 = z^3 = 1. (The first two equations imply that x,y,z are the three distinct complex cube roots of 1, but we don't need that to finish the problem.)
tenniskidperson319:56:37
raise each to the 669th power and ad
DPatrick19:56:39
Therefore, x^2007 = (x^3)^(669) = 1, and similarly y^2007 = z^2007 = 1.
Xantos C. Guin19:56:43
1^669+1^669+1^669=3
DPatrick19:56:47
So the answer is 3.
CatalystOfNostalgia19:57:07
shouldn't we first check that there actually is a solution?
DPatrick19:57:29
Indeed, you would have to show that there actually is a solution. But x=1, y=(1+sqrt(3)i)/2, and z=(1-sqrt(3)i)/2 works.
Laplace's_Demon19:57:38
We already showed the cube roots of unity solve it
Brut3Forc319:57:38
isn't it the cubic roots of unity?
DPatrick19:57:40
Exactly.
DPatrick19:57:48
Problem 3
DPatrick19:57:53
DPatrick19:58:14
Any quick observations?
CatalystOfNostalgia19:58:24
1=not possible
ifckoh19:58:24
can't use 1
McDutchy19:58:24
no 1
DPatrick19:58:29
1 cannot appear anywhere in the array, because all the numbers must be distinct.
tenniskidperson319:58:46
greedy algorithm
mathnerd31415919:58:46
Look at the lowest number.
VDLmath19:58:46
so then 2 is the smallest number in the array
DPatrick19:59:02
Intuitively it seems like we would want 2 at the bottom.
DPatrick19:59:14
Then, the smallest possibilities for the next row are 3 and 6:
DPatrick19:59:17
Aubrey Yang20:00:16
we see that whenever you move up 1 row, you have to add one more distinct facator
atennis20:00:16
the smallest number that could be in the 3rd row is 4
lingomaniac8820:00:16
The smallest possible value in the next row would be 4
DPatrick20:00:45
So we'll (again, just intuitively right now; we haven't proven anything rigorously yet) we'll want a 4 in the next row.
DPatrick20:00:55
DPatrick20:01:06
Finally, we should get a minimum number on the top row by making one of the entries a 5.
DPatrick20:01:15
This is the best we can do:
DPatrick20:01:20
DPatrick20:01:34
Is this the best we can do?
kostya20:01:46
put the 4 on the right would make 8 on the left and 24 in the middle
Brut3Forc320:01:50
but if you change the order of the third row to 8,24,4 a better solution is presented
DPatrick20:01:57
Right: a little more experimenting should give you this better configuration (with the 4 in the 3rd row on the end, not in the middle):
DPatrick20:02:01
DPatrick20:02:13
A bit more experimenting may convince you that we can't do any better than the above, which gives a value of 120.
Altheman20:02:21
We would hope that 2*3*4*5 is the largest element in the top row.
stanman20:02:21
120 is the best we can do, it is 2*3*4*5
DPatrick20:02:28
How can we prove that we must also have at least 120 somewhere in our triangle?
Note that 120 = 2*3*4*5. How can we use that?
Bernewtniz20:03:11
The greatest element of the nth row is >= the greatest element of the (n-1)th row, times the least element of the nth row.
DPatrick20:03:31
This is the simplest way to prove it I think.
DPatrick20:03:37
Let's elaborate.
DPatrick20:03:42
Let's look at the bottom two rows more abstractly:
DPatrick20:03:46
DPatrick20:03:55
Here a and b are distinct positive integers each at least 2.
DPatrick20:04:00
If c is the smallest element in the 3rd row, then how big is the largest element?
Brut3Forc320:04:21
abc
McDutchy20:04:21
abc
tenniskidperson320:04:21
abc
DPatrick20:04:26
The 3rd row must have at least one element of the form (something)*ab, since ab is the quotient of the two elements on the right side of the row (the way we're drawn it).
DPatrick20:04:55
Thus, if c is the smallest possible value of "something" in the line above, then the largest element of the 3rd row must be at least abc.
CatalystOfNostalgia20:05:04
then similarly above that theres an abcd
DPatrick20:05:24
Again, the largest row must contain an element at least of the form (something)*abc, and if the smallest that "something" can be is d, then the top row must contain an element of at least abcd.
stanman20:05:36
abcd must equal 2*3*4*5
Bernewtniz20:05:36
yes, and a,b,c and d are destinct integers greater than 1
VDLmath20:05:36
so the smallest value of abcd is 2*3*4*5 since a, b, c, and d are not equal
DPatrick20:05:42
To finish: the smallest choices for a,b,c,d are 2,3,4,5 (in some order), so the top row must contain an element of at least 2*3*4*5 = 120.
DPatrick20:05:53
We've constructed an example that has all elements at most 120, and we've proven that every triangle must have an element that's at least 120, so we're done. 120 is the answer.
DPatrick20:06:12
Problem 4
DPatrick20:06:16
DPatrick20:06:26
DPatrick20:06:36
I hope nobody tried to count them by hand from the picture. :)
apple pi20:06:47
Try a simpler problem.
DPatrick20:06:55
Sure -- we can try looking at smaller example to get a feel for the problem, and perhaps make a conjecture.
DPatrick20:07:02
Let's put 2 dots on each side:
DPatrick20:07:10
McDutchy20:07:19
trivial, 2 triangles
Xantos C. Guin20:07:19
There are 2 triangles
DPatrick20:07:30
Yeah, that's a bit *too* easy; there are clearly 2 triangles.
DPatrick20:07:34
Let's try 3 dots:
DPatrick20:07:42
DPatrick20:07:52
Already it's somewhat icky. I don't want to try to count them by hand.
DPatrick20:07:56
So let's try to count them more abstractly.
What types of triangles are there?
VDLmath20:08:08
we can split the kinds of triangles into two kinds: those with one vertex of the circle as a triangle, and those that have two points of the circle as vertices of the triangle
katrina2k259320:08:08
this can be split into 2 cases: triangles with one point on the circle as a vertex, and triangles with 2 points on the circle as vertices
DPatrick20:08:20
Right. There are basically two types: those with 1 of the points as a vertex, and those with 2 points as vertices.
DPatrick20:08:28
Let's try to count them separately.
DPatrick20:08:32
Case 1: Triangles with exactly 1 point as vertex.
DPatrick20:08:38
DPatrick20:08:46
How is such a triangle determined?
lingomaniac8820:09:07
three chords
Bernewtniz20:09:18
by a point in one are, a pair of points (one in each arc) and a pair of points in the opposing arc.
Criticalline185920:09:22
by the vertex on the arc, the two points oppositte, and the intersectind line segment
Xantos C. Guin20:09:28
3 chords intersecting at 5 points on the circle
DPatrick20:09:42
There are 5 points involved in determining such a triangle: 2 on one side and 3 on the other. The points are shown circled in the following picture:
DPatrick20:09:49
DPatrick20:10:02
These 5 points are the endpoints of the lines forming the 3 sides of the triangle.
DPatrick20:10:10
How many triangles does this particular set of 5 points determine?
Bernewtniz20:10:15
These 5 points also define a second such triangle (with one vertex on the circle)
alkjash20:10:20
2
DPatrick20:10:27
stanman20:10:36
So all you need to do is find all the possible combinations of such points
DPatrick20:10:54
And how many choices are there of 5 such points? (In the general problem, with 8 on each side.)
mathnerd31415920:11:27
8 choose 2 * 8 choose 3 * 2
alkjash20:11:27
8C3*8C2*2
DPatrick20:11:44
We must choose 2 on top and 3 on the bottom or vice versa, so that's 2 choices there (which side has 2 and which side has 3).
Then there are C(8,2) * C(8,3) choices for points.
DPatrick20:11:53
DPatrick20:12:06
(Remember, each set determines 2 triangles.)
DPatrick20:12:17
The other case is Case 2: Triangles with exactly 2 points as vertices.
DPatrick20:12:25
DPatrick20:12:33
How is such a triangle determined?
Criticalline185920:12:44
now we select 4 points
fireninja020:12:44
4 points
Laplace's_Demon20:12:44
determined by 2 points on one arc, 2 on the other
Xantos C. Guin20:12:44
2 points on each arc
DPatrick20:12:49
There are 4 points involved in determining such a triangle: 2 on each side. The points are shown circled in the following picture:
DPatrick20:12:57
krsattack20:13:14
2 triangles again
DPatrick20:13:30
Right...every such set of 4 points determines 2 triangles:
DPatrick20:13:36
VDLmath20:13:50
therefore, there will be 8 choose 2 squared times 2 such triangles
guiswim20:13:50
8C2 * 8C2 * 2
DPatrick20:13:58
DPatrick20:14:07
DPatrick20:14:41
Finally, Problem 5:
DPatrick20:14:48
firelight20:15:19
first, figure out what S looks like (a "twisted" triangular prism thingy)
DPatrick20:15:30
OK...what does S look like?
McDutchy20:15:49
triangular antiprism, a stretched octahedron
DPatrick20:16:02
S is an elongated regular octahedron. That is, it's an octahedron where the height between the faces ABC and XYZ has been stretched.
DPatrick20:16:29
Therefore, one approach is as follows: if we let T be a regular octahedron with side length 6, then the volume of S is:
DPatrick20:16:34
DPatrick20:16:53
Note that in this context, "height" means the distance from one face to the opposite face, NOT from a vertex to an opposite vertex.
DPatrick20:17:26
So let's work with T first. T is an octahedron with side length 6. What are the height and volume of T?
CatalystOfNostalgia20:17:57
split into pyramids
krsattack20:17:57
Volume of T = 1/3 (6)^2 (3 sqrt 2)
DPatrick20:18:05
T is the union of two pyramids, each with base 36 and height 3sqrt(2) (and here "height" is in the more usual sense for a pyramid.
DPatrick20:18:12
krsattack20:18:22
woops... times 2
DPatrick20:18:26
...yes. :)
DPatrick20:18:39
And what's the distance between opposite faces of T?
DPatrick20:19:05
An easy way to find the height of T is to use coordinates.
DPatrick20:19:14
For example, we can assign 4 vertices of T to the coordinates (3,3,0), (3,-3,0), (-3,3,0), and (-3,-3,0).
DPatrick20:19:29
Then the other two coordinates will be (0,0,z) and (0,0,-z) for some z.
DPatrick20:19:52
DPatrick20:20:03
Thus 18 + z^2 = 36, and z = sqrt(18) = 3*sqrt(2).
DPatrick20:20:20
So what is the distance between opposite faces?
DPatrick20:21:18
The distance from one face to the other is the distance from (0,2,sqrt(2)) to (0,-2,-sqrt(2)), which are the centers of opposite faces.
DPatrick20:21:32
stanman20:21:55
So now we have to find S
DPatrick20:22:00
Finally, to finish the problem, what is the height of S, or equivalently, of the icosahedron?
DPatrick20:22:14
(Again, here "height" is face-to-face distance, not vertex-to-vertex distance.)
stanman20:22:29
It is a line in a pentagon
VDLmath20:22:29
12sin54
DPatrick20:22:54
These are two ways to do it, but I won't discuss them here (you can check the message board if you want to discuss them further).
DPatrick20:23:09
I actually find coordinates easiest.
DPatrick20:23:24
DPatrick20:23:48
To verify that these coordinates work, we would need to check that each point is distance 6 from exactly 5 other points.
DPatrick20:24:04
For example, consider (3,3a,0). (I'll write "a" for "alpha" in text.)
It's clearly distance 6 from (-3,3a,0).
DPatrick20:24:22
It's also distance 6 from (0,3,3a), as we can easily compute:
DPatrick20:24:27
DPatrick20:24:46
DPatrick20:25:01
Therefore, we conclude that (3,3a,0) is distance 6 from all of:
(-3,3a,0), (0,3,3a), (0,3,-3a), (3a,0,3), (3a,0,-3).
And, by symmetry, every point is exactly distance 6 from 5 other points.
DPatrick20:25:10
(As a fun counting problem, see if you can find the 20 triangles in this icosahedron using the coordinates!)
timwu20:25:16
Do you prove it or use it as a fact?
mathfreak038820:25:16
did we have to prove this in our solution to get full credit
DPatrick20:25:36
I think you could cite the coordinates of an icosahedron.
DPatrick20:26:01
So now we can find two opposite sides. One side, for example, is determined by points (3,3a,0), (-3,3a,0), and (0,3,3a). (The opposite side is these points with all the coordinates multiplied by -1.)
DPatrick20:26:15
So the center of this side is the average of these three points, which is (0,1+2a,a).
DPatrick20:26:30
DPatrick20:26:35
This simplifies:
DPatrick20:26:40
DPatrick20:27:03
(I'm going kind of fast because the computations are not all that illuminating, plus we're just about out of time.)
krsattack20:27:12
Now, just plug in
DPatrick20:27:18
So we have all the data we need to answer the problem:
DPatrick20:27:22
stanman20:27:43
That is the answer!
DPatrick20:27:53
That's it for Round 2.
DPatrick20:28:10
Round 3 problems will be posted hopefully this week, or next Monday at the latest, on the USAMTS web site. The deadline for submitting your Round 3 solutions is January 8.
DPatrick20:28:32
We are currently processing the Round 2 submissions. We hope to have this completed this week (hopefully tomorrow, but no guarantees).
katrina2k259320:28:50
Woohoo for USAMTS! Q.E.D
stanman20:28:50
Thank you! I enjoyed the Math Jam!
SigmaTheta20:28:52
stanman20:28:56
The results are coming in tomorrow?
DPatrick20:29:07
No...I mean just processing them as received.
DPatrick20:29:14
They won't be graded for another couple of weeks.
Altheman20:29:25
Thanks for the math jam.
Bernewtniz20:29:25
Thanks!
fred36920:29:25
this was awesome
timwu_220:29:25
Thanks very much
VDLmath20:29:25
awesome Math Jam
katrina2k259320:29:25
On behalf of all us mathematicians, thank you DPatrick!
kostya20:29:31
for problem 1: if I just showed the case where n=8 does not work, and then went on to show that n=9 works, would that be acceptable?
nameless436520:29:31
It is not neccessary to prove that volume of S is equal to height of S divided by the height of T times the volume of T?
ccy_220:29:31
If we were braindead and we read "6" as "1" in 5, how much would we get marked down?
DPatrick20:29:51
We have not decided any of the grading criteria for any of the problems, so I cannot answer any of these questions, sorry.
stanman20:30:02
Thank you DPatrick!
fred36920:30:02
ty dpatrick
SigmaTheta20:30:02
thanks you so much
stanman20:30:04
Are you a grader?
DPatrick20:30:07
Nope.
DPatrick20:30:25
But I do help to decide the grading criteria, and I'm usually the person who deals with grading disputes.
electricj_220:30:28
can a calculus based answer ever be considered commended?
DPatrick20:30:36
Certainly, yes.
Altheman20:30:52
Have you guys considered some other sort of submission method, like something analogous to an attachment file on a post on the AoPS forum?
DPatrick20:31:06
We will be reconsidering our submission process next year.
stanman20:31:13
What is the definition of commended?
DPatrick20:31:31
We don't really have one...it's somewhat subjective based on the people grading the problem.
timwu_220:31:51
When will the Round 3 be posted
DPatrick20:32:00
Hopefully later this week; Monday Dec 3 at the latest.
DPatrick20:32:10
(We're still tweaking the problems!)
mathnerd31415920:32:21
If there are "too many" commended solutions, will some be dropped?
DPatrick20:32:24
No.
12345678920:32:51
By "sometime this week", is it possible to be posted tomorrow?
DPatrick20:33:12
Highly unlikely. Wednesday is the earliest likely date.
frank4420:33:29
what were the answer to the first 3 problems? sorry I missed that part of the mathjam =(
DPatrick20:33:41
9, 3, 120
DPatrick20:34:07
That's it for tonight -- thanks for coming and see you for Round 3!
