| Transcript
for the Math
Jam "USAMTS Round 1 Math Jam"
on Oct 12. |
| Math Jam hosted by DPatrick
(Dave Patrick ). |
DPatrick (19:30:31)
Hello and welcome to the first 2006-07 USA Math Talent Search Math Jam.
DPatrick (19:30:37)
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.
DPatrick (19:30:47)
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 class organized and on track. 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. 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.
DPatrick (19:31:15)
There will be images in this lecture. The images should appear directly in the classroom window, as in the example below:
DPatrick (19:31:21)
DPatrick (19:31:23)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/a7dcd8010feb2cc00e4a33dc6e5ef2cd.png
DPatrick (19:31:37)
If you click on the link, the image will also 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.
DPatrick (19:31:47)
You can view the first round problems as we discuss them by clicking on the following link:
DPatrick (19:31:51)
http://www.usamts.org/Tests/USAMTSProblems_18_1.pdf
DPatrick (19:31:59)
Let's get started!
DPatrick (19:32:05)
Problem 1
DPatrick (19:32:09)
DPatrick (19:32:11)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-237597135.gif
DPatrick (19:32:23)
How can we begin on this problem?
Elemennop (19:32:33)
algebracize it.
frodo (19:32:33)
x=number being slided
DPatrick (19:32:52)
OK, let's denote the original number by "x" and the new number (after the digit slide) by "y".
DPatrick (19:33:02)
What do we know about x and y?
13375P34K43V312 (19:33:07)
4x=y
Ro890Z (19:33:12)
x ends in a 4, and y starts with a 4
matt276eagles (19:33:13)
x has units digit 4
r_3weaver (19:33:14)
y=4x
DPatrick (19:33:33)
We know y = 4x.
DPatrick (19:33:46)
We also know that he units digit of x is 4. Therefore the leading (leftmost) digit of y is 4, since y is the digit slide of x.
DPatrick (19:33:58)
What else do we know?
bpms_2 (19:34:06)
Well we know that the units digit of the number is 1
perfectnumber628 (19:34:08)
the first digit of x must be 1
DPatrick (19:34:21)
Why do we know that?
13375P34K43V312 (19:34:36)
and they have he same number of digits
.cpp (19:34:37)
Because the number of digits can't change and 4x1 = 4
bpms_2 (19:34:45)
When we divide the number after the slide, we get that the first digit is 1
DPatrick (19:35:16)
Since y starts with a 4, and y=4x, and they have the same number of digits, the leading digit of x must be 1.
DPatrick (19:35:28)
This means that y = 41...
DPatrick (19:35:59)
At this point we know x = 1....4 and y = 41....
DPatrick (19:36:01)
Now what?
vishalarul (19:36:03)
Keep repeating this process.
frodo (19:36:17)
next digit of x is 0.
13375P34K43V312 (19:36:17)
we see that the 2nd digit of x must be a 0
DPatrick (19:36:41)
Right: if we divide 4 into y=41...., we get x=10....
DPatrick (19:37:05)
Another way to see it: if it were >= 1, then y = 4x >= 44...
DPatrick (19:37:13)
So now we know x = 10...4 and y = 410...
13375P34K43V312 (19:36:50)
and we do it again!
r_3weaver (19:37:21)
try doing this again
AdorableAddict (19:37:22)
next digit is 2
DPatrick (19:37:34)
Right, we can keep repeating the process.
DPatrick (19:37:41)
Now x = 102...4 and y = 4102...
pianist (19:37:46)
and the one after 2 is 5
ddkdolfin (19:37:52)
and the fourth digit of x is 5
DPatrick (19:38:02)
The 4th digit of x must be 5.
x = 1025...4 and y = 41025...
jetpilot55 (19:37:58)
5, then 6 come next
jamesdadeliverer (19:38:09)
now 6
DPatrick (19:38:18)
The 5th digit of x must be 6.
x = 10256....4 and y = 410256....
jamesdadeliverer (19:38:26)
and thats it
.cpp (19:38:27)
We're done here.
bpms_2 (19:38:28)
Next is 4
jetpilot55 (19:38:28)
lastly, 4
DPatrick (19:38:44)
The 6th digit of x must be 4.
So now we're done!
DPatrick (19:38:49)
x = 102564 and y = 410256. Check that 4(102564) = 410256.
DPatrick (19:38:56)
So the answer is
102564.
thinker (19:38:55)
but how do we know that this is the smallest possible result?
DPatrick (19:39:20)
Because at every step of the problem, the next digit in x was forced.
DPatrick (19:39:38)
We saw that the first digit of x had to be 1, then the next digit had to be 0, and so on.
oppenhejo (19:39:46)
I did it in reverse
DPatrick (19:40:15)
Sure, you could also do it by keeping track of the digits from right-to-left, as opposed to left-to-right like we just did.
DPatrick (19:40:31)
Let's move on to Problem 2.
DPatrick (19:40:41)
DPatrick (19:40:46)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-39204431.gif
DPatrick (19:40:52)
DPatrick (19:40:53)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/a7dcd8010feb2cc00e4a33dc6e5ef2cd.png
DPatrick (19:41:03)
Let me start off by labeling the circles so we can talk about them:
DPatrick (19:41:11)
DPatrick (19:41:14)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/6a1a8f09840d184a6310257dc93f4600.png
r_3weaver (19:41:02)
the amswer is 0
pianist (19:41:08)
number a) is impossible
ddkdolfin (19:41:09)
it's not possible is it?
DPatrick (19:41:30)
Yes: in fact there are no solutions.
DPatrick (19:41:33)
Why?
oppenhejo (19:41:27)
A is bigger than 1 so B is less than A, etcetera
SplashD (19:41:43)
1<A>B<C>D<E>F<1
jetpilot55 (19:41:42)
label the circles as being greater or less than the ones next to it, and you find that the last 2 circles must be simultaneously greater than and less than each other
DPatrick (19:41:52)
Right.
DPatrick (19:42:03)
We know that A is greater than both its neighbors, since clearly A must be greater than 1.
DPatrick (19:42:11)
Since A is greater than both its neighbors, we know that B is less than A. So, B is less than both its neighbors.
DPatrick (19:42:30)
As we move around the circle, they have to alternate "greater than neighbors" and "less than neighbors"
DPatrick (19:42:35)
Since B is less than its neighbors, C is greater than B, so C is greater than its neighbors. Therefore, D is less than C, so D is less than its neighbors. So, E is greater than D, which means E is greater than its neighbors. This means F is less than E and is therefore less than its neighbors.
DPatrick (19:42:49)
But one of F's neighbors is 1, and F can't possibly be less than 1. Therefore, it is impossible to arrange the numbers in this circle as described in the problem. There are
0 solutions.
DPatrick (19:43:02)
(There are, of course, lots of different ways that you could have worded this.)
DPatrick (19:43:15)
Now on to part (b):
DPatrick (19:43:18)
DPatrick (19:43:21)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-85847423.gif
DPatrick (19:43:26)
We again start by labeling the circles so we can talk about them.
DPatrick (19:43:31)
DPatrick (19:43:34)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/a005cd611b67a877e2e8979cf6b39853.png
AdorableAddict (19:43:31)
divide into cases
DPatrick (19:43:58)
It looks like we have to break this problem into cases.
DPatrick (19:44:12)
What can we observe first, before we get into the messy casework?
ddkdolfin (19:44:09)
i split the numbers into two groups of smaller and larger
Blaise_2 (19:44:17)
Label all of the numbers as "greater than neighbors" or "lesser than neighbors". Some can be both.
AdorableAddict (19:44:20)
set the values for BDF first then find the number of ways for each one
DPatrick (19:44:39)
That's a good approach (though others work too).
DPatrick (19:44:47)
We know that the 1 and cells B, D, and F must be less than their neighbors, and that cells A, C, E, G must be greater than their neighbors. Call 1,B,D,F "small" cells and A,C,E,G "large" cells.
DPatrick (19:44:55)
For ease in identifying the diagrams, let's make the small cells white and the large cells grey.
DPatrick (19:45:00)
DPatrick (19:45:03)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/376cb42da96c064f6b03a18d04b7abf0.png
Krazul (19:44:20)
2 must be in B, D, or F, while 7 and 8 must be in A, C, E, or G
captainclueless (19:44:23)
2 must be f, d, or b
DPatrick (19:45:19)
Right. Note that 2 must be in a small cell, and that 7 or 8 must be in large cells. The other numbers (3,4,5,6) might be either small or large.
DPatrick (19:45:43)
So one way to break up our count into cases is by which other two numbers are in the small cells. This will give us 6 possible cases.
AdorableAddict (19:45:32)
so we can do 234 for BDF first
DPatrick (19:46:18)
Case I: 3,4 are small cells (I'm going to omit writing "2" since it is small in all cases)
DPatrick (19:46:23)
Here's one example:
DPatrick (19:46:28)
DPatrick (19:46:31)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/3c6c886b1eb2c87803010465af41e7f7.png
DPatrick (19:46:34)
How many of these are there?
Krazul (19:46:39)
There are 3! times 4! to place it
oppenhejo (19:46:42)
144
krsattack (19:46:43)
144
DPatrick (19:46:58)
We can assign the small cells to be 2, 3, and 4 in any way we want, and the large cells to be 5 through 8 in any way we want. The first can be done in 3! = 6 ways and the second can be done in 4! = 24 ways, so there are (6)(24) = 144 such orderings.
DPatrick (19:47:17)
Case II: 3,5 are small cells
DPatrick (19:47:19)
Example:
DPatrick (19:47:28)
DPatrick (19:47:30)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/f44a54004e98265ce91acbef34a4d8df.png
DPatrick (19:47:33)
Are there any restrictions to how we assign the small cells to be 2, 3, and 5?
Krazul (19:47:44)
no
DPatrick (19:48:10)
We can place 2,3,5 in any order in the small cells. We can do them in 3!=6 ways.
DPatrick (19:48:21)
Do we have any restrictions on where we place the remaining 4, 6, 7, and 8?
captainclueless (19:47:47)
4 can't be next to 5
JerryS (19:47:48)
4 cannot be next to five
frodo (19:47:50)
4 can't be next to 5.
DPatrick (19:48:32)
We can't put the 4 next to the 5.
DPatrick (19:48:45)
So how many ways can we place 4,6,7,8 once we've placed 2,3,5?
frodo (19:48:46)
2*3! ways in gray
matt276eagles (19:48:53)
2(3!)
DPatrick (19:49:18)
Right. We have 2 choices for the 4, then once it's placed, there are 3!=6 choices left for 6,7,8.
DPatrick (19:49:23)
So, we have (3!)(2)(3!) = 72 total arrangements for this case.
DPatrick (19:49:34)
(3! for 2,3,5; 2 for 4; and 3! for 6,7,8)
DPatrick (19:49:47)
Next: Case III: 3,6 are small cells
DPatrick (19:49:58)
Now, we try letting the small cells be 2, 3, and 6. Are there any restrictions to how we do so? (Just the small cells first)
AdorableAddict (19:49:58)
still 3! = 6 ways for the placement of 2, 3, and 6.
DPatrick (19:50:14)
No restrictions, so there are 3! = 6 ways to assign these three numbers. Here's an example:
DPatrick (19:50:19)
DPatrick (19:50:22)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/bc6b9c54ff9c374b9004e6c32454099d.png
DPatrick (19:50:30)
We have 4,5,7,8 left to place. How do we count these?
frodo (19:50:23)
8 and 7 must be next to 6.
oppenhejo (19:50:32)
Yes 6 must be between 7 and 8
jetpilot55 (19:50:36)
4 and 5 cant be next to 6
krsattack (19:50:36)
4 and 5 can't be next to 6
DPatrick (19:50:59)
Right. 4 and 5 can't be next to the 6, so 7 and 8 must be.
DPatrick (19:51:05)
The two spaces next to the 6 must be 7 and 8. The other two spaces must be 4 and 5. We have two choices for each.
oppenhejo (19:50:59)
so 4
.cpp (19:51:04)
4 possibilities exist.
DPatrick (19:51:21)
Right, and there are a total of (3!)(4) = 24 arrangements in this case.
DPatrick (19:51:31)
So far, that's 144 in Case I, 72 in Case II, and 24 and Case III, for a total of 144+72+24 = 240 arrangements so far.
DPatrick (19:51:39)
Case IV: 4,5 are small.
DPatrick (19:51:50)
Are there any restrictions here for assigning 2,4,5 to small cells?
frodo (19:51:51)
2 is next to 1.
nallenusn (19:51:55)
1 and 2 must be adjacent
perfectnumber628 (19:51:56)
so 3 has to go between 1 and 2
DPatrick (19:52:30)
Right. The 3 has to be large, and the other way this can happen is if the 3 sits between the 1 and the 2. So 2 must be B or F (next to the 1); it can't be D.
DPatrick (19:52:35)
Here's an example:
DPatrick (19:52:41)
DPatrick (19:52:43)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/2246038a4f25a5da79d46d13c757be6a.png
DPatrick (19:52:54)
So, how many ways are there to assign 2, 4, and 5 to B, D, and F?
Elemennop_2 (19:52:59)
4
.cpp (19:53:01)
4.
nallenusn (19:53:05)
2*2
frodo (19:53:05)
2*2
DPatrick (19:53:17)
There are 2 ways to pick the slot for 2, then 2 ways to order 4 and 5, for a total of (2)(2) = 4 ways.
DPatrick (19:53:22)
3 is now forced.
DPatrick (19:53:28)
How many ways to place 6,7,8?
numberdance (19:53:30)
6,7,8 have no restrictions. There are 3! ways to place them.
frodo (19:53:31)
3! ways
DPatrick (19:53:41)
Right.
DPatrick (19:53:51)
So, we have a total of (4)(3!) = 24 orderings in this case.
DPatrick (19:53:59)
Our running total is now 240+24 = 264.
DPatrick (19:54:04)
Case V: 4,6 are small.
DPatrick (19:54:11)
How can we place 2,4,6 small?
jamesdadeliverer (19:54:17)
2 adjacent to 1
AdorableAddict (19:54:19)
only 4 ways
nallenusn (19:54:22)
1 and 2 must be adjacent
DPatrick (19:54:47)
It's the same as the previous case. The 2 must be in B or F, so the 3 can go between 1 and 2.
DPatrick (19:54:52)
Here's one example:
DPatrick (19:54:59)
DPatrick (19:55:02)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/d7876782b53622bcc99a3c7fae622291.png
DPatrick (19:55:10)
So, there are 2 ways to choose the slot for the 2, then there are 2 ways to choose where to put the 4 and 6. How about the number of ways to fill the remaining slots with 5, 7, and 8?
solafidefarms (19:55:12)
2 arrangements of the others
nallenusn (19:55:21)
5 cant be next to 6
bpms_2 (19:55:22)
2 ways
AdorableAddict (19:55:22)
5 can't be next to 6
DPatrick (19:55:45)
Right. Tthe 5 must go in the other slot that is not next to the 6, so the 5 is forced too.
DPatrick (19:55:58)
Our only choice is where 7 and 8 go.
DPatrick (19:56:10)
The 7 and 8 can be placed in either order next to the 6, so there are only 2 ways to complete the circle once the 2, 4, and 6 are placed in B, D, and F. Since there are (2)(2) = 4 ways to place 2, 4, and 6, and 2 ways to complete the circle for each of these arrangements, we have a total of (4)(2) = 8 arrangements in this case.
DPatrick (19:56:22)
Now our running total is 264 + 8 = 272.
DPatrick (19:56:30)
Last, Case VI: 5,6 are small.
DPatrick (19:56:37)
How can we place 2,5,6 all small?
JerryS (19:56:36)
There are no ways to arrange them
kostya (19:56:42)
none
DPatrick (19:56:57)
Why not?
matt276eagles (19:56:52)
no solutions, 3 and 4 must both be next to 1 and 2
nallenusn (19:56:56)
4 cant go anywhere
DPatrick (19:57:14)
Right. 3 and 4 would both have to go between 1 and 2, which is impossible.
DPatrick (19:57:22)
So adding all our cases, we have a total of
272 arrangements.
DPatrick (19:57:32)
Problem 2 was an example of doing
very careful casework. Many counting problems come down to this type of casework argument.
r_3weaver (19:57:43)
how do we know that there are no more possibilities?
DPatrick (19:58:01)
Because we
carefully covered all the possible cases.
DPatrick (19:58:19)
Let's more on to Problem 3.
DPatrick (19:58:21)
...move...
DPatrick (19:58:29)
DPatrick (19:58:36)
DPatrick (19:58:38)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/1248e7a0a0784462fb0790684e32eaaa.png
DPatrick (19:58:46)
Obviously there are some clever ways to do this -- which we'll look at in part (b) -- but supposing that we just wanted to grind this solution out, how could we do it?
krsattack (19:58:51)
realize that if we have an arithmetic sequence for every 3 consecutive vertices, the whole line segment forms an arithmetic sequence
AdorableAddict (19:58:58)
figure out numbers on teh edges
DPatrick (19:59:15)
Yes -- we note that "any three consecutive vertices" forming an arithmetic sequence implies that any line is also an arithmetic sequence.
DPatrick (19:59:23)
In particular, the three sides of the triangle are arithmetic sequences.
DPatrick (19:59:30)
So we can fill in the labels for all the vertices on the sides.
DPatrick (20:00:17)
For instance, the left side is a 6-term sequence: 1,_,_,_,_,4. It takes 5 steps to go from 1 to 4, so the common difference is 3/5.
jamesdadeliverer (19:59:59)
1.6, 2.2, 2.8, 3.4
2.6, 4.2, 5.8, 7.4
DPatrick (20:00:32)
Right, the same thing works on the right side.
DPatrick (20:00:39)
DPatrick (20:00:42)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/c1b92a07b356e8b56d7ce5c5cbaa6138.png
DPatrick (20:00:49)
We could continue and find all the vertices, but is there now a clever way to compute the sum of each row?
Aryth_2 (19:58:56)
sum of arithmetic sequence = (average of first and last term)(number of terms)
jamesdadeliverer (20:01:01)
averages of each row
DPatrick (20:01:14)
Each row's sum is equal to the number of vertices in that row times the average of the values of the vertices at the end of the row.
DPatrick (20:01:24)
So we can just compute from here:
DPatrick (20:01:29)
1 + 2(1.6+2.6)(1/2) + 3(2.2+4.2)(1/2) + 4(2.8+5.8)(1/2) + 5(3.4+7.4)(1/2) + 6(4+9)(1/2).
DPatrick (20:01:46)
This simplifies to
1 + 4.2 + 9.6 + 17.2 + 27 + 39 = 98.
DPatrick (20:01:59)
So the answer is (a) is
98.
DPatrick (20:02:09)
OK, that's the example, now on to part (b):
DPatrick (20:02:13)
AdorableAddict (20:02:21)
use the same approach in part (a), only in symbols
DPatrick (20:02:35)
We certainly could do that.
DPatrick (20:02:45)
Without loss of generality, let's set "a" to be the top vertex, and "b" and "c" to be the two bottom ones.
DPatrick (20:02:50)
DPatrick (20:02:52)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/2150c15fb19800e68e7a5445475d116d.png
DPatrick (20:03:00)
(I've drawn our picture with n = 5, but of course n could be anything.)
DPatrick (20:03:19)
We'll try the same approach as in part (a).
13375P34K43V312 (20:03:14)
fill the stuff in using arithmetic sequences
DPatrick (20:03:31)
The left side of the triangle is an arithmetic sequence. It takes n steps to get from a to b. Therefore, each step increases by (b-a)/n.
krsattack (20:03:30)
down the left, we have a, a+((b-a)/n), a+2((b-a)/n,...b
DPatrick (20:03:41)
Exactly.
DPatrick (20:03:53)
DPatrick (20:04:01)
DPatrick (20:04:13)
Also note that there are k+1 vertices in Row k (there is 1 in Row 0, and n+1 in Row n).
DPatrick (20:04:35)
Now we can sum the rows just like we did in part (a).
DPatrick (20:04:39)
DPatrick (20:04:54)
Note this is the same as in part (a): each row's sum is the number of terms in the row, times the average of the two endpoints.
DPatrick (20:05:01)
Yikes.
pianist (20:05:05)
is there any easier approach
DPatrick (20:05:13)
Let's hope so.
frodo (20:05:18)
Can it be further simplified?
DPatrick (20:05:32)
We can simplify it a bit:
DPatrick (20:05:37)
numberdance (20:05:48)
Not a whole lot better...
DPatrick (20:06:14)
There's one thing we did at the beginning, where I said "without loss of generality"...
DPatrick (20:06:30)
We arbitrarily chose a to be the top and b and c to be the two bottom vertices. We could have chosen any of the vertices to be the top.
DPatrick (20:06:40)
DPatrick (20:06:46)
Elemennop_2 (20:06:46)
Add up three cyclic sums of the sort, and divide by 3
DPatrick (20:07:09)
Right! When we add them up, the nasty stuff all cancels.
DPatrick (20:07:25)
bpms_2 (20:07:16)
What is a cyclic sum?
DPatrick (20:07:54)
"cyclic" basically means that we're "rotating" the vertices (letting b be where a was, c be where b was, and a be where c was)
DPatrick (20:08:05)
It's tantamount to rotating the triangle.
DPatrick (20:08:12)
chaostheory17 (20:08:26)
lots simpler
oppenhejo (20:08:28)
Still not in terms of n^2 however
vishalarul (20:08:30)
You can pull out the a+b+c.
AdorableAddict (20:08:34)
a+b+c can be factored out since its constant
DPatrick (20:08:52)
Right, we can pull out (a+b+c), and we just have the sum of the first n+1 positive integers.
frodo (20:08:50)
Now divide by 3.
DPatrick (20:09:05)
Right, don't forget to divide by 3!
DPatrick (20:09:12)
DPatrick (20:09:34)
...where we replace the sum of the first n+1 integers with its closed form (n+1)(n+2)/2.
DPatrick (20:09:45)
Let's check it: in part (a), a = 1, b = 4, c = 9, n = 5.
DPatrick (20:09:50)
Our formula then gives 6(7)(14)/6 = 98, so it agrees.
oppenhejo (20:09:37)
IF you simplift and seperate this, you get the avg of a,b,c times the number of labels
DPatrick (20:10:15)
Right -- it's just the number of vertices times the average value of the vertices, and the average value of the vertices is the average value of the three given numbers a, b, and c.
DPatrick (20:10:31)
Here's another solution, which is very quick, exploits symmetry even further, and shows exactly why this is the case.
nallenusn (20:07:10)
You could add the triangle to its 2 rotations, to get a triangle where every corner value is (a+b+c), then every value in the entire triangle is equal to (a+b+c), and that sum is 3 times the sum of the original triangle
DPatrick (20:11:00)
Yes! Here's a picture of this with the numbers from part (a):
DPatrick (20:11:04)
DPatrick (20:11:06)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/f8f397628bba45014ae39a0ab8913f2e.png
DPatrick (20:11:24)
Imagine taking the triangle, rotating it twice (as in the picture above), and adding all three triangles together.
DPatrick (20:11:37)
Then the labels at the corners of T are all 14. Furthermore, triangle T also has the arithmetic sequence property, since the sum of any two arithmetic sequences is also an arithmetic sequence.
.cpp (20:11:39)
I applaud that solution.
DPatrick (20:11:55)
Yeah, it's pretty cool.
chaostheory17 (20:11:48)
then they're all 14, and its lots easier
DPatrick (20:12:29)
Right, the triangle T on the right above has all interior vertices equal to 14, so its sum is just 14 times the number of vertices. Divide by 3 and you're done.
DPatrick (20:12:44)
It is clear that the argument generalizes, so that the answer to part (b) is indeed the average of a, b, and c times the number of labels, as derived above.
DPatrick (20:13:02)
Let's move on to Problem 4.
DPatrick (20:13:06)
DPatrick (20:13:18)
This problem is a lot more abstract than the others. There's nothing that we can sit down and grab onto and try to compute. So how should be proceed?
conartist (20:13:14)
pigeonhole!
Elemennop (20:13:18)
Sounds like Pigeonhole
AdorableAddict (20:13:20)
Pigeonhole Principle is my first reaction
DPatrick (20:13:55)
It certainly smacks of the Pigeonhole Principle, since we want to find a bunch of points (in this case, corners of a rectangle) that are all the same color.
SamE (20:13:46)
3 colors -> look at 4 points to start
DPatrick (20:14:20)
Sure, let's start with 4 points.
DPatrick (20:14:38)
By Pigeonhole, since we have 4 points and only 3 colors, at least 2 of the points must be the same color. (That's the Pigeonhole Principle)
DPatrick (20:14:52)
But how does this help us find a rectangle?
besttate (20:14:57)
then, expand this to 81 sets of 4 points
DPatrick (20:15:14)
Why 81? Where does 81 come from?
DPatrick (20:15:18)
Or maybe you meant 82? :)
jamesdadeliverer (20:15:18)
3^4
bpms_2 (20:15:19)
3^4
oppenhejo (20:15:19)
3^4
Ro890Z (20:15:24)
3^4 + 1 = 82
DPatrick (20:15:47)
Right idea. Let me back up a step.
DPatrick (20:16:01)
We need to find 4 points, corners of a rectangle, that are all the same color.
DPatrick (20:16:18)
So we need 2 points of one color on one line, and 2 points of that same color on another (parallel) line.
DPatrick (20:16:36)
We can get 2 points of one color on one line by picking any 4 points of the line, and using the Pigeonhole Principle.
DPatrick (20:16:58)
But we want to get the same color to also be two more points on a different parallel line.
bortreb (20:16:54)
so with 82 sets, teo sets will be exactly the same
chaostheory17 (20:17:04)
so 3 colors^4 points needed+1 to swing the balance?
DPatrick (20:17:33)
Right. We draw 82 = 3^4 + 1 parallel horizontal lines, and 4 parallel vertical lines. This gives us a 4x82 grid of points.
DPatrick (20:17:40)
For example, take all the points (i,j) with 1 <= i <= 4 and 1 <= j <= 82 integers.
DPatrick (20:18:08)
Since there are 3^4 possible patterns of colors at four points, we know that there must be two rows that have the same pattern at the 4 points.
DPatrick (20:18:23)
Keep those two rows and throw the rest away.
DPatrick (20:18:32)
Then, again by the Pigeonhole Principle, two of the points in each row must have the same color.
DPatrick (20:18:50)
Then we're done! These four points (two in each row) are the vertices of our rectangle.
Krazul (20:18:19)
Shouldn't we only need 19 parallel lines, not 82?
rgrgrg (20:18:50)
Do we need 82 lines, don't we only need 19 ?
DPatrick (20:19:18)
There are more clever ways to set up the Pigeonhole argument so that you need fewer lines. But I find 3^4+1 to be the simplest.
matt276eagles (20:19:11)
the same idea would work with any number of colors
DPatrick (20:19:51)
Indeed, you can generalize this argument both to an arbitrary number of colors, but also an arbitrary number of dimensions.
besttate (20:19:39)
82 lines would recieve full credit, right?
DPatrick (20:20:06)
Certainly! That's my solution! :)
DPatrick (20:20:38)
Finally, on to Problem 5:
DPatrick (20:20:42)
DPatrick (20:20:53)
We'll start with a drawing.
DPatrick (20:20:59)
DPatrick (20:21:01)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/9aab4d681cdc2ea92ecc778eb049c9e0.png
DPatrick (20:21:16)
Plane P intersects ABCD in triangle FXW, and plane Q intersects ABCD in triangle EZY. The intersection of these planes is line ST, and the problem asks us to find the length of ST.
jamesdadeliverer (20:21:23)
hard to interpret the drawing
DPatrick (20:21:33)
Best I could do.
DPatrick (20:21:40)
3D drawings are always tough.
oppenhejo (20:21:14)
We know ratio of volume to length is cube root
DPatrick (20:22:28)
Right, that's going to be a key fact. If the ratio of AEZY to ACDB in volume is 1:2, then the ratio in length is the cube root of that, 1:cubert(2).
DPatrick (20:22:55)
DPatrick (20:23:03)
So AY = r(AB), and so on.
DPatrick (20:23:28)
We're trying to find ST.
DPatrick (20:23:38)
What can we do?
oppenhejo (20:23:37)
Do we need to finsd the coordinates of all points?
DPatrick (20:23:46)
Let's hope not!
EunuchOmerta (20:23:57)
find the lengths of the triangle WST
AdorableAddict (20:24:04)
we can find the lines parallel to ST then use ratios
DPatrick (20:24:18)
Aha, you said the magic word!
DPatrick (20:24:22)
parallel
thinker_2 (20:24:19)
ratio?
DPatrick (20:24:34)
That's the next magic word!
DPatrick (20:24:38)
parallel -> similar triangles -> ratio
DPatrick (20:25:07)
We don't need coordinates, or trig, or any of that stuff: all we need are similar triangles.
SamE (20:24:51)
Or similar tetrahedrons...
DPatrick (20:25:15)
Right.
DPatrick (20:25:37)
Let me pop the picture back up.
DPatrick (20:25:47)
DPatrick (20:26:10)
DPatrick (20:26:28)
Somebody mentioned triangle SWT before.
DPatrick (20:26:34)
Note SWT ~ FWX.
DPatrick (20:26:44)
And we know FX = r(BC).
DPatrick (20:26:54)
So if we can find the ration between SWT and FWX, then we're done.
DPatrick (20:27:00)
(ratio, not ration!)
DPatrick (20:27:18)
How do we do that?
AdorableAddict (20:27:24)
use volumes again
vishalarul (20:27:29)
Similar triangles.
AdorableAddict (20:27:38)
use the tetrahedrons WSZT and WFDX
DPatrick (20:28:00)
Right. In particular, we can use SZW ~ FDW.
DPatrick (20:28:06)
We know DW = r(AD).
DPatrick (20:28:12)
What's ZW in terms of AD?
solafidefarms (20:28:35)
(2r-1)AD
DPatrick (20:28:43)
Exactly.
DPatrick (20:28:51)
Note DW = AZ = r(DA), so ZW = (DW+ZA-DA)=(2r-1)(DA).
DPatrick (20:29:13)
So the ratio ZW:DW = (2r-1):r.
DPatrick (20:29:24)
How does that help with ST?
AdorableAddict (20:29:38)
we can find SW : FW
AdorableAddict (20:29:44)
which is the same as ST : FX
DPatrick (20:30:11)
Right. SZW~FDW, so SW:FW = (2r-1):r.
DPatrick (20:30:21)
But then SZT ~ FDX, so ST:FX = (2r-1):r as well.
DPatrick (20:30:43)
So ST = (2r-1)/r (FX).
DPatrick (20:30:49)
But we know FX, right?
AdorableAddict (20:30:57)
FX = AB r
oppenhejo (20:31:03)
r(bc)
bpms_2 (20:31:05)
r(BC)
krsattack (20:31:05)
r(BC)
DPatrick (20:31:18)
Right, FX = r(BC).
DPatrick (20:31:28)
So ST = (2r-1)/r * r(BC) = (2r-1)*BC.
DPatrick (20:31:40)
And BC = 8.
DPatrick (20:31:50)
So the answer is 8*(2r-1) = 8(cubert(4) - 1).
mysmartmouth (20:31:57)
Nice.
DPatrick (20:32:11)
In particular, note that all the other lengths were irrelevant!
DPatrick (20:32:19)
Only BC=8 is important.
DPatrick (20:32:37)
And all it boiled down to was chasing similar triangles.
DPatrick (20:32:43)
That's it for the Math Jam.
DPatrick (20:32:52)
If you have questions about these solutions, you can now post them on the USAMTS message boad.
DPatrick (20:32:56)
...board.
DPatrick (20:33:10)
Round 2 should be available later this evening.
Aryth_2 (20:33:07)
do you know when we'll get our scores (approximately)?
DPatrick (20:33:21)
Probably early November.
mna851 (20:33:15)
Should I be worried if I have not received the little green check of confirmation yet?
DPatrick (20:33:30)
No.
DPatrick (20:33:34)
We're still processing the entries.
anhang (20:33:33)
i did some problems a different way, should i post them on the forum or ask you here?
DPatrick (20:33:49)
You can post them in the forum.
thinker_2 (20:33:43)
do you grade USAMTS entries?
DPatrick (20:34:02)
Yes, but many other people do as well.
ddkdolfin (20:33:47)
can you still get points for a wrong answer if you worked it out?
DPatrick (20:34:16)
Yes. We give a lot of partial credit.
solafidefarms (20:34:25)
If I think I sent in my entry form soon after they were released, but the site doesn't say it's been recorded, should I resend it?
DPatrick (20:34:48)
No, wait until Round 2. We may have it.
r_3weaver (20:34:26)
what do you look for in a solution?
DPatrick (20:35:19)
http://www.usamts.org/TipsFAQ/U_FAQ.php#complete
ddkdolfin (20:34:50)
can you still sign up if you didn't have a chance to participate in the first round?
DPatrick (20:35:41)
Yes, you can join at any point.
Varoon (20:35:22)
So round 2 problems will be out later tonight?
DPatrick (20:35:46)
Correct.
mna851 (20:35:27)
WHen will the transcript be up?
DPatrick (20:36:00)
As soon as I'm done answering questions and have a chance to post it.
bpms_2 (20:35:42)
If by some chance, when we sent in our solutions, they didn't make it, then is there no chance we can resend them?
DPatrick (20:36:10)
We treat this on a case-by-case basis.
.cpp (20:36:01)
Do the problems get increasingly hard with each round?
DPatrick (20:36:28)
Generally, a bit.
AdorableAddict (20:35:04)
When do we get a chance to see the official solutions and commended solutions?
DPatrick (20:36:45)
After all the grading is done, in about a month from now.
darthvarma (20:36:35)
where do you post the transcript
DPatrick (20:37:10)
Click on "Math Jams" from the Community menu on our website, then select "Transcripts"
kimby_102 (20:37:05)
I sent my solutions via email 12:30 a.m. on the due date... do you still accept my solutions
DPatrick (20:37:25)
Case-by-case basis.
DPatrick (20:37:56)
We won't reject any out of hand -- we'll always contact you if we're considering rejection a late submission, to give you a chance to present your side of the story.
bpms_2 (20:37:14)
If we email our solutions, do you guys print them anyways?
DPatrick (20:38:08)
Yes.
kimby_102 (20:37:57)
sorry, what do you mean by case-by-case basis?
DPatrick (20:38:36)
Meaning we don't automatically accept them, but we don't automatically reject them either.
SamE (20:38:26)
So are deadlines generally at midnight Pacific Time (CA) or Eastern Time (east coast)?
DPatrick (20:38:44)
Pacific.
DPatrick (20:38:57)
But
don't wait until the last minute, and we won't need to have this conversation.
DPatrick (20:39:06)
You have over 1 month to work on the problems!
DPatrick (20:39:26)
No, only once per year.
DPatrick (20:39:50)
(that was an answer to a question that I apparently deleted instead of posted, which was "do we have to send the entry form for each round?")
DPatrick (20:40:23)
That's all...have a good evening!
