Transcript for the Math Jam "USAMTS Round 2 Math Jam" on Nov 29.

Math Jam hosted by DPatrick (Dave Patrick ).

DPatrick (19:28:19)
Hello and welcome to the second 2006-07 USA Math Talent Search Math Jam.

DPatrick (19:28:28)
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.

DPatrick (19:28:38)
The classroom is moderated, meaning that students can type into the classroom, but only the moderator (that's me) can choose a comment to drop into the classroom. This helps keep the class organized and on track.

DPatrick (19:28:51)
This also means that only well-written comments will be dropped into the classroom, so please take time writing responses that are complete, easy to read, and on-topic.

DPatrick (19:29:05)
Also, only the moderator can enter into private chats with other users, although due to the size and nature of this Math Jam, it is unlikely that this will happen.

DPatrick (19:29:15)
There will be images in this lecture. The images should appear directly in the classroom window, as in the example below:

DPatrick (19:29:21)


DPatrick (19:29:26)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/a7dcd8010feb2cc00e4a33dc6e5ef2cd.png

DPatrick (19:29:35)
If you click on the link, the image will 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 (19:29:52)
You can view the 2nd round problems as we discuss them by clicking on the following link:

DPatrick (19:29:54)
http://www.usamts.org/Tests/USAMTSProblems_18_2.pdf

DPatrick (19:30:02)
Let's get started!

DPatrick (19:30:08)
Problem 1

DPatrick (19:30:11)


DPatrick (19:30:17)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-116500654.gif

DPatrick (19:30:25)
What can we first determine?

gregx007 (19:30:33)
bounds on n

vishalarul (19:30:39)
Try determining bounds on n.

bewareofdog (19:30:35)
it must have 4 digits

perfectnumber628 (19:30:37)
n has to have 4 digits

Xantos C. Guin (19:30:43)
n must contain 4 digits

cowpi (19:30:45)
n has 4 digits

DPatrick (19:31:00)
Clearly n must have at least 4 digits, otherwise n-2006 (the sum of the squares of the digits of n) is negative.

DPatrick (19:31:09)
How do we know that it can't have more than 4 digits?

vishalarul (19:31:19)
Otherwise n-2006 would be too big.

bewareofdog (19:31:38)
the number will be way more than the sum of squares, which is at most 405.

DPatrick (19:32:02)
We can compare n-2006 to the possible sums of the squares of the digits.

DPatrick (19:32:12)


DPatrick (19:32:24)


DPatrick (19:32:51)
Already if d=5, this gives 405 > 7994, no good. It's pretty clear that 81d grows much more slowly than 10^(d-1), so the left side can never be greater than the right side.

DPatrick (19:33:13)
So n must have 4 digits.

DPatrick (19:33:19)
Now what?

vfiroiu (19:31:43)
express n with digits

mna851 (19:33:32)
Assign variables to each digit

gregx007 (19:33:33)
first digit must be 2, since the sum of squares of digits is at most 4*81

bewareofdog (19:33:37)
the thousands digit must be 2

DPatrick (19:34:00)
Right, there are a couple of things we can do now.

DPatrick (19:34:13)
We can see right away that the first (thousands) digit of n must be 2.

DPatrick (19:34:26)
That's because the sum of the square of the digits of a 4-digit number is between 1 and 4*81=324. Therefore, n-2006 <= 324, hence n <= 2330.

DPatrick (19:34:59)
(and clearly n >= 2006 otherwise the problem doesn't make sense)

gregx007 (19:34:59)
so, write an equation relating the remaining three digits

DPatrick (19:35:18)
Right. We write n=2abc for unknown digits a,b,c. What equation can we write?

PI-Dimension (19:35:52)
100a+10b+c-6=a^2+b^2+c^2+4

gregx007 (19:36:03)
a^2 + b^2 + c^2 + 10 = 100 a + 10 b + c

DPatrick (19:36:19)
Right.

DPatrick (19:36:24)


DPatrick (19:36:31)
How do we solve this for digits a,b,c?

mna851 (19:36:50)
We can use a similar argument to show that $\ a $ must be 0 too

Andrew Kim (19:36:59)
We could simplify in terms of a

perfectnumber628 (19:37:01)
a can't be greater than 3

Xantos C. Guin (19:37:05)
a= 0, 1, 2, or 3, so solve for each case

DPatrick (19:37:53)
Sure. The 100a term on the right side limits a to 0, 1, and 2, since if a>2 the left side can't be that big.

DPatrick (19:38:15)
We can break it up into cases based on whether a is 0, 1, or 2.

DPatrick (19:38:37)
But a little experimentation will show you that a=1 and a=2 don't work (I'll leave those details out).

DPatrick (19:38:42)
So we only need to worry about a=0.

DPatrick (19:38:50)


VDLmath (19:39:12)
Now, can we just plug in numbers 1-9 for b?

Andrew Kim (19:39:16)
Then we could simplify this in terms of b

DPatrick (19:39:29)
Sure -- at this point, it's easy to just plug in numbers.

DPatrick (19:39:38)
One systematic way to do this is to write it as a quadratic equation, say in c.

DPatrick (19:39:43)


DPatrick (19:39:53)


DPatrick (19:40:17)
Now you just need to check which values of b give you a solution in c that's a integer digit.

PI-Dimension (19:40:29)
2,8

VDLmath (19:40:35)
I got 2 and 8

DPatrick (19:40:49)
Right. We check b=0,1,...,9, and find that only b=2 and b=8 make the discriminant a perfect square.

vishalarul (19:40:53)
In both cases, c=3.

DPatrick (19:41:01)
Correct.

vfiroiu (19:40:59)
so 2023 and 2083 are the answers

DPatrick (19:41:14)
So n=2023 and n=2083 are possible solutions. (We should also check these to make sure that they work; indeed they do.)

.cpp (19:41:17)
And we're done.

DPatrick (19:41:27)
:)

thinker (19:41:26)
how do you know there can't be another

DPatrick (19:41:50)
We checked all the cases -- none of the rest work. (I left out a couple of the cases here.)

DPatrick (19:42:01)
Problem 2

DPatrick (19:42:06)


DPatrick (19:42:11)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-264990061.gif

DPatrick (19:42:18)


DPatrick (19:42:19)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/cf9042f99e09d9fc88acb790de0ecafe.png

gregx007 (19:42:24)
extend AB and CD

thinker (19:42:28)
extend AB and DC until they intersect

solafidefarms (19:42:30)
Extend AB and CD to a point E :)

Xantos C. Guin (19:42:34)
Extend AB and DC to a point of intersection.

besttate (19:42:34)
Extend AB and CD to meet at a point E

rnwang2 (19:42:44)
extend AB and DC so that they intersect

DPatrick (19:43:06)
Most people did it this way:

DPatrick (19:43:16)


DPatrick (19:43:19)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/dfa0683c572b6c5f5bde840b732136fb.png

ragnarok23 (19:43:09)
notice 57+33=90

Andrew Kim (19:43:23)
We then get right triangle AXD

rnwang2 (19:43:25)
<AXD is a right angle

DPatrick (19:43:46)
That's the key. You have to notice that 57 and 33 are not just random numbers, but that 57+33 = 90.

mna851 (19:43:35)
Should we prove that X, M, and N are collinear?

DPatrick (19:44:12)
Absolutely! I drew MN going through point X as well, but that's definitely something that needs to be proven.

.cpp (19:44:21)
Midpoints!

Andrew Kim (19:44:21)
Right triangles hypoteneuse form diameter in a circumcircle

Andrew Kim (19:44:37)
N is the center with AN, DN radius

DPatrick (19:45:12)
Yes, there are different ways to do this.

DPatrick (19:45:45)
One is to note that the segment XN creates similar triangles XBM and XAN, so M must be the midpoint of BC (meaning that XN passes through M).

DPatrick (19:46:22)
So once we've drawn the above diagram, how do we finish the problem?

besttate (19:46:43)
Notice that ANX is isoceles, and solve for angle ANM (114)

mna851 (19:46:48)
The median to the hypotenuse of a right triangle is equal to half the hypotenuse.

ragnarok23 (19:46:52)
use medians on right triangles

DPatrick (19:47:39)
Right. We know that the median of a right triangle (from the right angle) is half the length of the hypotenuse. (If you don't know this, draw the circle around AXD; N is the center and XN is a radius.)

DPatrick (19:47:49)
So AN=NX, and ANX is isosceles.

DPatrick (19:48:13)
This lets us compute angle ANM = ANX = 180 - 33 - 33 = 114.

Xantos C. Guin (19:47:07)
XN=10/2=5, XM=6/2=3 MN=5-3=2

besttate (19:47:23)
Use that the median of a right triange to the hypotonuse is half the hypotonuse, so XN = 5, XM = 3, and thus MN = 2.

DPatrick (19:48:43)
Right, we also get the length MN: NX = NA = 5 and MX = MB = 3.

DPatrick (19:48:57)
So their difference is MN = NX - MX = 5-3 = 2.

DPatrick (19:49:08)
And we're done.

DPatrick (19:49:42)
The key, of course, was noticing that 57+33 = 90. You should always be on the lookout for complementary angles (angles that add to 90 degrees)!

DPatrick (19:49:59)
Problem 3

DPatrick (19:50:05)


DPatrick (19:50:08)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-264539342.gif

Xaero (19:50:15)
We could plug in a few numbers and see if we can find a pattern.

DPatrick (19:50:42)
I certainly would start by doing this.

DPatrick (19:50:50)
Let's try n=3.

DPatrick (19:50:58)
We'll make a chart of the floor and ceiling of k for all k between 1 and 9.

DPatrick (19:51:05)


DPatrick (19:51:24)
So the sum is 38, hence f(3)=38. (Let's hang on to this; we can use this to check our polynomial when we get finished.)

Andrew Kim (19:51:21)
Perfect square k's bring even numbers. Non perfect squares bring odd numbers!

PI-Dimension (19:51:39)
notice that between the sqaures its an odd and in them is even

DPatrick (19:51:52)
Right. That's definitely the pattern.

DPatrick (19:52:03)


kostya (19:52:07)
for (n-1)^2<k<n^2, the sum is 2n-1

DPatrick (19:52:28)
Also right.

Cmac89 (19:52:28)
we notice that the frequency of the odd numbers for every odd number 2n+1 is 2n

DPatrick (19:52:49)
Correct.

DPatrick (19:53:03)
We see that the "spaces" between the perfect squares are the positive even integers; in other words, in our example above, we have 2 3's and 4 5's. If we continued up to n=4, we'd get 6 7's and an 8.

spicychicken (19:53:19)
did we have to be rigorous about that step??

DPatrick (19:53:38)
Yes. Your solution would have to prove all these things.

DPatrick (19:54:11)
It's not enough to just observe these patterns, you'd need to prove them too; or, you can use the pattern to find the answer, but then prove that your answer works.

DPatrick (19:54:46)
Anyway, we we add up the even terms in our sum (which occur at perfect squares) and the odd terms (which occur in between), we get:

DPatrick (19:54:50)


DPatrick (19:55:24)
What a good way to simplify this?

VDLmath (19:55:47)
2i(2i+2)

Goistein (19:55:50)
Factor out the 2's.

Xaero (19:55:51)
Use formulas 1+...+n=n(n+1)/2

calc rulz (19:55:55)
Multiply out and use formulas for sums of k and k^2 from 1 to n.

DPatrick (19:56:08)
All good ideas.

DPatrick (19:56:18)
I would pull out the 2n term from the first sum, then combine them:

DPatrick (19:56:23)


Xantos C. Guin (19:56:17)
use 1+2+3+...+n = n(n+1)/2 and 1^2+2^2+3^2+...+n^2 = n(n+1)(2n+1)/6

DPatrick (19:56:53)
Now we can use the formulas for the sum of the first n-1 integers and squares:

DPatrick (19:56:57)


DPatrick (19:57:09)


DPatrick (19:57:36)


PI-Dimension (19:57:47)
and its done

thinker (19:57:48)
and we're done

DPatrick (19:58:03)
I would check it against our example first...

DPatrick (19:58:07)
We check that f(3) = 36+2 = 38, so we're pretty confident that we didn't make a mistake.

solafidefarms (19:58:13)
What if we had this in factored form? 2/3n (2n^2+1)?

DPatrick (19:58:44)
That would be fine.

DPatrick (19:59:29)
There are other, slightly more sophisticated ways to do this problems, such as using what are called "finite differences". (You can look it up if you're interested.)

DPatrick (19:59:36)
Problem 4

DPatrick (19:59:43)


DPatrick (19:59:52)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-179934029.gif

besttate (19:59:48)
Start by trying it with a real number to get a feel for the problem.

DPatrick (20:00:25)
Just like in #3, it probably pays to do some experimenting here to (and this also gives us a value to check out final answer at the end.)

DPatrick (20:00:35)
Let's try k=2 first. We'll try to color as big a set as possible so that (a) and (b) fail.

DPatrick (20:00:44)
Make 1 red (arbitrarily).

besttate (20:00:54)
2 has to be blue.

gregx007 (20:01:06)
then, 2 is blue

Xantos C. Guin (20:01:06)
1+1=2, so 2 must be blue

DPatrick (20:01:20)
Right. 1+1=2, so we don't want 2 red. 2 has to be blue.

besttate (20:01:21)
2+2=4, so 4 must be red.

gregx007 (20:01:23)
then 4 is red

JerryS (20:01:23)
So 4 is red

JAM (20:01:25)
2+2=4 so 4 has to be red

DPatrick (20:01:35)
4 then has to be red, since 2+2=4.

vishalarul (20:01:42)
So 3 is blue!

gregx007 (20:01:45)
and 1 + 3 = 4, so 3 must be blue

JAM (20:01:45)
Now 1+3 = 4 so 3 is blue

DPatrick (20:02:03)
Right. 1+3=4, so we can't have 3 red (since 1 and 4 both are). So 3 is blue.

besttate (20:01:58)
2+3=5 and 1+4=5, so what is 5???

kostya (20:02:08)
and 5 is the sum either way

DPatrick (20:02:33)
Yeah, now we're stuck. 1 and 4 are both red, 2 and 3 are both blue, but they both sum to 5.

VDLmath (20:02:21)
5 is n

Andrew Kim (20:02:30)
So is the answer here 5?

JerryS (20:02:41)
so n=5

DPatrick (20:02:53)
Right. When k=2, we get n=5.

DPatrick (20:03:15)
No matter how we color 1,2,3,4,5, we have two of the same color that sum to a third of the same color.

gregx007 (20:03:05)
a similar method can be employed in the general case

DPatrick (20:03:58)
We could try k=3 if we wanted another example, but let's go ahead and go to the general case.

perfectnumber628 (20:04:01)
for example, 1 is red, k is blue, k^2 is red

Xantos C. Guin (20:04:13)
WLOG, let 1 be red. Since 1+1+1+...+1=k, k must be blue. Since k+k+k+...+k=k^2, k^2 must be red

DPatrick (20:04:26)
Sure. We can start the same way.

DPatrick (20:04:33)
Arbitrarily, let's make 1 red.

DPatrick (20:04:46)
We're trying to build as big a set as possible that doesn't satisfy (a) and (b).

DPatrick (20:04:56)
Since k 1's can't add up to something red, we must have k blue.

DPatrick (20:05:09)
Then, since k k's can't add up to something blue, we must have k^2 red.

DPatrick (20:05:22)
Now what?

JAM (20:05:27)
1+(k-1)+(k-1)+....+(k-1) = k^2, so k+1 is blue

DPatrick (20:05:58)
Not quite, but right idea...

Scrambled (20:05:53)
k+1 is blue because

calc rulz (20:06:19)
you mean k+1

JAM (20:06:22)
Oops those should be k+1s

DPatrick (20:06:34)
Right.

DPatrick (20:06:57)
1 + (k-1) copies of (k+1) give k^2. Since 1 and k^2 are both red, we must have k+1 blue.

solafidefarms (20:05:36)
now k^2+k-1 must be blue

tjames_2 (20:05:49)
k^2+k-1 is the smallest red

DPatrick (20:07:33)
Where did this number come from?

besttate (20:07:36)
1+k+k+k+k....+k=k^2+k-1
(k+1)+(k+1)...+(k+1)+k = k^2 + k -1, so this is now a contradiction

DPatrick (20:08:03)
Right! We can add up to k^2+k-1 in two different ways.

DPatrick (20:08:33)
Wait a sec...

DPatrick (20:08:56)
The bottom one that you wrote is correct.

DPatrick (20:09:19)
k and k+1 are both blue, and if we add one k and (k-1) copies of k+1, we get k^2+k-1, so k^2+k-1 has to be red.

DPatrick (20:09:22)
On the other hand...

besttate (20:09:15)
Oops, the first addition should be k^2+1+1+1...+1=k^2+k - 1

perfectnumber628 (20:09:22)
1+1+1+...+1+k^2=k^2+k-1

DPatrick (20:09:30)
Right.

DPatrick (20:09:48)
1 and k^2 are both red, so if we add (k-1) copies of 1 and one k^2, we also get k^2+k-1, so it has to be blue.

DPatrick (20:09:56)
That's a contradiction.

DPatrick (20:10:17)
So we know that any coloring of {1,2,...,k^2+k-1} must satisfy the property. (We tried to construct one that didn't, and failed.)

Xantos C. Guin (20:09:59)
Now we must prove that we can color the elements when n=k^2+k-2 and avoid satisfying a and b

gregx007 (20:10:17)
now, we need to show that anything smaller than k^2 + k - 1 can be obtained

DPatrick (20:10:53)
Exactly. We need to show that we can color {1,2,...,k^2+k-2} without satisfying (a) and (b).

DPatrick (20:11:17)
(By the way, going back to our example, note that when k=2, that k^2+k-1 = 4+2-1 = 5, which is what we got, so we're probably correct!)

besttate (20:10:49)
Color [1,k-1] red; [k,k^2-1]blue; and [k^2,k^2+k-2] red.

gregx007 (20:11:09)
color 1, 2, ..., k-1 red; k, k+1, ..., k^2 - 1 blue; k^2, k^2 +1, ..., k^2 + k -2

Xantos C. Guin (20:11:21)
Let 1,2,3,...,k-1 and k^2,k^2+1,k^2+2,...k^2+k-2 be red and k,k+1,k+2,...,k^2-1 be blue

DPatrick (20:11:39)
Exactly.

DPatrick (20:11:52)
Playing around with the k=3 example might have helped you find this...

DPatrick (20:12:00)
We can color 1,2,up through k-1 all red.

DPatrick (20:12:32)
This means that all of k, k+1, ..., up through k^2-1 can be blue, and all the higher ones can be red.

DPatrick (20:13:09)
You then need to check that this coloring works: no sum of k reds is red, and no sum of k blues is blue. But that's fairly easy to check; I'll leave that for you to do.

Scrambled (20:12:47)
is it true that 1, 2, .... k-1 all have to be the same color for k^2+k-2 to work?

DPatrick (20:13:28)
I believe so, but I'm not entirely sure.

gregx007 (20:13:15)
out of question, is it possible to solve this problem generalized to m colors?

DPatrick (20:13:47)
I think so -- some people have posted on the message board about this.

DPatrick (20:14:08)
We thought the problem was interesting enough as is.

oppenhejo (20:13:59)
For full credit, was a proof necessary for this problem, becasue unlike problem 3, the question did not explicitly ask for one

DPatrick (20:14:29)
USAMTS answers always have to be justified.

DPatrick (20:14:51)
Let's move on to Problem 5.

DPatrick (20:14:57)


DPatrick (20:14:59)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-39971950.gif

DPatrick (20:15:05)


DPatrick (20:15:07)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/8d052d65ca937aa2bd41ceb30a4bf3ce.png

Xaero (20:15:23)
Connect EB to form a cyclic quadrilateral

kev015 (20:15:40)
The first step is to connect EB and EC

DPatrick (20:16:07)
This is going to be helpful later, but first, let's look at triangle ABC, the triangle that we have information about.

vfiroiu (20:15:11)
use law of cosines on angle BAC

rnwang2 (20:15:27)
<BAC is 60 degrees by the Law of Cosines

danielkiang (20:15:45)
law of cosines to find <BAC

JerryS (20:15:49)
law of cosines shows that BAC = 60

gregx007 (20:16:04)
you can find that angle BAC is 60 degrees

pbdude (20:16:12)
angle CAB=60

DPatrick (20:16:48)
Indeed, this is the key. We are given a triangle ABC in which we know all the lengths. It makes sense to try the Law of Cosines and see if any of the angles are nice.

DPatrick (20:16:59)


DPatrick (20:17:10)


DPatrick (20:17:23)
This seems useful!

besttate (20:16:46)
This means FAE and FAD must be 60 degrees as well.

kostya (20:17:10)
<BAE=60

Andrew Kim (20:17:53)
angle EAF is 60! 30 60 90

DPatrick (20:18:28)
Indeed, we now know that BAD = 120, so BAE = 60 and AEF is a 30-60-90 triangle.

Xantos C. Guin (20:18:37)
Triangle CEB is an equilateral triangle

DPatrick (20:18:52)
Cool! How do we know?

Cmac89 (20:18:56)
BEC is also 60 since it intercepts the same arc as BAC

Scrambled (20:19:03)
intercepted arcs

DPatrick (20:19:12)
Exactly.

DPatrick (20:19:17)
Let's draw in the segments that were suggested at the beginning.

DPatrick (20:19:25)


DPatrick (20:19:28)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/47071baeb0727646fa563ae0c006910a.png

DPatrick (20:19:46)
BAC and BEC intercept the same arc, so they're equal. Same for BCE and BAE.

DPatrick (20:19:59)
So EBC is an equilateral triangle!

gregx007 (20:19:17)
now, just use Ptolemy's theorem

rnwang2 (20:19:26)
and use Ptolemy's Theorem to finish it off

Xantos C. Guin (20:19:44)
Use Ptolemy's theorem and solve for EA

DPatrick (20:20:41)
That's one option. We know 3 sides of cyclic quadrilateral EACB and both diagonals, so we could use Ptolemy to get AE.

DPatrick (20:21:07)
We can also use the Law of Cosines on triangle AEC, or triangle EAB.

DPatrick (20:21:15)
Lots of ways to finish from here.

Goistein (20:21:11)
What's Ptolemy's thorom?

thinker (20:21:13)
What is a cyclic quadrilateral?

DPatrick (20:21:36)
A cyclic quadrilateral is one all of whose vertices are on a common circle, like EACB in this problem.

DPatrick (20:21:49)
Ptolemy's theorem then says:

DPatrick (20:21:54)


DPatrick (20:22:02)
or, in English:

DPatrick (20:22:15)
the sum of the product of the opposite sides equals the product of the diagonals.

DPatrick (20:22:28)
We can plug in the numbers:

DPatrick (20:23:00)
EB = CE = CB = 7, AB = 8, AC = 5, solve for AE.

DPatrick (20:23:16)
You should get AE = 3.

DPatrick (20:23:23)
And to finish...

besttate (20:21:03)
So we get EA is 3, and by the 30-60-90 ratio, AF is 3/2.

Andrew Kim (20:23:29)
AF= 3/2

DPatrick (20:23:47)
Right. AEF is 30-60-90, so AF = AE/2 = 3/2.

thinker (20:22:39)
is the proof for it simple? would you need to prove that to get full credit?

DPatrick (20:24:00)
You can cite Ptolemy's Theorem without proof.

DPatrick (20:24:11)
(Although the proof is a nice exercise if you've never done it before!)

DPatrick (20:24:21)
I'll leave you with an extra challenge: Suppose that ABC is any acute triangle with AB > AC, and the rest of the points in the problem are as defined. Show that AF = (AB - AC)/2. (As you can see, the law of cosines won't be as straightforward to apply here. But here's a hint: Ptolemy's Theorem still applies!)

rnwang2 (20:24:22)
is there a purely synthetic way to solve this problem?

DPatrick (20:25:01)
My extra challenge suggests that there is. [img id=em-0]

DPatrick (20:25:09)
That's it for Round 2!

DPatrick (20:25:17)
Round 3 problems should be available on Friday.

tjames_2 (20:25:16)
synthetic meaning?

DPatrick (20:25:30)
without using trig, I think he means.

sonyvaio22 (20:25:52)
will there be another classroom for round 3?

DPatrick (20:26:07)
Yes, we'll have one of these sessions after each round.

thinker (20:25:33)
would you have to prove theorems that you use to get your solutions in order to get full credit?

DPatrick (20:26:30)
Not if the theorems are "well known". Ptolemy's Theorem is considered "well known".

Andrew Kim (20:26:45)
When will we get the Round 2 scores?

DPatrick (20:27:19)
Late December.