| Transcript
for the Math
Jam "USAMTS Round 1 Math Jam"
on Oct 11. |
| Math Jam hosted by DPatrick
(Dave Patrick ). |
DPatrick19:29:19
Hello and welcome to the first 2007-08 USA Math Talent Search Math Jam.
DPatrick19:29:28
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.
DPatrick19:29:36
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.
DPatrick19:29:45
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.
DPatrick19:30:01
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.
DPatrick19:30:20
As there are a lot of people here 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.
DPatrick19:30:32
There will be images in this lecture. The images should appear directly in the classroom window, as in the example below:
DPatrick19:30:39
DPatrick19:30:51
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.)
DPatrick19:31:05
You can view the first round problems as we discuss them by clicking on the following link:
DPatrick19:31:22
Let's get started with Problem 1.
DPatrick19:31:27
DPatrick19:31:37
How can we begin on this problem?
cc119:32:17
find sum of the original pieces of paper first
frank4419:32:17
find the sum of all of the papers: n(2n+1)
p4fn2w19:32:17
the sum of 2n pieces is $\frac{(2n)(2n+1)}{2}$?
mathnerd31415919:32:17
Calculate the total sum of all the numbers on the pieces of paper.
DPatrick19:32:43
There are lots of different ways we could proceed. I would start by computing the sum the original 2n pieces of paper first.
DPatrick19:32:56
DPatrick19:33:11
vishalarul19:33:36
Set a variable for the first term.
Xantos C. Guin19:33:36
Find a formula for the sum in terms of n and the lowest numbered piece of paper
mthiyaga19:33:36
Let a be the first of the n numbers being removed.
cc119:33:36
then find the sum of the ones he removed
mthiyaga19:33:41
Let a be the first of the n numbers that are removed. Hence, the sum of the n numbers is n(2a+n-1)/2.
DPatrick19:33:48
Right. The next step is to consider the papers that he removed.
DPatrick19:33:59
DPatrick19:34:15
13375P34K43V31219:34:38
set it equal to 1615
Comatosis19:34:38
Subtract that from the original sum to get 1615
DPatrick19:34:46
DPatrick19:34:54
freezard773419:35:12
Factor n
nikki19:35:12
now solve for n
VDLmath19:35:17
simplify the equation
DPatrick19:35:25
Right, we can simplify it a bit and factor out an n:
DPatrick19:35:33
freezard773419:35:48
n has to be a factor of 1615 * 2 = 3230
Yuueesceri19:35:48
factor 3230
perfect62819:35:48
then let n equal eah of the factors of 3230
PI-Dimension19:35:48
n|3230
360_Fan19:35:48
and factor 3230 = 2*5*17*19
DPatrick19:36:07
Since n is an integer, we must have that n is a factor of 3230. We factor 3230 = 2(5)(17)(19).
DPatrick19:36:30
Do we need to check all 16 factors of 3230?
1=219:36:43
We don't want to try all 16 factors...
kstan01319:36:43
...but now a range must be determined for the possiblities of n
fireninja019:36:45
just first 8
Xantos C. Guin19:36:50
No, we can set bounds on n
ScoPI19:36:53
some are too extreme to work
DPatrick19:37:00
Right, we can save ourselves a bit of work first.
DPatrick19:37:13
For instance, note that 3n + 5 - 2a >= 3n + 3 - 2(n+1) = n+1, so we must have that n is the smaller factor and 3n+5-2a is the larger factor.
DPatrick19:37:36
(oops...that's 3n+3-2a in the line above)
360_Fan19:38:10
so find all the pairs of factors of 3230, and n is always the smaller one, and then try and see which ones work
DPatrick19:38:42
We can also eliminate some small values of n:
DPatrick19:38:55
3n + 3 - 2a <= 3n + 3 - 2 = 3n + 1, so the larger factor is at most 3 times the smaller factor, plus 1.
kstan01319:39:03
well 2, 5, 10, 17, 19 obviusly don't work (too small)
ductoan050319:39:03
we can set a lower bound for n
atennis19:39:08
n must be between 34 and 38
DPatrick19:39:24
Exactly. This rules out all factors n except for 34 and 38.
VDLmath19:39:30
n must be 34 and 38
applequest19:39:30
So n is 34 or 38.
ScoPI19:39:30
actually, 34 and 38 are the only ones that work
perfect62819:39:32
and both those work
DPatrick19:39:51
Yes...we're not quite done yet. We still have to check that these n give valid values of a.
DPatrick19:40:07
n=34 works: 3n+3-2a = 105 - 2a = 3230/34 = 95, so a=5. We check: the sum of 1 to 68 is 2346, the sum of 5 to 38 is 731, and their difference is 2346-731 = 1615.
DPatrick19:40:10
n=38 works: 3n+3-2a = 117 - 2a = 3230/38 = 85, so a=16. We check: the sum of 1 to 76 is 2926, the sum of 16 to 53 is 1311, and their difference is 2926-1311 = 1615.
DPatrick19:40:31
So the answer is n=34 and n=38.
krakola4519:40:52
So both are the answer
DPatrick19:41:08
Yes, to receive full credit, you'll need to show that both of these work and that no others values work.
DPatrick19:41:19
Let's move on to Problem 2.
DPatrick19:41:24
DPatrick19:41:37
DPatrick19:41:45
DPatrick19:41:56
Let's start with part (a). How do we find the angles?
ra524919:42:17
well (by the center), 6a = 360 so a = 60
eashwar1919:42:17
well 6A = 360
VDLmath19:42:17
we can figure out a =60 because 6 of it meet at the center
cyberspace19:42:17
6 A's make up 360 deg
360_Fan19:42:17
a is 60 because the 6 a's in the middle add up to 360
PI-Dimension19:42:17
A=60 because of the center
applequest19:42:17
A is easy. 6 meet at the center, so it's 60 degrees.
DPatrick19:42:25
Right. There are 6 copies of angle A around the center point, so 6A = 360, meaning that A=60. (All angles will be written in degrees.)
DPatrick19:42:36
Now what?
ShockDrake19:43:03
<B is an interior angle.
freezard773419:43:03
We can find B because it is the interior angle of the 18-gon
kstan01319:43:03
well b is an angle of the dodecagon... so that's easy
PI-Dimension19:43:03
B is (16)*180/18=160
fireninja019:43:03
B is 160 because the interior angle of 18-gon is 180-360/18
Spencer19:43:03
B is an angle of the 18-gon
DPatrick19:43:19
Right. B shows up in lots of positions as an interior angle of the 18-gon.
DPatrick19:43:24
But an interior angle of an 18-gon is 180(18-2)/18 = 160.
DPatrick19:43:32
So B = 160.
DPatrick19:43:38
How do we get the rest?
Twiz19:43:57
Drawing in segment BE gives a rhombus
darkprince19:43:57
draw BE
360_Fan19:43:59
Draw BE
DPatrick19:44:09
Good. Draw diagonal BE in our pentagon. This divides the pentagon into equilateral triangle ABE and rhombus BCDE.
DPatrick19:44:18
DPatrick19:44:52
Since all the outside sides are equal and angle A = 60, we have that ABE is equilaterial, and thus BCDE is a rhombus.
DPatrick19:44:57
So now we can just chase the rest of the angles.
mathnerd31415919:45:14
ABE is 60, so CBE is 160 - 60 = 100.
eashwar1919:45:24
and C = 80 and E = 80+60 = 140
ra524919:45:26
so then D = CBE = 100
DPatrick19:45:51
Right. ABE + EBC = 160, but ABE = 60, so EBC = 100. Thus D = 100 and C = 80, and finally E = 60+80 = 140.
DPatrick19:46:01
In summary, in the original pentagon, we have:
A = 60, B = 160, C = 80, D = 100, and E = 140.
DPatrick19:46:09
As a check, note that they sum to 60+160+80+100+140 = 540, which is the sum of the interior angles of a pentagon.
DPatrick19:46:26
Now for part (b). How do we prove that 3 points are collinear?
xw199219:46:46
to prove x, y, z to be collinear, i can prove angle ZYX to be 180 degres
fireninja019:46:46
prove XYZ is 180 degrees
darkprince19:46:46
180 degrees = angle XYZ
YellowMarker16119:46:46
if angle measure of xyz = 180 degrees
ShockDrake19:46:46
prove that <XYZ = 180
DPatrick19:46:59
Right. One way is to show that "angle" XYZ measures 180 degrees.
cc119:47:13
the problem will be easier to do if we add some points
1=219:47:13
Angle Chasing
DPatrick19:47:34
Right, we want to identify some smaller angles that make up XYZ, and chase angles so that we add up to 180 at point Y.
DPatrick19:47:42
There are a number of ways to go about this. Here's one possibility:
DPatrick19:47:48
DPatrick19:47:55
Remember, we cannot assume that XYZ is a line! We're trying to show that angle XYZ is 180 degrees.
DPatrick19:48:07
What's ZYQ?
VDLmath19:48:28
10
eashwar1919:48:28
10
krsattack19:48:28
10
13375P34K43V31219:48:28
10
Yuueesceri19:48:28
10
DPatrick19:48:37
Triangle ZQY is isosceles, and angles ZQY is the sum of our obtuse rhombus angle plus 60. So ZQY = 100+60 =160, hence ZYQ = 10.
DPatrick19:48:51
What's angle QYO?
13375P34K43V31219:49:08
60
1=219:49:08
60
360_Fan19:49:08
60
DPatrick19:49:14
About fifty of you said "60"! :)
DPatrick19:49:22
Triangle QOY is equilateral, so angle QYO = 60.
DPatrick19:49:33
Triangle YOP is also equilateral, so angle OYP = 60.
DPatrick19:49:39
Finally, what's PYX?
cyberspace19:49:52
50
360_Fan19:49:52
50
ra524919:49:52
half 100 so 50
Goldentide19:49:52
50
eashwar1919:49:52
50
DPatrick19:50:06
Finally, triangle YPX is isosceles, and base angle YPX = 80. So angle XYP = 50.
DPatrick19:50:12
Adding them up, we get:
ZYX = ZYQ + QYO + OYP + PYX = 10+60+60+50 = 180.
DPatrick19:50:21
Therefore, points X, Y, Z are collinear!
DPatrick19:50:37
Both parts of this problem were just "angle chasing". Remember, showing three points are collinear is equivalent to showing that the angle that they form is 180 degrees.
DPatrick19:50:56
Let's move on to Problem 3.
DPatrick19:51:02
apple pi19:51:22
Take the tangent of both sides, and use the tangent sum formula twice.
1=219:51:22
take the tangent of both sides
ShockDrake19:51:22
Take the tangent of both sides of the equation.
emilgouliev19:51:22
take the tan of both sides
DPatrick19:51:28
arctans are icky. Let's take the tangent of both sides:
DPatrick19:51:34
DPatrick19:51:47
Already we're happy: the right side is just 1.
13375P34K43V31219:51:59
tan(x+y)=(tanx+tany)/(1-tanxtany), let x=arctana, y=arctanb
DPatrick19:52:07
DPatrick19:52:13
What if we have 3 angles?
eashwar1919:52:22
wouldn't it be easier if u took teh arctan of 1/c over to the other side?
darkprince19:52:22
wouldn't moving one to the other side make it easier?
DPatrick19:52:44
That's actually another good option. It's not what I happened to do, though, but it works just as well.
Yuueesceri19:52:56
do Tan (A + B+ C) = Tan ((A+B)+C)
12345678919:52:56
perform the formula twice
cognos59919:52:56
add 2 at one time and add that to the third
McDutchy19:52:56
It doesn't matter, you can add x+y first then (x+y)+z
DPatrick19:53:02
Right, that's what I did.
DPatrick19:53:06
DPatrick19:53:15
Now we plug in our tan(x+y) formula from above:
DPatrick19:53:17
scissorblades19:53:33
Can we simplify that a bit?
DPatrick19:53:51
Sure -- let's get rid of the denominators (by multiplying numerator and denominator by 1-tan(x)tan(y)), then it looks a lot nicer:
DPatrick19:53:55
1=219:54:07
tanx=1/a, so on
mthiyaga19:54:07
=(ab+ac+bc-1)/(abc-a-b-c) plugging in 1/a, 1/b, and 1/c back in
DPatrick19:54:17
Right: in our problem, x = arctan(1/a), y = arctan(1/b), and z = acrtan(1/c).
DPatrick19:54:23
This means that tan(x) = 1/a, tan(y) = 1/b, and tan(z) = 1/c, so we can plug these in, and we get the formula:
DPatrick19:54:27
baozhale19:54:40
stiil want to simplify that
McDutchy19:54:40
simplify!
PI-Dimension19:54:42
multiply by abc/abc
DPatrick19:54:50
Indeed, multiply numerator and denominator by abc to get rid of the ugliness.
DPatrick19:54:56
360_Fan19:55:13
multiply by denominator
mathnerd31415919:55:17
now move the denominator to the other side.
DPatrick19:55:19
DPatrick19:55:31
Now what?
Twiz19:55:44
prove that a must be 2 or 3
perfect62819:55:44
find consraints on a-it must either be 2 or 3
napoleon633419:55:44
put bounds on a
Xantos C. Guin19:55:48
Set bounds for a
DPatrick19:56:20
There are certainly some slick algebraic manipulations that you can do here, but we can also just look for bounds on a. (Remember, a <= b <= c are positive integers.)
DPatrick19:56:26
Can a=1?
napoleon633419:56:55
no, because then b+c = 0
Xantos C. Guin19:56:55
no, since arctan1 = pi/4
3468Yz19:56:55
No. Arctan 1/1 = pi/4
DPatrick19:57:17
Right. We can see from the original arctan equation that a must be more than 1, since arctan(1/1) = pi/4 would make the left side too big.
DPatrick19:57:27
Or we plug in: if a=1, then we have
bc - (b+c+bc) - (1+b+c) + 1 = -2(b+c) = 0,
so there's no solution with a=1.
McDutchy19:57:49
also, a<4 since 3*arctan(1/4) < pi/4
13375P34K43V31219:57:53
if a>=4, then (4,4,4) makes RHS maximal, but it's too small
DPatrick19:58:09
We can also see that we cannot have a >= 4, in a number of ways.
DPatrick19:58:21
My favorite is: if we assume that 4 <= a <= b <= c, then we have
arctan(1/a) + arctan(1/b) + arctan(1/c) <= 3arctan(1/a) <= 3arctan(1/4).
DPatrick19:58:45
But tan(pi/12) = 2 - sqrt(3) > 1/4, so arctan(1/4) < arctan(2-sqrt(3)) = pi/12.
DPatrick19:59:07
So 3 arctan(1/4) < pi/4, and hence a >= 4 gives a contradiction.
DPatrick19:59:15
(You can show this algebraically too.)
baozhale19:59:23
so a=2 or a=3
DPatrick19:59:35
Right, so now we can just plug in a=2 and a=3, and see what solutions we get.
DPatrick19:59:45
If a=2, then we have
2bc - (2b + 2c + bc) - (2+b+c) + 1 = bc - 3b - 3c - 1 = 0.
DPatrick19:59:49
How do we solve this?
Xantos C. Guin20:00:19
SFFT
13375P34K43V31220:00:19
add 10 to both sides
tjhance20:00:19
simon
PI-Dimension20:00:19
SFFT
darkprince20:00:19
SFFT
perfect62820:00:19
sfft
1=220:00:19
Thank you, Simon!
DPatrick20:00:25
We now use Simon's Favorite Factoring Trick!
DPatrick20:00:29
bc - 3b - 3c -1 = bc - 3b - 3c + 9 - 10 = (b-3)(c-3) - 10.
DPatrick20:00:45
SFFT is adding a constant to make the expression factor.
DPatrick20:00:55
So (b-3)(c-3) = 10.
12345678920:01:11
find factors of 10
eashwar1920:01:23
1 2 5 10
McDutchy20:01:23
1,2,5,10
krsattack20:01:31
b=4, c=13 OR b=5, c=8
3468Yz20:01:31
Then b=5 and c=8, or b=4 and c=13.
DPatrick20:01:37
Right. The possible factorizations of 10 are (1)(10) or (2)(5), giving solutions (b,c) = (4,13) or (5,8).
DPatrick20:01:49
Thus (a,b,c) = (2,4,13) and (a,b,c) = (2,5,8) are possible solutions.
PI-Dimension20:02:01
do same for 3
atennis20:02:03
now plug 3 in for a
DPatrick20:02:13
If a=3, then we have
3bc - (3b+3c+bc) - (3+b+c) + 1 = 2bc - 4b - 4c - 2 = 0.
DPatrick20:02:27
Divide by 2 to give bc - 2b - 2c - 1 = 0.
Shadow Hunter20:02:39
Simon again
Goldentide20:02:39
sfft
12345678920:02:43
(b-2)(c-2)=5
DPatrick20:02:47
Again, Simon's Favorite Factoring Trick!
bc - 2b - 2c - 1 = bc - 2b - 2c + 4 - 5 = (b-2)(c-2) - 5.
So (b-2)(c-2) = 5.
PI-Dimension20:03:09
1,5
krakola4520:03:09
factors of 5 are 1,5
atennis20:03:09
find factors of 5
mathnerd31415920:03:11
only factors of 5 are 1 and 5
PI-Dimension20:03:15
b=3 and c=7
scissorblades20:03:15
b = 3, c = 7
3468Yz20:03:15
Thus b=3, c=7, as 5 is prime.
DPatrick20:03:21
This means that (b,c) = (3,7). So (a,b,c) = (3,3,7) is a solution.
DPatrick20:03:31
So we're done!
The solutions are (a,b,c) = (2,4,13), (2,5,8), and (3,3,7).
DPatrick20:04:01
Let's move on to Problem 4.
DPatrick20:04:09
Shadow Hunter20:04:28
diagram
uclabb20:04:28
diagram
scissorblades20:04:28
We should draw a sketch first.
DPatrick20:04:38
Let's start by sketching a picture.
DPatrick20:04:46
DPatrick20:04:55
We know that B = 77, C = 150, and AB=CD. How do we use the data about P?
emilgouliev20:05:21
draw the perpendicular bisectors
PI-Dimension20:05:21
perpendicular bisectors
cc120:05:21
draw bisectors
mathnerd31415920:05:23
Draw the perpendicular bisectors of AD, BC
DPatrick20:05:28
Let's add P to the picture. (Tip: use a ruler and protractor to make an accurate drawing; otherwise you might have P in the wrong spot!)
DPatrick20:05:34
cyberspace20:05:54
triangles APD and BPC are isosceles
apple pi20:05:54
it is equidistant from b and c and a and d
kstan01320:05:54
forms iscoeles triangles
napoleon633420:05:54
Now draw AP and AD, and BP and PC
DPatrick20:06:03
Right. We know that PB=PC and PA=PD.
DPatrick20:06:11
Let's draw these lines:
DPatrick20:06:17
DPatrick20:06:28
What do we have?
emilgouliev20:06:42
by SSS, we know ABP = DCP
electricj20:06:42
SSS congruency now
tjhance20:06:42
triangle ABP is congruent to triangle DCP
hsjangrun20:06:42
By given, AB = CD and triangles ABP and DCP are congruent
DPatrick20:06:52
Excellent. PB=PC, PA=PD, and AB=CD. So by SSS congruence, triangles PAB and PDC are congruent.
DPatrick20:07:03
Also PAD and PBC are isosceles.
DPatrick20:07:17
How do we use this information?
uclabb20:07:25
call angle PBC= x
monkeymatt20:07:27
find one of the base angles of PBC
DPatrick20:07:34
Good idea. Let x = angle PBC = angle PCB. What equation can we write for x?
monkeymatt20:08:26
x+77=360-150-x
cyberspace20:08:26
x+77+x+150=360
cc120:08:26
x+77=360-(150+x)
krsattack20:08:26
2x+77+150=360
DPatrick20:08:38
Yes. Angle PBA = 77 + x, and angle PCD = 360-(150+x) = 210-x. These must be equal (they're corresponding angles of PBA and PCD), so 77+x = 210-x.
DPatrick20:08:46
scissorblades20:09:04
2x+ bpc = 180
mathnerd31415920:09:04
2x + BPC = 180
phs194520:09:04
180 - 2x = angle BPC
DPatrick20:09:10
Right. We've got enough info to finish now.
DPatrick20:09:16
Thus 77+x = 210-x, so 2x = 133.
DPatrick20:09:22
Hence, angle BPC = 180 - 2x = 180 - 133 = 47 degrees.
DPatrick20:10:12
Let's finish up with a toughie: problem 5.
DPatrick20:10:18
DPatrick20:10:35
This problem is hard. (On the USAMTS, we generally try to make #5 on each round be fairly difficult.)
3468Yz20:10:53
Establish that c=-1/2 is valid.
uclabb20:10:53
notice that -.5 works
mathnerd31415920:10:53
Set a_2 = a_1 and solve for c.
DPatrick20:11:00
Right. One place to start is to look for a constant solution -- in other words, a solution where a_n = c for all n.
DPatrick20:11:12
DPatrick20:11:25
(The other solution to this quadratic is c=1, but clearly we can't have that since we want negative solutions.)
DPatrick20:11:39
So c = -1/2 is one solution. Are there any others?
calc rulz20:12:08
well let's get some bounds...1<c<0 to start
scissorblades20:12:08
Now set bounds
uclabb20:12:14
after quick examination, we conjecture that this is the only solution, so we call c= (-.5+a) and prove that a must be 0
DPatrick20:12:25
If you experimented with this problem (either by hand or on a computer), you might suspect that c = -1/2 is the only solution.
DPatrick20:13:00
There are a number of ways to proceed from here. One way is to try to set some sort of bounds on c,
DPatrick20:13:17
But a number of you have come up with a very clever observation.
junggi20:13:33
we can easily show 0>c>-1 so be like c=cos theta for some theta
cognos59920:13:33
you can notice that (2subn-1)^2 - 1 is in the form 2cos^2x-1 which is cos 2x.
Altheman20:13:33
We can make the substitution: a_1=cos t.
PI-Dimension20:13:39
cosine double angle formula
Xantos C. Guin20:13:41
cos2k = 2cos^2k-1
DPatrick20:13:51
Right!
DPatrick20:13:58
Our recursion looks a lot like the cosine double-angle formula:
DPatrick20:14:05
DPatrick20:14:18
So let's start with c = cos(theta) for some angle theta, and see where we get.
(Note: we can do this since if |c| > 1, then 2c^2 - 1 >= 1 > 0, so we'd fail right away if |c| > 1.)
panjia12320:14:43
cos(theta)
cos(2theta)
cos(4theta)
etc.
darkprince20:14:43
DPatrick20:14:51
Exactly!
DPatrick20:14:55
DPatrick20:15:03
DPatrick20:15:12
DPatrick20:15:26
How can we use this curious observation?
Altheman20:15:45
Now we must use the condition that a_k is negative.
DPatrick20:15:51
Right -- when is cosine negative?
Xantos C. Guin20:16:08
2^(n-1)*theta must be between 90 and 270
scissorblades20:16:08
when it is between 90 and 270
frank4420:16:08
2nd and 3rd quadrant
PI-Dimension20:16:08
90-270
DPatrick20:16:29
True...but our angle is going to increase without bound as we keep multiplying by 2.
DPatrick20:16:35
Is there another way we can describe it?
not_trig20:16:56
pi/2 < k < 3pi/2
McDutchy20:16:56
pi/2<x<3pi/2
monkeymatt20:16:56
multiple of pi
DPatrick20:17:04
OK, that's looking better.
DPatrick20:17:16
When is cos(k*pi) negative (if k is a real number)?
napoleon633420:17:51
when k is odd
ra524920:17:51
when k is odd?
bubble20:17:51
1/2<k<3/2
uclabb20:17:53
between pi/2 and 3pi/2
DPatrick20:18:05
True, but is there a statement that works for all values of k?
3468Yz20:18:13
Whenever 1/2<k<3/2, 5/2<k<7/2, etc.
not_trig20:18:15
1/2-3/2, 5/2-7/2, etc. etc.
Xantos C. Guin20:18:18
round(k)= some odd number?
DPatrick20:18:20
Aha!
DPatrick20:18:26
DPatrick20:18:46
DPatrick20:18:58
(Sorry, I used k to mean two different things there. Bad!)
DPatrick20:19:18
What real number(s) b satisfy this? How can we examine this?
Altheman20:19:51
Lets look at the binary representation of b.
DPatrick20:20:02
Indeed: all of the powers of 2, together with the rounding, suggest looking at a binary "decimal" representation. So let's write b as a binary real number:
DPatrick20:20:10
DPatrick20:20:16
Each digit is 0 or 1.
DPatrick20:20:31
Multiplying by a power of 2 just means moving the "binary" point to the right.
Altheman20:20:39
of course, b_0=0 since 0<b<1
DPatrick20:20:45
Right.
DPatrick20:21:02
How do we interpret rounding? How do we know if a binary decimal rounds to something even or odd?
1=220:21:53
rounds to a units digit of 0
mathnerd31415920:21:53
the digit in the units is 0 if the number is even.
bubble20:21:55
round up for 1 down for 0?
fireninja020:21:58
'ones' digit
ScoPI20:21:58
the ones digit?
DPatrick20:22:26
A binary decimal of the form .....0.1..... or .......1.0..... will round to something odd.
DPatrick20:22:39
There are two cases:
DPatrick20:22:43
(a) if the digit immediately to the left of the binary point is 1, then the digit immediately to the right of the binary point must by 0 (we "round down" to a binary integer ending in 1);
DPatrick20:22:48
or:
DPatrick20:22:52
(b) if the digit immediately to the left of the binary point is 0, then the digit immediately to the right of the binary point must by 1 (we "round up" to a binary integer ending in 1).
DPatrick20:23:18
So what can we conclude from all this?
applequest20:23:26
Digits alternate.
vishalarul20:23:32
Either 0.1010101... or 1.0101010..
krsattack20:23:33
b_0 = 0 => b_1 = 1 => b_2 = 0 => ...
DPatrick20:23:39
Bingo!
DPatrick20:23:45
The binary digits must alternate!
In other words, b = 0.101010101010....
DPatrick20:23:54
To put this another way, we can never have two consecutive zeros, since this will give us ...0.0... which will round down to a number ending in 0 (that is, an even number), which is bad.
Similarly, we can never have two consecutive ones, since this will gives us ...1.1... which will round up to a number ending in 0, again which is bad.
DPatrick20:24:09
So we must have b = 0.1010101010...
not_trig20:24:13
1/2+1/8+1/32+...
Xantos C. Guin20:24:15
thus b=2/3
DPatrick20:24:24
Yes. This is just 1/2 + 1/8 + 1/32 + ... = 1/2/(1-1/4) = 2/3.
DPatrick20:24:28
So b = 2/3, and c = cos(2/3 * pi) = -1/2.
DPatrick20:24:39
Thus c=-1/2 is the only solution.
DPatrick20:25:24
There are algebraic ways to do this problem too, without using trigonometry. You can post alternate solutions on the message board if you like.
DPatrick20:25:50
To wrap up: we are currently processing the Round 1 submissions. We hope to have this completed by Monday or Tuesday (there are a lot still to go). Your scores and feedback should be available in early November.
DPatrick20:25:58
Round 2 problems will be posted next week on the USAMTS web site. The deadline for submitting your Round 2 solutions is November 19.
DPatrick20:26:18
Feel free to discuss Round 1 on the message board now.
DPatrick20:26:23
Thanks for participating!
eashwar1920:27:47
for the third problem witht eh arctangents, wouldnt it have been easier to bring teh arctan of 1/c over to the other side?
eashwar1920:27:50
it gets rid of the messy algebra in teh middle
DPatrick20:28:16
It certainly is an acceptable way to do it. I happen to slightly prefer the triple-angle, but moving one arctan to the other side works very nicely too.
kstan01320:28:28
what if i got a problem right but the solution was kind of bad? 3?
DPatrick20:28:53
We'll be deciding on grading schemes for the various problems. There will definitely be partial credit for right answers with flawed solutions.
calc rulz20:28:59
Is the formula for tan(x+y+z) citable?
DPatrick20:29:02
Yes.
DPatrick20:29:17
that formula is sufficient common that you can cite it without proof.
DPatrick20:29:25
...sufficiently common...
smevawala20:29:29
what are algebraic ways?
DPatrick20:29:34
For #5 I assume you mean?
DPatrick20:30:05
You can very carefully place bounds on how big |a_n| must grow.
DPatrick20:30:19
Or, rather, how big |a_n + 1/2| must grow if it doesn't start at -1/2.
DPatrick20:30:37
There are a lot of little details to pay attention to, though :)
ra524920:30:46
is there a way to do the tangent problem without determining that a<4? (like by algebra)
DPatrick20:30:58
Sure.
DPatrick20:31:05
One nice way is: Let x=a-1, y=b-1, z=c-1, and plug them it. Things then simplify:
xyz - 2(x+y+z) = 4.
DPatrick20:31:22
Now the algebra is much easier to deal with, and you can use AM-GM for bounds too.
Altheman20:31:32
Hello, I don't think that we proved that the solutions for problem 3 actually worked.
DPatrick20:31:50
You do have to worry about all the angles being acute, so that all the tan/arctan stuff is valid.
napoleon633420:32:12
On problem 1, are the values of a (5 and 16) for 34 and 38 required to be shown in the solution? In other words, do you have to check those solutions to prove that they work?
DPatrick20:32:22
You don't have to explicitly find the values of a as long as you prove that your answers work.
uclabb20:32:38
if you prove that for (-.5+a)=a_n, a changes sign consistantly and abs value of a is always growing when it is positive and always increasing at an increasing rate when it is negative, then you have a proof, right?
DPatrick20:32:49
More or less, yes, as long as all the details are correct.
DPatrick20:33:21
darkprince20:33:52
what if you had a 10-line proof for 4? :)
DPatrick20:33:57
If it's correct, great!
DPatrick20:34:29
Any other questions?
3468Yz20:35:05
Does trig-bashing geometry problems yield full credit if correct?
DPatrick20:35:08
Certainly.
mathnerd31415920:35:30
What if we got the correct way to solve a problem but the wrong answer?
DPatrick20:35:50
We'll assign partial credit based on how much progress you made. We haven't decided on the specific grading schemes for this round yet.
uclabb20:36:11
when and where will the transcript be up?
DPatrick20:36:18
After we're finished, I'll put up the transcript.
DPatrick20:36:27
Click on "Math Jams" from the forum, then click "transcripts"
ccy20:36:42
Do you need to rigorously prove that c=-1/2 works for 5, or is it obvious enough?
DPatrick20:36:55
You'll need some statement at least like "c = -1/2 means a_n = -1/2 for all n".
DPatrick20:37:11
Saying "c = -1/2 works" and nothing else is not sufficient.
runsintights9920:37:40
If solutions are sent in a day late will they not be accepted in all cases?
DPatrick20:38:18
The due date was Tuesday. Unless you have a bona fide excuse, anything submitted after Tuesday is late and will not be accepted.
DPatrick20:39:03
Many of you are asking about alternate solutions. There are indeed alternate solutions to the problems. I encourage you to post them on the message board -- I obviously cannot cover them all here.
DPatrick20:39:21
When we post the "official" solutions on www.usamts.org after the grading, we'll try to post several different solution methods for each problem.
Ubemaya20:39:31
what if you emailed theh entry form?
DPatrick20:39:34
That's fine.
Yuueesceri20:39:46
What if I sent my solution through E-mail 10 minutes for 12 am of Wednesday?
DPatrick20:40:02
If it came before 12 am Pacific Wednesday, it's on time. If it came after, it's late.
Yuueesceri20:40:30
It is Eastern time here
DPatrick20:40:37
Then you're safe since it was only 9 p.m. here.
DPatrick20:40:40
:)
3468Yz20:40:53
What constitutes a bona fide excuse?
DPatrick20:41:12
If you think you have one, email usamts@usamts.org with your case. We decide on a case-by-case basis.
ShockDrake20:41:22
What if I sent a picture, but I didn't realize that I hadn't actually attached the picture to my tex/pdf file? Will it be seen?
DPatrick20:41:44
If there's a picture that's obviously missing from the completed file, we'll try to find it in your email.
DPatrick20:42:17
If it was an honest mistake in submitting a picture, and clearly a picture was there but was forgotten, we'll let you resubmit the picture.
3468Yz20:42:26
Something like "I tried to fax at 2:55 AM EDT but got a busy signal" is not a bona fide excuse, am I right?
DPatrick20:42:51
Correct. If you have a month to work on problems and wait until 5 minutes before it's due to submit it, then we're not going to be very sympathetic to that.
12345678920:43:05
how many points will i get for a solution for number 5 which only finds c=-0.5 and showing that a_n=-0.5 for all n?
DPatrick20:43:27
We haven't set the grading schemes for the problems yet. I would guess 1 or 2, but I don't unilaterally decide these things.
p4fn2w20:44:05
do you encourage using coordinates to solve geometry problem, even if they get really ugly, or do you dock points off for that?
DPatrick20:44:16
We will award full credit for any valid solution.
DPatrick20:44:23
Even if it's ugly.
DPatrick20:44:56
...unless it is poorly or improperly written. (In other words, the math may be correct, but it's not communicated clearly.)
#H34N120:45:12
I'm not sure if this is the question/answer part, but I inserted PNG files in my PDF. I forgot to send teh PNG files themselves in. Is that ok?
DPatrick20:45:43
It should be OK. Again, if things are obviously missing, and it was clearly a good-faith mistake that you forgot an image, then we'll give you a chance to resubmit what we need.
mathnerd31415920:45:51
Does "I punched it into my calculator and it came out right" count as a solution?
DPatrick20:45:58
It depends on the context, but usually no.
DPatrick20:46:31
As there are no more question, I'll end things here. Thanks again for participating!
