| Transcript
for the Math
Jam "2008 AMC 10/12 B Math Jam"
on Feb 29. |
| Math Jam hosted by DPatrick
(Dave Patrick ). |
DPatrick18:58:48
Welcome to the 2008 AMC 10B/12B Math Jam!
DPatrick18:58:58
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.
DPatrick18:59:18
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.
DPatrick18:59:28
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.
DPatrick18:59:41
When I post a problem, I will also post a link to the problem, like so:
DPatrick18:59:58
This allows you to click on the link and view the problem in a separate window as we work through the solution. Depending on your computer's configuration, you may have to hold down the Ctrl key while clicking on the link. If it still doesn't work, try disabling your pop-up browser. If it still doesn't work, sorry, I'm afraid you're out of luck for tonight.
DPatrick19:00:10
Also there will be some images in this session. The images should appear directly in the window, but I will also provide a link to images, like so:
DPatrick19:00:17
DPatrick19:00:46
There are a fair number of students here. As I said, only well-written comments will be passed to the entire group. Please do not take it personally if your comments do not get posted, and definitely do not complain about it.
DPatrick19:01:17
Please also remember that the purpose of this Math Jam is to work through the solutions to AMC problems, and not to merely present the answers. "Working through the solutions" includes at times discussing problem-solving tactics. So please, when a question is posted, do not simply respond with the answer. That's not why we're here. We're going to work through the problems step-by-step, and people who post comments that skip key steps or jump ahead in the problem, without providing explanation or motivation, are going to be ignored.
DPatrick19:01:48
Also, the goal here is not necessarily to present the "best" solution. We'll generally be presenting what I feel are the most "natural" solutions (at least to me). Some of these problems can be solved more quickly that I'm going to do them here, if you happen to know an advanced technique and/or notice a slick trick. Check out the AMC forum on our message board -- some people have posted some really nice solutions to some of these problems.
DPatrick19:02:10
Before we continue, I'd like to remind everyone about an upcoming Art of Problem Solving class.
DPatrick19:02:20
For those of you who will be taking the AIME, we will be hosting a weekend seminar for AIME preparation. This Special AIME Problem Seminar is on Saturday, March 8, and Sunday, March 9 from 3:30 - 6:30 PM ET (12:30 - 3:30 PM PT) each day. During this class we will discuss both general test-taking strategies and specific mathematical tactics for the AIME. We will discuss several problems both from past AIMEs and from other challenging competitions that exemplify some of the most common problem-solving tactics on the AIME.
DPatrick19:02:39
For more information about this class, please see the transcript from yesterday's informational Math Jam at:
http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=224
DPatrick19:03:01
...or you can always view the transcript by going to the "Math Jams" page on our website and clicking "Transcripts".
DPatrick19:03:13
There will also be a transcript of tonight's Math Jam posted as soon as we're done here.
miller4math19:03:24
how much does it cost?
jhangil19:03:24
Are we allowed in even though we do not take the AIME?
DPatrick19:03:31
The cost for the weekend seminar is $65, and it
DPatrick19:03:50
...and it's open to anyone, whether you're taking the AIME or not (though it is most appropriate for students taking the 2008 AIME).
DPatrick19:04:03
Tonight's Math Jam will proceed as follows:
DPatrick19:04:10
We will work the last 5 problems from the AMC 10B, then the last 5 problems from the AMC 12B. (Some of 21-25 on the 10B also appeared on the 12B.) After that, time permitting, I will take requests for some other problems for discussion.
DPatrick19:04:22
Let's get started with #21 from the 2008 AMC 10B:
DPatrick19:04:27
DPatrick19:04:55
Where can we begin?
karatemagic719:05:13
how can you place guys
DPatrick19:05:27
OK, let's just look at the men first. In how many ways can we seat them?
Fermat160119:06:15
2 to choose which chairs will be for men, that time 5! to arrange them = 5!*2 = 240.
karatemagic719:06:15
10*24
DPatrick19:06:30
There are 10 choices for where the first man seats. This fixes which seats will have men.
DPatrick19:06:37
Then there are 4! ways to seat the rest.
DPatrick19:06:57
Or we could do as Fermat1601 suggested, choose which seats will be the men (we have 2 choices), then seat them in 5! ways.
DPatrick19:07:14
It's the same thing, so we have 10*4! = 2*5! = 240 ways to seat the men.
DPatrick19:07:32
One thing to note: this means that the answer must be a multiple of 240. So it's got to be (A), (C), or (E).
DPatrick19:07:42
Now what?
miller4math19:07:53
Then for each man, there are only 2 spots possible for each his wife
karatemagic719:07:59
figure out how to place the women
DPatrick19:08:13
At this point, it might be helpful to make a quick sketch.
DPatrick19:08:23
DPatrick19:08:32
M0,...,M4 are the men and W0,...,W4 are their wives.
BOGTRO19:08:42
then W_0 only has 2 options
DPatrick19:09:11
Yes, if we look for example at W0, we see that there are only 2 possibilities: she can only sit between M1 and M2 or between M3 and M4.
DPatrick19:09:22
Suppose she sits between M1 and M2:
DPatrick19:09:28
d13579d19:10:03
W_3 must be between M_o and M_4
DPatrick19:10:09
Right, we can work through the rest of the wives.
DPatrick19:10:16
W3's position is now fixed. She can only sit between M0 and M4.
DPatrick19:10:23
DPatrick19:10:38
In the same way, the rest of the women are fixed too. W1 must be between M2 and M3, W2 must be between M3 and M4, and W4 must be between M0 and M1.
DPatrick19:10:45
worthawholebean19:10:54
This uniquely determines the seating.
DPatrick19:11:06
Right: once we seat the first wife, the rest of the wives' seats are determined.
BOGTRO19:11:22
so 240*2=480?
DPatrick19:11:39
Correct...recall we had 2 choices for the first wife's seat. So once we seat the men, we only have 2 ways to seat the women.
DPatrick19:11:49
Therefore the answer is 240*2 = 480. Answer (C).
DPatrick19:12:08
On to #22...also a counting problem.
DPatrick19:12:13
DPatrick19:12:44
To start, in how many ways can we arrange the beads?
DPatrick19:13:04
(Forgetting about the neighboring beads condition...in other words, we're getting our denominator.)
BOGTRO19:13:29
6!/3!2!=60
worthawholebean19:13:29
(6C3)(3C2)(1C1)
miller4math19:13:31
there are 6! ways, divided by 3! and 2!
DPatrick19:14:03
Right, that's one way to think of it. There are 6! arrangements of 6 items, but we divide by 3! (for 3 red beads) and 2! (for 2 blue beads), to get 6!/3!2! = 60 arrangements.
DPatrick19:14:19
Alternatively, we can think of having 6 choices for the spot for the blue bead, and then C(5,2) = 10 spots for the white beads. The reds go in the remaining slots.
DPatrick19:14:24
This also gives us 6*10 = 60 total.
DPatrick19:14:42
So now we need to count how many arrangements have no two beads of the same color together.
karatemagic719:15:02
list them?
DPatrick19:15:17
Sure, the numbers are small, so we can probably just brute-force list them
DPatrick19:15:22
What's the best way to begin?
BOGTRO19:15:29
r-r-r- > 3 ways
-r-r-r > 3 ways
r-r--r > 2 ways
r--r-r > 2 ways
(3+3+2+2)/60=10/60=1/6
so C
worthawholebean19:15:29
List possible red bead arrangements then fill in blanks
DPatrick19:15:45
When counting, we usually find it best to deal with the most restrictive condition first.
DPatrick19:15:52
In this problem, the position of the red beads is the most restrictive, because (intuitively) they're the "hardest" to keep apart.
DPatrick19:16:10
We can just list them (as BOGTRO did):
DPatrick19:16:15
R*R*R*
R*R**R
R**R*R
*R*R*R
(The *'s are beads of a different color)
miller4math19:16:27
then you fill in the non-red
DPatrick19:16:45
Exactly: we need to fill in 2 whites and 1 blue in each of these cases.
DPatrick19:16:52
We still have the condition that the two whites can't be next to each other.
bobbob010019:17:15
when 2 stars are together you have to have 1 white and 1 blue
DPatrick19:17:51
Right...in R*R**R and R**R*R, we must put the blue bead as one of the ** positions (otherwise the two whites are together), so we have 2 possibilities in each of these cases.
DPatrick19:18:05
What about R*R*R* and *R*R*R
DPatrick19:18:06
?
bobbob010019:18:23
in the others we have 3
worthawholebean19:18:23
no restrictions on W and B beads
miller4math19:18:27
then you can have any order be placed into them
hadasah19:18:27
3
chenhsi19:18:32
3 cause blue can go in any of them
DPatrick19:18:44
Right, we have no restriction, so the blue bead can go anywhere, and we have 3 possibilities for each.
DPatrick19:18:59
Adding up our cases, that's a total of 3+2+2+3 = 10 "legal" arrangements.
DPatrick19:19:10
So the probability is 10/60 = 1/6. Answer (C).
DPatrick19:19:41
Let's go on to #23, which is also a counting problem (in a way):
DPatrick19:19:46
DPatrick19:19:54
(This is also #16 on the AMC 12B)
DPatrick19:20:14
Wow, it's wordy.
DPatrick19:20:26
What observation makes this a lot simpler?
miller4math19:20:52
the smaller rectangle is (a-2) by (b-2)
kostya19:20:52
the painted part is also half the floot
karatemagic719:20:52
the center area is 1/2 of the total
ChrisBosh19:20:52
the small rectangle is half the area of the big rectangle?
DPatrick19:21:08
Right, you guys have collectively got all the relevant info.
DPatrick19:21:21
The painted part of the floor is half the area of the room, and it's a lot easier to compute than the unpainted area.
DPatrick19:21:33
The painted rectangle is (a-2) x (b-2).
miller4math19:21:51
and then you can make an equation!
miller4math19:21:51
namely 2(a-2)(b-2)=ab
karatemagic719:21:51
ab=2(a-2)(b-2)
DPatrick19:22:10
DPatrick19:22:31
How to we determine how many solutions this has with b>a?
worthawholebean19:22:40
Expand out and try to factor a different way.
DPatrick19:22:57
kostya19:23:19
this simplifies to (a-4)(b-4)=8
karatemagic719:23:19
0=ab-4a-4b+8 8=(a-4)(b-4)
miller4math19:23:19
then you can add 8 to each side!
worthawholebean19:23:19
SFFT!
DPatrick19:23:37
Yes...this is what we like to call around here Simon's Favorite Factoring Trick!
DPatrick19:23:47
DPatrick19:23:57
DPatrick19:24:03
DPatrick19:24:19
The "trick" is to add a constant to the equation so that it's easy to factor.
DPatrick19:24:36
(And Simon is the guy who first starting using it on our message boards.)
DPatrick19:24:48
How many solutions are there to this with b > a > 0 integers?
karatemagic719:25:00
a-4=1 b-4=8 or a-4=2 b-4=4 so the pairs are (5,13) and (6,8)
miller4math19:25:00
then you can break 8 up 2 different ways.. namely 2,4 and 1,8 giving you 6,8, and 5, 12
bobbob010019:25:05
so there is 2 solutions for b>a
johnfn19:25:08
We can just factor 8 and try things to find out
DPatrick19:25:30
Right...we need integers, and the only positive integer factorizations of 8 are 1*8 and 2*4.
DPatrick19:25:40
That's two solutions for (a,b): (5,12) and (6,8).
DPatrick19:26:04
We just need to quickly verify that negative factorizations don't give us a positive a and b, but they don't.
DPatrick19:26:23
(-2 * -4 gives us a=0, b=2, but that's not positive)
Fermat160119:26:44
8, 1 and 4, 2 don't work since b>a
miller4math19:26:44
thus the answer is 2, which is B
DPatrick19:26:49
So there are just two solutions. Answer (B).
DPatrick19:27:13
On to #24, where there's no counting at all!
DPatrick19:27:18
karatemagic719:27:34
draw a picture
infinity4ever_219:27:34
a lot of people got this with a protractor...
DPatrick19:27:43
Yes, I'd start by drawing a picture.
DPatrick19:27:53
DPatrick19:28:11
...and if you brought a ruler and protractor, you could draw an accurate picture, measure angle A, and be done.
DPatrick19:28:39
This is in no way cheating...as long as they let you bring a ruler and proctractor, you might as well use them!
DPatrick19:28:58
But of course we don't need to do that ... there is a solution.
miller4math19:29:20
start by drawing lines and maybe equilateral triangles?
karatemagic719:29:20
draw icsoscoles triangles
kostya19:29:20
then find angle BAC and BCA
DPatrick19:29:49
We've got lots of equal sides, so we can certainly draw in isosceles triangles and maybe even equilateral triangles if we get lucky.
DPatrick19:30:11
For example, we could draw in AC.
This makes ABC isosceles. So BAC = 110/2 = 55.
DPatrick19:30:38
You can go down this road, but this doesn't seem to help much. (At least I didn't get anywhere with it.)
miller4math19:31:28
draw an equilateral triangle at AB, since angle ABC is close to a 60 degree angle
DPatrick19:31:49
Sure...having lots of equal sides might get us to try to construct an equilateral triangle.
DPatrick19:31:57
Let's add point E so that ABE is equilateral.
DPatrick19:32:13
DPatrick19:32:35
EAB is 60. So if we can find DAE then we're done. So let's draw in segment ED.
Fermat160119:32:44
CBE is 10
karatemagic719:32:44
ebc=10
DPatrick19:32:57
miller4math19:33:11
EB is the same size as CB
BOGTRO19:33:11
Isn't EDCB a rhombus?
DPatrick19:33:19
Aha! EBCD is a rhombus...why?
miller4math19:33:46
and angle DEB =170, along with DEC = 10, so you can prove that it is a parallelogram
karatemagic719:33:46
equal sides
worthawholebean19:33:49
Parallelogram with consecutive sides congruent
DPatrick19:34:01
Right. From my picture, I only have 3 sides equal: BE = CB = CD.
DPatrick19:34:06
That's not enough to make a rhombus.
DPatrick19:34:27
But I also know that BE and CD are parallel, since angle C (170) and angle CBE (10) sum to 180.
DPatrick19:34:46
So it's indeed a rhombus.
DPatrick19:34:53
Why does that help?
kostya19:35:26
now all the sides except for AD are equal
kcn2rivers19:35:26
DE is also the same length
madshock19:35:26
DE=EB?
d13579d19:35:26
DE= AE
bubble19:35:26
DEA is isoscleses
chenhsi19:35:26
know measure od DEA
DPatrick19:35:35
Yes, it gives me two important new facts.
DPatrick19:35:38
Let me put the picture up again:
DPatrick19:35:46
DPatrick19:36:01
We know that DE is the same length as all the other marked lengths, so DEA is isosceles.
DPatrick19:36:25
We also have angle DEA. DEB is 170 (it's the same as angle C), and AEB is 60 (from the equilateral triangle), so DEA = 360-170-60 = 130.
miller4math19:36:41
Now we know that DEA =130, and EDA = 25, so then add that to DEC
karatemagic719:36:42
dae=25
DPatrick19:36:54
Exactly...we're home free now. This means that ADE and, more importantly, DAE are each 25 (since AED is isosceles).
DPatrick19:37:05
Add that to the EAB=60 that we had before...
kcn2rivers19:37:12
25+60 = 85
karatemagic719:37:12
85 is answer
Fermat160119:37:12
bad = 60+25 = 85 (c)
bubble19:37:12
25+60=85 C
DPatrick19:37:19
This gives us DAB = DAE + EAB = 25 + 60 = 85. Answer (C).
worthawholebean19:37:39
What are good strategies to try on magic construction problems like these?
DPatrick19:38:01
This was pretty tricky.
DPatrick19:38:15
The equal lengths lead us to try to make isosceles or (even better) equilateral triangles.
DPatrick19:38:33
I started by trying to construct isosceles triangles (by drawing in AC or BD), but didn't get anywhere.
DPatrick19:38:49
So then we have to try to "force" an equilateral triangle somewhere.
DPatrick19:39:20
At that point, it pays to notice the two given angles sum to 240, so when we use 60 of one angle in our equilateral triangle, we have 180 left, and it all kind of falls into place from there.
miller4math19:39:31
yes.. but NEXT TIME BRING PROTRACTERS PEOPLE!
ecommissioner19:39:31
is it a good idea to bring rulers and protractors? because they seem to save a lot of time.
DPatrick19:39:43
Indeed...the AMC lets you bring lots of stuff. It's too your advantage to bring it all.
DPatrick19:40:01
Let's finish the AMC 10B with #25:
DPatrick19:40:10
DPatrick19:40:34
(This is also #20 on the AMC 12B)
DPatrick19:40:44
My first reaction to reading this was "Ick".
Fermat160119:40:53
Let's draw a picture.
karatemagic719:40:53
draw a graph of the garbage trucks progress over time?
kostya19:40:53
i drew a picture
d13579d19:40:53
draw a graph?
chenhsi19:40:53
make a picture
DPatrick19:41:03
Certainly one solution is to draw an accurate picture.
DPatrick19:41:12
Here is what I drew:
DPatrick19:41:21
DPatrick19:41:31
(this graph is sped up to the time when Michael first meets the truck)
DPatrick19:41:44
The straight line is Michael's path. The jagged line is the truck's path. We can see the 5 intersection points clearly.
worthawholebean19:42:10
This type of problem is really easy to mess up.
DPatrick19:42:25
Indeed. So maybe we don't trust our graph; instead we can also carefully reason out why this graph is accurate.
madshock19:42:44
Figure out how much one moves when the other moves from one pail to the next
chenhsi19:42:44
every 50 seconds Michael moves 50 more feet than the truck
DPatrick19:42:50
Probably a good first step is to determine the length of a "cycle" for both Michael and the truck.
DPatrick19:43:05
Michael takes 200/ 5 = 40 seconds to get from one pail to the next.
DPatrick19:43:12
The truck takes 200 / 10 = 20 seconds to get from one pail to the next, but then it spends 30 seconds at the pail, for a total of 50 seconds.
DPatrick19:43:25
So every time Michael reaches the next pail, he's "gained" 10 seconds on the truck.
DPatrick19:43:50
At this point I just started listing stuff out.
DPatrick19:44:03
When Michael arrives at the 2nd pail, the truck will have been sitting at the pail in front of him for 20 seconds.
DPatrick19:44:12
When Michael arrives at the 3rd pail, the truck will have been sitting at the pail in front of him for 10 seconds.
DPatrick19:44:22
When Michael arrives at the 4th pail, the truck will just be arriving at the pail in front of him.
DPatrick19:44:38
(At no point so far has Michael caught up to the truck.)
ecommissioner19:44:43
then on the 5th the truck will be ten seconds behind? and so on?
DPatrick19:44:54
Right...when Michael arrives at the 5th pail, the truck will be halfway to the next pail.
DPatrick19:45:07
So when Michael arrives at the 6th pail, the truck will just be leaving from that same pail. This is the first time Michael meets the truck.
DPatrick19:45:15
(Another way to see this: Michael has traveled 5 pails, which takes him 200 seconds. In 200 seconds the truck goes through 4 complete cycles, so it's just leaving the 6th pail, since it started by just leaving the 2nd pail.)
DPatrick19:45:48
This is also the point at where by graph begins. (It was really hard to read the graph otherwise)
DPatrick19:45:56
...my graph...
DPatrick19:46:08
So now we continue the list...
DPatrick19:46:20
When Michael arrives at the 7th pail, the truck is already there and has been sitting there for 20 seconds. That's their second meeting.
DPatrick19:46:38
When Michael arrives at the 8th pail, the truck is already there and has been sitting there for 10 seconds. So they had to have met a third time when the truck passed Michael on its way between the 7th and 8th pails (since it was already at the 8th pail by the time Michael gets there). Then, when Michael arrives at the 8th pail, it's their FOURTH meeting.
DPatrick19:46:57
When Michael arrives at the 9th pail, the truck is just arriving too. This is their FIFTH meeting.
DPatrick19:47:06
That's it. Michael is now going to pull away. When he gets to the 10th pail, the truck is still halfway between pails 9 and 10. He'll pull away faster as he continues.
DPatrick19:47:19
So there are 5 meetings. Answer (B).
ecommissioner19:47:28
what do you suggest we should do for problems like this? make a graph or make a list? (which is faster? and which is safer?)
DPatrick19:47:43
I found the list better because it's easy to mess up the graph and not realize it.
DPatrick19:48:04
When I wrote out the list I felt like I really understood what was going on and it was hard for me to make a mistake.
Unemployed_219:48:16
safer: both
worthawholebean19:48:18
Do the graph to check, maybe?
DPatrick19:48:38
Right, once you've done a list, and feel like you understand the problem, then you can draw a graph and it will hopefully confirm your understanding.
DPatrick19:48:52
You might have tried to set this up with equations. I found that just reasoning it out as I did above was much simpler.
DPatrick19:49:20
Let's now switch over to the AMC 12B. We'll start with #21.
DPatrick19:49:30
DPatrick19:50:05
The first step is to make sure we're certain what the question means.
DPatrick19:50:18
We're choosing 0 <= a <= 2 and centering circle A at (a,0).
We're choosing 0 <= b <= 2 and centering circle B at (b,1).
And we need to determine the conditions on (a,b) such that they intersect.
DPatrick19:50:23
How can we proceed?
miller4math19:50:48
you can first draw a picture!!
BOGTRO19:50:48
Diagram
DPatrick19:51:14
You can do it algebraically or geometrically. I think this one is just as easy to do algebraically, so I actually didn't draw a picture at this step.
DPatrick19:51:33
What is the condition such that two circles of radius 1 do or don't intersect?
BOGTRO19:52:02
If the centers aren't more then 2 away from each other
redcomet4619:52:02
distance from centers greater than or equal to 2.
DPatrick19:52:18
The circles intersect as long as the centers are less than distance 2 apart.
DPatrick19:52:38
And since the centers are (a,0) and (b,1), we can easily write down the condition.
DPatrick19:53:00
karatemagic719:53:06
|a-b| is less than sqrt(3)
BOGTRO19:53:12
so its like a right triangle
BOGTRO19:53:12
with legs 1 and a-b?
Fermat160119:53:14
the ditance between y=0 and y=1 is 1. We have a right triangle with hypotenuse 2 and one leg 1
DPatrick19:53:22
miller4math19:53:45
then put a and b into a graph
DPatrick19:54:17
Right, now that we have the condition on a and b, we can graph it on the (a,b)-plane to compute the probability.
DPatrick19:54:26
DPatrick19:54:58
The big square has side length 2 (since 0 <= a <= 2 and 0 <= b <= 2), and the shaded area is the allowable region for (a,b) so that |a-b| <= sqrt(3).
worthawholebean19:55:14
What are the coordinates of the important points?
DPatrick19:55:34
The lengths of the long shaded sides are sqrt(3).
DPatrick19:55:54
So what are the sizes are areas of the little white triangles?
miller4math19:56:05
they are at (2-sqrt{3},sqrt{3}) and (sqrt{3},2-sqrt{3})
karatemagic719:56:18
(2-sqrt(3))^2/2
DPatrick19:56:48
Right. Probably the easiest thing to do is to smush the two little white triangle together form a square of side length (2-sqrt(3)).
DPatrick19:56:54
DPatrick19:57:12
That's the region where the circles don't intersect.
BOGTRO19:57:28
so 4-(7-4sqrt3)=-3+4sqrt3 ?
miller4math19:57:30
the darkened area is 3-(7-4sqrt{3})
DPatrick19:57:44
DPatrick19:58:05
DPatrick19:58:51
The next one, #22, was my co-favorite. (#25 was my other favorite, we'll get to it in a bit)
DPatrick19:58:56
DPatrick19:59:27
Where should we start?
Fermat160119:59:44
Let's count the complementary probablility(she isn't able to park)
miller4math19:59:44
find the probability that she isn't able to park
DPatrick20:00:16
Right, that will turn out to be a lot easier, because we can avoid casework (although you could certainly do it the direct way too, it just takes a lot longer).
DPatrick20:00:27
What's the denominator going to be?
lotsofmath20:00:48
16C12
kostya20:00:48
16 choose 12
bubble20:00:48
16C4
Fermat160120:00:48
the denominator is 16C12
DPatrick20:00:55
DPatrick20:01:17
So now we need to count the ways in which the 4 empty spaces are such that there aren't 2 spaces together.
DPatrick20:01:24
How do we do that?
erdogankerem12320:01:46
use balls and urns
erdogankerem12320:01:46
empty spaces are dividers
DPatrick20:02:15
That's the general idea, to take a "balls and urns" mindset...but it's not even that complicated in this case once we make a clever observation.
DPatrick20:02:44
We'll use the "dividers" approach that you just suggested: if there are 4 "isolated" empty spaces, then they divide the 12 cars into 5 groups, and the middle three groups are nonempty.
DPatrick20:02:56
For example:
*CCC*CC*CCCCC*CC
divides the 12 cars into groups of 0, 3, 2, 5, and 2. The two "outside" groups are allowed to be 0, but the three middle groups must be positive.
DPatrick20:03:15
How do we count this?
erdogankerem12320:03:52
cars must be between dividers, so subtract that from tot. number of objects
Fermat160120:03:52
isn't this the same as 13 objects, just the middle can be 0?
DPatrick20:04:00
Right, that's one clever way to do it.
DPatrick20:04:07
Remove a car from the three "middle" groups.
DPatrick20:04:29
Then we're simply arranging 9 cars and 4 *'s (empty space) in any order.
DPatrick20:04:35
DPatrick20:04:51
Another way to count this is to add a "phantom" car to each end.
DPatrick20:04:58
In our example above,
P*CCC*CC*CCCCC*CCP
we have groups of 1,3,2,5,3 cars (a total of 14), with a "phantom" car at each end.
DPatrick20:05:08
So to create such an arrangement, we have to insert 4 "*"'s (empty spaces) into the 13 slots between the cars (the 12 real cars and the 2 phantom cars).
DPatrick20:05:20
DPatrick20:05:33
It's essentially the same argument either way.
DPatrick20:06:09
Fermat160120:06:40
then the answer is 17/28.
miller4math20:06:40
then take 1-11/18
lotsofmath20:06:40
can=17/28
erdogankerem12320:06:40
subtract from 1 to get 17/28
DPatrick20:06:42
So the probability that she can park is 1-(11/28) = 17/28. Answer (E).
worthawholebean20:06:59
Balls and urns is really useful.
DPatrick20:07:14
Yes, this type of counting problem almost always appears on the AMC 12 somewhere.
DPatrick20:07:33
(in this problem, the "balls" are the cars and the "urns" are the slots between empty parking spaces)
DPatrick20:07:47
Let's move on to #23.
DPatrick20:07:52
DPatrick20:08:43
I've seen some clever, "hi-tech" solutions that use number-theory formulas, but we can do this in a low-tech way too.
BOGTRO20:08:51
10^n=2^n*5^n
Fermat160120:08:51
the divisors are of the form 2^a*5^b where a and b are between 0 and n inclusive.
DPatrick20:09:11
DPatrick20:09:25
And what's the log of this?
kostya20:10:01
alog2+blog5
Fermat160120:10:14
log(2^a*5^b) = log(2^a) + log(5^b) = alog(2) + blog(5).
DPatrick20:10:16
DPatrick20:10:31
How can we easily sum these?
kostya20:11:09
use summation then separate
worthawholebean20:11:09
sum all the as and all the bs
DPatrick20:11:19
Right, we can sum the log 2 terms and the log 5 terms separately.
DPatrick20:11:33
There are (n+1) of each value of a from 0 to n (one for each value of b).
DPatrick20:11:44
DPatrick20:12:05
The n+1 factor out front is for the n+1 different values of b.
DPatrick20:12:19
The 0+1+...+n term is for the n+1 different values of a.
DPatrick20:12:38
worthawholebean20:12:51
Factor out the n(n+1)^2.
kostya20:12:55
this mens that n*(n+1)^2/2=792
DPatrick20:13:03
Right, when we put them back together...
DPatrick20:13:11
DPatrick20:13:26
This is supposed to equal 792, so we have n(n+1)^2 = 2(792) = 1584.
kostya20:13:45
n=11
DPatrick20:13:55
Note that the right hand side is between 11^3 and 12^3, so we suspect 11 is the answer, and indeed we check that 11*(144) = 1584.
DPatrick20:14:18
So the answer is (A) 11.
DPatrick20:14:40
Honestly I thought this was a rather boring problem.
DPatrick20:14:50
Let's move on to #24.
DPatrick20:14:57
worthawholebean20:15:27
This looks like it will be ugly at first glance.
DPatrick20:15:37
Yes...to get a feel for it, I set up some notation.
DPatrick20:15:44
DPatrick20:16:00
What do we know about the relationships between these numbers?
redcomet4620:16:15
diagram (roughly)
Fermat160120:16:15
Draw a picture so it's easier to visualize.
miller4math20:16:15
lets try finding some points and drawing some diagrams
DPatrick20:16:44
To be honest, I didn't draw a diagram, because looking at the answer choices (all at least 13), I wasn't convinced that a diagram would help that much...
DPatrick20:16:53
...it would have to be extended a ways to see anything useful.
DPatrick20:17:24
However, this was probably not good intuition on my part.
DPatrick20:17:36
There are clever solutions that you can "see" from drawing the right diagram.
DPatrick20:17:43
But I went with an algebraic approach.
DPatrick20:18:29
My thought process (and that's part of what we're doing here...talking about the problem-solving process) was that I'd need to find the A's and B's algebraically to get all the way out to n=13.
DPatrick20:19:02
Fermat160120:19:07
why 13?
miller4math20:19:07
why n=13?
DPatrick20:19:14
Because 13 is the smallest answer choice.
DPatrick20:19:59
We can compare the x- and y- coordinates of the A's and B's since we have all those equilateral triangles.
BOGTRO20:20:09
But how did you know that you shouldn't work to get 15?
ecommissioner20:20:09
is the smallest answer choice always correct?
DPatrick20:20:13
Sorry, I'm not making myself clear.
DPatrick20:20:27
I was thinking that I'd have to compute out to at least 13.
DPatrick20:20:43
And that in order to do that, I'd want to know algebraically what a_n is.
DPatrick20:21:06
(It turns out, in the end, that it's not the case, but I'll come back to that later...)
DPatrick20:21:52
DPatrick20:21:58
What equations can I write?
Fermat160120:22:41
| a_n - a_(n-1) | other distances
DPatrick20:22:47
Right, that's the bottom side length.
DPatrick20:22:54
And the height is sqrt(3)/2 times that.
DPatrick20:23:06
kostya20:23:15
a_(n-1)+a_n=2b_n^2
DPatrick20:23:35
Right, we also have the x-coordinate of B_n lying halfway between A_{n-1} and A_n.
DPatrick20:23:51
Fermat160120:24:19
We can try to combine the two equatoins in some way.
kostya20:24:19
square the first equation
DPatrick20:24:24
DPatrick20:24:34
miller4math20:25:07
my approach was to find small values of an and see if there was a pattern
DPatrick20:25:25
Yes, at this point I didn't like the idea of trying to solve this explicitly, but now I feel like I can quickly calculate an's.
DPatrick20:25:37
a_0 = 0.
DPatrick20:25:47
DPatrick20:26:02
DPatrick20:26:14
DPatrick20:26:39
We might see the pattern now, but one more makes it clearer:
miller4math20:26:45
then you see that their differences are 2/3, 4/3, 6/3, 8/3 .........
DPatrick20:26:49
DPatrick20:27:04
The sequence is 0, 2/3, 2, 4, 20/3, ...
Better yet: the sequence is 0/3, 2/3, 6/3, 12/3, 20/3, ...
DPatrick20:27:26
It's just 2/3 times the triangular numbers 0,1,3,6,10,...
DPatrick20:27:46
(And if you don't see this right away, looking at the successive differences like miller4math suggested is helpful.)
DPatrick20:27:54
(Somebody must have been a fan of triangular numbers, since the last problem involved them too.)
Fermat160120:28:19
Great! so a_n = (2/3)n(n+1)/2
DPatrick20:28:45
You could prove it by induction if you like... but on the contest I'd probably take it on faith that it works.
DPatrick20:28:57
BOGTRO20:29:39
n=17
kostya20:29:39
17 or 18
miller4math20:29:40
n is about 17, since 17^2 is about 300
DPatrick20:29:44
300 lies between 17^2 and 18^2, and indeed 17*18 = 306, so the answer is (C) 17.
DPatrick20:30:38
There is a really nice diagram-based solution to this problem posted on the AMC message board (posted by rcv, who's here in attendance tonight).
DPatrick20:30:53
And there are other solutions too.
DPatrick20:31:11
Let's finish with another of my favorites, #25:
DPatrick20:31:18
DPatrick20:32:08
We'd of course want to sketch a crude picture (maybe not necessarily to scale):
DPatrick20:32:15
DPatrick20:32:34
Before we deal with the bisectors, is there anything we could do?
worthawholebean20:33:05
Draw in altitudes.
chenhsi20:33:05
draw two altitudes up from A and B to DC
madshock20:33:08
find the trapezoid's area
worthawholebean20:33:16
Draw in altitudes to get the area of the whole trapezoid.
DPatrick20:33:34
Right...it's probably going to be useful to have the height of the trapezoid, and thus also its area.
miller4math20:33:42
shouldn't you draw the bisectors too?
DPatrick20:33:54
we'll worry about them in a bit...right now they would clutter the picture.
DPatrick20:34:13
Let's call the height h. The area of the trapezoid will then be 15h.
DPatrick20:34:16
How do we find h?
DPatrick20:35:21
karatemagic720:35:37
sqrrt(49-h^2)+sqrrt(25-h^2)=8
kostya20:35:37
49-h^2=x^2
25-h^2=(8-x)^2
worthawholebean20:35:43
Pythagorean Theorem - draw one from A and B each.
DPatrick20:36:10
If we drop perpendiculars from A and B, what's "left over" of CD has length 19- 11 = 8.
DPatrick20:36:33
So we can apply the Pythagorean Theorem on the two little right triangles to find h.
DPatrick20:37:03
Kostya's way is slightly easier...let me label the points first.
DPatrick20:37:22
DPatrick20:37:42
We have AE=BF=h, ED = x, and CF = 8-x.
DPatrick20:38:11
DPatrick20:38:37
(I might have my x backwards from what I just said...sorry.)
DPatrick20:38:44
x = CF, 8-x = ED.
d13579d20:38:50
subtract equations
DPatrick20:39:22
Right, we just subtract them, the quadratics all cancel, and we have 64 - 16x = 24.
DPatrick20:39:34
This gives x = 5/2.
DPatrick20:39:51
And so h = 5*sqrt(3)/2...and we also note in passing that BCF is a 30-60-90 triangle!
DPatrick20:40:04
miller4math20:40:35
cut off those triangles you just drew! and stick them together to make another triangle.
miller4math20:40:35
you can use herons formula on that new triangle, and then find h!
DPatrick20:41:06
That's another way to find the height: you can smush AED and BCF together to form a 5-7-8 triangle, and from there you can use Heron's to find the area.
DPatrick20:41:17
...or Law of Cosine to find angle c = 60 degrees.
DPatrick20:41:37
(This problem has a lot of different approaches!)
DPatrick20:41:54
So we've got the height...now we can draw in the bisectors:
DPatrick20:41:59
DPatrick20:42:24
I gave a little bit away by extending AP and BQ to R and S. Why is that useful?
BOGTRO20:42:47
New Trapezoid?
DPatrick20:43:01
We've got a nice new trapezoid ABSR that's part of our hexagon.
DPatrick20:43:05
Do we know RS?
DPatrick20:43:41
My drawing is deceptive in a rather important feature.
DPatrick20:43:56
Do we know anything about angle APD?
karatemagic720:44:16
90?
madshock20:44:16
90 degrees
DPatrick20:44:18
How come?
DPatrick20:45:10
It's right! But why?
karatemagic720:45:21
AB and DC are parallel
madshock20:45:27
P comes from line that bisects DAB and CDA, which are supplementary
DPatrick20:45:41
Right. Angles A and D (in the original trapezoid) sum to 180, so half of their angles sum to 90.
DPatrick20:45:51
So DPA is 90 (so that the angles in APD sum to 180).
DPatrick20:46:20
So what? Why is it helpful that DPA is a right angle?
DPatrick20:46:28
DPatrick20:46:35
(same picture, just a repost)
BOGTRO20:47:14
DRP and APD are similar?
DPatrick20:47:21
Even better...they're congruent.
DPatrick20:47:36
They're right triangles with the same leg and the same angle (half of D).
DPatrick20:47:44
And the upshot is that DR = 7.
DPatrick20:48:05
We can do the same thing on the other side, so that CS = 6.
DPatrick20:48:09
...oops, bad typo, CS = 5.
DPatrick20:48:20
So RS = 19 - (7+5) = 7.
DPatrick20:48:54
We're really close now. How can we express the area of the hexagon in terms of the areas of ABCD (which is 15h) and ABSR (which is 9h)?
kostya20:49:45
(15h-9h)/2+9h
karatemagic720:49:45
(absr-abcd)/2+absr=12h
worthawholebean20:49:48
hexagon = ABSR+(ABCD-ABSR)/2
DPatrick20:49:51
Exactly.
DPatrick20:50:20
Our hexagon is the inner hexagon (which has area 9h), plus half of the outside triangles (which have total area 15h-9h = 6h, so that adds another 3h).
DPatrick20:50:30
So the area of the hexagon is 12h.
DPatrick20:50:38
DPatrick20:51:03
As I said, there are lots of ways to work through this problem...you can see some others (and maybe post your own!) on the message board.
madshock20:51:18
so we DID need h!
DPatrick20:51:35
Yeah, Richard Rusczyk and I talked about this problem a little this afternoon, and we didn't see a way to do it without finding the height.
DPatrick20:51:46
We found lots of ways to find h, then lots of ways to finish once we had h.
ecommissioner20:51:52
why did you like this problem?
DPatrick20:52:06
Mainly because it's interesting and there are lots of different, non-trivial ways to proceed with it.
DPatrick20:52:14
...and I don't usually like geometry! :)
DPatrick20:52:44
I've already run almost 25 minutes over, so I'm afraid we'll have to end it there.
DPatrick20:52:52
Most of the problems are posted on our message board.
DPatrick20:53:01
Thanks for coming!
kostya20:53:05
is the WIKI open??
DPatrick20:53:08
It should be.
worthawholebean20:53:25
I'd just like to say that I've posted all the problems on the 12B to the Contests section and am doing the 10B right now.
madshock20:53:28
whats the url for the WIKI?
DPatrick20:53:37
Just click "AoPSWiki" on the left-side menu of the website.
DPatrick20:53:59
Then just search for "AMC 10" or "AMC 12"
miller4math20:54:26
is AMC10A finished yet?
DPatrick20:54:58
The 10A/12A's have been scored and the cutoffs for the AIME were 117 and 97.5 respectively.
DPatrick20:55:23
The AMC score reports will be going out and your school should receive an email next week.
miller4math20:55:29
oh.. i meant on the wiki
DPatrick20:55:38
Don't know...what goes up on the wiki is up to the users of the wiki.
DPatrick20:55:54
Have a nice weekend and happy Leap Day! :)
