2016 AIME II Math Jam
Go back to the Math Jam ArchiveAoPS instructors discuss all 15 problems of the 2016 AIME II.
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Facilitator: Dave Patrick
DPatrick
2016-03-18 19:00:08
Welcome to the 2016 AIME II Math Jam!
Welcome to the 2016 AIME II Math Jam!
DPatrick
2016-03-18 19:00:14
I'm Dave Patrick, and I'll be leading our discussion tonight.
I'm Dave Patrick, and I'll be leading our discussion tonight.
DPatrick
2016-03-18 19:00:22
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.
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
2016-03-18 19:00:29
The classroom is moderated, meaning that students can type into the classroom, but these comments will not go directly into the room. These comments go to the instructors, who may choose to share your comments with the room.
The classroom is moderated, meaning that students can type into the classroom, but these comments will not go directly into the room. These comments go to the instructors, who may choose to share your comments with the room.
DPatrick
2016-03-18 19:00:43
This helps keep the session 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.
This helps keep the session 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.
DPatrick
2016-03-18 19:01:11
There are a lot of students here! As I said, only a relatively small fraction of the well-written comments will be passed to the entire group. Please do not take it personally if your responses do not get posted, and please do not complain about it. I expect this Math Jam to be much larger than our typical class, so please be patient with me---there will be quite a few of you here tonight!!
There are a lot of students here! As I said, only a relatively small fraction of the well-written comments will be passed to the entire group. Please do not take it personally if your responses do not get posted, and please do not complain about it. I expect this Math Jam to be much larger than our typical class, so please be patient with me---there will be quite a few of you here tonight!!
DPatrick
2016-03-18 19:01:28
Also, we won't be going through the math quite as thoroughly as we do in our classes -- I can't teach all the necessary material for every problem as we go. Another difference between tonight and our regular online classes is that it is very unlikely that we'll be able to answer every single question you ask. We always to try do so in our regular online classes, but we have a large number of students tonight! So, please go ahead and ask questions, but also please understand if we aren't able to answer them all!
Also, we won't be going through the math quite as thoroughly as we do in our classes -- I can't teach all the necessary material for every problem as we go. Another difference between tonight and our regular online classes is that it is very unlikely that we'll be able to answer every single question you ask. We always to try do so in our regular online classes, but we have a large number of students tonight! So, please go ahead and ask questions, but also please understand if we aren't able to answer them all!
DPatrick
2016-03-18 19:01:44
We do have two teaching assistants with us tonight to help answer your questions:
We do have two teaching assistants with us tonight to help answer your questions:
DPatrick
2016-03-18 19:01:55
Ben (hailstone) has a B.S. in Math from Carnegie Mellon University, with minors in Music, Philosophy, and Film Studies, and has an M.S. in Math from the University of Washington. He has spent almost as many hours working on his juggling and unicycling.
Ben (hailstone) has a B.S. in Math from Carnegie Mellon University, with minors in Music, Philosophy, and Film Studies, and has an M.S. in Math from the University of Washington. He has spent almost as many hours working on his juggling and unicycling.
DPatrick
2016-03-18 19:02:20
Ben and I went to college together, and in particular were in the killer problem solving class at CMU.
Ben and I went to college together, and in particular were in the killer problem solving class at CMU.
DPatrick
2016-03-18 19:02:29
Alina (laughinghead505) joined AoPS in 2006 as a young, impressionable mathematician. She is an alumna of MathPath, Canada/USA Mathcamp, the Illinois Math and Science Academy, and MIT. Currently, she is a graduate student in physics at the University of Illinois at Urbana-Champaign. In her scraps of free time, Alina enjoys eating and making sweets, reading science fiction, crafting, and exploring urban areas.
Alina (laughinghead505) joined AoPS in 2006 as a young, impressionable mathematician. She is an alumna of MathPath, Canada/USA Mathcamp, the Illinois Math and Science Academy, and MIT. Currently, she is a graduate student in physics at the University of Illinois at Urbana-Champaign. In her scraps of free time, Alina enjoys eating and making sweets, reading science fiction, crafting, and exploring urban areas.
hailstone
2016-03-18 19:02:37
Hi all!
Hi all!
DPatrick
2016-03-18 19:03:04
Ben or Alina can answer questions by whispering to you or by opening a window with you to chat 1-on-1. However, due to the large size of the session tonight, they may not be able to get to you right away (or at all). Repeating your question over and over is more likely to annoy us than to get it answered faster, so please, just ask your question once and be patient, and please understand that we may not be able to answer all the questions tonight.
Ben or Alina can answer questions by whispering to you or by opening a window with you to chat 1-on-1. However, due to the large size of the session tonight, they may not be able to get to you right away (or at all). Repeating your question over and over is more likely to annoy us than to get it answered faster, so please, just ask your question once and be patient, and please understand that we may not be able to answer all the questions tonight.
DPatrick
2016-03-18 19:03:34
Please also remember that the purpose of this Math Jam is to work through the solutions to AIME problems, and not to merely present the answers. "Working through the solutions" includes discussing problem-solving tactics. Also on occasion we may stop to prove things that you wouldn't necessary need to prove while doing the contest. So please, when a question is posted, do not simply respond with the final answer. That's not why we're here. We're going to work through the problems step-by-step.
Please also remember that the purpose of this Math Jam is to work through the solutions to AIME problems, and not to merely present the answers. "Working through the solutions" includes discussing problem-solving tactics. Also on occasion we may stop to prove things that you wouldn't necessary need to prove while doing the contest. So please, when a question is posted, do not simply respond with the final answer. That's not why we're here. We're going to work through the problems step-by-step.
DPatrick
2016-03-18 19:03:51
Let's get started! We're going to work through all 15 problems from the 2016 AIME II, in order.
Let's get started! We're going to work through all 15 problems from the 2016 AIME II, in order.
DPatrick
2016-03-18 19:03:57
1. Initially Alex, Betty, and Charlie had a total of 444 peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had form a geometric progression. Alex eats 5 of his peanuts, Betty eats 9 of her peanuts, and Charlie eats 25 of his peanuts. Now the three numbers of peanuts that each person has form an arithmetic progression. Find the number of peanuts Alex had initially.
1. Initially Alex, Betty, and Charlie had a total of 444 peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had form a geometric progression. Alex eats 5 of his peanuts, Betty eats 9 of her peanuts, and Charlie eats 25 of his peanuts. Now the three numbers of peanuts that each person has form an arithmetic progression. Find the number of peanuts Alex had initially.
DPatrick
2016-03-18 19:04:26
(You'll notice that I'll always put the current problem at the top. You can resize that top area by dragging the horizontal bar separating it from the discussion.)
(You'll notice that I'll always put the current problem at the top. You can resize that top area by dragging the horizontal bar separating it from the discussion.)
DPatrick
2016-03-18 19:05:33
We're not given too much info, but we are given information about the total number of peanuts. Which progression is it going to be easier to use that total in?
We're not given too much info, but we are given information about the total number of peanuts. Which progression is it going to be easier to use that total in?
whatsgucci17
2016-03-18 19:05:58
arithmetic
arithmetic
ilikepie2003
2016-03-18 19:05:58
arithmetic
arithmetic
mathlover3737
2016-03-18 19:05:58
arithmetic
arithmetic
blue8931
2016-03-18 19:05:58
arithmetic
arithmetic
celestialphoenix3768
2016-03-18 19:05:58
arithmetic
arithmetic
Liopleurodon
2016-03-18 19:05:58
Arithmetic
Arithmetic
maverick8
2016-03-18 19:05:58
arithmetic
arithmetic
DPatrick
2016-03-18 19:06:06
Yeah, the arithmetic progression. The total of an arithmetic progression is just the median term times the number of terms.
Yeah, the arithmetic progression. The total of an arithmetic progression is just the median term times the number of terms.
maverick8
2016-03-18 19:06:12
3b=444-5-9-25
3b=444-5-9-25
DPatrick
2016-03-18 19:06:30
Exactly. If $b$ is the middle term of the arithmetic sequence, then $3b$ is the total.
Exactly. If $b$ is the middle term of the arithmetic sequence, then $3b$ is the total.
DPatrick
2016-03-18 19:06:38
5+9+25 = 39 peanuts total were eaten to get to our arithmetic progression, so there are 444-39 = 405 peanuts remaining.
5+9+25 = 39 peanuts total were eaten to get to our arithmetic progression, so there are 444-39 = 405 peanuts remaining.
StoryScene
2016-03-18 19:06:52
b=135
b=135
whatsgucci17
2016-03-18 19:06:52
3b=405, b-135
3b=405, b-135
mattpi
2016-03-18 19:06:58
so 3b=405 => b=135
so 3b=405 => b=135
DPatrick
2016-03-18 19:07:08
Right. Our second progression is an arithmetic progression with total 405, so the middle term is 405/3 = 135.
Right. Our second progression is an arithmetic progression with total 405, so the middle term is 405/3 = 135.
mathlover3737
2016-03-18 19:07:23
135-a, 135, 135+a
135-a, 135, 135+a
DPatrick
2016-03-18 19:07:44
Indeed, more specifically, we can write the three terms as $135-d, 135, 135+d$ for some positive integer $d$. (It's traditional to use $d$ for "difference".)
Indeed, more specifically, we can write the three terms as $135-d, 135, 135+d$ for some positive integer $d$. (It's traditional to use $d$ for "difference".)
DPatrick
2016-03-18 19:07:56
Now what?
Now what?
AkashD
2016-03-18 19:08:25
Geometric series
Geometric series
Liopleurodon
2016-03-18 19:08:25
Do the geometric sequence now
Do the geometric sequence now
blue8931
2016-03-18 19:08:27
focus on the geometric progression
focus on the geometric progression
DPatrick
2016-03-18 19:08:39
Right...specifically, what should we do?
Right...specifically, what should we do?
DPatrick
2016-03-18 19:09:25
If $135-d$, $135$, $135+d$ is the arithmetic sequence after they eat some peanuts, then what was the original geometric sequence?
If $135-d$, $135$, $135+d$ is the arithmetic sequence after they eat some peanuts, then what was the original geometric sequence?
mathlover3737
2016-03-18 19:09:53
140-d, 144, 160+d
140-d, 144, 160+d
blue8931
2016-03-18 19:09:53
140-d, 144, 160+d
140-d, 144, 160+d
Liopleurodon
2016-03-18 19:09:53
140-d,144,160+d
140-d,144,160+d
qwerty733
2016-03-18 19:09:53
140-d, 144, 160+d
140-d, 144, 160+d
Peggy
2016-03-18 19:09:59
140-d, 144, 160+d
140-d, 144, 160+d
beanielove2
2016-03-18 19:09:59
$140 - d, 144, 160 + d.$
$140 - d, 144, 160 + d.$
DPatrick
2016-03-18 19:10:08
Right. Adding back the peanuts that were eaten, our original terms are $140-d, 144, 160+d$.
Right. Adding back the peanuts that were eaten, our original terms are $140-d, 144, 160+d$.
jwlw2014
2016-03-18 19:10:20
B^2=AC
B^2=AC
tarzanjunior
2016-03-18 19:10:20
$b^2=ac$
$b^2=ac$
DPatrick
2016-03-18 19:10:28
Aha, good: in a three-term geometric sequence, the outer two terms multiply to give the square of the middle term.
Aha, good: in a three-term geometric sequence, the outer two terms multiply to give the square of the middle term.
rlzhang
2016-03-18 19:10:36
(140-d)(160+d)=144^2
(140-d)(160+d)=144^2
DPatrick
2016-03-18 19:10:40
That is, $(140-d)(160+d) = 144^2$.
That is, $(140-d)(160+d) = 144^2$.
dragoncharm2b
2016-03-18 19:11:40
A square of a square. . .can we exploit that to make the arithmetic easier?
A square of a square. . .can we exploit that to make the arithmetic easier?
DPatrick
2016-03-18 19:11:49
True: you might be able to solve this "by inspection" by trying to make $140-d$ and $160+d$ be factors of $144^2$.
True: you might be able to solve this "by inspection" by trying to make $140-d$ and $160+d$ be factors of $144^2$.
mattpi
2016-03-18 19:11:51
We could plug into the quadratic formula
We could plug into the quadratic formula
DPatrick
2016-03-18 19:11:58
Alternatively, this quadratic simplifies to $d^2 + 20d - 1664 = 0$.
Alternatively, this quadratic simplifies to $d^2 + 20d - 1664 = 0$.
DPatrick
2016-03-18 19:12:15
Since we're looking for integer solutions, you might notice the factorization $(d-32)(d+52) = 0$, or you can use the quadratic formula (and some careful arithmetic) to see that $d=32$ is the only positive solution.
Since we're looking for integer solutions, you might notice the factorization $(d-32)(d+52) = 0$, or you can use the quadratic formula (and some careful arithmetic) to see that $d=32$ is the only positive solution.
maverick8
2016-03-18 19:12:24
$ d=32 $
$ d=32 $
AOPS12142015
2016-03-18 19:12:24
D=32
D=32
DPatrick
2016-03-18 19:12:34
So what's the final answer?
So what's the final answer?
24iam24
2016-03-18 19:12:51
so ans is 140-32=108
so ans is 140-32=108
beanielove2
2016-03-18 19:12:51
$108.$
$108.$
rechyyy
2016-03-18 19:12:51
108
108
blue8931
2016-03-18 19:12:51
so alex originally had 108
so alex originally had 108
IsabeltheCat
2016-03-18 19:12:51
so alex initially had 140-32=108
so alex initially had 140-32=108
ilikepie2003
2016-03-18 19:12:51
140-32=108
140-32=108
DPatrick
2016-03-18 19:12:58
The smallest term in the original sequence is $140 - d = 140 - 32 = \boxed{108}$.
The smallest term in the original sequence is $140 - d = 140 - 32 = \boxed{108}$.
StoryScene
2016-03-18 19:13:03
Quick question, how are we sure that 135 is Betty's though?
Quick question, how are we sure that 135 is Betty's though?
DPatrick
2016-03-18 19:13:26
That's an excellent point -- they might not necessarily have the same ordering after eating the peanuts.
That's an excellent point -- they might not necessarily have the same ordering after eating the peanuts.
DPatrick
2016-03-18 19:13:39
A quick check of our final answer is not a bad idea, to make sure this is the case.
A quick check of our final answer is not a bad idea, to make sure this is the case.
DPatrick
2016-03-18 19:14:06
The original amounts turned out to be 108, 144, and 192, which is indeed geometric.
The original amounts turned out to be 108, 144, and 192, which is indeed geometric.
DPatrick
2016-03-18 19:14:25
The final amounts then turned out to 103, 135, 167, which is indeed arithmetic.
The final amounts then turned out to 103, 135, 167, which is indeed arithmetic.
DPatrick
2016-03-18 19:14:32
So it worked out.
So it worked out.
DPatrick
2016-03-18 19:14:48
Onwards!
Onwards!
DPatrick
2016-03-18 19:14:52
2. There is a 40% chance of rain on Saturday and a 30% chance of rain on Sunday. However, it is twice as likely to rain on Sunday if it rains on Saturday than if it does not rain on Saturday. The probability that it rains at least one day this weekend is $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Find $a+b$.
2. There is a 40% chance of rain on Saturday and a 30% chance of rain on Sunday. However, it is twice as likely to rain on Sunday if it rains on Saturday than if it does not rain on Saturday. The probability that it rains at least one day this weekend is $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Find $a+b$.
DPatrick
2016-03-18 19:15:09
Ooh, conditional probability! My favorite!
Ooh, conditional probability! My favorite!
kbird
2016-03-18 19:15:18
This one was confusing.
This one was confusing.
DPatrick
2016-03-18 19:15:27
It is a little confusing. Let me introduce a little notation to make this problem easier to talk about.
It is a little confusing. Let me introduce a little notation to make this problem easier to talk about.
DPatrick
2016-03-18 19:15:43
As you may know, $P(X)$ is shorthand for the probability that $X$ happens.
As you may know, $P(X)$ is shorthand for the probability that $X$ happens.
DPatrick
2016-03-18 19:15:50
So we're given that $P(\text{rain Sat}) = \dfrac25$ and $P(\text{rain Sun}) = \dfrac{3}{10}$.
So we're given that $P(\text{rain Sat}) = \dfrac25$ and $P(\text{rain Sun}) = \dfrac{3}{10}$.
DPatrick
2016-03-18 19:16:01
Moreover, $P(X\;|\;Y)$ is shorthand for the probability that $X$ happens, assuming that $Y$ happens.
Moreover, $P(X\;|\;Y)$ is shorthand for the probability that $X$ happens, assuming that $Y$ happens.
DPatrick
2016-03-18 19:16:09
So what does the second sentence tell us?
So what does the second sentence tell us?
mathlover3737
2016-03-18 19:16:56
P(rain Sun | rain Sat) = 2 * P(rain Sun | no rain Sat)
P(rain Sun | rain Sat) = 2 * P(rain Sun | no rain Sat)
roseylily
2016-03-18 19:16:56
P(rain Sun | rain Sat) = 2P(rain Sun | no rain Sat)
P(rain Sun | rain Sat) = 2P(rain Sun | no rain Sat)
DPatrick
2016-03-18 19:17:09
Right. We have $P(\text{rain Sun}\;|\;\text{rain Sat}) = 2P(\text{rain Sun}\;|\;\text{no rain Sat})$.
Right. We have $P(\text{rain Sun}\;|\;\text{rain Sat}) = 2P(\text{rain Sun}\;|\;\text{no rain Sat})$.
DPatrick
2016-03-18 19:17:27
Let's set $p = P(\text{rain Sun} \;|\; \text{rain Sat})$ so we don't have to keep writing it all out.
Let's set $p = P(\text{rain Sun} \;|\; \text{rain Sat})$ so we don't have to keep writing it all out.
DPatrick
2016-03-18 19:17:57
How do we express the probability that it rains on Sun, in terms of what happens on Saturday?
How do we express the probability that it rains on Sun, in terms of what happens on Saturday?
jwlw2014
2016-03-18 19:18:58
P(Sun)=.4*p+.6*p/2
P(Sun)=.4*p+.6*p/2
qwerty733
2016-03-18 19:19:04
2/5 p + 3/10 p
2/5 p + 3/10 p
DPatrick
2016-03-18 19:19:13
That's right -- let's see where this comes from.
That's right -- let's see where this comes from.
DPatrick
2016-03-18 19:19:21
It either rains on Saturday (with probability $\dfrac25$) or it doesn't (with probability $\dfrac35$).
It either rains on Saturday (with probability $\dfrac25$) or it doesn't (with probability $\dfrac35$).
DPatrick
2016-03-18 19:19:32
If it does rain on Saturday, then the probability of rain on Sunday is $p$.
If it doesn't rain on Saturday, then the probability of rain on Sunday is $\dfrac12p$.
If it does rain on Saturday, then the probability of rain on Sunday is $p$.
If it doesn't rain on Saturday, then the probability of rain on Sunday is $\dfrac12p$.
DPatrick
2016-03-18 19:20:02
So when we combine these two cases, we get $P(\text{rain Sun}) = \dfrac25 \cdot p + \dfrac35 \cdot \dfrac12p$.
So when we combine these two cases, we get $P(\text{rain Sun}) = \dfrac25 \cdot p + \dfrac35 \cdot \dfrac12p$.
whatsgucci17
2016-03-18 19:20:22
.7p = P(Sun)
.7p = P(Sun)
DPatrick
2016-03-18 19:20:43
Right: this simpifies to $P(\text{rain Sun}) = \dfrac{7}{10}p$.
Right: this simpifies to $P(\text{rain Sun}) = \dfrac{7}{10}p$.
IsabeltheCat
2016-03-18 19:21:02
so p = 3/7
so p = 3/7
maverick8
2016-03-18 19:21:02
$ p= \frac {3}{7} $
$ p= \frac {3}{7} $
DPatrick
2016-03-18 19:21:17
Right -- the problem told us that $P(\text{rain Sun}) = \dfrac{3}{10}$.
Right -- the problem told us that $P(\text{rain Sun}) = \dfrac{3}{10}$.
DPatrick
2016-03-18 19:21:27
So $\dfrac{3}{10} = \dfrac{7}{10}p$, so $p = \dfrac{3}{7}$.
So $\dfrac{3}{10} = \dfrac{7}{10}p$, so $p = \dfrac{3}{7}$.
DPatrick
2016-03-18 19:21:42
So far, so good. How do we finish the problem from here?
So far, so good. How do we finish the problem from here?
IsabeltheCat
2016-03-18 19:22:25
so there's a 0.4 chance it rains on saturday, + 0.6*(3/7) chance it rains only on sunday
so there's a 0.4 chance it rains on saturday, + 0.6*(3/7) chance it rains only on sunday
blue8931
2016-03-18 19:22:34
complementary probability
complementary probability
Liopleurodon
2016-03-18 19:22:34
Find the probability of it not raining anyday?
Find the probability of it not raining anyday?
DPatrick
2016-03-18 19:22:44
Right, there are basically two ways to finish.
Right, there are basically two ways to finish.
DPatrick
2016-03-18 19:22:55
One way to is compute the probability of at least one day of rain directly.
One way to is compute the probability of at least one day of rain directly.
DPatrick
2016-03-18 19:23:05
To finish directly: if it rains on Saturday (with probability $\dfrac25$), then we're happy. Otherwise, if it doesn't rain on Saturday (with probability $\dfrac35$), we need it to rain on Sunday (with probability $\dfrac12p = \dfrac{3}{14}$).
To finish directly: if it rains on Saturday (with probability $\dfrac25$), then we're happy. Otherwise, if it doesn't rain on Saturday (with probability $\dfrac35$), we need it to rain on Sunday (with probability $\dfrac12p = \dfrac{3}{14}$).
DPatrick
2016-03-18 19:23:36
So the probability of rain on either day is $\dfrac25 + \dfrac35 \cdot \dfrac{3}{14} = \dfrac{37}{70}$.
So the probability of rain on either day is $\dfrac25 + \dfrac35 \cdot \dfrac{3}{14} = \dfrac{37}{70}$.
DPatrick
2016-03-18 19:24:02
Or -- we could compute $P(\text{no rain at all})$.
Or -- we could compute $P(\text{no rain at all})$.
blue8931
2016-03-18 19:24:12
or $1 - \frac{3}{5} * \frac{11}{14} = \frac{37}{70}$
or $1 - \frac{3}{5} * \frac{11}{14} = \frac{37}{70}$
Beast144
2016-03-18 19:24:20
P(No rain Sat) * P(No rain Sun) = 3/5 * 11/14 = 33/70
P(No rain Sat) * P(No rain Sun) = 3/5 * 11/14 = 33/70
DPatrick
2016-03-18 19:24:46
Right. It doesn't rain on Saturday with probability $\dfrac35$, and then given that, it also doesn't rain on Sunday with probability $\dfrac{11}{14}$.
Right. It doesn't rain on Saturday with probability $\dfrac35$, and then given that, it also doesn't rain on Sunday with probability $\dfrac{11}{14}$.
DPatrick
2016-03-18 19:25:14
So no rain at all is $\dfrac35 \cdot \dfrac{11}{14} = \dfrac{33}{70}$, and hence some rain has probability $1 - \dfrac{33}{70} = \dfrac{37}{70}$.
So no rain at all is $\dfrac35 \cdot \dfrac{11}{14} = \dfrac{33}{70}$, and hence some rain has probability $1 - \dfrac{33}{70} = \dfrac{37}{70}$.
dragoncharm2b
2016-03-18 19:25:25
so the answer is 107
so the answer is 107
jwlw2014
2016-03-18 19:25:25
107
107
DPatrick
2016-03-18 19:25:31
Either way, this is in lowest terms, so the final answer is $37 + 70 = \boxed{107}$.
Either way, this is in lowest terms, so the final answer is $37 + 70 = \boxed{107}$.
DPatrick
2016-03-18 19:26:12
If you'd like to learn more about conditional probability, may I recommend the book Intermediate Counting & Probability by (ahem) David Patrick.
If you'd like to learn more about conditional probability, may I recommend the book Intermediate Counting & Probability by (ahem) David Patrick.
DPatrick
2016-03-18 19:26:20
On to #3:
On to #3:
DPatrick
2016-03-18 19:26:25
3. Let $x$, $y$, and $z$ be real numbers satisfying the system\begin{align*}
\log_2(xyz - 3 + \log_5x) &= 5 \\
\log_3(xyz - 3 + \log_5y) &= 4 \\
\log_4(xyz - 3 + \log_5z) &= 4.
\end{align*}Find the value of $|\log_5x| + |\log_5y| + |\log_5z|$.
3. Let $x$, $y$, and $z$ be real numbers satisfying the system\begin{align*}
\log_2(xyz - 3 + \log_5x) &= 5 \\
\log_3(xyz - 3 + \log_5y) &= 4 \\
\log_4(xyz - 3 + \log_5z) &= 4.
\end{align*}Find the value of $|\log_5x| + |\log_5y| + |\log_5z|$.
DPatrick
2016-03-18 19:26:48
Ick, perhaps. What should we do first?
Ick, perhaps. What should we do first?
mathlover3737
2016-03-18 19:27:13
get rid of the outside logs
get rid of the outside logs
IsabeltheCat
2016-03-18 19:27:13
take out the outside logs
take out the outside logs
maverick8
2016-03-18 19:27:13
get rid of the left logarithms
get rid of the left logarithms
blue8931
2016-03-18 19:27:13
rewrite logs as exponential equations
rewrite logs as exponential equations
whatsgucci17
2016-03-18 19:27:13
take out the big logs?
take out the big logs?
DPatrick
2016-03-18 19:27:20
Sure, let's get rid of those outside logs. They're just window dressing.
Sure, let's get rid of those outside logs. They're just window dressing.
DPatrick
2016-03-18 19:27:56
For example, if $\log_2(\text{blah}) = 5$, then what's "blah"?
For example, if $\log_2(\text{blah}) = 5$, then what's "blah"?
legolego
2016-03-18 19:28:10
32
32
geogirl08
2016-03-18 19:28:10
32
32
InCtrl
2016-03-18 19:28:10
2^5
2^5
beanielove2
2016-03-18 19:28:10
$2^5 = 32.$
$2^5 = 32.$
JustinEli
2016-03-18 19:28:10
2^5
2^5
avchess
2016-03-18 19:28:10
32
32
DPatrick
2016-03-18 19:28:21
Right: by the definition of log, $\text{blah} = 2^5 = 32$.
Right: by the definition of log, $\text{blah} = 2^5 = 32$.
DPatrick
2016-03-18 19:28:32
We do this for all three equations, and we have \begin{align*}
xyz - 3 + \log_5x &= 2^5 = 32, \\
xyz - 3 + \log_5y &= 3^4 = 81, \\
xyz - 3 + \log_5z &= 4^4 = 256.\end{align*}
We do this for all three equations, and we have \begin{align*}
xyz - 3 + \log_5x &= 2^5 = 32, \\
xyz - 3 + \log_5y &= 3^4 = 81, \\
xyz - 3 + \log_5z &= 4^4 = 256.\end{align*}
DPatrick
2016-03-18 19:28:42
Now what?
Now what?
Dayranger
2016-03-18 19:28:52
add 3
add 3
Liopleurodon
2016-03-18 19:28:52
Add the 3s
Add the 3s
Beast144
2016-03-18 19:28:52
add 3 on both sides
add 3 on both sides
DPatrick
2016-03-18 19:29:02
I'd probably clean it up a little by adding 3 to all sides:
I'd probably clean it up a little by adding 3 to all sides:
DPatrick
2016-03-18 19:29:08
\begin{align*}
xyz + \log_5x &= 35, \\
xyz + \log_5y &= 84, \\
xyz + \log_5z &= 259. \end{align*}
\begin{align*}
xyz + \log_5x &= 35, \\
xyz + \log_5y &= 84, \\
xyz + \log_5z &= 259. \end{align*}
blue8931
2016-03-18 19:29:24
since they are symmetric we think about combining all of them (in this case, add them together)
since they are symmetric we think about combining all of them (in this case, add them together)
atmchallenge
2016-03-18 19:29:24
Then add all 3 equations together.
Then add all 3 equations together.
DPatrick
2016-03-18 19:29:36
I really want to add these together now, to try to exploit some symmetry, and also in part because a sum of logs looks a lot what we want.
I really want to add these together now, to try to exploit some symmetry, and also in part because a sum of logs looks a lot what we want.
DPatrick
2016-03-18 19:29:41
We get \[3xyz + \left(\log_5x + \log_5y + \log_5z\right) = 378.\]
We get \[3xyz + \left(\log_5x + \log_5y + \log_5z\right) = 378.\]
DPatrick
2016-03-18 19:29:52
But what is the sum of logs?
But what is the sum of logs?
geogirl08
2016-03-18 19:30:15
the log of their product
the log of their product
Saphira 7
2016-03-18 19:30:15
It's the log of the products
It's the log of the products
anixsarkar
2016-03-18 19:30:15
logxyz base 5
logxyz base 5
IsabeltheCat
2016-03-18 19:30:15
$log_5(xyz)$
$log_5(xyz)$
jfei2001
2016-03-18 19:30:15
log_5 xyz
log_5 xyz
whatsgucci17
2016-03-18 19:30:15
log5(xyz)
log5(xyz)
Dayranger
2016-03-18 19:30:15
log5 (xyz)
log5 (xyz)
DPatrick
2016-03-18 19:30:18
It's the log of the product!
It's the log of the product!
DPatrick
2016-03-18 19:30:22
So we have \[3xyz + \log_5xyz = 378.\]
So we have \[3xyz + \log_5xyz = 378.\]
DPatrick
2016-03-18 19:30:34
This also looks good, because now everything is in terms of the product $xyz$. In fact, for clarity, you might consider setting $t = xyz$, so that our equation now is just $3t + \log_5 t = 378$.
This also looks good, because now everything is in terms of the product $xyz$. In fact, for clarity, you might consider setting $t = xyz$, so that our equation now is just $3t + \log_5 t = 378$.
DPatrick
2016-03-18 19:30:49
Now what?
Now what?
maverick8
2016-03-18 19:31:08
$ t= 125 $
$ t= 125 $
IsabeltheCat
2016-03-18 19:31:08
$t=125$
$t=125$
blue8931
2016-03-18 19:31:08
note that $log_5 xyz$ needs to be an integer so xyz is a power a 5. simple testing gives us that xyz=125.
note that $log_5 xyz$ needs to be an integer so xyz is a power a 5. simple testing gives us that xyz=125.
tanishq1
2016-03-18 19:31:08
guess that t=125
guess that t=125
24iam24
2016-03-18 19:31:08
notice that t=125 works
notice that t=125 works
Saphira 7
2016-03-18 19:31:08
Solve by inspection
Solve by inspection
atmchallenge
2016-03-18 19:31:08
Let's try letting $t$ be a power of $5$.
Let's try letting $t$ be a power of $5$.
DPatrick
2016-03-18 19:31:21
I agree -- at this point I'm probably content to guess-and-check. Hopefully, $xyz$ is a power of 5 so that $\log_5 xyz$ is an integer.
I agree -- at this point I'm probably content to guess-and-check. Hopefully, $xyz$ is a power of 5 so that $\log_5 xyz$ is an integer.
DPatrick
2016-03-18 19:31:29
Indeed, it's apparent that $xyz = 125 = 5^3$ works: \[3(125) + 3 = 378.\]
Indeed, it's apparent that $xyz = 125 = 5^3$ works: \[3(125) + 3 = 378.\]
mssmath
2016-03-18 19:31:34
t=375, is a solution and the only one at the LHS increases with t
t=375, is a solution and the only one at the LHS increases with t
DPatrick
2016-03-18 19:32:12
Right: we can be confident that our guess-and-check worked, because the left-side expression strictly increases as we increase $xyz$ (or $t$ as I called it).
Right: we can be confident that our guess-and-check worked, because the left-side expression strictly increases as we increase $xyz$ (or $t$ as I called it).
DPatrick
2016-03-18 19:32:25
So there's only one value of $xyz$ that can possibly solve this equation, and we just found it.
So there's only one value of $xyz$ that can possibly solve this equation, and we just found it.
Beast144
2016-03-18 19:32:34
Now we can plug xyz back into the above equations in order to solve for log {x,y,z} base 5
Now we can plug xyz back into the above equations in order to solve for log {x,y,z} base 5
blue8931
2016-03-18 19:32:42
now, plug back into our original exponential equations and solve the problem
now, plug back into our original exponential equations and solve the problem
DPatrick
2016-03-18 19:32:47
Right. We go back to our earlier system and substitute $xyz = 125$: \begin{align*}
125 + \log_5x &= 35, \\
125 + \log_5y &= 84, \\
125 + \log_5z &= 259. \end{align*}
Right. We go back to our earlier system and substitute $xyz = 125$: \begin{align*}
125 + \log_5x &= 35, \\
125 + \log_5y &= 84, \\
125 + \log_5z &= 259. \end{align*}
DPatrick
2016-03-18 19:33:00
So: \begin{align*}
\log_5x &= -90, \\
\log_5y &= -41, \\
\log_5z &= 134.\end{align*}
So: \begin{align*}
\log_5x &= -90, \\
\log_5y &= -41, \\
\log_5z &= 134.\end{align*}
blue8931
2016-03-18 19:33:14
$90 + 41 + 134 = \boxed{265}$
$90 + 41 + 134 = \boxed{265}$
calculus_riju
2016-03-18 19:33:14
265
265
DPatrick
2016-03-18 19:33:18
And the sum of the absolute value of these is $90 + 41 + 134 = \boxed{265}$.
And the sum of the absolute value of these is $90 + 41 + 134 = \boxed{265}$.
DPatrick
2016-03-18 19:34:00
This was probably my least-favorite problem: if you know logs (and aren't scared of them) it was pretty easy algebra, but if you don't know logs you had nowhere to go.
This was probably my least-favorite problem: if you know logs (and aren't scared of them) it was pretty easy algebra, but if you don't know logs you had nowhere to go.
DPatrick
2016-03-18 19:34:18
Plus the guess-and-check step is slightly annoying.
Plus the guess-and-check step is slightly annoying.
DPatrick
2016-03-18 19:34:23
Anyway...
Anyway...
DPatrick
2016-03-18 19:34:27
4. An $a \times b \times c$ rectangular box is built from $a \cdot b \cdot c$ unit cubes. Each unit cube is colored red, green, or yellow. Each of the $a$ layers of size $1 \times b \times c$ parallel to the $(b \times c)$-faces of the box contains exactly 9 red cubes, exactly 12 green cubes, and some yellow cubes. Each of the $b$ layers of size $a \times 1 \times c$ parallel to the $(a \times c)$-faces of the box contains exactly 20 green cubes, exactly 25 yellow cubes, and some red cubes. Find the smallest possible volume of the box.
4. An $a \times b \times c$ rectangular box is built from $a \cdot b \cdot c$ unit cubes. Each unit cube is colored red, green, or yellow. Each of the $a$ layers of size $1 \times b \times c$ parallel to the $(b \times c)$-faces of the box contains exactly 9 red cubes, exactly 12 green cubes, and some yellow cubes. Each of the $b$ layers of size $a \times 1 \times c$ parallel to the $(a \times c)$-faces of the box contains exactly 20 green cubes, exactly 25 yellow cubes, and some red cubes. Find the smallest possible volume of the box.
maverick8
2016-03-18 19:35:00
set up equations for different colors
set up equations for different colors
DPatrick
2016-03-18 19:35:12
That seems like a good strategy. Can we count the total number of cubes of each color in the whole box?
That seems like a good strategy. Can we count the total number of cubes of each color in the whole box?
StoryScene
2016-03-18 19:35:35
total green cubes: $20b=12a$
total green cubes: $20b=12a$
24iam24
2016-03-18 19:35:35
well we know the total greens are 12a=20b
well we know the total greens are 12a=20b
DPatrick
2016-03-18 19:36:04
Right. Each of $a$ layers has 12 greens, and each of $b$ layers has 20 greens. So there are $12a = 20b$ green cubes total.
Right. Each of $a$ layers has 12 greens, and each of $b$ layers has 20 greens. So there are $12a = 20b$ green cubes total.
DPatrick
2016-03-18 19:36:20
How about red?
How about red?
mathlover3737
2016-03-18 19:36:51
9a
9a
DPatrick
2016-03-18 19:37:07
Each of the $a$ layers has 9 red, so that's $9a$ total.
Each of the $a$ layers has 9 red, so that's $9a$ total.
DPatrick
2016-03-18 19:37:13
And each of the $b$ layers has...how many red?
And each of the $b$ layers has...how many red?
maverick8
2016-03-18 19:37:26
ac-45
ac-45
DPatrick
2016-03-18 19:37:45
Right: each layer has $ac$ total, and 45 of them are green or yellow, so the remaining $ac - 45$ are red.
Right: each layer has $ac$ total, and 45 of them are green or yellow, so the remaining $ac - 45$ are red.
DPatrick
2016-03-18 19:38:04
So there are $b(ac-45)$ red total as well.
So there are $b(ac-45)$ red total as well.
DPatrick
2016-03-18 19:38:39
So we have $9a = b(ac - 45)$. What might we do with that?
So we have $9a = b(ac - 45)$. What might we do with that?
whatsgucci17
2016-03-18 19:39:03
substitute a for b or vice versa
substitute a for b or vice versa
mochi
2016-03-18 19:39:03
9a=abc-45b
9a=abc-45b
DPatrick
2016-03-18 19:39:13
I like both of these suggestions!
I like both of these suggestions!
DPatrick
2016-03-18 19:39:30
Since $12a = 20b$, we can replace $9a$ with $15b$ on the left side of our latest equation.
Since $12a = 20b$, we can replace $9a$ with $15b$ on the left side of our latest equation.
DPatrick
2016-03-18 19:39:51
And, when we then expand out the right side, we get $15b = abc - 45b$, or $60b = abc$.
And, when we then expand out the right side, we get $15b = abc - 45b$, or $60b = abc$.
DPatrick
2016-03-18 19:40:07
In terms of what the problem is asking for, what does that tell us?
In terms of what the problem is asking for, what does that tell us?
DPatrick
2016-03-18 19:40:16
(Try to never forget what the problem is asking for!)
(Try to never forget what the problem is asking for!)
thatindiankid55
2016-03-18 19:40:21
The volume in terms of b.
The volume in terms of b.
atmchallenge
2016-03-18 19:40:27
We want to minimize $60b$, or minimize $b$.
We want to minimize $60b$, or minimize $b$.
whatsgucci17
2016-03-18 19:40:27
that the smallest volume is 60b
that the smallest volume is 60b
DPatrick
2016-03-18 19:40:39
Right: in order to minimize the volume $abc$, we want to minimize $b$.
Right: in order to minimize the volume $abc$, we want to minimize $b$.
DPatrick
2016-03-18 19:40:54
And then $60b$ will be our answer.
And then $60b$ will be our answer.
lucylou
2016-03-18 19:41:09
and make sure the lengths are integers
and make sure the lengths are integers
DPatrick
2016-03-18 19:41:23
Right: that's important too. $a$, $b$, $c$ all have to be integers.
Right: that's important too. $a$, $b$, $c$ all have to be integers.
DPatrick
2016-03-18 19:41:40
So we can try to get lucky and find the smallest positive integer solution to $12a = 20b$. What is it?
So we can try to get lucky and find the smallest positive integer solution to $12a = 20b$. What is it?
blue8931
2016-03-18 19:42:08
b=3
b=3
InCtrl
2016-03-18 19:42:08
a = 5, b = 3
a = 5, b = 3
legolego
2016-03-18 19:42:08
(5, 3)
(5, 3)
jwlw2014
2016-03-18 19:42:08
b=3
b=3
roseylily
2016-03-18 19:42:08
a = 5, b = 3
a = 5, b = 3
Oleksenko
2016-03-18 19:42:08
5,3
5,3
tanishq1
2016-03-18 19:42:08
(5,3)
(5,3)
Liopleurodon
2016-03-18 19:42:08
b=3,a=5,c=12
b=3,a=5,c=12
kapilak
2016-03-18 19:42:08
a=5 ,b=3
a=5 ,b=3
rechyyy
2016-03-18 19:42:08
a=5, b=3?
a=5, b=3?
calculus_riju
2016-03-18 19:42:08
(5,3)
(5,3)
StoryScene
2016-03-18 19:42:08
$5,3$ for $a,b$
$5,3$ for $a,b$
DPatrick
2016-03-18 19:42:25
Right. Dividing by 4 leaves $3a = 5b$, so the smallest positive integer solution is clearly $a=5$, $b=3$.
Right. Dividing by 4 leaves $3a = 5b$, so the smallest positive integer solution is clearly $a=5$, $b=3$.
DPatrick
2016-03-18 19:42:39
And then using $60b = abc$, we get $c = 12$ and $abc = 180$.
And then using $60b = abc$, we get $c = 12$ and $abc = 180$.
DPatrick
2016-03-18 19:42:43
But does this solution actually work?
But does this solution actually work?
DPatrick
2016-03-18 19:43:13
Can we actually have such a box with $a=5$, $b=3$, $c=12$, and the required numbers of blocks in each layer?
Can we actually have such a box with $a=5$, $b=3$, $c=12$, and the required numbers of blocks in each layer?
kbird
2016-03-18 19:43:42
yeah
yeah
maverick8
2016-03-18 19:43:42
yes
yes
legolego
2016-03-18 19:43:42
yes
yes
blue8931
2016-03-18 19:43:42
yes
yes
DPatrick
2016-03-18 19:43:49
Well, we can, but how would you convince me?
Well, we can, but how would you convince me?
DPatrick
2016-03-18 19:44:17
If you plug in the numbers:
The $b \times c$ layer is a $3 \times 12$ layer of 9 red, 12 green, and 15 yellow.
The $a \times c$ layer is a $5 \times 12$ layer of 20 green, 25 yellow, and 15 red.
If you plug in the numbers:
The $b \times c$ layer is a $3 \times 12$ layer of 9 red, 12 green, and 15 yellow.
The $a \times c$ layer is a $5 \times 12$ layer of 20 green, 25 yellow, and 15 red.
DPatrick
2016-03-18 19:44:28
Can we satisfy both of these conditions simultaneously?
Can we satisfy both of these conditions simultaneously?
DPatrick
2016-03-18 19:45:27
The key is that both of our types of layers have red : green : yellow in the ratio 3 : 4 : 5.
The key is that both of our types of layers have red : green : yellow in the ratio 3 : 4 : 5.
DPatrick
2016-03-18 19:45:38
So we just need to make sure that each $1 \times 1 \times 12$ line of blocks (perpendicular to the $a \times b$ faces) has 3 red, 4 green, and 5 yellow, and we'll be all set.
So we just need to make sure that each $1 \times 1 \times 12$ line of blocks (perpendicular to the $a \times b$ faces) has 3 red, 4 green, and 5 yellow, and we'll be all set.
DPatrick
2016-03-18 19:45:57
So the minimum volume is $abc = \boxed{180}$.
So the minimum volume is $abc = \boxed{180}$.
DPatrick
2016-03-18 19:46:30
On the actual contest, with the clock ticking, you might not take the time to convince yourself that the configuration actually works. But it is part of a complete solution.
On the actual contest, with the clock ticking, you might not take the time to convince yourself that the configuration actually works. But it is part of a complete solution.
DPatrick
2016-03-18 19:46:52
OK, on to our first geometry problem!
OK, on to our first geometry problem!
DPatrick
2016-03-18 19:46:56
5. Triangle $ABC_0$ has a right angle at $C_0$. Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$. Let $C_1$ be the foot of the altitude to $\overline{AB}$, and for $n \ge 2$, let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$. The sum $\sum_{n=1}^\infty C_{n-1}C_n = 6p$. Find $p$.
5. Triangle $ABC_0$ has a right angle at $C_0$. Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$. Let $C_1$ be the foot of the altitude to $\overline{AB}$, and for $n \ge 2$, let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$. The sum $\sum_{n=1}^\infty C_{n-1}C_n = 6p$. Find $p$.
calculus_riju
2016-03-18 19:47:13
diagram plz!
diagram plz!
Dayranger
2016-03-18 19:47:13
Diagram please!!!
Diagram please!!!
whatsgucci17
2016-03-18 19:47:13
draw a diagram'
draw a diagram'
idomath12345
2016-03-18 19:47:13
Diagram
Diagram
DPatrick
2016-03-18 19:47:29
Certainly, we'd start with a sketch of a picture of what's going on:
Certainly, we'd start with a sketch of a picture of what's going on:
DPatrick
2016-03-18 19:47:33
DPatrick
2016-03-18 19:47:40
What do you notice in this picture?
What do you notice in this picture?
legolego
2016-03-18 19:48:01
lots of similar triangles
lots of similar triangles
maverick8
2016-03-18 19:48:01
similar triangles
similar triangles
thatindiankid55
2016-03-18 19:48:01
Parallel lines which means similar triangles.
Parallel lines which means similar triangles
.
blue8931
2016-03-18 19:48:01
smaller and smaller with similar triangles... infinite geo series
smaller and smaller with similar triangles... infinite geo series
idomath12345
2016-03-18 19:48:01
SImilar trinagles.
SImilar trinagles.
fredgauss
2016-03-18 19:48:01
the triangles are all similar
the triangles are all similar
mochi
2016-03-18 19:48:01
Similar triangles.
Similar triangles.
DPatrick
2016-03-18 19:48:29
Sooo many similar triangles here!
Sooo many similar triangles here!
DPatrick
2016-03-18 19:48:36
All of the little triangles are similar to the big triangle.
All of the little triangles are similar to the big triangle.
DPatrick
2016-03-18 19:49:13
Just to be clear what's going on, let's set $a = BC_0$, $b = AC_0$, and $c = AB$ to be the sides of the big triangle $ABC$ (so that $a$ is opposite $A$, as usual). And let's focus on one of the little triangles, say $C_1C_2C_3$:
Just to be clear what's going on, let's set $a = BC_0$, $b = AC_0$, and $c = AB$ to be the sides of the big triangle $ABC$ (so that $a$ is opposite $A$, as usual). And let's focus on one of the little triangles, say $C_1C_2C_3$:
DPatrick
2016-03-18 19:49:16
DPatrick
2016-03-18 19:49:25
What is the ratio $\dfrac{C_2C_3}{C_1C_2}$?
What is the ratio $\dfrac{C_2C_3}{C_1C_2}$?
Shaddoll
2016-03-18 19:49:49
they have ratio $\dfrac{a}{c}$
they have ratio $\dfrac{a}{c}$
tanishq1
2016-03-18 19:49:49
a/c
a/c
maverick8
2016-03-18 19:49:49
a/c
a/c
Dayranger
2016-03-18 19:49:49
a/c
a/c
blue8931
2016-03-18 19:49:49
a/c
a/c
DPatrick
2016-03-18 19:49:57
$C_1C_2$ is the hypotenuse of the red triangle, and $C_2C_3$ corresponds to the $a$ side of the big triangle (it's the longer leg the way I drew it).
$C_1C_2$ is the hypotenuse of the red triangle, and $C_2C_3$ corresponds to the $a$ side of the big triangle (it's the longer leg the way I drew it).
DPatrick
2016-03-18 19:50:02
So $\dfrac{C_2C_3}{C_1C_2} = \dfrac{a}{c}$.
So $\dfrac{C_2C_3}{C_1C_2} = \dfrac{a}{c}$.
DPatrick
2016-03-18 19:50:20
There's nothing special about the example I picked, so in fact, this is true for any ratio $\dfrac{C_nC_{n+1}}{C_{n-1}C_n}$.
There's nothing special about the example I picked, so in fact, this is true for any ratio $\dfrac{C_nC_{n+1}}{C_{n-1}C_n}$.
DPatrick
2016-03-18 19:50:28
So what does that tell us about $\sum_{n=1}^\infty C_{n-1}C_n$?
So what does that tell us about $\sum_{n=1}^\infty C_{n-1}C_n$?
blue8931
2016-03-18 19:50:58
infinite geo series
infinite geo series
24iam24
2016-03-18 19:50:58
its an infinite geometric series
its an infinite geometric series
maverick8
2016-03-18 19:50:58
geo series with ratio a/c
geo series with ratio a/c
StoryScene
2016-03-18 19:50:58
It's the sum of a geometric series
It's the sum of a geometric series
atmchallenge
2016-03-18 19:50:58
It's an infinite geometric series.
It's an infinite geometric series.
DPatrick
2016-03-18 19:51:02
It's a geometric series with ratio $\dfrac{a}{c}$!
It's a geometric series with ratio $\dfrac{a}{c}$!
DPatrick
2016-03-18 19:51:17
To compute its sum, we just need the first term $C_0C_1$. What is it?
To compute its sum, we just need the first term $C_0C_1$. What is it?
maverick8
2016-03-18 19:51:50
ab/c
ab/c
blue8931
2016-03-18 19:51:50
ab/c
ab/c
24iam24
2016-03-18 19:51:50
ab/c
ab/c
mathlover3737
2016-03-18 19:51:50
ab/c
ab/c
idomath12345
2016-03-18 19:51:50
altitude
altitude
DPatrick
2016-03-18 19:52:18
Right. $C_0C_1$ is an altitude, so $c(C_0C_1) = ab = 2(\text{area of } ABC_0)$.
Right. $C_0C_1$ is an altitude, so $c(C_0C_1) = ab = 2(\text{area of } ABC_0)$.
DPatrick
2016-03-18 19:52:28
Thus $C_0C_1 = \dfrac{ab}{c}$.
Thus $C_0C_1 = \dfrac{ab}{c}$.
DPatrick
2016-03-18 19:52:36
Therefore, what's the sum of the series?
Therefore, what's the sum of the series?
24iam24
2016-03-18 19:53:12
ab/(c-a)
ab/(c-a)
geogirl08
2016-03-18 19:53:12
ab/(c-a)
ab/(c-a)
tanishq1
2016-03-18 19:53:12
$\frac{ab/c}{1-a/c}=\frac{ab}{c-a}$
$\frac{ab/c}{1-a/c}=\frac{ab}{c-a}$
calculus_riju
2016-03-18 19:53:12
$\frac{ab}{c-a}$
$\frac{ab}{c-a}$
Shaddoll
2016-03-18 19:53:12
$\dfrac{ab}{c-a}$
$\dfrac{ab}{c-a}$
DPatrick
2016-03-18 19:53:25
Right. It's $\dfrac{ab}{c} \cdot \dfrac{1}{1-\frac{a}{c}}$, which simplifies to $\dfrac{ab}{c-a}$.
Right. It's $\dfrac{ab}{c} \cdot \dfrac{1}{1-\frac{a}{c}}$, which simplifies to $\dfrac{ab}{c-a}$.
DPatrick
2016-03-18 19:53:35
Now what?
Now what?
idomath12345
2016-03-18 19:53:49
=6p
=6p
jwlw2014
2016-03-18 19:53:49
Set equal to 6p
Set equal to 6p
tanishq1
2016-03-18 19:53:49
set it equal to 6(a+b+c)?
set it equal to 6(a+b+c)?
blue8931
2016-03-18 19:53:49
compare this to the perimeter: a+b+c
compare this to the perimeter: a+b+c
DPatrick
2016-03-18 19:54:06
Right: we're told that this is 6 times the perimeter. Therefore, we have \[\frac{ab}{c-a} = 6(a+b+c).\]
Right: we're told that this is 6 times the perimeter. Therefore, we have \[\frac{ab}{c-a} = 6(a+b+c).\]
DPatrick
2016-03-18 19:54:15
That's good -- now what?
That's good -- now what?
Shaddoll
2016-03-18 19:54:41
just guess and check?
just guess and check?
Shaddoll
2016-03-18 19:54:41
there aren't many triangles with relatively pairwise prime positive integers that are small enough
there aren't many triangles with relatively pairwise prime positive integers that are small enough
DPatrick
2016-03-18 19:55:01
It's not a terrible idea to start guessing-and-checking even at this point, but we could also try to work the algebra a bit further first.
It's not a terrible idea to start guessing-and-checking even at this point, but we could also try to work the algebra a bit further first.
kbird
2016-03-18 19:55:08
get rid of the fraction
get rid of the fraction
DPatrick
2016-03-18 19:55:18
OK, let's clear the denominator: \[ab = 6(a+b+c)(c-a).\]
OK, let's clear the denominator: \[ab = 6(a+b+c)(c-a).\]
DPatrick
2016-03-18 19:55:31
The right side expands: \[ab = 6(ac - a^2 + bc - ab + c^2 - ac).\]
The right side expands: \[ab = 6(ac - a^2 + bc - ab + c^2 - ac).\]
DPatrick
2016-03-18 19:55:37
Those $ac$ terms cancel; any other simplification?
Those $ac$ terms cancel; any other simplification?
Shaddoll
2016-03-18 19:55:58
$c^2-a^2=b^2$
$c^2-a^2=b^2$
tanishq1
2016-03-18 19:55:58
c^2-a^2=b^2
c^2-a^2=b^2
calculus_riju
2016-03-18 19:56:07
c^2-a^2=b^2
c^2-a^2=b^2
DPatrick
2016-03-18 19:56:10
Nice: the Pythagorean Theorem lets us replace $-a^2 + c^2$ with $b^2$: \[ab = 6(b^2 + bc - ab).\]
Nice: the Pythagorean Theorem lets us replace $-a^2 + c^2$ with $b^2$: \[ab = 6(b^2 + bc - ab).\]
tbob
2016-03-18 19:56:19
then divide by b
then divide by b
whatsgucci17
2016-03-18 19:56:19
then factor out b from all terms
then factor out b from all terms
24iam24
2016-03-18 19:56:21
divide by b
divide by b
DPatrick
2016-03-18 19:56:23
Ooh, now $b$ divides out: \[a = 6(b+c-a).\]
Ooh, now $b$ divides out: \[a = 6(b+c-a).\]
DPatrick
2016-03-18 19:56:34
And maybe I want to collect terms and write this as $7a = 6(b+c)$ nice and neatly.
And maybe I want to collect terms and write this as $7a = 6(b+c)$ nice and neatly.
DPatrick
2016-03-18 19:56:46
Great -- so now all we have to do is find a primitive Pythagorean triple $(a,b,c)$ that satisfies this relationship.
Great -- so now all we have to do is find a primitive Pythagorean triple $(a,b,c)$ that satisfies this relationship.
DPatrick
2016-03-18 19:57:07
It's definitely guess-and-checkable from here, but we can also continue to finish it a little more systematically.
It's definitely guess-and-checkable from here, but we can also continue to finish it a little more systematically.
DPatrick
2016-03-18 19:57:30
There's one more algebraic manipulation to try with this...any ideas?
There's one more algebraic manipulation to try with this...any ideas?
DPatrick
2016-03-18 19:57:55
Say, if I wanted to get rid a variable?
Say, if I wanted to get rid a variable?
whatsgucci17
2016-03-18 19:58:28
oh a^2 + b^2 = c^2
oh a^2 + b^2 = c^2
jfei2001
2016-03-18 19:58:28
a=sqrt(c^2-b^2)
a=sqrt(c^2-b^2)
DPatrick
2016-03-18 19:58:32
Right. I'd square both sides: \[ 49a^2 = 36(b+c)^2. \] How does this help?
Right. I'd square both sides: \[ 49a^2 = 36(b+c)^2. \] How does this help?
24iam24
2016-03-18 19:58:51
a^2=c^2-b^2
a^2=c^2-b^2
atmchallenge
2016-03-18 19:58:51
$a^2=c^2-b^2$.
$a^2=c^2-b^2$.
pie314159265
2016-03-18 19:58:51
difference of squares!!
difference of squares!!
DPatrick
2016-03-18 19:58:55
We know that $a^2 = c^2 - b^2$ from the right triangle. So this is \[49(c^2-b^2) = 36(b+c)^2.\]
We know that $a^2 = c^2 - b^2$ from the right triangle. So this is \[49(c^2-b^2) = 36(b+c)^2.\]
JonXu101
2016-03-18 19:59:14
factor b+c from both sides
factor b+c from both sides
whatsgucci17
2016-03-18 19:59:21
divide both sides by b+c
divide both sides by b+c
DPatrick
2016-03-18 19:59:22
And now we can divide by $b+c$ to get \[49(c-b) = 36(b+c).\]
And now we can divide by $b+c$ to get \[49(c-b) = 36(b+c).\]
DPatrick
2016-03-18 19:59:35
This gives us $13c = 85b$.
This gives us $13c = 85b$.
DPatrick
2016-03-18 20:00:00
So just like in the last problem, what's our guess for the simplest solution?
So just like in the last problem, what's our guess for the simplest solution?
24iam24
2016-03-18 20:00:06
so now let b=13 and c=85
so now let b=13 and c=85
idomath12345
2016-03-18 20:00:06
so 13 and 85.
so 13 and 85.
whatsgucci17
2016-03-18 20:00:06
c= 85, b= 13
c= 85, b= 13
Shaddoll
2016-03-18 20:00:06
by the relatively prime condition ,we must have $(b,c)=(13,85)$.
by the relatively prime condition ,we must have $(b,c)=(13,85)$.
DPatrick
2016-03-18 20:00:23
In fact, to get a relatively prime solution, that's what we must have.
In fact, to get a relatively prime solution, that's what we must have.
DPatrick
2016-03-18 20:00:30
So $(b,c) = (13,85)$ is a solution. Does this solution work?
So $(b,c) = (13,85)$ is a solution. Does this solution work?
mathlover3737
2016-03-18 20:00:42
13, 84, 85
13, 84, 85
Liopleurodon
2016-03-18 20:00:42
13,84,85
13,84,85
GOMATHmath
2016-03-18 20:00:42
13, 84, 85
13, 84, 85
JonXu101
2016-03-18 20:00:42
a-84
a-84
blue8931
2016-03-18 20:00:42
yes, a=84
yes, a=84
DPatrick
2016-03-18 20:00:47
Yes! We get $7a = 6(b+c) = 6(98)$, so $a = 6(14) = 84$, and we can check that 13-84-85 is a Pythagorean triple.
Yes! We get $7a = 6(b+c) = 6(98)$, so $a = 6(14) = 84$, and we can check that 13-84-85 is a Pythagorean triple.
Liopleurodon
2016-03-18 20:01:08
p=182
p=182
pramodmitikiri
2016-03-18 20:01:08
so.. 13+84+85= 182
so.. 13+84+85= 182
DPatrick
2016-03-18 20:01:09
So our triangle has sides 13, 84, and 85, and its perimeter is $13 + 84 + 85 = \boxed{182}$.
So our triangle has sides 13, 84, and 85, and its perimeter is $13 + 84 + 85 = \boxed{182}$.
DPatrick
2016-03-18 20:01:49
Certainly like some of you suggested, there were many points at which you could have guessed-and-checked the answer if you know your Pythagorean triples well, but it was also possible to use algebra all the way to the end.
Certainly like some of you suggested, there were many points at which you could have guessed-and-checked the answer if you know your Pythagorean triples well, but it was also possible to use algebra all the way to the end.
mochi
2016-03-18 20:01:56
Are Math Jams always about 2 hours long?
Are Math Jams always about 2 hours long?
DPatrick
2016-03-18 20:02:07
More like 3 tonight. We're only 1/3 of the way through, and it's been an hour.
More like 3 tonight. We're only 1/3 of the way through, and it's been an hour.
DPatrick
2016-03-18 20:02:20
6. For polynomial $P(x) = 1 - \frac13x + \frac16x^2$, define \[ Q(x) = P(x)P(x^3)P(x^5)P(x^7)P(x^9) = \sum_{i=0}^{50} a_ix^i. \] Then $\sum_{i=0}^{50} |a_i| = \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
6. For polynomial $P(x) = 1 - \frac13x + \frac16x^2$, define \[ Q(x) = P(x)P(x^3)P(x^5)P(x^7)P(x^9) = \sum_{i=0}^{50} a_ix^i. \] Then $\sum_{i=0}^{50} |a_i| = \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
DPatrick
2016-03-18 20:02:59
If we didn't have the absolute value condition, how would we find the sum of the coefficients of $Q(x)$?
If we didn't have the absolute value condition, how would we find the sum of the coefficients of $Q(x)$?
maverick8
2016-03-18 20:03:19
let x=1
let x=1
Shaddoll
2016-03-18 20:03:19
Find $Q(1)$
Find $Q(1)$
Aopser1221
2016-03-18 20:03:19
x=1
x=1
mssmath
2016-03-18 20:03:19
Q(1)
Q(1)
idomath12345
2016-03-18 20:03:19
x=1
x=1
blue8931
2016-03-18 20:03:19
let x=1?
let x=1?
JonXu101
2016-03-18 20:03:19
make x=1
make x=1
DPatrick
2016-03-18 20:03:25
Right. We'd just plug in $x=1$, and we'd get $\sum_{i=0}^{50} a_i = Q(1)$.
Right. We'd just plug in $x=1$, and we'd get $\sum_{i=0}^{50} a_i = Q(1)$.
DPatrick
2016-03-18 20:03:33
So maybe we can manipulate things so that we get a polynomial whose coefficients are the $|a_i|$, and then plug in $x=1$.
So maybe we can manipulate things so that we get a polynomial whose coefficients are the $|a_i|$, and then plug in $x=1$.
Shaddoll
2016-03-18 20:03:43
Note that all the even degree coefficients are positive and odd degree coefficients are negative in Q
Note that all the even degree coefficients are positive and odd degree coefficients are negative in Q
Aopser1221
2016-03-18 20:03:43
Note that all even powers of Q(x) are positive and all odd powers are negative
Note that all even powers of Q(x) are positive and all odd powers are negative
DPatrick
2016-03-18 20:03:53
Aha, that's the key observation. The even-power terms have positive coefficients, and the odd-power terms have negative coefficients.
Aha, that's the key observation. The even-power terms have positive coefficients, and the odd-power terms have negative coefficients.
DPatrick
2016-03-18 20:04:11
That seems annoying -- it'd be easier to deal with if they had all positive coefficients. Then we wouldn't have to worry about absolute values. How can we make that happen?
That seems annoying -- it'd be easier to deal with if they had all positive coefficients. Then we wouldn't have to worry about absolute values. How can we make that happen?
Oleksenko
2016-03-18 20:04:37
Consider P(-x)
Consider P(-x)
maverick8
2016-03-18 20:04:37
let x=-1
let x=-1
tanishq1
2016-03-18 20:04:37
so you can actually just plug in x=-1
so you can actually just plug in x=-1
calculus_riju
2016-03-18 20:04:37
so put x=-1
so put x=-1
Aopser1221
2016-03-18 20:04:37
Plug in x=-1
Plug in x=-1
mssmath
2016-03-18 20:04:37
P(-x)?
P(-x)?
idomath12345
2016-03-18 20:04:37
x=-1!
x=-1!
poptower99
2016-03-18 20:04:37
Make x=-1
Make x=-1
24iam24
2016-03-18 20:04:37
plug in x=-1
plug in x=-1
tanishq1
2016-03-18 20:04:37
plug in x=-x?
plug in x=-x?
DPatrick
2016-03-18 20:04:43
Right! We can replace $x$ with $-x$. Note that $P(-x) = 1 + \dfrac13x + \dfrac16x^2$ has all positive coefficients!
Right! We can replace $x$ with $-x$. Note that $P(-x) = 1 + \dfrac13x + \dfrac16x^2$ has all positive coefficients!
DPatrick
2016-03-18 20:04:54
So $Q(-x) = P(-x)P(-x^3)P(-x^5)P(-x^7)P(-x^9)$ has all positive coefficients too. But that's exactly what we want: the (positive) coefficients of $Q(-x)$ are all the absolute values of the coefficients of $Q(x)$.
So $Q(-x) = P(-x)P(-x^3)P(-x^5)P(-x^7)P(-x^9)$ has all positive coefficients too. But that's exactly what we want: the (positive) coefficients of $Q(-x)$ are all the absolute values of the coefficients of $Q(x)$.
idomath12345
2016-03-18 20:05:13
Yayy we're done!
Yayy we're done!
DPatrick
2016-03-18 20:05:28
I agree: we can get the sum of the coefficients of $Q(-x)$ in the usual way, by plugging in $x=1$. So $Q(-1)$ is our answer.
I agree: we can get the sum of the coefficients of $Q(-x)$ in the usual way, by plugging in $x=1$. So $Q(-1)$ is our answer.
Shaddoll
2016-03-18 20:05:42
thus, the desired $\dfrac{m}{n}$ is just $Q(-1)=P(-1)^{5}$
thus, the desired $\dfrac{m}{n}$ is just $Q(-1)=P(-1)^{5}$
DPatrick
2016-03-18 20:05:44
Fortunately, that's easy to compute: $Q(-1) = P(-1)P(-1)P(-1)P(-1)P(-1) = \left(P(-1)\right)^5$.
Fortunately, that's easy to compute: $Q(-1) = P(-1)P(-1)P(-1)P(-1)P(-1) = \left(P(-1)\right)^5$.
DPatrick
2016-03-18 20:05:59
Since $P(-1) = 1 + \dfrac13 + \dfrac16 = \dfrac96 = \dfrac32$, we get that $Q(-1) = \left(\dfrac32\right)^5 = \dfrac{243}{32}.$
Since $P(-1) = 1 + \dfrac13 + \dfrac16 = \dfrac96 = \dfrac32$, we get that $Q(-1) = \left(\dfrac32\right)^5 = \dfrac{243}{32}.$
blue8931
2016-03-18 20:06:16
275 is our answer
275 is our answer
Dayranger
2016-03-18 20:06:19
243 + 32 = 275
243 + 32 = 275
DPatrick
2016-03-18 20:06:22
Thus our final answer is $243 + 32 = \boxed{275}$.
Thus our final answer is $243 + 32 = \boxed{275}$.
blue8931
2016-03-18 20:06:37
that was quick
that was quick
DPatrick
2016-03-18 20:06:51
Indeed, this was a classic AIME problem that looked a lot harder than it actually was.
Indeed, this was a classic AIME problem that looked a lot harder than it actually was.
DPatrick
2016-03-18 20:07:07
More geometry (sort of) next:
More geometry (sort of) next:
DPatrick
2016-03-18 20:07:11
7. Squares $ABCD$ and $EFGH$ have a common center and $\overline{AB} \parallel \overline{EF}$. The area of $ABCD$ is 2016, and the area of $EFGH$ is a smaller positive integer. Square $IJKL$ is constructed so that each of its vertices lies on a side of $ABCD$ and each vertex of $EFGH$ lies on a side of $IJKL$. Find the difference between the largest and smallest possible integer values for the area of $IJKL$.
7. Squares $ABCD$ and $EFGH$ have a common center and $\overline{AB} \parallel \overline{EF}$. The area of $ABCD$ is 2016, and the area of $EFGH$ is a smaller positive integer. Square $IJKL$ is constructed so that each of its vertices lies on a side of $ABCD$ and each vertex of $EFGH$ lies on a side of $IJKL$. Find the difference between the largest and smallest possible integer values for the area of $IJKL$.
DPatrick
2016-03-18 20:07:24
Here's a picture of what's going on:
Here's a picture of what's going on:
DPatrick
2016-03-18 20:07:28
DPatrick
2016-03-18 20:07:56
What do we (once again) see in the diagram?
What do we (once again) see in the diagram?
blue8931
2016-03-18 20:08:16
similar triangles
similar triangles
pramodmitikiri
2016-03-18 20:08:16
similarities
similarities
mochi
2016-03-18 20:08:16
similar triangles!
similar triangles!
bluephoenix
2016-03-18 20:08:16
similar trianggles
similar trianggles
DPatrick
2016-03-18 20:08:24
There are lots of similar triangles here: the four smaller triangles (like $IEF$) are similar to the four larger triangles (like $BJI$). How does that help us relate the areas of the squares?
There are lots of similar triangles here: the four smaller triangles (like $IEF$) are similar to the four larger triangles (like $BJI$). How does that help us relate the areas of the squares?
pie314159265
2016-03-18 20:09:21
ratio of sides
ratio of sides
DPatrick
2016-03-18 20:09:24
Some of you have it -- but if you don't see what to do right away, a good starting place is to simply write down the ratios that the similar triangles give us.
Some of you have it -- but if you don't see what to do right away, a good starting place is to simply write down the ratios that the similar triangles give us.
DPatrick
2016-03-18 20:09:30
\[ \frac{IE}{BJ} = \frac{IF}{BI} = \frac{FE}{IJ}. \]
\[ \frac{IE}{BJ} = \frac{IF}{BI} = \frac{FE}{IJ}. \]
DPatrick
2016-03-18 20:09:38
The last ratio looks especially helpful: it's the ratio of the side lengths of the smaller two squares.
The last ratio looks especially helpful: it's the ratio of the side lengths of the smaller two squares.
DPatrick
2016-03-18 20:09:43
But what can we do with those first two ratios?
But what can we do with those first two ratios?
brian22
2016-03-18 20:10:40
Add numerators and denominators!
Add numerators and denominators!
DPatrick
2016-03-18 20:10:44
Good idea!
Good idea!
DPatrick
2016-03-18 20:10:49
We can add the ratios. That is, we can use the identity $\displaystyle\frac{w}{x} = \frac{y}{z} = \frac{w+y}{x+z}$. Think of it as an addition of ratios.
We can add the ratios. That is, we can use the identity $\displaystyle\frac{w}{x} = \frac{y}{z} = \frac{w+y}{x+z}$. Think of it as an addition of ratios.
DPatrick
2016-03-18 20:11:04
So the first two ratios combine to give $\dfrac{IE+IF}{BJ+BI} = \dfrac{FE}{IJ}$.
So the first two ratios combine to give $\dfrac{IE+IF}{BJ+BI} = \dfrac{FE}{IJ}$.
DPatrick
2016-03-18 20:11:11
But that's a huge win...because?
But that's a huge win...because?
calculus_riju
2016-03-18 20:11:37
side of squares
side of squares
mathwhiz16
2016-03-18 20:11:44
ratios of square side lengths
ratios of square side lengths
DPatrick
2016-03-18 20:11:47
Right. $IE + IF = IJ$, a side of the middle square, and $BJ + BI = AB$, a side of the big square.
Right. $IE + IF = IJ$, a side of the middle square, and $BJ + BI = AB$, a side of the big square.
DPatrick
2016-03-18 20:11:57
So we have $\dfrac{IJ}{AB} = \dfrac{FE}{IJ}$. What does that tell us?
So we have $\dfrac{IJ}{AB} = \dfrac{FE}{IJ}$. What does that tell us?
Shaddoll
2016-03-18 20:12:14
and we can use similar triangles to prove that the 3 square's areas form a geometric sequence
and we can use similar triangles to prove that the 3 square's areas form a geometric sequence
Aopser1221
2016-03-18 20:12:14
they are a geometric sequence!
they are a geometric sequence!
maverick8
2016-03-18 20:12:18
geo series
geo series
pramodmitikiri
2016-03-18 20:12:25
geo series
geo series
DPatrick
2016-03-18 20:12:32
The three side lengths of the squares are in geometric progression! So their areas are too.
The three side lengths of the squares are in geometric progression! So their areas are too.
DPatrick
2016-03-18 20:12:51
We can write our geometric progression in the usual way with a common ratio $r$, with $0 < r < 1$, so that we have $2016,\, 2016r,\, 2016r^2$ as the three areas.
We can write our geometric progression in the usual way with a common ratio $r$, with $0 < r < 1$, so that we have $2016,\, 2016r,\, 2016r^2$ as the three areas.
DPatrick
2016-03-18 20:13:00
Quick check: is $0 < r < 1$ the right bound on $r$?
Quick check: is $0 < r < 1$ the right bound on $r$?
DPatrick
2016-03-18 20:13:34
$r<1$ just reflects that the three areas are decreasing, which is correct. (2016 is the largest area.)
$r<1$ just reflects that the three areas are decreasing, which is correct. (2016 is the largest area.)
DPatrick
2016-03-18 20:13:39
But...
But...
DPatrick
2016-03-18 20:13:54
How small can $r$ be and still make the geometry work?
How small can $r$ be and still make the geometry work?
maverick8
2016-03-18 20:14:12
1/2
1/2
calculus_riju
2016-03-18 20:14:12
it cant be less than 1/2 i guess
it cant be less than 1/2 i guess
1023ong
2016-03-18 20:14:12
1/2
1/2
Aopser1221
2016-03-18 20:14:12
1/2
1/2
magicflippinertle
2016-03-18 20:14:12
half
half
DPatrick
2016-03-18 20:14:18
We get the smallest possible area of $IJKL$ when those points are the midpoints of the sides of the big square, like so:
We get the smallest possible area of $IJKL$ when those points are the midpoints of the sides of the big square, like so:
DPatrick
2016-03-18 20:14:27
DPatrick
2016-03-18 20:14:42
The middle square is half the area of the big square, so $r = \frac12$ in this picture, and that's as small as we can go.
The middle square is half the area of the big square, so $r = \frac12$ in this picture, and that's as small as we can go.
DPatrick
2016-03-18 20:14:58
Thus, our possibilities are values of $r$ such that $\frac12 \le r < 1$, giving areas of $2016,\,2016r,\,2016r^2$.
Thus, our possibilities are values of $r$ such that $\frac12 \le r < 1$, giving areas of $2016,\,2016r,\,2016r^2$.
DPatrick
2016-03-18 20:15:16
But what else do we need to ensure?
But what else do we need to ensure?
mathlover3737
2016-03-18 20:15:28
all integers
all integers
roseylily
2016-03-18 20:15:28
integer areas
integer areas
magicflippinertle
2016-03-18 20:15:32
all positive integers
all positive integers
DPatrick
2016-03-18 20:15:41
Right: all these areas need to be integers.
Right: all these areas need to be integers.
Aopser1221
2016-03-18 20:15:55
a perfect square that goes into 2016
a perfect square that goes into 2016
DPatrick
2016-03-18 20:16:09
Indeed, if we write $r^2 = \dfrac{p^2}{q^2}$ in lowest terms, then we need $q^2$ to be a divisor of 2016.
Indeed, if we write $r^2 = \dfrac{p^2}{q^2}$ in lowest terms, then we need $q^2$ to be a divisor of 2016.
mathwhiz16
2016-03-18 20:16:15
prime factorize 2016
prime factorize 2016
DPatrick
2016-03-18 20:16:20
The prime factorization is $2016 = 2^5 \cdot 3^2 \cdot 7$. (Hopefully you had this memorized before the contest and didn't have to compute it on the fly.)
The prime factorization is $2016 = 2^5 \cdot 3^2 \cdot 7$. (Hopefully you had this memorized before the contest and didn't have to compute it on the fly.)
blue8931
2016-03-18 20:16:37
largest is 144
largest is 144
bluephoenix
2016-03-18 20:16:37
12 = q
12 = q
DPatrick
2016-03-18 20:16:49
Right, the largest possible value of $q$ is 12.
Right, the largest possible value of $q$ is 12.
Aopser1221
2016-03-18 20:16:52
11/12 is the largest factor
11/12 is the largest factor
DPatrick
2016-03-18 20:17:06
So correct, the largest possible value of $r$ is thus $\dfrac{11}{12}$.
So correct, the largest possible value of $r$ is thus $\dfrac{11}{12}$.
DPatrick
2016-03-18 20:17:26
So to finish, we need to compare the smallest and largest possible values of $2016r$.
So to finish, we need to compare the smallest and largest possible values of $2016r$.
magicflippinertle
2016-03-18 20:18:08
So smallest area for ILKJ is 1008
So smallest area for ILKJ is 1008
calculus_riju
2016-03-18 20:18:08
1008 is the smallest area of IJKL
1008 is the smallest area of IJKL
magicflippinertle
2016-03-18 20:18:08
largest area is 1848
largest area is 1848
idomath12345
2016-03-18 20:18:12
1008, 1848
1008, 1848
Aopser1221
2016-03-18 20:18:15
or just 5/12*2016
or just 5/12*2016
DPatrick
2016-03-18 20:18:18
Right.
Right.
DPatrick
2016-03-18 20:18:23
The largest area is $2016 \cdot \dfrac{11}{12}$ and the smallest is $2016 \cdot \dfrac12$.
The largest area is $2016 \cdot \dfrac{11}{12}$ and the smallest is $2016 \cdot \dfrac12$.
DPatrick
2016-03-18 20:18:28
So their difference is $2016\left(\dfrac{11}{12}-\dfrac12\right) = 2016 \cdot \dfrac{5}{12} = \boxed{840}$.
So their difference is $2016\left(\dfrac{11}{12}-\dfrac12\right) = 2016 \cdot \dfrac{5}{12} = \boxed{840}$.
blue8931
2016-03-18 20:18:46
very nice problem (although I couldn't solve it earlier lol)
very nice problem (although I couldn't solve it earlier lol)
DPatrick
2016-03-18 20:19:07
Yeah, I liked the nice blend of geometry and number theory. A blend of two different areas of math is a hallmark of a good AIME problem.
Yeah, I liked the nice blend of geometry and number theory. A blend of two different areas of math is a hallmark of a good AIME problem.
DPatrick
2016-03-18 20:19:30
On to the median problem (#8):
On to the median problem (#8):
DPatrick
2016-03-18 20:19:33
8. Find the number of sets $\{a,b,c\}$ of three distinct positive integers with the property that the product of $a$, $b$, and $c$ is equal to the product of 11, 21, 31, 41, 51, and 61.
8. Find the number of sets $\{a,b,c\}$ of three distinct positive integers with the property that the product of $a$, $b$, and $c$ is equal to the product of 11, 21, 31, 41, 51, and 61.
Shaddoll
2016-03-18 20:19:52
prime factorize the product
prime factorize the product
maverick8
2016-03-18 20:19:52
prime factorize product
prime factorize product
Aopser1221
2016-03-18 20:19:52
Factor it first
Factor it first
DPatrick
2016-03-18 20:20:04
Sure. We need $abc = 11 \cdot 21 \cdot 31 \cdot 41 \cdot 51 \cdot 61$. A better way to write that product is in terms of primes.
Sure. We need $abc = 11 \cdot 21 \cdot 31 \cdot 41 \cdot 51 \cdot 61$. A better way to write that product is in terms of primes.
DPatrick
2016-03-18 20:20:28
$21 = 3 \cdot 7$ and $51 = 3 \cdot 17$, and the rest are already prime.
$21 = 3 \cdot 7$ and $51 = 3 \cdot 17$, and the rest are already prime.
DPatrick
2016-03-18 20:20:39
Thus, $abc = 3^2 \cdot 7 \cdot 11 \cdot 17 \cdot 31 \cdot 41 \cdot 61$.
Thus, $abc = 3^2 \cdot 7 \cdot 11 \cdot 17 \cdot 31 \cdot 41 \cdot 61$.
DPatrick
2016-03-18 20:20:52
How do we count the solution sets $\{a,b,c\}$.
How do we count the solution sets $\{a,b,c\}$.
DPatrick
2016-03-18 20:20:53
?
?
Aopser1221
2016-03-18 20:21:12
We can count ordered triplets first and then divide by 6
We can count ordered triplets first and then divide by 6
Shaddoll
2016-03-18 20:21:15
we count how to distribute each factors while making sure no 2 are the same, and divide by $6$ since we counted each way $6$ times
we count how to distribute each factors while making sure no 2 are the same, and divide by $6$ since we counted each way $6$ times
mathlover3737
2016-03-18 20:21:19
find the number of ordered ones?
find the number of ordered ones?
blue8931
2016-03-18 20:21:19
and then divide by 6
and then divide by 6
atmchallenge
2016-03-18 20:21:21
First, let's suppose order counts.
First, let's suppose order counts.
DPatrick
2016-03-18 20:21:48
Sounds like a good plan. It seems easier to count ordered triples $(a,b,c)$ that work, then divide by 6 to account for the $3! = 6$ different ordering of sets versus ordered triples.
Sounds like a good plan. It seems easier to count ordered triples $(a,b,c)$ that work, then divide by 6 to account for the $3! = 6$ different ordering of sets versus ordered triples.
idomath12345
2016-03-18 20:22:04
Each prime besides three can be distributed to exactly one of a,b,c
Each prime besides three can be distributed to exactly one of a,b,c
DPatrick
2016-03-18 20:22:32
Indeed, this is now a problem of "distributing" the primes to $a$, $b$, or $c$. Kind of like putting balls into boxes, except two of the balls (the two 3's) are identical.
Indeed, this is now a problem of "distributing" the primes to $a$, $b$, or $c$. Kind of like putting balls into boxes, except two of the balls (the two 3's) are identical.
DPatrick
2016-03-18 20:22:54
Each prime (other than 3) has 3 choices of $a$, $b$, or $c$ to go to. So the other primes get distributed in $3^6 = 729$ ways.
Each prime (other than 3) has 3 choices of $a$, $b$, or $c$ to go to. So the other primes get distributed in $3^6 = 729$ ways.
DPatrick
2016-03-18 20:23:05
But then what about the 3's?
But then what about the 3's?
bluephoenix
2016-03-18 20:23:20
6 ways to do so
6 ways to do so
lucylou
2016-03-18 20:23:20
6 ways?
6 ways?
legolego
2016-03-18 20:23:26
they go in the same number or different numbers
they go in the same number or different numbers
DPatrick
2016-03-18 20:23:41
Right. Either both 3's can go to one number (in 3 ways), or we can give a 3 to two numbers (again in 3 ways). So that's 6 choices for the 3's.
Right. Either both 3's can go to one number (in 3 ways), or we can give a 3 to two numbers (again in 3 ways). So that's 6 choices for the 3's.
DPatrick
2016-03-18 20:23:49
That gives a total of $6 \cdot 729$ possibilities for ordered triples $(a,b,c)$.
That gives a total of $6 \cdot 729$ possibilities for ordered triples $(a,b,c)$.
DPatrick
2016-03-18 20:23:54
But...?
But...?
bluephoenix
2016-03-18 20:24:14
but we overcount when 2 of them are 1 or 3
but we overcount when 2 of them are 1 or 3
InCtrl
2016-03-18 20:24:14
1,1 and 3,3 in any one set
1,1 and 3,3 in any one set
24iam24
2016-03-18 20:24:14
subtract off 1, 1, ... and 3, 3, ...
subtract off 1, 1, ... and 3, 3, ...
calculus_riju
2016-03-18 20:24:14
(3,3,p) is not valid
(3,3,p) is not valid
DPatrick
2016-03-18 20:24:26
Right. We only want to count triples of distinct integers.
Right. We only want to count triples of distinct integers.
DPatrick
2016-03-18 20:24:36
There are the 3 triples with terms 1,1,9N (where $N = 7 \cdot 11 \cdot 17 \cdot 31 \cdot 41 \cdot 61$) and the 3 triples with terms 3,3,N, in which the numbers are not distinct.
There are the 3 triples with terms 1,1,9N (where $N = 7 \cdot 11 \cdot 17 \cdot 31 \cdot 41 \cdot 61$) and the 3 triples with terms 3,3,N, in which the numbers are not distinct.
Shaddoll
2016-03-18 20:24:46
we counted (1,1,x) and (3,3,x) which gives 6 bad cases
we counted (1,1,x) and (3,3,x) which gives 6 bad cases
DPatrick
2016-03-18 20:25:05
So we have to throw out those 6 possibilities, to get $6 \cdot 729 - 6 = 6 \cdot 728$ possible ordered triples of distinct numbers.
So we have to throw out those 6 possibilities, to get $6 \cdot 729 - 6 = 6 \cdot 728$ possible ordered triples of distinct numbers.
bluephoenix
2016-03-18 20:25:18
so we subtract 2*3 = 6, then divide by 6 to get $728$ yay done
so we subtract 2*3 = 6, then divide by 6 to get $728$ yay done
poptower99
2016-03-18 20:25:18
Divide by 3!=6 for unordered sets
Divide by 3!=6 for unordered sets
DPatrick
2016-03-18 20:25:24
And to finish, we divide by 6 to account for the fact that sets are unordered, and we end up with $\boxed{728}$ possible sets.
And to finish, we divide by 6 to account for the fact that sets are unordered, and we end up with $\boxed{728}$ possible sets.
DPatrick
2016-03-18 20:26:02
Next!
Next!
DPatrick
2016-03-18 20:26:06
9. The sequences of positive integers $1,a_2,a_3,\ldots$ and $1,b_2,b_3,\ldots$ are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let $c_n = a_n + b_n$. There is an integer $k$ such that $c_{k-1} = 100$ and $c_{k+1} = 1000$. Find $c_k$.
9. The sequences of positive integers $1,a_2,a_3,\ldots$ and $1,b_2,b_3,\ldots$ are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let $c_n = a_n + b_n$. There is an integer $k$ such that $c_{k-1} = 100$ and $c_{k+1} = 1000$. Find $c_k$.
DPatrick
2016-03-18 20:26:22
How can we work with these sequences?
How can we work with these sequences?
idomath12345
2016-03-18 20:26:46
Use general formula?
Use general formula?
mathlover3737
2016-03-18 20:26:46
write them in terms of differece and ratio
write them in terms of differece and ratio
JonXu101
2016-03-18 20:26:46
write them in a short equation
write them in a short equation
DPatrick
2016-03-18 20:26:57
Right: we can do the usual thing where we write the elements in terms of their common difference or ratio.
Right: we can do the usual thing where we write the elements in terms of their common difference or ratio.
atmchallenge
2016-03-18 20:27:03
Express them in the forms $1,1+d,1+2d,...$ and $1,r,r^2,r^3,...$.
Express them in the forms $1,1+d,1+2d,...$ and $1,r,r^2,r^3,...$.
DPatrick
2016-03-18 20:27:08
Exactly: if the common difference of the arithmetic sequence is $d$, then we can write the sequence as \[ 1, 1+d, 1+2d, 1+3d, \ldots.\]
Exactly: if the common difference of the arithmetic sequence is $d$, then we can write the sequence as \[ 1, 1+d, 1+2d, 1+3d, \ldots.\]
DPatrick
2016-03-18 20:27:15
Similarly, the geometric sequence with common ratio $r$ is \[1, r, r^2, r^3, \ldots.\]
Similarly, the geometric sequence with common ratio $r$ is \[1, r, r^2, r^3, \ldots.\]
DPatrick
2016-03-18 20:27:20
Then what is $c_n$?
Then what is $c_n$?
blue8931
2016-03-18 20:27:53
$1+(n-1)d+r^{n-1}$
$1+(n-1)d+r^{n-1}$
24iam24
2016-03-18 20:27:53
r^(n-1)+1+(n-1)d
r^(n-1)+1+(n-1)d
atmchallenge
2016-03-18 20:27:53
$c_n=1+(n-1)d+r^{n-1}$.
$c_n=1+(n-1)d+r^{n-1}$.
Oleksenko
2016-03-18 20:27:53
r^n-1 + 1 + (n-1)d
r^n-1 + 1 + (n-1)d
walnutwaldo20
2016-03-18 20:27:58
$1+d\cdot(n-1)+r^{n-1}$
$1+d\cdot(n-1)+r^{n-1}$
Liopleurodon
2016-03-18 20:27:58
1+d(n-1)+r^n-1
1+d(n-1)+r^n-1
pramodmitikiri
2016-03-18 20:27:58
the sum of the two sequesnces, or 1+(n-1)d+r^n-1
the sum of the two sequesnces, or 1+(n-1)d+r^n-1
kbird
2016-03-18 20:27:58
1 + (n-1)d + r^{n-1)
1 + (n-1)d + r^{n-1)
DPatrick
2016-03-18 20:28:15
Right. We have $c_n = r^{n-1} + (n-1)d + 1$. (Resist the temptation to write $n$ where $n-1$ should be!)
Right. We have $c_n = r^{n-1} + (n-1)d + 1$. (Resist the temptation to write $n$ where $n-1$ should be!)
DPatrick
2016-03-18 20:28:28
Of course, we want to keep in mind that $d$ is a positive integer, and $r$ is a positive integer greater than 1.
Of course, we want to keep in mind that $d$ is a positive integer, and $r$ is a positive integer greater than 1.
DPatrick
2016-03-18 20:28:49
Now we can write the given data: \[\begin{array}{rcrcl}
c_{k-1} &=& r^{k-2} + (k-2)d + 1 &=& 100, \\
c_{k+1} &=&r^k + kd + 1 &=& 1000. \end{array}\]
Now we can write the given data: \[\begin{array}{rcrcl}
c_{k-1} &=& r^{k-2} + (k-2)d + 1 &=& 100, \\
c_{k+1} &=&r^k + kd + 1 &=& 1000. \end{array}\]
DPatrick
2016-03-18 20:29:02
Now what?
Now what?
idomath12345
2016-03-18 20:29:14
Subtract.
Subtract.
pramodmitikiri
2016-03-18 20:29:14
we subtract?
we subtract?
JonXu101
2016-03-18 20:29:20
subtract the equations?
subtract the equations?
mochi
2016-03-18 20:29:22
subtract
subtract
Oleksenko
2016-03-18 20:29:22
Subtract the two equations
Subtract the two equations
DPatrick
2016-03-18 20:29:23
Subtracting those seems useful: \[ r^k - r^{k-2} + 2d = 900. \]
Subtracting those seems useful: \[ r^k - r^{k-2} + 2d = 900. \]
DPatrick
2016-03-18 20:29:35
I might even factor it a little bit now: \[r^{k-2}(r^2-1) + 2d = 900.\]
I might even factor it a little bit now: \[r^{k-2}(r^2-1) + 2d = 900.\]
DPatrick
2016-03-18 20:29:46
What now?
What now?
kbird
2016-03-18 20:30:01
trial and error
trial and error
mathlover3737
2016-03-18 20:30:01
just casework on r now? it's at least 2
just casework on r now? it's at least 2
DPatrick
2016-03-18 20:30:11
Yeah, to be honest, in real time I pretty much guessed-and-checked from here. I looked at the system \begin{align*}
r^{k-2} + (k-2)d &= 99, \\
r^{k-2}(r^2-1) + 2d &= 900. \end{align*}
Yeah, to be honest, in real time I pretty much guessed-and-checked from here. I looked at the system \begin{align*}
r^{k-2} + (k-2)d &= 99, \\
r^{k-2}(r^2-1) + 2d &= 900. \end{align*}
DPatrick
2016-03-18 20:30:40
$r$ and $k$ are going to have to be pretty small to make this system work.
$r$ and $k$ are going to have to be pretty small to make this system work.
DPatrick
2016-03-18 20:31:08
I just tried $k=3$ (trying values of $k$ is more appealing to me because it gets rid of the variable exponent): \begin{align*}
r+d &= 99, \\
r(r^2-1) + 2d &= 900. \end{align*}
I just tried $k=3$ (trying values of $k$ is more appealing to me because it gets rid of the variable exponent): \begin{align*}
r+d &= 99, \\
r(r^2-1) + 2d &= 900. \end{align*}
DPatrick
2016-03-18 20:31:28
And now what?
And now what?
Liopleurodon
2016-03-18 20:31:55
multiply the first one by 2 and subtract to get rid of d?
multiply the first one by 2 and subtract to get rid of d?
idomath12345
2016-03-18 20:31:55
Multiply top by 2 and subtract.
Multiply top by 2 and subtract.
BobThePotato
2016-03-18 20:31:55
multiply the first equation by 2 and subtract?
multiply the first equation by 2 and subtract?
DPatrick
2016-03-18 20:32:05
We could then double the top one and subtract, to eliminate $d$ and get $r(r^2-3) = 702$.
We could then double the top one and subtract, to eliminate $d$ and get $r(r^2-3) = 702$.
magicflippinertle
2016-03-18 20:32:17
guess and check
guess and check
DPatrick
2016-03-18 20:32:34
Now it's really guess and check. $r^3 - 3r = 702$, so $r^3 > 702$ and...?
Now it's really guess and check. $r^3 - 3r = 702$, so $r^3 > 702$ and...?
blue8931
2016-03-18 20:32:44
r has to be 9
r has to be 9
JonXu101
2016-03-18 20:32:44
something around...9-10?
something around...9-10?
maverick8
2016-03-18 20:32:44
r=9
r=9
mathlover3737
2016-03-18 20:32:47
r=9
r=9
mathwhiz16
2016-03-18 20:32:47
r=9!
r=9!
DPatrick
2016-03-18 20:32:54
Right, that's just a little smaller than $r^3$, and hey, $9^3 = 729$ is close. And indeed, it works: $9(78) = 702$.
Right, that's just a little smaller than $r^3$, and hey, $9^3 = 729$ is close. And indeed, it works: $9(78) = 702$.
kbird
2016-03-18 20:33:04
it's r = 9 and d = 90.
it's r = 9 and d = 90.
24iam24
2016-03-18 20:33:04
r=9 d=90
r=9 d=90
DPatrick
2016-03-18 20:33:15
Right, backing up gives $d=90$ from the first equation, and indeed the sequences are: \[\begin{array}{rrrrl}
1,& 91,& 181,& 271,& \ldots \\
1,& 9,& 81,& 729,& \ldots \end{array}\]
Right, backing up gives $d=90$ from the first equation, and indeed the sequences are: \[\begin{array}{rrrrl}
1,& 91,& 181,& 271,& \ldots \\
1,& 9,& 81,& 729,& \ldots \end{array}\]
DPatrick
2016-03-18 20:33:28
This works: the sum of the 2nd terms is 100, and the sum of the 4th terms is 1000.
This works: the sum of the 2nd terms is 100, and the sum of the 4th terms is 1000.
blue8931
2016-03-18 20:33:38
262 is our answer
262 is our answer
kbird
2016-03-18 20:33:38
so it's 181 + 81 = 262
so it's 181 + 81 = 262
Liopleurodon
2016-03-18 20:33:38
262
262
DPatrick
2016-03-18 20:33:41
So our answer is the sum of the 3rd terms, which is $181 + 81 = \boxed{262}$.
So our answer is the sum of the 3rd terms, which is $181 + 81 = \boxed{262}$.
tanishq1
2016-03-18 20:33:55
i wish there was a better way
i wish there was a better way
DPatrick
2016-03-18 20:34:06
We could solve this a little more systematically. For example, we can go back to $r^{k-2}(r^2-1) + 2d = 900$. Is there any modulus which is fruitful to look at this equation?
We could solve this a little more systematically. For example, we can go back to $r^{k-2}(r^2-1) + 2d = 900$. Is there any modulus which is fruitful to look at this equation?
poptower99
2016-03-18 20:34:20
3 must divide d
3 must divide d
DPatrick
2016-03-18 20:34:27
Yeah -- mod 3 helps. Either $r \equiv 0 \pmod{3}$ or $r^2 \equiv 1 \pmod{3}$. Either way, the first term is a multiple of 3, and so is 900 of course. That means $2d$ is a multiple of 3, and hence so is $d$.
Yeah -- mod 3 helps. Either $r \equiv 0 \pmod{3}$ or $r^2 \equiv 1 \pmod{3}$. Either way, the first term is a multiple of 3, and so is 900 of course. That means $2d$ is a multiple of 3, and hence so is $d$.
poptower99
2016-03-18 20:34:50
And then 3 must divide r from the original equations
And then 3 must divide r from the original equations
DPatrick
2016-03-18 20:34:53
And now if we go back to $r^{k-2} + (k-2)d = 99$, we see that $r$ must be a multiple of 3 as well.
And now if we go back to $r^{k-2} + (k-2)d = 99$, we see that $r$ must be a multiple of 3 as well.
DPatrick
2016-03-18 20:35:01
So now $r$ must be 3, 6, or 9 (because we must have $r^3 < 1000$), and using the fact that $k \ge 3$ and $r^k < 1000$ really limits the possibilities -- we can just check them all.
So now $r$ must be 3, 6, or 9 (because we must have $r^3 < 1000$), and using the fact that $k \ge 3$ and $r^k < 1000$ really limits the possibilities -- we can just check them all.
DPatrick
2016-03-18 20:35:34
OK, I'm going to take a quick break before we go to the double-digit problems. Let's resume at :40 past.
OK, I'm going to take a quick break before we go to the double-digit problems. Let's resume at :40 past.
DPatrick
2016-03-18 20:40:32
OK, we're back and ready to go on to #10!
OK, we're back and ready to go on to #10!
DPatrick
2016-03-18 20:40:37
10. Triangle $ABC$ is inscribed in circle $\omega$. Points $P$ and $Q$ are on side $\overline{AB}$ with $AP < AQ$. Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$), respectively. If $AP = 4$, $PQ = 3$, $QB = 6$, $BT = 5$, $AS = 7$, then $ST = \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
10. Triangle $ABC$ is inscribed in circle $\omega$. Points $P$ and $Q$ are on side $\overline{AB}$ with $AP < AQ$. Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$), respectively. If $AP = 4$, $PQ = 3$, $QB = 6$, $BT = 5$, $AS = 7$, then $ST = \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
DPatrick
2016-03-18 20:40:47
Here's a sketch with all the info. Note that it's pretty hard to get a picture to scale properly, and I didn't really try.
Here's a sketch with all the info. Note that it's pretty hard to get a picture to scale properly, and I didn't really try.
DPatrick
2016-03-18 20:40:52
maverick8
2016-03-18 20:41:14
cyclic quadrilaterals
cyclic quadrilaterals
DPatrick
2016-03-18 20:41:32
Certainly. We need to find $ST$. We have a lot of data about all the other sides of $ABTS$. And $ABTS$ is cyclic.
Certainly. We need to find $ST$. We have a lot of data about all the other sides of $ABTS$. And $ABTS$ is cyclic.
JonXu101
2016-03-18 20:41:52
ptolemy's theorem? (later on)
ptolemy's theorem? (later on)
DPatrick
2016-03-18 20:42:03
This suggests Ptolemy's Theorem might help, which states that the product of the diagonals of a cyclic quadrilateral is equal to the sum of the products of the opposite sides.
This suggests Ptolemy's Theorem might help, which states that the product of the diagonals of a cyclic quadrilateral is equal to the sum of the products of the opposite sides.
DPatrick
2016-03-18 20:42:06
That is, \[ (AT)(BS) = (AB)(ST) + (AS)(BT).\]
That is, \[ (AT)(BS) = (AB)(ST) + (AS)(BT).\]
DPatrick
2016-03-18 20:42:55
There are certainly other ways to do this problem: you can chase the Law of Sines around and around, and it works. But I managed to get Ptolemy's to work nicely, so that's the solution I'm going to lead us through here.
There are certainly other ways to do this problem: you can chase the Law of Sines around and around, and it works. But I managed to get Ptolemy's to work nicely, so that's the solution I'm going to lead us through here.
calculus_riju
2016-03-18 20:43:10
we need BS AB AT
we need BS AB AT
DPatrick
2016-03-18 20:43:14
But this means we'll have to find $AT$ and $BS$. Let me draw them in so we can see them:
But this means we'll have to find $AT$ and $BS$. Let me draw them in so we can see them:
DPatrick
2016-03-18 20:43:18
DPatrick
2016-03-18 20:43:29
Now what? We have to use $P$ and $Q$ somehow...
Now what? We have to use $P$ and $Q$ somehow...
maverick8
2016-03-18 20:43:44
similar triangles
similar triangles
JonXu101
2016-03-18 20:43:46
ratios?
ratios?
DPatrick
2016-03-18 20:43:56
They do create lots of similar triangles, which means lots of ratios.
They do create lots of similar triangles, which means lots of ratios.
DPatrick
2016-03-18 20:44:15
In particular, around point $P$ we have $PAS \sim PCB$ and $PSB \sim PAC$.
And around point $Q$ we have $QAT \sim QCB$ and $QTB \sim QAC$.
In particular, around point $P$ we have $PAS \sim PCB$ and $PSB \sim PAC$.
And around point $Q$ we have $QAT \sim QCB$ and $QTB \sim QAC$.
DPatrick
2016-03-18 20:44:40
So let's just plow in by trying to find $AT$.
So let's just plow in by trying to find $AT$.
DPatrick
2016-03-18 20:44:45
It only appears in $QAT \sim QCB$ so let's write those ratios down: \[ \frac{AT}{CB} = \frac{QA}{QC} = \frac{QT}{QB}. \]
It only appears in $QAT \sim QCB$ so let's write those ratios down: \[ \frac{AT}{CB} = \frac{QA}{QC} = \frac{QT}{QB}. \]
DPatrick
2016-03-18 20:45:16
What can we do with this?
What can we do with this?
mochi
2016-03-18 20:45:32
simplify
simplify
DPatrick
2016-03-18 20:45:40
Certainly we know some of those numbers at least.
Certainly we know some of those numbers at least.
DPatrick
2016-03-18 20:45:49
I can fill the known lengths in: \[ \frac{AT}{CB} = \frac{7}{QC} = \frac{QT}{6}. \]
I can fill the known lengths in: \[ \frac{AT}{CB} = \frac{7}{QC} = \frac{QT}{6}. \]
Liopleurodon
2016-03-18 20:46:13
42=QC(QT)
42=QC(QT)
tanishq1
2016-03-18 20:46:24
power of a point?
power of a point?
DPatrick
2016-03-18 20:46:41
Right...one thing to note is that the second equals sign here is just the power of the point $Q$.
Right...one thing to note is that the second equals sign here is just the power of the point $Q$.
DPatrick
2016-03-18 20:46:55
But we can't solve for $AT$ yet.
But we can't solve for $AT$ yet.
DPatrick
2016-03-18 20:47:11
But maybe we don't have to! Ptolemy only requires us to know the product $(AT)(BS)$.
But maybe we don't have to! Ptolemy only requires us to know the product $(AT)(BS)$.
DPatrick
2016-03-18 20:47:34
So let's write down the same equation for $BS$ using $PSB \sim PAC$: \[ \frac{BS}{AC} = \frac{PS}{PA} = \frac{PB}{PC}. \] Again, we know some of these: \[ \frac{BS}{AC} = \frac{PS}{4} = \frac{9}{PC}. \]
So let's write down the same equation for $BS$ using $PSB \sim PAC$: \[ \frac{BS}{AC} = \frac{PS}{PA} = \frac{PB}{PC}. \] Again, we know some of these: \[ \frac{BS}{AC} = \frac{PS}{4} = \frac{9}{PC}. \]
DPatrick
2016-03-18 20:47:54
Maybe we can multiply them together to get an $(AT)(BS)$ term, and see what happens.
Maybe we can multiply them together to get an $(AT)(BS)$ term, and see what happens.
DPatrick
2016-03-18 20:48:29
Let's just pick one of the ratio equalities to use. We probably want to pick the same one in each. Let's use the one with the numbers in the denominator. That is, let's try multiplying $\dfrac{AT}{CB} = \dfrac{QT}{6}$ with $\dfrac{BS}{AC} = \dfrac{PS}{4}$.
Let's just pick one of the ratio equalities to use. We probably want to pick the same one in each. Let's use the one with the numbers in the denominator. That is, let's try multiplying $\dfrac{AT}{CB} = \dfrac{QT}{6}$ with $\dfrac{BS}{AC} = \dfrac{PS}{4}$.
DPatrick
2016-03-18 20:48:49
We get \[ \frac{(AT)(BS)}{(AC)(BC)} = \frac{(PS)(QT)}{24}.\]
We get \[ \frac{(AT)(BS)}{(AC)(BC)} = \frac{(PS)(QT)}{24}.\]
DPatrick
2016-03-18 20:49:07
And now we can isolate what we want: \[(AT)(BS) = \frac{1}{24}(AC)(BC)(PS)(QT).\] But what now?
And now we can isolate what we want: \[(AT)(BS) = \frac{1}{24}(AC)(BC)(PS)(QT).\] But what now?
DPatrick
2016-03-18 20:50:06
When stalled out in a geometry problem, it's often helpful to go back and see what we haven't used yet.
When stalled out in a geometry problem, it's often helpful to go back and see what we haven't used yet.
tanishq1
2016-03-18 20:50:45
QT/5=7/AC so QT=35/AC
QT/5=7/AC so QT=35/AC
DPatrick
2016-03-18 20:50:59
Aha! We have another two pairs of similar triangles that we can use!
Aha! We have another two pairs of similar triangles that we can use!
DPatrick
2016-03-18 20:51:15
That is, we can go to the other two pairs of similar triangles: $PAS \sim PCB$ and $QTB \sim QAC$.
That is, we can go to the other two pairs of similar triangles: $PAS \sim PCB$ and $QTB \sim QAC$.
DPatrick
2016-03-18 20:51:42
The second pair gives the ratio $\dfrac{QT}{QA} = \dfrac{TB}{AC}$, which (using known lengths) gives $\dfrac{QT}{7} = \dfrac{5}{AC}$.
The second pair gives the ratio $\dfrac{QT}{QA} = \dfrac{TB}{AC}$, which (using known lengths) gives $\dfrac{QT}{7} = \dfrac{5}{AC}$.
I_m_ram
2016-03-18 20:51:50
QT*AC =35
QT*AC =35
DPatrick
2016-03-18 20:52:13
So $(AC)(QT) = 35$, and we substitute this in: \[ (AT)(BS) = \dfrac{35}{24} (BC)(PS).\]
So $(AC)(QT) = 35$, and we substitute this in: \[ (AT)(BS) = \dfrac{35}{24} (BC)(PS).\]
JonXu101
2016-03-18 20:52:31
*BC/9=7/PS
*BC/9=7/PS
tanishq1
2016-03-18 20:52:31
PS/7=9/BC so PS=63/BC
PS/7=9/BC so PS=63/BC
DPatrick
2016-03-18 20:52:41
Unless we're really cursed, this should work exactly the same way on the remaining pair of similar triangles.
Unless we're really cursed, this should work exactly the same way on the remaining pair of similar triangles.
DPatrick
2016-03-18 20:53:12
$PAS \sim PCB$ gives us $\dfrac{PS}{9} = \dfrac{7}{BC}$.
$PAS \sim PCB$ gives us $\dfrac{PS}{9} = \dfrac{7}{BC}$.
kjw3704
2016-03-18 20:53:18
BC*PS=63
BC*PS=63
DPatrick
2016-03-18 20:53:24
So we get $(BC)(PS) = 63$.
So we get $(BC)(PS) = 63$.
DPatrick
2016-03-18 20:53:52
And then \[(AT)(BS) = \frac{35}{24} \cdot 63 = \frac{735}{8}.\]
And then \[(AT)(BS) = \frac{35}{24} \cdot 63 = \frac{735}{8}.\]
JonXu101
2016-03-18 20:53:59
we can substitute into ptolemy's to find ST pretty quickly...
we can substitute into ptolemy's to find ST pretty quickly...
DPatrick
2016-03-18 20:54:05
Right: and now we have all the data we need to plug in Ptolemy: \[ \frac{735}{8} = (AB)(ST) + (AS)(BT) = 13(ST) + 35. \]
Right: and now we have all the data we need to plug in Ptolemy: \[ \frac{735}{8} = (AB)(ST) + (AS)(BT) = 13(ST) + 35. \]
DPatrick
2016-03-18 20:54:24
This simplifies to $\dfrac{455}{8} = 13(ST)$, so $ST = \dfrac{35}{8}$, and our final answer is $35 + 8 = \boxed{043}$.
This simplifies to $\dfrac{455}{8} = 13(ST)$, so $ST = \dfrac{35}{8}$, and our final answer is $35 + 8 = \boxed{043}$.
DPatrick
2016-03-18 20:55:08
This was not easy. In fact this was the last problem I solved when I worked through the contest.
This was not easy. In fact this was the last problem I solved when I worked through the contest.
DPatrick
2016-03-18 20:55:43
You have to know Ptolemy for this solution, and you have to be persistent about chasing all the similar triangle ratios, and you have to not make a mistake.
You have to know Ptolemy for this solution, and you have to be persistent about chasing all the similar triangle ratios, and you have to not make a mistake.
JonXu101
2016-03-18 20:55:56
how do we do it using sine law? im curious...
how do we do it using sine law? im curious...
DPatrick
2016-03-18 20:56:19
I'm not sure -- I didn't try it, but some of my colleagues here at AoPS did -- but you can probably find some discussion (or start one!) on our community.
I'm not sure -- I didn't try it, but some of my colleagues here at AoPS did -- but you can probably find some discussion (or start one!) on our community.
calculus_riju
2016-03-18 20:56:29
onto #11
onto #11
DPatrick
2016-03-18 20:56:33
11. For positive integers $N$ and $k$, define $N$ to be $\textit{$k$-nice}$ if there exists a positive integer $a$ such that $a^k$ has exactly $N$ positive divisors. Find the number of positive integers less than 1000 that are neither 7-nice nor 8-nice.
11. For positive integers $N$ and $k$, define $N$ to be $\textit{$k$-nice}$ if there exists a positive integer $a$ such that $a^k$ has exactly $N$ positive divisors. Find the number of positive integers less than 1000 that are neither 7-nice nor 8-nice.
DPatrick
2016-03-18 20:56:53
So to start, how do we count the positive divisors of an integer?
So to start, how do we count the positive divisors of an integer?
brian22
2016-03-18 20:57:17
Prime factorize, multiply the exponents + 1
Prime factorize, multiply the exponents + 1
JonXu101
2016-03-18 20:57:17
add one to the exponents of prime factors and then multiply them together
add one to the exponents of prime factors and then multiply them together
Jack11
2016-03-18 20:57:23
prime factorization exponent's product adding one to each exponent
prime factorization exponent's product adding one to each exponent
DPatrick
2016-03-18 20:57:25
We take its prime factorization, and multiply together each exponent incremented by 1.
We take its prime factorization, and multiply together each exponent incremented by 1.
DPatrick
2016-03-18 20:57:30
That is, if $b = p_1^{e_1} p_2^{e_2} \cdots p_m^{e_m}$, then the number of positive divisors of $b$ is \[ (e_1+1)(e_2+1)\cdots(e_m+1). \]
That is, if $b = p_1^{e_1} p_2^{e_2} \cdots p_m^{e_m}$, then the number of positive divisors of $b$ is \[ (e_1+1)(e_2+1)\cdots(e_m+1). \]
DPatrick
2016-03-18 20:57:36
(You can think of this as counting the ways to "choose" a divisor of $b$, by choosing $0,1,2,\ldots,e_1$ powers of $p_1$, then $0,1,2,\ldots,e_2$ powers of $p_2$, and so on.)
(You can think of this as counting the ways to "choose" a divisor of $b$, by choosing $0,1,2,\ldots,e_1$ powers of $p_1$, then $0,1,2,\ldots,e_2$ powers of $p_2$, and so on.)
DPatrick
2016-03-18 20:57:45
So if $a = p_1^{e_1}p_2^{e_2}\cdots p_m^{e_m}$, then how many positive divisors does $a^k$ have?
So if $a = p_1^{e_1}p_2^{e_2}\cdots p_m^{e_m}$, then how many positive divisors does $a^k$ have?
tanishq1
2016-03-18 20:58:22
(e_1k+1)(e_2k+1)...
(e_1k+1)(e_2k+1)...
calculus_riju
2016-03-18 20:58:22
$\prod (e_i\cdot k +1)$
$\prod (e_i\cdot k +1)$
DPatrick
2016-03-18 20:58:30
Right: we have $a^k = p_1^{ke_1}p_2^{ke_2}\cdots p_m^{ke_m}$ as the prime factorization, so the number of divisors is \[ (ke_1+1)(ke_2+1)\cdots(ke_m+1).\]
Right: we have $a^k = p_1^{ke_1}p_2^{ke_2}\cdots p_m^{ke_m}$ as the prime factorization, so the number of divisors is \[ (ke_1+1)(ke_2+1)\cdots(ke_m+1).\]
DPatrick
2016-03-18 20:58:37
So $N$ is $k$-nice if and only if $N = (ke_1+1)(ke_2+1)\cdots(ke_m+1)$ for some $m$ and some positive integers $e_1,e_2,\ldots,e_m$.
So $N$ is $k$-nice if and only if $N = (ke_1+1)(ke_2+1)\cdots(ke_m+1)$ for some $m$ and some positive integers $e_1,e_2,\ldots,e_m$.
DPatrick
2016-03-18 20:58:45
Is there an easier way to say this?
Is there an easier way to say this?
poptower99
2016-03-18 20:59:06
1 mod k
1 mod k
flaminpotato
2016-03-18 20:59:10
1 mod k
1 mod k
DPatrick
2016-03-18 20:59:22
All those terms in the product are 1 more than a multiple of $k$.
All those terms in the product are 1 more than a multiple of $k$.
DPatrick
2016-03-18 20:59:29
So when we multiply them together, the product is still 1 more than a multiple of $k$.
So when we multiply them together, the product is still 1 more than a multiple of $k$.
DPatrick
2016-03-18 20:59:38
And the conclusion is: $N$ is $k$-nice if and only if $N \equiv 1 \pmod{k}$.
And the conclusion is: $N$ is $k$-nice if and only if $N \equiv 1 \pmod{k}$.
DPatrick
2016-03-18 20:59:52
(You might worry that we've only proved the "only if" direction. But we can easily construct an $N$ to prove the "if" direction: just let $e_1 = \dfrac{N-1}{k}$ and then take $a = p^{e_1}$ for any prime $p$.)
(You might worry that we've only proved the "only if" direction. But we can easily construct an $N$ to prove the "if" direction: just let $e_1 = \dfrac{N-1}{k}$ and then take $a = p^{e_1}$ for any prime $p$.)
DPatrick
2016-03-18 21:00:08
So, happily, now our problem is much more simply stated:
So, happily, now our problem is much more simply stated:
DPatrick
2016-03-18 21:00:11
11. Find the number of positive integers $N$ less than 1000 such that $N \not\equiv 1 \pmod{7}$ and $N \not\equiv 1 \pmod{8}$.
11. Find the number of positive integers $N$ less than 1000 such that $N \not\equiv 1 \pmod{7}$ and $N \not\equiv 1 \pmod{8}$.
24iam24
2016-03-18 21:00:29
complementary counting
complementary counting
blue8931
2016-03-18 21:00:29
complementary counting
complementary counting
mdu
2016-03-18 21:00:30
Inclusion-Exclusion
Inclusion-Exclusion
DPatrick
2016-03-18 21:00:34
This is a job for complementary counting and PIE!
This is a job for complementary counting and PIE!
DPatrick
2016-03-18 21:00:49
That is, we find the number of integers that are 1 (mod 7) or 1 (mod 8), and subtract from the total of 999 positive integers less than 1000.
That is, we find the number of integers that are 1 (mod 7) or 1 (mod 8), and subtract from the total of 999 positive integers less than 1000.
DPatrick
2016-03-18 21:01:16
How many positive integers less than 1000 are equivalent to 1 (mod 7)?
How many positive integers less than 1000 are equivalent to 1 (mod 7)?
mdu
2016-03-18 21:01:44
143 btw
143 btw
mathlover3737
2016-03-18 21:01:44
143
143
DPatrick
2016-03-18 21:01:50
Note that $1000 = 142 \cdot 7 + 6$.
Note that $1000 = 142 \cdot 7 + 6$.
DPatrick
2016-03-18 21:01:58
So the positive integers that are 1 (mod 7) are \[0 \cdot 7 + 1,\; 1 \cdot 7 + 1,\; 2 \cdot 7 + 1,\; \ldots,\; 142 \cdot 7 + 1.\] There are 143 of them.
So the positive integers that are 1 (mod 7) are \[0 \cdot 7 + 1,\; 1 \cdot 7 + 1,\; 2 \cdot 7 + 1,\; \ldots,\; 142 \cdot 7 + 1.\] There are 143 of them.
DPatrick
2016-03-18 21:02:14
How many are 1 (mod 8)?
How many are 1 (mod 8)?
24iam24
2016-03-18 21:02:28
125
125
Shaddoll
2016-03-18 21:02:28
$125$
$125$
pramodmitikiri
2016-03-18 21:02:28
125
125
blue8931
2016-03-18 21:02:28
125
125
poptower99
2016-03-18 21:02:28
125
125
DPatrick
2016-03-18 21:02:31
We have $1000 = 125 \cdot 8$, so all of $0 \cdot 8 + 1$ through $124 \cdot 8 + 1$ are 1 (mod 8). There are 125 of them.
We have $1000 = 125 \cdot 8$, so all of $0 \cdot 8 + 1$ through $124 \cdot 8 + 1$ are 1 (mod 8). There are 125 of them.
blue8931
2016-03-18 21:02:41
but consider the overlap (1 (mod 56))
but consider the overlap (1 (mod 56))
DPatrick
2016-03-18 21:02:46
Right. We've double-counted those that are both -- that is, those that are 1 (mod 56). How many are there?
Right. We've double-counted those that are both -- that is, those that are 1 (mod 56). How many are there?
24iam24
2016-03-18 21:03:13
18
18
pie314159265
2016-03-18 21:03:13
18
18
pramodmitikiri
2016-03-18 21:03:13
18
18
Oleksenko
2016-03-18 21:03:13
18?
18?
DPatrick
2016-03-18 21:03:16
Note that $1000 = 17 \cdot 56 + 48$. So all of $0 \cdot 56 + 1$ through $17 \cdot 56 + 1$ are double-counted. That's 18 double-counted numbers.
Note that $1000 = 17 \cdot 56 + 48$. So all of $0 \cdot 56 + 1$ through $17 \cdot 56 + 1$ are double-counted. That's 18 double-counted numbers.
DPatrick
2016-03-18 21:03:32
To finish, there are 143 + 125 - 18 = 250 positive integers less than 1000 that are 7-nice or 8-nice.
To finish, there are 143 + 125 - 18 = 250 positive integers less than 1000 that are 7-nice or 8-nice.
DPatrick
2016-03-18 21:03:48
Hence, the final answer is...?
Hence, the final answer is...?
blue8931
2016-03-18 21:04:04
999-250 = 749
999-250 = 749
Shaddoll
2016-03-18 21:04:04
We have to subtract from $999$ since we counted the complement, so the answer is 749
We have to subtract from $999$ since we counted the complement, so the answer is 749
poptower99
2016-03-18 21:04:04
999-250=749
999-250=749
mdu
2016-03-18 21:04:04
$999-250=749$
$999-250=749$
pramodmitikiri
2016-03-18 21:04:04
749
749
DPatrick
2016-03-18 21:04:08
Hence, there are $999 - 250 = \boxed{749}$ that are neither 7-nice nor 8-nice.
Hence, there are $999 - 250 = \boxed{749}$ that are neither 7-nice nor 8-nice.
DPatrick
2016-03-18 21:04:33
It is really, really, really easy to be off by 1 in problems like this. You have to be extra-careful in your counting!
It is really, really, really easy to be off by 1 in problems like this. You have to be extra-careful in your counting!
DPatrick
2016-03-18 21:04:56
On to a stretch of three problems that I really liked, starting with #12:
On to a stretch of three problems that I really liked, starting with #12:
DPatrick
2016-03-18 21:04:59
12. The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
12. The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
DPatrick
2016-03-18 21:05:02
avchess
2016-03-18 21:05:43
casework
casework
Shaddoll
2016-03-18 21:05:44
We use recursion
We use recursion
tanishq1
2016-03-18 21:06:00
i honestly just unraveled it in and used casework when the ends were same
i honestly just unraveled it in and used casework when the ends were same
Goobersmith
2016-03-18 21:06:00
casework for the colors
casework for the colors
DPatrick
2016-03-18 21:06:13
It is possible to solve this directly using very very careful casework. I did it for fun(!) just to see how messy it was.
It is possible to solve this directly using very very careful casework. I did it for fun(!) just to see how messy it was.
DPatrick
2016-03-18 21:06:20
But there's a clever solution using recursion.
But there's a clever solution using recursion.
DPatrick
2016-03-18 21:06:38
For example, let's suppose the top two pieces are red and blue, and consider the piece marked X:
For example, let's suppose the top two pieces are red and blue, and consider the piece marked X:
DPatrick
2016-03-18 21:06:41
DPatrick
2016-03-18 21:06:49
Of course, X cannot be blue. What if X is any color but red?
Of course, X cannot be blue. What if X is any color but red?
DPatrick
2016-03-18 21:07:21
How can I make a smaller ring? (That's the goal with recursion: try to relate the problem to a smaller version of the same problem.)
How can I make a smaller ring? (That's the goal with recursion: try to relate the problem to a smaller version of the same problem.)
legolego
2016-03-18 21:07:59
get rid of the blue
get rid of the blue
calculus_riju
2016-03-18 21:08:04
take 1 segment out
take 1 segment out
DPatrick
2016-03-18 21:08:19
Right. If X is not red, then I can cut out the blue piece, and glue the rest together to make a 5-segment ring.
Right. If X is not red, then I can cut out the blue piece, and glue the rest together to make a 5-segment ring.
DPatrick
2016-03-18 21:08:54
And that's reversible: if I have a 5-segment ring, I can "splice in" a 6th piece like the above.
And that's reversible: if I have a 5-segment ring, I can "splice in" a 6th piece like the above.
DPatrick
2016-03-18 21:09:14
But what if X is red? I can't cut the blue piece out, because the two red pieces won't legally connect.
But what if X is red? I can't cut the blue piece out, because the two red pieces won't legally connect.
DPatrick
2016-03-18 21:09:18
tanishq1
2016-03-18 21:09:32
cut out the blue AND x (OMG OMG)
cut out the blue AND x (OMG OMG)
DPatrick
2016-03-18 21:09:51
Right! If X is red, then Y can't be red -- so I can cut out both the blue and X, and get a 4-segment ring.
Right! If X is red, then Y can't be red -- so I can cut out both the blue and X, and get a 4-segment ring.
DPatrick
2016-03-18 21:10:11
So reversing these two cases, we can form a 6-piece ring in one of two ways:
(a) we can take a 5-piece ring, and splice in a piece whose color doesn't match either endpoint; or
(b) we can take a 4-piece ring, splice in a piece whose color matches the other endpoint (this is the red piece marked "X" in my example above), and then splice in a 6th piece whose color doesn't match the two new endpoints.
So reversing these two cases, we can form a 6-piece ring in one of two ways:
(a) we can take a 5-piece ring, and splice in a piece whose color doesn't match either endpoint; or
(b) we can take a 4-piece ring, splice in a piece whose color matches the other endpoint (this is the red piece marked "X" in my example above), and then splice in a 6th piece whose color doesn't match the two new endpoints.
DPatrick
2016-03-18 21:10:37
How many choices for a new color do we have in the above cases?
How many choices for a new color do we have in the above cases?
I_m_ram
2016-03-18 21:10:56
3
3
mathlover3737
2016-03-18 21:10:56
2
2
24iam24
2016-03-18 21:10:56
3
3
DPatrick
2016-03-18 21:11:01
You're both right.
You're both right.
DPatrick
2016-03-18 21:11:21
In case (a), we have 2 choices for the new piece's color: it can't match either endpoint. (Which are the red piece of the piece marked "X" in the first example.)
In case (a), we have 2 choices for the new piece's color: it can't match either endpoint. (Which are the red piece of the piece marked "X" in the first example.)
DPatrick
2016-03-18 21:11:34
In case (b), we have no choice for the first added piece (the piece labeled "X" -- it has to be red by construction), but then 3 choices for the final piece (it just can't match X's color).
In case (b), we have no choice for the first added piece (the piece labeled "X" -- it has to be red by construction), but then 3 choices for the final piece (it just can't match X's color).
DPatrick
2016-03-18 21:11:49
So if we let $c_n$ be the number of ways to color a ring with $n$ pieces, then what is the recursive formula?
So if we let $c_n$ be the number of ways to color a ring with $n$ pieces, then what is the recursive formula?
tanishq1
2016-03-18 21:12:03
F_(n)=2F_(n-1)+3F_(n-2)
F_(n)=2F_(n-1)+3F_(n-2)
DPatrick
2016-03-18 21:12:17
(You wrote that before I said "c" I think!)
(You wrote that before I said "c" I think!)
DPatrick
2016-03-18 21:12:23
We get $2c_{n-1}$ colorings from case (a) and $3c_{n-2}$ colorings from case (b).
We get $2c_{n-1}$ colorings from case (a) and $3c_{n-2}$ colorings from case (b).
DPatrick
2016-03-18 21:12:31
So the formula is $c_n = 2c_{n-1} + 3c_{n-2}$.
So the formula is $c_n = 2c_{n-1} + 3c_{n-2}$.
DPatrick
2016-03-18 21:12:45
And what do we do with this formula?
And what do we do with this formula?
InCtrl
2016-03-18 21:13:05
plug n chug
plug n chug
Oleksenko
2016-03-18 21:13:05
we know the first terms
we know the first terms
DPatrick
2016-03-18 21:13:18
We can compute the base cases.
We can compute the base cases.
DPatrick
2016-03-18 21:13:30
$c_1$ doesn't really make sense so I'd avoid it.
$c_1$ doesn't really make sense so I'd avoid it.
DPatrick
2016-03-18 21:13:39
$c_2 = 12$: just choose any two colors.
$c_3 = 24$: just choose any three colors.
$c_2 = 12$: just choose any two colors.
$c_3 = 24$: just choose any three colors.
DPatrick
2016-03-18 21:14:00
And now we just compute:
And now we just compute:
calculus_riju
2016-03-18 21:14:05
$c_4=84
$c_4=84
DPatrick
2016-03-18 21:14:08
$c_4 = 2c_3 + 3c_2 = 2(24) + 3(12) = 48 + 36 = 84$
$c_4 = 2c_3 + 3c_2 = 2(24) + 3(12) = 48 + 36 = 84$
DPatrick
2016-03-18 21:14:21
$c_5 = 2c_4 + 3c_3 = 2(84) + 3(24) = 168 + 72 = 240$
$c_5 = 2c_4 + 3c_3 = 2(84) + 3(24) = 168 + 72 = 240$
blue8931
2016-03-18 21:14:38
732 is our answer
732 is our answer
Liopleurodon
2016-03-18 21:14:38
732!
732!
DPatrick
2016-03-18 21:14:39
And finally $c_6 = 2c_5 + 3c_4 = 2(240) + 3(84) = 480 + 252 = \boxed{732}$
And finally $c_6 = 2c_5 + 3c_4 = 2(240) + 3(84) = 480 + 252 = \boxed{732}$
DPatrick
2016-03-18 21:15:13
Like I said, you can certainly do very, very careful casework and count this directly. But the recursion is way more fun.
Like I said, you can certainly do very, very careful casework and count this directly. But the recursion is way more fun.
DPatrick
2016-03-18 21:15:26
Three more!
Three more!
DPatrick
2016-03-18 21:15:30
13. Beatrix is going to place six rooks on a $6 \times 6$ chessboard where both the rows and columns are labeled 1 to 6; the rooks are placed so that no two rooks are in the same row or the same column. The value of a square is the sum of its row number and column number. The score or an arrangement of rooks is the least value of any occupied square. The average score over all valid configurations is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.
13. Beatrix is going to place six rooks on a $6 \times 6$ chessboard where both the rows and columns are labeled 1 to 6; the rooks are placed so that no two rooks are in the same row or the same column. The value of a square is the sum of its row number and column number. The score or an arrangement of rooks is the least value of any occupied square. The average score over all valid configurations is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.
DPatrick
2016-03-18 21:15:41
To be clear, here's the picture of the chessboard, with the squares' values labeled:
To be clear, here's the picture of the chessboard, with the squares' values labeled:
DPatrick
2016-03-18 21:15:45
DPatrick
2016-03-18 21:15:57
How many configurations are there in total?
How many configurations are there in total?
Shaddoll
2016-03-18 21:16:11
$6!$
$6!$
24iam24
2016-03-18 21:16:11
6!=720
6!=720
Peggy
2016-03-18 21:16:11
6!
6!
maverick8
2016-03-18 21:16:11
720
720
DPatrick
2016-03-18 21:16:18
There are 6 choices for the rook in the first column, then 5 choices remaining for the rook in the second column, etc., for a total of $6! = 720$ legal arrangements.
There are 6 choices for the rook in the first column, then 5 choices remaining for the rook in the second column, etc., for a total of $6! = 720$ legal arrangements.
DPatrick
2016-03-18 21:16:30
How do we count how many arrangements have each particular score?
How do we count how many arrangements have each particular score?
DPatrick
2016-03-18 21:16:41
Or is there a better way to approach the problem?
Or is there a better way to approach the problem?
blue8931
2016-03-18 21:17:02
we could do casework, but I feel like there's something more efficient
we could do casework, but I feel like there's something more efficient
DPatrick
2016-03-18 21:17:14
Yeah, if you try to count how many arrangements have a score of $k$, it quickly bogs down in really messy casework.
Yeah, if you try to count how many arrangements have a score of $k$, it quickly bogs down in really messy casework.
DPatrick
2016-03-18 21:17:31
However, there is a clever insight that makes the problem much more doable.
However, there is a clever insight that makes the problem much more doable.
DPatrick
2016-03-18 21:18:03
Counting the number of arrangements with a score of at least $k$ is fairly easy!
Counting the number of arrangements with a score of at least $k$ is fairly easy!
DPatrick
2016-03-18 21:18:22
For instance, how many arrangements have a score of at least 3?
For instance, how many arrangements have a score of at least 3?
24iam24
2016-03-18 21:18:47
600
600
Shaddoll
2016-03-18 21:18:47
$5*5*4*3*2*1$
$5*5*4*3*2*1$
Liopleurodon
2016-03-18 21:18:47
600
600
DPatrick
2016-03-18 21:18:51
We only have to avoid the lower-left square:
We only have to avoid the lower-left square:
DPatrick
2016-03-18 21:18:54
DPatrick
2016-03-18 21:19:01
There are now only 5 choices for the rook in the first column.
There are now only 5 choices for the rook in the first column.
DPatrick
2016-03-18 21:19:11
But once that rook is placed, there are no more additional restrictions (except to avoid that first rook's row, as usual), so there are $5!$ ways to finish the arrangement from here.
But once that rook is placed, there are no more additional restrictions (except to avoid that first rook's row, as usual), so there are $5!$ ways to finish the arrangement from here.
DPatrick
2016-03-18 21:19:25
So there are $5 \cdot 5! = 600$ configurations with a score of at least 3.
So there are $5 \cdot 5! = 600$ configurations with a score of at least 3.
DPatrick
2016-03-18 21:19:32
How many with a score of at least 4?
How many with a score of at least 4?
Shaddoll
2016-03-18 21:20:02
$4*4*4*3*2*1$
$4*4*4*3*2*1$
flaminpotato
2016-03-18 21:20:02
4*4*4*3*2*1
4*4*4*3*2*1
Oleksenko
2016-03-18 21:20:02
4*4*4*3*2*1
4*4*4*3*2*1
pramodmitikiri
2016-03-18 21:20:07
384
384
DPatrick
2016-03-18 21:20:09
Now we have to avoid the three squares in the bottom left:
Now we have to avoid the three squares in the bottom left:
DPatrick
2016-03-18 21:20:13
DPatrick
2016-03-18 21:20:18
There are 4 choices for the rook in the first column.
There are 4 choices for the rook in the first column.
DPatrick
2016-03-18 21:20:24
This will leave 4 squares for the rook in the second column.
This will leave 4 squares for the rook in the second column.
DPatrick
2016-03-18 21:20:29
And then there are no more additional restrictions, so there are $4!$ ways to finish.
And then there are no more additional restrictions, so there are $4!$ ways to finish.
DPatrick
2016-03-18 21:20:35
So the number of arrangements with score at least 4 is $4 \cdot 4 \cdot 4! = 16 \cdot 24 = 384$.
So the number of arrangements with score at least 4 is $4 \cdot 4 \cdot 4! = 16 \cdot 24 = 384$.
tanishq1
2016-03-18 21:20:49
3*3*3*3*2*1
3*3*3*3*2*1
DPatrick
2016-03-18 21:20:54
Right, the pattern continues.
Right, the pattern continues.
DPatrick
2016-03-18 21:20:58
If we need our score to be at least 5, then there are 3 choices for each of the first three columns, and then 3! = 6 ways to finish.
If we need our score to be at least 5, then there are 3 choices for each of the first three columns, and then 3! = 6 ways to finish.
DPatrick
2016-03-18 21:21:06
So there are $3^3 \cdot 3! = 27 \cdot 6 = 162$ ways with a score of at least 5.
So there are $3^3 \cdot 3! = 27 \cdot 6 = 162$ ways with a score of at least 5.
DPatrick
2016-03-18 21:21:17
Similarly, there are $2^4 \cdot 2! = 32$ ways with a score of at least 6.
Similarly, there are $2^4 \cdot 2! = 32$ ways with a score of at least 6.
DPatrick
2016-03-18 21:21:21
And there's just 1 way with a score of at least 7 -- we use all the "7" squares. (And, no ways to have a score of 8 or more.)
And there's just 1 way with a score of at least 7 -- we use all the "7" squares. (And, no ways to have a score of 8 or more.)
DPatrick
2016-03-18 21:21:25
Here's the summary of our data: \[ \begin{array}{r|c|c|c|c|c|c|l}
\text{Score at least} & 2 & 3 & 4 & 5 & 6 & 7 & 8+ \\ \hline
\text{Number} & 720 & 600 & 384 & 162 & 32 & 1 & 0 \end{array}\]
Here's the summary of our data: \[ \begin{array}{r|c|c|c|c|c|c|l}
\text{Score at least} & 2 & 3 & 4 & 5 & 6 & 7 & 8+ \\ \hline
\text{Number} & 720 & 600 & 384 & 162 & 32 & 1 & 0 \end{array}\]
DPatrick
2016-03-18 21:21:28
What do we do now?
What do we do now?
24iam24
2016-03-18 21:21:53
subtract off stuff to get exactly 2, 3, 4, ...
subtract off stuff to get exactly 2, 3, 4, ...
InCtrl
2016-03-18 21:21:56
1 thats i exactly 7, 31 thats is exactly 6, 130 that is exactly 5 etc et
1 thats i exactly 7, 31 thats is exactly 6, 130 that is exactly 5 etc et
DPatrick
2016-03-18 21:22:03
Right. The number of positions with an exact score of $k$ is just the difference of (score at least $k$) - (score at least $k+1$).
Right. The number of positions with an exact score of $k$ is just the difference of (score at least $k$) - (score at least $k+1$).
DPatrick
2016-03-18 21:22:21
So the differences give us the counts we want. We can make this chart too: \[ \begin{array}{r|c|c|c|c|c|c|l}
\text{Score exact} & 2 & 3 & 4 & 5 & 6 & 7 & 8+ \\ \hline
\text{Number} & 120 & 216 & 222 & 130 & 31 & 1 & 0 \end{array}\]
So the differences give us the counts we want. We can make this chart too: \[ \begin{array}{r|c|c|c|c|c|c|l}
\text{Score exact} & 2 & 3 & 4 & 5 & 6 & 7 & 8+ \\ \hline
\text{Number} & 120 & 216 & 222 & 130 & 31 & 1 & 0 \end{array}\]
Shaddoll
2016-03-18 21:22:45
Now we just multiply out and divide by $720$ to find the expected value
Now we just multiply out and divide by $720$ to find the expected value
DPatrick
2016-03-18 21:22:50
And now, to finish, we can compute the expected value:
And now, to finish, we can compute the expected value:
DPatrick
2016-03-18 21:23:06
\[\frac{2(120) + 3(216) + 4(222) + 5(130) + 6(31) + 7(1)}{720}\]
\[\frac{2(120) + 3(216) + 4(222) + 5(130) + 6(31) + 7(1)}{720}\]
DPatrick
2016-03-18 21:23:15
\[= \frac{240 + 648 + 888 + 650 + 186 + 7}{720} = \frac{2619}{720}.\]
\[= \frac{240 + 648 + 888 + 650 + 186 + 7}{720} = \frac{2619}{720}.\]
tanishq1
2016-03-18 21:23:27
cancel factors
cancel factors
DPatrick
2016-03-18 21:23:32
This reduces to lowest terms (by dividing out a 9) as $\dfrac{291}{80}$, so our answer is $291 + 80 = \boxed{371}$.
This reduces to lowest terms (by dividing out a 9) as $\dfrac{291}{80}$, so our answer is $291 + 80 = \boxed{371}$.
DPatrick
2016-03-18 21:23:56
Kind of a cute problem, I thought.
Kind of a cute problem, I thought.
DPatrick
2016-03-18 21:24:05
On to our last geometry problem of the night:
On to our last geometry problem of the night:
DPatrick
2016-03-18 21:24:10
14. Equilateral triangle $\triangle ABC$ has side length 600. Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on the opposite sides of the plane. Furthermore, $PA = PB = PC$, and $QA = QB = QC$, and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^\circ$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A$, $B$, $C$, $P$, and $Q$ is $d$. Find $d$.
14. Equilateral triangle $\triangle ABC$ has side length 600. Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on the opposite sides of the plane. Furthermore, $PA = PB = PC$, and $QA = QB = QC$, and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^\circ$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A$, $B$, $C$, $P$, and $Q$ is $d$. Find $d$.
DPatrick
2016-03-18 21:24:18
Ooh, 3D!
Ooh, 3D!
mattpi
2016-03-18 21:24:34
diagram?
diagram?
pie314159265
2016-03-18 21:24:34
diagram?
diagram?
mochi
2016-03-18 21:24:34
PICTURE!!!
PICTURE!!!
DPatrick
2016-03-18 21:24:36
We'll want to try to draw a picture. We need to deduce some facts first. What do we know about $P$ and $Q$?
We'll want to try to draw a picture. We need to deduce some facts first. What do we know about $P$ and $Q$?
mathlover3737
2016-03-18 21:24:51
PQ intersects center of ABC
PQ intersects center of ABC
flaminpotato
2016-03-18 21:24:56
on the line perpendicular to the center
on the line perpendicular to the center
bluephoenix
2016-03-18 21:25:02
Note that P is on the line perpendicular to ABC plane and passsing thtough the circumcneter, same with !
Note that P is on the line perpendicular to ABC plane and passsing thtough the circumcneter, same with !
DPatrick
2016-03-18 21:25:10
Right. They both lie on a line that's perpendicular to the plane containing $\triangle ABC$ (let's call that plane $X$), one on either side of the plane, and that passes through the center of $\triangle ABC$, which I'll call $Z$.
Right. They both lie on a line that's perpendicular to the plane containing $\triangle ABC$ (let's call that plane $X$), one on either side of the plane, and that passes through the center of $\triangle ABC$, which I'll call $Z$.
DPatrick
2016-03-18 21:25:22
DPatrick
2016-03-18 21:25:32
To see this, note that $Z$ is the only point in $X$ that's equidistant from $A$, $B$, and $C$. Then we can move "up" or "down" away from plane $X$ as much as we like, as long as we stay directly above or below $Z$.
To see this, note that $Z$ is the only point in $X$ that's equidistant from $A$, $B$, and $C$. Then we can move "up" or "down" away from plane $X$ as much as we like, as long as we stay directly above or below $Z$.
flaminpotato
2016-03-18 21:25:43
also point O is the midpoint of PQ
also point O is the midpoint of PQ
DPatrick
2016-03-18 21:25:51
Since $OP = OQ$, it must be the midpoint of $\overline{PQ}$. Let's leave it off the diagram for now.
Since $OP = OQ$, it must be the midpoint of $\overline{PQ}$. Let's leave it off the diagram for now.
DPatrick
2016-03-18 21:26:01
How are we going to use the dihedral angle data?
How are we going to use the dihedral angle data?
DPatrick
2016-03-18 21:26:37
Where is that angle in our picture?
Where is that angle in our picture?
treemath
2016-03-18 21:26:42
drop altitudes of PAB and QAB
drop altitudes of PAB and QAB
mathlover3737
2016-03-18 21:26:42
perpendiculars to AB from P and Q
perpendiculars to AB from P and Q
DPatrick
2016-03-18 21:26:51
Right. The two planes "hinge" across the line containing $\overline{AB}$.
Right. The two planes "hinge" across the line containing $\overline{AB}$.
DPatrick
2016-03-18 21:26:56
We drop perpendiculars from $P$ and $Q$ down to $\overline{AB}$. The angle they form there is the dihedral angle.
We drop perpendiculars from $P$ and $Q$ down to $\overline{AB}$. The angle they form there is the dihedral angle.
DPatrick
2016-03-18 21:27:08
Where do those perpendiculars intersect $\overline{AB}$?
Where do those perpendiculars intersect $\overline{AB}$?
Peggy
2016-03-18 21:27:19
midpoint
midpoint
blue8931
2016-03-18 21:27:19
midpoint
midpoint
DPatrick
2016-03-18 21:27:22
At its midpoint. Let's call it $M$.
At its midpoint. Let's call it $M$.
DPatrick
2016-03-18 21:27:26
Here's an updated picture, with the two new segments in blue:
Here's an updated picture, with the two new segments in blue:
DPatrick
2016-03-18 21:27:29
DPatrick
2016-03-18 21:27:38
The dihedral angle data is that $\angle PMQ = 120^\circ$.
The dihedral angle data is that $\angle PMQ = 120^\circ$.
DPatrick
2016-03-18 21:27:49
Now what?
Now what?
maverick8
2016-03-18 21:27:59
Draw MZ
Draw MZ
pramodmitikiri
2016-03-18 21:28:12
draw MZ to make a triangle
draw MZ to make a triangle
DPatrick
2016-03-18 21:28:26
I might want to draw a little more...there's one point we haven't used at all really yet.
I might want to draw a little more...there's one point we haven't used at all really yet.
Liopleurodon
2016-03-18 21:28:49
C
C
tanishq1
2016-03-18 21:28:52
MC?
MC?
DPatrick
2016-03-18 21:28:59
Right. We notice that $\overline{CM}$ passes through $Z$. In particular, $P$, $Q$, $Z$, $C$, $M$, and $O$ all lie in the same plane.
Right. We notice that $\overline{CM}$ passes through $Z$. In particular, $P$, $Q$, $Z$, $C$, $M$, and $O$ all lie in the same plane.
DPatrick
2016-03-18 21:29:17
So we can just look at that plane, and now we've got a 2-D problem! (This is often a goal in 3-D geometry problems: get down to 2 dimensions.)
So we can just look at that plane, and now we've got a 2-D problem! (This is often a goal in 3-D geometry problems: get down to 2 dimensions.)
DPatrick
2016-03-18 21:29:24
Here's the side view of the plane we care about:
Here's the side view of the plane we care about:
DPatrick
2016-03-18 21:29:27
DPatrick
2016-03-18 21:29:39
We know that the blue angle is $120^\circ$.
We know that the blue angle is $120^\circ$.
DPatrick
2016-03-18 21:29:50
We also know that $\overline{MC}$ is a median from the original triangle, where $Z$ is the center. What does that tell us?
We also know that $\overline{MC}$ is a median from the original triangle, where $Z$ is the center. What does that tell us?
flaminpotato
2016-03-18 21:30:08
MZ=100sqrt(3)
MZ=100sqrt(3)
DPatrick
2016-03-18 21:30:16
The original triangle had side length 600. So $CM = 300\sqrt3$, meaning $MZ = 100\sqrt3$ and $CZ = 200\sqrt3$. (Remember that $Z$ divides the median $\overline{CM}$ of $\triangle ABC$ by a $2:1$ ratio.)
The original triangle had side length 600. So $CM = 300\sqrt3$, meaning $MZ = 100\sqrt3$ and $CZ = 200\sqrt3$. (Remember that $Z$ divides the median $\overline{CM}$ of $\triangle ABC$ by a $2:1$ ratio.)
DPatrick
2016-03-18 21:30:30
To make this easier to write, let's set $a = 100\sqrt3$, so that $MZ = a$ and $CZ = 2a$.
To make this easier to write, let's set $a = 100\sqrt3$, so that $MZ = a$ and $CZ = 2a$.
DPatrick
2016-03-18 21:30:34
DPatrick
2016-03-18 21:31:04
What else? I slyly added $O$ to the diagram as the midpoint of $\overline{PQ}$, but what else do we know about $O$?
What else? I slyly added $O$ to the diagram as the midpoint of $\overline{PQ}$, but what else do we know about $O$?
tanishq1
2016-03-18 21:31:07
OC=OQ=OP too
OC=OQ=OP too
DPatrick
2016-03-18 21:31:11
Aha.
Aha.
DPatrick
2016-03-18 21:31:37
You might even say that $O$ is the circumcenter of $\triangle PQC$.
You might even say that $O$ is the circumcenter of $\triangle PQC$.
DPatrick
2016-03-18 21:31:46
DPatrick
2016-03-18 21:32:06
What seems natural to do now?
What seems natural to do now?
DPatrick
2016-03-18 21:32:39
How can we use the $120^\circ$ angle?
How can we use the $120^\circ$ angle?
mathlete478
2016-03-18 21:32:43
Extend MC
Extend MC
DPatrick
2016-03-18 21:32:53
I think that's a good idea.
I think that's a good idea.
DPatrick
2016-03-18 21:32:57
Let's extend $\overline{MC}$ out to meet the circle again at $D$. Note that $MD = a$:
Let's extend $\overline{MC}$ out to meet the circle again at $D$. Note that $MD = a$:
DPatrick
2016-03-18 21:33:00
DPatrick
2016-03-18 21:33:18
I don't know if that really "helps" but for some reason it did help me in real time to find the next step.
I don't know if that really "helps" but for some reason it did help me in real time to find the next step.
DPatrick
2016-03-18 21:33:36
(That's often the goal in geometry problems: just keep looking for the next step. Hopefully you end up somewhere nice.)
(That's often the goal in geometry problems: just keep looking for the next step. Hopefully you end up somewhere nice.)
DPatrick
2016-03-18 21:33:57
We still haven't explicitly used $\angle PMQ = 120^\circ$. Maybe now is the time...
We still haven't explicitly used $\angle PMQ = 120^\circ$. Maybe now is the time...
Shaddoll
2016-03-18 21:34:42
we can use the tangent addition formula to find angle PMZ and MZQ
we can use the tangent addition formula to find angle PMZ and MZQ
DPatrick
2016-03-18 21:35:08
I like that general idea...though I want to suggest something slightly different (but essentially the same idea).
I like that general idea...though I want to suggest something slightly different (but essentially the same idea).
DPatrick
2016-03-18 21:35:28
If we let $PZ = p$ and $QZ = q$, then $\tan \angle PMZ = \dfrac{p}{a}$ and $\tan \angle QMZ = \dfrac{q}{a}$.
If we let $PZ = p$ and $QZ = q$, then $\tan \angle PMZ = \dfrac{p}{a}$ and $\tan \angle QMZ = \dfrac{q}{a}$.
DPatrick
2016-03-18 21:35:58
But these tangents can be combined to give us $\tan \angle PMQ = \tan 120^\circ = -\sqrt3$, using the tangent addition formula.
But these tangents can be combined to give us $\tan \angle PMQ = \tan 120^\circ = -\sqrt3$, using the tangent addition formula.
DPatrick
2016-03-18 21:36:16
Specifically, \[ -\sqrt3 = \tan\angle PMQ = \frac{\tan\angle PMZ + \tan\angle QMZ}{1 - \tan\angle PMZ \cdot \tan\angle QMZ}.\]
Specifically, \[ -\sqrt3 = \tan\angle PMQ = \frac{\tan\angle PMZ + \tan\angle QMZ}{1 - \tan\angle PMZ \cdot \tan\angle QMZ}.\]
DPatrick
2016-03-18 21:36:29
Substituting the known tangents gives us \[ -\sqrt3 = \frac{\frac{p}{a} + \frac{q}{a}}{1 - \frac{p}{a}\cdot\frac{q}{a}}.\]
Substituting the known tangents gives us \[ -\sqrt3 = \frac{\frac{p}{a} + \frac{q}{a}}{1 - \frac{p}{a}\cdot\frac{q}{a}}.\]
DPatrick
2016-03-18 21:36:52
And simplifying by multiplying through by $a^2$ gives \[ - \sqrt3 = \frac{a(p+q)}{a^2 - pq}.\]
And simplifying by multiplying through by $a^2$ gives \[ - \sqrt3 = \frac{a(p+q)}{a^2 - pq}.\]
mathlover3737
2016-03-18 21:36:56
pq=4a^2
pq=4a^2
I_m_ram
2016-03-18 21:37:16
Pq=4a²
Pq=4a²
tanishq1
2016-03-18 21:37:16
PoP!
PoP!
DPatrick
2016-03-18 21:37:17
Right: that's the last key step. Power of a point at $Z$ tells us $pq = (2a)^2 = 4a^2$.
Right: that's the last key step. Power of a point at $Z$ tells us $pq = (2a)^2 = 4a^2$.
DPatrick
2016-03-18 21:37:25
So we have $-\sqrt3 = \dfrac{a(p+q)}{a^2 - 4a^2} = \dfrac{p+q}{-3a}$.
So we have $-\sqrt3 = \dfrac{a(p+q)}{a^2 - 4a^2} = \dfrac{p+q}{-3a}$.
DPatrick
2016-03-18 21:37:44
And that gives us $p+q = 3\sqrt3a = 3\sqrt3(100\sqrt3) = 900$.
And that gives us $p+q = 3\sqrt3a = 3\sqrt3(100\sqrt3) = 900$.
DPatrick
2016-03-18 21:37:50
Are we done?
Are we done?
Oleksenko
2016-03-18 21:38:09
now we just need that over 2
now we just need that over 2
Peggy
2016-03-18 21:38:09
no! divide by 2
no! divide by 2
tarzanjunior
2016-03-18 21:38:09
Almost
Almost
flaminpotato
2016-03-18 21:38:09
yes 900/2=450
yes 900/2=450
treemath
2016-03-18 21:38:09
so OP = OQ = 450
so OP = OQ = 450
DPatrick
2016-03-18 21:38:21
We're basically done. The distance we want is $OP = \frac12(PQ) = \frac12(900) = \boxed{450}$.
We're basically done. The distance we want is $OP = \frac12(PQ) = \frac12(900) = \boxed{450}$.
DPatrick
2016-03-18 21:38:54
I thought that was pretty too. Lots of different aspects of geometry going on.
I thought that was pretty too. Lots of different aspects of geometry going on.
DPatrick
2016-03-18 21:39:09
By contrast to 12-14, I thought #15 was a bit of a letdown. But let's see, perhaps you'll disagree:
By contrast to 12-14, I thought #15 was a bit of a letdown. But let's see, perhaps you'll disagree:
DPatrick
2016-03-18 21:39:17
15. For $1 \le i \le 215$ let $a_i = \frac{1}{2^i}$ and $a_{216} = \frac{1}{2^{215}}$. Let $x_1,x_2,\ldots,x_{216}$ be positive real numbers such that \[ \sum_{i=1}^{216} x_i = 1 \quad\text{and}\quad \sum_{1 \le i < j \le 216} x_ix_j = \frac{107}{215} + \sum_{i=1}^{216} \frac{a_ix_i^2}{2(1-a_i)}.\] The maximum possible value of $x_2 = \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
15. For $1 \le i \le 215$ let $a_i = \frac{1}{2^i}$ and $a_{216} = \frac{1}{2^{215}}$. Let $x_1,x_2,\ldots,x_{216}$ be positive real numbers such that \[ \sum_{i=1}^{216} x_i = 1 \quad\text{and}\quad \sum_{1 \le i < j \le 216} x_ix_j = \frac{107}{215} + \sum_{i=1}^{216} \frac{a_ix_i^2}{2(1-a_i)}.\] The maximum possible value of $x_2 = \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
DPatrick
2016-03-18 21:39:41
It's a little tiny in the problem statement: $a_{216} = \dfrac{1}{2^{215}}$.
It's a little tiny in the problem statement: $a_{216} = \dfrac{1}{2^{215}}$.
DPatrick
2016-03-18 21:39:59
Yikes. How do we possibly approach this?
Yikes. How do we possibly approach this?
DPatrick
2016-03-18 21:40:17
Note that the $a_i$'s are $\dfrac12$, $\dfrac14$, $\dfrac18$, etc., down to $\dfrac{1}{2^{215}}$, and then $\dfrac{1}{2^{215}}$ is repeated.
Note that the $a_i$'s are $\dfrac12$, $\dfrac14$, $\dfrac18$, etc., down to $\dfrac{1}{2^{215}}$, and then $\dfrac{1}{2^{215}}$ is repeated.
DPatrick
2016-03-18 21:40:21
What's interesting about that?
What's interesting about that?
Oleksenko
2016-03-18 21:40:29
sum of 1
sum of 1
flaminpotato
2016-03-18 21:40:29
sum is 1
sum is 1
DPatrick
2016-03-18 21:40:33
They all sum to 1. That is, $\displaystyle\sum_{i=1}^{216} a_i = 1$.
They all sum to 1. That is, $\displaystyle\sum_{i=1}^{216} a_i = 1$.
DPatrick
2016-03-18 21:40:43
This is probably important, but it's not clear how it's useful yet. The term with $a_i$ in it is a mess.
This is probably important, but it's not clear how it's useful yet. The term with $a_i$ in it is a mess.
mattpi
2016-03-18 21:40:52
Try to incorporate the first sigma expression into the second one
Try to incorporate the first sigma expression into the second one
DPatrick
2016-03-18 21:40:58
I agree -- we've got to try to relate the linear constraint to the quadratic constraint.
I agree -- we've got to try to relate the linear constraint to the quadratic constraint.
mathlover3737
2016-03-18 21:41:13
square first expression
square first expression
Shaddoll
2016-03-18 21:41:19
we can square the linear constraint
we can square the linear constraint
DPatrick
2016-03-18 21:41:22
One idea is to square the linear expression. That is, \[ 1 = \left(\sum_{i=1}^{216} x_i\right)^2. \]
One idea is to square the linear expression. That is, \[ 1 = \left(\sum_{i=1}^{216} x_i\right)^2. \]
DPatrick
2016-03-18 21:41:27
What do we get when we expand that square?
What do we get when we expand that square?
Oleksenko
2016-03-18 21:41:49
sum of the squares + 2*(sum taken two at a time)
sum of the squares + 2*(sum taken two at a time)
DPatrick
2016-03-18 21:41:55
Exactly. We get the sum of the $x_i^2$, and we get $2\displaystyle\sum_{1 \le i < j \le 216} x_ix_j$.
Exactly. We get the sum of the $x_i^2$, and we get $2\displaystyle\sum_{1 \le i < j \le 216} x_ix_j$.
DPatrick
2016-03-18 21:42:16
But the problem statement tells us what that $\sum x_ix_j$ expression is.
But the problem statement tells us what that $\sum x_ix_j$ expression is.
DPatrick
2016-03-18 21:42:29
So we end up with \[ 1 = \sum_{i=1}^{216} x_i^2 + 2\left(\frac{107}{215}\right) + \sum_{i=1}^{216} \frac{a_ix_i^2}{1-a_i}.\]
So we end up with \[ 1 = \sum_{i=1}^{216} x_i^2 + 2\left(\frac{107}{215}\right) + \sum_{i=1}^{216} \frac{a_ix_i^2}{1-a_i}.\]
DPatrick
2016-03-18 21:42:47
Hmmm...what can we do with this?
Hmmm...what can we do with this?
mathlover3737
2016-03-18 21:42:56
combine the two sigmas
combine the two sigmas
bluephoenix
2016-03-18 21:42:56
combine first and third terms of RHS
combine first and third terms of RHS
tanishq1
2016-03-18 21:42:56
add the two sigmas together
add the two sigmas together
DPatrick
2016-03-18 21:43:10
I agree, but first, let me move the constant to the left side: note that we end up subtracting $\dfrac{214}{215}$ from 1, leaving just $\dfrac{1}{215}$ on the left side.
I agree, but first, let me move the constant to the left side: note that we end up subtracting $\dfrac{214}{215}$ from 1, leaving just $\dfrac{1}{215}$ on the left side.
DPatrick
2016-03-18 21:43:23
And now we can combine the two sums: each term of each sum is something times $x_i^2$.
And now we can combine the two sums: each term of each sum is something times $x_i^2$.
DPatrick
2016-03-18 21:43:29
So, we're at \[ \frac{1}{215} = \sum_{i=1}^{216} \left(1 + \frac{a_i}{1-a_i}\right)x_i^2.\]
So, we're at \[ \frac{1}{215} = \sum_{i=1}^{216} \left(1 + \frac{a_i}{1-a_i}\right)x_i^2.\]
DPatrick
2016-03-18 21:43:39
Now what?
Now what?
tanishq1
2016-03-18 21:43:53
the sum inside the parantheses simplifies
the sum inside the parantheses simplifies
Shaddoll
2016-03-18 21:44:01
$(1+\dfrac{a_i}{1-a_i}=\dfrac{1}{1-a_i}$
$(1+\dfrac{a_i}{1-a_i}=\dfrac{1}{1-a_i}$
DPatrick
2016-03-18 21:44:07
It sure does: \[ 1 + \frac{a_i}{1-a_i} = \frac{(1-a_i)+a_i}{1-a_i} = \frac{1}{1-a_i}.\]
It sure does: \[ 1 + \frac{a_i}{1-a_i} = \frac{(1-a_i)+a_i}{1-a_i} = \frac{1}{1-a_i}.\]
DPatrick
2016-03-18 21:44:15
So now we have \[ \frac{1}{215} = \sum_{i=1}^{216} \frac{1}{1-a_i}x_i^2.\]
So now we have \[ \frac{1}{215} = \sum_{i=1}^{216} \frac{1}{1-a_i}x_i^2.\]
DPatrick
2016-03-18 21:44:58
Do you notice anything interesting here? How do we relate the 215 on the left side to some stuff that resembles the right side?
Do you notice anything interesting here? How do we relate the 215 on the left side to some stuff that resembles the right side?
DPatrick
2016-03-18 21:45:26
Hint: we made an observation at the beginning that I moved up top, because it was probably important.
Hint: we made an observation at the beginning that I moved up top, because it was probably important.
calculus_riju
2016-03-18 21:45:39
216-1
216-1
DPatrick
2016-03-18 21:46:07
The sum of the $a_i$'s is 1.
The sum of the $a_i$'s is 1.
DPatrick
2016-03-18 21:46:17
So the sum of the $(1-a_i)$'s is...
So the sum of the $(1-a_i)$'s is...
Shaddoll
2016-03-18 21:46:31
$215$
$215$
DPatrick
2016-03-18 21:46:35
215!!!
215!!!
DPatrick
2016-03-18 21:46:42
That is, $\displaystyle\sum_{i=1}^{216} (1-a_i) = 216 - \sum_{i=1}^{216} a_i = 216 - 1 = 215$.
That is, $\displaystyle\sum_{i=1}^{216} (1-a_i) = 216 - \sum_{i=1}^{216} a_i = 216 - 1 = 215$.
DPatrick
2016-03-18 21:46:56
That cannot possibly be a coincidence.
That cannot possibly be a coincidence.
DPatrick
2016-03-18 21:47:01
So our equation becomes \[ \frac{1}{\displaystyle \sum_{i=1}^{216} (1-a_i)} = \sum_{i=1}^{216} \frac{1}{1-a_i} x_i^2.\]
So our equation becomes \[ \frac{1}{\displaystyle \sum_{i=1}^{216} (1-a_i)} = \sum_{i=1}^{216} \frac{1}{1-a_i} x_i^2.\]
DPatrick
2016-03-18 21:47:16
Blech -- let's not have a sum in a denominator -- let's write it as a product instead: \[ 1 = \sum_{i=1}^{216} (1-a_i) \sum_{i=1}^{216} \frac{1}{1-a_i}x_i^2. \]
Blech -- let's not have a sum in a denominator -- let's write it as a product instead: \[ 1 = \sum_{i=1}^{216} (1-a_i) \sum_{i=1}^{216} \frac{1}{1-a_i}x_i^2. \]
idomath12345
2016-03-18 21:47:20
Multiply.
Multiply.
tanishq1
2016-03-18 21:47:20
multiply then
multiply then
Oleksenko
2016-03-18 21:47:20
CAUCHY!
CAUCHY!
DPatrick
2016-03-18 21:47:37
Oleksenko said the magic word: Cauchy!
Oleksenko said the magic word: Cauchy!
DPatrick
2016-03-18 21:47:44
If you've seen a few problems with products of equal-length sums like this, you probably know what to try next.
If you've seen a few problems with products of equal-length sums like this, you probably know what to try next.
DPatrick
2016-03-18 21:47:55
Let's try applying the Cauchy-Schwarz inequality. (It's legal because everything in sight is positive.)
Let's try applying the Cauchy-Schwarz inequality. (It's legal because everything in sight is positive.)
DPatrick
2016-03-18 21:48:04
The Cauchy-Schwarz inequality tells us that if we have two sequences of positive numbers $x_1,\ldots,x_n$ and $y_1,\ldots,y_n$, then \[ (x_1^2 + \cdots + x_n^2)(y_1^2 + \cdots + y_n^2) \ge (x_1y_1 + \cdots + x_ny_n)^2.\]
The Cauchy-Schwarz inequality tells us that if we have two sequences of positive numbers $x_1,\ldots,x_n$ and $y_1,\ldots,y_n$, then \[ (x_1^2 + \cdots + x_n^2)(y_1^2 + \cdots + y_n^2) \ge (x_1y_1 + \cdots + x_ny_n)^2.\]
DPatrick
2016-03-18 21:48:25
So for our equation, we get \[ 1 = \sum_{i=1}^{216} (1-a_i) \sum_{i=1}^{216} \frac{1}{1-a_i}x_i^2 \ge \left(\sum_{i=1}^{216} \sqrt{(1-a_i) \cdot \frac{1}{1-a_i}x_i^2}\right)^2.\]
So for our equation, we get \[ 1 = \sum_{i=1}^{216} (1-a_i) \sum_{i=1}^{216} \frac{1}{1-a_i}x_i^2 \ge \left(\sum_{i=1}^{216} \sqrt{(1-a_i) \cdot \frac{1}{1-a_i}x_i^2}\right)^2.\]
DPatrick
2016-03-18 21:49:01
Wow, this is super-useful.
Wow, this is super-useful.
Shaddoll
2016-03-18 21:49:23
the RHS reduces to 1 because of the sum of $x_i$s equal to 1 condition
the RHS reduces to 1 because of the sum of $x_i$s equal to 1 condition
DPatrick
2016-03-18 21:49:58
The $(1-a_i)$'s cancel, and all that's left is \[ 1 = \sum_{i=1}^{216} (1-a_i) \sum_{i=1}^{216} \frac{1}{1-a_i}x_i^2 \ge \left(\sum_{i=1}^{216} x_i\right)^2 = 1. \]
The $(1-a_i)$'s cancel, and all that's left is \[ 1 = \sum_{i=1}^{216} (1-a_i) \sum_{i=1}^{216} \frac{1}{1-a_i}x_i^2 \ge \left(\sum_{i=1}^{216} x_i\right)^2 = 1. \]
idomath12345
2016-03-18 21:50:07
so 1>=1???
so 1>=1???
blue8931
2016-03-18 21:50:13
oh... it's the equality case, right?
oh... it's the equality case, right?
Shaddoll
2016-03-18 21:50:13
so it must be the equality case, when the sequences are proportional to each other
so it must be the equality case, when the sequences are proportional to each other
tarzanjunior
2016-03-18 21:50:13
Equality occurs when...
Equality occurs when...
DPatrick
2016-03-18 21:50:29
$1 \ge 1$ isn't exactly shocking, but what this tells us is that we must be in the equality case of Cauchy-Schwarz.
$1 \ge 1$ isn't exactly shocking, but what this tells us is that we must be in the equality case of Cauchy-Schwarz.
DPatrick
2016-03-18 21:50:36
We only get equality if one list of terms is a constant multiple of the other.
We only get equality if one list of terms is a constant multiple of the other.
DPatrick
2016-03-18 21:50:50
In particular, we must have some constant $k$ such that $\dfrac{1}{1-a_i} x_i^2 = k(1-a_i)$ for all $i$.
In particular, we must have some constant $k$ such that $\dfrac{1}{1-a_i} x_i^2 = k(1-a_i)$ for all $i$.
DPatrick
2016-03-18 21:51:13
That is, $x_i^2 = k(1-a_i)^2$ for all $i$, or $x_i = k'(1-a_i)$ for all $i$ (where $k' = \sqrt{k}$).
That is, $x_i^2 = k(1-a_i)^2$ for all $i$, or $x_i = k'(1-a_i)$ for all $i$ (where $k' = \sqrt{k}$).
DPatrick
2016-03-18 21:51:20
How do we determine $k'$?
How do we determine $k'$?
Shaddoll
2016-03-18 21:51:37
We know that the $x_i$s sum to $1$
We know that the $x_i$s sum to $1$
DPatrick
2016-03-18 21:51:45
Right, we compare the sums: \[ 1 = \sum_{i=1}^{216} x_i = k' \sum_{i=1}^{216} (1-a_i) = k'(215). \]
Right, we compare the sums: \[ 1 = \sum_{i=1}^{216} x_i = k' \sum_{i=1}^{216} (1-a_i) = k'(215). \]
DPatrick
2016-03-18 21:51:55
So $k' = \dfrac{1}{215}$, and thus $x_i = \dfrac{1}{215}(1-a_i)$ for all $i$.
So $k' = \dfrac{1}{215}$, and thus $x_i = \dfrac{1}{215}(1-a_i)$ for all $i$.
Oleksenko
2016-03-18 21:52:13
Now we can find x_2
Now we can find x_2
DPatrick
2016-03-18 21:52:15
FINALLY, $x_2 = \dfrac{1}{215}\left(1 - \dfrac14\right) = \dfrac{1}{215} \cdot \dfrac34 = \dfrac{3}{860}$. (The phrase "maximum possible value" in the problem statement is a bit deceiving: there is only one possible value!)
FINALLY, $x_2 = \dfrac{1}{215}\left(1 - \dfrac14\right) = \dfrac{1}{215} \cdot \dfrac34 = \dfrac{3}{860}$. (The phrase "maximum possible value" in the problem statement is a bit deceiving: there is only one possible value!)
DPatrick
2016-03-18 21:52:31
Hence, our final answer is $3 + 860 = \boxed{863}$.
Hence, our final answer is $3 + 860 = \boxed{863}$.
DPatrick
2016-03-18 21:53:05
I was a little dissatisfied with the problem, because it mainly tests experience rather than problem-solving skills.
I was a little dissatisfied with the problem, because it mainly tests experience rather than problem-solving skills.
DPatrick
2016-03-18 21:53:18
If you've seen this sort of problem before and know Cauchy-Schwarz, it's not tremendously hard.
If you've seen this sort of problem before and know Cauchy-Schwarz, it's not tremendously hard.
DPatrick
2016-03-18 21:53:22
If you haven't, it's near impossible.
If you haven't, it's near impossible.
DPatrick
2016-03-18 21:53:47
But it's a nice problem despite that, if a little contrived.
But it's a nice problem despite that, if a little contrived.
DPatrick
2016-03-18 21:53:58
That's it for the 2016 AIME II! Thanks for participating tonight.
That's it for the 2016 AIME II! Thanks for participating tonight.
DPatrick
2016-03-18 21:54:04
Have a nice weekend!
Have a nice weekend!
blue8931
2016-03-18 21:54:14
thanks for the great math jam, as usual
thanks for the great math jam, as usual
Oleksenko
2016-03-18 21:54:18
Which problem do you think was the hardest?
Which problem do you think was the hardest?
DPatrick
2016-03-18 21:54:33
I think #15 was technically the hardest: you had to have the tools and experience to do it.
I think #15 was technically the hardest: you had to have the tools and experience to do it.
DPatrick
2016-03-18 21:54:47
#10 took me the longest.
#10 took me the longest.
DPatrick
2016-03-18 21:55:09
(Any problem with loads of similar triangles, it's hard to keep all the data straight.)
(Any problem with loads of similar triangles, it's hard to keep all the data straight.)
tanishq1
2016-03-18 21:55:21
what is your guess for the USAMO cutoff?
what is your guess for the USAMO cutoff?
idomath12345
2016-03-18 21:55:21
Do you think there will be split cutoffs?
Do you think there will be split cutoffs?
DPatrick
2016-03-18 21:55:33
No idea whatsoever. I didn't really look at AIME I so I can't compare.
No idea whatsoever. I didn't really look at AIME I so I can't compare.
kbird
2016-03-18 21:55:37
what's your favorite
what's your favorite
DPatrick
2016-03-18 21:55:48
I liked all of 12-14 -- it'd be hard to pick a favorite of those three.
I liked all of 12-14 -- it'd be hard to pick a favorite of those three.
idomath12345
2016-03-18 21:55:58
Are u coming to mc nats!
Are u coming to mc nats!
DPatrick
2016-03-18 21:56:05
Yes! rrusczyk and jbatterson too!
Yes! rrusczyk and jbatterson too!
Liopleurodon
2016-03-18 21:56:33
Yay! Hope to meet you there!
Yay! Hope to meet you there!
bluephoenix
2016-03-18 21:56:42
Was this AIME harder compared to recent years (maybe even 2011 AIME I?)
Was this AIME harder compared to recent years (maybe even 2011 AIME I?)
Oleksenko
2016-03-18 21:56:42
Overall, do you think this was a hard AIME?
Overall, do you think this was a hard AIME?
DPatrick
2016-03-18 21:57:05
It's so hard for me to tell -- I've been looking at AIMEs for literally 30 years so they tend to blur together at this point.
It's so hard for me to tell -- I've been looking at AIMEs for literally 30 years so they tend to blur together at this point.
DPatrick
2016-03-18 21:57:20
Having said that, this one didn't strike me as overly easy or overly hard.
Having said that, this one didn't strike me as overly easy or overly hard.
tarzanjunior
2016-03-18 21:58:16
Which do you think was the easiest?
Which do you think was the easiest?
DPatrick
2016-03-18 21:58:20
#3 by a mile.
#3 by a mile.
DPatrick
2016-03-18 21:58:53
logs are scary-looking but they're all bark and no bite
logs are scary-looking but they're all bark and no bite
DPatrick
2016-03-18 21:59:44
Well...I actually have to help teach a class tomorrow morning, so I should go. Have a good night!
Well...I actually have to help teach a class tomorrow morning, so I should go. Have a good night!
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