Mandelbrot Competition Math Jam
Go back to the Math Jam ArchiveWarmup for the Mandelbrot. Medium to hard problems discussed.
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Facilitator: Mathew Crawford
MANDELBROT COMPETITION MATH JAM
MCrawford (19:30:25)
MCrawford (19:30:29)
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MCrawford (19:30:34)
What can we do to help us evaluate this series?
bmsiv (19:31:01)
first thing, separate the terms
Monkey (19:31:09)
We can split it up into two series: 3^n/8^n and 5^n/8^n.
bmsiv (19:31:22)
then figure out each term
Neutron (19:31:26)
it's the sum of two geometric series
MCrawford (19:31:33)
MysticTerminator (19:30:52)
64/15
MCrawford (19:31:54)
MCrawford (19:32:13)
This problem was not all that difficult, but helps us recognize the importance of mathematical notation. So long as we are comfortable with writing series in sigma (sum) notation, we can quickly recognize ways of working with many of them that are harder to see when examining just the terms alone.
MCrawford (19:32:28)
MCrawford (19:32:32)
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MCrawford (19:32:36)
We all know the Pythagorean theorem, but I'm guessing none of you are any more wild about squaring 3- and 5-digits number than I am, which is to say I'd rather be watching Regis Philbin lecture about the hygiene of minor celebrities.
MCrawford (19:32:43)
Okay, it's not quite that bad, but what can we do that would make this problem easier?
sleepsta (19:32:56)
factor
MCrawford (19:33:16)
Does that help us get started?
bmsiv (19:33:21)
guess thattheanswer will be an integer
MCrawford (19:33:39)
Heh, sometimes such guess give us a hand.
Aryth (19:33:24)
the legs are easier
Aryth (19:33:41)
then later, just multiply it back into the hypotenuse length
samath (19:33:50)
we can only factor out 2 though
MCrawford (19:34:12)
True. We may have to look for something else.
sirkuku (19:34:12)
its still a start
MCrawford (19:34:43)
MCrawford (19:34:49)
How can we make use of this equation without having to square everything?
MysticTerminator (19:35:00)
diff of squares
MCrawford (19:35:13)
MCrawford (19:35:18)
We can now factor a difference of squares. This is particularly useful to use as we are working with integers and algebraic factors can often be matched up with prime factorizations as we cross the bridge between algebra and number theory.
MCrawford (19:35:22)
What are the factorizations of both sides of the equation?
Neutron (19:35:28)
a^2 - c^2 = (a +c)(a-c)
bmsiv (19:35:37)
left side: c-408, c+408
Aryth (19:35:42)
(c+408)(c-408)=20806^2
MCrawford (19:35:55)
How does the RHS factor?
bmsiv (19:35:57)
right side: 2^2, 101^2, 103^
MCrawford (19:36:10)
MCrawford (19:36:19)
Now what?
Teddy (19:36:41)
split rhs into 2 factors of difference 916
Teddy (19:36:44)
816
MCrawford (19:37:04)
We are looking for two factors of the right hand side that differ by 816.
How can we find them?
bmsiv (19:36:46)
2*101^1, 2*103^2
Monkey (19:36:57)
c+408=2x103^2, c-408=2x101^2
MCrawford (19:37:32)
There are a limited number of ways we could divide up the right hand side into the product of two integers. Since the difference between those factors must be 916, which is not a multiple of 101 or 103, we know we must keep those factors separate. If we don't divide up the two powers of 2, the difference cannot be even.
MCrawford (19:37:38)
MCrawford (19:37:43)
What is c?
MysticTerminator (19:37:02)
20810
Monkey (19:38:04)
20810
MCrawford (19:38:14)
MCrawford (19:38:31)
Notice how I used algebraic factorization to make the computations easier. We squared and added integers after all, but these were not much of a chore given the application of algebra to arithmetic.
MCrawford (19:38:40)
Rearranging sums of squares to highlight differences of squares is not an uncommon problem solving technique, particularly when working with the Pythagorean theorem, integers, or both.
MCrawford (19:38:57)
MCrawford (19:39:02)
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MCrawford (19:39:06)
We could just multiply everything out, but that would take a lot of time.
How can we make this problem easier?
bmsiv (19:39:22)
see that each one cubed is a real number
bmsiv (19:39:32)
factor out the 3rd roots of unity
MCrawford (19:39:44)
We can still do much better.
Monkey (19:39:18)
The two numbers are conjugates
agolsme (19:39:36)
(a + b)^8 + (a - b)^8 ?
MCrawford (19:40:03)
This will help us in some sense, but there s a powerful approach we have not considered.
Teddy (19:40:03)
convert to r cis pheta
MCrawford (19:40:16)
There are different ways in which we can express complex numbers which stem from the fact that we can represent complex numbers geometrically in a plane where the x-axis represents real numbers and the y-axis represents imaginary numbers. In this complex plane we call the axes the real axis and the imaginary axis.
MCrawford (19:40:24)
MCrawford (19:40:26)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classes/Complex/Images/argandplane.gif
MCrawford (19:40:30)
In the picture above, O is the origin of the complex plane, which corresponds to the number 0. Our old x-axis is now called the 'Real Axis', and our old y-axis is the 'Imaginary Axis'.
MCrawford (19:40:41)
So in the above picture, A is the point (2,5) which corresponds to the complex number 2+5i, and B is the point (7,6) which corresponds to the complex number 7+6i.
Often we will be a bit looser with our terminology, and say that the point A "is" the complex number 2+5i.
MCrawford (19:41:10)
We can now view each complex number (each point in the complex plane) in terms of different variables -- namely, their distance from the origin (which we call the magnitude) and the angle that the ray to the point makes with the real axis (going counterclockwise).
MCrawford (19:41:16)
How can we write the two complex numbers in the original problem?
MysticTerminator (19:41:49)
factor out a sqrt(3) out of each to make the norm 1
bmsiv (19:41:57)
as root 3 radius, 2pi/3 angle and root 3 radius, 4pi/3 angle
MCrawford (19:42:20)
Your angles are slightly off.
MCrawford (19:42:42)
What angles should we use?
bmsiv (19:42:44)
oh...right. pi/3, 5pi/3
MCrawford (19:42:50)
MCrawford (19:43:00)
I could have written the angle to the second complex number as 330 degrees instead of -30 degrees, but it's all the same and for our purposes, -30 is easier to work with.
MCrawford (19:43:19)
MCrawford (19:43:22)
This notation summarizes more compactly the value of a complex number.
MCrawford (19:43:45)
There is in fact an even more useful notation involving Euler's constant that we will not use here, but is worth noting.
MCrawford (19:43:51)
MysticTerminator (19:43:49)
MCrawford (19:44:05)
MCrawford (19:44:20)
Though MysticTerminator shows us the other form.
MCrawford (19:44:26)
Moving on...
How can we now continue with our computations?
bmsiv (19:44:40)
we can notice that cis mustiplies the angle by the exponent
bmsiv (19:44:51)
that is, it becomes root3 cis 240 + root3 cis 120
variable (19:44:59)
DeMoivre's Theorem
MCrawford (19:45:10)
bmsiv'
MCrawford (19:45:21)
bmsiv's explanation is close. Something else happens.
bmsiv (19:45:33)
except the radius is still put to the eighth
MCrawford (19:45:40)
MCrawford (19:46:01)
DeMoivre's theorem can be proven using trigonometry identities involving angle addition. Those of you unfamiliar with this proof are encouraged to try it on your own and discuss it in the Intermediate High School forum.
MCrawford (19:46:19)
MCrawford (19:46:22)
Now, how can we finish?
MysticTerminator (19:46:27)
draw a picture!
MCrawford (19:46:40)
If that helps, sure.
MysticTerminator (19:46:36)
(much better than doing it algebraically)
bmsiv (19:46:36)
notice that the sin parts cancel out, and that both the cosines are -1/2
MysticTerminator (19:46:41)
and be like ha there's an equilateral triangle there
Teddy (19:46:42)
i sin 240 + i sin 120 = 0
MysticTerminator (19:40:21)
-81 or something
bmsiv (19:46:45)
which means that it totals to 81*-1 = -81
MCrawford (19:47:02)
bmsiv (19:47:18)
um...could you keep the equations a bit shorter? they go off the screen here
MCrawford (19:47:32)
You can drag your window wider.
MCrawford (19:47:37)
If this result is surprising at all, then it is recommended that you work more with complex numbers until this problem seems simple and natural. An understanding of DeMoivre's theorem and also complex numbers known as roots of unity lay the basis for a level of understanding of complex numbers makes a very powerful toolset. This toolset is not simply one for performing calculations, but also solving polynomial problems, geometry problems, and a host of other kinds of problems.
MCrawford (19:47:51)
The AoPS Intermediate Complex Numbers/Trigonometry class covers these topics in detail, carrying on to even more difficult problems.
MCrawford (19:48:09)
MCrawford (19:48:20)
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MCrawford (19:48:38)
How can we begin exploring this problem?
uclabb (19:48:29)
draw
MCrawford (19:48:49)
It makes sense to start with a diagram. I'll let you all draw your own starting diagrams.
Once we have our initial diagrams, what kinds of information can we fill in?
Teddy (19:48:50)
now i actually have to use paper
MCrawford (19:49:08)
Sorry Teddy. We all know how you struggle with geometric diagrams.
Monkey (19:49:00)
the stuff given
MCrawford (19:49:29)
Well...how do we interpret the given statements?
MCrawford (19:49:44)
There is both qualitative and quantitative information to incorporate here.
MCrawford (19:50:00)
Not recognizing this fact is precisely what leads many students to a quick dead end.
bmsiv (19:49:48)
first, we find that we are given the lengths of two medians
Aryth (19:50:21)
The two medians intersect where the area closer to the point is twice the length than the length closer to the side
Monkey (19:50:31)
Then we use the fact that medians divide each other into 2/3-1/3 parts.
MCrawford (19:50:38)
Since we have medians to two of the sides, we know that the medians will divide each other up into segments with ratios 2:1:
MCrawford (19:50:44)
MCrawford (19:50:55)
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MCrawford (19:50:58)
Does our diagram help us with anything?
Monkey (19:51:00)
medians are perpedicular
samath (19:51:08)
AEG is a right triangle
Aryth (19:51:10)
EGA is a right triangle!
MCrawford (19:51:25)
We see that AEG is a 3-4-5 triangle, meaning that the angles at G are all right angles:
MCrawford (19:51:34)
MCrawford (19:51:37)
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Teddy (19:50:58)
let the intersection of the medians be X, we know AX = 4, XE = 3, and AE = 5 so the area of ABC = 6 [AXE] = 6 * 6 = 36
bmsiv (19:51:32)
each medial triangle has the same area, and there are 6 of them, and since we have a 3-4-5 medial triangle, the area of each medial triangle is 6, and so the toal area is 36
MysticTerminator (19:51:47)
ehh 36
Aryth (19:52:02)
good point...
MCrawford (19:52:16)
Nicely done.
MCrawford (19:52:37)
Now, let's suppose a student did not recognize that the medial triangles had the same area...
MCrawford (19:52:47)
Would they be dead in the water?
Neutron (19:53:02)
Use Heron's Formula
MCrawford (19:53:13)
Ouch, please, no.
Teddy (19:53:17)
[ABC] = AG * EB
MCrawford (19:53:50)
Explain.
bmsiv (19:53:59)
the ADE triangle is a quarter of ABC in area
MCrawford (19:54:35)
We have a couple of ways, yes. Explain them both please.
bmsiv (19:54:33)
A(ABC) = 27 + A(ADE) = 27 + A(ABC)/4
MCrawford (19:55:01)
Where do you get that equation?
Neutron (19:55:03)
[ABE] = [ACE]
MCrawford (19:55:28)
I think you have one letter wrong Neutron
Teddy (19:55:20)
AG is perpendicular to EB and A and C are the same distance to EB since AE = EC and AEC is a line so [abc] = AG * EB
MCrawford (19:55:58)
There are a number of ways we can compare the areas of smaller parts of the triangle.
MCrawford (19:56:09)
We now see that AG is the altitude to base BE of triangle ABE.
MCrawford (19:56:14)
Neutron (19:56:14)
I meant [ABE] = [BCE]. Equal base, shared height
MCrawford (19:56:28)
MCrawford (19:56:30)
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MCrawford (19:56:35)
It is important to recognize the relationships between areas of triangles even when either the base or altitude is not obvious. In this case we see that the bases AE and CE of triangles ABE and CBE are the same and even though we have not identified the altitude to those bases, we know that it must also be the same since AE and CE are on the same line.
Monkey (19:56:57)
[ABC]=36
MCrawford (19:57:08)
MCrawford (19:57:28)
The key to getting off the ground with this problem was starting with the information available and scanning our diagram for new information. It's important to recognize the relationships between medians, even if they are not identified as medians to us. From there we had to recognize the relationships between the areas of pieces that made up the whole triangle. If that step seemed difficult, it might have helped to have labeled the altitudes to the sides. Recognition of such relationships can become second nature with experience.
MysticTerminator (19:57:11)
yay I finished doing it by heron's formula :-)
MCrawford (19:57:51)
1. I don't believe you because I know how smart you are.
MCrawford (19:58:13)
2. If you did, you're just punishing yourself. [img id=em-0]
Aryth (19:57:47)
lol
Monkey (19:58:22)
lol
MysticTerminator (19:58:24)
well I'm hyper. more hyper than normal ...
Sabre_X (19:58:29)
rofl, owned
MCrawford (19:58:38)
Okay, okay.
MCrawford (19:58:42)
I've only got one more today.
MCrawford (19:58:50)
I wasn't expecting a crowd this large.
MCrawford (19:58:52)
MCrawford (19:58:59)
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MCrawford (19:59:18)
A few highly ambitious students might look at this expression and endeavor to estimate the computations, but the answer is a 5-digit integer, so I wouldn't suggest banging your head against that particular wall.
MCrawford (19:59:28)
What kinds of tools does this problem seem to invite?
Monkey (19:59:37)
Binomial Theorem
solafidefarms (19:59:47)
combinations and binomial expansions ;o
Sabre_X (19:59:51)
binomial theorem?
MCrawford (20:00:00)
Since have a sum of two numbers raised to an exponent, binomial expansion comes to mind.
MCrawford (20:00:06)
Can we just expand here and calculate nicely?
Monkey (20:00:19)
No
Aryth (20:00:21)
perhaps not
MCrawford (20:00:28)
Why not?
bmsiv (19:59:36)
hmm...I personally like doingthis with the binomial theorem, and then separating the square root terms and the integers
Aryth (20:00:38)
the square roots interfere
Monkey (20:00:40)
Still way too many square roots
MCrawford (20:00:54)
bmsiv (20:01:01)
7^3+6*7^2*sqrt(35)+15*7^2*5+20*7*5*sqrt(35)+ ...
MCrawford (20:01:12)
Unfortunately, plain ole binomial expansion still leaves us with square roots that would require estimation. That might make the super-crazy-ambitious-head-to-wall-banging-number-crunches a little happier, but it's hardly inspiring.
MCrawford (20:01:22)
Is there any nice way of getting rid of the messy stuff, or are we barking up the wrong tree?
MysticTerminator (20:01:27)
try squaring it first
MysticTerminator (20:01:30)
bit less square roots
MCrawford (20:01:54)
That might have made things easier. Good call.
MCrawford (20:01:58)
But we're down this path already...
MysticTerminator (19:59:10)
add (sqrt(7) - sqrt(5))^6
Teddy (20:00:00)
that expression + (rt 7 - rt 5)^6 since (rt 7 - rt 5)^6 < 1
MCrawford (20:02:40)
MCrawford (20:03:08)
What's nice about this?
MysticTerminator (20:00:49)
13536
bmsiv (20:03:13)
add the two
Aryth (20:03:19)
cancel out terms with square roots
Monkey (20:03:19)
It cancels out the square roots when we add
MCrawford (20:03:34)
MCrawford (20:03:56)
What is our answer?
MysticTerminator (20:03:55)
but! we added something
MysticTerminator (20:01:17)
13535
Teddy (20:04:08)
13535
bmsiv (20:04:10)
we added something less than 1
bmsiv (20:04:15)
so we must subtract 1
Sabre_X (20:04:16)
13535
MCrawford (20:04:42)
It's not the closest integer! But we know that we rounded up to 13536, so we must go down 1 to 13535.
Monkey (20:04:23)
13535, since (\sqrt {7}-\sqrt {5})^6>0, but less than 1.
yangzhi (20:04:30)
13535
MCrawford (20:04:49)
E^(pi*i)=-1 (20:05:15)
MCrawford (20:05:21)
It's good to recognize the ways that we can combine binomial expansions to rid us of messy parts of otherwise nasty calculations. This method is also useful when dealing with expansions of some complex numbers. This method is discussed in detail in the AoPS Intermediate Algebra class.
MCrawford (20:05:36)
MCrawford (20:05:48)
I'll leave the rest of this solution to all you young geniuses.
MCrawford (20:06:00)
That was our last problem for the night.
Thank you for coming to this Mandelbrot Math Jam. I hope you enjoyed it.
Are there any questions about Math Jams or AoPS classes?
MCrawford (20:06:08)
Sorry that was shorter than I planned.
MCrawford (20:06:34)
I didn't expected a reigning Mandelbrot champion and some other high scorers to show up.
Teddy (20:06:17)
we blazed too fast?
Aryth (20:06:21)
is Mandelbrot only for High schools?
MCrawford (20:06:54)
Middle schoolers can take it too.
MCrawford (19:30:25)
MCrawford (19:30:29)
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MCrawford (19:30:34)
What can we do to help us evaluate this series?
bmsiv (19:31:01)
first thing, separate the terms
Monkey (19:31:09)
We can split it up into two series: 3^n/8^n and 5^n/8^n.
bmsiv (19:31:22)
then figure out each term
Neutron (19:31:26)
it's the sum of two geometric series
MCrawford (19:31:33)
MysticTerminator (19:30:52)
64/15
MCrawford (19:31:54)
MCrawford (19:32:13)
This problem was not all that difficult, but helps us recognize the importance of mathematical notation. So long as we are comfortable with writing series in sigma (sum) notation, we can quickly recognize ways of working with many of them that are harder to see when examining just the terms alone.
MCrawford (19:32:28)
MCrawford (19:32:32)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-152039150.gif
MCrawford (19:32:36)
We all know the Pythagorean theorem, but I'm guessing none of you are any more wild about squaring 3- and 5-digits number than I am, which is to say I'd rather be watching Regis Philbin lecture about the hygiene of minor celebrities.
MCrawford (19:32:43)
Okay, it's not quite that bad, but what can we do that would make this problem easier?
sleepsta (19:32:56)
factor
MCrawford (19:33:16)
Does that help us get started?
bmsiv (19:33:21)
guess thattheanswer will be an integer
MCrawford (19:33:39)
Heh, sometimes such guess give us a hand.
Aryth (19:33:24)
the legs are easier
Aryth (19:33:41)
then later, just multiply it back into the hypotenuse length
samath (19:33:50)
we can only factor out 2 though
MCrawford (19:34:12)
True. We may have to look for something else.
sirkuku (19:34:12)
its still a start
MCrawford (19:34:43)
MCrawford (19:34:49)
How can we make use of this equation without having to square everything?
MysticTerminator (19:35:00)
diff of squares
MCrawford (19:35:13)
MCrawford (19:35:18)
We can now factor a difference of squares. This is particularly useful to use as we are working with integers and algebraic factors can often be matched up with prime factorizations as we cross the bridge between algebra and number theory.
MCrawford (19:35:22)
What are the factorizations of both sides of the equation?
Neutron (19:35:28)
a^2 - c^2 = (a +c)(a-c)
bmsiv (19:35:37)
left side: c-408, c+408
Aryth (19:35:42)
(c+408)(c-408)=20806^2
MCrawford (19:35:55)
How does the RHS factor?
bmsiv (19:35:57)
right side: 2^2, 101^2, 103^
MCrawford (19:36:10)
MCrawford (19:36:19)
Now what?
Teddy (19:36:41)
split rhs into 2 factors of difference 916
Teddy (19:36:44)
816
MCrawford (19:37:04)
We are looking for two factors of the right hand side that differ by 816.
How can we find them?
bmsiv (19:36:46)
2*101^1, 2*103^2
Monkey (19:36:57)
c+408=2x103^2, c-408=2x101^2
MCrawford (19:37:32)
There are a limited number of ways we could divide up the right hand side into the product of two integers. Since the difference between those factors must be 916, which is not a multiple of 101 or 103, we know we must keep those factors separate. If we don't divide up the two powers of 2, the difference cannot be even.
MCrawford (19:37:38)
MCrawford (19:37:43)
What is c?
MysticTerminator (19:37:02)
20810
Monkey (19:38:04)
20810
MCrawford (19:38:14)
MCrawford (19:38:31)
Notice how I used algebraic factorization to make the computations easier. We squared and added integers after all, but these were not much of a chore given the application of algebra to arithmetic.
MCrawford (19:38:40)
Rearranging sums of squares to highlight differences of squares is not an uncommon problem solving technique, particularly when working with the Pythagorean theorem, integers, or both.
MCrawford (19:38:57)
MCrawford (19:39:02)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-455849.gif
MCrawford (19:39:06)
We could just multiply everything out, but that would take a lot of time.
How can we make this problem easier?
bmsiv (19:39:22)
see that each one cubed is a real number
bmsiv (19:39:32)
factor out the 3rd roots of unity
MCrawford (19:39:44)
We can still do much better.
Monkey (19:39:18)
The two numbers are conjugates
agolsme (19:39:36)
(a + b)^8 + (a - b)^8 ?
MCrawford (19:40:03)
This will help us in some sense, but there s a powerful approach we have not considered.
Teddy (19:40:03)
convert to r cis pheta
MCrawford (19:40:16)
There are different ways in which we can express complex numbers which stem from the fact that we can represent complex numbers geometrically in a plane where the x-axis represents real numbers and the y-axis represents imaginary numbers. In this complex plane we call the axes the real axis and the imaginary axis.
MCrawford (19:40:24)
MCrawford (19:40:26)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classes/Complex/Images/argandplane.gif
MCrawford (19:40:30)
In the picture above, O is the origin of the complex plane, which corresponds to the number 0. Our old x-axis is now called the 'Real Axis', and our old y-axis is the 'Imaginary Axis'.
MCrawford (19:40:41)
So in the above picture, A is the point (2,5) which corresponds to the complex number 2+5i, and B is the point (7,6) which corresponds to the complex number 7+6i.
Often we will be a bit looser with our terminology, and say that the point A "is" the complex number 2+5i.
MCrawford (19:41:10)
We can now view each complex number (each point in the complex plane) in terms of different variables -- namely, their distance from the origin (which we call the magnitude) and the angle that the ray to the point makes with the real axis (going counterclockwise).
MCrawford (19:41:16)
How can we write the two complex numbers in the original problem?
MysticTerminator (19:41:49)
factor out a sqrt(3) out of each to make the norm 1
bmsiv (19:41:57)
as root 3 radius, 2pi/3 angle and root 3 radius, 4pi/3 angle
MCrawford (19:42:20)
Your angles are slightly off.
MCrawford (19:42:42)
What angles should we use?
bmsiv (19:42:44)
oh...right. pi/3, 5pi/3
MCrawford (19:42:50)
MCrawford (19:43:00)
I could have written the angle to the second complex number as 330 degrees instead of -30 degrees, but it's all the same and for our purposes, -30 is easier to work with.
MCrawford (19:43:19)
MCrawford (19:43:22)
This notation summarizes more compactly the value of a complex number.
MCrawford (19:43:45)
There is in fact an even more useful notation involving Euler's constant that we will not use here, but is worth noting.
MCrawford (19:43:51)
MysticTerminator (19:43:49)
MCrawford (19:44:05)
MCrawford (19:44:20)
Though MysticTerminator shows us the other form.
MCrawford (19:44:26)
Moving on...
How can we now continue with our computations?
bmsiv (19:44:40)
we can notice that cis mustiplies the angle by the exponent
bmsiv (19:44:51)
that is, it becomes root3 cis 240 + root3 cis 120
variable (19:44:59)
DeMoivre's Theorem
MCrawford (19:45:10)
bmsiv'
MCrawford (19:45:21)
bmsiv's explanation is close. Something else happens.
bmsiv (19:45:33)
except the radius is still put to the eighth
MCrawford (19:45:40)
MCrawford (19:46:01)
DeMoivre's theorem can be proven using trigonometry identities involving angle addition. Those of you unfamiliar with this proof are encouraged to try it on your own and discuss it in the Intermediate High School forum.
MCrawford (19:46:19)
MCrawford (19:46:22)
Now, how can we finish?
MysticTerminator (19:46:27)
draw a picture!
MCrawford (19:46:40)
If that helps, sure.
MysticTerminator (19:46:36)
(much better than doing it algebraically)
bmsiv (19:46:36)
notice that the sin parts cancel out, and that both the cosines are -1/2
MysticTerminator (19:46:41)
and be like ha there's an equilateral triangle there
Teddy (19:46:42)
i sin 240 + i sin 120 = 0
MysticTerminator (19:40:21)
-81 or something
bmsiv (19:46:45)
which means that it totals to 81*-1 = -81
MCrawford (19:47:02)
bmsiv (19:47:18)
um...could you keep the equations a bit shorter? they go off the screen here
MCrawford (19:47:32)
You can drag your window wider.
MCrawford (19:47:37)
If this result is surprising at all, then it is recommended that you work more with complex numbers until this problem seems simple and natural. An understanding of DeMoivre's theorem and also complex numbers known as roots of unity lay the basis for a level of understanding of complex numbers makes a very powerful toolset. This toolset is not simply one for performing calculations, but also solving polynomial problems, geometry problems, and a host of other kinds of problems.
MCrawford (19:47:51)
The AoPS Intermediate Complex Numbers/Trigonometry class covers these topics in detail, carrying on to even more difficult problems.
MCrawford (19:48:09)
MCrawford (19:48:20)
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MCrawford (19:48:38)
How can we begin exploring this problem?
uclabb (19:48:29)
draw
MCrawford (19:48:49)
It makes sense to start with a diagram. I'll let you all draw your own starting diagrams.
Once we have our initial diagrams, what kinds of information can we fill in?
Teddy (19:48:50)
now i actually have to use paper
MCrawford (19:49:08)
Sorry Teddy. We all know how you struggle with geometric diagrams.
Monkey (19:49:00)
the stuff given
MCrawford (19:49:29)
Well...how do we interpret the given statements?
MCrawford (19:49:44)
There is both qualitative and quantitative information to incorporate here.
MCrawford (19:50:00)
Not recognizing this fact is precisely what leads many students to a quick dead end.
bmsiv (19:49:48)
first, we find that we are given the lengths of two medians
Aryth (19:50:21)
The two medians intersect where the area closer to the point is twice the length than the length closer to the side
Monkey (19:50:31)
Then we use the fact that medians divide each other into 2/3-1/3 parts.
MCrawford (19:50:38)
Since we have medians to two of the sides, we know that the medians will divide each other up into segments with ratios 2:1:
MCrawford (19:50:44)
MCrawford (19:50:55)
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MCrawford (19:50:58)
Does our diagram help us with anything?
Monkey (19:51:00)
medians are perpedicular
samath (19:51:08)
AEG is a right triangle
Aryth (19:51:10)
EGA is a right triangle!
MCrawford (19:51:25)
We see that AEG is a 3-4-5 triangle, meaning that the angles at G are all right angles:
MCrawford (19:51:34)
MCrawford (19:51:37)
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Teddy (19:50:58)
let the intersection of the medians be X, we know AX = 4, XE = 3, and AE = 5 so the area of ABC = 6 [AXE] = 6 * 6 = 36
bmsiv (19:51:32)
each medial triangle has the same area, and there are 6 of them, and since we have a 3-4-5 medial triangle, the area of each medial triangle is 6, and so the toal area is 36
MysticTerminator (19:51:47)
ehh 36
Aryth (19:52:02)
good point...
MCrawford (19:52:16)
Nicely done.
MCrawford (19:52:37)
Now, let's suppose a student did not recognize that the medial triangles had the same area...
MCrawford (19:52:47)
Would they be dead in the water?
Neutron (19:53:02)
Use Heron's Formula
MCrawford (19:53:13)
Ouch, please, no.
Teddy (19:53:17)
[ABC] = AG * EB
MCrawford (19:53:50)
Explain.
bmsiv (19:53:59)
the ADE triangle is a quarter of ABC in area
MCrawford (19:54:35)
We have a couple of ways, yes. Explain them both please.
bmsiv (19:54:33)
A(ABC) = 27 + A(ADE) = 27 + A(ABC)/4
MCrawford (19:55:01)
Where do you get that equation?
Neutron (19:55:03)
[ABE] = [ACE]
MCrawford (19:55:28)
I think you have one letter wrong Neutron
Teddy (19:55:20)
AG is perpendicular to EB and A and C are the same distance to EB since AE = EC and AEC is a line so [abc] = AG * EB
MCrawford (19:55:58)
There are a number of ways we can compare the areas of smaller parts of the triangle.
MCrawford (19:56:09)
We now see that AG is the altitude to base BE of triangle ABE.
MCrawford (19:56:14)
Neutron (19:56:14)
I meant [ABE] = [BCE]. Equal base, shared height
MCrawford (19:56:28)
MCrawford (19:56:30)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Community/MJImages/MandMJg.gif
MCrawford (19:56:35)
It is important to recognize the relationships between areas of triangles even when either the base or altitude is not obvious. In this case we see that the bases AE and CE of triangles ABE and CBE are the same and even though we have not identified the altitude to those bases, we know that it must also be the same since AE and CE are on the same line.
Monkey (19:56:57)
[ABC]=36
MCrawford (19:57:08)
MCrawford (19:57:28)
The key to getting off the ground with this problem was starting with the information available and scanning our diagram for new information. It's important to recognize the relationships between medians, even if they are not identified as medians to us. From there we had to recognize the relationships between the areas of pieces that made up the whole triangle. If that step seemed difficult, it might have helped to have labeled the altitudes to the sides. Recognition of such relationships can become second nature with experience.
MysticTerminator (19:57:11)
yay I finished doing it by heron's formula :-)
MCrawford (19:57:51)
1. I don't believe you because I know how smart you are.
MCrawford (19:58:13)
2. If you did, you're just punishing yourself. [img id=em-0]
Aryth (19:57:47)
lol
Monkey (19:58:22)
lol
MysticTerminator (19:58:24)
well I'm hyper. more hyper than normal ...
Sabre_X (19:58:29)
rofl, owned
MCrawford (19:58:38)
Okay, okay.
MCrawford (19:58:42)
I've only got one more today.
MCrawford (19:58:50)
I wasn't expecting a crowd this large.
MCrawford (19:58:52)
MCrawford (19:58:59)
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MCrawford (19:59:18)
A few highly ambitious students might look at this expression and endeavor to estimate the computations, but the answer is a 5-digit integer, so I wouldn't suggest banging your head against that particular wall.
MCrawford (19:59:28)
What kinds of tools does this problem seem to invite?
Monkey (19:59:37)
Binomial Theorem
solafidefarms (19:59:47)
combinations and binomial expansions ;o
Sabre_X (19:59:51)
binomial theorem?
MCrawford (20:00:00)
Since have a sum of two numbers raised to an exponent, binomial expansion comes to mind.
MCrawford (20:00:06)
Can we just expand here and calculate nicely?
Monkey (20:00:19)
No
Aryth (20:00:21)
perhaps not
MCrawford (20:00:28)
Why not?
bmsiv (19:59:36)
hmm...I personally like doingthis with the binomial theorem, and then separating the square root terms and the integers
Aryth (20:00:38)
the square roots interfere
Monkey (20:00:40)
Still way too many square roots
MCrawford (20:00:54)
bmsiv (20:01:01)
7^3+6*7^2*sqrt(35)+15*7^2*5+20*7*5*sqrt(35)+ ...
MCrawford (20:01:12)
Unfortunately, plain ole binomial expansion still leaves us with square roots that would require estimation. That might make the super-crazy-ambitious-head-to-wall-banging-number-crunches a little happier, but it's hardly inspiring.
MCrawford (20:01:22)
Is there any nice way of getting rid of the messy stuff, or are we barking up the wrong tree?
MysticTerminator (20:01:27)
try squaring it first
MysticTerminator (20:01:30)
bit less square roots
MCrawford (20:01:54)
That might have made things easier. Good call.
MCrawford (20:01:58)
But we're down this path already...
MysticTerminator (19:59:10)
add (sqrt(7) - sqrt(5))^6
Teddy (20:00:00)
that expression + (rt 7 - rt 5)^6 since (rt 7 - rt 5)^6 < 1
MCrawford (20:02:40)
MCrawford (20:03:08)
What's nice about this?
MysticTerminator (20:00:49)
13536
bmsiv (20:03:13)
add the two
Aryth (20:03:19)
cancel out terms with square roots
Monkey (20:03:19)
It cancels out the square roots when we add
MCrawford (20:03:34)
MCrawford (20:03:56)
What is our answer?
MysticTerminator (20:03:55)
but! we added something
MysticTerminator (20:01:17)
13535
Teddy (20:04:08)
13535
bmsiv (20:04:10)
we added something less than 1
bmsiv (20:04:15)
so we must subtract 1
Sabre_X (20:04:16)
13535
MCrawford (20:04:42)
It's not the closest integer! But we know that we rounded up to 13536, so we must go down 1 to 13535.
Monkey (20:04:23)
13535, since (\sqrt {7}-\sqrt {5})^6>0, but less than 1.
yangzhi (20:04:30)
13535
MCrawford (20:04:49)
E^(pi*i)=-1 (20:05:15)
MCrawford (20:05:21)
It's good to recognize the ways that we can combine binomial expansions to rid us of messy parts of otherwise nasty calculations. This method is also useful when dealing with expansions of some complex numbers. This method is discussed in detail in the AoPS Intermediate Algebra class.
MCrawford (20:05:36)
MCrawford (20:05:48)
I'll leave the rest of this solution to all you young geniuses.
MCrawford (20:06:00)
That was our last problem for the night.
Thank you for coming to this Mandelbrot Math Jam. I hope you enjoyed it.
Are there any questions about Math Jams or AoPS classes?
MCrawford (20:06:08)
Sorry that was shorter than I planned.
MCrawford (20:06:34)
I didn't expected a reigning Mandelbrot champion and some other high scorers to show up.
Teddy (20:06:17)
we blazed too fast?
Aryth (20:06:21)
is Mandelbrot only for High schools?
MCrawford (20:06:54)
Middle schoolers can take it too.
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