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Math for fun and glory
Course: Math for fun and glory > Unit 4
Lesson 1: 2003 AIME- 2003 AIME II problem 1
- 2003 AIME II problem 3
- Sum of factors of 27000
- Sum of factors 2
- 2003 AIME II problem 4 (part 1)
- 2003 AIME II problem 4 (part 2)
- 2003 AIME II problem 5
- 2003 AIME II problem 5 minor correction
- Area circumradius formula proof
- 2003 AIME II problem 6
- 2003 AIME II problem 7
- 2003 AIME II problem 8
- Sum of polynomial roots (proof)
- Sum of squares of polynomial roots
- 2003 AIME II problem 9
- 2003 AIME II problem 10
- Trig challenge problem: area of a triangle
- 2003 AIME II problem 12
- 2003 AIME II problem 13
- Trig challenge problem: area of a hexagon
- 2003 AIME II problem 15 (part 1)
- 2003 AIME II problem 15 (part 2)
- 2003 AIME II problem 15 (part 3)
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2003 AIME II problem 4 (part 2)
Created by Sal Khan.
Want to join the conversation?
- I've got a faster way to do this problem.
Let's look at one face of the tetrahedron. The front one with the blue line for example.
Blue line = √(2^2-1^2)=√3
Draw a line bisecting a bottom corner of the triangle. It goes through the center of the face.
To calculate length of the blue line from the bottom to the center of the face do:
tan(30) x 1 = √(3)/3
smaller/larger = ((√(3)/3)/√(3))^3=1/27
1+27=28(5 votes) - I worked this out using a relatively easy method that didn't involve coordinate bashing.
Let's call the side length of the larger tetrahedron x.
The centroid of one of the larger faces would be at one third of the height, or √(3)/6 x. Therefore, we can form an isosceles triangle with two segments going from the centroids of two faces to the midpoint of their common side and a segment between them (that segment between the centroids is one of the sides of the smaller tetrahedron). We now have an isosceles triangle with one missing side length of the smaller tetrahedron and two side lengths of √(3)/6 x. This triangle is a dilation of the triangle formed with one side of the tetrahedron, x, and two heights meeting at the middle of the opposite side, each √(3)/2 x.
That makes the smaller tetrahedron's side length equal 1/3 x.
And since the ratio of the larger tetrahedron's side length to the smaller tetrahedron's is 3 : 1, the ratio of their volumes would be 27 : 1. Therefore, the two relatively prime numbers would be 27 and 1 and their sum would be 28.(2 votes) - This solution is good!!
But I want to solve it faster, not drawing.
Can anyone explain it to me?(2 votes)- Simply find the similarity ratio. The centroid is 1/3 up the face, so the answer is simply (1/3)^3=1/27.
http://artofproblemsolving.com/community/c4h125800p713207(1 vote)
- I have an alternate solution: Look at the triangle constructed at time. It is known that the blue side of the triangle is the same length as the other blue side (both are vertical while the other red one is horizontal). It is a known fact that the center of a triangle occurs at a height 1/3 of the triangle's height (for equilaterals), so the center of the face on the larger tetrahedron is 1/3 it's slant height. Therefore, the sides of that small triangle are 1/3 and 1/3. That 1/3 is one of the side lengths of the smaller tetrahedron, and since we know that is regular, all of its sides are 1/3. Because of the relationship of volume to length, any ratio of 1 : a in length causes a ratio of 1 : a^3 in volume. Thus, the smaller tetrahedron is (1/3)^3 times the size of the larger, or 1/27. 2:38(2 votes)
- What do he mean at? 0:33(1 vote)
- How many right angles are on a hexagon?(1 vote)
- It depends on the shape of the hexagon. It could have one, two, or none.(1 vote)
- Honestly there is an easier way to do this. You split each of the big tetrahedron's faces into 3 layers of 9 smaller triangles, then you can easily find the centre of each face and know that Lb/Ls is 3. 3^3 is 27. I calculated that in like 30 seconds.(1 vote)
- why is there 2 parts to 2003 AIME II problem 4(1 vote)
- I think it takes more than 15 minutes so Sal made two videos.(1 vote)
- I found Sal's work to be very complicating. Is there a faster/easier way to do this?(1 vote)
- Yes, definitely. Since the centroid is 1/3 up from the base of the tetrahedron, the height of the smaller tetrahedron is 1/3 of the height of the larger tetrahedron. So, our answer is (1/3)^3=1/27.(0 votes)
Video transcript
Where we left off
in the last video, we were able to figure
out the coordinates of all of the vertices for this face
right here of the tetrahedron. And so now we can use those
coordinates to figure out the coordinates of the center
of that face, which is also one of the vertices of
the smaller tetrahedron. When we figure out
that coordinate, then we can figure
out the distance between that point
and that point, or between this
point and this point. And that will give
us the length of one of the sides of the
small tetrahedron. And then we can compare
that to the length of the large tetrahedron and
get this ratio right over here. So let's get the
coordinates of that point or this point right over here. Let's get the coordinates
of this thing. To do that, we just have to
take the average of the x, y, and z-coordinates the
vertices of the face. Now for x, you
can see, hopefully maybe you see that this
is sitting on the yz axis. Actually, let me draw the z-axis
here just for visualization. So z could go up like
that, and then this would be the negative z
direction would be going down like that. So this is clearly
lying on the zy plane. But even if you
didn't see that, you could just average
the x points here. So we have negative
1 plus 1 plus 0 over 3, which is
going to be 0 over 3. And then you could
average the y points. 0 plus 0 plus square root of
3 over 3, all of that over 3. And then you can
average the z points. Let me do this in
a different color. Average the z's. 0 plus 0 plus 2 square roots
of 2 over the square root of 3, all of that over 3. Now this first coordinate right
over here, clearly just 0. Second coordinate
over here is going to be the square root of-- this
is the square root of 3 over 3. This was hard to read. Square root of 3
over 3 divided by 3. So it's the square
root of 3 over 9. And then we have 2 times
the square root of 2 over the square root
of 3 divided by 3. So it's going to be 2
times the square root of 2 over 3 times the
square root of 3. This right here
are the coordinates for the center of this space
face, this coordinate right over here. Now using that, we can
now find the distance between that point
and that point. We already know this
point's coordinates. We already know the
center of the base. We already know that
right over here. So let's figure out this
distance or this distance, the length of a side of
the smaller tetrahedron. So let's write it. Let me do it over here. The length of the side of a
smaller tetrahedron squared is going to be our difference
in x value squared. So this has an x value of 0. This has an x value of 0. So 0 minus 0 squared plus
the difference in y values. Square root of 3 over 9
minus square root of 3 over 3 minus-- Square root
of 3 over 3, just so we have a common
denominator, that's the same thing as 3
square roots of 3 over 9. I just wanted to
rewrite it this way so that I have a common
denominator, squared. Plus 2 square roots of 2. It's getting messy because I'm--
plus 2 square roots of 2 over 3 square roots of 3 minus
this guy's z-coordinate. Well, this guy was
on the xy plane, so it's minus 0,
minus 0 squared. So this is going to be equal
to-- let me write it here. Let me scroll to the
left a little bit so I have some space. So we have a length of the
small tetrahedron squared is going to be equal to-- well,
this is just going to be 0. This right here is negative
2 times the square root of 3 over 9 squared. So negative 2 times the square
root of 3 over 9 squared is 4 times 3. Negative 2 times square
root of 3 squared is going to be 4 times 3
over 9 squared, over 81. So we did that part
right over here. We had negative 2 times
square root of 3 over 9. Square that, you
get this over here. So that turns into that. And then this term
right over here, this is just going to be 2
root 2 over 3 root 3 squared. So it's going to be plus-- so
this is going to be 4 times 2 over 9 times 3. Or another way of
saying it, this is going to be equal
to-- let's see. Well, the numerator and
denominator are divisible by 3. So this is the same
thing as 4 over 27. So this first term is
4 over 27 plus what we have over here, 4 times 2 is
8 over 27, which is equal to, which is the same
thing as 12 over 27. You divide the numerator
and denominator by 3. This is equal to 4 over 9. So the length of the small side
squared is equal to 4 over 9. You take the principal
root on both sides, positive square root. We only care about
positive values because we're talking
about distances. So the length of
one side is going to be equal to the
square root of that, which is going to be 2 over 3. So this distance right
here, this length right here is going to be 2/3, or
this length right here is going to be 2/3. So what's the ratio of
a length of a big side to the length of a small side? So the length of a big side
we already figured out the. Length of a big side is 2. This thing is 2. What's the length
of a small side? Well, based on all the math
we just did, it is 2/3. Now before we even
cube it, we can divide the numerator and
the denominator by 2. Actually, we can divide the
numerator and the denominator by 2, so this becomes 1/2 1/3. And 1 over 1/3 is
the same thing as 3. So 3 cubed is 27. Or I could write it
as 27 to 1, since we want to write it as a ratio. So when you simplify all this,
you really will just get 27. But if you want to express
it as a ratio, it is 27 to 1, the volume of the
big tetrahedron to the volume of the small. So that ratio is m over n. This is equal to m over
n where m and n are relatively prime
positive integers. These are definitely
relatively prime. Their only common factor is 1. Find m plus n? So 27 plus 1 is 28,
so our answer is 28.