2003 AIME II problem 6
2003 AIME II problem 6
- In triangle ABC, AB is equal to 13, BC is equal to 14, and AC is equal to 15.
- Let me just draw that. It looks like a pretty involved problem
- So it looks something like that
- Let's say this is triangle ABC
- They tell us AB is equal to 13, BC is equal to 14 and AC is equal to 15
- and then they tell us point G is the intersection of the medians.
- If you've watched the video on medians and centroids you might recognize
- that the intersection of the medians is just the centroid.
- Let me just show that here.
- The line from B to the midpoint of AC is a median.
- Let me draw that a little cleaner.
- So that is a median.
- Now the line from A to the midpoint of BC is also a median.
- And finally the line from C to the midpoint of AB is a median.
- We learned -- if you've watched the video on medians and centroids --
- that these three medians always intersect at a point called the centroid.
- They even imply that in the question. So that right there is point G.
- Point A', B', and C' are the iamges of A, B, and C after a 180 degree rotation about G.
- What is the area of the nion of the two regions enclosed by the triangles ABC and A'B'C'?
- Let's draw the second triangle.
- What we want to do is rotate each of these about G 180 degrees.
- So if A is here and we rotate it 180 degrees about G,
- it's going to end up just as far from G on that side.
- So A' is going to be right over there.
- B is this far right now, if you rotate it 180 degrees,
- B' will then be about that far.
- Same for C. If you rotate it 180 degrees,
- it's going to go about here. So you have C'.
- So if we wanted to draw A'B'C', it's going to look like this.
- Here's my best attempt at it.
- This line should be parallel to that line
- since it's been rotated by 180 degrees.
- And this line [C'B'] should be parallel to this line [BC],
- and this line [C'A'] should be parallel to that line [AC].
- It seems like it's going to be relevant.
- So they're asking, they want the area of the union of the two triangles.
- So it's the area of this star. This slightly skewed or tilted star of David
- I'm not doing this in real time; it took me a while before I did this problem
- before it jumped out at me.
- Kind of knowing that most of these problems don't require
- really hard mathematics.
- A lot of the time you just kind of have to see something
- or rearrange things and the answer jumps out.
- It's also the case with this one here.
- And the thing that jumps out (and the motivation for making this video)
- I showed you in the last video that if you have the median of a triangle
- and the centroid of the trianlge is over here,
- this distance [GA] is twice the distance from the centroid to the midpoint [of BC]
- and the same on all 3 sides.
- But this distance [AG] is also the same as this whole entire distance [A'G]
- This distance has been rotated 180 and degrees
- and if this distance [G-midpoint of BC] is half of that one [AG]
- this is exactly halfway between this point [G] and that point [A']
- Another way to think about it:
- we know that this length over here is equal to this length over here,
- which is equal to this length over here, which is equal to this length over here
- And so you start seeing, maybe, some triangles will jump out.
- Let's see if I can draw these triangles.
- I have this triangle up here.
- This is actually going to be a median of this triangle over here
- It's going to be the same as the median or equivalent line of this triangle over here
- It's actually going to be an altitude -- we're kind of drawing a rhombus
- and the diagonals of a rhombus are perpendicular.
- So this altitude right here is going to be the same as this altitude over here.
- This side is the exact same thing as this side.
- Let me be careful here.
- We already know that this side is going to be the same thing as this side
- We know these two sides are the same because this is acting as a median of this triangle
- We already know that this yellow side I'll draw two marks is the same thing
- as that yellow side down there.
- If this is the same as this, and this is the same in both triangles,
- we know that this side must be the same thing as this side,
- which also must be the same thing as this side and this side over here.
- Another way to think about it, this triangle is going to have the exact same area
- as this triangle.
- This triangle is going to have the same exact area as this triangle.
- Or, you could view it another way.
- This purple triangle right over here is going to be a congruent triangle to this triangle over here.
- All of its sides are exactly the same. They share one of them.
- We could use the same argument to say that this triangle right over here
- is completely congruent to this triangle right over there.
- They have all of the same sides.
- In fact, we even know that they are congruent to these two up here.
- So these are all congruent.
- Let's think about this triangle right over here.
- We know that this distance, the distance when started talking about medians,
- from the centroid to the base, that's the exact same distance
- over here , if you were to draw a parallel line.
- It's just that this line up here and this line over here are parallel
- If this is parallel to this, and they're intersecting at the same point
- I won't go into the depth of proving they're parallel -- it's not too hard to prove.
- You can show that this distance is the same as this distance.
- We share this magenta side right over here,
- so we know that this side is going to
- be the same as that side
- Or we know that this triangle is going to be
- the same thing as that triangle.
- And then we can take the mirror image.
- Using essentially the same argument,
- that's going to be the same thing as this triangle over here.
- And it's a similar argument to show that that's going
- to be the same thing as that triangle over there.
- And then do it on the other side.
- I think you see where this is going!
- All of these triangles are congruent.
- There's actually multiple ways you can make that argument.
- The same argument we made to show that these magenta triangles are congruent,
- we could make to say that this distance is exactly half of this entire distance.
- Same argument to tell you that this triangle's congruent
- to that that triangle.
- And that one to that one.
- So that entire shape --
- and actually I was a little handwavy about it --
- I think you can see it actually wouldn't be hard to do a rigorous proof.
- But this shape, this slanted star of david
- is made up of [counts to 12] congruent triangles.
- 12 congruent triangles.
- Now, how many of these congruent triangles originally made up our triangle ABC?
- [counts to 9]
- So we have 12 congruent triangles in the star
- we had 9 congruent triangles in ABC,
- so the area of the star is just going to be 12/9 the area of ABC.
- So if we can figure out the area of ABC,
- we know that each triangle will have 1/9 that area.
- Multiplying by 12 we'll get the area of the actual star.
- So let's figure out the area of triangle ABC.
- Let me redraw the triangle.
- There are formulas for figuring out the areas of triangles if you know the sides.
- I always forget them, but I do remember the law of cosines.
- It helps me derive those formulas.
- Area of a triangle is 1/2 base * height,
- so if we could figure out this height,
- we could multiply 1/2 and 14, or multiply it by 7 and
- then we would be able to figure out the area of this entire triangle.
- How can we do that?
- Let's just break out a little trigonometry.
- Let's take this to be theta over here.
- The law of cosines tells us that
- the opposite side of the angle squared, so 15 squared
- is equal to the sum of the other sides squared
- (13 squared + 14 squared)
- minus 2 times the product of these two sides
- so times 13 times 14
- times the cosine of theta
- Now, what is the cosine of theta?
- Cosine is adjacent over hypotenuse,
- so that's this side over our hypotenuse.
- So let's just call this length over here x.
- So cosine of theta is x over 13.
- So if you could solve for x you could then solve for h using the Pythogorean Theorem
- These two characters cancel out, and we are left with
- 15 squared is 225, 13 squared is 169
- 14 squared -- we have 140 + 40 + 16 = 196
- minus 28 x
- So let's solve for x.
- We'll do a couple of things. We'll subtract 225 from both sides
- and I'm going to add 28x to both sides
- On the left hand side we are left with 28 x is equal to
- and we have to do a little bit of math
- [does arithmetic aloud] 365. Did I do that right?
- And from that we want to subtract 225
- It is going to be 140
- so x is going to be 140 divided by 28
- Let's think about it. It goes 5 times. 5 times 20 is 100,
- 5 times 8 is 40, so 5 times 28 is 140
- So x in this situation is equal to 5.
- So we can use that information to figure out this height.
- We know from the pythagorean theorem that
- 5 squared plus h squared is equal to the hypotenuse squared.
- It's equal to 169
- or h squared = 169 -25 = 144.
- h is equal to 12
- Now we can figure out the area of the triangle.
- The area sometimes denoted by brackets [ABC] is equal to
- 1/2 times 14 times 12
- which is the same as 7 times 12, which is equal to 84
- That's just the area of these 9 congruent triangles.
- We know that the area of the whole thing will be 12/9 of that
- So the area of the star is going to be equal to 12/9 times 84
- 12/9, dividing numerator and denominator by 3 is the same as 4/3
- 3 does go into 84. How many times? Is it 28?
- Let me make sure. 2*3 = 6, so yup, 28 times.
- So this is equal to 4 times 28
- Which is going to be 80 plus 32
- which is equal to 112.
- Let me make sure I did that right
- 4 times 20 is 80 plus 32 is 112.
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