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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 6

Created by Sal Khan.

## Want to join the conversation?

- is this 8th grade work because i am in 6th grade(3 votes)
- lol, this is AIME stuff. AIME is a huge competition for math geniuses. :)(2 votes)

- Why is the download link not provide like it is , for the other videos ?(3 votes)
- Maybe there is a bug affecting you to watch the video(2 votes)

- Around the time from6:00to7:15, Sal talks about how all four of the triangles he highlights at the top are congruent. However, I don't really understand why that is true. I see the congruent segments that he marks with the ticks (the bisections of the intersecting segments), but I don't understand why he says that he's forming a rhombus because the diagonals are perpendicular. If he has a rhombus, doesn't that mean (by similar triangles) that triangle ABC was isosceles with base BC?(1 vote)
- I get that all four triangles will have the same areas because A=1/2 ab sin(C), and also sin(C)=sin(180-C). However, I don't get why all four of them are congruent. The diagonally opposite pairs are congruent, I know, by SAS.

In short, my question is this:

Why did Sal say that the thing was a rhombus and its diagonals were perpendicular?

BTW: I used two posts because I needed more characters in my original question.(4 votes)

- You could also use the 6 congruent triangles to find the area of the hexagon in the center. Then, using Heron's Formula, find the area of the triangle that you started with. Multiply this by 2, subtract the area of the central hexagon, and you would get the same answer with a lot less work.(2 votes)
- Couldn't Sal have shown that the magenta rhombuses were congruent to the green by first assuming that the 2 angles are x and y in the rhombus. Then you know x + y = 180. When the star is cut in half there are 180 degrees on one side, but the angles x/2 and x/2 take up some space. Therefore one of the angles is y, which means the other is y, which means the other 2 are x.It may seem like you could just scale that up and down, but since the 2 rhombuses share 2 sides you can't. Also, a general triangle with lengths a,b, and c has an area equal to square root(s(s-a)(s-b)(s-c)) where s is equal to (a+b+c)/2(1 vote)
- This is wrong! in a scalene triangle, all medians are of different length. Using this logic we end up proving that all medians are of the same lenght which wouldnt be trueee(1 vote)
- Why doesn't he just use Heron's? Heron's formula states that the area, A, is equal to the square root of the semiperimeter ((a+b+c)/2) times (S (semi perimeter) -a) times (S-b) times (S-c). I know he forgot, but it seems so much easier. Also, the specific equation is A = sqrt(S*(S-a)*(S-b)*(S-c)) where A is the area, a, b, and c are the side lengths, and S = (a+b+c)/2(1 vote)
- how does he write that so good(1 vote)
- Why does the cosine of theta and 13 cancel out?(1 vote)
- I hate this add a little fun(0 votes)

## Video transcript

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. And we are done!