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Current time:0:00Total duration:17:42

2003 AIME II problem 6

Video transcript

in triangle ABC a B is equal to 13 BC is equal to 14 and AC is equal to 15 let me just draw that because it seems like a pretty involved problem so let me see if how well I can draw triangle ABC so let's say it looks like that like that and then like about to draw let me draw it like that pretty well I could draw a slightly straighter line there you go so let's say that this is triangle a B C they tell us that a B is equal to 13 B C is equal to 14 and then a C is equal to 15 and then they tell us Point G is the intersection of the medians and if you've watched the video on medians and centroids you might recognize that the intersection of the medians is just the centroid so let me show you that so this is the midpoint of AC so the line connecting B to that midpoint is a median let me draw a little bit cleaner than that shaky hand so that is a median the line collecting vertex a with the midpoint of BC so maybe it's gonna go right about there is also another median drawing the straight lines is the hard part and then finally you have your third median here connecting C to the midpoint of a B so it will connect it just like just like that and we learned if you watch the video on medians and centroids that these three means they always intersect in a point called the centroid and they even imply that on this vid on this on this question right here that it's going to intersect at Point G so that right there is Point G points a prime B Prime and C prime are the images of a B and C respectively after 180 this must be 180 degree rotation about G what is the area of the union of the two regions and closed by the triangles ABC and a prime B prime C prime so let's draw the second triangle a prime B prime C prime so what we want to do is rotate each of these each of these points about G 180 degrees so if a is here if we rotate it if we rotate it 180 degrees about G it's going to end up so if it's this far from G right now it's going to be that far again on this side so a is going to be a is going to be right over there or I should say a prime it's the image of a after being rotated 180 degrees B is this far right now if you rotate it if you rotate it 180 degrees the B will then be about that far the prime will be that far and same thing for C if it's this far right now if you've rotated 180 degrees it's going to get it's going to get about this looks about right it's going to go about right over there so you have C Prime and so if we wanted to draw a triangle a prime B prime C prime it would look like this let me see how well I can so let me draw my best attempt at it so let me so it would look something something like something like this and this line should be proud to that line because it's now going to be it's the line connecting a and B it's been rotated 180 degrees this line should be parallel to this one so I just want to make drew my best attempt at drawing it because it seems like it's going to be relevant and then this line should be parallel to that one since it's just rotated 180 degrees so it's gonna look something it's going to look something like this so what they're asking they want the area of the union of the two regions enclosed by the triangles so it's essentially the area of this entire star is the six pointed star kind of a slightly skewed or slightly tilted star of david' over here so this is not you know this just I'm not doing it in real time now I made sure I could do this problem before I did this video but it took me a while to for it to kind of jump out at me kind of knowing that most of these problems don't require to do some really hard mathematics they a lot of times you just kind of have to see something or rearrange things and the answer jumps out and it's also the case with this one here and the thing that jumps out and this was actually my motivation for making the video I showed you in the I showed you in the last well if last video you watched that if you have a median of a triangle if you have a median of a triangle and the centroid of the triangle is over here this distance the distance from the vertex to the centroid is twice the distance from the centroid to the midpoint of the other side or another way if you look at this diagram right over here this distance this distance is going to be twice this distance but this distance is also the same thing as this whole entire distance right this distance and this distance are going to be the same it's been reflected 180 degrees or it's been rotated 180 degrees I should say and if this distance is half of that then we know that this is exactly halfway between this point and this point or this is exactly halfway between this point and this point up there or another way to think about it we know that this length right over here is equal to this length right over here which is equal to this length over here which is equal to this length over here and so you start seeing maybe maybe some triangles will start jumping out at you and let me see if I can draw these triangles and have it respectably neat so I have this triangle up here I have this triangle up here we know it's we could view it as this is actually going to be a median of this triangle right over here is going to be is going to be the same same as the median I guess we could say or the equivalent I guess line of this triangle it's actually going to be an altitude as well because we're going to see we're kind of drawing a rhombus and the diagonals of a rhombus intersect they're perpendicular to each other so this actually these are altitudes so this altitude right here is going to be the same as this altitude right over here this this side is the exact same thing as this side over here for both triangles so this side is the same thing as that side this side let me be careful here we already know we already know that this side is going to be the same thing as this side we know that these two sides are the same because this is acting as a media of this triangle over here we already know that this side this pink side up here or actually I say this yellow side I'll direct two marks it's the same thing as that yellow side down there so this side is this if this is the same as this and this is the same in both of these triangles we know that this side we know we know that this side must be must be the same thing as this side which also must be the same thing by the same exact argument as this side which also must be the same thing as this side right over here or 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 exact same area as this triangle or you could view it another way this purple triangle right over here is going to have this purple triangle right over here it's going to have the exact same area is actually going to be a congruent triangle to this triangle over here all of its sides are exactly the same they actually share one of them well we can use that exact same argument to say that this triangle right over here this triangle right over here is completely congruent to this triangle right over there they have all of the same size even share side in fact we even know that these two triangles are congruent to these two triangles up here so these these are all these are all congruent now let's think about let's think about another triangle let's think about this triangle right over here let me draw some more triangles over here let me draw some more triangles over here now we know we know that this distance this distance over here that's the distance that we started about talking about the medians this distance right over here from a centroid to a base or from a centroid to the base that's the exact same thing as this distance right over here if you were to draw a parallel line right over there this obviously it's just these this line up here and this line over here are parallel so if you have if if you are interest is parallel to this and they're intersecting at the same points and I won't go into the depths of proving that they're parallel it's actually not too hard to do you can set show that this distance is the same as this distance well those two distances are the same and we share this point right we share this side right over here we share this magenta side right over here then we know we know that this side is going to be the same thing we know that this side is going to be the same thing as that side or we know that this triangle right over here is going to be the same thing as that triangle and then we could take the mirror image we could say well that uses essentially the same argument that's going to be the same thing as this triangle over here and you can I guess it's a similar argument to show that that's going to be the same thing as that triangle over there which is going to be the same thing as that triangle over there so and then you do it on the other side that's gonna be the same thing there same thing there I think you see where this is going all of these triangles are congruent this try those magenta triangles are congruent to this triangle right over here which is congruent to this triangle right over here which is congruent to this triangle right over there which is congruent to that triangle right over there there's actually multiple ways you can make that argument I won't go into the depths of it but the same argument we said to say these magenta triangles are equivalent you can make that same argument so this this distance is exactly half of this entire distance same argument will tell you that this triangle is congruent to that triangle and that triangle is congruent to that one and that one's congruent to that one and that one's congruent to that one so that entire shape and it was kind of a tedious way of doing it actually was a little hand wavy about it I didn't do a very rigorous proof about it but I think you can see it's it actually would not be hard to do a rigorous proof that this shape this kind of slanted star of david' is made up of 1 2 3 4 5 6 7 8 9 10 11 12 congruent triangles 12 congruent triangles now how many of these congruent triangles was our original triangle ABC made up of well we have 1 we have 2 3 4 5 six seven eight nine so we have twelve congruent triangles in in the star in the star we had nine congruent triangles congruent triangles in a b c so the area of the star is just going to be twelve ninth of the area of triangle a b c so if we can figure out the area of triangle ABC we know that each of those triangles are going to be one ninth of that and then you multiply that by twelves you'll get the area of the actual star so let's actually figure out the area of triangle ABC so if we do ABC I'll do it I don't want to lose the problem here so our triangle looked something like this now there are formulas for figuring out the areas of triangles if you just know the sides I always forget those formulas but I do remember the law of cosines and I always so I always break out the law of cosines which actually helps you derive those formulas to figure out the area of an arbitrary triangle you can figure out the area of a triangle area of a triangle is one-half base times height so if we were able to figure out this height right over here if we could figure out this height over here we can multiply it by 1/2 and 14 or multiply it by 7 and then we would be able to figure out the area of this entire triangle so how can we do that well let's just break out let's just break out a little bit of a little bit of trigonometry let's take let's take this to be theta right 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 13 squared plus 14 squared minus minus 2 times the product of these two sides so times 13 times 14 times the cosine of theta times the cosine of theta now what is the cosine of theta theta is so cosine is adjacent over potenuse so it's this length right over here over our hypotenuse so let's just call this length over here X let's call this X this obviously this distance over here would obviously be 14 minus X so let's just call this X so cosine of theta is equal to x over 13 so we can replace this with x over 13 and obvious if you could solve for X you can then solve for H using the Pythagorean theorem but these two characters cancel out and we are left with we are left with let's see 15 squared is 225 is equal to 169 14 squared 14 squared we have 140 plus 40 plus 16 is 196 so 196 minus minus 2 times 14 X so minus 28 X let's see if we can solve for X now so let's add 28 X to both sides let's subtract let's do a couple of things let's subtract 225 from both sides so I'm going to subtract 225 from both sides and I'm going to add 28 X to both sides so add 28 X to both sides that should help us with the sign so on the left hand side we are left with we are left with 28 X is equal to now we just have to do a little bit of math 169 169 plus 196 6 plus 9 is 15 this is going to be 16 365 did I do that right 15 and then use 68 and 365 and then from that we want to subtract 225 it's going to be a 0 this is going to be a 4 it is going to be 140 so 28 X is going to be 140 or X is going to be equal to you divide both by 28 let's see 28 goes into 140 let's think about it it goes well it goes 5 times 5 times 20 is 100 5 times 8 is 40 so it goes 5 times so X in this situation is equal is equal to 5 so we can use that information figure out this height we know from the Pythagorean theorem that five squared plus h squared is equal to 13 squared is equal to the hypotenuse squared is equal to 169 or we have H squared is equal to 169 minus 25 so it's equal to 144 or H is equal to H is equal to 12 so now we can figure out the area of the triangle the area and sometimes denoted by brackets the area of triangle ABC is equal to 1/2 times 14 times times 12 which is the same thing as well this is the same thing as 7 times 12 which is equal to 84 now that's just the area of these 9 that's just the area of these 9 congruent triangles we know that the area of the whole thing will be 12 ninths of that because this is 9 congruent triangles we want 12 congruent triangles so the area of the whole shape so the area of the star the area of the star I'll put it in quotes is going to be equal to 12 over 9 times times 84 and let's see 12 over nine this is the same thing as if you divide the numerator in the denominator by 3 this is the same thing as 4 over 3 times 84 let's see 84 3 does go into 84 how many times it looks like it goes 28 times let me make sure so 3 goes into 84 2 times 3 is 2 times 3 is 6 subtract you get a 20 yep 28 times so this is equal to 4 times 4 times 28 which is going to be 80 plus 32 which is equal to 112 let me make sure I did that right don't want to make a careless mistake 18 minutes into this video 4 times 20 is 80 plus 32 is 112 and we are and we are done