Geometry (all content)
2003 AIME II problem 7
Find the area of
rhombus ABCD given that the radii of the circles
circumscribed about triangles ABD and ACD. Created by Sal Khan.
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- Aime is for high school right?(15 votes)
- Mostly, yes. AIME is short for American Invitational Mathematics Examination. The "invitational" part is because you have to qualify to take it, either by scoring highly (usually around 98th percentile or higher - there is a set score, but it varies year to year) on the AMC 10/12 (which are open to anyone who is not above that grade level - that is, you can take it in middle school, although it is intended for people in high school) or by getting the highest score in your state on the AMCs. If you score highly on the AIME, which has fifteen questions of increasing difficulty, you take the USAMO, United States of America Mathematical Olympiad, which consists of 6 problems given in two 4.5 - hour sets of three. Of the top twelve high scorers from this, six go on to represent the USA in the International Mathematical Olympiad, or IMO. AIME problems and solutions can be found at http://www.artofproblemsolving.com/Wiki/index.php/Aime_problems(6 votes)
- does AIME stand for somthing(11 votes)
- It stands for American Invitational Mathematics Examination.(13 votes)
- What is a rhombus?(2 votes)
- According to my on-line dictionary, a "RHOMBUS" is "an equilateral parallelogram, including the square as a special case." In other words, a rhombus is a 4-sided figure, with all 4 sides of the same length, where the angles do not have to be (but can be) right angles. Hope this helps. Good Luck.(14 votes)
- at1:30Sal says that the diameter of the circles circumscribed around the triangles abd and acd are 12.5 and 25 but the text says radii(3 votes)
- Yeah, he did the wording wrong, but he still did indeed solve the problem correctly.(2 votes)
- Ths video refers to several other videos about triangles and competition geometry. It seems to have been set up from another video, which wasn't directly before the video. I would be great if there were video links to videos to whihc the lesson refers. It was very confusing and startling to have something with such an increased level of complexity with very little preparation.(3 votes)
- i think he is referring to his competition math playlist where this vid actually used to be.
- Misran Rehman and I have the same question. The "10 = a" should be "10 = a^2", which would make the end steps a little more complicated.(1 vote)
- No. He says around12:29that 50a=5a^2. 5a^2 is 5*a*a. So dividing by 5a on both sides, 50a/5a = 10. 5*a*a/5*a=a so 10=a, not a^2(3 votes)
- what is a circumscribed circle?(2 votes)
- Wiki said: " the circumscribed circle or circumcircle of a polygon is a circle which passes through all the vertices of the polygon"
In another word, all corners of a polygon are on its circumscribed circle.
Each Triangle, rectangle and regular polygon have its one and only one circumscribed circle. Other polygons may or may not have circumscribed circles.
I hope that helps.(2 votes)
- at12:23, he said 50=5a^2 , then he divided by 5 and got 10=a. what about that squared symbol?(2 votes)
- The equation was 50a=5a^2. He then took each side divided by 5a, which came out to be 10=a, or a=10.(3 votes)
- What does AIME stand for?(1 vote)
- AIME stands for the American Invitational Mathematics Examination.(4 votes)
- This is pretty cool! I didn't like the idea of cutting the diagonals and making use of them to get the answer, but i have to admit it's a lot neater than my way. I went on to use the concepts of the cosine rule to evaluate this solution. long and taxing no doubt. great question!!(2 votes)
Find the area of rhombus ABCD given that the radii of the circles circumscribed about triangles ABD and ACD are 12.5 and 25, respectively. So let's draw ourselves rhombus ABCD. So let's draw a rhombus. So let me draw it. Here we go. That's a decent rhombus right over there. We know that all the sides of a rhombus are equal. And let's label the vertices. So vertex A, B, C, D. So there we go, rhombus ABCD. And then they say the radii of the circle circumscribed about triangles ABD-- so triangle ABD, that's A, B, D. That is triangle ABD. So let's draw out the circle. Let's draw-- its circumcircled. It's circumscribed circle, or the circle that passes through the vertices A, B, and D. So let me do my best job at that. So it would look something-- this is not a trivial thing to do. It's not always easy. So let's do it like that. There we go. That's its circumscribed circle, or it's the circle circumscribed about ABD right over there. Now they're telling us that its diameter is 12.5-- so they're saying that this diameter right over here, so if I were to draw a diameter of this circle right over here, it is 12.5. Now, the other circle-- the circumcircle for triangle ACD, so let's draw ACD. So let's draw a circle that can go through these three points. It looks like it would have to be something like this. It looks like it would have to be a somewhat bigger circle, and that gels with the information that they gave us, the way I drew it. So the circle would look something like that. I don't want to spend too much time trying to draw that circle. But they're telling us that it's-- I should be very careful. They're saying that the radius is 12.5, not the diameter. So let me make it very clear. Actually, let me delete that circle since it's just so messy. And I can delete that 12.5 too. Let me get-- there you go. So the 12.5 is the radius, the radii of the circle. So this first circle around ABD-- around triangle ABD-- this distance right over here is 12.5. This distance over here is also 12.5. Now, let's focus on triangle ACD. Let's focus on that triangle. Its circumcircle will look something like this. Let me draw. There you-- oh, that doesn't look too good. There you go. Something like that. The whole point here isn't trying to draw a circumcircle. But it's a circle that will go through those three points. And it has a radius of 25. So if you had its center, if I were to draw a diameter of it, it is 25. Fair enough. Now, we would need to figure out the area of rhombus ABC and D. Now, if you've been seeing the videos that I've been uploading lately, I've actually been uploading a few of the prerequisites for this. Because there is a formula, and we proved the formula in the geometry and the competition math playlist. We proved the formula that relates the area of a triangle to the radius of its circumcircle. And let me just rewrite the formula right over here. The formula is the radius of a triangle's circumcircle is equal to the product of the triangle's sides. All of that over 4 times the area of the triangle. So let's see if we can use this formula that we have proved in a previous video to figure out areas of triangle ABD-- or express them somehow-- and triangle ACD and then see if we can use that information to figure out the area of the entire rhombus. So let me redraw it a little bit because I think my diagram's gotten kind of messy. So I'll redraw the rhombus. We actually won't even have to draw the circumcircles, or the circumscribed circles, because we know this formula right over here. So this is A, B, C, and D. Now, let's think first about triangle ABD. Actually, let me just draw the diagonals here. BD is one of the diagonals. AC is another one of the diagonals. We know that the diagonals of a rhombus are perpendicular bisectors. We know that that's a right angle. That's a right angle. That's a right angle. That's a right angle. And we know that this length is equal to this length, and we also know that that length is equal to that length. Now, if we knew this green length here or this blue length here, we would be able to figure out the area of the rhombus. Let's label them. Let's call this lowercase a, and let's call this length over here lowercase b. a times b times 1/2 would be the area of this triangle right over there. a times b times 1/2 times 2 would give us this area and that area. Or another way to think about it, this triangle is completely congruent. It has sides a, b, and this side over here. All of these four triangles have those three sides, so all four of these triangles are congruent. So you could take the area of this triangle, multiply it by 4-- you have the area of the rhombus. Let me write this down. The rhombus area is equal to 4 times 1/2ab. 1/2ab gives us just this triangle right over here. 4 times that, so 4 times 1/2ab is 2ab, is going to be the area of the rhombus. If we can somehow figure out a and b, we can figure out the rhombus' area. So let's focus on this first piece of information. Let's focus on triangle AB and D. Now they tell us that its circumradius is 12.5. So let's just use this formula right over here. We get 12.5. It's circumradius is 12.5 is equal to the product of the length of the sides. So what's the length of the sides here? So we have this side right over here, side bd. That's just going to be 2a, right? That's an a plus another a, so it's going to be 2a, times this side right over here. What's this side, which is just one of the sides of the rhombus. Well, this is the hypotenuse of this right triangle right over here, right? This is a right angle. So it's going to be the square root of a squared plus b squared. But all of the sides are going to be that. It's a rhombus. All the sides are the same-- a squared plus b squared. They're all going to have that exact same length. So the product of the sides-- you have 2a-- that's the length of bd-- times the length of ba, which is going to be the square root of a squared plus b squared, times the length of ad, which is the square root of a squared plus b squared. All of that over 4 times the area. 4 times the area of ABD. Now, what's the area of a, b, and d? Well, ABD is just two of these triangles right over here. This guy right over here is 1/2ab. This guy over here is also 1/2ab. So the entire area is going to be two of these guys, so it's just going to be a times b. It gives you the area of both of these triangles. Each of them are 1/2ab. So instead of writing area right here, I could write ab. Now let's see. This simplifies to 12.5, is equal to. Divide the numerator and the denominator by 2. So that becomes a 1. That becomes a 2, divided by a. That becomes a 1. That becomes a 1. Square root of a squared plus b squared times square root of a squared plus b squared is just a squared plus b squared. And the denominator, we're just left with a 2b. So this first piece of information, the circumradius for ABD being 12.5, gives us this equation right over here. Now, let's do the same thing for triangle ACD. It's circumradius is 25. 25 is equal to the length of this side. This is a b. This is also a b, so it's going to be 2b. 2b times the length of this side, which is just the square root of a squared plus b squared, times the length of this side, which is, again, just the square root of a squared plus b squared. All of that over 4 times the area. Now, the area, once again-- it's this triangle, which is 1/2ab, plus this triangle, which is another 1/2ab. You add them together, you just get ab. You just get ab, 2 divided by 2, you get a 1 there. You get a 2 here. Divide by b, get a 1. That just becomes an a. And so you get 25 is equal to the numerator, square root of a squared b squared times itself is just going to be a squared plus b squared over 2a. So that second triangle, its circumradius being 25 gives us this equation right over here. Now we can use both of this. We have two equations with two unknowns. Let's solve for a and b. If we know a and b, we can then go back here and figure out the rhombus's area. So over here, we get-- let's multiply both sides by 2b. We get 25b is equal to a squared plus b squared. Over here, if we multiply both sides by 2a, we get 50a is equal to a squared plus b squared. So 50a is equal to a squared plus b squared. 25b is equal to a squared plus b squared. So 25b must be the same thing as 50a. They're both a squared plus b squared. So we get 25b must be equal to 50a. They're both equal to a squared plus b squared. Now, divide both sides by 25, you get b is equal to 2a. b is equal to-- actually I wanted to do that in the magenta. b is equal to 2a. So we can take this information and then now substitute back into either one of these equations to solve for b, and then we can solve for a. So let's go back into this one. So we get 50a-- actually we'll solve for a first. 50a is equal to a squared plus b squared. Instead of writing b squared, we know b is the same thing as 2a, so let's write 2a squared. So we get 50a is equal to a squared plus 4a squared. Or we get 50a is equal to 5a squared. We could divide both sides by 5a. If we divide this side by 5a, we get 10. And if we divide this side by 5a, we get a. So a is equal to 10. And then we could just substitute back here to figure out b. 2 times a is equal to b. b is equal to 2 times 10, which is equal to 20. So we know a is 10. b is 20. We just have to go right back here to figure out the area of the rhombus. The area of the rhombus is equal to 2 times-- a was 10-- 2 times 10 times 20. This is 20 times 20. This is equal to 400. And we're done. The area of rhombus ABCD is 400.