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## Perpendicular bisectors

Current time:0:00Total duration:13:13

# 2003 AIME II problem 7

## Video transcript

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.