More trig examples
2003 AIME II Problem 14 Trigonometry and geometry to find the area of an equilateral (but not regular) hexagon
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- Let A=(0,0) and B=(b,2) be points on the coordinate plane.
- Let ABCDEF be a convex Equilateral hexagon.
- So convex means it's not concave,a concave hexagon would look like this
- so that's two sides, three, four, five, six,
- this would be a concave hexagon,
- this is going to be popped out
- and all the sides are going to be equal,
- so it's equilateral hexagon.
- They're not telling us it's a regular hexagon,
- so we don't know that all of the angles are going to be the same
- but all the sides will be the same.
- then they show us a bunch of sides that are parallel to each other
- and that the y-coordinates of its vertices are
- distinct elements of the set 0,2,4,6,8,10.
- The area of the hexagon can be written in the form m square root of n
- where m and n are positive integers
- and n is not divisible by the square of any prime.
- That's just a fancy way of saying that
- we've simplified this radical as much as possible.
- Find m plus n.
- So really the first part let's just make sure
- we can visualize this hexagon.
- So let me draw, we know one point one vertex for sure zero zero,
- so let me draw my x-axis, that is my x-axis right over there
- and then my y axis, my y axis will look like that.
- y axis, we know that the vertex A sits at the point (0,0),
- that is the vertex A.
- Now we know that all the vertices have y-coordinates that
- are either zero two four six eight ten,
- and they're distinct members of the set
- which means no two of the vertices share the same y-coordinates.
- They are not going to be on the same horizontal line.
- So let me draw these words on lines.
- With the x-axis is a zero
- then you have Y is equal to two,
- then you have four, and you have six, and then you have eight,
- and then you have ten up here.
- Now B we already know. So first we've already used up the zero for A,
- A is already using up the zero.
- B uses up the two they tell us, that the y-coordinate of B is two.
- The y-coordinate of B is two, so we use that as well
- let me see if I can draw a B over here.
- Let me see if I can draw a B sits on the horizontal some place,
- and we - the hexagon has side length s
- we don't know what that length is but they're all the same,
- so let's call this s, it can help me think about it,
- now that I know that this is equilateral hexagon,
- all the sides are going to be the same length,
- and so we're going to go out here,
- we're going to go out here the (b,2),
- we don't know what b is but that is our vertex B.
- Now F is the other vertex that is connected to A.
- F cannot sit on this horizontal, cannot sit on y is equal to two,
- it can't sit on y is equal to six
- because this distance would be super far,
- clearly much further than this distance over here
- where actually you could have that,
- but then you wouldn't be able to draw a really a convex hexagon.
- so y,the next vertex is just going to have to sit on this horizontal,
- so it's going to be s away,
- so it's going to be s away, maybe it will be something like that,
- something like that, so let me draw it,
- so that is the next vertex, that is vertex F Right?
- Because we're going ABCDEF and then back to A.
- Fair enough, now what about vertex C.
- well vertex C can't be on the four horizontal
- so it gonna have to be on the six horizontal.
- So vertex C is going to have to be some place like that,
- Some place like that, that's vertex C.
- Once again that length is s, this length is s,
- now what about vertex E?
- can't be on the six horizontal now already taken up by vertex C
- so the four and the six are already taken up,
- so has to be at the at the eight horizontal.
- And so this is length s.
- And we also know that we're going back to the origin now.
- so this is vertex E right here,
- we know that we're going back to the origin,
- not to the origin we are going back to the same x value,
- this is going to be on the y intercept
- and the reason why we know that is this is length s,
- and this is length s and they both have to,
- both of these diagonals travel the same,
- they travel the same vertical distance, this base is four,
- this base is four,so you can kind of view this is two right triangles,
- both of them have base four and hypotenuse s,
- and so they share this side right over here
- so they both - this one goes out to the left that distance
- and this one is going to come back that distance.
- Now by the same logic over here,
- this guy's going to have to come back
- so he's - we can now use the ten coordinate,
- the ten y horizontal or the y coordinate of ten
- that's the only one we haven't used yet for D,
- and since we came out,
- when we had a diagonal of length s,
- traveling four up,
- this time same logic, we had a diagonal length s,
- it traveled up four over here and moved out this distance
- then we go back at the other direction and traveling up four,
- and you're gonna go back in the same direction
- so this is going to be directly on top of B.
- so the coordinate for D is actually going to be b comma ten,
- the y coordinate here is ten.
- And there we have our hexagon,
- we're done drawing our actual hexagon.
- and all this parallel line information they told us is that
- AB is parallel to DE,
- so AB is parallel to DE and this is kind of obvious here.
- BC is parallel to EF,
- BC is parallel to EF.
- and they say CD is parallel to FA, so CD
- is parallel to FA and the way we drew it looks pretty clear
- that is the case.
- Now we define the area,
- we defined the area of this hexagon
- and it seems like a good starting point,
- would be to figure out what s is
- and to figure out what s is,
- it's really going to be a function of
- how much we've inclined this thing. So let's -
- let's draw, you can see that this is an equiangular hexagon
- that this is kind of skewed, is kind of, we distorted a little bit
- but all the sides are of the same length.
- So let's just call this theta,
- let's call that angle right over there theta,
- and they tell us that the angle FAB is 120 degrees,
- FAB is a hundred and twenty degrees
- that is a hundred and twenty degrees.
- So this angle over here on the left,
- this angle over here on the left is going to be 180 minus one twenty,
- minus theta. So 180 minus one twenty is sixty
- So this single over here sixty minus theta.
- The reason I did that is because we have some information,
- we know that we traveled up four over here,
- and we know that we traveled up two over here
- and maybe we could use that information to solve for s
- because s is the hypotenuse of both of these right triangles
- that I just constructed. Let me draw them.
- So this right triangle right over here I could draw like this,
- I could draw like this. So I have s,
- I have theta and I have 2,that's this right triangle right over here.
- This right triangle looks like this,
- it looks like this, this is,
- this angle is sixty minus theta,
- and this height over here is four
- So let's see what we can do to solve for s.
- This triangle on the left or is on the right over here,
- this triangle says that
- If we take the sine of theta,
- the sine of theta is equal to the opposite over the hypotenuse,
- is equal to two over s, this triangle tells us that the sine,
- remember this hypotenuse over here is also s,
- the sine of sixty minus theta,
- the sine of sixty minus theta is equal to four over s.
- And if you want set these equal to each other
- we could multiply this guy two on both sides.
- You can see two sine of theta is equal to four over s,
- Sine of sixty minus theta is also equal to four over s,
- we can set them equal to each other.
- So we have two sine of theta, two,
- we have two sine two sine of theta is
- equal to sine of sixty minus theta.
- and then we could use some of our trig identities
- we know the sine of a minus b is the same thing,
- the sine of a minus b this is equal to
- the sine of a times the cosine of b,
- or I should say theta in this case,
- So sine of sixty cos theta, minus,
- this is just a standard trig identity
- So it's a difference, sum and difference identity,
- minus cosine of sixty,
- minus cosine of sixty times the sine of theta,
- times the sine of theta and all this is equal to two sine of theta.
- two sine of theta. well sine of sixty degrees
- this is the square root of three over two,
- square root of three over two.
- Cosine of sixty degrees is one-half, is one-half.
- So we could add one half sine theta to both sides of this
- and what are we going to get.
- So we are going to add one-half sine theta,
- then this guy's going to go away
- and then you add one-half sine of theta to two sine of theta
- which is really four half sine of theta,
- so that's going to be five halves sine of theta,
- so let me add one-half sine of theta, that's five -
- five halves sine of theta
- is equal to square root of three over two cosine of theta,
- square root of three over two cosine of theta.
- right? I added one half sine of theta to both sides of this
- to get this I can multiply both sides by two just to simplify it.
- So I get five sine of theta
- five sine of theta is equal to the square root of three cosine of theta
- Now I want to use that identity sine squared theta plus
- cosine squared theta is equal to one.
- So let me just square both sides that
- will also help us with this radical
- So we'll get twenty five sine squared of theta is
- equal to square three cosine squared of theta,
- Instead of writing cosine squared of theta,
- let's just write that's one minus sine squared of theta. Right?
- cosine squared theta is one minus sine squared of theta
- Just squared both sides.
- So let me just write what I just did,
- I just squared both sides and so we get twenty five sine squared theta
- is equal to three minus three sine squared theta
- and we can add three sine squared theta to both sides
- we get twenty eight sine squared theta is equal to three,
- or that the sine squared theta,home stretch,to three over twenty eight
- Or we could even write that sine of theta,
- sine of theta is equal to the square root of three over twenty eight.
- So it's equal to the square root of three over twenty eight
- now we can simplify that twenty eight is four times seven
- so we can take it out,
- but that's good enough for now that'll,
- that may be you know we'll simplify it
- later if we have to sometimes it's easier to deal with.
- So let's see, over here we have we have the sine of theta
- now we can relate that actually to s over here we know that,
- we know that before I messed with this thing,
- we know that the sine of theta is equal to two over s,
- or that s over two is equal to one over sine of theta,
- or that s is equal to two over the sine of theta.
- Well we know what sine of theta is,
- it is square root of three over twenty eight.
- so s is equal to two divided by sine of theta,
- that's like multiplying by the inverse of sine of theta,
- so that's two times the square root of twenty eight over three,
- twenty eight over three so that is we figured out our s,
- two times this thing over here.
- Now given that we know s let's see how we can figure out the area.
- Well what immediately pops out
- is that we have this triangle over here,
- that has height or I should say maybe it's base
- if you view it sideways, its base is eight,
- its base is eight,
- and this distance right over here we should be able to figure out,
- we should be able to figure out using the Pythagorean theorem,
- because we know that this distance right over here is four,
- We know that this distance is four.
- We know that this distance, the hypotenuse is s,
- so we could call this the height of it right over here
- we could say that h squared,
- h squared plus four squared plus sixteen is equal to
- the hypotenuse squared, equal to s squared.
- S squared - s is this thing over here so if we want to square s,
- it becomes four times twenty-eight,
- four times twenty eight over three,
- and we'd subtract sixteen from both sides so h is equal to
- four times twenty-eight over three minus
- if I want to write sixteen over three
- or if I want to write sixteen something over three
- it's going to be minus forty eight,
- minus forty eight over three
- let's see I don't wanna multiply,
- I don't want to have to multiply four times twenty-eight,
- so you can write forty-eight as four times twelve,
- so this numerator is going to be
- four times twenty-eight minus twelve over -
- remember that's h squared I should say,
- H squared is going to be
- four times twenty-eight minus twelve over three
- which is equal to four times sixteen over three,
- which is equal to sixty-four over three, that's h squared
- so h is going to be the square root of that which is eight
- over the square root of three,
- so h right over here is eight over the square root of three
- so if I want to find the area of this whole thing over here.
- well first let's find the area of the small thing right over here
- that's just going to be h times four,
- so it's going to be,
- well,I could do it either way,
- but let's just say this is h times four, times one-half,
- so it's going to be two times so the area of this triangle,
- Let me do it in blue right here,this triangle's area is going to be h,
- which is eight, over the square root of three, times four,
- times four times one-half,
- so this guy right over here is this going to be
- two times eight over the square root of three,
- or that's going to be sixteen over the square root of three
- so this guy over here is sixteen over the square root of three.
- So that is sixteen over the square root of three.
- Now we have a bunch that, we have
- this guy and this guy is going to have
- the exact same area right over there,
- and then you have this guy
- who's going to have the exact same area again,
- same exact logic, same exact logic, same base, same height,
- they're actually congruent,
- so you have four of these triangles,
- so you are gonna multiply by four
- if you want the area of this
- or this area I've already shaded in, four times this
- where it's sixty four over
- where it's sixty four over the square root of three.
- now the only area we have left to figure out
- is the area of this parallelogram,
- the area of the parallelogram in the middle.
- Now we know the base of the parallelogram,
- the base of this parallelogram is eight,
- the base of the parallelogram is eight.
- We just have to figure out its height.
- We just have to figure out its height
- and once again we can use the Pythagorean theorem.
- So I'll call this, I don't know, we've already used h,
- I'll use h again but this is,
- you'd have to remember this is a different height over here,
- this base over here is of length two,
- two and it's hard to read now,
- so we can now write,
- we can now write that h squared plus four plus two squared is equal to
- is equal to s squared
- now we already figured out what s squared was in the past,
- it's four times twenty-eight over three,
- four times twenty eight over three.
- Let's subtract four from both sides.
- Let's subtract four over there so minus twelve over three.
- And now let's see twelve is the same thing as four times three,
- so this is, this is equal to four times twenty-eight,
- four times twenty-eight minus three,
- so that's four times twenty-five over three,
- which is equal to one hundred over three, that's h squared,
- so this h is going to be equal to the square root of this,
- which is ten over the square root of three.
- This is ten over the square root of three,
- so this distance right over here is ten over the square root of three.
- So we want the area of this parallelogram is going to be
- that height times the base of the parallelogram,
- so the area of parallelogram is going to be
- eight times ten square root of three, or eighty -
- eighty square roots of three.
- Eighty, oh no, let me be very careful.
- Let me be very careful.
- This was ten over -
- this is ten over the square root of three is this height.
- So the whole area this parallelogram is
- eight times ten over the square root of three
- so it's eighty over the square root of three.
- So the entire area now, if we add everything together,
- If we add everything together
- we have sixty-four square root of three for these four triangles
- plus eighty over square root of three, so let's add it together.
- So we have eighty over square root of three for the parallelogram
- plus the sixty-four over square root of three for the triangle parts,
- and this is equal to one hundred forty-four
- over the square root of three,
- we can rationalize the denominator,
- so times the square root of three over the square root of three,
- and in the denominator now we're going to get a
- three one forty-four over three is going to be,
- what, that's forty-eight, right?
- three times forty's a hundred twenty,
- three times eight's twenty-four
- so it's going to be forty-eight square roots of three
- for the area of our entire our entire hexagon
- and so we have in the form forty eight square roots of three,
- so if you want to find m plus n,
- it's forty-eight plus three which is fifty-one.
- That was a tiring problem I just find my brain started to
- fry near the end of that I had trouble keeping track of things,
- anyway, hopefully you enjoyed that.
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