You people are so lazy!

[A$$Bandit]

Oh come on... I can't believe none of you geniuses can solve this problem...

"Find the area x"

Come on people, are you THAT lazy?
I tried and I couldn't do it...

[MrPolite]

impossible need more info

[snakeeyes1000]

Just a guess.. 16cm^2

[A$$Bandit]

It can't be impossible or I wouldn't have got set it... and I need an answer because I have to hand it in on Wednesday!!! Help!

[DavidHooper]

Is it equilateral?
Is it to scale?

[Destined Soul]

I know this problem is solvable, I remember doing one just like it in Euclidian Geometry. However, I don't have the book with me (I know a problem very similar to this was actually quite easy to solve if you knew a couple of important theorums.)

However, I don't have the book with me - it's at home. I keep forgetting to look this up when I go home. I'll send myself an email to check into this.

[EDIT]
Yes, I am bit lazy, but not in a bad way.

Oh, and yes, you do have enough information. This is more intuitive for me, but I feel that I'm right. Given the shape of the combined triangles, I think there's only one solution that could result in a larger triangle. (Area = 5+8+10+x)
[/EDIT]

Destined

[A$$Bandit]

Thanks dude

Even if, for any reason, you are unable or really can't be arsed to try this out, could I have what you think would be the relevant formulae anyway?

And we don't know if it's equilateral or not so we have to assume it's not, and it certainly isn't to scale, so drawing a scale diagram won't help. I think it's purely algebraic.

[Destined Soul]

Originally posted by A$$Bandit
Even if, for any reason, you are unable or really can't be arsed to try this out, could I have what you think would be the relevant formulae anyway?

Heh.. if I knew that, I'd have figured it out for ya.

In Euclid. Geometry, they often use the relationships between triangles and circles, and how the radii of these circles are related.

The second option is about lines intersecting with a corner point of a triangle, as well as intersections of where THREE of these lines (you have two drawn) all intersect at the same point. There's something special about this case. (Some of them assume splitting corner by half, which won't work in this case.)

The third thing that comes to mind, and has been what I've been attempting to use, is the relation between the triangle and the formula for its area (1/2 * base * height) While there's no direct way of applying this, I've been debating with myself as to if an answer can be found with the help from some linear algebra.

I'm curious, what level of homework question is this? Usually, the teacher should have gone over something that you can use as a tool against this. If not, it SHOULD BE coming up in the course pretty soon. If I were you, I'd look there if you already haven't. It's interesting that you did find the area of that part of x earlier. I think you might have been on to something. (possibly.) I haven't thought about how that answer might help, though.. it may or may not help anymore than what you've done.

Destined

[A$$Bandit]

Well if this helps, I am in the 3rd top pupil in the top maths set in the 7th best school in the UK for my age group, so the homework is quite demanding. In the UK we take 2 sets of public exams, one at 16 (just dun 'em), and one at 18. Before the beginning of the next school year, my new teacher thought he'd set us some preparation work so I guess that's what this is. I can give you the formulae about triangles that I'm expected to know:

1) The area of a triange using two sides and the angle between them.
2) The length of a side or the size of an internal angle, given any combination of sides and angles (so long as it's physically possible to do).

I know a lot more than I'm supposed to so you don't need to explain anything which I haven't mentioned here

This one has really got me stuck!

[bugzpodder]

do you know hero (heron)'s formula for an area of the triangle?

btw as i said i believe there are more than one value for x. if you are looking for any value such that the the other 3 areas are as follows on the graph, i could probably give you an answer.

[bugzpodder]

yea ok, i got it. a little bit too easy.

lets assume the right side of the triangle are in the ratio of 1

then since the area of the portions of the triangle are portional to its base:

(x+5)/(10+8)=1/p
so:

(x+5)/18=1/p

p=18/(x+5)

p=3 yields:

3x+15=18
x=1

p=2 yields: x=4
p=1 yields x=13
...
so on.

similarly on the other side,

1/q=(x+8)/15
q=15/(x+8)

when q=1,x=7
...
thus all you need to find is a value for x that generate both positive p and q. note this MAY generate some cases where its not a valid triangle.

[snakeeyes1000]

if that is really in actual centimeters, measure it with a ruler. Then you can get the value for X very easily. If the drawing is not to scale, there is no way I'm aware of to solve this.

[A$$Bandit]

Mr Podder, I'm not sure exactly what you're trying to do, could you draw a diagram or something? I got stuck understanding what you meant by "lets assume the right side of the triangle are in the ratio of 1". Ratio of what to what?

[bugzpodder]

u can call me Bugz

hmm, i don't have softwares to draw the diagram. what i mean by

I got stuck understanding what you meant by "lets assume the right side of the triangle are in the ratio of 1 to p". Ratio of what to what?

is

ok, first in ur diagram, the 3 sides of the triangle are: left side, right side, and bottom side. what i mean by that is the length of the top segment of the right side to the length of the bottom segment of the right side is 1. is that better? and the area of the area of two triangles formed using these two segments as its base are in the ratio of 1 also.

[A$$Bandit]

Aren't you assuming they are in that ratio? Do you know for sure?

[bugzpodder]

well they are in some ratio right? 1/p can be anything! p doesn't have to be an integer!

[A$$Bandit]

Oh sorry yeah, I misunderstood in the first place, I see exactly what you mean... I thought you meant the two lengths were in a ratio of 1:1, which would be a little on the fishy side...

Thanks

[Destined Soul]

Originally posted by bugzpodder
lets assume the right side of the triangle are in the ratio of 1 : p

then since the area of the portions of the triangle are portional to its base:

(x+5)/(10+8)=1/p

Heh.. I was going to ask where this came from, but I just figured it out. Here's a breakdown for those who didn't get it like I first did.

area = 1/2 * base * height of the triangle, thus:

if a = length of one of the bases (of triangle with area = x+5), and b = length of the base of the other triange (with area = 10+8), we know that these have the same heights. So far we have:

area_x+5 = 1/2 * a * height
area₁₀₊₈ = 1/2 * b * height

Even though we don't know the formulas, dividing these two equations results in:

(x+5)/(10+8) = a/b. (BugZ uses a = 1, b = p)

Thus x = 18a/b - 5

Similarly for the other side, (c = base of x+8, d = base of 5+10):

x = 15c/d - 8

I'm not sure how well this will work for finding the actual value.

Destined

[NotLKH]

So,
We've got till wednesday?

I think the method 'll be figured by then.


I've been trying to use coordinate geometry, but the equations got hairy.

So, I'm now trying to use equations based on ratios of the lengths of sides, with 1 perpindicular bisector defined, whose origin is (0, 0), Height is H1, the base of the triangle is defined as {m1h1,0}, {M2h1,0}, and so on...{The intersection point would be {0, h1}


Mind you, I hate triangles. Given a search on Euler, You might find what your looking for.

[bugzpodder]

No this is all of it. you guys are thinking that there is one single value for x, but what i am saying that there are multiple values for x that would satisfy the given conditions. if you choose any value for x such that p and q are positive real numbers, then that value is legit. similarly, if you choose any value for p such that x>0 and q>0, then that value for x is also legit. thats what i meant in the first place by multiple values.

btw the proof Destined Soul gave of my method is correct.

[Destined Soul]

Okay, I sifted through my math book (at 12.30am ) and I couldn't find anything directly applicable to this. I don't feel like getting into the nitty gritty stuff, so I'm gonna say that I couldn't finish it tonight. I have to work tomorrow. (Last day before going back to school! )

One thing I do believe is that there's only one unique solution, but I can't really derive a proper proof for this. BugZ says that there's more than one - I'd like to see two solutions and still see if it's a triangle. Either way, we still need one complete solution, regardless if it's unique or not.

I'm working on a proof of this, which might result in an answer, but I haven't worked it out, as I know if there are any holes in it, BugZ (and others) will find it.

Destined

[bugzpodder]

we've established that when q=1,x=7.

since p=18/(x+5)

p=18/12=3/2

so x=7 is a solution.

so how do you construct such a triangle?

i think this would work: first draw any triangle with an area of 30 (5+8+10+7), then bisect the sides to their specific ratio of 1:q and 1:p (left and right) using diagonals and you'll get your desired triangle.

i'll try this on geometry sketchpad.

[bugzpodder]

a little further look-in reveals that my solution implies (x,b,c,d are areas of triangle, and we are looking for x):

x+b=12
c+d=18
x+c=15
b+d=15

the solutions to this system is not unique since while x=7,b=5,c=8,d=10 is a solution, so is x=1,b=11,c=14,d=4 is also a solution to the system of equations. that means if we guarentee area of one of the triangles, we guarentee it to be a solution. unfortunately, following my contruction methods, i can't guarentee the area of any triangle, however, i am confident that x=7 will result in a valid triangle (hey the question didn't ask for the sides, just ask for the value of x!).

btw i believe that for any value of x>0, there is a valid triangle such that fits the given conditions. maybe its hard to believe but i certainly think that its true.

[A$$Bandit]

I have been informed by a teacher that there is only one solution to the problem. Just one, so you'll all have to do a bit more thinking!

[bugzpodder]

no way... unless x is an integer or somethin'

[A$$Bandit]

Oh it's pretty much guaranteed that x is an integer.

[A$$Bandit]

Latest development from me is that if you consider the other side of the whole big triangle to be in ratio 1(top part):q(bottom part), then q = 5p/(6+p)...

[bugzpodder]

agreed but that doesn't help much. i'll try to find a x such that all 3 sides are valid. i think i'll stick with x=7 and i'll let u know if i get anything

[A$$Bandit]

You're right. It can only help if we find ANOTHER two equations containing only p's q's and x...

This problem is turning out to be a royal pain in the arse...

[bugzpodder]

Originally posted by A$$Bandit
You're right. It can only help if we find ANOTHER two equations containing only p's q's and x...

if such an equation does not exist, it simiply implies that there are multiple solutions for x since any value for x will work (provided that gives you a valid value for p and q)

btw you only need 1 more equation with p,q, and/or x and no other new variables

[A$$Bandit]

Yeah, sorry, I actually did mean to write "one" not "TWO", but for some reason the hands thought otherwise...

I do think you're right, it is 7, BUT we need some solid proof!

[bugzpodder]

well, i just subbed a random value in. as i said x could be anything. i just solved x such that q=1. really if you think about it, construct ANY triangle with an area of 10. extend the sides and connect them to the vertices such that area of one section is 5 and the other is 8. connect these two new line segements so they meet and the enclosed area is x. x could be literally anything. i am sure of this!

[A$$Bandit]

Well I tell you this: we will all find out from the Mathematics God on Wednesday or Thursday, because he'll be explaining this one in class. I think you're right, you've pretty much confirmed what I thought to be the case all along. The maths teacher I asked must have been wrong. Thanks everyone, especially Bugz, because that "7" will be enough to get me the marks!

Continue at your for your own pleasure only!

[bugzpodder]

make sure you post here what his solution after wednesday. i am very interested in seeing how to solve this problem by the "mathematical god". ty

[NotLKH]

I think its 22, but I could have made a mistake.

Care to check this out?

[bugzpodder]

how did u generate the jpeg file?

[NotLKH]

MSWord, Screen Capture, And Paint.

[bugzpodder]

how did you type it in the first place?

[NotLKH]

I typed the whole thing out in MSWord.

After spending 3 days scratching on paper!
I just figured this method out in the last 2 hours.

So Far, I think its good!

[bugzpodder]

did u use any plug-ins in word?