extremely hard math bonus problem

[earlz]

Hello! I really am needing bonus points for my math class.. and this is one of the bonus problems..

I am not saying I want the answer with no work.. I would just like some kind of hint as to how to do this... I've tried everything but all I end up with are equations to solve length that are just 1=1 type things...

this is the text:
"A Trisected Sqare"

A square is divided into three pieces of equal area by two parallel cuts, as shown. The distance between the parallel lines is 6 inches. what is the area of the square in square inches?

attached is my remake of the diagram...

(oops, in my diagram I put x... replace X with 6)

[JohnnyTheDon]

What level math course is this? Because I have a feeling my solution is too simple.

Since the only constraints on the lines were in the square, parallel to each other, and 6 inches apart, I just placed them like so (please excuse my horrible attempt as ASCII drawing):

|````````````````|
|----------------|
|----------------|
|_________|

As shown in the drawing, I made the parallel lines perpendicular to the sides of the square that they intersect.
So, the center becomes a rectangle, with a height of 6. The problem also said it divided the square into 3 pieces of equal area.
That means the other remaining parts of the square (also rectangles) should also have a height of 6, because their width is constant. That means each side of the square equals 3*6 or 18 inches.

I might be wrong. I'll check in more in depth later.

[stephenj]

At what level do you need to prove your solution?

I think I have a solution, but I haven't proved 2 properties that I would if it were my problem.

I'll tell you that the trick of this problem is breaking it down into sub-triangles, rebuilding them into sub-squares, then finally a big square.

Again though, this method may be incorrect, and you'll need to determine that on your own. But I'm not doing the tedious part for someone else's homework (I'm done with school).

[michael1234]

Hi,
I THINK I solved it, this would seem to be well above my level I get taught in school though, not that that makes a difference
Anyway, here's a bit of my working...

Area of square = x^2
Area of one section = (x^2)/3
Area of a any triangle = (w^2)/2 (make w triangle height and width, x already means something else)
Area of shape A (or B) = (x^2)/3 = (w^2)/2 (where shape A is the top left triangle)
(x^2)/3 = (w^2)/2
(x^2) = (w^2) * 3/2
X = W * sqrt(3/2)
X/sqrt(3/2) = W

Distance from top left corner of triangle A to centre of hypotenuse of triangle A = sqrt(2*(w^2))/2
Distance from top left corner to bottom left corner of square = sqrt(2*(x^2)) = (sqrt(2*(w^2))* 2/2) + 6
sqrt(2*(x^2)) - (sqrt(2*(w^2))* 2/2) = 6

sqrt(2*(x^2)) - (sqrt(2*((x/sqrt(3/2))^2))* 2/2) = 6
simplify and solve x.

[JohnnyTheDon]

as shown

Whoops. Ignore my answer, I completely ignored a very important part of the problem *facepalm*

[JohnnyTheDon]

Area of a any triangle = (w^2)/2 (make w triangle height and width, x already means something else)

How do you know the sides of the triangle are equal to each other? Actually, its impossible, unless one side of the square is longer than the other (making it not a square).

[Combuster]

From the description alone the problem is indeterminate - what michael and johnny solved are two 'trivial' cases of the problem, and each has a different solution

So, do each of the cuts intersect a corner, if so there is a triangle similarity you can use to build a system of equations.

[earlz]

Wow! I am suprised by the amount of responses... yes, the cuts intersect 1 corner each(but I don't think they bisect it)

The 3 rectangle thing makes a lot of sense... I might try to prove that that is the answer... the only thing I have to have is work to show my answer is correct and that I understand the solution... (not a formal proof or anything)

[JohnnyTheDon]

My previous answer cannot be the answer, if it is supposed to be done "as shown". As shown, the lines do not bisect, and the whole 3 rectangles thing isn't possible. I tried to solve the problem as shown, but I ended up with a system of equations with 1 equation but 2 variables. I just got that one from setting the area of the triangles equal to the area of the parallelogram in the center.

[earlz]

oh.. I see that now.. lol...
I keep getting the same 2 variable problem too...

I hate this problem...

[stephenj]

With the corner intersect, one of my pre-conditions is wrong (I probably should have checked the drawing closer).

Anyway... There are essentially 2 lengths in this problem l and l-6.

The area of one triangle is: A = b*h/2, A = l*(l-6)/2

But we know that the area of the three objects are not only equal to themselves, but if multiplied together have a product of l^2.

3A = l^2

From there, I did the algebra to get l = 18. But I leave that to you.

[JohnnyTheDon]

Sorry about my OCD, but this problem was seriously bothering me

I attatched two pictures of my solution (taken on my crappy old webcam XD )

This solution works under the constaints set by the pictures.

I'll write out my equations and try to explain them.

At = Area of Triangle
Ap = Area of Parallelogram
Atr = Area of Trapezoid
x = Length on one side of Square
y = Find it on the picture.
At = Ap
Atr = 2At = 2Ap = At+Ap
(x^2)/3 = At = Ap

At = (x(x-y))/2
Ap = 6 sqrt(x^2+(x-y)^2 )
Atr = (x(x+y))/2

Actually, the trapezoid stuff was completely useless. In fact, I only used the third identity (x^2)/3 = At = Ap.

You better get that extra credit

[JohnnyTheDon]

The area of one triangle is: A = b*h/2, A = l*(l-6)/2

The distance of 6 is only the length of line segments that are perpendicular to the parallel lines. The line segment you used is not, so its length is not 6.

[stephenj]

Damn, I should have looked at the drawing clearer. Twice in one day.

Our proofs work the same, just prove a different thing (mine assumes your y=6, then simplifies).

hckr83, you may want to state the assumption that Johnny mentioned.

[Combuster]

The cuts can not possibly bisect (bonus points from me if you prove that)