A math challenge problem

You know the total area of the square = 36 (6 squared) so adding up all of the regions inside the square will equal 36.

So:
4x + 4y + z = 36

You also know that the area of a quarter circle is 1/4(pi*r-squared), and the radius of these quarter cirlces is 6 so add up the regions within one quarter circle and they should equal 1/4(pi*36) or 9pi.

For the third equation, you could take the difference between the area of the square and the area of the quarter circle -- 36 - 9pi. Adding up the regions in the space left (2 x's and 1 y) should equal that difference.

Hope that helps -- I haven't worked them all out to know if it's correct, but that does give you three equations.

Yes, but the equations have to be independent of themselves. So you cant just say z= 36- (y+x). Like for the formula for x i have gotten:
x=(36-(9√3)-6π)cm^2
So i also need to find one for y and z in this form.

I think you want to use the approach suggested by FD. You state in your original question that you want three independent equations in X, Y and Z. To me that means that you do as FD suggests. What you presented is not an equation for X; it is the value of X.

Of course, you are the one who sat in class. You could use the approach above, then substitute into various equations to get a simple equation for each variable if you feel that is what is required.

FreeDuck's first two equations are independent. The third is a linear combination of the first two.

Yep, you are correct. The third is just the first minus second. Maybe that equation for x that dimatt came up with will work as the third.

Yes but the question is asking for an independent equation for each value that determines the 'area of each region'. That to me tells me that they are asking for an equation that will get you a value for each region. Which is how i got x=(36-(9√3)-6π)cm^2. I understand what freeduck is saying i just dont think that its the way im am supposed to complete the question.

I don't interpret "independent equations in x, y, and z" to mean each equation must contain only one of the three variables. Freeduck's first two equations are independent equations in x, y, and z.

Where did you get that equation for x?

Hey, i have this EXACT same problem and i'm wondering if you figured it out? i cannot get the x equation easily... if i could it's possible that i may be able to figure it out, i'm currently trying to make points for the centres of the circles and set the equations equal to each other to find the overlap but then i still dont know how to use that to get area...stuck if you know anything that may help please!!

Like dimatt above, you can find the equation for the circle (x^2 + (y-6)^2 = 36) and integrate from 0 to 3 to find the value of half of x. You get

y = 6 - sqrt(36-x^2)

That is a bear to integrate, but when you do and evaluate from 0-3 then multiply by 2, you get what dimatt posted above:

x = 36- 9sqrt(3) - 6pi

From there, you can use FreeDuck's equations to get Y and Z.

engineer you are my hero...that's amazing, i can get it completely if i can get the value of x and then sub it into other equations i have made, however, i don't fully understand the integration of 0-3...am i supposed to make x 0 and y 3? a little further explaination would be amazing. Thanks!

^^ it actually turned out that i was going about it totally the wrong way. You can use 2 of the equations that freeduck mentioned and the third is representative of the equilateral triangle formed within the diagram. You can then take those and put them into a matirces to find x,y,z formulas :p

Cool. I was wondering where the third equation could come from. Totally didn't see that. Thanks for the update, dimatt.

And thank you for putting me on the right track to getting the answer

Anytime. It's an interesting problem.

Hey one last question. I looked at the diagram again and the triangle for the third equation didn't pop out at me. Can you help me see it?

This did take me a while to find

Aha. Thank you so much.