Wednesday, April 8, 2015

Simplifying a Proof

I found this problem online yesterday (briefly) via a post on the google+ k-12 education.  In the figure below

  • $\bigtriangleup$ BCX has an area of 5 
  • $\bigtriangleup$ AXY has an area of 3 
  • $\bigtriangleup$ CDY has an area of 4
What is the area of $\bigtriangleup$ CXY in the middle?  So I have a reasonable solution for the problem but I'm still trying to see if I can find a way to eliminate the quadratic equation  from the process.

My solution

First add in some perpendicular lines and divide the big rectangle into 4 quadrants.

Conveniently we know most of the quadrants areas from the  triangles.

  • $\Box$ AXOY has an area of 6 since its two $\bigtriangleup$ AXY. 
  • $\Box$ BCXZ has an area of 10 since its two $\bigtriangleup$ BCX. 
  • $\Box$ CDYW has an area of 8 since its two $\bigtriangleup$ CDY.
Notice the last two boxes overlap in the upper right corner and I split that out into x. Now we have 2 sets of proportional rectangles going either horizontal or vertical. I'll just choose the vertical ones but you can go either way.

$\frac{\Box BWOX}{\Box AXOY} = \frac{\Box WCZO}{\Box OZDY}$


$\frac{10-x}{6} = \frac{x}{8-x}$ which once simplified produces the following quadratic:

$x^2 - 24x  + 80 = 0$ or $(x - 4)(x - 20) = 0$  We choose the first solution where x = 4 since the second produces negative areas in the other boxes.   That means the entire rectangle has a size of 

6 + 6 + 4 + 4 = 20 and 

$\bigtriangleup$ CXY =  Area of the big rectangle minus the 3 triangles or  20 - ( 3 + 4 + 5) = 8.


The presence of the quadratic means I couldn't give this one to the kids but it seems like there should be  a refinement that lets me remove it. I thought about it a bit yesterday and couldn't come up with any simplifications. Jump in if you have any ideas.

1 comment:

  1. You can certainly do this by guess and check.

    Take advantage of the fact the aspect ratio is under-specified, so you can just pick them. Make the rectangle 5 units wide. That tells you the height of the 5 triangle is 2. Then you can pick the 3 triangles is 2 units high and 3 wide. The 4 triangle works out at 4 units high and 2 wide. This is a guess & check solution - you are essentially guessing the answer on the quadratic.