A rectangle is divided into four small rectangles as shown. The perimeters of three of the four small rectangles are 24, 32, and 42, respectively. The remaining small rectangle has the shortest perimeter among the four. What is the perimeter of remaining rectangle?

2a + 2b  =  24     →     2b  =  24 - 2a

2c + 2d  =  32     →     2d  =  32 - 2c

2a + 2c  =  42

perimeter of remaining rectangle

  =  2b + 2d                      Substitute  24 - 2a  in for  2b ,   and substitute  32 - 2c  in for  2d .

  =  24 - 2a + 32 - 2c

  =  24 + 32 - 2a - 2c

  =  24 + 32 -1(2a + 2c)         Substitute  42  in for  2a + 2c .

  =  24 + 32 -1(42)

  =  24 + 32 - 42

  =  14

Id solved #1 and got 15

All triangles are under the same height

If AWX  = 30......and AYZ  = 20....we have that

(1/2)WX * height   =  30

(1/2)YZ * height  = 20

Which implies that    WX =  (3/2)YZ

Which implies that

WX + XY = 9       and

YZ +  XY = 7        so

(3/2)YZ + XY  = 9

YZ        + XY = 7     subtract these two equations

(1/2)YZ  =  2      ⇒  YZ  =  4       ⇒ XY = 3

And the height is   (1/2)YZ  * height = 20    ⇒  (1/2)4 * height = 20  ⇒ height = 10

So.....the area of AXY  =  (1/2) XY * height  = (1/2) * 3 * 10   =  15 units^2

Very nice, hectictar.....this one stumped me, too!!!!

I happened to notice something else interesting about this last problem.....the sums of the two intermediate perimeters = the sum of the two "extreme" perimeters......

Probably  just a coincidence  ???

But maybe not  ???

Note .....perimeter of  AEJF = 18......perimeter of JHDG  = 8

Perimeter of ECHJ  = 10  ......perimeter of FJGB = 16

So.....the sum of the extreme perimeters  = 26

And the sum of the intermediate perimeters  = 26

Does anyone know if this is a "theorem"  ????

If not.....I posit it as "The CPHill  Conjecture of Mean and Extreme Perimeters".....LOL!!!!!!

Actually......here's the reason for this.........

Let's suppose that  we have a rectangle divided into 4 smaller rectangles of equal  sides of  "m" length  and "n" height.....

And the exact center of the rectangle will be at  (n, m).....

Now...for instance...let us shift  this point down x units and to the right by  y  units

Drawing two lines parallel to the sides of our original rectangle through this point  will produce four new rectangles.......

So.......the perimeter of the original top left rectangle will increase by 2x + 2y  units

And the perimeter of the original bottom right rectangle will decrease by 2x + 2y units

And the height of the original rectangle on the top right will increase by y units but the width will decrease by x units.......so......the new perimeter is 2(m+n) - 2x + 2y units

Finally, the height of the original rectangle on the bottom left will decrease by y units but the width will increase by x units..........so......the new perimeter is 2(m+n) + 2x - 2y units

So.....the sum  of the perimeters of the top left and bottom right rectangles will be :

2(m +n) + 2x + 2y + 2(m + n) -2x - 2y  =   4 (m + n)

And.....the sum  of the perimeters of the top right and bottom left rectangles will be :

2(m +n) - 2x + 2y + 2(m + n) +2x - 2y  =   4 (m + n)

So........the sum of the perimeters of the diagonally situated rectangles will be exactly the same !!!