| * |
| A * |
| O |
| * * * B |
| * * * |
| * * * |
| * * * |
| * D * C * |
What is the minimum area of rectangle WXYZ if all
lengths are whole numbers, as are the areas of the similar triangles, denoted by A, B, C & D?
The guess is that each triangle is similar to a 3,4,5 right triangle.
If QY were 3, QO were 4 and OY 5, then OX would be 3*5/4=15/4, and YX would be 25/4, and ZX would be 125/12.
So if we scale that up by a factor of 12, everything would be integral, with ZX at 125 units and XY at 75 and ZY at 100, making the area of WXYZ = 7500.
I don't have a proof that this is a minimum, however, so the following program checks for solutions with an area of 7500 or less.
FOR h = 1 TO SQR(7500)
FOR w = h + 1 TO 7500 / h
sumSq = h * h + w * w
diag = INT(SQR(sumSq) + .5)
IF diag * diag = sumSq THEN
OY = h * w / diag
OX = h * h / diag
IF OY = INT(OY) AND OX = INT(OX) THEN
OQ = OY * w / diag
QY = OY * h / diag
IF OQ = INT(OQ) AND QY = INT(QY) THEN
PRINT h, w, diag
PRINT OY, OX, OQ, QY
PRINT h * w, h * w / 2, OX * OY / 2, QY * OQ / 2, OQ * (w - QY) / 2
where h and w are the height and width of the large rectangle.
In the above, diag represents the length of ZX, and OY = h * w / diag since
OY/XY = ZY/ZX due to triangle B's similarity to triangle A.
Similarly for OX, as OX/YX = YX/ZX.
Then OQ/OY = ZY/ZX and QY/OY = YX/ZX are used to find OQ and QY.
If everything is integral, the height, width and diagonal are shown, followed on the next line by OY, OX, OQ and QY. As the difference between integers is also an integer, we don't need to verify ZQ and OZ. The last line shows the area: h*w, and the area A, area B, area C and area D. The result is:
ZW=YX=75 WX=ZY=100 ZX=125
OY=60 OX=45 OQ=48 QY=36
Area of WXYZ=7500 Area A=3750 Area B=1350 Area C=864 Area D=1536
So there is nothing smaller than the initial guess.
Posted by Charlie
on 2008-03-18 10:50:27