Calculate largest rectangle in a rotated rectangle

Question

I\'m trying to find the best way to calculate the biggest (in area) rectangle which can be contained inside a rotated rectangle.

Some pictures should help (I hope) i

Answer 1

Here is the easiest way to do this... :)

Step 1
//Before Rotation

int originalWidth = 640;
int originalHeight = 480;

Step 2
//After Rotation
int newWidth = 701;  //int newWidth = 654;  //int newWidth = 513;
int newHeight = 564; //int newHeight = 757; //int newHeight = 664;

Step 3
//Difference in height and width
int widthDiff ;
int heightDiff;
int ASPECT_RATIO = originalWidth/originalHeight; //Double check the Aspect Ratio

if (newHeight > newWidth) {

    int ratioDiff = newHeight - newWidth;
    if (newWidth < Constant.camWidth) {
        widthDiff = (int) Math.floor(newWidth / ASPECT_RATIO);
        heightDiff = (int) Math.floor((originalHeight - (newHeight - originalHeight)) / ASPECT_RATIO);
    }
    else {
        widthDiff = (int) Math.floor((originalWidth - (newWidth - originalWidth) - ratioDiff) / ASPECT_RATIO);
        heightDiff = originalHeight - (newHeight - originalHeight);
    }

} else {
    widthDiff = originalWidth - (originalWidth);
    heightDiff = originalHeight - (newHeight - originalHeight);
}

Step 4
//Calculation
int targetRectanleWidth = originalWidth - widthDiff;
int targetRectanleHeight = originalHeight - heightDiff;

Step 5
int centerPointX = newWidth/2;
int centerPointY = newHeight/2;

Step 6
int x1 = centerPointX - (targetRectanleWidth / 2); 
int y1 = centerPointY - (targetRectanleHeight / 2);
int x2 = centerPointX + (targetRectanleWidth / 2);
int y2 = centerPointY + (targetRectanleHeight / 2);

Step 7
x1 = (x1 < 0 ? 0 : x1);
y1 = (y1 < 0 ? 0 : y1);