Question

这部分是一个数学问题，部分是一个java实现问题。

我正在尝试使用Java Graphics中的fillPolygon(Polygon p)方法绘制三角形。

有一个类Triangle.java，它扩展了Polygon并采用了两个参数：Point location和int height。 height参数的目的是轻松调整三角形的高度，然后通过求解30,60,90三角形来计算边长。

public class Triangle extends Polygon {

    private final static int numberOfSides = 3;

    // Class creates a polygon for use in fillPolygon() method call in
    // Piece.java
    public Triangle(Point p, int height) {

        // Pass in the hieght of the desired triangle and the side lengths will
        // be calculated by solving a 30, 60, 90 triangle
        super(new int[] { p.x,
                (p.x + ((int) ((2 * height) / Math.sqrt(3.0)) / 2)),
                p.x + (int) ((2 * height) / Math.sqrt(3.0)) }, new int[] {
                (p.y + (int) ((2 * height) / Math.sqrt(3.0))), p.y,
                (p.y + (int) ((2 * height) / Math.sqrt(3.0))) }, numberOfSides);
    }

我需要做的是创建一个同心三角形的方法，无论三角形的大小或位置如何。显然，内三角形的椎体必须与外三角形的所有顶点等距（如果外三角形是等边的，则特别容易）。三角形之间的差距是无关紧要的，因为我希望可以作为参数传递。

就像我说这主要是一个数学问题，虽然我还没有找到一个与同心三角形一般等距的方程。

感谢您的高级帮助。

编辑：我解决了这个问题。我不得不重做我接近三角形的方法（使用Path2D.Double而不是Polygon以获得更好的准确性），但它确实有效。我就这样做了：

根据@ajb的建议，我计算了等边三角形的Circumcenter（与其他质心相同），然后计算三角形的坐标，用于Path2D.Double：

public class Triangle extends Path2D.Double {

private double[] xCoordinates, yCoordinates, innerXCoordinates,
        innerYCoordinates;
private Point location;
private double height, circumcenterRadius, sideLength;

public Triangle(Point p, double h) {

    if (p != null && h > 0) {
        this.location = p;
        this.height = h;

        this.circumcenterRadius = (2.0 / 3.0) * (this.height);
        this.sideLength = (2 * this.height)/Math.sqrt(3.0);
    }

    // Set values for the x coordiantes using location and height
    this.xCoordinates = new double[] { p.x,
            p.x + sideLength/ 2,
            p.x + sideLength };

    // Set values for the x coordinates using location and height
    this.yCoordinates = new double[] { p.y + this.height, p.y,
            p.y + this.height };
}

然后使用正确的坐标，迭代然后将它们添加到路径：

 public void constructPath() {
    // Sets the path to the first coordinates of the triangle (bottom left).
    this.moveTo(this.xCoordinates[0], this.yCoordinates[0]);

    // Iterates through the coordinates and adds them to the path
    for (int i = 1; i < this.xCoordinates.length; i++) {
        this.lineTo(this.xCoordinates[i], this.yCoordinates[i]);
    }

    // Closes the path and compeletes the Polygona
    this.closePath();
}

然后我创建了一个方法，它将在Circumcenter Radii中获得所需的差异（以确定缩小外接三角形的程度）。使用它，我计算了x和y偏移量，以便重新定位内三角形，使其正确地刻在外三角形内部

public Path2D constructInscribedTriangle(double radiusDifference) {

    // Calculate the new circumcenterRadius of the inscribed Triangle
    double innerCircumcenterRadius = (this.circumcenterRadius - radiusDifference);
    double innerHeight = (3.0 / 2.0) * innerCircumcenterRadius;
    double innerSideLength = (2 * innerHeight)/Math.sqrt(3.0);

    double xOffset = (radiusDifference/2) * (Math.sqrt(3.0)); 

    // Recalculate coordintes for inner Triangle
    this.innerXCoordinates = new double[] {
            (this.location.x + xOffset),
            (this.location.x + xOffset)
                    + innerSideLength/ 2,
            (this.location.x + xOffset)
                    + innerSideLength };

    this.innerYCoordinates = new double[] {
            (this.location.y + radiusDifference) + innerHeight,
            (this.location.y + radiusDifference),
            (this.location.y + radiusDifference) + innerHeight };

    Path2D.Double inscribedTriangle = new Path2D.Double();

    inscribedTriangle.moveTo(this.innerXCoordinates[0],
            this.innerYCoordinates[0]);
    for (int h = 1; h < this.innerXCoordinates.length; h++) {
        inscribedTriangle.lineTo(this.innerXCoordinates[h],
                this.innerYCoordinates[h]);
    }

    inscribedTriangle.closePath();

    return inscribedTriangle;
}

在Java图形中绘制同心三角形

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