电脑
Mathematica12
绘制一个经过ABC三点的圆。这是一个随机的几何图形:RandomInstance[ GeometricScene[{a, b, c}, {Point[{a, b, c}], CircleThrough[{a, b, c}]}]]字母用小写字母表示。
加上线段BC的中点:RandomInstance[ GeometricScene[{a, b, c, d}, {Point[{a, b, c, d}], d == Midpoint[{b, c}], CircleThrough[{a, b, c}]}]]
绘制三角形ABC:RandomInstance[ GeometricScene[{a, b, c, d}, {Point[{a, b, c, d}], Triangle[{a, b, c}], d == Midpoint[{b, c}], CircleThrough[{a, b, c}]}]]
限定∠ABC的度数是36°:RandomInstance[ GeometricScene[{a, b, c, d}, {Point[{a, b, c, d}], Triangle[{a, b, c}], d == Midpoint[{b, c}], CircleThrough[{a, b, c}], PlanarAngle[{a, b, c}] == 36 Degree}]]
再强加一个限制条件:RandomInstance[ GeometricScene[{a, b, c, d}, {Point[{a, b, c, d}], Triangle[{a, b, c}], d == Midpoint[{b, c}], CircleThrough[{a, b, c}], PlanarAngle[{a, b, c}] == 36 Degree, PlanarAngle[{a, d, c}] == 99 Degree}]]此时的三角形ABC的形状已经确定了。
如果条件自相矛盾,作图的程序就会报错,什么陷入死循环。