Finding Circle Center via Extremum Points
2020-12-08 08:40
This example demonstrates using GeoGebra's Minimize command to find the optimal solution and determine a circle's center. By constructing polygons, midpoints, and arcs, it illustrates extremum problems in geometric construction. Involving circles, inscribed circles, and similar right triangles, it visually presents solving complex geometric compositions via optimization methods.
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正方形ABCD边长100，
（1）圆K与圆F和圆B都相切，求最大的圆K的半径。
（2）圆O与圆F、圆B和CD都相切，求圆O的半径。""/>
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(1) 在两弧外围时，圆心到B点的距离为r+a，到F点的距离为r+\frac{a}{2}\\
(2) 在两弧相交区域时，圆心到B点的距离为a-r，到F点的距离为\frac{a}{2}-r\\
故两者之差都为 \frac{a}{2},即有\\
abs((x, y) - B) - abs((x, y) - F) = ad / 2 + 0 / (y(A) < y < y(D))\\

为了排除正方形之外的情形，采用了逻辑表达\\
0/(y(A)<y<y(D)), 不要用 (x(A)<x<x(B)) 后者达不到要求""/>
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<layer val="0"/>
<labelMode val="0"/>
<lineStyle thickness="5" type="0" typeHidden="1"/>
<eigenvectors x0="1.0" y0="0.0" z0="1.0" x1="-0.0" y1="1.0" z1="1.0"/>
<matrix A0="1.0" A1="1.0" A2="2500.000008715183" A3="0.0" A4="-20.7106781276036" A5="-45.71067820209985"/>
<eqnStyle style="specific"/>
</element>
<expression label="horizonCurve" exp="horizonCurve: abs((x, y) - B) + abs((x, y) - F) = (3 * ad / 2) + 0 / (y(F) < y < y(D)) + 0 / (x(A) < x < x(B))" />
<element type="implicitpoly" label="horizonCurve">
<show object="true" label="false" ev="4"/>
<condition showObject="horizoncc"/>
<objColor r="255" g="127" b="0" alpha="0.0"/>
<layer val="0"/>
<labelMode val="0"/>
<fixed val="true"/>
<lineStyle thickness="10" type="30" typeHidden="1"/>
<userinput show="false"/></element>
<element type="boolean" label="horizoncc">
<value val="true"/>
<show object="true" label="true"/>
<objColor r="0" g="0" b="0" alpha="0.0"/>
<layer val="0"/>
<labelOffset x="371" y="281"/>
<labelMode val="4"/>
<checkbox fixed="true"/>
<ggbscript onUpdate=""/>
<caption val="圆心轨迹-横式"/>
</element>
</construction>
</geogebra>