[color=#999999]This activity belongs to the [i]GeoGebra book[/i] [url=https://www.geogebra.org/m/sw2cat9w]GeoGebra Principia[/url].[/color][br][br][br]The following is one of the images generated using the dynamic color scanner that I particularly like. The scanner has tremendous versatility, it can create a heat map for virtually any situation [[url=https://www.geogebra.org/m/sw2cat9w#material/er8nf4qt]3[/url], [url=https://www.geogebra.org/m/sw2cat9w#material/er8nf4qt]19[/url], [url=https://www.geogebra.org/m/sw2cat9w#material/er8nf4qt]26[/url], [url=https://www.geogebra.org/m/sw2cat9w#material/er8nf4qt]27[/url], [url=https://www.geogebra.org/m/sw2cat9w#material/er8nf4qt]31[/url]].[br][br][img]https://www.geogebra.org/resource/aE2jc9Rh/QwFAk0chdCWt2FaD/material-aE2jc9Rh.png[/img][br] [br]In this case, the first isogonic point I[sub]1[/sub] is visualized by intersecting the loci that see each pair of sides of the triangle under the same angle.[br][list][*][color=#808080]Note: I[sub]1[/sub] coincides with the Fermat point when the triangle's largest angle is not greater than 120º; otherwise, the Fermat point coincides with the vertex corresponding to that angle. It can be calculated directly as the triangle's center X(13) [/color][url=https://faculty.evansville.edu/ck6/encyclopedia/ETC.html][img]https://www.geogebra.org/resource/scjbyz2p/0tuzuVw455vxurEw/material-scjbyz2p.png[/img][/url] [color=#808080]:  [/color][/*][/list][color=#808080][center]I[sub]1[/sub] = TriangleCenter(O, A, B, 13).[/center][/color]However, constructing the scanner takes some effort. But we can use CAS to define not only distances but also angles. If someone still thinks that using the expression [b]XA[/b] instead of sqrt((x − x(A))² + (y − y(A))²) doesn't save much work, perhaps they'll reconsider now when they can use the expression [b]OXA[/b], defined in the CAS view as:[br]  [br] [color=#CC3300]OXA(x,y):= Angle(O, X, A)[/color][br]  [br]instead of its algebraic equivalent (with O=(a,b) and A=(c,d)):[br][list][*]cos[sup]–1[/sup]((a c – a x + b d – b y – c x – d y + x² + y²) sqrt(a² c² – 2a² c x + a² d² – 2a² d y + a² x² + a² y² – 2a c² x + 4a c x² – 2a d² x + 4a d x y – 2a x³ – 2a x y² + b² c² – 2b² c x + b² d² – 2b² d y + b² x² + b² y² – 2b c² y + 4b c x y – 2b d² y + 4b d y² – 2b x² y – 2b y³ + c² x² + c² y² – 2c x³ – 2c x y² + d² x² + d² y² – 2d x² y – 2d y³ + x⁴ + 2x² y² + y⁴) / (a² c² – 2a² c x + a² d² – 2a² d y + a² x² + a² y² – 2a c² x + 4a c x² – 2a d² x + 4a d x y – 2a x³ – 2a x y² + b² c² – 2b² c x + b² d² – 2b² d y + b² x² + b² y² – 2b c² y + 4b c x y – 2b d² y + 4b d y² – 2b x² y – 2b y³ + c² x² + c² y² – 2c x³ – 2c x y² + d² x² + d² y² – 2d x² y – 2d y³ + x⁴ + 2x² y² + y⁴))[br][/*][/list]Naturally, this expression is merely a development deduced from the dot product of two vectors:[br] [br] vO(x,y):= Vector(X, O)[br] vA(x,y):= Vector(X, A)[br] OXA(x,y):= acos((vO vA)/(|vO|*|vA|))[br] [br]The significant advantage, besides convenience, is that the Angle command allows us to explore angular relationships without needing to know even the basic operations with vectors, like the dot product.[br][br]Here, we can see, for example, the locus corresponding to points that form an angle (in radians) equal to the distance to point A with segment OA:[br][br][color=#CC3300] OXA – XA = 0[/color][br][list][*][color=#808080]Note: Circles whose arcs span an angle OXA equivalent to XA radians have centers: [/color][color=#808080] [/color][/*][/list][color=#808080][center](O + A)/2 ± PerpendicularVector(OA)/(2 tan(XA))[/center][/color]And those that see segments OA and OB from the same angle:[br][br][color=#CC3300] OXA – OXB = 0[/color][br] [br]Finally, the intersection of this latter locus with the one corresponding to the equation OXA – AXB = 0 is the sought-after Fermat point.

[color=#999999]Author of the construction of GeoGebra: [url=https://www.geogebra.org/u/rafael]Rafael Losada[/url].[/color]