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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click on the figure below to enlarge it.
Angela Drei has an excellence solution for this equal incircles problem.See link below for his solution.http://www.cut-the-knot.org/triangle/EqualIncirclesTheorem.shtml
Since I prefer basic geometrical argument, here is my proposal... http://bleaug.free.fr/gogeometry/935.pngFirst, forget about the 3 equal circles, it's a strawman ;-) Let's simplify the picture and keep A, B, C, D, O1 and Q2 (which I renamed Q1 for simplicity in this post). Let's add S as the center of ABC incircle. By construction O1 lies on AS and Q1 lies on CS.Bisectors of ACxBD in D are orthogonal, therefore O1-D-Q1 is a right triangle. Let's build P1 such as O1-P1-Q1-D is a rectangle. It has the following properties:- PS is orthogonal to AC (simple angle calculation)- altitudes relative to AC obey: alt(P1) = alt(O1)+alt(Q1) because of rectangle projection propertyConclusion: alt(O1) and alt(Q1) play a symmetric role. Setting O1' on CS such as alt(O1')=alt(O1) would imply corresponding alt(Q1')=alt(Q1). Therefore in original problem wordings: alt(O1)=alt(O3) implies alt(Q1)=alt(Q2) irrespective of alt(O2).bleaug