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Euclidean geometry is a mathematical system attributed to the Alexandrian Greek mathematician Euclid, which he described in his textbook on geometry: the Elements. Euclid's method consists in assuming a small set of intuitively appealing axioms, and deducing many other propositions (theorems) from these.

4 votes

ten concurrent lines

It is possible to give a syhtnetic proof using the following result: The perpendiculars dropped from the midpoints of a cyclic quadrilateral to the opposite sides are concurrent. Furthermore, thei …
Beni Bogosel's user avatar
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4 votes

Minimum distance between two arbitrary circles in space?

I didn't manage to solve the problem (edit: in the meantime an answer was posted which says a precise formula using radicals cannot be found), but I can post a proof that the line joining the points w …
Beni Bogosel's user avatar
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4 votes
3 answers
3k views

How to solve geometry problems using involutions

Some geometry problems ( like this and this ) have short solutions if we use involutions. What references are there for solving geometry problems using involutions? I am particularly interested in the …
Beni Bogosel's user avatar
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1 vote

Strengthened version of Isoperimetric inequality with n-polygon

The keyword you are looking for is "quantitative isoperimetric inequalities". The case of polygons was solved in the following paper: https://arxiv.org/abs/1402.4460 The "quantitative" term in the pap …
Beni Bogosel's user avatar
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1 vote

Intersection point of three circles

Switching the roles of $ABC$ and $A'B'C'$, consider the circles passing through the vertices of a triangle $A,B,C$, midpoints of the opposite sides $A',B',C'$ and the circumcenter $O$. It is straightf …
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0 votes

A generalization of Napoleon's theorem

Notice that triangles $ACD, AEB, FCB$ are similar. Working out the ratios of the sides and the angles one can see that: Triangles $AI_1I_3$ and $ACE$ are similar. Rotating $I_1I_3$ around $A$ with an …
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0 votes

Three circles intersecting at one point

It is straightforward to see that $A',B',C'$ are reflections of the circumcenter $O$ with respect to $BC, CA,AB$. Therefore, the center of $(AOA')$ is just the intersection of the mediatrix of $OA$ wi …
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