Centroid of six equilateral triangle lie on a circle
Let ABC be a triangle, F is the first or secon Fermat point. Construct
FA1A2; FB1B2;FC1C2 are three equilateral triangles
FAcAb; FCbCa; FBaBc are three equilateral triangles
A3,B3,C3 are the centroids of three equilateral triangles FA1A2; FB1B2;FC1C2
A4,B4,C4 are the centroids of three equilateral triangles FAcAb; FCbCa; FBaBc
Problem 1: Then A3,C4,B3,A4,C3,B4 lie on a circle
Problem 2: A3B3C3 is an equilateral triangle
Problem 3: AA3;BB3;CC3 at are concurrent (figure 1)
Figure 1
Problem 4: A5,B5,C5 respectively are midpoints of B1C2; C1A2; A1B2 then A5B5C5 is an
equilateral triangle (figure 2)
Problem 5: Perpendicular bisector of B1C2; C1A2; A1B2 concurrent at X15 (or X16) (Figure 2)
Figure 2
Problem 6: Denote A6,B6,C6 respectively are midpoint of AcAb, BaBc, CaCb. Prove that A6B6C6
is a equilateral triangle(Figure 3)
Problem 7: AA6,BB6, CC6 are concurrent (Figure 3)
Figure 3
Problem 8: A7,B7,C7 respectively are circumcenter of of three circle (XB1C2); (XC1A2);(XA1B2)
then A7B7C7 is an equilateral triangle(X is X15 or X16 respectively with F is the first or secon
Fermart point) (Figure 4)
Figure 4
Problem 9: Denote a point A’c lie on AB such that \angleC5XA’c=\angleBCA-300, circle with
center X and radii XA’c meets three sides of the triangle at A’b; C’b, C’a,B’a (show in figure).
Then XA’cA’b; XC’bC’a; XB’aB’c are equal equilateral triangle. Figure 5
(I think this is logo of http://www.xtec.cat/~qcastell/ttw/ttweng/portada.html)
I also mention this problem here:
http://groups.yahoo.com/neo/groups/AdvancedPlaneGeometry/conversations/messages/253
Figure 5
Problem 10: Circumcenter of three circle FA’cB’c; FB’aC’a; FC’bA’b is an equilateral triangle
Figure 6
Figure 6
Dao Thanh Oai
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