Construct a triangle if you are given the positions of A,O,K
- The problem on the subject was posted by Frans Gremmen to geometry-college.
(Frans wants the vertex to be C, but I prefer A :-)
Nice problem! I myself tried it for ~1/2 hour but without success.
Then I had had a look in my geometry textbooks. In a Greek one, namely
N. Kiskyras: Theorems and Problems of Geometry. Book IV - V. p. 137,
I located the problem but as asterisked: *1049. The asterisk means that
the problem is difficult. However, the author gives a hint:
See the figure of the Theorem paragraph 312, The problem of parag. 315, etc.
Aha...... The theorem of par312 is on the First Lemoine Circle, while
The Problem of par315 is on its radius and the distance OK.
Now I can solve it!
N M | P
The 1st Lemoine circle is centered at L, midpoint of KO.
Let N,P be its intersections with AB, AC, resp. nearest to A.
NP is perpendicular to LM, where M is the midpoint of AK.
(==> LM =// AO/2 = R/2)
Now, OK^2 = AO^2 - 3(----------------)^2 (1)
a^2 + b^2 + c^2
NP = --------------- (2)
a^2 + b^2 + c^2
(1) & (2) ==> NP = sqrt((AO^2 - OK^2)/3).
The above analysis leads to this construction:
Find the midpoints L, M of OK,AK, resp.
Draw a line l perpendicular to LM at M.
Draw the circle with center L and radius sqrt((AO^2 - OK^2)/3),
intersecting the line l at N, P.
Draw the lines AN, AP intersecting the circle (O,OA) [=circumcircle of ABC]
at B,C. The triangle ABC is the triangle in question.
From (1) we see that OK is necessarily < than AO (= circumradius).