In Spanish
Francisco
García (with Mathematica)
After placing some approximate points, one can see that the solution is a circle by two vertex.
I draw then the perpendicular bisector
of a side, a point on it, and the circle by two vertex. I drag the center
to make de symetric-lateral of one of its points right.
Looking for some relations with the triangle, I have found that the angle formed by the center and the side is aproximatly the same as the opposite angle.
So, I rotate the line AC about A with angle ABC. The intersection of this line with the perpendicular bisector of AC is the center of the solution circle.
I draw first a right triangle ADE
and found that all points on hypotenuse have a right symetric-lateral.
I thoguht it should be the same for
a general triangle, and P should be on a line.
Then I worked with Pablo, who said me that some points were on a circle, and this infrimed my hypothesis.
We decided to draw a general triangle and to verifiy the hypothesis.
The three first points seemed collinear, and we draw some more points to confirm this hypothesis.
Surprisingly, the points were not on a line, but on a curve.
We have redrawn a smaller triangle
We draw then a circle and adjusted it to fit the points. We were sure
the locus was a circle, but we had to construct it.
We construct the circumcenter C of
ADE, and noted that the perpendicular to AD by C, it passed by the center
of the solution circle.
Then I draw the midpoint of C and
the center of the solution circle and we tried unsuccessfully to relate
it to the central points of the triangle. We decided to draw a circle with
center on the midpoint and passing by a vertex, and we saw it passed by
C
So we draw the triangle ADC and
its circumcenter C´
The intersection point of the perpendicular and the circle of center C' by C is the center of the solution center.
The hole solution is formed by three circles which centers K1, K2 and K3 are the intersection of the perpendicular bisectors of the sides with lines by the opposite vertex and the symedian point. The radii are the distances from K1, K2 and K3 to the vertex.
PROCES OF SOLUTION
1 – I draw a triangle ABC and a Point P
2 – I draw A' and B', reflections of P on AB and AC, and the segment A'B'
3 – Then the perpendicular line to A'B' by B'.
4 – C' the reflection of P on BC
We should see the solution when C' is on the perpendicular to A'B'. In this way i draw some solution points.
I saw that the points were not collinear and thought they could be on a circle.
I draw the perpendicular bissectors of the solution points, who seemed cross on a single point.
Then I made the same in inverse order:
1 - I draw triangle ABC, the perpendicular bisector of AB and a point K on it
2 – The circle of center K and a point on it.
3 – The symetric-lateral of P. then I dragged K and the radius to obtain the symetric-lateral to be right angled.
4 – I saw that when the circle passes by a vertex, an angle of the symetricl-lateral is constant.
5 – I tried to relate the center of the circle with some central points, and I saw that it is on the intersection of the perpendicular bisector with the line through the opposite vertex and the symmedian point.
I draw a triangle a point M, the
symetric-lateral triangle, and I mesure one angle. I drag M to obtain 90°.
I put a point at this position;
I repeat this to some positions
of M and see that the locus seems to be a circle through two vertex of
the reference triangle.
I deduce that the solution is formed by three circles through two vertex of the triangle.
I put points on the perpendicular bisectors of the reference triangle, and draw the ciercles through them and two vertex; I drag the points to make the symetric-lateral right-angled.
When I draw the triangle formed by the three circle centers, its sides seem to pass by the vertex of the reference triangle.
I draw the circles through M and two vertex. Nothing interesting, but I draw also the circumcircle of the reference triangle who will give me the solution....
I take the idea Sol2, find a method to deduce two circles from the first one. When I draw the lines through the centers, I see they are tangent to the circumcircle: that is my solution.
Solution :
Circumcircle of the reference triangle
Tangents to the circle through the
vertex
These tangents cut by pairs on the
solution circle's centers.
these circles pass through the vertex
of the tirangle.