Perpendicular bisector passes through a fixed point
by liberator, Sep 1, 2014, 7:50 PM
Problem: Let 
be two circles of different radii, centers 
respectively, which intersect at 
. Let 
be a line through 
meeting at 
respectively. Prove that
a) For any choice of
, the perpendicular bisector of passes through a fixed point 
, and
b)
is an isosceles trapezium.
See: http://www.geogebratube.org/student/mn7v1UIdK
be two circles of different radii, centers 
respectively, which intersect at 
. Let 
be a line through 
meeting at 
respectively. Prove thata) For any choice of
, the perpendicular bisector of passes through a fixed point 
, andb)
is an isosceles trapezium.See: http://www.geogebratube.org/student/mn7v1UIdK
![[asy]
/* Free script by liberator, 1 September 2014 */
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
/* Original Geogebra file created by liberator, geogebratube.org/student/mn7v1UIdK */
unitsize(1.5cm);
pointpen=black;
pen dsp = rgb(0.4,0.6,0.8);
pen rcp = rgb(0.9,0,0);
/* Initialize objects */
pair A = (2.1,2.31);
pair B = (2.1,-0.31);
pair M = (2.9,2.31);
pair N = (3.28,-0.12);
pair O1 = (1,1);
pair O2 = (4,1);
pair P = (0.35,-0.58);
pair Q = (6.21,0.33);
pair X = rotate(180,O1)*B;
pair Y = rotate(180,O2)*B;
/* Draw objects */
draw(B--X--Y--cycle, dsp+linewidth(1));
draw(O1--A--M--O2--cycle, rgb(0,0.6,0)+linewidth(1));
draw(P--Q, dsp);
draw(O1--M, dsp);
draw(P--X, dsp);
draw(Q--Y, dsp);
draw(M--N, dsp);
draw(A--O2, dsp);
draw(circumcircle(A,B,P), rcp);
draw(circumcircle(A,B,Q), rcp);
draw(circumcircle(A,M,O1), rgb(0.8,0,0.8)+dotted+linewidth(1));
/* Draw angle marks */
draw(anglemark(O1,M,O2,7), dsp);
draw(anglemark(O2,B,O1,7), dsp);
draw(anglemark(O1,A,O2,7), dsp);
draw(rightanglemark(Y,Q,N,5), dsp);
draw(rightanglemark(Q,N,M,5), dsp);
draw(rightanglemark(N,P,X,5), dsp);
/* Place dots on and label each point */
Drawing("A", A, 1.5*dir(90));
Drawing("B", B, dir(-95));
Drawing("M", M, dir(90));
Drawing("N", N, dir(-90));
Drawing("O_1", O1, dir(-150));
Drawing("O_2", O2, dir(-30));
Drawing("P", P, dir(-110));
Drawing("Q", Q, dir(0));
Drawing("X", X, dir(150));
Drawing("Y", Y, dir(20));
/* Label circles */
label("$\mathcal{S}_1$", O1, 21*dir(90));
label("$\mathcal{S}_2$", O2, 28*dir(90));
[/asy]](http://latex.artofproblemsolving.com/9/a/3/9a370f5082729b4c8a986997acea6f33fa0ea77c.png)
be antipodal to 
;
,
is the midpoint of PQ, then 
,
are collinear. Considering 
as the medial triangle of 
,
and 
,
is the nine-point circle of This post has been edited 1 time. Last edited by liberator, Sep 4, 2014, 5:34 PM
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