Ken's POTW


Crossing Chords 
        +++++++++
     +++         +++
   +S               ++
  + |\                +
 +  | \                +
 +  |  \               +
+   |   \              _Q
+   |    \  O       _- /+
+   |     \      _-   / +
 +  |      \  _-     / +
 A--X-------C-------Y--B
  + |    _-  \     /  +
   ++ _-      \   / ++
    P+++       \ /++
        ++++++++R
In circle O, a chord AB is drawn, with midpoint C. Through C, two more chords are drawn, PQ and RS. Connecting the endpoints, PS passes through AC at point X, and QR passes through BC at point Y. Show that CX = CY.

First try showing:

  • If RS is a diameter of O, that (CR)(CS) = (CA)(CB)
  • that (CP)(CQ) = (CR)(CS).

Source: The rec.puzzles newsgroup long ago, though I can't find it in the rec.puzzles archive.

I received the following elegant solution from Radu Ionescu. In only a few lines, it completes the proof:

  1. Let M be on SP and OM perpendicular to SP ==>OMXC is a cyclic quadrilateral ==> angle(CMX)=angle(XOC) (1)
  2. Let N be on RQ and ON perpendicular to RQ ==>ONYC is a cyclic quadrilateral ==> angle(CNY)=angle(YOC) (2)
  3. Triangles (SPC) and (QRC) are similar (have the same angles) and OM & ON are the medians ==> angle(CMP)=angle(CNR) (3)
  4. (1),(2)&(3) ==> angle(XOC)=angle(YOC) (4)
  5. OC=OC, (4) & angle(OCX)=angle(OCY)=90 ==> triangle(XOC)=triangle(YOC) ==> CX=CY
I have received a solution from Philippe Fondanaiche, Paris, France, but I haven't had time to work through it. I put it here verbatim temporarily until I can help add some clarification to it - or perhaps someone else could send me some. I believe that Philippe's point M is my point C. 
From Philippe Fondanaiche, Paris, France


I have lost the recollection of the basical theorems of geometry.I have 
found a solution of your problem, I admit, not very elegant:


1) RS is  a diameter of the circle
 Let OM=d and OR=r
The law of sines in the triangle MPR gives:
sin(MPR)/MR = sin(PMR)/PR
therefore sin(MPR)/(d+r) = sin(MPR+MRP)/(2rcos(MRP))
hence (r+d) tg(MRP) = (r-d) tg(MPR)
as MPR = MSY then MX =MY


2) any R on the circle
Let AMB the x-axis and MO the y-axis
Let OR = r = 1(to simplify)
Let OM = d <1
Equations of line PQ y=ax a>0
               of line RS  y=bx b<0
               of circle O x^2 + (y-d)^2 = 1
Let A = sqroot(1-d^2+a^2) and B=sqroot(1-d^2+b^2)
The coordinates of the vertices P,Q,R,S are the following:
P x_p = (da+A)/(1+a^2)    y_p = ax_p
Q x_q = (da-A)/(1+a^2)     y_q = ax_q
R x_r = (db-B)/(1+b^2)      y_r = bx_r
S x_s = (db+B)/(1+b^2)    y_s = bx_s


MX= - MY is equivalent to x_p.x_s/(y_p - y_s) = - x_q.x_r/(y_q - y_r)
So we hane to demonstrate the following relation (R):
a.x_p.x_q.(x_r+x_s) = b.x_r.x_s.(x_p+x_q)
Whereas x_p.x_q = (1-d^2)/(1+a^2)        x_p + x_q = 2da/(1+a^2)
      and   x_r.x_s = (1-d^2)/(1+ b^2)       x_r  + x_s = 2db/(1+b^2)
the relation (R) is easily proved.
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