Tangent Lines and Circles in Convex Quadrilateral

Let A0, A1, A2, and A3 list in order the vertices of a convex quadrilateral Q. Treat all indices as integers modulo 4. Let Lk denote the line through Ak and Ak+1, and let Ck be the circle tangent to Lk-1, Lk, and Lk+1 outside Q. Let Mk be the line through the points where Ck is tangent to Lk-1 and to Lk+1. Let Ek be the intersection of Mk with Mk+1. Prove that E1E3 bisects A1A3.

 

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

(The problem was proposed by Chu Cheng, Yan An Middle School, Shanghai, China. The solution is by Marius Stefan, Los Angeles, CA. American Mathematical Monthly, v 114 (December 2007), 926-927).

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny

For distinct points A, B,C, write d(A, BC) for the distance from point A to the line through B and C. Denote by Tlk the point where the line Lk touches the circle Cl. Let M be the midpoint of the line segment A1A3. Let α be half the angle between L1 and L3, lines that enclose C0 and C2. Let β be half the angle between L0 and L2, the lines enclosing C1 and C3.

 

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

With these definitions, we have

 d(M, E3E0)= (d(A1, E3E0) + d(A3, E3E0)) / 2
  = (A1T01 cos(α) + A3T03 cos(α)) / 2
  = (A1T01 + A3T03) cos(α) / 2
  = (A1T00 + A3A0 + A0T00) cos(α) / 2
  = (A1A0 + A3A0) cos(α) / 2

Similarly,

 d(M, E2E3)= (d(A3, E2E3) + d(A2, E2E3)) / 2
  = (A3T32 cos(β) + A1T30 cos(β)) / 2
  = (A3T32 + A1T30) cos(β) / 2
  = (A3T33 + A1A0 + A0T33) cos(β) / 2
  = (A1A0 + A3A0) cos(β) / 2

It follows that

 d(M, E3E0) / d(M, E2E3)= cos(α) / cos(β).

The lines E3E0 and E1E2 are parallel because they are both perpendicular to the bisector of 2α. The lines E0E1 and E2E3 are also parallel, since they are perpendicular to the bisector of 2β. Hence E0E1E2E3 is a parallelogram. The distance d0 between E3E0 and E1E2 is

 d0= T01T21 cos(α) = T03T23 cos(α)
  = (T01T21 + T03T23) cos(α)) / 2
  = [(A1T00 + A1A2 + A2T22) + (A0T00 + A0A3 + A3T22)] cos(α) / 2
  = (A0A1 + A1A2 + A2A3 + A3A0) cos(α) / 2.

Similarly, the distance d1 between E0E1 and E2E3 is

 d1= (A0A1 + A1A2 + A2A3 + A3A0) cos(β) / 2.

Therefore,

 d0 / d1= cos(α) / cos(β)= d(M, E3E0) / d(M, E2E3),

which implies that M ∈ E1E3.

The configuration has additional properties:

  • The centers of the four circles C0, ..., C3 are concyclic.

  • If s is the semiperimeter of the quadrilateral A0A1A2A3 then

      T01T21 = T03T23 = T10T30 = T12T32 = s.

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny

 62075750

Search by google: