Equilic Quadrilateral I: What is this about?
A Mathematical Droodle


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?

Discussion

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

Copyright © 1996-2017 Alexander Bogomolny

The applet may suggest the following statement due to Jack Garfunkel [Honsberger]:

If an equilateral triangle DCQ is drawn on the side DC (away from AB) of an equilic quadrilateral ABCD (AD = BC), then the triangle ABQ is also equilateral.

(To remind, a quadrilateral ABCD is called equilic if it has a pair of equal opposite sides inclined 60° to each other. In the problem, AD = BC.)


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?

Triangles ADQ and BCQ are equal. Indeed, by construction, AD = BC and DQ = CQ. Now, on the surface, the proof proceeds smoothly. We have to consider two possible cases as shown the diagrams below:

We denote the angles in the quadrilateral by a single letter as in the corresponding vertex so that, by construction, ∠A + ∠B = 120°. Therefore, ∠C + ∠D = 240°. Taking up the left diagram:

∠ADQ= ∠D + 60°
 = 240° - ∠C + 60°
 = 300° - ∠C
 = 360° - ∠C - 60°
 = ∠BCQ.

It follows that ΔADQ = ΔBCQ. In particular, ∠AQD = ∠BQC. Thus, by removing and adding equal angles from and to ∠CQD we still get an angle of 60°. Therefore, ∠AQB = 60°. Since also AQ = BQ (as the corresponding sides in equal triangles), ΔABQ is indeed equilateral.

In the case of the right diagram above we similarly arrive at the same conclusion. Q.E.D.

A Flaw in the Proof

Now, this proof contains a subtle flaw. For the convenience sake, to simplify the process of drawing an equilic quadrilateral, I used the fact that the two sides AD and BC make a 60° angle so that their intersection S is bound to a 240° arc above AB. The statement we have attempted to prove is equivalent to stating that Q lies at the middle of that arc. The argument was based on the fact that their were just two possibilities: either C or D is inside ΔABQ whereas the other one is outside. However, if Q does not lie on the arc, two other possibilities have to be considered: either both C and D are inside or outside ΔABQ. It appears that we based our proof on an assumption equivalent to the statement to be proved.

A valid proof

References

  1. R. Honsberger, Mathematical Gems III, MAA, 1985, pp. 32-35

Equilic Quadrilateral

  1. Equilic Quadrilateral I
  2. Equilic Quadrilateral II
  3. Equilateral Triangles on Segments of Equilic Quadrilateral
  4. Equilic Quadrilateral I, A Variation
  5. Equilateral Triangles on Diagonals of Antiequilic Quadrilateral

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

Copyright © 1996-2017 Alexander Bogomolny

A Valid Proof

A rotation around Q through 60° maps D to C and A to, say, A' such that CA' and AD form a 60° angle. But the same is true of AD and CB. Therefore, A' = B. It follows that that rotation maps the whole ΔADQ on top of ΔBCQ. The triangle are then equal, their angles at Q are also equal, and we can proceed as before to conclude that ∠AQB = 60°.

Besides establishing the fact it was intended to prove, the proof has a virtue of suggesting a generalization.

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

Copyright © 1996-2017 Alexander Bogomolny

 62687581

Search by google: