# Around the Incircle

In a curious twist several geometric constructs emerge (or merge) in a single dynamic diagram:

- H. Eves' problem of getting 6 concyclic points by rotating the side lines of a triangle around its incircle.
- Bui Quang Tuan's closed chain of 6 touching circles.
- Adams' circle is centered at the incenter and passes through the points of intersection of Gergonne's lines with the side lines of the given triangle.
- Conway's circle passes through the six points obtained by producing the sides of ΔABC a units beyond A, b units beyond B, and c units beyond C.
- A construction of shapes of constant width.

Do you see how?

What if applet does not run? |

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

Copyright © 1996-2018 Alexander Bogomolny

Let's consider the problems one at a time.

### Howard Eves' Quickie

Starting at a point P on the side BC of DABC, mark Q on AB with

### Solution

Let D, E, F be the points of tangency of the incircle of DABC with the sides BC, AC, and AB, respectively. Then, say,

(1) | FQ = ER = DP' = FQ' = ER' = DP. |

What if applet does not run? |

The process indeed is getting closed. (This is a generalization of the process wherewith a line is rotated around a triangle.)

Now, points D, E, F are equidistant from the center I of the incircle. The radius-vectors ID, IE, IF are perpendicular to the sides of DABC. Additionally, D, E, F serve as the midpoints of equal segments PP', RR', QQ'. Which implies that the six points P, P', Q, Q', R, R' are equidistant from I.

What if applet does not run? |

Note that if DP = 0, to start with, then after three rotations DP' = 0. Which means that point D is fixed by successive rotations around B, A and C.

(It is worth noting that the construction can be expanded to n-gons, with

### Adams' Circle

What if applet does not run? |

This is a particular case of the above situation where P is chosen so that the lines PR', RQ, and P'Q' are concurrent. Wonderfully, they meet at the Gergonne point of DABC.

### Conway's Circle

Conway's circle are obtained by producing the sides AB and AC beyond A by a (= BC), etc. Concentric circles are also obtained if all the extension are modified by a fixed amount, say, x. ("Beyond" should be understood here metaphorically as "away from".)

What if applet does not run? |

### Proof

We know that

CE = CD = (a + b - c)/2,

BD = BF = (a - b + c)/2,

AE = EF = (-a + b + c)/2.

If AC is produced beyond C by (c+x) to, say, C', and beyond A by (a+x) to, say, A', then

EA' | = (a + x) + (-a + b + c)/2 |

= (a + b + c)/2 + x | |

= (a + b - c)/2 + (c + x) | |

= EC'. |

The situation is exactly as before for the points R, E, R'. Conway's construction leads to exactly the circles in Eves' problem.

### Curves of Constant Width

Move point P away so that all six points D, E, F, D', E', F' are outside DABC.

What if applet does not run? |

You should get something like

Eves' arcs form a convex shape with an interesting property: In whatever direction it is squeezed between two parallel lines the distance between the lines remains the same. Figures with this property are known as shapes (or curves) of constant width. Shapes of constant width can be constructed starting with equilateral but not necessarily equiangular stars. Eves' arcs suggest a different approach that generalizes the *crossed-lines method* discussed in [Gardner, p. 217, Honsberger, p. 158, Rademacher, p. 167]. The crossed-lines method no longer requires that we start with an equilateral triangle.

Note that three segments PP', QQ', RR' all have the same length

### References

- M. Gardner,
*The Unexpected Hanging and Other Mathematical Diversions*, The University of Chicago Press, 1991 - R. Honsberger,
*Ingenuity in Mathematics*, MAA, New Math Library, 1970 - H. Rademacher and O.Toeplitz,
*The Enjoyment of Mathematics*, Dover Publications, 1990. - C. W. Trigg,
*Mathematical Quickies*, Dover, 1985, #181 (*Am Math Monthly*,**50**(June, 1943), 391)

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

Copyright © 1996-2018 Alexander Bogomolny

68360325