Two Circles in an Angle: What is this about?
A Mathematical Droodle

A simple sangaku deals with two circles inside an angle. The circles are related in two ways which imply each other.


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Solution

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Copyright © 1996-2006 Alexander Bogomolny

The applet presents two circles inscribed in an angle. One of the circles is incident with the center of the other. The latter is tangent to the chord in the former that joins the points of tangency with the angle.


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?

More accurately, let there be two circles - (O) centered at O, (Q) centered at Q - inscribed into an angle with vertex C. (Obviously, OQC is the bisector of the angle.) Let MN be a chord in (O) perpendicular to OC. Then any two of the three conditions

  1. M and N are the points of tangency of (O) with the angle,
  2. (Q) is tangent to MN,
  3. Q lies on (O)

imply the third. There are two cases depending on whether O is father from C than Q or nearer. I shall treat only the former.

1 and 2 imply 3

Let Q' and V be the points where OC meets (O), V farthest from C. The tangent to (O) at V meets CM and CN in U and W, respectively such that (O) is inscribed into ΔCUW. O then is the incenter of that triangle and is located at the intersection of its internal angle bisectors. It follows that

(1) ∠CUO = ∠OUV.

In addition, quadrilateral OMUV is cyclic (because both angles OMU and OVU are right) which shows that

(2) ∠MUV = 180° - ∠OUV = ∠MOQ'.

The latter is the central angle in (O) subtended by chord MQ' and thus is twice the angle formed by the chord and a tangent at one of its ends:

(3) ∠MOQ' = 2·∠CMQ'.

From (1)-(3),

(4) ∠CUO = 2·∠CMQ',

UO||MQ'. Since also UW||MN, this means that MQ' is the bisector of ∠CMN in ΔCMN. Q' is then the incenter of that triangle. Since (Q) is tangent to all three side lines of the latter, so is Q. We conclude that Q' = Q.

1 and 3 imply 2

Let (Q') be the incircle of ΔCMN. As we just seen this assumption implies that Q' lies on (O). Since the same holds for Q and both are on the same side from O as C, they coincide: Q = Q'. (Q) is the same as (Q') and is therefore tangent to MN.

2 and 3 imply 1

Let M' and N' be the points where (O) touches CU and CV, respectively. Let O(Q') be the incircle of ΔCM'N'. Then, as we already observed, Q' is bound to lie on (O); it thus coincides with Q: Q = Q'.

References

  1. H. Fukagawa, D. Pedoe, Japanese Temple Geometry Problems, The Charles Babbage Research Center, Winnipeg, 1989

    Write to:

    Charles Babbage Research Center
    P.O. Box 272, St. Norbert Postal Station
    Winnipeg, MB
    Canada R3V 1L6

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

Copyright © 1996-2006 Alexander Bogomolny

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