### Tangent Circles and an Isosceles Triangle

Solution by N. Bowler

The applet presents a solution to the following Sangaku problem: Given a circle S with center O and diameter AC and point B on AC. Form circle G with center P and diameter AB and an isosceles triangle BCE with E on the circle S. Circle W with center Q is inscribed in the *curvilinear* triangle formed by circles S and G and the line BE. Prove that QB is perpendicular to AC.

What if applet does not run? -->

I shall use inversion. I shall denote the inverse of any object by the name of that object followed by a '.

First note that the problem is equivalent to showing that the circle with centre on the perpendicular from B, and tangent to BE and to G is also tangent to S. I shall still call this circle W.

Invert with respect to B, i.e., in a circle centered at B and some radius r, say.

**Step 1:**C'BE' will be similar to EBC and so isosceles. This is becauseBC·BC' = r such that^{2}= BE·BE',BC/BE = BE'/BC'. It follows that alsoBC'/C'E' = CE/BE, but in DBCE we haveCE = BE. **Step 2:**S' will be the circle on A'C' as diameter, and G' the tangent to it at A'.**Step 3:**Let U be the intersection of G' with E'B. Then A'UE' = pi/2 - UBA' = pi/2 - E'BC' = pi/2 - C'E'B = UE'A' so A'UE' is isosceles.**Step 4:**Let V be the centre of W'. Since W' is tangent to both G' and BE', V lies on the bisector of A'UB. SoVUB = A'UV = BVU and so also BVU is isosceles.**Step 5:**Let L be the centre of S', and call the radius of this circle R. Call the radius of W' r.There are right angles everywhere, so we may now indulge ourselves with a Pythagorean orgy:

LV ^{2}= LB ^{2}+ BV^{2}= (R - A'B) ^{2}+ BU^{2}= (R-r) ^{2}+ A'B^{2}+ A'U^{2}= (R-r) ^{2}+ r^{2}+ A'E'^{2}= (R-r) ^{2}+ r^{2}+ A'C'^{2}- E'C'^{2}= (R-r) ^{2}+ r^{2}+ 4R^{2}- BC'^{2}= (R-r) ^{2}+ r^{2}+ 4R^{2}- (2R-r)^{2}= R ^{2}+ 2Rr + r^{2}= (R + r) ^{2}.So that W' is tangent to S' and so W is tangent to S as required.

Another solution makes use of inversion with negative power.

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

Copyright © 1996-2018 Alexander Bogomolny