# Square Inscribed in Triangle II:

What is this about?

A Mathematical Droodle

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Copyright © 1996-2018 Alexander Bogomolny

Here's a problem:

Construct on AB an associated Vecten square AC_{a}C_{b}B. Next, join C to both C_{a} and C_{b} and consider the points of intersection of CC_{a} and CC_{b} with AB. (In the applet they are denoted P and Q.)

I claim that PQ is the base of the sought square. To complete the square find points P' and Q' on AC and BC, respectively, such that PP'⊥ AB and QQ'⊥ AB.

There are several ways to view the configuration. An important thing to note is that there are three pair of similar triangles: ACC_{a} and P'CP, C_{a}CC_{b} and PCQ, and BCC_{b} and Q'CQ. The segments CP, CC_{a}, as well as, the segments CQ, CC_{b} are shared between two of those pairs. Therefore, the relevant side lengths ratios are all equal, which means that PP'Q'Q is indeed a square homothetic to C_{a}ABC_{b} from C.

Under this homothety, the centers of the two squares correspond to each other and, therefore, are collinear with C. As a consequence, they are also collinear with one of the Vecten points. The problem has in general two solutions depending on whether we start with an outwardly or inwardly constructed Vecten square.

(Homothety can be also used to produce an entirely different solution.)

### References

- D. M. Bradley,
__On Shutting up and Listening__,*MAA Focus*, (Jan 2009, v 29, n 1), pp. 20-21 - F. van Lamoen,
__Inscribed Squares__,*Forum Geometricorum*, Volume 4 (2004) 207-214

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

Copyright © 1996-2018 Alexander Bogomolny