# Chaitanya's Random Pages

## January 30, 2013

### Similar triangles fitting together in two ways

Filed under: mathematics — ckrao @ 11:58 am

In my previous mathematical post we observed that if three triangles fit together to form a triangle or quadrilateral, then they also can fit together in a second way. I thought this point was noteworthy enough to form a blog post of its own.

One of the examples from the previous post is reproduced below.

More generally, we have the following result.

Given points $A, B, C, D$ in the plane there exist points $P, Q, R, S$ in the plane so that

• $\displaystyle \triangle ABD \sim \triangle PQS$
• $\displaystyle \triangle ACD \sim \triangle RQS$
• $\displaystyle \triangle BCD \sim \triangle RPS$ To prove this, first construct $\triangle PQS \sim ABD$ and choose $R$ so that $\triangle ACD \sim \triangle RQS$ and the triangles have the same orientation (i.e. vertices labelled in the same direction clockwise or anticlockwise). This can be done in one way only. We now wish to show that in this configuration we have the third condition $\triangle BCD \sim \triangle RPS$. We shall use complex number geometry to show this.

Let $z_1$ represent the spiral similarity that maps $DA$ to $DB$.

Let $z_2$ represent the spiral similarity that maps $DB$ to $DC$.

Let $z_3$ represent the spiral similarity that maps $DC$ to $DA$.

Then the composition of these three maps is the identity so we have $z_3 z_2 z_1 = 1$. But since $\triangle ABD \sim \triangle PQS$, $z_1$ also maps $SP$ to $SQ$. Similarly $z_3$ maps $SQ$ to $SR$.

Therefore $z_3z_1 = 1/z_2$ maps $SP$ to $SR$, or in other words $z_2$ maps $SR$ to $SP$. This shows that $\triangle BCD \sim \triangle RPS$, as we wished to show.

It’s fascinating to me that non-trivial geometrical ideas can follow from the arithmetic of complex numbers (in this case, the simple fact that $z_1 z_2 z_3 = z_2 z_3 z_1$).

Advertisements

## Leave a Comment »

No comments yet.

Create a free website or blog at WordPress.com.