Iterative Tangents

A curiosity journal of math, physics, programming, astronomy, and more.

Geometry

Right triangle legs

I was thinking about right triangles the other day and wondered what constraints on a triangle's geometry preserve the right angle under rotation. In more visual terms, given the diagram below, what are the relationships between a, b, c, and d that make the angle θ a right angle?

abcdθ
A right triangle embedded in a rectangle.

The legs adjacent to θ have the lengths

StartRoot a squared plus c squared EndRoot StartRoot d squared plus left-parenthesis a plus b right-parenthesis squared EndRoot

and the hypotenuse opposite θ has the length

StartRoot b squared plus left-parenthesis c plus d right-parenthesis squared EndRoot

To find the constraints on a, b, c, and d that make the inner triangle a right triangle, we can put these lengths into the Pythagorean theorem:

StartLayout 1st Row 1st Column StartRoot a squared plus c squared EndRoot squared plus StartRoot d squared plus left-parenthesis a plus b right-parenthesis squared EndRoot squared 2nd Column equals StartRoot b squared plus left-parenthesis c plus d right-parenthesis squared EndRoot squared 2nd Row 1st Column a squared plus c squared plus d squared plus left-parenthesis a plus b right-parenthesis squared 2nd Column equals b squared plus left-parenthesis c plus d right-parenthesis squared 3rd Row 1st Column a squared plus c squared plus d squared plus a squared plus 2 a b plus b squared 2nd Column equals b squared plus c squared plus 2 c d plus d squared 4th Row 1st Column 2 a squared plus 2 a b 2nd Column equals 2 c d 5th Row 1st Column a squared plus a b 2nd Column equals c d 6th Row 1st Column a left-parenthesis a plus b right-parenthesis 2nd Column equals c d 7th Row 1st Column StartFraction a Over c EndFraction 2nd Column equals StartFraction d Over left-parenthesis a plus b right-parenthesis EndFraction EndLayout

That is, a is to c as d is to a + b.

If that relationship is a bit abstract, look back at the diagram and note that a and c form the legs of an outer right triange, as do d and a + b. The ratio we found tells us that to make the inner triangle a right triangle, those outer right triangles have to be similar to each other.

That seemed too simple until I remembered the middle-school algebra fact that, given a line, any line perpendicular to it will have a slope that's the negative reciprocal of the slope of the original line. We've lost the negative sign (since directions are implicit in the diagram), but we still have the reciprocal.