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Right Triangle

A right triangle is triangle with an angle of ( radians). The sides , , and of such a triangle satisfy the Pythagorean theorem

(1)

where the largest side is conventionally denoted and is called the hypotenuse. The other two sides of lengths and are called legs, or sometimes catheti.

The favorite A-level math exam question of the protagonist Christopher in the novel The Curious Incident of the Dog in the Night-Time asks for proof that a triangle with sides of the form , , and where is a right triangle, and that the converse does not hold (Haddon 2003, pp. 214 and 223-226).

The side lengths of a right triangle form a so-called Pythagorean triple. A triangle that is not a right triangle is sometimes called an oblique triangle. Special cases of the right triangle include the isosceles right triangle (middle figure) and 30-60-90 triangle (right figure).

For any three similar shapes of area on the sides of a right triangle,

(2)

which is equivalent to the Pythagorean theorem.

For a right triangle with sides , , and hypotenuse , the area is simply

(3)

The inradius can be found by equating the area of the triangle with the sum of the areas of the three triangles , , and having the inradii as altitudes, giving

(4)

Solving for then gives

(5)

This can also be written in the equivalent forms

			(6)
			(7)

The hypotenuse of a right triangle is a diameter of the triangle's circumcircle, so the circumradius is given by

(8)

A primitive right triangle is a right triangle having integer sides , , and such that , where is the greatest common divisor. The set of values is then known as a primitive Pythagorean triple.

For a right triangle with integer side lengths, any primitive Pythagorean triple can be written

			(9)
			(10)
			(11)

Using these, equation (6) becomes

			(12)
			(13)

which is an integer whenever and are integers (Ogilvy and Anderson 1988, p. 68).

Given a right triangle , draw the altitude from the right angle . Then the triangles and are similar.

In a right triangle, the midpoint of the hypotenuse is equidistant from the three polygon vertices (Dunham 1990). This can be proved as follows. Given , let be the midpoint of (so that ). Draw , then since is similar to , it follows that . Since both and are right triangles and the corresponding legs are equal, the hypotenuses are also equal, so we have and the theorem is proved.

In addition, the triangle median and altitude of a triangle are reflections about the angle bisector of iff is a right triangle (G. McRae, pers. comm., May 1, 2006).

Fermat showed how to construct an arbitrary number of equiareal nonprimitive right triangles. An analysis of Pythagorean triples demonstrates that the right triangle generated by a triple has common area

(14)

(Beiler 1966, pp. 126-127). The only extremum of this function occurs at . Since for , the smallest area shared by three nonprimitive right triangles is given by , which results in an area of 840 and corresponds to the triplets (24, 70, 74), (40, 42, 58), and (15, 112, 113) (Beiler 1966, p. 126). One can also find quartets of right triangles with the same area. The quartet having the smallest known area is (111, 6160, 6161), (231, 2960, 2969), (518, 1320, 1418), (280, 2442, 2458), with area (Beiler 1966, p. 127). Guy (1994) gives additional information.

It is also possible to find sets of three and four right triangles having the same perimeter (Beiler 1966, pp. 131-132).

In a given right triangle, an infinite sequence of squares that alternately lie on the hypotenuse and longest leg can be constructed, as illustrated above. These create a sequence of increasingly smaller similar right triangles. Let the original triangle have legs of lengths and and hypotenuse of length . Also define

			(15)
			(16)

Then the sides of the square are of length

(17)

Number the upper left triangle as 1, and then the remainder by following the "strip" of triangles at adjoining vertices. Then the side lengths of these triangles are

		${s_((n+1)/2) for n odd; (ab)/cx^(n/2) for n even$	(18)
		${(b^2)/ax^((n+1)/2) for n odd; (b^2)/cx^(n/2) for n even$	(19)
		${(bc)/ax^((n+1)/2) for n odd; s_(n/2) for n even.$	(20)

The inradii of the corresponding triangles can be found from

(21)

giving

$r_n={b/ayx^((n+1)/2) for n odd; b/cyx^(n/2) for n even.$

(22)

A Sangaku problem from 1913 in the Miyagi Prefecture asks for the relationships between the first, third, and fifth inradii (Rothman 1998). This can be solved using elementary trigonometry as well as the explicit equations given above, and has solution

(23)

Right Triangle

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