Talk:Tangential trapezoid

Dubious uncited formulas

One formula given for the area is

$K={\frac {a+b}{2|b-a|}}{\sqrt {ab(a-c)(c-a)}},$

and one formula given for the inradius is

$r={\frac {K}{a+b}}={\frac {\sqrt {ab(a-c)(c-a)}}{2|b-a|}}.$

Both of these involve the square root of a negative number, so they can't be right. Unfortunately no citation is given, so I can't look up the correct version. Does anyone know it? Duoduoduo (talk) 23:42, 18 May 2012 (UTC)

You are correct! There was in fact two errors, both the letters and the factor 2 in the denominator. I wrote the first draft for this article and the fault was mine. I didn't find a reference for these formulas, so I used the formula for the area of a trapezoid

$K={\frac {a+b}{4|b-a|}}{\sqrt {(-a+b+c+d)(a-b+c+d)(a-b+c-d)(a-b-c+d)}}$

and simplified it using Pitot's theorem. The problem is that the sides in a trapezoid are in consecutive order: a, c, b, d and not a, b, c, d as in most (all) other quadrilaterals. Thus Pitot's theorem in a tangential trapezoid states a + b = c + d instead of the usual a + c = b + d. Hence the problem you address. It would of course be prefarable to be able to cite a reference giving these formulas, but this is a special case that does not seem to appear often in the litterature. Please check my algebra so it is correct this time, and lets be on the look out for a reference. Circlesareround (talk) 09:23, 19 May 2012 (UTC)

Looks good to me now! Duoduoduo (talk) 17:35, 19 May 2012 (UTC)

Characterization(s)

Here this characterization of tangential trapezoid is given:

A convex quadrilateral is a tangential trapezoid if and only if opposite sides satisfy Pitot's theorem (so it is tangential) and it has two adjacent angles that are supplementary (then this is also true for the other two angles) (so it is a trapezoid). Hence AB and CD are the bases in a tangential trapezoid ABCD if and only if $\displaystyle AB+CD=BC+DA$ and $\displaystyle A+D=B+C=\pi$.

At bicentric quadrilateral a similar one (if not the same) characterization is given:

A convex quadrilateral ABCD with sides a, b, c, d is bicentric if and only if opposite sides satisfy Pitot's theorem and opposite angles are supplementary, that is

$\displaystyle a+c=b+d\quad {\text{and}}\quad A+C=B+D=\pi .$

But not every tangential trapezoid is (also) bicentric (only isosceles (tangential) ones). So, what is wrong here? A quadrilateral with these two properties is not automatically (also) bicentric - am I right? I guess something is missing in 1st characterization for bicentric quadrilateral. --xJaM (talk) 23:28, 11 July 2012 (UTC)

There is nothing wrong. As you said yourself, they are similar but not the same! Pitot's theorem must hold in both so the quadrilateral is tangential (has an incircle). The other condition is a little different. For a tangential trapezoid it is that adjacent angles are supplementary, whereas in a bicentric quadrilateral it is that opposite angles are supplementary. That is the difference between that opposite sides are parallel (trapezoid) or that the quadrilateral has a circumcircle (bicentric).

Further, you are correct that a tangential trapezoid is not always bicentric. If a convex quadrilateral shall be a bicentric trapezoid, then it shall satisfy three conditions: Pitot's, adjacent angles supplementary and opposite angles supplementary. Then it is a tangential isosceles trapezoid, as in the second figure. Circlesareround (talk) 09:50, 12 July 2012 (UTC)

Oh, yes. I didn't read well and haven't seen words 'opposite' and 'adjacent' in both characterizations; and also both second formuale for angles, which of course differ. For sure - it makes sense then. Perhaps what should also can be noted is that in tangential trapezoid (or in any other trapezoid) adjacent angles lie on legs ($\displaystyle A+D=B+C=\pi$ holds), not perhaps on bases, so relation for other pair of adjacent angles $\displaystyle A+B\neq C+D\neq \pi$ does not hold - this can be perhaps well seen in image for right tangential trapezoid. Also I think that now citations for area and inradius formulae are not needed anymore, as they can be derived from area formula for trapezoid. --xJaM (talk) 13:15, 12 July 2012 (UTC)