Let $ABCD$ be a cyclic quadrilateral with sides $a$, $b$, $c$ and $d$. Denote $s$ the semiperimeter and let $\angle{DAB}=\alpha$, $\angle{ABC}=\beta$, $\angle{BCD}=\gamma$ and $\angle{CDA}=\delta$. Then the following inequality holds
$$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\gamma}{2}}\tan{\frac{\delta}{2}}+\tan{\frac{\delta}{2}}\tan{\frac{\alpha}{2}}\geq4.\tag{1}$$GeoDom
domingo, 19 de junio de 2022
$\sum_{cyc}\tan\frac\alpha2\tan\frac\beta2\geq4$ for a cyclic quadrilateral
lunes, 23 de mayo de 2022
The theoretical importance of the half-angle formulas
Unlike the laws of sines, cosines and tangents, which are very well known, the half-angle formulas seem (although they appear timidly in the mathematical literature) not to enjoy the same popularity. Thus, while there are entire chapters devoted to the law of sines, cosines, and tangents and their applications, there is not even a Wikipedia article on half-angle formulas. Right now you may be imagining this version of the half angle formulas
$$\sin{\frac{\alpha}{2}}=\pm\sqrt{\frac{1-\cos{\alpha}}{2}}\qquad\qquad\cos{\frac{\alpha}{2}}=\pm\sqrt{\frac{1+\cos{\alpha}}{2}},$$
that do appear in the textbooks of the first trigonometry courses (at least with the one I studied). But actually I mean these
$$\sin^2{\frac{\alpha}{2}} = \frac{(s-b)(s-c)}{bc}\qquad\qquad\cos^2{\frac{\alpha}{2}}= \frac{s(s-a)}{bc},\tag{1}$$
where $a$, $b$, and $c$ are the sides of a triangle, $\alpha$ is the angle opposite side $a$, and $s$ is the semiperimeter. I found the furthest reference to these formulas in a conversation posted online between Conway and Doyle, where Conway uses them to prove Heron's formula and later claims to have taken it from a sequel by Casey.
I discovered $(1)$ independently trying to prove the law of cosines by contradiction. When I realized that they were known, I tried to generalize them and I got this
$$\sin^2{\frac{\alpha}{2}}=\frac{(s-a)(s-d)}{ad+bc}\qquad\qquad \cos^2{\frac{\alpha}{2}}=\frac{(s-b)(s-c)}{ad+bc},\tag{2}$$
where $a$, $b$, $c$ and $d$ are the sides of a cyclic quadrilateral, $s$ is semiperimeter and $\angle{DAB}=\alpha$.
Before discovering the conversation between Conway and Doyle, I had been excited that I had found an original proof of Heron's formula using $(1)$. When I found $(2)$, I thought that by analogous reasoning I could prove Brahmagupta's formula. So it was. But in a geometry forum someone referred me to an ancient Greek book that contained $(2)$. I then sent my proof of Brahmagupta's formula to Martin Josefsson who referred me to Casey's book "A Treatise On Plane Trigonometry" where my proof already appeared. But I didn't give up and tried to generalize $(2)$ getting this
$$ad\sin^2{\frac{\alpha}{2}}+bc\cos^2{\frac{\gamma}{2}}=(s-a)(s-d)\qquad\qquad bc\sin^2{\frac{\gamma}{2}}+ad\cos^2{\frac{\alpha}{2}}=(s-b)(s-c),\tag{3}$$
where $a$, $b$, $c$ and $d$ are the sides of a general quadrilateral, $s$ is semiperimeter, $\angle{DAB}=\alpha$ and $\angle{BCD}=\gamma$.
Bretschneider's formula is known to be a generalization of Heron's and Brahmagupta's formulas. Naturally, I wondered if I could generalize Casey's proof of Brahmagupta's formula using $(3)$ and thus derive Bretschneider's formula. And I did it. I sent my formulas in $(3)$ and my proof of the Bretschneider's formula to Josefsson (among many other mathematicians) and he told me this:
"I like your paper, especially how you put these important formulas in a single framwork. I cannot say that I remember seeing the identities (4) and (5) anywhere else before."
Where identities (4) and (5) are the identities $(3)$ in this post.
And then he said:
"Even though much has already been written about these formulas, the ideas for proving Bretschneider' formula and the area of a bicentric quadrilateral are novel as far as I know. I hope you get your paper published."
I decided to write an article about these formulas called "Two Identities and their Consequences" which was published in MATINF.
In almost three years of exploring possible applications of $(1, 2, 3)$, this is what I have found:
Using $(1, 2)$ we can also derive (you can see most of the proofs here):
- The law of cosines
- The law of sines
- The law of tangents
- Stewart's theorem
- Compound angle formulas
- Mollweide's formula
- The product $AI\cdot{BI}\cdot{CI}$
- The bisector length formula
- Mahavira's formulas
- Zelich's lemma
- Yun’s Inequality (Josefsson)
Other unnamed identities and inequalities:
- $\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\alpha}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}=1$
- $r=4R\sin{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}$
- $s=4R\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}$
- $\sin{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}\le\frac{1}{8}$
- $\sum_{cyc}\tan\frac\alpha2\tan\frac\beta2\geq4$ for a cyclic quadrilateral
- I could go on...
The formulas $(1, 2, 3)$ explain the Heron-Brahmagupta-Bretschneider development better than I have seen anywhere else. This made me wonder what would happen if I analogously applied the half-angle formulas to formulas where half-angles explicitly appeared, such as Mollweide's (rather Newton's) formula or the law of tangents. This is how these two generalizations arose:
When questioning Martin Josefsson about the originality of these generalizations, this is what he said:
"As far as I can recall, I have not seen any of them, at least not in modern books or papers, and even if some of them where to be found in an old text, they are at least not well known, and deserve to be wider known."
And Alexander Mednykh said:
"I never saw these results before. Certainly, it is interesting to find a no-Euclidean generalization of these results."Apart from the proof of the Bretschneider's formula, I haven't found any other applications for $(3)$.
Interestingly, half angles seem to be everywhere: from circle angle theorems to the Weierstrass substitution in Integral Calculus. Even when Viète derived his formula for $\pi$ using an infinite product, he started by writing $\sin{x}=2\sin{\frac12x}\cos{\frac12x}$.
Some comments:
Thibaut Demaerel, from Leuven University, commented:
James Cook, from the University of West Alabama, commented:
domingo, 22 de mayo de 2022
La importancia teórica de las fórmulas de medio ángulo
A diferencia de las leyes de senos, de cosenos y de las tangentes, que son muy bien conocidas, las fórmulas de medio ángulo parecen (aunque aparecen tímidamente en la literatura matemática) no gozar de la misma popularidad. Así, mientras hay capítulos enteros dedicados a la ley de senos, de cosenos, de tangentes y a sus aplicaciones, no hay ni siquiera un artículo de Wikipedia sobre las fórmulas de medio ángulo. En estos momentos a lo mejor te estés imaginando esta versión de las fórmulas de medio ángulo
$$\sin{\frac{\alpha}{2}}=\pm\sqrt{\frac{1-\cos{\alpha}}{2}}\qquad\qquad\cos{\frac{\alpha}{2}}=\pm\sqrt{\frac{1+\cos{\alpha}}{2}},$$- La ley de cosenos
- La ley de senos
- La ley de las tangentes
- El teorema de Stewart
- Las fórmulas para ángulos compuestos
- La fórmula de Mollweide
- El producto $AI\cdot{BI}\cdot{CI}$
- La fórmula de la longitud del bisector
- Las fórmulas de Mahavira
- El lema de Zelich
- $\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\alpha}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}=1$
- $r=4R\sin{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}$
- $s=4R\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}$
- Podría seguir...
viernes, 29 de abril de 2022
Falacia lógica en Bosch - Composición Social Dominicana
Una falacia de división es una falacia informal que ocurre cuando uno razona que algo que es cierto para un todo también debe ser cierto para todas o algunas de sus partes.
La falacia de división fue abordada por Aristóteles en sus Refutaciones sofísticas.
Ejemplos:
1. El sabor del guacamole es sabroso, por lo que los componentes con que se elabora (aguacate, sal, limón, chiles) también lo son.
2. El auto es azul; por lo tanto, sus neumáticos también lo son.
¿Ves el fallo? Ahora veamos el argumento de Bosch reescrito para que lo notes más fácilmente (ten en cuenta que Bosch hablaba de la época en que Santo Domingo era colonia española):
Premisa 1: España no asimilaba la oligarquía esclavista.
Premisa 2: Santo Domingo era parte de España.
Conclusión: Santo Domingo no asimilaba la oligarquía esclavista.
¡Claramente una falacia de división y apenas estamos en el preámbulo del libro! Cabe señalar que la conclusión podría ser verdadera, pero mi objeción va dirigida a la manera en que Bosch la infiere, una falacia registrada desde los tiempos de Aristóteles. La conclusión ha sido mal defendida y eso le resta rigor a la obra, precisamente de lo que se cuidan los pensadores de verdad.
Esta falacia está en el preámbulo de la obra Bosch-Composición Social Dominicana, página 10, párrafo 1. ¿Crees que sea la única?
miércoles, 27 de abril de 2022
The compound angle formulas from the half angle formulas
My goal is not necessarily to give the simplest derivation. I've been trying to rescue the half angle formulas from oblivion and give them the status they deserve by showing its many applications.
Various proofs of the formulas of compound angles are given here. The cosine of the sum of two angles is given by the formula$$\cos{(\alpha+\beta)}=\cos{\alpha}\cos{\beta}-\sin{\alpha}\sin{\beta}\tag{1}$$
Proof. We take advantage of the cyclic nature of the half angle formulas whose proof can be found here. The proof of the half angle formulas given in the link makes use of the Pythagorean trigonometric identity and the double-angle formula for sine whose proofs (I hope you clicked on the links) are independent of the compound angle formulas.
Suppose $\triangle{ABC}$ is a triangle with sides $|BA|=a$, $|AC|=b$ and $|AB|=c$. Let $\angle{BAC}=2\alpha$, $\angle{CBA}=2\beta$ and $\angle{ACB}=2\gamma$. Let's start with the right-hand side of formula $(1)$. Substituting from the half angle formulas we have
$$\begin{aligned} \cos{\alpha}\cos{\beta}-\sin{\alpha}\sin{\beta} &= \sqrt{\frac{s(s-a)}{bc}\cdot{\frac{s(s-b)}{ac}}}-\sqrt{\frac{(s-b)(s-c)}{bc}\cdot{\frac{(s-a)(s-c)}{ac}}}\\&=\frac{s}{c}\cdot{\sqrt{\frac{(s-a)(s-b)}{ab}}}-\frac{s-c}{c}\cdot{\sqrt{\frac{(s-a)(s-b)}{ab}}}\\&=\sqrt{\frac{(s-a)(s-b)}{ab}}\cdot{\left(\frac{s}{c}-\frac{s-c}{c}\right)}\\&=\sin{\gamma}\\&=\sin{\left(\frac{\pi}{2}-(\alpha+\beta)\right)}\\&=\cos{(\alpha+\beta)}\end{aligned}$$
$\square$
The other formulas of compound angles can be obtained similarly.
domingo, 27 de marzo de 2022
A generalization of the law of tangent for a cyclic quadrilateral
$$\frac{\tan\frac12(\alpha-\beta)}{\tan\frac12(\alpha+\beta)}=\frac{(a-c)(b-d)}{(a+c)(b+d)}\tag{2}$$
$$\frac{\tan\frac12(\alpha-\beta)}{\tan\frac12(\alpha+\beta)}=\frac{\sin\frac12(\alpha-\beta)\cos\frac12(\alpha+\beta)}{\cos\frac12(\alpha-\beta)\sin\frac12(\alpha+\beta)}=\frac{\sin{\alpha}-\sin{\beta}}{\sin{\alpha}+\sin{\beta}}$$
The area of a cyclic quadrilateral can be expressed as $\Delta=\frac12(ad+bc)\sin{\alpha}$ (see $(12)$ at Killing three birds with one stone) and similarly for the other angles. Then substituting, simplifying and factorizing we have
$$\begin{align*}\frac{\tan\frac12(\alpha-\beta)}{\tan\frac12(\alpha+\beta)}&=\frac{\frac{2\Delta}{ad+bc}-\frac{2\Delta}{ab+cd}}{\frac{2\Delta}{ad+bc}+\frac{2\Delta}{ab+cd}}=\frac{ab-ad+cd-bc}{ab+ad+cd+bc}=\frac{(a-c)(b-d)}{(a+c)(b+d)}\end{align*}$$
The formula $(2)$ reduces to the law of tangent for a triangle when $c=0$.
sábado, 15 de enero de 2022
A generalization of Mollweide's formula (rather Newton's)
$$\frac{a+b}{c}=\frac{\cos{\left(\frac{\alpha-\beta}{2}\right)}}{\sin{\left(\frac{\gamma}{2}\right)}}\tag{1}$$
and
$$\frac{a-b}{c}=\frac{\sin{\left(\frac{\alpha-\beta}{2}\right)}}{\cos{\left(\frac{\gamma}{2}\right)}}\tag{2}$$