A generalization of Mollweide's formula (rather Newton's)

Mollweide’s formula, sometimes also referred to as Mollweide’s equation, is a set of two relationships between sides and angles in a triangle. This equation is particularly useful in checking one’s result after solving an oblique triangle since all six components of the triangle are involved.

Let $a$, $b$, and $c$ be the lengths of the three sides of a triangle. Let $\alpha$, $\beta$, and $\gamma$ be the measures of the angles opposite those three sides respectively. Mollweide's formula states that
$$\frac{a+b}{c}=\frac{\cos{\left(\frac{\alpha-\beta}{2}\right)}}{\sin{\left(\frac{\gamma}{2}\right)}}\tag{1}$$
and
$$\frac{a-b}{c}=\frac{\sin{\left(\frac{\alpha-\beta}{2}\right)}}{\cos{\left(\frac{\gamma}{2}\right)}}\tag{2}$$

The equations adopt their name from a German mathematician and astronomer Karl Brandan Mollweide. Nonetheless, this pair of equations was discovered earlier by Isaac Newton and fully developed by Thomas Simpson. An excellent overview of the history of Mollweide’s formula is given by Wu (2007).

For proofs of the Mollweide's formula we invite the readers to see Karjanto's article on the subject (see here). 

Generalization. Let $a$, $b$, $c$ and $d$ be the sides of a cyclic convex quadrilateral. Let $\angle{DAB}=\alpha$, $\angle{ABC}=\beta$, $\angle{BCD}=\gamma$ and $\angle{CDA}=\delta$. Denote $\theta$ the angle between the diagonals (see figure below), then the following identity holds

$$\frac{\sin{\left(\frac{\alpha+\beta}{2}\right)}}{\cos{\left(\frac{\gamma-\delta}{2}\right)}}=\frac{a+c}{b+d}\cot{\left(\frac{\theta}{2}\right)}\tag{3}$$

Proof. We take advantage of the cyclic nature of the half-angle formulas (see $(1)$ here) in combination with the formulas of compound angles. 

$$\frac{\sin{\left(\frac{\alpha+\beta}{2}\right)}}{\cos{\left(\frac{\gamma-\delta}{2}\right)}}=\frac{\sin{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}+\cos{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}}{\cos{\frac{\gamma}{2}}\cos{\frac{\delta}{2}}+\sin{\frac{\gamma}{2}}\sin{\frac{\delta}{2}}}$$

Substituting from the half-angle formulas

$$\frac{\sin{\left(\frac{\alpha+\beta}{2}\right)}}{\cos{\left(\frac{\gamma-\delta}{2}\right)}}=\frac{\sqrt{\frac{(s-a)(s-d)}{ad+bc}}\sqrt{\frac{(s-c)(s-d)}{ab+cd}}+\sqrt{\frac{(s-b)(s-c)}{ad+bc}}\sqrt{\frac{(s-a)(s-b)}{ab+cd}}}{\sqrt{\frac{(s-a)(s-d)}{ad+bc}}\sqrt{\frac{(s-a)(s-b)}{ab+cd}}+\sqrt{\frac{(s-b)(s-c)}{ad+bc}}\sqrt{\frac{(s-c)(s-d)}{ab+cd}}}$$

Simplifying and factorizing

$$\frac{\sin{\left(\frac{\alpha+\beta}{2}\right)}}{\cos{\left(\frac{\gamma-\delta}{2}\right)}}=\sqrt{\frac{(s-a)(s-c)}{(s-b)(s-d)}}\cdot{\frac{(s-d)+(s-b)}{(s-a)+(s-c)}}$$

It is well-known that $\tan{\left(\frac{\theta}{2}\right)}=\sqrt{\frac{(s-b)(s-d)}{(s-a)(s-c)}}$ (see here), so the formula reduces to 

$$\frac{\sin{\left(\frac{\alpha+\beta}{2}\right)}}{\cos{\left(\frac{\gamma-\delta}{2}\right)}}=\frac{a+c}{b+d}\cot{\left(\frac{\theta}{2}\right)}$$

$\square$

Note that as with Mollweide's formulas, this version for a cyclic quadrilateral not only relates the four sides to the four angles, but also includes the angle between the diagonals. I have to admit I was skeptical as to whether it was really a generalization of Mollweide's formula (it was not clear to me). I have published the result in a forum where someone whose nickname is Blue has shown that indeed the formula $(3)$ generalizes Mollweide's formula, more specifically, it generalizes Newton's version (see here).

You can download the entire article here. And here, you can download comments from reviewers of the Mathematical Association of America.

Update. I am pleased to announce that this generalization was published by the Romanian journal MATINF.

This is another example of what the half angle formulas can do for us. See also

• A generalization of the law of tangents
• Proofs and applications of two well-known formulae involving sine, cosine and the semiperimeter of a triangle
• Two Identities and their Consequences

PAGMO 2021 - Problemas 2, 6

Estas son mis soluciones a los problemas $2$ y $6$ de la Olimpiada Panamericana Femenil de Matemáticas.

Problema 2. Considere el triángulo rectángulo isósceles $\triangle{ABC}$ con $\angle{BAC}=90^\circ$. Sea $l$ la recta que pasa por $B$ y el punto medio del lado $AC$. Sea $\tau$ la circunferencia con diámetro $AB$. La recta $l$ y la circunferencia $\tau$ se intersecan en el punto $P$, diferente de $B$. Muestre que la circunferencia que pasa por los puntos $A$, $C$ y $P$ es tangente a la recta $BC$ en $C$.

Demostración. Considera una línea, $m$, paralela a $AB$ que pase por $C$. Denota con $D$ la intersección de $l$ y $m$. Por el Teorema de Tales, $\angle{APD}=90^\circ$. Como $\angle{ACD}$ es recto, tenemos que $APCD$ es cíclico. Llamemos $M$ al punto medio de $AC$. Claramente, $\triangle{ABM}\cong{\triangle{CDM}}$ (por el criterio $ALA$) con $CD=AB=AC$, por lo que $\triangle{ACD}$ es también un triangulo rectángulo isósceles con $\angle{ADC}=\angle{ACB}$. Por la conversa del Teorema del Segmento Alterno, la circunferencia que pasa por los puntos $A$, $P$ y $C$ debe ser tangente a $BC$ en $C$.

Problema 6. Sea $ABC$ un triángulo con incentro $I$ y sea $\tau$ el excírculo opuesto al vértice $A$. Suponga que $\tau$ es tangente a las rectas $BC$, $AC$ y $AB$ en los puntos $A_1$, $B_1$ y $C_1$, respectivamente. Suponga que las rectas $IA_1$, $IB_1$ e $IC_1$ intersectan nuevamente a $\tau$ en los puntos $A_2$, $B_2$ y $C_2$, respectivamente. Sea $M$ el punto medio del segmento $AA_1$. Si las rectas $A_1B_1$ y $A_2B_2$ se intersectan en $X$ y las rectas $A_1C_1$ y $A_2C_2$ se intersectan en $Y$, demuestre que $MX=MY$.

Demostración. Claramente, $\angle{ABI}=\angle{BC_1A_1}$, por lo que  $BI\parallel{A_1C_1}$. Por propiedad de ángulos entre paralelas, $\angle{BIC_1}=\angle{IC_1A_1}$ y por propiedad del ángulo semiinscrito, $\angle{BC_1I}=\angle{C_1A_1C_2}$. Esto implica que $\triangle{IBC_1}\sim{\triangle{A_1C_1C_2}}$. Luego sigue que $\angle{IC_2A_1}=180^\circ-\angle{C_1C_2A_1}=180^\circ-\angle{C_1BI}=\angle{ABI}=\angle{A_1BI}=\frac{\angle{ABC}}{2}$. De donde se deduce que $IBC_2A_1$ es cíclico. Denotemos con $N$ la segunda intersección de $BC_2$ con $\tau$. Por el Teorema de Reim, $BI\parallel{A_2N}$. Note que $A_1NC_1C_2$ es un cuadrilátero armónico, por lo que si proyectamos desde $A_2$ sobre la recta $A_1C_1$, también paralela a $A_2N$ (recuerde que $BI\parallel{A_1C_1}$), tenemos

$$-1=(N, C_2; C_1, A_1)\stackrel{A_2}{=}(P_\infty, Y; C_1, A_1).$$

Esto implica que $Y$ es el punto medio de $A_1C_1$. Análogamente, $X$ es el punto medio de $A_1B_1$.  Por el Teorema del Segmento Medio, $MY=\frac{AC_1}{2}$ y $MX=\frac{AB_1}{2}$. Pero $AC_1=AB_1$ por ser tangentes comunes, por lo tanto $MY=MX$.

Generalized half-angle formulas - Hyperbolic version

In Mathoverflow I wonder about the possibility of finding non-Euclidean versions of the generalized half-angle formulas $[1]$. The Russian mathematician Alexander Mednykh has kindly answered my question. Specifically, he has derived the hyperbolic version. Look at the following link:

Note: Generalized half-angle formula - Hyperbolic version

Reference
$[1]$ E. A. José García, Two Identities and their Consequences, MATINF, 6 (2020) 5-11.

Using the half-angle formula for cosine to derive Zelich's lemma on mixtilinear incircles

Lemma (Ivan Zelich). Let $w$ be the $A$-mixtilinear incircle in $\triangle{ABC}$ touching side $AB$ at $E$, side $AC$ at $F$. Then

$$AE=AF=\frac{bc}{s}\tag{1}$$

where $s$ is the semiperimeter.

Proof 1. The radius of $w$ inscribed in $\angle{CAB}=\alpha$ is given by

$$\rho_a=r\sec^2{\frac{\alpha}{2}}\tag{2}$$

where $r$ is the inradius of the reference triangle and $\rho_a$ is the radius of $w$ (Durell and Robson 1935).

A proof of $(2)$ can be found here (see pp. 13).

The half-angle formula for cosine states that 

$$\cos^2{\frac{\alpha}{2}}=\frac{s(s-a)}{bc}\tag{3}$$

See here for a proof of $(3)$.

Call $I$ the Incenter of $\triangle{ABC}$ and $D$ the touchpoint between the incircle and $AC$. Denote $K$ the center of $w$. Notice that $\triangle{AID}\sim\triangle{AKF}$. Now, by similarity of triangles we have

$$\frac{r}{s-a}=\frac{\rho_a}{AF}\tag{4}$$

Combining $(2)$ and $(3)$ in $(4)$ we get $(1)$.

$\square$

Proof 2. We can also derive $(1)$ using the relationships $\Delta=rs$ and $\Delta=\frac{bc\sin{\alpha}}{2}$. Indeed, since 

$$r=\frac{\Delta}{s}=\frac{\frac{bc\sin{\alpha}}{2}}{s}=\frac{bc\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}}{s}.$$

Substituting in $(2)$ and simplifying we get 

$$\rho_a=r\sec^2{\frac{\alpha}{2}}=\frac{bc}{s}\tan{\frac{\alpha}{2}}=\frac{bc\rho_a}{s\cdot{AF}}$$

from which the result holds. 

$\square$

A proof using inversion can be found here. 

Related material

Conic in Mixtilinear Incircles

Using the half-angle formulas to derive Mahavira's identities

In a cyclic quadrilateral $ABCD$, let $a$, $b$, $c$, $d$ denote the lengths of sides $AB$, $BC$, $CD$, $DA$, and $m$, $n$ the lengths of the diagonals $BD$ and $BC$. Then Mahavira's result is expressed as
$$m^2=\frac{(ab+cd)(ac+bd)}{ad+bc}\tag{1}$$
$$n^2=\frac{(ac+bd)(ad+bc)}{ab+cd}\tag{2}.$$

Proof. By the Law of Cosines, 

$$\begin{align*}m^2&=a^2+d^2-2ad\cos{A}\\&=a^2+d^2-2ad(2\cos^2{\frac{A}{2}}-1)\\&=(a+d)^2-4ad\cos^2{\frac{A}{2}}\end{align*}$$

Substituting from the half-angle formula (see formula $(5)$ in this page) we get

$$\begin{align*}m^2&=(a+d)^2-\frac{ad[(a+d)^2-(b-c)^2]}{ad+bc}\\&=\frac{bc(a+d)^2+ad(b-c)^2}{ad+bc}\\&=\frac{a^2bc+bcd^2+ab^2d+ac^2d}{ad+bc}\\&=\frac{(ab+cd)(ac+bd)}{ad+bc}.\end{align*}$$

$\square$

Similarly we can get $(2)$.

Related material 

Brahmagupta-Mahavira Identities

Ptolemy's Theorem

Length of angle bisector: yet another application of the half-angle formulas

Let $\triangle{ABC}$ be a triangle. Let $AD$ be the angle bisector of $\angle{BAC}$ in $\triangle{ABC}$. Let $d$ be the length of $AD$. Then $d$ is given by:

$$d^2=\frac{bc}{(b+c)^2}\left[(b+c)^2-a^2\right]\tag{1}$$

where $a$, $b$, and $c$ are the sides opposite $A$, $B$ and $C$ respectively.

Lemma 1 (Tehebycheff). Let $\triangle{ABC}$ be a triangle. Let $AD$ be the angle bisector of $\angle{BAC}$ in $\triangle{ABC}$. Let $d$ be the length of $AD$ and $\angle{BAD}=\angle{CAD}=\frac{\alpha}{2}$. Then $d$ is given by:

$$d=\frac{2bc\cos{\frac{\alpha}{2}}}{b+c}\tag{2}$$

Proof. Let $BD=m$ and $CD=n$. We make use of the Angle Bisector Theorem and the Law of Cosines. Indeed, by the Angle Bisector Theorem we have

$$\frac{b}{c}=\frac{n}{m}$$

which can also be written as

$$\frac{b^2}{c^2}=\frac{n^2}{m^2}\tag{3}$$

Now, by the Law of Cosines, 

$$n^2=b^2+d^2-2bd\cos{\frac{\alpha}{2}}\tag{4}$$

$$m^2=c^2+d^2-2cd\cos{\frac{\alpha}{2}}\tag{5}$$

Dividing $(4)$ by $(5)$ and substituting $\frac{n^2}{m^2}$ by $\frac{b^2}{c^2}$ we have

$$\frac{b^2}{c^2}=\frac{b^2+d^2-2bd\cos{\frac{\alpha}{2}}}{c^2+d^2-2cd\cos{\frac{\alpha}{2}}}$$

From which we get

$$d^2(b^2-c^2)=2bcd\cos{\frac{\alpha}{2}}(b-c)$$

Isolating $d$, factoring and simplifying we obtain $(2)$.

$\square$

Another proof of Lemma 1 can be found here.

Lemma 2. Let $\triangle{ABC}$ be a triangle and let $a$, $b$ and $c$ be the lengths of sides $BC$, $AC$ and $BC$, respectively. Then the following identity holds

$$2\cos{\frac{\alpha}{2}}=\sqrt{\frac{(b+c)^2-a^2}{bc}}\tag{6}$$

A proof of Lemma 2 can be found here.

$\square$

Main proof

Combining $(2)$ and $(6)$ we obtain

$$d=\frac{bc}{(b+c)}\cdot{\sqrt{\frac{(b+c)^2-a^2}{bc}}}$$

Raising both sides to the power of 2 we get $(1)$.

$\square$

Related material

Length of Angle Bisector

Proving the length of angle bisector

Publicaciones, reportajes y entrevistas

En arXiv (Cornell)

• A Unified Substitution Method for Integration 

En Mathematics Magazine

• A Generalization of the Law of Tangents.

En The College Mathematics Journal

• A Variant of the Eyeball Theorem

Este artículo inspiró la portada del CMJ, Volumen 53, 2022 - 2. Mire aquí.

En Forum Geometricorum (ahora disponible aquí)

• A note on reflections, Forum Geometricorum, 14 (2014) 155--161.
• Another Archimedean circle in an arbelos, Forum Geometricorum, 15 (2015) 127--128.
• (With Paul Yiu) Golden sections of triangle centers in the golden triangles, Forum Geometricorum, 16 (2016) 119--124.
• (With Le Viet An) Some Archimedean circles in an arbelos, Forum Geometricorum 19 (2019) 53--58.

En MATINF

• A Generalization of the Law of Cotangents, pp. 16-19.
• A generalization of Mollweide's formula (rather Newton's), pp. 19-22.
• Two Identities and their Consequences, pp. 5-11.

En Sangaku Journal of Mathematics

• (With Le Viet An) A pair of Archimedean incircles, pp.1-2.
• Three Archimedean circles arising from equilateral triangles, pp.86-87.

Otras publicaciones no indexadas

En Encyclopedia of Triangle Centers - ETC

• García-Feuerbach Point (vea X(5550))
• Inner/Outer García Triangles
• García-Reflection Triangle

En Online Catalogue of Archimedean Circles

• Archimedean circles 59a and 59b: First García pair
• Archimedean circles 60a and 60b: Second García pair
• Archimedean circle 62: Sanchez-García triplet circle

En Schnittpunkte 701–800 de Hans Walser

• Schnittpunkt 756 - 760
• Schnittpunkt 770 - 773

Artículos donde se citan mis trabajos

• Lopes, A. von B., dos Santos, R. C., & dos Reis, P. R. A. (2025). Bretschneider’s Formula Via Complex Numbers. The American Mathematical Monthly, 132(8), 818–821.
  
• Roger C. Alperin, Reflections on Poncelet's pencil, Forum Geometricorum, 15 (2015) 93--98.
• Hiroshi Okumura, A remark on Archimedean incircles of an isosceles triangle, Forum Geometricorum, 19 (2019) 59—61.
• M. Dalcín, S. N. Kiss Some Properties of the García Reflection Triangles 119--126.
• Masa Nahtigal, Zlati Rez V Geometriji, Univerza V Ljubljani, 2017.
• Michell F. Marques, Alguns círculos congruentes aos círculos gêmeos de Arquimedes, Universidade de Brasília, 2019.
• HORČIČKOVÁ, Klára. Arbelos. Bakalářská práce, vedoucí Škorpilová, Martina. Praha: Univerzita Karlova, Matematicko-fyzikální fakulta, Katedra didaktiky matematiky, 2021.
  
• NOVÁKOVÁ, Klára. Kružnice v širších souvislostech. Diplomová práce, vedoucí Škorpilová, Martina. Praha: Univerzita Karlova, Matematicko-fyzikální fakulta, Katedra didaktiky matematiky, 2024.

Reportajes

• Diario Libre: El dominicano que ha hecho aportes a la geometría
  
• Periódico Hoy: La geometría, lengua universal para resolver problemas
  
• Diario Libre: En RD podemos hacer matemáticas
  
• El Nuevo Diario: Dominicano asegura haber superado fórmula matemática de Newton
  
• RD Futuro: El profesor dominicano que descubre nuevos caminos para el álgebra y la geometría mundial
  

Entrevistas

• Paneo Semanal: El dominicano que superó una fórmula de Newton
• Diario Libre: En RD podemos hacer matemáticas
  

Reconocimientos / Distinciones / Premios

• Hijo Distinguido y Meritorio del Municipio Santo Domingo Este, otorgado mediante Resolución No. 05-2025 por el Ayuntamiento de Santo Domingo Este.