sábado, 5 de diciembre de 2020

The Pythagorean Theorem from the Half-Angle Formula for Cosine

I've been trying to rescue the Half-Angle Formulas from oblivion and give them the status they deserve by showing its many applications and equivalence with other fundamental theorems (i.e., the Law of Cosines or the Pythagorean Theorem). In this page the Pythagorean Theorem is presented as a mere corollary of the aforementioned formulas.

Let $\triangle{ABC}$ be a right-triangle with $BC=a$, $AC=b$, $AB=c$ and $\angle{ACB}=\gamma=90^\circ$. Also, let $s=\frac{a+b+c}{2}$. The proof is based on the formula

$$\cos^2{\frac{\gamma}{2}}=\frac{s(s-c)}{ab}=\frac{(a+b)^2-c^2}{4ab}\tag{1}$$

This is true for any triangle and can be proven independently of the Pythagorean theorem or the Pythagorean Identity. The proof of $(1)$ can be found here.

There has been controversy about whether it is possible to prove the Pythagorean Theorem using trigonometry. Elisha Loomis and many others believed and still believe that no trigonometric proof of the Pythagorean theorem is possible. This belief stems from the assumption that any such proof would rely on the most fundamental of trigonometric identities $\sin^2{\theta} + \cos^2{\theta} = 1$ which is nothing but a reformulation of the Pythagorean theorem proper. However, Jason Zimba showed it is possible to prove the Pythagoren Identity without a recourse to the Pythagorean theorem (see $[1]$). He then has used the Pythagorean Identity to prove the Pythagorean theorem. Another trigonometric proof is given by Luc Gheysens (see here).

Pythagorean Theorem. Given a right-triangle $\triangle{ABC}$ with $BC=a$, $AC=b$, $AB=c$ and $\angle{ACB}=\gamma=90^\circ$. Then

$$a^2+b^2=c^2$$

Proof. From the Double-Angle Formula for cosine we know

$$\cos{2\theta}=2\cos^2{\theta}-1\tag{2}$$


The proof of $(2)$ does not rely on the Pythagorean Theorem (see this video). Using the unit circle, the proof of $\cos{90^\circ}=0$ only depends on the definition of Cosine (see here). So applying $(2)$ we can calculate $\cos{45^\circ}=\frac{\sqrt{2}}{2}$ without risk of Petitio Principii. Hence


$$\cos^2{\frac{\gamma}{2}}=\frac{s(s-c)}{ab}=\frac{(a+b)^2-c^2}{4ab}=\frac{1}{2}$$


from which we have


$$\begin{align*}(a+b)^2-c^2&=2ab\\a^2+2ab+b^2-c^2&=2ab\\a^2+b^2-c^2&=0\end{align*}$$

I have to admit that this proof is not as satisfying as I would have hoped. Discussion of this proof along with geometric proofs (provided by Blue) of the Half-Angle Formulas can be seen here (pay no attention to the votedown, some intolerant people still believe in the impossibility of trigonometric proofs of the Pythagorean Theorem).


References
$[1]$ A. Bogomolny, More Trigonometric Proofs of the Pythagorean Theorem, from Cut-the-Knot.org

domingo, 29 de noviembre de 2020

An Alternative Form of Bretschneider's Formula

If you are already familiar with Bretschneider's Formula, have you ever wonder how would it like if we interchange the cosine-part by a sine-part?

There are three different forms of expressing the Bretschneider's Formula in MathWorld. In this note we will give another one which is almost as simple as the original one.

Given a general convex quadrilateral with sides $a$, $b$, $c$ and $d$, its area is given by the formula

$$K=\sqrt{abcd\sin^2\left({\frac{\alpha+\gamma}{2}}\right)-s(s-c-d)(s-b-d)(s-b-c)}\tag{1},$$

where $s$ is the semiperimeter and  $\alpha$ and $\gamma$ are opposite angles.

The proof is based on the following unexpected simplification lemma.

Lemma 1. Given a general quadrilateral with sides $a$, $b$, $c$ and $d$, then

$$(s-a)(s-b)(s-c)(s-d)+s(s-c-d)(s-b-d)(s-b-c)=abcd,\tag{2}$$

where $s$ is the semiperimeter.

Proof. Let's focus on the left side of the identity. Substituting and rewriting as difference of squares,

$$\begin{align*}\frac{\left[(c+d)^2-(a-b)^2\right]\left[(a+b)^2-(c-d)^2\right]}{16}+\frac{\left[(a+b)^2-(c+d)^2\right]\left[(a-b)^2-(c-d)^2\right]}{16}&=\\\frac{(a+b)^2\left[(c+d)^2-(c-d)^2\right]+(a-b)^2\left[(c-d)^2-(c+d)^2\right]}{16}&=\\\frac{4cd(a+b)^2-4cd(a-b)^2}{16}&=abcd.\end{align*}$$

$\square$

Now, consider the original Bretschneider's Formula, 

$$K=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left(\frac{\alpha+\gamma}{2}\right)}.\tag{3}$$

Using the Pythagorean Identity $\sin^2{\left(\frac{\alpha+\gamma}{2}\right)}+\cos^2{\left(\frac{\alpha+\gamma}{2}\right)}=1$ in combination with $(2)$, you get $(1)$.

Remark. Assume $d=0$. Then $(2)$ reduces to 

$$s(s-a)(s-b)(s-c)+s(s-b)(s-c)(s-b-c)=0$$

From which we get the following alternative form of Heron's Formula:

$$K=\sqrt{-s(s-b)(s-c)(s-b-c)}$$

or

$$K=\sqrt{-s(s-a)(s-c)(s-a-c)}$$

$$K=\sqrt{-s(s-a)(s-b)(s-a-b)}$$

jueves, 26 de noviembre de 2020

La importancia de tener un amigo matemático

La mayoría de nosotros, enfrentados ante un problema matemático de esos de “cuánto tardarían dos trenes en chocar si uno sale de la estación a las tal y tal…”, pensamos en nuestra juventud que de qué demonios nos serviría eso en la vida, al menos si no queríamos ser ingenieros. Hasta a un servidor, que siempre le gustó la ciencia de los números, le parecía una pérdida de tiempo averiguar la hora del trágico desenlace sabiendo que alguna señal impediría que los convoyes ocuparan la misma vía al mismo tiempo. Para eso ya habría expertos, mientras que los de letras nos dedicábamos a otras cosas. Pero conforme crecemos, nos damos cuenta de que muchas de las situaciones mencionadas en los exámenes sí se presentan en nuestras vidas, y aunque no lo hagan, las matemáticas son necesarias para prácticamente cualquier carrera que decidamos seguir, aunque sea para calcular los impuestos que debamos pagar al insaciable estado o para que no nos hagan trampa en algún negocio. Algunos más listos, como los señores que inventaron Google, supieron sacarle jugo a los números, sus ecuaciones y algoritmos para forrarse superlativamente; otros aprendieron a usarlos para ganar la lotería, apostando a lo seguro.

Probablemente los nombres de Charles Marie de la Condamine y Francois-Marie Arouet no digan nada a la mayoría, pero las cosas cambian si os digo que este segundo utilizó en su vida decenas de seudónimos, entre ellos el de Voltaire, uno de los pensadores más célebres de la Ilustración y un crítico constante de las autoridades francesas, su carencias y abusos en las décadas previas a la Revolución. Por su parte, de la Condamine no alcanzó la fama de Voltaire en los libros de historia, pero sí en los de matemáticas y en los de geografía, a través de sus valiosas investigaciones en Sudamérica, patrocinadas, como no, por sus ganancias con la lotería. Esta había sido instituida en la segunda década del siglo XVIII con la intención de revalorizar los bonos del tesoro expedidos anteriormente y, de paso, conseguir fondos nuevos para el estado. El problema es que el encargado de diseñarla, Le Pelletier Desforts, Vice Ministro de finanzas, cometió un grave error.

Los futuros socios de la Condamine y Voltaire se conocieron en una cena cuando el segundo acababa de volver a París después de uno de sus múltiples exilios. Sus finanzas no eran muy boyantes y el matemático aprovechó para contarle el fallo que había encontrado en la lotería. Él no tenía problemas de dinero, pero quería probar su teoría y sabía que Voltaire contaba con ciertas conexiones que les ayudarían en la empresa. El truco era el siguiente: todos los dueños de bonos podían comprar un billete de la lotería, uno por bono, sin importar su valor, al precio de una milésima del valor del bono. El premio equivaldría al valor del bono, más 500.000 libras extras. Así, los que comparaban billetes a 1 libra con bonos de 1000 libras, tenían la misma oportunidad de ganar el premio que alguien que había comprado el billete a 100 libras por tener un bono de 100.000 libras. Lo único que tenían que hacer los socios era comprar la mayoría de los bonos más baratos, de los que había más, y así se aseguraban ganar un premio invirtiendo menos. No todo era fácil, pues se necesitaba un capital para empezar, y aquí fue donde las conexiones sociales de Voltaire entraban en juego, sino que los billetes podían comprarse exclusivamente a través de un notario. El filósofo no tuvo problemas en encontrar uno que aceptara ser parte del cartel.

Así, cada mes Voltaire acudía a su amigo el notario a comprar los billetes y poco después volvía a cobrar sus premios. El truco funcionó durante varios meses en los que todos los participantes se llevaron pingües beneficios a costa del estado, hasta que fueron descubiertos. Resulta que la gente de aquella época acostumbraba escribir mensajes en cada cupón de la lotería, normalmente deseándose suerte a sí mismos, pero Voltaire no pudo evitar burlarse del mismo gobierno que le estaba haciendo rico. En muchas ocasiones, escribió en sus billetes cosas como “¡A la buena idea de Marie de la Condamine!”, o incluso burlas directas, “¡Brindo por el Vice-Ministro de Finanzas!”. Este, que ya se había dado cuenta de que un mismo grupo de personas ganaba casi todos los sorteos, no tardó en descubrir su identidad. Desforts demandó a Voltaire y a de la Condamine por fraude, pero como en realidad no habían hecho nada ilegal, el tribunal los exoneró y pudieron quedarse lo ganado. Eso sí, el ministro tuvo la sensata idea de cancelar la lotería.

Charles Marie de la Condamine, miembro de la Academia de las Ciencias, utilizó sus ganancias para pagarse un par de expediciones a Sudamérica, donde confirmó la teoría de Isaac Newton de que la Tierra no era completamente esférica, sino ligeramente achatada y abultada a la altura del Ecuador. También fue el primero en conocer el caucho, y quien lo introdujo a Europa junto con un documento científico demostrando sus propiedades, y realizó el primer mapa topográfico basado en observaciones astronómicas de la Cuenca del Amazonas. Por su parte, Voltaire, rico de por vida con el medio millón de libras obtenido de la lotería, se dedicó a lo que mejor sabía hacer, criticar al gobierno y promover las libertades de expresión y religión y la separación de estado e iglesia en más de dos mil libros y folletos y 20.000 cartas que aún se preservan. De la Condamine y Voltaire probablemente no utilizaron los medios más éticos para enriquecerse, pero al menos dieron buen uso al dinero.


Fuente: Ciencia Histórica

The product AI*BI*CI

Consider a triangle $\triangle{ABC}$ and its Incenter, $I$. Denote $R$ and $r$ the circumradius and inradius, respectively. Also let $AI=k$; $BI=l$; $CI=m$. Then, the following identity holds

$$klm=4Rr^2$$


Proof
. We make use of the semiperimeter-half-angle formula, 

$$\cos^2{\frac{\gamma}{2}}= \frac{s(s-c)}{ab}\tag{1}$$ 

where $s$ is the semiperimeter and $\gamma$ denotes the angle $\angle{ACB}$. The proof for this formula can be found here.

Notice that $\cos{\frac{\gamma}{2}}=\frac{(s-c)}{m}$. Also, because of $(1)$ we have $\cos{\frac{\gamma}{2}}=\sqrt{\frac{s(s-a)}{ab}}$. Equating both expressions and solving for $m^2$, 


$$m^2=\frac{ab(s-c)}{s}$$

Similarly you get $k^2=\frac{bc(s-a)}{s}$ and $l^2=\frac{ac(s-b)}{s}$. Hence, 

$$(klm)^2=\frac{a^2b^2c^2(s-a)(s-b)(s-c)}{s^3}=\frac{a^2b^2c^2s(s-a)(s-b)(s-c)}{s^4}$$

Substituting from Heron's formula,

$$(klm)^2=\frac{a^2b^2c^2\Delta^2}{s^4}$$

Simplifying and using the well-known formulas $abc=4R\Delta$ and $\Delta=rs$ you get the desired result. 

$$klm=\frac{abc\Delta}{s^2}=\frac{4R\Delta^2}{s^2}=4Rr^2$$

sábado, 21 de noviembre de 2020

Still very Bretschneider, isn't it?

Given a general  convex quadrilateral with sides of lengths $a$, $b$, $c$, and $d$, the area is given by

$$\begin{align*} K&=\frac{1}{4}\sqrt{4p^2q^2-(b^2+d^2-a^2-b^2)^2}\tag{1}\\&=\sqrt{(s-a)(s-b)(s-c)(s-d)-\frac{1}{4}(ac+bd+pq)(ac+bd-pq)}\tag{2}\end{align*}$$

where $p$ and $q$ are the diagonal lengths and $s$ is the semiperimeter.

In MathWorld the American mathematician Julian Coolidge is credited with giving the second form of this formula, stating "here is one [formula] which, so far as I can find out, is new," while at the same time crediting Bretschneider and Strehlke with "rather clumsy" proofs of the related formula

$$\begin{align*}K&=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left(\frac{\alpha+\gamma}{2}\right)}\tag{3}\end{align*}$$

where $\alpha$ and $\gamma$ are two opposite angles of the quadrilateral.

The author of this note has realized that Coolidge's Formula is a easy consequence of Bretschneider's results (i.e., the Bretschneider's formula and Bretschneider's generalization of Ptolemy's Theorem).

In 1842 Bretschneider derived the following generalization of Ptolemy's theorem, regarding the product of the diagonals in a convex quadrilateral.

Theorem 1 (Bretschneider). Given a general  convex quadrilateral with sides of lengths $a$, $b$, $c$, and $d$, then

$$p^2q^2=a^2c^2+b^2d^2-2abcd\cos{(\alpha+\gamma})\tag{4}$$

where $p$ and $q$ are the diagonal lengths and $\alpha$ and $\gamma$ are two opposite angles of the quadrilateral.

We avoid proving Theorem 1; however, you can consult $[1]$ for that purpose. 

Using the cosine double angle formula and substituting in $(4)$,

$$\begin{align*}p^2q^2&=a^2c^2+b^2d^2-2abcd\left[2\cos^2{\left(\frac{\alpha+\gamma}{2}\right)}-1\right]\tag{5}\\&=(ac+bd)^2-4abcd\cos^2{\left(\frac{\alpha+\gamma}{2}\right)}\tag{6}\end{align*}$$

Substituting from Bretschneider's Formula, 

$$\begin{align*}p^2q^2&=(ac+bd)^2-4\left[K^2-(s-a)(s-b)(s-c)(s-d)\right]\tag{7}\end{align*}$$

Isolating $K$ and factorizing you get $(2)$. Still very Bretschneider, isn't it? Coolidge's proof can be found in $[2]$.


References
$[1]$ Andreescu, Titu & Andrica, Dorian, Complex Numbers from A to...Z, Birkhäuser, 2006, pp. 207–209.

$[2]$ Coolidge, J. L. "A Historically Interesting Formula for the Area of a Quadrilateral." Amer. Math. Monthly 46, 345-347, 1939.

sábado, 31 de octubre de 2020

Two Identities and their Consequences (draft)

Renowned British mathematician John Conway, in correspondence with Peter Doyle, used two trigonometric formulae to prove Heron's formula. John Casey, in his book "A Treatise of Plane Trigonometry" used two other trigonometric formulae to prove the Brahmagupta's formula. It turns out that the two trigonometric formulae used by Casey for a cyclic quadrilateral generalize the two used by Conway for a triangle. No one seems to have wondered if the two formulae used by Casey could be generalized to a general quadrilateral and use it to prove the Bretschneider's formula. The last link in this chain is precisely what my last article is about!

Article: Two Identities and their Consequences

Abstract. In this note we prove the Heron’s formula (although known, see Conway’s dicussion in $[7]$), the Brahmagupta’s formula (also known, see $[6]$) and the formula for the area of a bicentric quadrilateral (possibly new, see $[12, 13]$), $\sqrt{abcd}$, based on two lesser-known trigonometric formulae $[6, 16]$ involving sine, cosine, the semiperimeter and the side lenghts of a cyclic quadrilateral. Once the two trigonometric formulae have been established (and the necessary adjustments made), the proofs of these area theorems are greatly simplified. Furthermore, we present a generalization of the two aforementioned trigonometric formulae and use it to give an alternative proof of Bretschneider’s formula. Since all these area theorems can be derived from this new generalization, the approach presented in this note, unlike others, provides a more holistic view of these theorems. Our main result for a general convex quadrilateral are the identities
\[ad\sin^2{\frac{\alpha}{2}}+bc\cos^2{\frac{\gamma}{2}}=(s-a)(s-d)\]

and

\[bc\sin^2{\frac{\gamma}{2}}+ad\cos^2{\frac{\alpha}{2}}=(s-b)(s-c),\]
where $a$, $b$, $c$, $d$ are the sides lengths, $s$ is the semiperimeter, and  $\alpha$ and $\gamma$ are opposite angles.

Regarding the novelty of the results and proofs presented in this article, I have consulted Martin Josefsson (whom I consider an expert on these issues) and this was the message he sent me:

"Dear Emmanuel,
 
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.
 
I have seen (somewhere on the Internet) that proof of Heron's formula using half angle triangle formulas before - but not from your point of view of using cyclic quadrilateral formulas and setting one side = 0. These cyclic quadrilateral half angle formulas are, as you say, not so well known, but both them and your proof of Brahmagupta's formula can be found - more or less in the same way, but with fewer details - in Casey's 1888 book "A Treatise on Plane Trigonometry", see the attachment.
 
About the structure of the article I have not much to say, except perhaps to eliminate the penultimate step in the proofs of Theorems 1 and 5 where you explain how to introduce the semiperimeter in the formulas.
 
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.
 
Best regards,
Martin"

The identities $(4)$ and $(5)$ mentioned by Martin are now identities $(5)$ and $(6)$ in the present version of the paper. Martin's work on these issues can be found in the following link:


Below I've added an incomplete concept map of identities $(5)$ and $(6)$ so you can better appreciate the way they relate to other well-known identities.


Update. The article has been published by MATINF. See here.

lunes, 6 de julio de 2020

Generalization of two formulae and an alternative proof of Bretschneider's formula

"If we do not succeed in solving a mathematical problem, the reason frequently consists in our failure to recognize the more general standpoint from which the problem before us appears only as a single link in a chain of related problems. After finding this standpoint, not only is this problem frequently more accessible to our investigation, but at the same time we come into possession of a method which is applicable also to related problems."  David Hilbert
 

The following formulae generalize $(1)$ in my previous post Killing three birds with one stone. For implications in a triangle see also Proofs and applications of two well-known formulae involving sine, cosine and the semiperimeter of a triangle

Here, $a$, $b$, $c$, $d$ are the sides of a general convex quadrilateral, $s$ is the semiperimeter, and $\alpha$ and $\gamma$ are two opposite angles. Then



$$\sin^2{\frac{\alpha}{2}}=\frac{(s-a)(s-d)-bc\cos^2{\frac{\gamma}{2}}}{ad}\quad and \quad \cos^2{\frac{\alpha}{2}}=\frac{(s-b)(s-c)-bc\sin^2{\frac{\gamma}{2}}}{ad}\tag{1}$$

Proof. By the Law of Cosines,

$$a^2+d^2-2ad\cos{\alpha}=b^2+c^2-2bc\cos{\gamma}\tag{2}$$

Yielding $\cos{\alpha}=\frac{a^2+d^2-b^2-c^2+2bc\cos{\gamma}}{2ad}$. Now, making use of the half angle formula for cosine,

$$\begin{align*} \cos^2{\frac{\alpha}{2}}&=\frac{a^2+d^2+2ad-b^2-c^2+2bc\cos{\gamma}}{4ad}\tag{3}\\ &=\frac{a^2+d^2+2ad-b^2-c^2+2bc(1-2\sin^2{\frac{\gamma}{2}})}{4ad}\tag{4}\\&=\frac{(a+d)^2-(b-c)^2-4bc\sin^2{\frac{\gamma}{2}}}{4ad}\tag{5}\\&=\frac{(a+d+b-c)(a+d-b+c)-4bc\sin^2{\frac{\gamma}{2}}}{4ad}\tag{6}\\&=\frac{1}{ad}\left(\frac{a+b+c+d}{2}-c\right)\left(\frac{a+b+c+d}{2}-b\right)-\frac{bc\sin^2{\frac{\gamma}{2}}}{ad}\tag{7}\\&=\frac{(s-b)(s-c)-bc\sin^2{\frac{\gamma}{2}}}{ad}\tag{8}\end{align*}$$

$\square$

The other formula can be obtained similarly by replacing $\cos^2{\frac{\alpha}{2}}$ by $1 - \sin^2{\frac{\alpha}{2}}$ in $(3)$.

A proof of Bretschneider's formula
The formulae in $(1)$ can be rewritten as follows

$$ad\sin^2{\frac{\alpha}{2}}+bc\cos^2{\frac{\gamma}{2}}=(s-a)(s-d)\tag{9}$$

and

$$bc\sin^2{\frac{\gamma}{2}}+ad\cos^2{\frac{\alpha}{2}}=(s-b)(s-c)\tag{10}$$

Multiplying $(9)$ and $(10)$ we get

$$\begin{align*}\left(ad\sin^2{\frac{\alpha}{2}}+bc\cos^2{\frac{\gamma}{2}}\right)\left(bc\sin^2{\frac{\gamma}{2}}+ad\cos^2{\frac{\alpha}{2}}\right) &= (s-a)(s-b)(s-c)(s-d)\tag{11}\end{align*}$$

Expanding, factorizing, completing the squares and keeping in mind some well-known trigonometric identities, 

$$\begin{align*}abcd\cos^2\left({\frac{\alpha+\gamma}{2}}\right)+\left(ad\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}+bc\sin{\frac{\gamma}{2}}\cos{\frac{\gamma}{2}}\right)^2 &=(s-a)(s-b)(s-c)(s-d)\tag{12}\\abcd\cos^2\left({\frac{\alpha+\gamma}{2}}\right)+\left(\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{\gamma}}{2}\right)^2 &=(s-a)(s-b)(s-c)(s-d)\tag{13}
\end{align*}$$

Since the area of $ABCD$ can be expressed as the sum of the areas of $\triangle{ABD}$ and $\triangle{CBD}$, which in turn can be written as $\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{\gamma}}{2}$, then we are done.
$\square$

An alternative form of Bretschneider's formula
We encourage readers to prove the following formula for themselves.

Prove that the area of a general convex quadrilateral is given by the following formula:

$$K=\sqrt{abcd\sin^2\left({\frac{\alpha+\gamma}{2}}\right)-s(s-c-d)(s-b-d)(s-b-c)},$$

where $a$, $b$, $c$ and $d$ are the sides lengths, $s$ is the semiperimeter, and  $\alpha$ and $\gamma$ are opposite angles.

A concept map of identities $(9)$ and $(10)$
Below you can find a concept map of the identities $(9)$ and $(10)$ so you can see clearly what's going on here (click on the image to have a better view). 


It would be interesting to investigate whether identities $(9)$ and $(10)$ can be generalized to other geometries.

I have organized my ideas presented in this page (and related links) and put it in a draft paper which you can download here.

sábado, 4 de julio de 2020

Killing three birds with one stone

In this note we derive Heron's formula, Brahmagupta's formula and the bicentric quadrilateral's area formula, $\sqrt{abcd}$, from two formulae involving sine, cosine, semiperimeter and the side lenghts of a cyclic quadrilateral. 

Let $ABCD$ be a cyclic quadrilateral with $AB=a$, $BC=b$, $CD=c$, $DA=d$ and $s=\frac{a+b+c+d}{2}$. If $\angle{BAD}=\alpha$, then



$$\sin^2{\frac{\alpha}{2}}=\frac{(s-a)(s-d)}{ad+bc}\quad and \quad \cos^2{\frac{\alpha}{2}}=\frac{(s-b)(s-c)}{ad+bc}\tag{1}$$

or 

$$sin^2{\frac{\alpha}{2}}=\frac{(a+b+c-d)(-a+b+c+d)}{4(ad+bc)}\quad and \quad cos^2{\frac{\alpha}{2}}=\frac{(a+b-c+d)(a-b+c+d)}{4(ad+bc)}\tag{2}$$

Proof. First we will find an expression for $\cos{\alpha}$ in terms of $a$, $b$, $c$ and $d$. Let $\angle{BCD}=\gamma$. By the Law of Cosines and keeping in mind that $\alpha$ and $\gamma$ are supplementary, we have

$$a^2+d^2-2ad\cos{\alpha}=b^2+c^2-2bc\cos{(180^\circ-\alpha)}\tag{3}$$

Yielding $\cos{\alpha}=\frac{a^2+d^2-b^2-c^2}{2(ad+bc)}$. Now, making use of the half angle formula for cosine,

$$\begin{align*} \cos^2{\frac{\alpha}{2}}&=\frac{2ad+2bc+a^2+d^2-b^2-c^2}{4(ad+bc)}\tag{4}\\ &=\frac{(a+d)^2-(b-c)^2}{4(ad+bc)}\tag{5}\\&=\frac{(a+b-c+d)(a-b+c+d)}{4(ad+bc)}\tag{6}\\&=\frac{1}{ad+bc}\left(\frac{a+b+c+d}{2}-c\right)\left(\frac{a+b+c+d}{2}-b\right)\tag{7}\\&=\frac{(s-b)(s-c)}{ad+bc}\end{align*}$$

$\square$

The other formulae can be obtained similarly by replacing $\cos^2{\frac{\alpha}{2}}$ by $1 - \sin^2{\frac{\alpha}{2}}$.

A generalization of $(1)$ together with a proof of Bretschneider's formula can be found here.

Remark$(1)$ appears as exercise 400 in V. Panagiotis' 1000 General Trigonometry Exercises, Volume B.

A proof of Heron's Formula
For a triangle, if in $(1)$ we assume $c=0$, then we have

$$\sin^2{\frac{\alpha}{2}}=\frac{(s-a)(s-d)}{ad}\quad and \quad \cos^2{\frac{\alpha}{2}}=\frac{s(s-b)}{ad}\tag{8}$$


Let $\Delta_0$ be the area of $\triangle{ABD}$. Making use of the double-angle identity for sine we have

$$\sin{\alpha}=2\sqrt{\frac{s(s-b)}{ad}}\sqrt{\frac{(s-a)(s-d)}{ad}}=2\frac{\sqrt{s(s-a)(s-b)(s-d)}}{ad}\tag{9}$$

Since $\Delta_0=\frac{ad\sin{\alpha}}{2}$, it follows 

$$\Delta_0=\sqrt{s(s-a)(s-b)(s-d)}\tag{10}$$

$\square$

For more implications in a triangle see here.

A proof of the Brahmagupta's formula
Denote $\Delta_1$ the area of the cyclic quadrilateral, $ABCD$. Then

$$\begin{align*}\Delta_1&=\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{(180^\circ-\alpha)}}{2}\tag{11}\\&=\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{\alpha}}{2}\tag{12}\\&=\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}(ad+bc)\tag{13}\\&=\sqrt{\frac{(s-a)(s-d)}{ad+bc}}\sqrt{\frac{(s-b)(s-c)}{ad+bc}}(ad+bc)\tag{14}\\&=\sqrt{(s-a)(s-b)(s-c)(s-d)}\tag{15}\end{align*}$$

$\square$

A proof of the bicentric quadrilateral's area formula
Since $a+c=b+d$ in a bicentric quadrilateral, the formulae in $(2)$ reduce to 

$$sin^2{\frac{\alpha}{2}}=\frac{bc}{ad+bc}\quad and \quad cos^2{\frac{\alpha}{2}}=\frac{ad}{ad+bc}\tag{16}$$

Assume $ABCD$ is a bicentric quadrilateral and let $\Delta_2$ be its area, then

$$\begin{align*}\Delta_2&=\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{(180^\circ-\alpha)}}{2}\tag{17}\\&=\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{\alpha}}{2}\tag{18}\\&=\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}(ad+bc)\tag{19}\\&=\sqrt{\frac{bc}{ad+bc}}\sqrt{\frac{ad}{ad+bc}}(ad+bc)\tag{20}\\&=\sqrt{abcd}\tag{21}\end{align*}$$

$\square$

sábado, 20 de junio de 2020

Proofs and applications of two well-known formulae involving sine, cosine and the semiperimeter of a triangle

In Cut-the-knot's Relations between various elements of a triangle, the formulae (a generalization can be found here)

$$\sin^2{\frac{\gamma}{2}} = \frac{(s-a)(s-b)}{ab}\quad and\quad\cos^2{\frac{\gamma}{2}}= \frac{s(s-c)}{ab}$$ 

are derived using Heron's formulaHere we give an alternative proof without using Heron's Formula and we demonstrate several well-known theorems based on these formulae as a sample of its power. We will be using standard notation: $BC=a$, $AC=b$, $AB=c$, $\Delta$ for the area, $s$ for the semiperimeter, $R$ for the circumradius and $r$ for the inradius. Let $D$, $E$ and $F$ be the contact points of the incircle with $AC$, $AB$ and $BC$, respectively. Also, let $AE=AD=x$; $BE=BF=y$; $CD=CF=z$.



Notice that $\frac{\cot{\frac{\gamma}{2}}}{s-c} = \frac{1}{r}$. Also, We know $Δ = rs$ and $Δ =\frac {ab\sin{\gamma}}{2}$, hence

$$\frac{\cot{\frac{\gamma}{2}}}{s-c}=\frac{1}{r}=\frac{s}{\Delta}=\frac{2s}{ab\sin{\gamma}}$$

But, $\sin{\gamma} = 2\sin{\frac{\gamma}{2}}\cos{\frac{\gamma}{2}}$, so 
 
$$\frac{\cos{\frac{\gamma}{2}}}{\sin{\frac{\gamma}{2}}}\cdot{\sin{\frac{\gamma}{2}}\cos{\frac{\gamma}{2}}} = \frac{s(s-c)}{ab}$$

from which we get $\cos^2{\frac{\gamma}{2}} = \frac{s(s-c)}{ab}$.
 
The other formula can be obtained replacing $\cos^2{\frac{\gamma}{2}}$ by $1 - \sin^2{\frac{\gamma}{2}}$. Indeed, 

$$\begin{aligned} 1-\sin^2{\frac{\gamma}{2}} &= \frac{s(s-c)}{ab} \\ \sin^2{\frac{\gamma}{2}} &= 1-\frac{s(s-c)}{ab} \\  &= \frac{\left(ab-s(s-c)\right)}{ab} \\ &= \frac{\left((y+z)(x+z)-(x+y+z)(z)\right)}{ab}\\ &= \frac{xy}{ab} \\ &=\frac{(s-a)(s-b)}{ab} \end{aligned}$$

$\square$

Purely geometrical proofs can be found at Trigonography.com.

1. A proof of Heron's Formula
Making use of the formulae proven above and the double angle identity for sine we have

$$\sin{\gamma}=2\sqrt{\frac{s(s-c)}{ab}}\sqrt{\frac{(s-a)(s-b)}{ab}}=2\frac{\sqrt{s(s-a)(s-b)(s-c)}}{ab}$$

Since $\Delta=\frac{ab\sin{\gamma}}{2}$, it follows 

$$\Delta=\sqrt{s(s-a)(s-b)(s-c)}$$

$\square$

2. A proof of the Law of Cosines
Since $(s-a)=x$, $(s-b)=y$ and $(s-c)=z$, then the following identity holds:
$$ab\cos{\gamma}=ab\cos^2{\frac{\gamma}{2}}-ab\sin^2{\frac{\gamma}{2}}=s(s-c)-(s-a)(s-b)$$
Substituting and multiplying by 4, 
$$4ab\cos{\gamma}=(a+b+c)(a+b-c)-(b+c-a)(a+c-b)$$
Expanding and Simplifying,
$$2ab\cos{\gamma}=a^2+b^2-c^2$$
$\square$

A similar reasoning must show that $a^2=b^2+c^2-2bc\cos{\alpha}$ and $b^2=a^2+c^2-2ac\cos{\beta}$.


3. Proofs for some trigonometric identities associated to a triangle
a) $\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\alpha}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}=1$.

As a consequence of the formulae proven at the beginning of the note,

$$\tan{\frac{\alpha}{2}}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}, \quad\tan{\frac{\beta}{2}}=\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}\quad and \quad \tan{\frac{\gamma}{2}}=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$$

 So, by canceling and simplifying you get

$$\begin{aligned}\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\alpha}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}} &=\frac{s-c}{s}+\frac{s-b}{s}+\frac{s-a}{s}\\ &=\frac{z+y+x}{s}\\ &=\frac{s}{s}=1\end{aligned}$$

$\square$

b) $r=4R\sin{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}$.

We make use of the well-known relationship $abc=4R\Delta$ (see here for a proof) and Heron's Formula.

$$\begin{aligned}r&=4R\sin{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}\\ &=4R\sqrt{\frac{(s-b)(s-c)}{bc}}\sqrt{\frac{(s-a)(s-c)}{ac}}\sqrt{\frac{(s-a)(s-b)}{ab}}\\&=4R\sqrt{\frac{(s-a)^2(s-b)^2(s-c)^2}{a^2b^2c^2}}\\&=4R\sqrt{\frac{\frac{\Delta^4}{s^2}}{a^2b^2c^2}}\\&=4R\sqrt{\frac{\frac{\Delta^4}{s^2}}{16R^2\Delta^2}}\\&=\frac{\Delta}{s}\\&=\frac{rs}{s}\\&=r\end{aligned}$$

$\square$

c) $s=4R\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}$.

$$\begin{aligned}s&=4R\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}\\ &=4R\sqrt{\frac{s(s-a)}{bc}}\sqrt{\frac{s(s-b)}{ac}}\sqrt{\frac{s(s-c)}{ab}}\\&=4R\sqrt{\frac{s^2\Delta^2}{a^2b^2c^2}}\\&=4R\frac{s\Delta}{abc}\\&=4R\frac{s\Delta}{4R\Delta}\\&=s\end{aligned}$$

$\square$


Consequently, the following relationship also holds

$$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}=\frac{r}{s}$$

or

$$\cot{\frac{\alpha}{2}}\cot{\frac{\beta}{2}}\cot{\frac{\gamma}{2}}=\frac{s}{r}$$

d) $\cot{\frac{\alpha}{2}}\cot{\frac{\beta}{2}}\cot{\frac{\gamma}{2}}=\cot{\frac{\alpha}{2}}+\cot{\frac{\beta}{2}}+\cot{\frac{\gamma}{2}}$.

To prove the above identity we will show that the right hand side equals $\frac{s}{r}$.

$$\begin{aligned}\frac{s}{r}&= \cot{\frac{\alpha}{2}}\cot{\frac{\beta}{2}}\cot{\frac{\gamma}{2}}=\cot{\frac{\alpha}{2}}+\cot{\frac{\beta}{2}}+\cot{\frac{\gamma}{2}}\\&=\frac{\sqrt{s(s-a)}}{\sqrt{(s-b)(s-c)}}+\frac{\sqrt{s(s-b)}}{\sqrt{(s-a)(s-c)}}+\frac{\sqrt{s(s-c)}}{\sqrt{(s-a)(s-b)}}\\&=\frac{\Delta(s-a)+\Delta(s-b)+\Delta(s-c)}{(s-a)(s-b)(s-c)}\\&=\frac{\Delta(x+y+z)}{xyz}\\&=\Delta\frac{s^2}{\Delta^2}\\&=\frac{s^2}{rs}\\&=\frac{s}{r}\end{aligned}$$

$\square$


We invite the reader to prove the following identity (possibly new) on their own.
$$\frac{\cot{\frac{\alpha}{2}}\cot{\frac{\beta}{2}}+\cot{\frac{\alpha}{2}}\cot{\frac{\gamma}{2}}+\cot{\frac{\beta}{2}}\cot{\frac{\gamma}{2}}}{(s-a)(s-b)+(s-a)(s-c)+(s-b)(s-c)}=\frac{1}{r^2}$$


4. The product $AI\cdot{BI}\cdot{CI}$
Consider a triangle $\triangle{ABC}$ and its Incenter, $I$. Denote $R$ and $r$ the circumradius and inradius, respectively. Also let $AI=k$; $BI=l$; $CI=m$. Then, the following identity holds

$$klm=4Rr^2$$

Proof. We make use of the half-angle formula, 

$$\cos^2{\frac{\gamma}{2}}= \frac{s(s-c)}{ab}$$
 
Notice that $\cos{\frac{\gamma}{2}}=\frac{(s-c)}{m}$. Also, because of half-angle formula we have $\cos{\frac{\gamma}{2}}=\sqrt{\frac{s(s-a)}{ab}}$. Equating both expressions and solving for $m^2$, 

$$m^2=\frac{ab(s-c)}{s}$$

Similarly you get $k^2=\frac{bc(s-a)}{s}$ and $l^2=\frac{ac(s-b)}{s}$. Hence, 

$$(klm)^2=\frac{a^2b^2c^2(s-a)(s-b)(s-c)}{s^3}=\frac{a^2b^2c^2s(s-a)(s-b)(s-c)}{s^4}$$

Substituting from Heron's formula,

$$(klm)^2=\frac{a^2b^2c^2\Delta^2}{s^4}$$

Simplifying and using the well-known formulas $abc=4R\Delta$ and $\Delta=rs$ you get the desired result. 

$$klm=\frac{abc\Delta}{s^2}=\frac{4R\Delta^2}{s^2}=4Rr^2$$

See also

viernes, 5 de junio de 2020

Yet Another Proof of the Law of Cosines

The law of cosines relates the lengths of the sides of a triangle to the cosine of one of its angles. Using standard notation, the law of cosines states

$$c^2=a^2+b^2-2ab\cos{\gamma},$$

where $\gamma$ denotes the angle contained between sides of lengths $a$ and $b$ and opposite the side of length $c$. For the same figure, the other two relations are analogous:

$$a^2=b^2+c^2-2ac\cos{\alpha},$$
$$b^2=a^2+c^2-2ac\cos{\beta}.$$

Proof. Let $D$, $E$ and $F$ be the contact points of the incircle with $AC$, $AB$ and $BC$, respectively. Also, let $AE=AD=x$; $BE=BF=y$; $CD=CF=z$. We start from two well-known relationships of a triangle: $$\sin^2{\frac{\gamma}{2}}=\frac{(s-a)(s-b)}{ab} \qquad\text{and}\qquad \cos^2{\frac{\gamma}{2}}=\frac{s(s-c)}{ab}$$  
(See Cut-the-knot's Relations between various elements of a triangle for proofs), where $s$ denotes the semiperimeter of $\triangle{ABC}$. Since $(s-a)=x$, $(s-b)=y$ and $(s-c)=z$, then the following identity holds:
$$ab\cos{\gamma}=ab\cos^2{\frac{\gamma}{2}}-ab\sin^2{\frac{\gamma}{2}}=sz-xy$$
Substituting and multiplying by 4, 
$$4ab\cos{\gamma}=(a+b+c)(a+b-c)-(b+c-a)(a+c-b)$$
Simplifying,
$$2ab\cos{\gamma}=a^2+b^2-c^2$$
$\square$

A similar reasoning must show that $a^2=b^2+c^2-2bc\cos{\alpha}$ and $b^2=a^2+c^2-2ac\cos{\beta}$.

AcknowledgementMy sincerest thanks to Angina Seng for giving helpful comments which allowed me to simplify the proof.

Related material.

miércoles, 20 de mayo de 2020

Another Simple Proof of Johnson's Theorem

Other proofs can be found in cut-the-knot.org. See also Johnson's Three Circles Theorem Revisited and Johnson's theorem proof.

Johnson's theorem: Let three equal circles with centers $J_a$, $J_b$, and $J_c$ intersect in a single point $H$ and intersect pairwise in the points $A$, $B$, and $C$. Then the circumcircle of the triangle $\triangle{ABC}$ is congruent to the original three.


Proof. Since $\odot{ABH}$ and $\odot{BCH}$ are congruent and as $\angle{BAH}$ and $\angle{BCH}$ are angles subtended by the same arc, it follows that $\angle{BAH}=\angle{BCH}$. Analogously, $\angle{CAH}=\angle{CBH}$. Let $O$, $J_a$ be the centers of $\odot{ABC}$ and $\odot{BCH}$, respectively, then

$$\angle{COB}=2\angle{CAB}=2\angle{CAH}+2\angle{BAH}=2\angle{CBH}+2\angle{BCH}=\angle{BJ_aC}.$$

It follows that $\triangle{BCO}$ and $\triangle{BCJ_a}$ are similar isosceles triangles sharing a common side, $BC$, so by $ASA$ we deduce that $\triangle{BCO}\cong{\triangle{BCJ_a}}$. Consequently, $\odot{ABC}$ is congruent to the original three. 

$\square$

Note: The point $H$ may cross the side lines of the triangle $\triangle{ABC}$ in points either interior or exterior to the sides. The reasoning in cases other than that considered above requires only minor adjustments.

sábado, 9 de mayo de 2020

The Pythagorean Theorem by Reductio ad Absurdum

Other proofs by contradiction can be found in cut-the-knot.org (see proof #122) and in Loomis' collection (see proofs # 16/2 and # 32).

Let $\triangle{ABC}$ be a right-triangle with $\angle{ACB}=90^\circ$. Let $D$, $E$ and $F$ be the contact points of the incircle with $BC$, $AC$ and $AB$, respectively. Also, let $AE=AF=x$; $BD=BF=y$; $CD=CE=r$, where $r$ is the inradius of $\triangle{ABC}$.




Assume to the contrary that $a^2+b^2>c^2$. Then, 

$$(r+y)^2+(x+r)^2>(x+y)^2.$$
Expanding, collecting like terms and simplifying we get
$$ry+rx+r^2>xy.$$

Notice that $ry+rx+r^2=r(y+x+r)=rs=\Delta$, where $\Delta$ denotes the area of $\triangle{ABC}$ and $s$ its semiperimeter. Moreover, it is well-known that $xy=\Delta$. So $ry+rx+r^2>xy$ is equivalent to write $\Delta>\Delta$, which is a contradiction. A similar situation arise if you assume $a^2+b^2<c^2$.

There is a simple direct proof starting from $\Delta=rs$ and $\Delta=xy$, can you find it? My sincerest thanks to Andrius Navas and José Hernández for pointing me out this. 

jueves, 7 de mayo de 2020

Yet Another Proof of the Pythagorean Theorem - Simpler version

In cut-the-knot.org you can find more than a hundred proofs of the Pythagorean theorem. Here I give another proof which I hope to be new.

Let $\triangle{ABC}$ be a right-triangle with $\angle{ACB}=90^\circ$. Below I will be using standard notations: $a$, $b$, $c$, for the side length, $s$ for the semiperimeter, $\Delta$ for the area and $r$ for the inradius.

$$s=c+r$$
$$s-\frac{c}{2}=\frac{c}{2}+r$$
$$\frac{a}{2}+\frac{b}{2}=\frac{c}{2}+r$$
$$a+b=c+2r$$

Squaring both sides, 

$$a^2+2ab+b^2=c^2+4cr+4r^2$$

It is well-known $\Delta=rs$, which implies $2ab=4r(c+r)$. So the equation $a^2+2ab+b^2=c^2+4cr+4r^2$ can be rewritten like this $a^2+b^2=c^2$, which is the Pythagorean theorem.

Below is the comment by John Molokach, whom I consider an expert on this topic (he has proved the theorem in more than ten different ways).



miércoles, 6 de mayo de 2020

Yet Another Proof of the Pythagorean Theorem

In cut-the-knot.org you can find more than a hundred proofs of the Pythagorean theorem. Here we give another proof which we hope to be new.


Let $\triangle{ABC}$ be a right-triangle with $\angle{ACB}=90^\circ$. Denote $D$, $E$ the contact points of the incircle and sides $BC$ and $AC$, respectively. Below I shall be using standard notations: $a$, $b$, $c$, for the side length, $p$ for the semiperimeter and $r$ for the inradius.

$$p=c+r$$
$$p-\frac{c}{2}=\frac{c}{2}+r$$
$$\frac{a}{2}+\frac{b}{2}=\frac{c}{2}+r$$
$$a+b=c+2r$$

Squaring both sides, 

$$a^2+2ab+b^2=c^2+4cr+4r^2$$

Construct a parallel line, $l$, to $AB$ passing through $I$, the incenter of $\triangle{ABC}$. Call $H$ and $G$ the orthogonal projections of $B$ and $A$ onto $l$, respectively. Then, $[ABC]=[CDIE]+[ABHG]$. Indeed, let $K$, $J$ be the intersections of $l$ with $AC$, $BC$, respectively. Notice that $[ABC]=[ABJK]+[CDIE]+[EIK]+[DJI]$. Clearly, $\triangle{AGK}\cong{\triangle{EIK}}$ and $\triangle{BHJ}\cong{\triangle{DJI}}$, hence $[ABC]=[CDIE]+[ABGH]=r^2+cr$. As a consequence, $2ab=4r^2+4cr$. Now, the equation $a^2+2ab+b^2=c^2+4cr+4r^2$ can be rewritten like this $a^2+b^2=c^2$, which is the Pythagorean theorem.

$\square$

I realized the construction was not necessary, so I have published a simpler version of this proof here.

lunes, 4 de mayo de 2020

Galileo's Genius: The Experiment That Never Was.

Stolen from Clueless Fundatma.

Many of us have this mental picture of Galileo standing on top of the Leaning Tower of Pisa and dropping a feather and stone to watch them fall on the ground simultaneously, thus upsetting the Aristotelian assertion that "heavier objects fall faster".


Of course, we know Galileo couldn't possibly have done that experiment, because if he actually had, the presence of air would have reaffirmed the Aristotelian world-view, and the feather would have fallen later.

So what exactly did Galileo do?

I haven't read his original work, but my PhD advisor - Ron Larson - told us this charming story during a class I took with him, and its stuck with me since.

Apparently Galileo did a thought experiment.

He used the idea of "reductio ad absurdum", which begins by assuming something that we wish to prove as false, and hitting a contradiction.

The proof is beautiful, and here is a rough outline.

Consider two objects, with the heavier object of mass M, and the lighter object of mass m. Let us assume that "heavier objects fall faster" - which is opposite of what we wish to prove.

Further consider a really tall tower. If you can think of an infinitely tall tower, that is even better.

Now we've assumed that mass M falls faster than mass m. Here's the beautiful construction: Consider tying the two masses with a piece of string - to create a new object of mass M + m, as shown below.


Let us drop this composite object from the top of our tall tower, and think about what may happen. Because of what we've assumed, the mass M is going to fall faster than m, and in a short while the composite object (that is still falling, remember) will have adopted a configuration that looks more like the following.


Let us now consider the tension inside the string (red line in my diagram). The big mass tugs on the small mass, while the small mass tugs back.

Thus, the small mass slows the big mass down.

Consequently, the composite object of mass (M+m) falls slower than the unfettered object of mass M. But the composite object is heavier, and should fall faster.

Reductio ad absurdum!

Reductio ad absurdum or the contrapositive?

Response by Joel David Hamkins in mathoverflow.
Although the other answers correctly explain the basic logical equivalence of the two proof methods, I believe an important point has been missed:

With good reason, we mathematicians prefer a direct proof of an implication over a proof by contradiction, when such a proof is available. (all else being equal)

"What is the reason? The reason is the fecundity of the proof, meaning our ability to use the proof to make further mathematical conclusions. When we prove an implication (p implies q) directly, we assume p, and then make some intermediary conclusions r1, r2, before finally deducing q. Thus, our proof not only establishes that p implies q, but also, that p implies r1 and r2 and so on. Our proof has provided us with additional knowledge about the context of p, about what else must hold in any mathematical world where p holds. So we come to a fuller understanding of what is going on in the p worlds."

Similarly, when we prove the contrapositive (¬q implies ¬p) directly, we assume ¬q, make intermediary conclusions r1, r2, and then finally conclude ¬p. Thus, we have also established not only that ¬q implies ¬p, but also, that it implies r1 and r2 and so on. Thus, the proof tells us about what else must be true in worlds where q fails. Equivalently, since these additional implications can be stated as (¬r1 implies q), we learn about many different hypotheses that all imply q.

These kind of conclusions can increase the value of the proof, since we learn not only that (p implies q), but also we learn an entire context about what it is like in a mathematial situation where p holds (or where q fails, or about diverse situations leading to q).

With reductio, in contrast, a proof of (p implies q) by contradiction seems to carry little of this extra value. We assume p and ¬q, and argue r1, r2, and so on, before arriving at a contradiction. The statements r1 and r2 are all deduced under the contradictory hypothesis that p and ¬q, which ultimately does not hold in any mathematical situation. The proof has provided extra knowledge about a nonexistent, contradictory land. (Useless!) So these intermediary statements do not seem to provide us with any greater knowledge about the p worlds or the q worlds, beyond the brute statement that (p implies q) alone.

I believe that this is the reason that sometimes, when a mathematician completes a proof by contradiction, things can still seem unsettled beyond the brute implication, with less context and knowledge about what is going on than would be the case with a direct proof.

Edit: For an example of a proof where we are led to false expectations in a proof by contradiction, consider Euclid's proof that there are infinitely many primes. In a common proof by contradiction, one assumes that p1, ..., pn are all the primes. It follows that since none of them divide the product-plus-one p1...pn+1, that this product-plus-one is also prime. This contradicts that the list was exhaustive. Now, many beginner's falsely expect after this argument that whenever p1, ..., pn are prime, then the product-plus-one is also prime. But of course, this isn't true, and this would be a misplaced instance of attempting to extract greater information from the proof, misplaced because this is a proof by contradiction, and that conclusion relied on the assumption that p1, ..., pn were all the primes. If one organizes the proof, however, as a direct argument showing that whenever p1, ..., pn are prime, then there is yet another prime not on the list, then one is led to the true conclusion, that p1...pn+1 has merely a prime divisor not on the original list.

viernes, 24 de abril de 2020

Selectivo Argentina Cono Sur - Problema 3

Sea $\triangle{ABC}$  un triángulo e $I$ el punto de intersección de sus bisectrices. Sea $\tau$ la circunferencia con centro $I$ que es tangente a los tres lados del triángulo y sean $D$ en $BC$ y $E$ en $AC$ los puntos de tangencia de $\tau$ con $BC$ y $AC$. Sean $M$ y $N$ los puntos medios de $BC$ y $AB$, respectivamente. Demostrar que $AI$, $DE$ y $MN$ concurren en un punto común. 


Demostración. Supongamos que $ED$ y $MN$ se cortan en $P$. Si llamamos $X$ a la intersección de $EF$ con $MN$, donde $F$ es el punto de contacto de $\tau$ con $AB$, sabemos que $BXFD$ es cíclico y $NX=NB$ (ver demostración aquí). Claramente $\triangle{FMX}\sim{\triangle{AEF}}$ con $\angle{FXP}=\angle{AFE}=\angle{FDP}$, implicando que $P$ yace sobre $(BXFD)$. Al ser $\triangle{NXB}$ isósceles, $\angle{PFB}=\angle{PXB}=\frac{180^\circ-\angle{ACB}}{2}$. Dicho esto, no es difícil darse cuenta que $FP=EP$, significando que $PFAE$ es un deltoide y, como consecuencia, $AP$ es la bisectriz de $\angle{FAE}$. Esto demuestra que $MN$, $ED$ y $AI$ son concurrentes y hemos terminado.

jueves, 23 de abril de 2020

Un lema sobre segmentos congruentes

Considera un triángulo $\triangle{ABC}$. Los puntos $E$ y $F$ son los puntos donde el incírculo toca los lados $AB$ y $AC$, respectivamente. $M$ y $N$ son los puntos medios de los lados $AB$ y $BC$, respectivamente. Llamemos $P$ es la intersección de $EF$ y $MN$.

Lema.
a) $BN=CN=NP$.
b) $CP$ es la bisectriz de $\angle{ACB}$.
c) $BPEID$ es cíclico ($I$ es el incentro de $\triangle{ABC}$).



Demostración a). Denotemos con $s$ el semiperímetro de $\triangle{ABC}$. Entonces tenemos

$$s=AE+BC$$
$$\frac{AB}{2}+\frac{BC}{2}+\frac{AC}{2}=AE+BC$$
$$\frac{AB}{2}+\frac{AC}{2}-AE=\frac{BC}{2}$$

Pero $\frac{AB}{2}-AE=EM$ y $\frac{AC}{2}=MN$. Además, es fácil notar que $\triangle{AEF}$ y $\triangle{EMP}$ son triángulos isósceles semejantes con $EM=MP$, de modo que  $\frac{AB}{2}+\frac{AC}{2}-AE=\frac{BC}{2}$ puede reescribirse como $MP+MN=\frac{BC}{2}$ y hemos terminado.

Demostración b)Note que $AC$ es paralela con $MN$ y $\triangle{PNC}$ es isósceles, de donde resulta que  $\angle{NCP}=\angle{NPC}=\frac{180^\circ-(180-\angle{ACB})}{2}=\frac{\angle{ACB}}{2}$. 

Demostración c). El cuadrilátero $BEID$ es cíclico puesto que es un deltoide recto. Note que como consecuencia de a) y b) $\angle{EPI}=\angle{EPM}-\frac{\angle{ACB}}{2}=\frac{180^\circ-\angle{BAC}}{2}-\frac{\angle{ACB}}{2}=\frac{\angle{ABC}}{2}$. Pero $\angle{EBI}=\frac{\angle{ABC}}{2}$, por lo tanto, $BPEID$ es cíclico.

sábado, 18 de abril de 2020

Another Tran Viet Hung's problem




Lemma 1. Let $CD$ cuts circle $(D, DB)$ in $G$. Then, $BGIC$ is cyclic.

Proof. Notice that $\angle{BDC}=\angle{BAC}$, so 

$$\angle{GBI}=\frac{\angle{ABC}}{2}-(\angle{DBG}-\angle{DBA})=$$ 
$$\angle{GBI}=\frac{\angle{ABC}}{2}-(90^\circ-\frac{\angle{BAC}}{2})+\angle{DCA}.$$

Replacing $90^\circ$ by $\frac{\angle{ABC}}{2}+\frac{\angle{BAC}}{2}+\frac{\angle{BCA}}{2}$ and simplifying, we get $\angle{GBI}=\angle{DCA}-\frac{\angle{BCA}}{2}$. But $\angle{DCA}=\angle{BCA}-\angle{BCD}$, thus, $\angle{GBI}=\frac{\angle{BCA}}{2}-\angle{BCD}$. Moreover, $\angle{DCI}=\angle{GCI}=\frac{\angle{BCA}}{2}-\angle{BCD}$, hence,$\angle{GBI}=\angle{GCI}$ implying $BGIC$ is cyclic.


$\square$

Lemma 2. $IE=IG$.

Proof. Notice that $\angle{BEG}=\frac{\angle{BDG}}{2}=\frac{\angle{BAC}}{2}=\angle{BAI}$ which means that $EG\parallel{AI}$ implying that $ID\perp{EG}$ and as $DI$ passes through the center of circle $(D, DB)$ we deduce $DI$ is the perpendicular bisector of $EG$ meaning $IE=IG$.

$\square$

Back to our main problem.

$$\angle{IEG}=\angle{IGE}=360^\circ-\angle{EGF}-\angle{DGB}-\angle{BGC}-\angle{CGI}.$$

Combining lemmas 1 and 2 we get  $\angle{EGF}=\frac{\angle{BAC}}{2}+\angle{DCA}$; $\angle{DGB}=90^\circ-\frac{\angle{BAC}}{2}$; $\angle{BGC}=\angle{BIC}=180^\circ-\frac{\angle{ABC}}{2}-\frac{\angle{BCA}}{2}$; $\angle{CGI}=\frac{\angle{ABC}}{2}$.

Replacing and simplifying we get $\angle{IEG}=90^\circ+\angle{BCD}-\frac{\angle{BCA}}{2}$. Finally, $\angle{FEI}+\angle{FCI}=180^\circ+\angle{BCD}-\frac{\angle{BCA}}{2}+\frac{\angle{BCA}}{2}-\angle{BCD}=180^\circ$ and we are done.


domingo, 12 de abril de 2020

Solution to problem 4975 in Romantics of Geometry

This problem was proposed by Tran Viet Hung (Vietnam). Here is my proof. 



Proof for a). 


Perform an inversion around circle, $w$, centred at $A$ with radius $AE=AD$. The incircle is fixed since it is orthogonal with $w$. Touch points $E$ and $D$ are also fixed since they are on the circumference of $w$. The inverted image of $I$ is the intersection of $AI$ with $ED$, denoted $I'$. $B$ and $C$ are transformed into $B'$ and $C'$ on lines $AB$ and $AC$, respectively. The circumcircle $(ABC)$ is sent to the line passing through $B'$, $C'$. Line $BC$ is sent to the circle $(AB'C')$ and is tangent to the incircle at $P'$, the inverted image of $P$. The point $M$ is sent to $M'$, the intersection of $AI$ and $B'C'$. The line $MP$ is sent to the circle passing through $A$, $M'$ and $P'$. Finally, the inverted image of $Q$ is the second intersection of line $B'M'$ with $(AM'P')$ (See the inverted diagram above).

As angles are preserved under inversion, we want to show that $Q'$ lies on $E'D'$, taking advantage of the fact that $\angle{AI'E'}=90^\circ$. Let's supose that $E'D'$ cuts $B'C'$ in $Q''$. It suffices to show $AM'P'Q''$ is cyclic. Indeed, if $N$ is the intersection of $AI'$ with arc $B'C'$,  then, $N$ is the midpoint of arc $B'C'$, and it is known that $Q''$, $P'$ and $N$ are collinear (notice that the incircle turns out to be the $A$-mixtilinear excircle of $\triangle{AB'C'}$). A simple angle chase shows that $\angle{M'AP'}=\angle{M'Q''P'}$, implying $AM'P'Q''$ is cyclic and that $Q''=Q'$, as desired.

$\square$


Proof for b).

Under the same inversion we perfom in part a), the $A$-mixtilinear excircle is sent to the incircle of $\triangle{AB'C'}$ which touches $B'C'$ in $X''$, the inverted image of $X'$. The intersection of $AI$ with $BC$ is sent to $N$, the same midpoint of arc $B'C'$ (not containing $A$) we described in part a). It suffices to show that $Q'NX''A$ is cyclic. Indeed, we proved already that $Q'$ lies on $B'C'$, so if we prove that $\angle{X''Q'N} = \angle{X''AN}$ we are done. But it is well known that $AX''$, $AP'$ are isogonal and $Q'$, $P'$ and $N$ are collinear, so $\angle{X''AN} = \angle{M'AP'} = \angle{M'Q'P'} = \angle{X''Q'N}$, implying that $Q'NX''A$ is cyclic, as desired.

$\square$