The half-angle formulas are central!

 As the title suggests, the half-angle formulas are central. Even more central than the law of cosines, which is nothing more than the half-angle formulas in disguise. Virtually, every metric relationship characterizing the triangle can be derived from the triangular half-angle formulas. The mind map below can give you an idea of the pivotal role that the half-angle formulas play in relation to the other most important metric relationships in classical geometry. I have also written an essay titled "The Theoretical Importance of Half-Angle Formulas" where you can see the details of the proofs, as well as the new generalizations I have managed to derive from this novel approach.

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Identifying the central theorems of an area is important because it helps streamline the process of understanding it. By knowing the basic principles, you can, with logic and a bit of ingenuity, understand (by proving) the rest of the discipline without having to memorize as much.

Solution to Problem 1315 of Gogeometry

 The following problem was proposed by my esteemed friend, Ángel Mejía. However, this is problem 1315 from Gogeometry. Here I give a trigonometric proof.

The figure shows a triangle $ABC$ with the inscribed circle $O$ ($D$, $E$, and $T$ are the tangency points). $OB$ cuts chord $DE$ and arc $DE$ at $M$ and $F$, respectively. $AF$ and $CF$ cut chord $DE$ at $G$ and $N$, respectively. Prove that $DG = MN$.

Proof. Let's denote $\angle{BAC}=\alpha$, $\angle{ABC}=\beta$, $\angle{ACB}=\gamma$, $\angle{AFM}=\angle{AFO}=\delta$, and $\angle{DFA}=\epsilon$. Applying the sine law in triangles $GFM$ and $DFG$, we obtain

$$GM = FG \cdot \sin{\delta},\tag{1}$$

$$DG = \frac{FG \cdot \sin{\epsilon}}{\sin{\frac12\left(\frac{\pi}{2}-\frac{\beta}{2}\right)}}.\tag{2}$$

Dividing $(1)$ by $(2)$ and then squaring, we get

$$\left(\frac{GM}{DG}\right)^2=\frac{\sin^2{\delta}}{\sin^2{\epsilon}}\cdot{\sin^2{\frac12\left(\frac{\pi}{2}-\frac{\beta}{2}\right)}}=\frac{\sin^2{\delta}}{\sin^2{\epsilon}}\left(\frac{1-\sin{\frac12\beta}}{2}\right).\tag{3}$$

Now, applying the sine law once more, this time in triangles $AFO$ and $ADF$, we obtain

$$\sin{\delta}=\frac{AO\cdot{\sin{\frac12(\alpha+\beta)}}}{AF},\tag{4}$$

$$\sin{\epsilon}=\frac{AD\cdot{\sin{\frac14(3\pi+\beta})}}{AF}.\tag{5}$$

Dividing $(4)$ by $(5)$ and then squaring, we have

$$\frac{\sin^2{\delta}}{\sin^2{\epsilon}}=\frac{\sec^2{\frac12\alpha}\cos^2{\frac12\gamma}}{\sin^2{\frac14(3\pi+\beta})}=\sec^2{\frac12\alpha}\cos^2{\frac12\gamma}\left(\frac{2}{1-\sin{\frac12\beta}}\right).\tag{6}$$

Substituting $(6)$ in $(3)$ and taking square roots,

$$\frac{GM}{DG}=\frac{\cos{\frac12\gamma}}{\cos{\frac12\alpha}}.$$

Analogously,

$$\frac{MN}{EN}=\frac{\cos{\frac12\alpha}}{\cos{\frac12\gamma}}.$$

This implies $\frac{GM}{DG}=\frac{EN}{MN}$ and, since $DM=ME$, we conclude that $DG=MN$.

$\square$

Another Identity on Triangle Areas Arising from Reflection

In a previous post, we proved an identity about triangle areas arising from reflection. In this occasion, we prove another interesting identity associated with triangle areas.

Theorem. Let $ABC$ be a triangle. Denote $D$, $E$ and $F$ arbitrary points on sides $BC$, $AC$ and $AB$, respectively. Let $A'$ be the reflection of $A$ with respect to $D$. Define $B'$ and $C'$ similarly. Denote $A''$ the reflection of $A'$ with respect to the midpoint of $BC$. Define $B''$ and $C''$ similarly. Then 

$$[A'B'C']-[A''B''C''] = 3[ABC].$$

The brackets $[\, ]$ represent the area of the enclosed figure.

Lemma 1. $[A'B'C]=[ABA''B'']$, $[A'C'B]=[ACA''C'']$ and $[B'C'A]=[BCB''C'']$.

Proof. Observe that the diagonals of $BCB'C'$ and $BCEF$ are in a 2:1 ratio, and the angle between them remains unchanged. Hence, $[BCB'C']=4[BCEF]$. This is easily deduced from the formula for the area of a quadrilateral $K=\frac12pq\sin{\theta}$, where the lengths of the diagonals are $p$ and $q$, and the angle between them is $\theta$.

Also, observe that $AF$ and $BF$ are medians of triangles $\triangle{ACC'}$ and $\triangle{BCC'}$, respectively. It follows that $[ACF]=[AC'F]$ and $[BCF]=[BC'F]$, and then $[ABC]=[ABC']$. Analogously, $[ABC]=[ACB']=[BCA']$. Now, it turns out that

$$[AB'C']=[BCB'C']-3[ABC]=4[BCEF]-3[ABC]=4[ABC]-4[AEF]-3[ABC]=[ABC]-4[AEF].\tag{1}$$

Let $H$ and $I$ be the midpoints of $AC$ and $AB$, respectively. Then

$$[ABC]=4[AHI]=4[AEF]+4[HEFI].\tag{2}$$

From $(1)$ and $(2)$ follows that $[AB'C']=4[HEFI]$. Now, observe that $BC$, $BB''$, and $CC''$ are the homothetic images with a scale factor of 2 of $IH$, $EH$, and $FI$ with respect to $A$, $B'$, and $C'$, respectively. It follows that $BCC''B''$ and $HEFI$ are homothetic with $[BCC''B'']=4[HEFI]$, therefore,

$$[AB'C']=[BCC''B''].$$

$\square$

A similar reasoning must show that $[A'B'C]=[ABB''A'']$, and for the case of $\triangle{A'C'B}$ and $ACA''C''$ where $F$ and $D$ are on opposite sides of the line $GI$, where $G$ is the midpoint of the side $BC$, only a minor adjustment will be required.

$\color{blue}{[A'B'C]=[ABA''B'']}$, $\color{red}{[A'C'B]=[ACA''C'']}$ and $\color{green}{[B'C'A]=[BCB''C'']}$.

Remark. Notice lemma 1 gives us an alternative way to construct a triangle with an area equal to a given quadrilateral, as long as it is not a rectangle or a square.

Back to the main problem

Notice that 

$$[A'B'C']=[A'C'B'C]-[A'B'C]=4[ABC]+[AB'C']+[BA'C']-[A'B'C].$$

But $[AB'C']=[ABC]-4[AEF]$ and similarly $[BA'C']=[ABC]-4[BDF]$ and $[A'B'C]=4[CDE]-[ABC]$, so

$$[A'B'C']= 7[ABC]-4[AEF]-4[BDF]-4[CDE].\tag{3}$$

As $HIFE$ and $BCC''B''$ are inversely homothetic with scale factor $-2$, then so are segments $EF$ and $B''C''$ and similarly for $ED$ and $A''C''$ and $FD$ and $A''C''$, meaning that triangles $\triangle{DEF}$ and $\triangle{A''B''C''}$ are also inversely homothetic with factor $-2$ and then

$$[A''B''C'']=4[DEF]=4[ABC]-4[AEF]-4[BDF]-4[CDE].\tag{4}$$

Substracting $(4)$ from $(3)$ we get 

$$[A'B'C']-[A''B''C'']=3[ABC].$$

$\square$

A curious family of integrals that give rational multiples of $\pi$

 I have noticed experimentally that:

$$\int_0^1 \frac{\color{red}{x}}{x^4+2x^3+2x^2-2x+1} \,dx=\color{blue}{\frac{\pi}{8}},\tag{1}$$

$$\int_0^1 \frac{\color{red}{1-x^2}}{x^4+2x^3+2x^2-2x+1} \, dx=\color{blue}{\frac{\pi}{4}},\tag{2}$$

$$\int_0^1 \frac{\color{red}{1+x-x^2}}{x^4+2x^3+2x^2-2x+1} \, dx =\color{blue}{\frac{3\pi}{8}}.\tag{3}$$

So slight variations in the numerator always seem to produce something like $n\pi$, where $n$ is a rational number.

At MathSE I have asked what the exact relationship is between $n$ and the numerator of the integrand, to which Quanto has responded with a general formula:

$$\int_{0}^{1} \frac{ax^2 +b x + c}{x^4+2x^3+2x^2-2x+1} \, dx=\frac\pi8\color{green}{(c+b-a)}+\frac\pi{3\sqrt3}\color{green}{(a+c)}.\tag{4}$$

However, is it necessary for the denominator to remain fixed? Not really. The following integral is formula $(34)$ in this list of $\pi$ formulas:

$$\int_0^1 \frac{\color{red}{16x-16}}{x^4-2x^3+4x-4}\,dx=\color{blue}{\pi}.\tag{5}$$

Notice that the denominator is different. But again, a slight variation in the numerator and it still produces something like $n\pi$:

$$\int_0^1 \frac{\color{red}{x^2-x-1}}{x^4-2x^3+4x-4}\,dx=\color{blue}{\frac{3\pi}{16}}.\tag{6}$$

More generally, 

$$\int_{0}^{1} \frac{ax^2+bx+c}{x^4-2x^3+4x-4}\,dx=\frac{\pi}{16}\color{green}{(2a-c)}+\frac{\ln{(3-\sqrt8)}}{\sqrt32}\color{green}{(b+c)}+\frac{\ln{(3-\sqrt8)}}{\sqrt8}\color{green}{a}.\tag{7}$$

From $(7)$, we can deduce that the integral will yield rational multiples of $\pi$ when $b$ and $c$ are opposite to each other and $a=0$. However, this formula is still far from being the ultimate generalization since it does not take into account the coefficients of the denominator. The fact that the denominator is not the same in $(4)$ and $(7)$ suggests that a further generalization is possible.

A more intimidating integral

Doing some arithmetic with the integrands, we can obtain more intimidating integrands that still yield $\pi$. The following one comes from adding integrands (2) and (5) with different denominators:

$$\int_{0}^{1} \frac{2(1-x)(x^5-5x^4-10x^3-4x^2+8x-8)}{x^8-2x^6-2x^5+9x^4-2x^3-16x^2+12x-4}=\pi.$$

Addendum. I wonder if it is possible to characterize the integrand $\frac{P(x)}{Q(x)}$ in such a way that by simple inspection we can say $I=n\pi$? Or in other words, what should be the relationship between the coefficients of the numerator and the denominator for the integral to yield $n\pi$?

Sine half-angle substitution

 If you are a student of integral calculus, it is highly likely that you have come across or will come across the famous Weierstrass substitution, which is very useful for converting rational expressions involving trigonometric functions into ordinary rational expressions involving $t$, where $t=\tan{\frac12{x}}$. The general transformation formula is as follows:

$$\int f(\sin{x},\cos{x})\,dx =  \int f\left(\frac{2t}{1+t^2}, \frac{1-t^2}{1+t^2}\right)\frac{2dt}{1+t^2}.$$

In this note, we introduce another substitution as a companion to the Weierstrass substitution that can transform certain rational expressions of trigonometric functions into simpler ordinary rational expressions than the Weierstrass substitution. For instance, it would be useful when a common linear factor of $\sin{x}$ appears in the numerator or denominator of the integrand. The general transformation formula is given by:

$$\int f(\sin{x},\cos{x})\,dx =  \int f(2\sqrt{s-s^2}, 1-2s)\frac{ds}{\sqrt{s-s^2}}.$$

Here $s=\sin^2{\frac12x}$. 

Derivation

Using the double-angle formulas and the Pythagorean identity, one gets
$$\sin{x}=2\sin{\frac12x}\cos{\frac12x}=2\sin{\frac12x}\sqrt{1-\sin^2{\frac12x}}=2\sqrt{\sin^2{\frac12x}-\sin^4{\frac12x}}=2\sqrt{s-s^2},$$
$$\cos{x}=1-2\sin^2{\frac12x}=1-2s.$$
Finally, since $s=\sin^2{\frac12x}$, differentiation rules imply
$$ds=\sin{\frac12x}\cos{\frac12x}\,dx=\frac{\sin{x}}{2}\,dx,$$
and thus,
$$dx=\frac{ds}{\sqrt{s-s^2}}.$$

Example 1

By applying the sine half-angle substitution and simplifying,

$$\int \frac{\sin{x}}{\sin^2{x}+2\cos{x}}\,dx=\int \frac{1}{1-2s^2}\,ds.$$

Using standard integrals,

$$\int \frac{1}{1-2s^2}\,ds=\frac{\tanh^{-1}{(\sqrt{2}s)}}{\sqrt{2}}+C=\frac{\tanh^{-1}{(\sqrt{2}\sin^2{\frac12x})}}{\sqrt{2}}+C.$$

The advantage of this substitution is evident when comparing it to the solution provided in this integral calculator (which solution, by the way, is equivalent to the one given here) or when using the Weierstrass substitution.

Example 2 

By applying the sine half-angle substitution,

$$\int \frac{1}{\sin^3{x}}\,dx=\frac18\int \frac{1}{(s-s^2)^2}\,ds.$$

Using partial fraction decomposition, 

$$\frac18\int \frac{1}{(s-s^2)^2}\,ds=\frac18\int \left(\frac{1}{s}+\frac{1}{s^2}-\frac{2}{s-1}+\frac{1}{(s-1)^2}\right)\,ds=\frac14\ln{\left|\frac{s}{1-s}\right|}+\frac18\left(\frac{1}{1-s}-\frac{1}{s}\right)+C.$$

Substituting $s$ by $\sin^2{\frac12x}$, we have

$$\frac18\int \frac{1}{(s-s^2)^2}\,ds=\frac14\ln{\left|\tan^2{\frac12x}\right|}+\frac18\left(\frac{1}{\cos^2{\frac12x}}-\frac{1}{\sin^2{\frac12x}}\right)+C=\frac12\left(\ln{|\tan{\frac12x}|}-\frac{\cos{x}}{\sin^2{x}}\right)+C.$$

A solution using integration by parts is given at Integrals For You. 

A Generalization of the Law of Sines

 The following is an extension of the law of sines for cyclic quadrilaterals that comes to accompany the generalization of Mollweide's formula (rather Newton's) and the generalization of the law of tangent.

Generalization. Consider a cyclic quadrilateral, $ABCD$, with side lengths $AB=a$, $BC=b$, $CD=c$, and $DA=d$. Let $\angle{DAB}=\alpha$, $\angle{ABD}=\beta$, $\angle{BCD}=\gamma$, and $\angle{CDA}=\delta$. Then the following identity holds

$$\frac{ab+cd}{\sin{\alpha}}=\frac{ad+bc}{\sin{\beta}}=\frac{ab+cd}{\sin{\gamma}}=\frac{ad+bc}{\sin{\delta}}.$$

At first, I was reluctant to publish this result because the proof is very straightforward (hence, I leave it as an exercise to the reader). However, I have shared it on my social media platforms (see here, here and here) and the audience's response has been more favorable than my generalizations of Mollweide's formula and the law of tangents combined.

Related material

Generalizing Lami's theorem

Triangle Proofs of the Sine and Cosine Addition Formulas

Introduction
The addition formulas for sine and cosine admit many proofs in the mathematical literature. Standard textbook treatments use coordinates on the unit circle and the distance formula; see Abramson. Other presentations use geometric arrangements of triangles and area decompositions (Bogomolny), or Ptolemy's theorem for cyclic quadrilaterals (Bogomolny). This note presents two short triangle proofs based on the area formula and the cosine rule. The hypotheses needed for a nondegenerate triangle are stated explicitly, and the boundary cases are verified separately.

Notation and hypotheses
Throughout the two proofs, assume
$$\alpha>0,\qquad \beta>0,\qquad \alpha+\beta<\pi.\tag{1}$$
Choose a triangle $ABC$ with
$$a=BC,\qquad b=CA,\qquad c=AB,\qquad \angle BAC=\alpha,\qquad \angle CBA=\beta,\qquad \angle ACB=\gamma.$$
Then $a,b,c>0$ and
$$\gamma=\pi-(\alpha+\beta)>0.$$
Write $\Delta$ for its area. We use
$$2\Delta=bc\sin\alpha=ac\sin\beta=ab\sin\gamma\tag{2}$$
and the cosine rule
$$\cos\alpha=\frac{b^2+c^2-a^2}{2bc},\qquad \cos\beta=\frac{a^2+c^2-b^2}{2ac},\qquad \cos\gamma=\frac{a^2+b^2-c^2}{2ab}.\tag{3}$$
These facts can be established using altitudes and the Pythagorean theorem, independently of the addition formulas. The unit-circle definition also gives
$$\sin(\pi-t)=\sin t,\qquad \cos(\pi-t)=-\cos t,\qquad \sin^2 t+\cos^2 t=1.$$

The sine addition formula
Theorem 1. Under the hypotheses (1),
$$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta.$$

Proof. Using (2) and then (3), we obtain
$$\begin{aligned}\sin\alpha\cos\beta+\cos\alpha\sin\beta&=\frac{2\Delta}{bc}\cos\beta+\frac{2\Delta}{ac}\cos\alpha\\&=\frac{2\Delta}{c}\left(\frac{\cos\beta}{b}+\frac{\cos\alpha}{a}\right)\\&=\frac{2\Delta}{c}\left(\frac{a^2+c^2-b^2}{2abc}+\frac{b^2+c^2-a^2}{2abc}\right)\\&=\frac{2\Delta}{ab}\\&=\sin\gamma\\&=\sin\bigl(\pi-(\alpha+\beta)\bigr)\\&=\sin(\alpha+\beta).\end{aligned}$$

The cosine addition formula
Theorem 2. Under the hypotheses (1),
$$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.$$

Proof. The first two identities in (3) give
$$c-b\cos\alpha=c-\frac{b^2+c^2-a^2}{2c}=\frac{a^2+c^2-b^2}{2c}=a\cos\beta.$$
Together with (2), this yields
$$\cos\beta=\frac{c-b\cos\alpha}{a},\qquad \sin\beta=\frac{b\sin\alpha}{a}.$$
Consequently,
$$\begin{aligned}\cos\alpha\cos\beta-\sin\alpha\sin\beta&=\frac{c\cos\alpha-b\cos^2\alpha-b\sin^2\alpha}{a}\\&=\frac{c\cos\alpha-b}{a}\\&=\frac{c^2-a^2-b^2}{2ab}\\&=-\cos\gamma\\&=\cos(\alpha+\beta).\end{aligned}$$
The last equality uses
$$\gamma=\pi-(\alpha+\beta)$$
and the unit-circle identity for the cosine of a supplementary angle.

Boundary cases
Corollary. Both addition formulas hold for
$$\alpha\ge0,\qquad \beta\ge0,\qquad \alpha+\beta\le\pi.$$

Proof. The strict inequalities were covered above. If $\alpha=0$ or $\beta=0$, both formulas follow immediately from $\sin 0=0$ and $\cos 0=1$.
If $\alpha+\beta=\pi$, then
$$\beta=\pi-\alpha,$$
so
$$\sin\beta=\sin\alpha\qquad\text{and}\qquad \cos\beta=-\cos\alpha.$$
Hence
$$\begin{aligned}\sin\alpha\cos\beta+\cos\alpha\sin\beta&=-\sin\alpha\cos\alpha+\cos\alpha\sin\alpha=0=\sin\pi,\\cos\alpha\cos\beta-\sin\alpha\sin\beta&=-\cos^2\alpha-\sin^2\alpha=-1=\cos\pi.\end{aligned}$$
This covers every boundary case without using a degenerate triangle.