Extending the Newton-Mollweide Formulas and the Law of Tangents to General Convex Quadrilaterals

 Introduction and notation

For a triangle with sides $a,b,c$ opposite angles $\alpha,\beta,\gamma$, the classical Mollweide pair is

$$\frac{a+b}{c}=\frac{\cos((\alpha-\beta)/2)}{\sin(\gamma/2)}, \qquad \frac{a-b}{c}=\frac{\sin((\alpha-\beta)/2)}{\cos(\gamma/2)}. \tag{1}$$

The triangle notation in the classical Mollweide formulas.

In 2022, the present author established a cyclic-quadrilateral generalization of Newton's form of Mollweide's formula; in 2024, the author established a cyclic-quadrilateral law of tangents. The present article extends these two earlier results to arbitrary strictly convex quadrilaterals using all four sides and both diagonals. It also derives a signed Newton–Mollweide companion identity. 

Let $ABCD$ be a strictly convex quadrilateral with distinct vertices in boundary order. Set

$$AB=a,\quad BC=b,\quad CD=c,\quad DA=d,\qquad AC=e,\quad BD=f,$$

$$\alpha=\angle DAB,\quad \beta=\angle ABC,\quad \gamma=\angle BCD,\quad \delta=\angle CDA.$$

The diagonals meet at $E$, and $\theta=\angle CED\in(0,\pi)$. All six lengths are strictly positive, all interior angles belong to $(0,\pi)$, and $\alpha+\beta+\gamma+\delta=2\pi$.

A strictly convex quadrilateral with $AC=e$, $BD=f$ and $E=AC\cap BD$; the diagonal angle is $\theta=\angle CED$.

The six-length Newton–Mollweide pair

Define the quadratic and quartic expressions

$$\begin{aligned} L_+&=e^2+f^2-a^2-c^2+2bd, &R_+&=(e^2-d^2)(f^2-b^2)+c^2((b+d)^2-a^2),\\ L_-&=a^2+c^2+2bd-e^2-f^2, &R_-&=(e^2-d^2)(b^2-f^2)+c^2(a^2-(b-d)^2). \end{aligned} \tag{2}$$

To resolve the companion's sign using lengths alone, also set

$$S=b(a^2+d^2-f^2)+d(a^2+b^2-e^2),\qquad T=d(b^2+c^2-f^2)-b(c^2+d^2-e^2). \tag{3}$$

We use $\operatorname{sgn}(0)=0$.

Theorem 1. For every strictly convex $ABCD$, $L_+>0$ and $R_+>0$, and

$$\boxed{\displaystyle \frac{\sin\frac{\alpha+\beta}{2}}{\cos((\gamma-\delta)/2)} =c\sqrt{\frac{L_+}{R_+}}.} \tag{4}$$

Moreover, $T=0$ if and only if $\gamma=\delta$. If $T\ne0$, then $R_->0$, $L_-\ge0$, and the signed six-length companion is

$$\boxed{\displaystyle \frac{\cos\frac{\alpha+\beta}{2}}{\sin\frac{\delta-\gamma}{2}} =\operatorname{sgn}(ST)\,c\sqrt{\frac{L_-}{R_-}}.} \tag{5}$$

In particular its square equals $c^2L_-/R_-$. Also, $S=0$ if and only if $\alpha+\beta=\pi$; in that case the right side of (5) is zero whenever $T\ne0$.

Proof. Write $\nu=\cos(\alpha+\beta)$. The vector identity

$$2\overrightarrow{AD}\cdot\overrightarrow{BC} =e^2+f^2-a^2-c^2$$

and the directed-turn relation $\cos(\alpha+\beta)=-\overrightarrow{AD}\cdot\overrightarrow{BC}/(bd)$ give

$$\nu=\frac{a^2+c^2-e^2-f^2}{2bd}.$$

The cosine rule yields

$$\cos\gamma=\frac{b^2+c^2-f^2}{2bc},\qquad \cos\delta=\frac{c^2+d^2-e^2}{2cd}.$$

Since $\gamma+\delta=2\pi-(\alpha+\beta)$, $\cos(\gamma-\delta)=2\cos\gamma\cos\delta-\nu$. A direct substitution into (2) now gives

$$\begin{aligned} 1-\nu&=\frac{L_+}{2bd}, &1+\cos(\gamma-\delta)&=\frac{R_+}{2bdc^2},\\ 1+\nu&=\frac{L_-}{2bd}, &1-\cos(\gamma-\delta)&=\frac{R_-}{2bdc^2}. \end{aligned}$$

Because $0<\alpha+\beta<2\pi$ and $|\gamma-\delta|<\pi$, $1-\nu>0$ and $1+\cos(\gamma-\delta)>0$. Taking the positive square root of their ratio proves (4). When $\gamma\ne\delta$, the denominator $1-\cos(\gamma-\delta)$ is positive and the second ratio yields

$$\left(\frac{\cos\frac{\alpha+\beta}{2}}{\sin\frac{\delta-\gamma}{2}}\right)^2 =c^2\frac{L_-}{R_-}. \tag{6}$$

It remains to determine the sign, which cannot be inferred from (6) alone. By the cosine rule at $A,B$, and again at $C,D$,

$$S=2abd(\cos\alpha+\cos\beta),\qquad T=2bcd(\cos\gamma-\cos\delta).$$

The identity $\cos\alpha+\cos\beta =2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$ and $\cos\frac{\alpha-\beta}{2}>0$ show that $\operatorname{sgn} S=\operatorname{sgn}\cos\frac{\alpha+\beta}{2}$. As cosine is strictly decreasing on $(0,\pi)$, $\operatorname{sgn} T=\operatorname{sgn}(\delta-\gamma)=\operatorname{sgn}\sin\frac{\delta-\gamma}{2}$. Equation (5) follows, including the zero case $S=0$. $\square$

Specialization to cyclic quadrilaterals

Put

$$\begin{gathered} s=\tfrac12(a+b+c+d),\quad X=(s-a)(s-c),\quad Y=(s-b)(s-d),\\ P=ab+cd,\quad Q=ad+bc,\quad M=ac+bd. \end{gathered} \tag{7}$$

For a strictly convex quadrilateral $s$ exceeds each side, so $X,Y,P,Q,M>0$.

Corollary 1. If $ABCD$ is cyclic, then

$$\frac{\sin\frac{\alpha+\beta}{2}}{\cos((\gamma-\delta)/2)} =\frac{a+c}{b+d}\cot\frac{\theta}{2}, \tag{8}$$

which is the cyclic Newton–Mollweide identity previously established by the present author. If $\delta\ne\gamma$ (equivalently, $b\ne d$ in the cyclic case), the signed companion is

$$\frac{\cos\frac{\alpha+\beta}{2}}{\sin\frac{\delta-\gamma}{2}} =\frac{a-c}{b-d}\tan\frac{\theta}{2}. \tag{9}$$

Proof. For a cyclic quadrilateral, the classical side–diagonal identities, which also follow from the cosine rule and the supplementary-opposite-angle relations, give

$$e^2=\frac{MQ}{P},\qquad f^2=\frac{MP}{Q},\qquad ef=M. \tag{10}$$

The definitions in (7) imply the elementary identities

$$X+Y=M,\qquad X-Y=\frac{b^2+d^2-a^2-c^2}{2}. \tag{11}$$

For any quadrilateral, the cosine rule in the four triangles with vertex $E$ gives

$$2ef\cos\theta=b^2+d^2-a^2-c^2. \tag{12}$$

Together with (10) and (11), this implies

$$\cot^2\frac{\theta}{2}=\frac{X}{Y},\qquad \tan^2\frac{\theta}{2}=\frac{Y}{X}. \tag{13}$$

Both half-angle functions are positive. Direct substitution of (10) into (2) yields

$$\begin{aligned} L_+&=\frac{4bd(a+c)^2X}{PQ},& R_+&=\frac{4bdc^2(b+d)^2Y}{PQ},\\ L_-&=\frac{4bd(a-c)^2Y}{PQ},& R_-&=\frac{4bdc^2(b-d)^2X}{PQ}. \end{aligned} \tag{14}$$

Thus (4) and (13) give (8), whereas (6) gives only the square of (9). To verify the sign, substitution into (3) further factors as

$$S=\frac{4abd(a-c)(b+d)Y}{PQ},\qquad T=\frac{4bcd(a+c)(b-d)X}{PQ}.$$

Consequently $\operatorname{sgn}(ST)=\operatorname{sgn}((a-c)(b-d))$; combining (5), (14) and (13) proves the signed identity (9). In particular $b=d\iff T=0\iff\gamma=\delta$ in the cyclic case. $\square$

A rational six-length law of tangents

The cyclic law of tangents established by the present author reads

$$\frac{\tan\frac{\alpha-\beta}{2}}{\tan\frac{\alpha+\beta}{2}} =\frac{(a-c)(b-d)}{(a+c)(b+d)}. \tag{15}$$

When $\alpha+\beta=\pi$, the tangent quotient is interpreted by its continuous extension $ (\sin\alpha-\sin\beta)/(\sin\alpha+\sin\beta)$. Define the degree-two and degree-four expressions

$$H=b^2+d^2-a^2-c^2,\quad U=e^2+a^2-b^2,\quad V=f^2+a^2-d^2, \quad P_t=2f^2U+HV,\quad Q_t=2e^2V+HU.$$

Theorem 2. For every strictly convex $ABCD$, $P_t>0$, $Q_t>0$, and

$$\boxed{\displaystyle \frac{\tan\frac{\alpha-\beta}{2}}{\tan\frac{\alpha+\beta}{2}} =\frac{bP_t-dQ_t}{bP_t+dQ_t},} \tag{16}$$

with the preceding continuous-extension convention when $\alpha+\beta=\pi$.

Proof. The sine addition/subtraction identities give, whenever the tangent quotient is defined,

$$\frac{\tan\frac{\alpha-\beta}{2}}{\tan\frac{\alpha+\beta}{2}} =\frac{\sin\alpha-\sin\beta}{\sin\alpha+\sin\beta}. \tag{17}$$

The right side is continuous for all $\alpha,\beta\in(0,\pi)$ and so defines the stated extension.

Let $x=AE$, $y=BE$, and $k=\cos\theta$. From the areas of triangles $ABD$ and $ABC$, respectively,

$$\sin\alpha=\frac{xf\sin\theta}{ad},\qquad \sin\beta=\frac{ye\sin\theta}{ab}.$$

Hence (17) equals $(bfx-dey)/(bfx+dey)$. The cosine rule in $AEB,BEC,AED$ gives

$$a^2=x^2+y^2-2xyk,\quad b^2=(e-x)^2+y^2+2(e-x)yk,\quad d^2=x^2+(f-y)^2+2x(f-y)k.$$

Subtracting the first identity from each of the others yields

$$x-ky=\frac{U}{2e},\qquad y-kx=\frac{V}{2f}. \tag{18}$$

By (12), $k=H/(2ef)$. Since $|k|<1$, solving (18) gives

$$x=\frac{2f^2U+HV}{4ef^2(1-k^2)} =\frac{P_t}{4ef^2(1-k^2)},\qquad y=\frac{2e^2V+HU}{4e^2f(1-k^2)} =\frac{Q_t}{4e^2f(1-k^2)}.$$

Because $x,y>0$, both quartics are positive. Multiplication by $bf$ and $de$, respectively, puts $bfx$ and $dey$ over the common positive denominator $4ef(1-k^2)$, proving (16). $\square$

Corollary 2. For a cyclic convex quadrilateral, the six-length law of tangents reduces to the author's previously published cyclic identity (15).

Proof. Let $K$ be the quadrilateral's area. The cyclic supplementary-angle relations imply $\sin\gamma=\sin\alpha$ and $\sin\delta=\sin\beta$. Splitting along $BD$ and along $AC$, respectively, gives

$$2K=(ad+bc)\sin\alpha=(ab+cd)\sin\beta.$$

Therefore (17) equals

$$\frac{(ab+cd)-(ad+bc)}{(ab+cd)+(ad+bc)} =\frac{(a-c)(b-d)}{(a+c)(b+d)},$$

as required. $\square$

Remark 1 (Domains and triangle degeneration). The first ratio (4) and the rational law (16) are defined on the entire strictly convex class (with the stated convention for the latter). The companion is not defined when $\delta=\gamma$; in particular, the rectangle has a $0/0$ companion and must not be included in an unqualified statement. In a cyclic degeneration $D\to C$ (so $c\to0$), use the limiting triangle $ABC$ with opposite-side lengths $\widetilde a=b$, $\widetilde b=d$, $\widetilde c=a$. The author's earlier article explains the geometric limit of the first cyclic formula: $\theta\to\angle ACB$. Under this relabelling, (8) reduces to the first formula in (1). The companion (9), when $b\ne d$, reduces to the reciprocal of the second formula in (1); equivalently, the second classical formula follows after cross-multiplication. The tangent formula (15) becomes the classical triangle law of tangents. These are limiting statements, not claims about a degenerate quadrilateral.

The Ubiquitous Half-Angle Formulas and the Riemann Hypothesis

 For some time I have been studying the recurring role of half-angle formulas in geometry, rational parametrization, and integration. Some of that background is collected in The Ubiquitous Half-Angle Formulas (PDF link). Henning Dathe’s Die Halbwinkelsubstitution und ihre Anwendungen (PDF link) gives a broad treatment of the tangent half-angle substitution and its applications. A related paper, A Unified Substitution Method for Integration (see arXiv:2505.03754) (see also the trigonometric/hyperbolic formulas for the roots of quadratic equations that gave rise to the core identities of this method, and how the tangent of half-angle emerges here as well!), develops a branch-conscious framework in which several classical substitutions and identities arise from a common pair of principal-branch exponential identities. It recovers Euler’s first and second substitutions, after the necessary scaling, sign choices, and component-dependent reciprocal reparametrizations and obtains the classical Weierstrass substitution $t=\tan(\omega/2)$ as a local unit-radius corollary of Transform 5. The same identities also contain two notable geometric specializations. For the Cayley/tangent-half-angle form, take the positive hyperbolic parametrization $y=\cosh u$. Then $\arccos(\cosh u)=iu$ and

$$\tan\!\left(\tfrac12\operatorname{arcsec}(\cosh u)\right)=\tanh(u/2),$$

so the exterior identity becomes

$$e^{-u}=\frac{1-\tanh(u/2)}{1+\tanh(u/2)}.$$

Continuing by $u=i\phi$, with $\tanh(i\phi/2)=i\tan(\phi/2)$, gives the classical Cayley half-angle representation

$$e^{-i\phi}=\frac{1-i\tan(\phi/2)}{1+i\tan(\phi/2)},\qquad e^{i\phi}=\frac{1+i\tan(\phi/2)}{1-i\tan(\phi/2)}.$$

This follows directly from Theorems 3.1–3.2 of the arXiv draft and the branch computation used in §4; the derivation is isolated explicitly in this Cayley-type half-angle note. For Lobachevsky’s angle-of-parallelism formula, the same positive hyperbolic parametrization $y=\cosh u$ in Theorem 3.1 identifies $\beta=\operatorname{arccsc}(\cosh u)=\Pi(u)$ and yields

$$\tan\!\left(\frac{\Pi(u)}2\right)=e^{-u};$$

equivalently, Theorem 3.2 gives the same result using the reciprocal parametrization $y=\operatorname{sech} u$, for which $\psi=\arcsin(\operatorname{sech} u)=\Pi(u)$. Both derivations are given explicitly in §4, “Lobachevsky’s formula as a geometric specialization,” of the arXiv draft.

The transforms also simultaneously rationalize the associated quadratic radicals and inverse-trigonometric half-angle composites, and for several natural families cancellation between the parametrized radical and the Jacobian reduces the transformed integrand all the way to a finite Laurent polynomial, allowing term-by-term integration (in particular, some quotient-radical integrals collapse directly to Laurent polynomials and require no partial-fraction decomposition; see the arXiv draft, especially Example 4 and the termwise-integration remark). 

Recent benchmarks against Mathematica’s native `Integrate` show the strongest gains in the most difficult half-angle families: on mutually solved Transform 1 cases, native `Integrate` was about $194\times$ slower in cold timing and produced expressions about $753\times$ larger on average; for Transform 4 it was about $110\times$ slower cold, with mean closed-form output more than $100\times$ larger and one native result reaching roughly $2.96$ MB. Across the five benchmark families, USM solved and verified every tested example in Transforms 2–5 and produced 95 non-timeout results out of 100 in Transform 1. 

The point of this background is not that any of it is needed for the argument below, but rather to illustrate the remarkable range and structural power of half-angle formulas. Against that background, it is especially intriguing to encounter $\tan(\theta/2)$ once again, this time in a formulation equivalent to the Riemann Hypothesis. That recurrence is what made the observation below worth recording.

I want to emphasize that I am not claiming a proof of the Riemann Hypothesis. The RH argument below was generated by ChatGPT 5.6 (OpenAI) in response to my prompts (so take it with a grain of salt).

Let

$$\xi(s)=\frac12s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s),$$

and let $\lambda_n$ denote the Li coefficients. Li’s criterion states

$$\mathrm{RH}\iff \lambda_n\ge 0\qquad\text{for every }n\ge1.$$

The Li coefficients also have, in the standard symmetric limiting sense, the zero-sum representation

$$\lambda_n=\sum_\rho\left[1-\left(1-\frac1\rho\right)^n\right].$$

Consider the Möbius transformation

$$t(s)=1-\frac1s=\frac{s-1}{s}.$$

For finite $s$,

$$\operatorname{Re}(s)=\frac12 \iff |t(s)|=1.$$

Thus, assuming RH and writing

$$\rho=\frac12+i\gamma,\qquad \gamma>0,$$

we may write

$$1-\frac1\rho=e^{i\theta_\gamma},\qquad 0<\theta_\gamma<\pi.$$

A direct calculation gives

$$\cos\theta_\gamma=\frac{4\gamma^2-1}{4\gamma^2+1},\qquad\sin\theta_\gamma=\frac{4\gamma}{4\gamma^2+1},$$

and hence

$$\tan\frac{\theta_\gamma}{2}=\frac{\sin\theta_\gamma}{1+\cos\theta_\gamma}=\frac1{2\gamma}.$$

Equivalently,

$$\gamma=\frac1{2\tan(\theta_\gamma/2)}.$$

Under RH, a conjugate pair $\frac12\pm i\gamma$ contributes

$$2[1-\cos(n\theta_\gamma)]$$

to the Li zero sum. If $N(T)$ is the usual Riemann–von Mangoldt zero-counting function, counting upper-half-plane zeros with multiplicity, define

$$\theta(T)=2\arctan\frac1{2T},\qquad g_n(T)=2[1-\cos(n\theta(T))].$$

Then the proposed derivation is

$$\lambda_n=\int_0^\infty g_n(T)\,dN(T).$$

Stieltjes integration by parts gives

$$\lambda_n=-\int_0^\infty N(T)g_n'(T)\,dT,$$

since the boundary term vanishes using

$$N(T)=O(T\log T),\qquad g_n(T)=O_n(T^{-2}).$$

Since

$$\theta'(T)=-\frac4{4T^2+1},$$

this becomes

$$\lambda_n=8n\int_0^\infty\frac{N(T)\sin(n\theta(T))}{4T^2+1}\,dT.$$

Finally, with

$$T=\frac1{2\tan(\theta/2)}=\frac12\cot\frac\theta2,$$

we have

$$\frac{dT}{4T^2+1}=-\frac14\,d\theta,$$

and therefore, under RH,

$$\boxed{\lambda_n=2n\int_0^\pi N\left(\frac1{2\tan(\theta/2)}\right)\sin(n\theta)\,d\theta}\tag{*}$$

for every $n\ge1$.

The integral appears to converge: as $\theta\to0$,

$$\frac1{2\tan(\theta/2)}\sim\frac1\theta,$$

so

$$N\left(\frac1{2\tan(\theta/2)}\right)=O\left(\frac{\log(1/\theta)}{\theta}\right),$$

while $\sin(n\theta)\sim n\theta$. Thus the integrand is

$$O_n(\log(1/\theta)),$$

which is integrable at $0$.

The proposed converse is the following. Define

$$F(\theta)=N\left(\frac1{2\tan(\theta/2)}\right),\qquad 0<\theta<\pi.$$

Because $N(T)$ is nonnegative and nondecreasing while $\theta\mapsto1/(2\tan(\theta/2))$ is decreasing, $F$ is nonnegative and nonincreasing.

The elementary lemma being used is:

If $F:(0,\pi)\to[0,\infty)$ is nonincreasing and the improper integral converges, then

$$\int_0^\pi F(\theta)\sin(n\theta)\,d\theta\ge0$$

for every $n\ge1$.

Indeed, putting $h=\pi/n$ and translating the $n$ intervals of length $h$ back to $(0,h)$ gives

$$\int_0^\pi F(\theta)\sin(n\theta)\,d\theta=\int_0^h\left[\sum_{j=0}^{n-1}(-1)^jF(x+jh)\right]\sin(nx)\,dx.$$

For fixed $x\in(0,h)$,

$$F(x)\ge F(x+h)\ge F(x+2h)\ge\cdots\ge0,$$

so the alternating sum is nonnegative, and $\sin(nx)\ge0$ on $(0,h)$.

Consequently, if $(*)$ holds for the actual Li coefficients for every $n\ge1$ without assuming RH, then

$$\lambda_n\ge0$$

for every $n$, and Li’s criterion implies RH. Thus

$$\boxed{\mathrm{RH}\iff\left[\lambda_n=2n\int_0^\pi N\left(\frac1{2\tan(\theta/2)}\right)\sin(n\theta)\,d\theta\quad\text{for every }n\ge1\right].}$$

An Elegant Collinearity Problem Involving the Incircle

A problem from a vietnamese Facebook group.

Problem. Let $ABC$ be a triangle, and let its incircle, with center $I$, be tangent to $BC$ at $D$. Let $M$ and $N$ be the midpoints of $AD$ and $BC$, respectively. Prove that $M$, $I$, and $N$ are collinear. 

Proof 1. Let $BC=a$, $CA=b$, $AB=c$, let $s$ be the semiperimeter, $\Delta=[ABC]$, and let $r$ be the inradius. Assume $b>c$; the case $b<c$ is symmetric, while $b=c$ is immediate. Since $BD=s-b$ and $N$ is the midpoint of $BC$,
$$DN=\frac a2-(s-b)=\frac{b-c}{2}.$$
As $ID\perp BC$ and $ID=r$,
$$\tan\angle DNI=\frac{r}{DN}=\frac{2r}{b-c}.$$
Let $M'$ be the foot of the perpendicular from $M$ to $BC$. Since $M$ is the midpoint of $AD$,
$$[CDM]=\frac12[CDA]=\frac14\,b(s-c)\sin C.$$
On the other hand,
$$[CDM]=\frac12(s-c)MM',$$
hence
$$MM'=\frac{b\sin C}{2}.$$
By the cosine law in $\triangle ACD$,
$$AD^2=b^2+(s-c)^2-2b(s-c)\cos C.$$
Since $MD=AD/2$,
$$DM'^2=DM^2-MM'^2=\frac14\bigl(b\cos C-(s-c)\bigr)^2.$$
Using
$$\sin^2\frac C2=\frac{(s-a)(s-b)}{ab},\qquad\cos^2\frac C2=\frac{s(s-c)}{ab},$$
we get
$$\cos C=\cos^2\frac C2-\sin^2\frac C2=\frac{s(s-c)-(s-a)(s-b)}{ab},$$
so
$$b\cos C-(s-c)=\frac{(s-a)(b-c)}{a}.$$
Therefore
$$DM'=\frac{(s-a)(b-c)}{2a}.$$
Since $M',D,N$ occur in this order,
$$NM'=ND+DM'=\frac{b-c}{2}+\frac{(s-a)(b-c)}{2a}=\frac{s(b-c)}{2a}.$$
Thus
$$\tan\angle DNM=\frac{MM'}{NM'}=\frac{ab\sin C}{s(b-c)}=\frac{2\Delta}{s(b-c)}=\frac{2r}{b-c}.$$
Hence
$$\tan\angle DNM=\tan\angle DNI,$$
and therefore $M,I,N$ are collinear.

Proof 2. This problem is essentially a degenerate case of Newton’s theorem for tangential quadrilaterals: in every tangential quadrilateral, the center of the inscribed circle lies on the line joining the midpoints of the two diagonals. As one of the vertices of the quadrilateral approaches the point of tangency $D$ on $BC$, the quadrilateral degenerates into triangle $ABC$, and the Newton line becomes precisely the line containing the midpoint $M$ of $AD$, the incenter $I$, and the midpoint $N$ of $BC$.