# A generalization of Pythagoras'

Consider a quadrilateral inscribed in a semicircle of diameter $$d$$, as in the picture below, then $$d^2 = a^2 + b^2 + c^2 + \frac{2 a b c}{d}$$

Notes:

1. If one of the $$a$$, $$b$$, $$c$$ equals $$0$$, we get Pythagoras'

2. An older problem on this site asks (essentially) to determine $$d$$ if $$(a,b,c) = (2,7,11)$$.

3. The set $$\{ (a,b,c,d) \ | d^3 = (a^2 + b^2 + c^2)d + 2 a b c\}$$ can be parametrized by $$\begin{eqnarray} a &=& p(-p^2 + q^2 + r^2)\\ b&=&q(p^2 - q^2 + r^2)\\ c&=& r(p^2 + q^2 - r^2)\\ d&=& 2 p q r \end{eqnarray}$$

4. Provided as reference.

Any feedback would be appreciated!

$$\bf{Added:}$$

1. will write an answer tomorrow, perhaps some users would like to solve it on their own.

2. In the answer of @Cristhian Camacho to the older problem linked, we find this formula.

$$\bf{Added:}$$

A possible solution : the relation is equivalent to

$$1 = x^2 + y^2 + z^2 + 2 x y z$$ where $$x = \frac{a}{d}, \ldots$$, but then also $$x = \sin \frac{\alpha}{2}, \ldots$$, with $$\alpha+ \beta + \gamma = \pi$$. Now we have to show that if $$u + v+w = \frac{\pi}{2}$$, then

$$1- (\sin^2 u + \sin^2 v + \sin^2 w) - 2 \sin u \sin v \sin w$$ In fact we have a factoring.

Equivalently, one can show that if $$\phi+ \psi + \theta = \pi$$ ( angles of a triangle) then

$$1 = \cos^2 \phi + \cos^2 \psi + \cos^2 \theta + 2 \cos \phi \cos \psi \cos \theta$$

( $$\phi = \frac{\pi}{2} - \frac{\alpha}{2}, \ldots$$). This also provides the parametrization of the set, using formulas for the $$\cos$$ in a triangle.

$$\bf{Added:}$$ Just an idea how to handle more general things: say we want to determine a relation between the cosines of some angles with a sum a given rational multiple of $$\pi$$. Write $$x_i = \cos \phi_i$$, with $$\sum \phi_i = \frac{ k \pi}{l}$$. Then $$\cos \phi_i = \frac{1}{2}( t_i + \frac{1}{t_i})$$ and $$(t_1 \cdots t_n)^l = (-1)^k$$. Eliminating $$t_i$$ from the above gets a relation between the $$x_i$$. It could be rather involved, but our case is simple( $$\phi_1 + \phi_2 + \phi_3 = \pi$$, so $$t_1 t_2 t_3 + 1 = 0$$, and we get $$1 = x_1^2 + x_2^2 + x_3^2 + 2 x_1 x_2 x_3$$).

• Do you have a proof? if so include it to "show work" in hope to avoid question close votes. Sep 14, 2023 at 3:22
• @coffeemath: will write one soon, just in case somebody wants to solve it, it's not difficult :-) Sep 14, 2023 at 3:45
• Sorry, just noticed "provided as challenge" [still a proof maybe along with asking for other proofs would be good. Sep 14, 2023 at 4:16

Clearly, due to $$90^\circ$$ subtended by the diameter on the semicircle, the diagonals are $$\sqrt{d^2-a^2}$$ and $$\sqrt{d^2-c^2}$$. By Ptolemy theorem, which says that product of diagonals of a cyclic quadrilateral equals the sum of products of opposite sides, we have $$\sqrt{d^2-a^2}\cdot\sqrt{d^2-c^2}=ac+bd$$ On squaring and simplifying, this rearranges into $$d^2=a^2+b^2+c^2+\frac{2abc}{d}$$

• Great solution. Remarkably efficient! Sep 14, 2023 at 5:37

In the figure,

$$x=y$$ $$\because \sin y = \frac{c}{d}$$ $$\therefore \sin x = \frac{c}{d}$$

Using Cosine Rule on $$\Delta ABC$$,

\begin{align} AC^2 &= a^2+b^2-2ab\cos(90^{\text {o}}+x) \\ &= a^2+b^2+2ab\sin x \\ &= a^2+b^2+2ab\left(\frac{c}{d} \right) \\ &= a^2+b^2+\frac{2abc}{d} \end{align}

In $$\Delta ACD$$, $$AD^2 = CD^2+AC^2$$ $$d^2=c^2+a^2+b^2+\frac{2abc}{d}$$

• Nice and clean! +1 Sep 14, 2023 at 17:33

As shown in the figure below, I've added lines joining the $$2$$ points on the upper part of the circle to the center (labeled $$O$$), and labeled the points around the circle, starting from the left and going clockwise, as $$A$$, $$B$$, $$C$$ and $$D$$.

With $$r$$ being the circle's radius, we have

$$\lvert AO\rvert = \lvert BO\rvert = \lvert CO\rvert = \lvert DO\rvert = r \tag{1}\label{eq1A}$$

Also, label the angles in the the upper semi-circle around the point $$O$$ as

$$\measuredangle AOB = \alpha, \; \measuredangle BOC = \beta, \; \measuredangle COD = \gamma, \;\;\; \alpha + \beta + \gamma = \pi \tag{2}\label{eq2A}$$

Next, using the law of cosines on $$\triangle AOB$$, $$\triangle BOC$$ and $$\triangle COD$$ gives

$$a^2 = r^2 + r^2 - 2r^2\cos\alpha = 2r^2(1 - \cos\alpha) = \frac{d^2}{2}(1 - \cos\alpha) \tag{3}\label{eq3A}$$

$$b^2 = r^2 + r^2 - 2r^2\cos\beta = 2r^2(1 - \cos\beta) = \frac{d^2}{2}(1 - \cos\beta) \tag{4}\label{eq4A}$$

$$c^2 = r^2 + r^2 - 2r^2\cos\gamma = 2r^2(1 - \cos\gamma) = \frac{d^2}{2}(1 - \cos\gamma) \tag{5}\label{eq5A}$$

Adding the above $$3$$ equations together results in

$$a^2 + b^2 + c^2 = \frac{d^2}{2}(3 - (\cos\alpha + \cos\beta + \cos\gamma)) \tag{6}\label{eq6A}$$

As shown in Prove trigonometry identity for $\cos A+\cos B+\cos C$, we have

$$\cos\alpha + \cos\beta + \cos\gamma = 1 + 4\sin\left(\frac{\alpha}{2}\right)\sin\left(\frac{\beta}{2}\right)\sin\left(\frac{\gamma}{2}\right) \tag{7}\label{eq7A}$$

By drawing perpendicular bisector lines from $$O$$ to each of $$AB$$, $$BC$$ and $$CD$$, we see that

$$\sin\left(\frac{\alpha}{2}\right) = \frac{\left(\frac{a}{2}\right)}{r} = \frac{\left(\frac{a}{2}\right)}{\left(\frac{d}{2}\right)} = \frac{a}{d} \tag{8}\label{eq8A}$$

Similarly, $$\sin\left(\frac{\beta}{2}\right) = \frac{b}{d}$$ and $$\sin\left(\frac{\gamma}{2}\right) = \frac{c}{d}$$. Substituting these into \eqref{eq7A}, and then that result into \eqref{eq6A}, gives

\begin{aligned} a^2 + b^2 + c^2 & = \frac{d^2}{2}\left(3 - 1 - \frac{4abc}{d^3}\right) \\ a^2 + b^2 + c^2 & = d^2 - \frac{2abc}{d} \\ d^2 & = a^2 + b^2 + c^2 + \frac{2abc}{d} \end{aligned}\tag{9}\label{eq9A}

• Nice! ($2$ is in the numerator, look at half of a regular hexagon) Sep 14, 2023 at 4:26
• @orangeskid You're right, I forgot the factor of $4$ in the equation for the sum of cosines. Thanks for letting me know. I've now corrected my answer accordingly. Sep 14, 2023 at 4:28