Talk:Holditch's theorem

Proof
Does anyone have a proof for this theorem? N Shar 02:33, 25 December 2005 (UTC)

Here is a proof. Let $x(t)$ denote a parametrization of the curve and let $v(t)$ a vector of constant length $v$ along the curve. Let $\lambda$ be a real number and consider the curve $x_\lambda (t):= x(t) + \lambda v(t)$. A well-known application of Green's theorem gives a formula for the area $A_\lambda$ enclosed by $\x_\lambda$. Namely, \[ A_\lambda = 1/2 \int  x'_\lambda (t) \wedge x''_\lambda (t) dt \] Assuming that $x_1(t)$ parametrizes the same curve as $x_0(t)$ we have \[ A_1 = A_0 \] which implies \[ \int x' \wedge v + v' \wedge x dt  = -\int v' \wedge v'' dt\] and so \[ A_\lambda - A_0 = -\lambda \int v' \wedge v + \lambda^2 \int v' wedge v dt =  (1/2) \lambda\cdot(\lambda -1) \int v' \wedge v'' \, dt and since $v(t)$ goes once around a circle of radious $v$ the last integral is $\pi v^2$. Thus \[ A_0 - A_\lambda = pq \pi \] where $p=\lambda. v$ and $q= \lambda. v $. QED — Preceding unsigned comment added by Holonomia (talk • contribs) 16:23, 24 April 2012 (UTC)

Here is a trascription of "Holonomia" 's comments: Here is a proof. Let x(t) denote a parametrization of the curve and let v(t) a vector of constant length v along the curve. Let &lambda; be a real number and consider the curve x&lambda;(t):= x(t) + &lambda;v(t). A well-known application of Green's theorem gives a formula for the area A&lambda; enclosed by x&lambda;. Namely,
 * $$ A_\lambda = \frac 1 2 \int x'_\lambda (t) \wedge x''_\lambda (t) \, dt $$
 * $$ A_\lambda = \frac 1 2 \int x'_\lambda (t) \wedge x''_\lambda (t) \, dt $$

Assuming that x1(t) parametrizes the same curve as x0(t) we have
 * $$ A_1 = A_0 \, $$
 * $$ A_1 = A_0 \, $$

which implies
 * $$ \int x' \wedge v + v' \wedge x \, dt = -\int v' \wedge v'' \, dt $$

and so
 * $$ A_\lambda - A_0 = -\lambda \int v' \wedge v + \lambda^2 \int v' \wedge v \, dt = \frac 1 2 \lambda\cdot(\lambda -1) \int v' \wedge v'' \, dt $$
 * $$ A_\lambda - A_0 = -\lambda \int v' \wedge v + \lambda^2 \int v' \wedge v \, dt = \frac 1 2 \lambda\cdot(\lambda -1) \int v' \wedge v'' \, dt $$

and since v(t) goes once around a circle of radius v the last integral is $\pi$v2.

Thus
 * $$ A_0 - A_\lambda = pq \pi \, $$
 * $$ A_0 - A_\lambda = pq \pi \, $$

where p = &lambda; &middot; v and q = &lambda; &middot; v. QED Michael Hardy (talk) 19:05, 17 May 2017 (UTC)

Pat Ballew's blog
It seems too bad that policy dictates immediate removal of an external link to Pat Ballew's blog entry on Hamnett Holditch: In this instance, at least, the blog entry gives helpful information of a strictly factual nature not in the article. An alternative is to duplicate this information in the article, for example in a new section on Holditch's biography, but that would not give due credit to Ballew as source. — Preceding unsigned comment added by 129.180.1.224 (talk) 04:12, 22 September 2013 (UTC)
 * Pat's blog.