Posted on Mar 20, 2026
I have a loop of string whose total length is 10. I place it around a unit disk (radius 1) and pull a point on the string away from the disk until the string is taut.
I drag this point around the disk in all directions, always keeping the string taut, tracing out a loop. What is the area inside this resulting loop?
Let \(P\) be the pulled point and \(O\) the center of the disk. The string is tangent at two points and wraps around the far side of the circle.
Let \(d=|OP|\). If the angle between \(OP\) and the radius to a tangency point is \(\alpha\), then
So the total length condition is
\[ 10=2\tan\alpha+2\pi-2\alpha \quad\Longrightarrow\quad \tan\alpha-\alpha=5-\pi. \]
This gives \(\alpha\approx1.260529\) radians, so
\[ d=\sec\alpha\approx 3.2753267. \]
The dragged point stays at distance \(d\) from \(O\), so the traced loop is a circle of radius \(d\).
Area:
\[ A=\pi d^2=\pi\sec^2\alpha\approx 33.7022675. \]
Now I have a loop of string whose total length is 14, and I place it around two adjacent unit disks. I pull a point on the string away from the disks until the string is taut, and I drag this point around the disks in all directions, keeping the string taut.
What is the area inside the resulting loop?
Drag the angle slider to move the pulled point around the two-disk stadium and watch the taut string update.
The two disks are tangent, so the convex hull is a stadium: two semicircles of radius 1 joined by two line segments of length 2. Its perimeter is
\[ P_0=2\pi+4. \]
When the string is taut, it consists of two tangent segments from \(P\) to the stadium plus the far-side boundary arc between the tangency points.
For points on the right side with \(|y|\le 1\), both tangency points lie on the right semicircle. The same computation as the main problem gives
\[ 14=P_0-2\alpha+2\tan\alpha \quad\Longrightarrow\quad \tan\alpha-\alpha=5-\pi, \]
so those parts of the loop are circular arcs centered at \((\pm1,0)\) with radius \(d=\sec\alpha\approx 3.2753267\).
For the top and bottom portions (\(|y|>1\)), tangency points lie on opposite semicircles. If \(P=(x,y)\) with \(y>1\), let
\[ d_r=\sqrt{(x-1)^2+y^2},\quad d_l=\sqrt{(x+1)^2+y^2}, \]
then the tangent lengths are \(\sqrt{d_r^2-1}\) and \(\sqrt{d_l^2-1}\), and the wrapped arc length is
\[ P_0-\Bigl(2+(\theta_l-\theta_r)\Bigr), \]
where \(\theta_r,\theta_l\) are the tangency angles on the right/left circles. This yields an implicit equation for the boundary curve.
Numerically, I solve for the polar radius \(r(\phi)\) (measured from the midpoint between the circles) that satisfies the taut-string condition at each angle \(\phi\). Then I compute the area by
\[ A=\frac12\int_{0}^{2\pi} r(\phi)^2\,d\phi \]
More explicitly, let \(P=(r\cos\phi,r\sin\phi)\). Define
\[ d_r=\sqrt{r^2-2r\cos\phi+1},\quad d_l=\sqrt{r^2+2r\cos\phi+1} \]
and
\[ \phi_r=\operatorname{atan2}(r\sin\phi,\ r\cos\phi-1),\quad \phi_l=\operatorname{atan2}(r\sin\phi,\ r\cos\phi+1), \]
\[ \alpha_r=\arccos\!\left(\frac{1}{d_r}\right),\quad \alpha_l=\arccos\!\left(\frac{1}{d_l}\right). \]
The tangency angles on each circle are \(\theta_r^{\pm}=\phi_r\pm\alpha_r\) and \(\theta_l^{\pm}=\phi_l\pm\alpha_l\). Choose the two tangency points that are visible from \(P\). Let \(L_{\text{far}}(r,\phi)\) be the length of the stadium boundary on the far side between those two points (perimeter minus the visible arc). Then the taut-string condition is
\[ L(r,\phi)=|P-T_1|+|P-T_2|+L_{\text{far}}(r,\phi)=14. \]
For each \(\phi\), I solve \(L(r,\phi)=14\) for \(r\) by bisection, then plug \(r(\phi)\) into the area integral above. The result:
Numerical code used for the integration
Extra credit area (numerical):
\[ A_{\text{two disks}}\approx 52.2851. \]
Ellipse fit approximation (separate page)