{"id":843,"date":"2014-06-01T01:00:57","date_gmt":"2014-06-01T01:00:57","guid":{"rendered":"http:\/\/blogs.ams.org\/visualinsight\/?p=843"},"modified":"2015-07-29T00:51:28","modified_gmt":"2015-07-29T00:51:28","slug":"grace-danielsson-inequality","status":"publish","type":"post","link":"https:\/\/blogs.ams.org\/visualinsight\/2014\/06\/01\/grace-danielsson-inequality\/","title":{"rendered":"Grace–Danielsson Inequality"},"content":{"rendered":"
\n
\"Grace-Danielsson<\/a>

Grace-Danielsson Inequality – Antony Milne<\/p><\/div>\n<\/div>\n

 <\/p>\n

When can you fit a tetrahedron between two nested spheres? <\/p>\n

Here’s the answer. Suppose the radius of the large sphere is $R$ and the radius of the small one is $r$. Suppose the distance between their centers is $d$. Then you can fit a tetrahedron between these spheres if and only if the Grace–Danielsson inequality<\/b> holds:<\/p>\n

$$ d^2 \\le (R + r)(R – 3r) $$<\/p>\n

This was independently proved by Grace in 1917 and Danielsson in 1949. But Antony Milne has found a new proof of this inequality using quantum information theory:<\/p>\n

• Antony Milne, The Euler and Grace-Danielsson inequalities for nested triangles and tetrahedra: a derivation and generalisation using quantum information theory<\/a>.<\/p>\n

The rough idea is this. Suppose Alice and Bob have entangled qubits. By choosing which measurements to perform, Bob can ‘steer’ the results of Alice’s measurements. Given all possible measurements by Bob, the set of results Alice can get forms an ellipsoid… and studying this ellipsoid leads to the Grace–Danielsson inequality.<\/p>\n

A similar puzzle in 2 dimensions was solved by Chapple in 1746 and Euler in 1765. Namely: suppose you have a circle contained in a larger circle. Suppose the radius of the large circle is $R$ and the radius of the small one is $r$. Suppose the distance between their centers is $d$. Then you can fit a triangle between these two circles if and only if<\/p>\n

$$ d^2 \\le R(R – 2r) $$<\/p>\n

Here is Milne’s picture of the situation:<\/p>\n

\n
\"The<\/a>

The Euler Equation – by Antony Milne<\/p><\/div>\n<\/div>\n

In the comments on this blog, Greg Egan has proposed a generalized Grace–Danielsson inequality in $n$ dimensions. More precisely, he has shown the following. Suppose we have a smaller ball of radius $r$ inside a larger ball of radius $R$ in $\\mathbb{R}^n$. Suppose the distance between the centers of these balls is $d$. Then we can fit an $n$-simplex inside the larger ball and outside the smaller one if<\/p>\n

$$d^2 \\le (R + (n-2) r) (R – n r),$$<\/p>\n

where if the inequality is an equation the simplex may touch the surface of these balls. He conjectures that this sufficient condition is also necessary. <\/p>\n

Egan writes:<\/p>\n

\nHere\u2019s a sketch of a result leading towards a higher-dimensional Grace–Danielsson inequality.<\/p>\n

Given a small sphere with radius $r$ in $\\mathbb{R}^n$, inside a larger sphere of radius $R$, and a distance of $d$ between their centres, there\u2019s an $n$-simplex we can try to construct that\u2019s very symmetrical with respect to these spheres.<\/p>\n

First, pick the point $P$ on the small sphere that lies on the axis joining the centres of the spheres, and which is the furthest of the two such points from the larger sphere. The hyperplane tangent to the small sphere at $P$ intersects the large sphere in an sphere of dimension one less—an $(n-2)$-sphere. Construct a regular $(n-1)$-simplex for which that $(n-2)$-sphere is its circumsphere.<\/p>\n

Now, extend the $(n-1)$-simplex into an $n$-simplex by adding the point where the axis joining the centres of the original spheres intersects the large sphere, and which lies on the opposite side of the small sphere from P. By construction, one face of this $n$-simplex is tangent to the small sphere, and by symmetry if any other face avoids intersecting the small sphere, or is tangent to it, then the $n$-simplex will enclose the small sphere.<\/p>\n

It\u2019s not too hard to grind through the algebra and show that the condition that allows this is:<\/p>\n

$$d^2 \\le (R + (n-2)r)(R \u2013 n r)$$<\/p>\n

So this condition is certainly sufficient for an $n$-simplex to fit between the nested spheres.\n<\/p><\/blockquote>\n

And then:<\/p>\n

\nTo flesh out my conjecture a little more, it’s helpful to start with an equation that is satisfied by the inradius $r$ of an isosceles triangle with base $2b$ and height $h$:<\/p>\n

$$b^2 (h-2r) – h r^2 = 0$$<\/p>\n

This is easy to prove from Pythagoras and similar triangles. The inradius is the radius of a circle that is tangent to all three sides of the triangle.<\/p>\n

Now in the context of two nested spheres in $R^n$, the large one with radius $R$, the small one with radius $r$, and a distance of $d$ between their centres, suppose we place a regular $(n-1)$-simplex tangent to the small sphere, with its centre at one of the points of intersection with the axis that runs between the centres of both spheres. If we choose a scale for this $(n-1)$-simplex so that its vertices lie on the large sphere, its circumradius will be given by the radius of the sphere of intersection between the large sphere and the hyperplane tangent to the small sphere, which is:<\/p>\n

$$\\sqrt{R^2-(r\\pm d)^2}$$<\/p>\n

Here the sign depends on which point of intersection with the axis we use as the point of tangency. The inradius of a regular $(n-1)$-simplex is smaller than the circumradius by a factor of $n-1$, so if we call that inradius $b$, we will have:<\/p>\n

$$b^2 = \\frac{R^2-(r\\pm d)^2}{(n-1)^2}$$<\/p>\n

Now, suppose we turn the regular $(n-1)$-simplex into an $n$-simplex by adding a point at the intersection of the axis and the large sphere that lies on the opposite side of the small sphere to the $(n-1)$-simplex. The “altitude” of this $n$-simplex measured from the $(n-1)$-simplex as its “base” will be given by:<\/p>\n

$$h = R + (r\\pm d)$$<\/p>\n

We can construct an isosceles triangle by taking the centre of the regular $(n-1)$-simplex as the centre of the triangle’s base, the centre of any of the $(n-2)$-simplex faces of the regular $(n-1)$-simplex as an endpoint of the triangle’s base, and the added point we use to create the $n$-simplex as the triangle’s apex. This isosceles triangle with have a half-base of $b$ and a height of $h$. What’s more, if every face of the $n$-simplex is tangent to the small sphere, then the radius of the small sphere, $r$, will be equal to the inradius of the isosceles triangle. So we have:<\/p>\n

$$b^2 (h-2r) – h r^2 = 0$$<\/p>\n

It’s convenient to divide this through by $h$ and multiply by $(n-1)^2$, to obtain:<\/p>\n

$$\\frac{b^2 (n-1)^2 (h-2r)}{h} – (n-1)^2 r^2 = 0$$<\/p>\n

Substituting in for $b^2$ and $h$ we get:<\/p>\n

$$(R – (r\\pm d)) (R + (r\\pm d) – 2r) – (n-1)^2 r^2 = 0$$<\/p>\n

If we solve this for $d^2$, we get:<\/p>\n

$$d^2 = (R + (n-2) r) (R – n r)$$<\/p>\n

So, this is the condition for an $n$-simplex with a regular $(n-1)$-simplex as its “base” to have all its faces tangent to the small sphere and all its vertices lying on the large sphere, given that we choose to make the “base” orthogonal to the axis running between the centres of the spheres.<\/p><\/blockquote>\n

For more details, see the comments.<\/p>\n

\n

Visual Insight<\/i> is a place to share striking images that help explain advanced topics in mathematics. I\u2019m always looking for truly beautiful images, so if you know about one, please drop a comment here<\/a> and let me know!\t<\/p>\n

<\/div>","protected":false},"excerpt":{"rendered":"

When can you fit a tetrahedron between two nested spheres? Suppose the radius of the large sphere is $R$ and the radius of the small one is $r$. Suppose the distance between their centers is $d$. Then you can fit a tetrahedron between these spheres if and only if the Grace–Danielsson inequality<\/b> $ d^2 \\le (R + r)(R – 3r) $ holds. This was independently proved by Grace in 1917 and Danielsson in 1949. But Antony Milne has found a new proof of this inequality using quantum information theory!<\/p>\n

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