## Problem 3

$A, B, C, D, X$ are five points in space, such that $A B, B C, C D, D A$ all subtend the acute angle $\theta$ at $X$. Find the maximum and minimum possible values of $\angle \mathrm{AXC}+\angle \mathrm{BXD}$ (for all such configurations) in terms of $\theta$.


## Solution

Answer: minimum 0 (see below); maximum $2 \cos ^{-1}(2 \cos \theta-1)$, achieved by a square pyramid.

If we take $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ to lie in a plane so that $\mathrm{AC}$ and $\mathrm{BD}$ meet at $\mathrm{X}$ with the angle $\mathrm{AXC}=\theta$, then $\mathrm{AB}, \mathrm{BC}$, $\mathrm{CD}$, DA all subtend the angle $\theta$ at $\mathrm{X}$, but $\mathrm{AC}$ and $\mathrm{BD}$ subtend the angle zero. So the minimum is 0 .

A little playing around suggests that we should take $\mathrm{ABCD}$ to be a square, with $\mathrm{X}$ on the normal through the center, so that $\mathrm{XABCD}$ is a square pyramid. We calculate the result for this case. Suppose $\mathrm{XA}=\mathrm{XB}=\mathrm{XC}=$ $\mathrm{XD}=1, \angle \mathrm{AXB}=\angle \mathrm{BXC}=\angle \mathrm{CXD}=\angle \mathrm{DXA}=\theta$. Then $\mathrm{AB}=2 \sin \theta / 2$, so $\mathrm{AC}=2 \sqrt{2} \sin \theta / 2$. So $\sin$ $\mathrm{AXC} / 2=\sqrt{2} \sin \theta / 2$. Hence $\cos \mathrm{AXC}=1-2 \sin ^{2} A X C / 2=1-4 \sin ^{2} \theta / 2=2 \cos \theta-1$. Hence $\angle \mathrm{AXC}+\angle$ BXD $=2 \cos ^{-1}(2 \cos \theta-1)$. We note that this increases monotonically from 0 (at $\theta=0$ ) to $2 \pi$ (at $\theta=\pi / 2$ ). For $\theta>0$, we have $\cos \theta2 \theta$. In other words, except for $\theta=0$, we can certainly do better than $2 \theta$.

Take vectors origin $\mathrm{X}$. Write $\mathrm{XA}=\mathbf{A}$ etc. Since we are only interested in the angles, it is convenient to take a to be the unit vector in the direction $\mathbf{A}$. Then we have $\mathbf{a . b}=\mathbf{b} . \mathbf{c}=\mathbf{c} . \mathbf{d}=\mathbf{d} . \mathbf{a}=\cos \theta$. So $(\mathbf{a}-\mathbf{c}) .(\mathbf{b}-\mathbf{d})=$ 0 . So either $\mathbf{a}=\mathbf{c}$, or $\mathbf{b}=\mathbf{d}$ or $\mathbf{a}-\mathbf{c}$ is perpendicular to $\mathbf{b}-\mathbf{d}$. Suppose $\mathbf{a}=\mathbf{c}$. Then $X$ lies on the line $A C$. In this case we certainly have $A D$ and $C D$ subtending the same angle at $X$, and $A B$ and $B C$ subtending the same angle at $\mathrm{X}$. But $\angle \mathrm{AXC}=0 . \mathrm{XABD}$ is a tetrahedron with $\angle \mathrm{AXD}=\angle \mathrm{AXB}=\theta . \angle \mathrm{BXD} \leq \angle \mathrm{AXB}+\angle$ AXD (with equality only if A, B, D, $X$ lie in a plane). So $\angle B X D+\angle A X C \leq 2 \theta$ and we have just seen that this is not maximal. Similarly if $\mathbf{b}=\mathbf{d}$. So we may assume that $\mathbf{a}-\mathbf{c}$ is perpendicular to $\mathbf{b}-\mathbf{d}$ and both are non-zero.

We also have $(\mathbf{a}+\mathbf{c}) \cdot(\mathbf{a}-\mathbf{c})=\mathrm{a}^{2}-\mathbf{c}^{2}=0$ and $(\mathbf{a}+\mathbf{c}) .(\mathbf{b}-\mathbf{d})=0$. So $\mathbf{a}+\mathbf{c}$ is normal to the plane containing $\mathbf{a}-$ $\mathbf{c}$ and $\mathbf{b}-\mathbf{d}$. Similarly, $\mathbf{b}+\mathbf{d}$. Hence $\mathbf{a}+\mathbf{c}$ and $\mathbf{b}+\mathbf{d}$ are multiples of each other and $(\mathbf{a}+\mathbf{c}) .(\mathbf{b}+\mathbf{d})=|\mathbf{a}+\mathbf{c}|$ $|\mathbf{b}+\mathbf{d}| \cdot$ But lhs $=4 \cos \theta$ and rhs $=(2 \cos \mathrm{AXC} / 2)(2 \cos \mathrm{BXD} / 2)$. So $\cos \theta=\cos \mathrm{AXC} / 2 \cos \mathrm{BXD} / 2=1 / 2$ $\cos (\mathrm{AXC} / 2+\mathrm{BXD} / 2)+1 / 2 \cos (\mathrm{AXC} / 2-\mathrm{BXD} / 2)$. We want to maximise $\mathrm{AXC} / 2+\mathrm{BXD} / 2$ and hence to minimise $\cos (\mathrm{AXC} / 2+\mathrm{BXD} / 2)$, so we must maximise $\cos (\mathrm{AXC} / 2$ - $\mathrm{BXD} / 2)$ and hence take $\angle \mathrm{AXC}=\angle$ BXD. That then gives $\cos (\mathrm{AXC} / 2+\mathrm{BXD} / 2)=2 \cos \theta-1$, so $\mathrm{AXC}+\mathrm{BXD}=2 \cos ^{-1}(2 \cos \theta-1)$. So the square pyramid is indeed maximal.

