About four points lying on a sphere

Question

This seems to be a curious problem:

Four points $A,B,C,D$ are chosen on a sphere of radius $1$ such that the centre $O$ of the sphere lies inside the tetrahedron $ABCD$. Prove that $|\vec{OA}+\vec{OB}+\vec{OC}+\vec{OD}|<2$.

An analogous problem in the plane can be formulated like this:

If three points $A,B,C$ are chosen on a circle of radius $1$ such that the centre $O$ of the circle lies inside triangle $ABC$, then $|\vec{OA}+\vec{OB}+\vec{OC}|<1$.

I don't know how the first problem is solved, but I also don't have a good solution to the second problem, just some thoughts:

We can assume that the angles $AOB$, $BOC$, $AOC$ satisfy the inequalities $$ 180^\circ\geq\ \angle AOB\ \geq\ \angle BOC\ \geq\ \angle AOC. $$ Then $AOB\geq120^\circ$ and point $C$ lies inside arc $A'B'$ (see figure). If $\vec{OP}=\vec{OA}+\vec{OB}$ and $\vec{OQ}=\vec{OP}+\vec{OC}$, then $|OP|<1$ and it is easy to see that $\angle POQ\>\angle OPQ$. (I wrote easy to see, but to be honest I could not strictly justify this point. These angles are equal if $C=A'$ or $C=B'$; if $C$ lies on arc $A'BAB'$ then the opposite inequality holds; if $C$ lies on arc $A'B'$ (as in the figure) then the above inequality holds.) Consider $\triangle OCQ$, $\angle CQO=\angle POQ\geq90^\circ$ $\Longrightarrow$ $OQ<OC=1$.

Added 2024/01/07.
Patrick suggested another solution to the problem, but unfortunately his explanations are very vague.

I will try to present his solution here, it seems to be the best one that is presented here.

So, let $\vec{OS}=\vec{OA}+\vec{OB}+\vec{OC}+\vec{OD}$. Let us assume that $\vec{OS}\neq0$, otherwise our statement is obvious. Consider an axis $L$ passing through $O$ and parallel to $\vec{OS}$. We can look at $L$ as a number line, $1$ corresponds to the point of intersection of the ray $OS$ and the sphere. Let $\pi$ be the orthogonal projection of $\mathbb{R}^3$ onto the $L$. If $\pi(A)=a,\pi(B)=b,\pi(C)=c,\pi(D)=d,\pi(S)=s$, then $s=|\vec{OS}|$ and $a+b+c+d=s$. We can assume that $$ -1\leq a\leq b\leq c\leq d\leq1. $$ It is clear, if $d=1$, then $c<d=1$. (In the picture below, the named projections are marked in red.)

1. If $a>0$, then the tetrahedron $ABCD$ lies on one side of the plane perpendicular to $\vec{OS}$ and passing through the center $O$ and so $O$ is outside the tetrahedron.

2. If $b>-a$ and $a\neq-1$ and $P$ is the plane passing through the three points $A$, $O$, $S$ (in this plane is the figure below), then the tetrahedron $ABCD$ is located on one side of the plane perpendicular to $P$ and passing through $O$ and $A$.

3. If $a=-1$, then $s=a+b+c+d=(a+d)+b+c=0+b+c<2$.

4. If $b\leq-a$, then $s=a+b+c+d=(a+b)+c+d\leq0+c+d<2$.

The proof is now complete.

But that still leaves the question. Are there any simpler considerations?

Answer 1

Not a bad idea given by Reza Rajaei in his comment. Here is another solution to the problem for the circle based on the dot product.

Let $\angle AOB=2\alpha$, $\angle AOC=2\beta$, and $\angle BOC=2\gamma$. First we prove a well-known trigonometric identity $$ \cos2\alpha+\cos2\beta+\cos2\gamma=-1-4\cos\alpha\cos\beta\cos\gamma. $$ Indeed, since $\alpha+\beta+\gamma=180^\circ$ we get \begin{align*} \cos2\alpha+\cos2\beta+\cos2\gamma & \\ = &\ 2\cos(\alpha+\beta)\cos(\alpha-\beta)+\cos2\gamma \\ = & -2\cos\gamma\cos(\alpha-\beta)+2\cos^2\gamma-1\\ = & -1-2\cos\gamma(\cos(\alpha-\beta)-\cos(\alpha+\beta))\\ = & -1 -4\cos\gamma\cos\alpha\cos\beta. \end{align*} Then since $0\leq\alpha,\beta,\gamma<90^\circ$ we have $$ \cos2\alpha+\cos2\beta+\cos2\gamma<-1.\tag1 $$

Now consider the dot product \begin{align*} (\vec{OA}+\vec{OB}+\vec{OC})\cdot(\vec{OA}+\vec{OB}+\vec{OC})&\\ = &\ 3+2\vec{OA}\cdot\vec{OB}+2\vec{OA}\cdot\vec{OC}+2\vec{OB}\cdot\vec{OC}\\ = &\ 3+2\cos2\alpha+2\cos2\beta+2\cos2\gamma. \end{align*} Hence and from $(1)$ it follows that $$ |\vec{OA}+\vec{OB}+\vec{OC}|=3+2(\cos2\alpha+\cos2\beta+\cos2\gamma)<3-2=1. $$

PS. Although this is not a solution to the main problem, but maybe someone will find a similar solution for the 3D problem. On the other hand I am sure that there must be a purely geometric solution for both circle and sphere.

Addition. (2023/12/31)

Here is another proof, of this assertion, which D S says in his commentary.

Let $O$ be the circumcenter of a triangle $\triangle ABC$ and let $Q$ denote the orthocenter of $\triangle ABC$, then $$ \vec{OQ}=\vec{OA}+\vec{OB}+\vec{OC}.\tag2 $$ This vector equality is sometimes called Sylvester's triangle problem.

As we know, the orthocentre is inside an acute triangle. In our case, the triangle $\triangle ABC$ is an acute triangle, as follows from the condition. Hence $|OQ|<1$ and according to $(2)$ we get $\vec{OA}+\vec{OB}+\vec{OC}<1$.

For the sake of completeness, I will give a short proof of $(2)$ (see the figure from my question). Let $O$ be the circumcenter of a triangle $\triangle ABC$, and let $M$ be the midpoint of side $AB$, $S$ the median intersection, and $\vec{SQ}=2\vec{OS}$. So $MS=2CS$ and the triangles $MOS$ and $CQS$ are similar.
Hence $OM\|CQ$, $CQ$ is an altitude of the triangle, and $Q$ is the orthocenter of the triangle. Now we get \begin{align*} &\vec{OA}+\vec{OB}=2\vec{OM}=\vec{CQ} \text{ and } \vec{OC}+\vec{CQ}=\vec{OQ}\\ \Longrightarrow\ & \vec{OA}+\vec{OB}+\vec{OC}=\vec{OQ}. \end{align*}

Answer 2

I think the idea mentioned by @D S in the comments might work. WLOG, we may assume:

$$B=(x_1,y_1,-z_1), \\ C=(x_2,y_2,-z_1), \\D=(x_3,y_3,-z_1),$$ where $z_1 \geq 0.$ Moreover, we assume $A=(x_4,y_4,z_4)$ with $z_4>0$ and $O=(0,0,0).$ Note that if $z_4 \leq z_1$, then $O$ does not lie inside the tetrahedron. Hence, we suppose $z_4>z_1.$

I think at this point we need to make use of the $2$D case. Let $H$ be the projection of $O$ onto the plane $z=-z_1.$ Since $O$ lies inside the tetrahedron,$|\vec{HB}+\vec{HC}+\vec{HD}|$ is exactly the $2$D case where radius $1$ is replaced with radius $\sqrt{1-z_1^2}$. To see that, just consider the circle shaped by the intersection of the sphere and the plane $z=-z_1.$ Thus, we will have:

$$|\vec{HB}+\vec{HC}+\vec{HD}|^2=(x_1+x_2+x_3)^2+(y_1+y_2+y_3)^2<1-z_1^2. \ \ \ \ \ \ (*)$$

Now, notice that our main goal is to prove:

$$(x_1+x_2+x_3+x_4)^2+(y_1+y_2+y_3+y_4)^2+(-3z_1+z_4)^2<4.$$

Considering $(*)$, it is enough to show:

\begin{align*} (1-z_1^2)+x_4^2+2x_4(x_1+x_2+x_3)+y_4^2+2y_4(y_1+y_2+y_3)+9z_1^2+z_4^2-6z_1z_4 <4\\ \iff 4z_1^2-3z_1z_4+x_4(x_1+x_2+x_3)+y_4(y_1+y_2+y_3)<1. \end{align*}

However, using $(*)$ once again together with $z_4>z_1 \geq 0$, we have:

\begin{align*} 4z_1^2-3z_1z_4+x_4(x_1+x_2+x_3)+y_4(y_1+y_2+y_3)&\\ \leq 4z_1^2-3z_1z_4+ \frac{x_4^2+(x_1+x_2+x_3)^2}{2}+\frac{y_4^2+(y_1+y_2+y_3)^2}{2} & \\ <4z_1^2-3z_1z_4+\frac{2-z_1^2-z_4^2}{2} &\\ <4z_1^2-3z_1^2-\frac{z_1^2+z_4^2}{2}+1 &\\ =\frac{z_1^2-z_4^2}{2}+1<1. \end{align*}

We are done.

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Another proof for the $2$D case:

WLOG, we may assume $A=(x_1, y_1)$ and $B=(-x_1,y_1)$ where $y_1 > 0$. In order to have $O=(0,0) $ inside the triangle, we must have $C=(x_3,-y_3)$ such that $y_3>y_1> 0$.

Now, we need to prove that:

$$|\vec{OA}+\vec{OB}+\vec{OC}|=|(x_3,-y_3+2y_1)|<1.$$

However, $$x_3^2+(-y_3+2y_1)^2=x_3^2+y_3^2-4y_3y_1+4y_1^2=1-4y_1(y_3-y_1)<1.$$

Answer 3

Suppose that instead of maximizing the length of the sum of the four vectors we instead want to maximize the x-coordinate of the sum of the four vectors. If we can prove that there is a strict upper bound for this then this solves the stated problem. To see this note that we can choose the system of coordinates to that the sum of the vectors end up on the positive x-axis.

For the origin to be in the interior of the tetrahedron then one of the four points without loss of generality A has to have a negative x-coordinate. Otherwise the tetrahedron would not intersect the plane x=0 that contains the origin.

Now project onto the plane spanned by A and the x-axis. We have now 4-gon in the circle disk so that the origin is in the interior and one of the vertices is A.

One of the points obtained by projecting B,C,D has to have x-coordinate at most the same as the point -A for the origin to be in the interior (otherwise the line segments from A to these points would all be on the same side of the line trough A and the origin). The other points has x-coordinate at most the same as (1,0,0) as we are on a circle centered on the origin with radius 1.

Now the rather deranged tetrahedron spanned by the points A,-A,(1,0,0),(1,0,0) gives a strict upper bound of 2 for the x-coordinate of the sum.

Answer 4

Reza Rajaei's solution to the sphere problem is wonderful, but requires some computation. After some thought about his solution, I have found that the computations required is not that much.

I think it is reasonable to post this reasoning as a new answer. Circle case.

Let the three points of $A,B,C$ be placed on a circle with a radius of $1$. If the center $O$ of the circle lies inside $\triangle ABC$, then $$ |\vec{OA}+\vec{OB}+\vec{OC}|<1. $$

Proof. Let $OQ\perp AB$. We have $\vec{OA}+\vec{OB}=2\vec{OQ}$.

Let $A'$ and $B'$ be points centrally symmetric to points $A$ and $B$ respectively with respect to the center $O$. The center $O$ lies inside $\triangle ABC$ if and only if $C$ lies inside the arc $A'UB'$ (see figure on the left).

It is almost obvious that $$ |2\vec{OQ}+\vec{OC}|\leq|2\vec{OQ}+\vec{OA'}| $$ for any point $C$ on arc $A'UB'$. (You can for example compute $(\vec{OC}+2\vec{OQ})^2$ then use the geometric definition of dot product, the properties of the $\cos$, and the fact that $\angle A'OQ>90^\circ$.)

We have $2\vec{OQ}+\vec{OA'}=2\vec{OQ}-\vec{OA}=\vec{OB}$.

Therefore we get $$ |\vec{OA}+\vec{OB}+\vec{OC}|= |2\vec{OQ}+\vec{OC}|\leq |2\vec{OQ}+\vec{OA'}|= |\vec{OB}|\leq1. $$

Case sphere.

Let $A,B,C,D$ be four points on a sphere of radius $1$. If the centre $O$ of the sphere lies inside the tetrahedron $ABCD$, then
$$|\vec{OA}+\vec{OB}+\vec{OC}+\vec{OD}|<2.$$

Proof. Let $P$ be the plane passing through the three points $A,B,C$, and let $Q$ be the projection of the point $O$ onto the plane $P$. Consider a plane $P'$ that passes through the points $O$ and $D$ and is perpendicular to the plane $P$. The figure shows the circle lying in the intersection of the plane $P'$ and the sphere and the line $XY$ lying in the intersection of the planes $P$ and $P'$. Let $X'$ and $Y'$ be points centrally symmetric to points $X$ and $Y$ respectively with respect to the center $O$. We get exactly the same figure as in the case of a circle (in the figure on the right).

We know that $\vec{QA}+\vec{QB}+\vec{QC}=\vec{QH}$ and $|QH|<|QA|$ (|QA| is the radius of the circle lying in the intersection of the sphere with the plane $P$). Hence $$ |\vec{OH}| =|\vec{OQ}+\vec{QH}| \leq|\vec{OQ}+\vec{QX}| =|\vec{OX}| \leq1. $$

Just as in the case of a circle we have $$ |2\vec{OQ}+\vec{OD}|<1. $$

Consequently, \begin{align*} |\vec{OA}+\vec{OB}+\vec{OC}+\vec{OD}|\\ &=|\vec{QA}+\vec{QB}+\vec{QC}+3\vec{OQ}+\vec{OD}|\\ &=|\vec{QH}+3\vec{OQ}+\vec{OD}|\\ &=|\vec{OH}+2\vec{OQ}+\vec{OD}|\\ &<|\vec{OH}|+|2\vec{OQ}+\vec{OD}|=2. \end{align*}

That was what we had to prove.

PS. I hope that everything is all right here. But once again, all of the credit goes to Reza. It is too bad that he does not have a lot of votes for his answer.