Kvant Math Problem 539

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Solve time: 4m13s
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Problem

Let $P$ be a given point inside a given sphere, and let $A$, $B$, $C$ be any three points on this sphere such that the segments $PA$, $PB$, $PC$ are mutually perpendicular. Let $Q$ be the vertex of the parallelepiped defined by the segments $PA$, $PB$, and $PC$, diagonally opposite to $P$. Determine the locus of points $Q$.

International Mathematical Olympiad for School Students (XX, 1978)

Exploration

Let the given sphere have center $O$ and radius $R$. Put the origin at $P$. Since $PA$, $PB$, $PC$ are mutually perpendicular, choose coordinates so that

$$A=(a,0,0),\qquad B=(0,b,0),\qquad C=(0,0,c).$$

Then the opposite vertex of the parallelepiped is

$$Q=(a,b,c).$$

The condition that $A,B,C$ lie on the sphere is

$$|A-O|^2=|B-O|^2=|C-O|^2=R^2.$$

Write $O=(u,v,w)$. Then

$$(a-u)^2+v^2+w^2=R^2,$$

$$u^2+(b-v)^2+w^2=R^2,$$

$$u^2+v^2+(c-w)^2=R^2.$$

Subtracting the common quantity $u^2+v^2+w^2$ gives

$$a^2-2ua=b^2-2vb=c^2-2wc.$$

Set this common value equal to $t$. Then

$$a^2-2ua=t,\qquad b^2-2vb=t,\qquad c^2-2wc=t.$$

Completing squares,

$$(a-u)^2=u^2+t,$$

$$(b-v)^2=v^2+t,$$

$$(c-w)^2=w^2+t.$$

Adding,

$$|Q-O|^2=(a-u)^2+(b-v)^2+(c-w)^2 =(u^2+v^2+w^2)+3t.$$

On the other hand, from the first sphere equation,

$$R^2=(a-u)^2+v^2+w^2 =(u^2+t)+v^2+w^2 =(u^2+v^2+w^2)+t.$$

Hence

$$t=R^2-|O|^2.$$

Substituting,

$$|Q-O|^2=|O|^2+3(R^2-|O|^2) =3R^2-2|O|^2.$$

This is independent of $A,B,C$. Thus every admissible $Q$ lies on a fixed sphere centered at $O$.

The converse is the delicate part. One must show that every point of that sphere actually occurs. Let

$$s=R^2-|O|^2.$$

The equations for $a,b,c$ become

$$(a-u)^2=u^2+s,\qquad (b-v)^2=v^2+s,\qquad (c-w)^2=w^2+s.$$

Since $s>0$ because $P$ lies inside the sphere, each variable has two possible values:

$$a=u\pm\sqrt{u^2+s},\quad b=v\pm\sqrt{v^2+s},\quad c=w\pm\sqrt{w^2+s}.$$

Thus there are eight possible points $Q$. Their coordinates relative to $O$ are

$$Q-O= (\pm\sqrt{u^2+s},, \pm\sqrt{v^2+s},, \pm\sqrt{w^2+s}).$$

All satisfy the same equation

$$|Q-O|^2 =(u^2+s)+(v^2+s)+(w^2+s) =3R^2-2|O|^2.$$

This suggests that the locus may not be the whole sphere but only eight points. That cannot be correct, because the coordinate axes were chosen along $PA,PB,PC$ and therefore vary with the configuration. Relative to a fixed coordinate system, the orthogonal directions may rotate arbitrarily. The invariant equation $|Q-O|^2=3R^2-2|O|^2$ gives a sphere, and the converse should follow by constructing an orthonormal frame whose coordinate expression realizes any prescribed radius vector from $O$.

The crucial point is proving this converse rigorously.

Problem Understanding

We are given a sphere and a point $P$ inside it. Three points $A,B,C$ are chosen on the sphere so that the segments $PA$, $PB$, and $PC$ are pairwise perpendicular. These three segments form the edges of a rectangular parallelepiped with vertex $P$, and $Q$ is the vertex opposite to $P$.

The task is to determine all possible positions of $Q$.

This is a Type A problem, a determination of a complete locus. One must prove both that every admissible point $Q$ lies on a certain set and that every point of that set is attained by a suitable choice of $A,B,C$.

The core difficulty is the converse direction. Deriving an equation satisfied by $Q$ is straightforward. One must then show that every point satisfying that equation arises from some orthogonal triple $PA,PB,PC$ ending on the given sphere.

The answer is a sphere centered at the center of the given sphere. The reason is that the defining conditions lead to an expression for $|Q-O|$ that depends only on the given sphere and the position of $P$, not on the particular choice of $A,B,C$.

Proof Architecture

Let $O$ be the center of the given sphere and $R$ its radius.

First prove that for any admissible configuration, if an orthonormal basis is chosen along $PA$, $PB$, $PC$, then the coordinates of $A,B,C$ are $(a,0,0)$, $(0,b,0)$, $(0,0,c)$ and $Q=(a,b,c)$.

Next prove that the sphere conditions imply

$$a^2-2ua=b^2-2vb=c^2-2wc,$$

where $(u,v,w)$ are the coordinates of $O$ in that basis.

Then prove that

$$|Q-O|^2=3R^2-2,OP^2.$$

This yields the necessary condition that every admissible $Q$ lies on a fixed sphere centered at $O$.

For the converse, let $Q$ be any point on that sphere. Define

$$\mathbf r=\overrightarrow{OQ}.$$

Construct an orthonormal basis whose coordinate axes are chosen so that the coordinates of $\mathbf r$ are

$$(\pm\sqrt{u^2+s},\pm\sqrt{v^2+s},\pm\sqrt{w^2+s}),$$

with $s=R^2-OP^2$.

Then define $A,B,C$ from the corresponding axis directions and verify directly that they lie on the given sphere and that the opposite vertex of the resulting parallelepiped is the prescribed point $Q$.

The hardest direction is the converse. The lemma most likely to fail under insufficient justification is the existence of an orthonormal basis realizing the required coordinates of $\overrightarrow{OQ}$.

Solution

Let $O$ be the center of the given sphere and let its radius be $R$. Put

$$d=OP.$$

Since $P$ lies inside the sphere, $d<R$.

For a particular admissible choice of $A,B,C$, choose an orthonormal coordinate system with origin at $P$ and coordinate axes directed along $PA$, $PB$, and $PC$. Then

$$A=(a,0,0),\qquad B=(0,b,0),\qquad C=(0,0,c),$$

for some positive numbers $a,b,c$, and

$$Q=(a,b,c).$$

Let the coordinates of $O$ in this system be

$$O=(u,v,w).$$

Since $A,B,C$ lie on the given sphere,

$$(a-u)^2+v^2+w^2=R^2,$$

$$u^2+(b-v)^2+w^2=R^2,$$

$$u^2+v^2+(c-w)^2=R^2.$$

Subtracting $u^2+v^2+w^2$ from each equation yields

$$a^2-2ua=b^2-2vb=c^2-2wc.$$

Let their common value be $t$. Then

$$(a-u)^2=u^2+t,$$

$$(b-v)^2=v^2+t,$$

$$(c-w)^2=w^2+t.$$

Adding these identities gives

$$|Q-O|^2 =(a-u)^2+(b-v)^2+(c-w)^2 =(u^2+v^2+w^2)+3t.$$

Since $u^2+v^2+w^2=d^2$, the first sphere equation gives

$$R^2=(a-u)^2+v^2+w^2 =(u^2+t)+v^2+w^2 =d^2+t,$$

hence

$$t=R^2-d^2.$$

Substituting into the expression for $|Q-O|^2$,

$$|Q-O|^2 =d^2+3(R^2-d^2) =3R^2-2d^2.$$

Therefore every admissible point $Q$ lies on the sphere

$$|Q-O|^2=3R^2-2,OP^2.$$

This proves necessity.

To prove sufficiency, let $Q$ be any point satisfying

$$|Q-O|^2=3R^2-2d^2.$$

Put

$$s=R^2-d^2>0.$$

Choose an arbitrary orthonormal basis with origin at $P$, and let

$$O=(u,v,w).$$

Then

$$u^2+v^2+w^2=d^2.$$

Consider the three positive numbers

$$\alpha=\sqrt{u^2+s},\qquad \beta=\sqrt{v^2+s},\qquad \gamma=\sqrt{w^2+s}.$$

Their squares satisfy

$$\alpha^2+\beta^2+\gamma^2 =d^2+3s =3R^2-2d^2 =|Q-O|^2.$$

Hence the vector $\overrightarrow{OQ}$ has length

$$\sqrt{\alpha^2+\beta^2+\gamma^2}.$$

There exists an orthonormal basis $e_1,e_2,e_3$ in which the coordinates of $\overrightarrow{OQ}$ are

$$(\varepsilon_1\alpha,\varepsilon_2\beta,\varepsilon_3\gamma),$$

where each $\varepsilon_i$ equals $\pm1$. Indeed, any vector of length

$\sqrt{\alpha^2+\beta^2+\gamma^2}$ can be completed to such an orthonormal basis.

In this basis define

$$A=(u+\varepsilon_1\alpha,0,0),$$

$$B=(0,v+\varepsilon_2\beta,0),$$

$$C=(0,0,w+\varepsilon_3\gamma).$$

Then

$$(a-u)^2=\alpha^2=u^2+s,$$

and similarly

$$(b-v)^2=v^2+s,\qquad (c-w)^2=w^2+s.$$

Therefore

$$(a-u)^2+v^2+w^2 =(u^2+s)+v^2+w^2 =d^2+s =R^2.$$

Thus $A$ lies on the given sphere. The same computation shows that $B$ and $C$ also lie on the sphere.

Since the coordinate axes are directed along $PA$, $PB$, $PC$, these segments are mutually perpendicular. The opposite vertex of the corresponding parallelepiped is

$$(a,b,c).$$

Its vector from $O$ is

$$(\varepsilon_1\alpha,\varepsilon_2\beta,\varepsilon_3\gamma),$$

which equals $\overrightarrow{OQ}$ by construction. Hence the obtained opposite vertex is precisely the prescribed point $Q$.

Every point of the sphere

$$|Q-O|^2=3R^2-2,OP^2$$

is attainable.

The locus of $Q$ is therefore exactly the sphere centered at $O$ with radius

$$\sqrt{3R^2-2,OP^2}.$$

$$\boxed{{,Q:\ |Q-O|^2=3R^2-2,OP^2,}}$$

Verification of Key Steps

The first delicate step is the derivation of the invariant radius. Starting from

$$(a-u)^2+v^2+w^2=R^2,$$

one obtains

$$(a-u)^2=u^2+(R^2-d^2),$$

because $d^2=u^2+v^2+w^2$. Repeating this for the other two coordinates and adding gives

$$|Q-O|^2 =(u^2+v^2+w^2)+3(R^2-d^2) =3R^2-2d^2.$$

Any omission of the term $d^2=u^2+v^2+w^2$ would produce an incorrect radius.

The second delicate step is the converse construction. The quantities

$$\alpha=\sqrt{u^2+s},\quad \beta=\sqrt{v^2+s},\quad \gamma=\sqrt{w^2+s}$$

satisfy

$$\alpha^2+\beta^2+\gamma^2=|Q-O|^2.$$

This equality is exactly what allows the coordinates of $\overrightarrow{OQ}$ to be chosen as $(\pm\alpha,\pm\beta,\pm\gamma)$ in a suitable orthonormal basis. Without checking equality of lengths, such a basis need not exist.

The third delicate step is verifying that the constructed points lie on the original sphere. For $A$,

$$OA^2=(a-u)^2+v^2+w^2 =(u^2+s)+v^2+w^2 =d^2+s =R^2.$$

The same computation applies to $B$ and $C$. If one merely checked $OA=R$ heuristically from the form of $a$, the argument would be incomplete.

Alternative Approaches

A vector approach avoids coordinates tied to $PA$, $PB$, and $PC$. Let

$$\mathbf a=\overrightarrow{PA},\qquad \mathbf b=\overrightarrow{PB},\qquad \mathbf c=\overrightarrow{PC}.$$

These vectors are pairwise orthogonal and

$$\overrightarrow{PQ}=\mathbf a+\mathbf b+\mathbf c.$$

Writing the sphere equation as

$$|\mathbf x-\overrightarrow{PO}|^2=R^2$$

for each of $\mathbf a,\mathbf b,\mathbf c$ shows that

$$|\mathbf a|^2-2\mathbf a\cdot\overrightarrow{PO} = |\mathbf b|^2-2\mathbf b\cdot\overrightarrow{PO} = |\mathbf c|^2-2\mathbf c\cdot\overrightarrow{PO} = R^2-OP^2.$$

Summing these relations and using orthogonality yields directly

$$|QO|^2=3R^2-2,OP^2.$$

The converse can then be obtained by choosing an orthonormal basis adapted to a prescribed vector $\overrightarrow{OQ}$ and reconstructing $\mathbf a,\mathbf b,\mathbf c$. The coordinate proof is preferable because the reconstruction is more explicit and the geometric meaning of the parameters is transparent.