Kvant Math Problem 1373

Let the sphere have radius $R$, and let the given plane be $\pi$.

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Problem

A plane intersects a sphere with center $O$ along a circle. Points $A$ and $B$ are taken on the sphere on opposite sides of the plane, with radius $OA$ perpendicular to the given plane. Through the line $AB$, an arbitrary plane is drawn. It intersects the circle at points $X$ and $Y$. Prove that the product $BX\cdot BY$ does not depend on the choice of such a plane.

B. Chinik

Interrepublican Mathematical Olympiad 1992

Exploration

Let the sphere have radius $R$, and let the given plane be $\pi$. Since $OA\perp\pi$, the center $O$ lies on the line through $A$ perpendicular to $\pi$. The circle cut from the sphere by $\pi$ has center at the foot $C$ of the perpendicular from $O$ to $\pi$.

Every plane through $AB$ intersects $\pi$ in some line through the point $P=AB\cap\pi$. Since $X$ and $Y$ belong both to the chosen plane and to the circle in $\pi$, the chord $XY$ is exactly the intersection of the circle with a line through $P$. Thus the freedom in choosing the plane is equivalent to choosing an arbitrary chord of the fixed circle through the fixed point $P$.

The quantity to study is $BX\cdot BY$. Since $B,P,X,Y$ all lie in the chosen plane and $P,X,Y$ are collinear, perhaps the power of a point with respect to the circle is relevant. For the circle in $\pi$,

$PX\cdot PY=\operatorname{Pow}_{\Gamma}(P),$

which is independent of the chosen chord through $P$.

To connect this with $BX\cdot BY$, it is natural to compare triangles having vertex $B$ and base on the line $PXY$. Let $A$ be placed above $\pi$ and $B$ below it. Because $A,O,B$ are collinear and $A$ lies on the sphere, the line $AB$ passes through the center, hence $AB$ is a diameter. Then $O$ is the midpoint of $AB$.

Choose coordinates. Let

$\pi:\ z=h,\qquad O=(0,0,0),\qquad A=(0,0,R),\qquad B=(0,0,-R).$

Then $P=(0,0,h)$, because $AB$ is the $z$-axis. Any point of the circle has coordinates $(u,v,h)$ with $u^2+v^2=R^2-h^2$.

For such a point $X$,

$BX^2=u^2+v^2+(h+R)^2=(R^2-h^2)+(h+R)^2=2R(R+h).$

The same holds for every point of the circle. Hence $B$ is equidistant from all points of the circle. This is the crucial observation. Then in any chosen plane, $BX=BY$ equals a constant.

Thus it suffices to prove that $BX^2$ is constant and combine this with the power of $P$ in the circle. Along the line $XY$,

$BX^2=BP^2+PX^2,\qquad BY^2=BP^2+PY^2,$

because $BP$ is perpendicular to $\pi$, hence perpendicular to every line in $\pi$, in particular to $XY$.

Since $BX=BY$, we get $PX^2=PY^2$, so $P$ is the midpoint of $XY$. Then

$PX\cdot PY=PX^2=BX^2-BP^2.$

Therefore

$BX\cdot BY=BX^2=BP^2+PX\cdot PY.$

The first term is fixed, and the second is the power of $P$ with respect to the fixed circle. Hence the product is constant.

The delicate point is showing that $P$ is the midpoint of every chord through $P$. That follows from $BX=BY$ together with the right triangles $BPX$ and $BPY$.

Problem Understanding

A sphere is cut by a plane $\pi$ in a circle $\Gamma$. Points $A$ and $B$ lie on the sphere on opposite sides of $\pi$, and the radius $OA$ is perpendicular to $\pi$. Through the line $AB$ we draw an arbitrary plane. This plane meets the circle $\Gamma$ at points $X$ and $Y$. The task is to prove that the product $BX\cdot BY$ is independent of the chosen plane.

This is a Type B problem. The statement is to be proved.

The core difficulty is to exploit simultaneously the geometry of the sphere and the geometry of the circle section. The key observation is that every point of the circle is at the same distance from $B$.

Proof Architecture

First, prove that $AB$ is a diameter of the sphere, because $A$, $O$, and $B$ are collinear and both $A$ and $B$ lie on the sphere.

Second, let $P=AB\cap\pi$ and prove that every point of the section circle is equidistant from $B$; this follows from a right triangle computation involving the distance from $B$ to the plane.

Third, show that for any chosen plane through $AB$, the chord $XY$ lies on a line of $\pi$ passing through $P$, and that $BP\perp XY$.

Fourth, use the equality $BX=BY$ together with the right triangles $BPX$ and $BPY$ to deduce $PX=PY$, so $P$ is the midpoint of the chord $XY$.

Fifth, apply the power-of-a-point theorem to the fixed circle $\Gamma$ and the fixed point $P$, obtaining that $PX\cdot PY$ is constant.

Finally, combine $BX^2=BP^2+PX^2$ with $PX^2=PX\cdot PY$ and the constancy of the power to conclude that $BX\cdot BY$ is constant.

The lemma most likely to fail under scrutiny is the claim that $PX=PY$ for every chord through $P$; it requires the prior proof that $BX=BY$ for all points of the circle.

Solution

Let $\pi$ be the given plane, let $\Gamma$ be the circle in which $\pi$ intersects the sphere, and let

$P=AB\cap\pi.$

Since $OA$ is a radius perpendicular to $\pi$, the center $O$ lies on the line $AB$. The points $A$ and $B$ lie on the sphere, hence

$OA=OB.$

Since $A$ and $B$ are on opposite sides of $\pi$, they lie on opposite sides of $O$ on the same line. Thus $O$ is the midpoint of $AB$, and $AB$ is a diameter of the sphere.

Let $C$ be the center of the circle $\Gamma$. Since $\Gamma$ is obtained by intersecting the sphere with the plane $\pi$, the point $C$ is the orthogonal projection of $O$ onto $\pi$. Denote by $r$ the radius of $\Gamma$.

The line $AB$ is perpendicular to $\pi$, because it contains the radius $OA$ perpendicular to $\pi$. Hence

$BP\perp\pi.$

Take any point $T\in\Gamma$. Since $T$ lies in $\pi$, the triangle $BPT$ is right-angled at $P$. Therefore

BT^{2}=BP^{2}+PT^{2}. \tag{1}

The point $P$ lies on the line through $O$ perpendicular to $\pi$, and $C$ also lies on that line. Consequently $P=C$. Hence $PT=r$ for every point $T$ of the circle $\Gamma$. From (1),

$BT^{2}=BP^{2}+r^{2},$

which is independent of the choice of $T$. Thus every point of $\Gamma$ is at the same distance from $B$.

In particular,

BX=BY. \tag{2}

Now choose any plane through $AB$. Its intersection with $\pi$ is a line through $P$. Since $X$ and $Y$ belong both to this plane and to $\pi$, the points $P,X,Y$ are collinear.

Because $BP\perp\pi$, we have

$BP\perp XY.$

Hence the triangles $BPX$ and $BPY$ are right triangles. Using (2) and the Pythagorean theorem,

$BP^{2}+PX^{2}=BX^{2}=BY^{2}=BP^{2}+PY^{2},$

so

$PX^{2}=PY^{2}.$

Since $P,X,Y$ are collinear, this gives

PX=PY. \tag{3}

Therefore $P$ is the midpoint of the chord $XY$.

Apply the power-of-a-point theorem to the fixed circle $\Gamma$ and the fixed point $P$. For every line through $P$ meeting $\Gamma$ at $X$ and $Y$,

PX\cdot PY=\operatorname{Pow}_{\Gamma}(P). \tag{4}

The right-hand side depends only on $\Gamma$ and $P$, hence it is constant.

Using (3),

PX^{2}=PX\cdot PY. \tag{5}

From the right triangle $BPX$,

$BX^{2}=BP^{2}+PX^{2}.$

Combining this with (5) and (4),

$BX^{2}=BP^{2}+\operatorname{Pow}_{\Gamma}(P).$

The right-hand side is independent of the chosen plane. Since $BX=BY$, we obtain

$BX\cdot BY=BX^{2},$

and therefore $BX\cdot BY$ is independent of the choice of the plane through $AB$.

This completes the proof.

∎

Verification of Key Steps

The first delicate point is the claim that every point of the circle is equidistant from $B$. Let $T\in\Gamma$. Since $BP$ is perpendicular to the whole plane $\pi$, the triangle $BPT$ is right. The distance $PT$ equals the fixed radius $r$ of the circle. Hence

$BT^{2}=BP^{2}+r^{2},$

which is the same for all $T$. No property of the chosen plane through $AB$ is involved.

The second delicate point is the deduction that $PX=PY$. A careless argument might confuse equality of chords with equality of distances from $P$. The correct reasoning is

$$BY^{2}=BP^{2}+PY^{2},$$

and $BX=BY$. Subtracting gives $PX^{2}=PY^{2}$. Since distances are nonnegative, $PX=PY$.

The third delicate point is the use of power of a point. The theorem applies because $P$ is fixed and every admissible chord $XY$ lies on a line through $P$. Thus

$PX\cdot PY$

is the same for every choice of the line, which is exactly the invariant needed in the final computation.

Alternative Approaches

A coordinate solution is very short. Let the sphere be

$x^{2}+y^{2}+z^{2}=R^{2},$

and let the cutting plane be $z=h$. Then

$A=(0,0,R),\qquad B=(0,0,-R).$

The circle is

$$$$

For every point $T=(x,y,h)$ on this circle,

$$=(R^{2}-h^{2})+(R+h)^{2} =2R(R+h),$$

so all points of the circle are at the same distance from $B$. Hence $BX=BY=\sqrt{2R(R+h)}$, and

$$$$

which is manifestly independent of the chosen plane.

The synthetic proof is preferable because it explains geometrically why the quantity is invariant and uses only elementary properties of sphere sections and powers of points.