Kvant Math Problem 1263

Let the outer circle have center $O$ and radius $R$.

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Problem

Inside a circle, there are two more circles, each tangent to the outer circle at points $A$ and $B$ respectively, and intersecting each other. Prove that if one of the points of intersection lies on the segment $AB$, then the sum of the radii of the smaller circles equals the radius of the larger circle. Is the converse true?

A. Veselov

Exploration

Let the outer circle have center $O$ and radius $R$. Let the two inner circles have centers $O_1,O_2$ and radii $r_1,r_2$. Since each inner circle is internally tangent to the outer circle, the points $A,O_1,O$ are collinear and $OO_1=R-r_1$; similarly $B,O_2,O$ are collinear and $OO_2=R-r_2$.

Let $P$ be the intersection point of the two inner circles lying on the segment $AB$.

The condition $P\in AB$ suggests using coordinates. Place $O=(0,0)$ and let the line $AB$ be the $x$ axis. Write

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

with $a,b>0$. Since $O$ lies on the perpendicular bisector of $AB$, one has

$$a^2+h^2=b^2+h^2=R^2$$

for some $h$, hence $a=b$. Thus if $AB$ is chosen as the $x$ axis, the outer circle has equation

$$x^2+y^2=R^2,\qquad A=(-R,0),\quad B=(R,0).$$

Because $O_1$ lies on $OA$ and $OO_1=R-r_1$,

$$O_1=(-R+r_1,0).$$

Similarly

$$O_2=(R-r_2,0).$$

The crucial observation is that all four points $A,P,B$ and the centers $O_1,O_2$ lie on the same line. Since $P$ belongs to both small circles, its distances from the centers are exactly $r_1$ and $r_2$. Along the line there are only two possibilities:

$$P=-R+2r_1=R-2r_2.$$

This immediately yields

$$r_1+r_2=R.$$

The converse should also be tested. If $r_1+r_2=R$, then the same coordinate description gives

$$(-R+2r_1)=(R-2r_2),$$

so there is a point on the line $AB$ at distance $r_1$ from $O_1$ and distance $r_2$ from $O_2$. Hence it belongs to both circles. This suggests that the converse is true.

The step most likely to conceal an error is the determination of the correct intersection point on the line. A circle centered at $O_1$ meets the line $AB$ in two points, namely $A$ and $-R+2r_1$. One must check carefully that the common intersection cannot be $A$ or $B$, because the circles are distinct and intersect inside the outer circle.

Problem Understanding

We have a circle of radius $R$ containing two intersecting circles of radii $r_1$ and $r_2$. The first is tangent to the outer circle at $A$, the second at $B$. One of their intersection points lies on the chord $AB$.

The problem asks first to prove that this geometric condition forces

$$r_1+r_2=R.$$

It then asks whether the converse implication is valid.

This is a Type B problem. The main difficulty is to translate the tangency condition into a simple relation between the positions of the centers and then exploit the fact that an intersection point lies on the line $AB$.

Proof Architecture

The first claim is that the centers of the two inner circles lie on the line $AB$ together with the center of the outer circle. This follows from internal tangency, because the center of a circle and the point of tangency with another circle lie on the common normal.

The second claim is that, after choosing coordinates with the outer circle given by $x^2+y^2=R^2$ and $A=(-R,0)$, $B=(R,0)$, the centers of the inner circles are

$$(-R+r_1,0),\qquad (R-r_2,0).$$

This is a direct consequence of the first claim and the distances from the outer center.

The third claim is that if a common point $P$ of the two inner circles lies on $AB$, then

$$P=-R+2r_1=R-2r_2.$$

This comes from computing the second intersection of each inner circle with the line $AB$.

The final claim is that the preceding equality is equivalent to

$$r_1+r_2=R,$$

and conversely that this relation produces a common point on $AB$.

The hardest direction is proving that the existence of $P\in AB$ forces $r_1+r_2=R$. The most delicate lemma is the identification of the correct intersection point of each small circle with the line $AB$.

Solution

Let the outer circle have center $O$ and radius $R$. Let the inner circles have centers $O_1,O_2$ and radii $r_1,r_2$.

Since the first inner circle is internally tangent to the outer circle at $A$, the points $O,A,O_1$ are collinear and

$$OO_1=R-r_1.$$

Similarly, the points $O,B,O_2$ are collinear and

$$OO_2=R-r_2.$$

Choose coordinates so that

$$O=(0,0),$$

and the line $AB$ is the $x$ axis. Because $A$ and $B$ lie on the outer circle and $O$ lies on the perpendicular bisector of $AB$, we may write

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

From the tangency relations,

$$O_1=(-R+r_1,0),\qquad O_2=(R-r_2,0).$$

Let $P$ be a common point of the two inner circles lying on the segment $AB$.

The first inner circle intersects the line $AB$ at two points. One of them is the tangency point $A=(-R,0)$. Since the center is $O_1=(-R+r_1,0)$ and the radius is $r_1$, the other intersection point is

$$x=-R+r_1+r_1=-R+2r_1.$$

Likewise, the second inner circle intersects the line $AB$ at $B=(R,0)$ and at

$$x=R-r_2-r_2=R-2r_2.$$

Since $P$ belongs to both circles and lies on the line $AB$, it must be this second intersection point for each circle. Hence

$$-R+2r_1=R-2r_2.$$

Rearranging,

$$2(r_1+r_2)=2R,$$

and therefore

$$r_1+r_2=R.$$

This proves the first statement.

For the converse, assume

$$r_1+r_2=R.$$

Then

$$-R+2r_1 = R-2r_2.$$

Let

$$P=(-R+2r_1,0).$$

Its distance from $O_1=(-R+r_1,0)$ equals

$$|(-R+2r_1)-(-R+r_1)|=r_1,$$

so $P$ lies on the first inner circle.

Its distance from $O_2=(R-r_2,0)$ equals

$$|(R-r_2)-(-R+2r_1)| = |2R-r_2-2r_1| = |R-r_1-r_2+r_2| = r_2,$$

because $r_1+r_2=R$.

Thus $P$ also lies on the second inner circle. Since $P$ lies on the line $AB$, the two circles have an intersection point on $AB$.

Hence the converse is true.

This completes the proof.

∎

Verification of Key Steps

The first delicate step is locating the centers. Internal tangency at $A$ means that the common tangent at $A$ is perpendicular to both radii through $A$. Consequently $O,A,O_1$ are collinear. The same argument gives $O,B,O_2$ collinear. Without this fact, the coordinate description of the centers would be unjustified.

The second delicate step is finding the second intersection of the first inner circle with the line $AB$. The center is at $x=-R+r_1$. Points of the circle on the $x$ axis satisfy

$$|x-(-R+r_1)|=r_1.$$

The solutions are

$$x=-R,\qquad x=-R+2r_1.$$

The first is the tangency point $A$; the second is the only other point of the circle on the line. Confusing these two solutions would invalidate the argument.

The third delicate step is the converse. One must prove that the point obtained from

$$-R+2r_1=R-2r_2$$

actually lies on both circles. This requires checking both distance equalities explicitly. Merely observing that the coordinates coincide does not establish membership in the circles.

Alternative Approaches

A synthetic proof can be obtained by projecting everything onto the line $AB$. Because the centers lie on $AB$, the first small circle cuts the line at $A$ and at a second point $P_1$ with

$$AP_1=2r_1.$$

Similarly, the second circle cuts the line at $B$ and at a second point $P_2$ with

$$BP_2=2r_2.$$

If a common intersection lies on $AB$, then $P_1=P_2$. Since

$$AB=2R,$$

one gets

$$2r_1+2r_2=2R,$$

hence $r_1+r_2=R$. The converse is obtained by reversing the argument.

The coordinate method is preferable because it makes the location of the relevant points completely explicit and reduces the entire problem to a single linear equation.