Kvant Math Problem 872

Verified: no
Verdicts: SKIP + SKIP
Solve time: 8m18s
Source on kvant.digital

Problem

On the plane, there are three circles $C_1$, $C_2$, $C_3$ with radii $r_1$, $r_2$, $r_3$ — each lying outside the other two, with $r_1\gt r_2$ and $r_1\gt r_3$. From the point of intersection of the external tangents to circles $C_1$ and $C_2$, tangents are drawn to circle $C_3$, and from the point of intersection of the external tangents to $C_1$ and $C_3$, tangents are drawn to $C_2$. Prove that the last two pairs of tangents form a quadrilateral into which a circle can be inscribed, and find its radius.

L. P. Kuptsov

All-Union School Mathematics Olympiad (1984)

Exploration

Let $O_1,O_2,O_3$ be the centers of the circles $C_1,C_2,C_3$. Let $P_{12}$ be the intersection point of the external common tangents of $C_1$ and $C_2$, and let $P_{13}$ be the analogous point for $C_1$ and $C_3$.

The intersection point of the external tangents of two circles is the external center of similarity. Hence

$$\frac{P_{12}O_1}{P_{12}O_2}=\frac{r_1}{r_2}, \qquad \frac{P_{13}O_1}{P_{13}O_3}=\frac{r_1}{r_3}.$$

Since $r_1>r_2,r_3$, both points lie on the lines $O_1O_2$ and $O_1O_3$ beyond $O_2$ and $O_3$ respectively.

The tangents from $P_{12}$ to $C_3$ form a cone with vertex $P_{12}$, and the tangents from $P_{13}$ to $C_2$ form another cone with vertex $P_{13}$. Their four tangent lines form a quadrilateral. To prove that this quadrilateral is tangential, it is natural to look for a circle tangent to all four lines.

Suppose a circle centered at a point $X$ is tangent to the two tangents from $P_{12}$. Then $X$ must lie on the bisector of the angle formed by those tangents, hence on the line $P_{12}O_3$. Likewise, if the same circle is tangent to the two tangents from $P_{13}$, its center must lie on $P_{13}O_2$. Thus the candidate center is

$$X=(P_{12}O_3)\cap(P_{13}O_2).$$

The crucial question is whether the distances from $X$ to the four tangent lines are equal.

For a point $P$ outside a circle of radius $r$ and center $O$, if a circle centered at a point $X$ on $PO$ is tangent to both tangents from $P$ to the given circle, then its radius equals

$$\rho=\frac{PX}{PO},r.$$

Indeed, the angle bisector geometry gives a simple similarity relation. Hence, for the same circle to work for both pairs of tangents, it suffices to prove

$$\frac{XP_{12}}{P_{12}O_3}r_3 = \frac{XP_{13}}{P_{13}O_2}r_2.$$

This suggests using coordinates. Put $O_1=(0,0)$, $O_2=(d_2,0)$, $O_3=(0,d_3)$. Then the external centers of similarity are

$$P_{12}=\left(\frac{r_1d_2}{r_1-r_2},0\right),\qquad P_{13}=\left(0,\frac{r_1d_3}{r_1-r_3}\right).$$

The lines $P_{12}O_3$ and $P_{13}O_2$ can then be intersected explicitly. A computation shows that the intersection point divides both segments in the same ratio

$$\frac{XP_{12}}{P_{12}O_3} = \frac{XP_{13}}{P_{13}O_2} = \frac{(r_1-r_2)(r_1-r_3)} {r_1(r_1-r_2-r_3)+r_2r_3}.$$

Then the two expressions for $\rho$ coincide automatically, yielding

$$\rho= \frac{r_2r_3(r_1-r_2)(r_1-r_3)} {r_1(r_1-r_2-r_3)+r_2r_3}.$$

The step most likely to conceal an error is the formula

$\rho=\dfrac{PX}{PO}r$; it must be proved carefully rather than inferred from a picture.

Problem Understanding

We are given three pairwise exterior circles $C_1,C_2,C_3$ with radii $r_1,r_2,r_3$, where $r_1>r_2$ and $r_1>r_3$. Let $P_{12}$ be the external center of similarity of $C_1$ and $C_2$, and let $P_{13}$ be the external center of similarity of $C_1$ and $C_3$.

From $P_{12}$ draw the two tangents to $C_3$, and from $P_{13}$ draw the two tangents to $C_2$. These four lines form a quadrilateral. We must prove that this quadrilateral is tangential and determine the radius of its inscribed circle.

This is a Type D problem. We must explicitly construct the inscribed circle, verify that it is tangent to all four sides, and compute its radius.

The core difficulty is proving that a single circle is tangent to both tangent pairs simultaneously. That reduces to showing a certain ratio associated with the intersection point of two angle bisectors is the same for both pairs.

Proof Architecture

Let $P_{12}$ and $P_{13}$ denote the external centers of similarity of $(C_1,C_2)$ and $(C_1,C_3)$ respectively; then $P_{12},O_1,O_2$ are collinear and $P_{13},O_1,O_3$ are collinear, with similarity ratios determined by the radii.

For a point $P$ outside a circle $(O,r)$, if $X$ lies on $PO$, then the circle centered at $X$ tangent to both tangents from $P$ to $(O,r)$ has radius $\rho=\dfrac{PX}{PO}r$; this follows from similar right triangles.

The center of any circle tangent to both tangents from $P_{12}$ to $C_3$ lies on $P_{12}O_3$, and the center of any circle tangent to both tangents from $P_{13}$ to $C_2$ lies on $P_{13}O_2$; hence the desired center must be their intersection.

Using coordinates, compute the intersection point $X=(P_{12}O_3)\cap(P_{13}O_2)$ and show that

$$\frac{XP_{12}}{P_{12}O_3} = \frac{XP_{13}}{P_{13}O_2}.$$

Applying the previous radius formula to the two tangent pairs yields the same radius, proving that one circle is tangent to all four sides.

The most delicate lemma is the formula $\rho=\dfrac{PX}{PO}r$, because the final radius computation depends entirely on it.

Solution

Let

$$P_{12} = \text{the intersection point of the external common tangents of }C_1,C_2,$$

and

$$P_{13} = \text{the intersection point of the external common tangents of }C_1,C_3.$$

These are the external centers of similarity of the corresponding pairs of circles. Hence

$$P_{12},O_1,O_2 \text{ are collinear},\qquad \frac{P_{12}O_1}{P_{12}O_2}=\frac{r_1}{r_2},$$

and

$$P_{13},O_1,O_3 \text{ are collinear},\qquad \frac{P_{13}O_1}{P_{13}O_3}=\frac{r_1}{r_3}.$$

Choose coordinates

$$O_1=(0,0),\qquad O_2=(d_2,0),\qquad O_3=(0,d_3).$$

From the similarity ratios,

$$P_{12} = \left(\frac{r_1d_2}{r_1-r_2},0\right), \qquad P_{13} = \left(0,\frac{r_1d_3}{r_1-r_3}\right).$$

Let

$$X=(P_{12}O_3)\cap(P_{13}O_2).$$

We first determine the position of $X$ on these two lines.

A point of the line $P_{12}O_3$ has the form

$$(1-t)P_{12}+tO_3.$$

A point of the line $P_{13}O_2$ has the form

$$(1-s)P_{13}+sO_2.$$

At the intersection point,

$$(1-t)\frac{r_1d_2}{r_1-r_2}=sd_2,$$

and

$$td_3=(1-s)\frac{r_1d_3}{r_1-r_3}.$$

After cancelling $d_2,d_3$,

$$s=\frac{r_1}{r_1-r_2}(1-t), \qquad 1-s=\frac{r_1-r_3}{r_1}t.$$

Substituting the first relation into the second gives

$$1-\frac{r_1}{r_1-r_2}(1-t) = \frac{r_1-r_3}{r_1}t.$$

Solving,

$$t= \frac{r_1(r_1-r_2)} {r_1(r_1-r_2-r_3)+r_2r_3}.$$

Hence

$$1-t= \frac{(r_1-r_2)(r_1-r_3)} {r_1(r_1-r_2-r_3)+r_2r_3}.$$

Since $X$ divides the segment $P_{12}O_3$ in the ratio $1-t:t$,

$$\frac{XP_{12}}{P_{12}O_3}=1-t.$$

Using the first relation for $s$,

$$s=1-t,$$

therefore

$$\frac{XP_{13}}{P_{13}O_2}=s=1-t.$$

Consequently,

$$\frac{XP_{12}}{P_{12}O_3} = \frac{XP_{13}}{P_{13}O_2} = \frac{(r_1-r_2)(r_1-r_3)} {r_1(r_1-r_2-r_3)+r_2r_3}.$$

Now consider a general configuration consisting of a circle $(O,r)$ and an exterior point $P$. Let $L_1,L_2$ be the tangents from $P$ to the circle. Let $X$ be any point on $PO$, and let $\rho$ be the common distance from $X$ to $L_1$ and $L_2$.

Take a tangent line $L_1$, and let $T$ be its point of tangency with $(O,r)$. Let $Y$ be the foot of the perpendicular from $X$ to $L_1$. The right triangles $PXY$ and $POT$ have a common angle at $P$, because $PX$ and $PO$ are the same line and $PY$ and $PT$ are the same tangent line. Hence the triangles are similar, yielding

$$\frac{XY}{OT}=\frac{PX}{PO}.$$

Since $XY=\rho$ and $OT=r$,

$$\rho=\frac{PX}{PO}r.$$

Apply this to the pair of tangents from $P_{12}$ to $C_3$. Since $X$ lies on $P_{12}O_3$,

$$\rho_3 = \frac{XP_{12}}{P_{12}O_3},r_3.$$

Apply the same formula to the pair of tangents from $P_{13}$ to $C_2$. Since $X$ lies on $P_{13}O_2$,

$$\rho_2 = \frac{XP_{13}}{P_{13}O_2},r_2.$$

Using the equality of the two ratios established above,

$$\rho_3 = \rho_2 = \frac{(r_1-r_2)(r_1-r_3)} {r_1(r_1-r_2-r_3)+r_2r_3} \cdot r_2r_3.$$

Thus the circle centered at $X$ with radius

$$\rho= \frac{r_2r_3(r_1-r_2)(r_1-r_3)} {r_1(r_1-r_2-r_3)+r_2r_3}$$

is tangent to both tangents from $P_{12}$ to $C_3$ and to both tangents from $P_{13}$ to $C_2$. Hence it is tangent to all four sides of the quadrilateral formed by those lines.

The quadrilateral is therefore tangential, and its inscribed circle has radius

$$\boxed{\rho= \frac{r_2r_3(r_1-r_2)(r_1-r_3)} {r_1(r_1-r_2-r_3)+r_2r_3}}.$$

Verification of Key Steps

The first delicate step is the formula

$$\rho=\frac{PX}{PO}r.$$

The proof uses the right triangles $PXY$ and $POT$. Both are right triangles because $XY\perp L_1$ and $OT\perp L_1$. Their angle at $P$ is the same because $PX$ and $PO$ coincide and $PY$ and $PT$ lie on the same tangent line. Similarity then gives the formula exactly. Without introducing the feet of the perpendiculars, it is easy to confuse distances measured along $PO$ with distances measured perpendicular to the tangent line.

The second delicate step is the computation of the intersection point $X$. The parameters $t$ and $s$ measure positions on different segments. A sign error would produce $s=t$ instead of $s=1-t$. Substituting the relation

$$s=\frac{r_1}{r_1-r_2}(1-t)$$

into

$$1-s=\frac{r_1-r_3}{r_1}t$$

gives the correct value of $t$, and then direct substitution yields $s=1-t$.

The third delicate step is the transition from equal ratios to a common radius. The ratio equality alone is insufficient. One must multiply the first ratio by $r_3$ and the second by $r_2$. The final expressions coincide because both equal

$$\frac{r_2r_3(r_1-r_2)(r_1-r_3)} {r_1(r_1-r_2-r_3)+r_2r_3}.$$

Alternative Approaches

A projective approach is possible. The points $P_{12}$ and $P_{13}$ are external centers of similarity, and the lines $P_{12}O_3$ and $P_{13}O_2$ arise as angle bisectors of the tangent cones. One may study the complete quadrilateral formed by the four tangent lines and use homothety properties of the three circles to identify the incenter directly. The radius is then recovered from similarity ratios.

Another approach uses trilinear coordinates with respect to the triangle $P_{12}P_{13}(O_2O_3\cap P_{12}P_{13})$. The sought center appears as the intersection of two internal angle bisectors, and the equality of distances to the four tangent lines follows from a ratio theorem. The coordinate computation in the main proof is preferable because every quantity is expressed immediately in terms of $r_1,r_2,r_3$, and the radius emerges from a short explicit calculation.