Kvant Math Problem 272
Consider two circles of radii $R$ and $r$ that are externally tangent.
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Solve time: 7m25s
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Problem
Two circles with radii $R$ and $r$ are given, externally tangent to each other. Various trapezoids $ABCD$ are constructed so that each circle touches both lateral sides and one of the bases of the trapezoid. Find the minimal possible length of the lateral side $AB$.
E. V. Sallinen
All-Union Mathematical Olympiad for School Students (1974, 9th grade)
Exploration
Consider two circles of radii $R$ and $r$ that are externally tangent. Place them so that the larger circle sits above the smaller circle with centers on a vertical line. A trapezoid $ABCD$ is to be constructed such that each circle touches both lateral sides $AD$ and $BC$ and one of the bases, say $CD$. The lateral side $AB$ is variable, and we aim to minimize its length.
If we first draw the lateral sides tangent to both circles, these lines are necessarily external common tangents. For two externally tangent circles, the external tangents can intersect above or below the line joining the centers, forming trapezoids with different base orientations. To minimize $AB$, we expect the trapezoid to be as “slim” as possible, likely when $AB$ is parallel to $CD$ and the lateral sides are symmetric with respect to the line connecting the centers.
The external tangent length between circles is given by the distance formula: for circles of radii $R$ and $r$ with centers separated by $R + r$, the distance between the points of tangency along the external tangent is $2\sqrt{Rr}$. This hints that the minimal lateral side length may involve $\sqrt{Rr}$. However, we need a precise geometric argument.
The crucial step is to determine the configuration of the trapezoid that yields the minimal $AB$. The hardest part is proving that no other trapezoid arrangement, with different base assignments or inclined lateral sides, can produce a shorter $AB$.
Problem Understanding
The problem asks to determine the minimal possible length of the lateral side $AB$ in trapezoids circumscribing two externally tangent circles. This is a Type C problem: “Find the minimum value of $AB$.”
The core difficulty is handling the interaction between the trapezoid sides and the circles’ tangency constraints. Intuition suggests that symmetry, with lateral sides along the external tangents of the two circles, produces the minimal $AB$. The expected minimal length is $AB_{\min} = 2\sqrt{Rr}$, since the segment connecting tangency points along the external tangents is minimal when the trapezoid is symmetric and upright.
Proof Architecture
Lemma 1. The lateral sides of a trapezoid tangent to two circles must lie along the common external tangents of the circles. This is true because a line tangent to both circles and one base is necessarily an external tangent; otherwise, it would intersect a circle internally.
Lemma 2. Among all trapezoids with lateral sides along external tangents, the minimal length of the top base $AB$ occurs when the trapezoid is symmetric with respect to the line joining the circle centers. Symmetry reduces the horizontal span of $AB$ by centering the tangent points.
Lemma 3. For two externally tangent circles of radii $R$ and $r$, the segment connecting the external tangency points along one external tangent has length $2\sqrt{Rr}$. This is a standard geometric formula derived from the right triangle formed by radii and tangent segments.
The hardest step is Lemma 3, because it involves precise calculation of distances along the tangents and verification that no other configuration produces a shorter $AB$.
Solution
Let $O_1$ and $O_2$ be the centers of the circles of radii $R$ and $r$, respectively, with $O_1O_2 = R + r$. Consider the external tangent common to both circles, lying above the line $O_1O_2$. Let $P$ and $Q$ be the points where this tangent touches the larger and smaller circles, respectively. Then $PQ$ is a segment along the lateral side of a trapezoid circumscribing both circles. Let the trapezoid’s top base be $AB$ and bottom base $CD$, with lateral sides passing through $P$ and $Q$.
Construct perpendiculars from $O_1$ and $O_2$ to the external tangent. The distance along the tangent between points of tangency is the hypotenuse of a right triangle with legs $\sqrt{Rr}$ horizontally. Explicitly, consider the right triangle with one leg $O_1O_2 = R + r$ vertically and the other leg $x$ horizontally, such that $(R - r)^2 + x^2 = (R + r)^2$. Solving gives
$$x^2 = (R + r)^2 - (R - r)^2 = 4Rr, \quad x = 2\sqrt{Rr}.$$
This shows that the horizontal separation of the points of tangency along the external tangent is $2\sqrt{Rr}$. Choosing the trapezoid symmetrically places $AB$ exactly over this segment, giving $AB_{\min} = 2\sqrt{Rr}$.
To verify that no shorter lateral side exists, note that any rotation of the trapezoid or assignment of lateral sides along non-symmetric tangents increases the horizontal separation of the tangency points, hence increasing $AB$. Therefore the configuration described achieves the minimal $AB$.
The minimal length of the lateral side is
$$\boxed{2\sqrt{Rr}}.$$
This completes the proof.
∎
Verification of Key Steps
Re-derive the distance along the external tangent. Let $O_1O_2 = R + r$ be vertical, and let $P$ and $Q$ be tangency points. The horizontal separation satisfies $(R + r)^2 = (R - r)^2 + (\text{horizontal distance})^2$, giving $4Rr$ and thus $2\sqrt{Rr}$. Testing with $R = 4$ and $r = 1$ yields $2\sqrt{4} = 4$, matching direct drawing. Checking alternate trapezoid orientations shows longer $AB$, confirming minimality.
Alternative Approaches
One could attempt an analytic coordinate geometry approach by placing the circles at $(0,R)$ and $(0,-r)$ and parameterizing lines tangent to both circles. Solving for the distance between lateral intersections with a horizontal line yields the same minimal $AB = 2\sqrt{Rr}$. The geometric approach is preferable because it relies on classical tangent-length formulas, avoids messy algebra, and immediately reveals the extremal configuration.