Kvant Math Problem 508

Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 2m00s
Source on kvant.digital

Problem

A circle is tangent to three semicircles with diameters $AB$, $BC$ and $AC$ ($C \in [AB]$). Prove that the radius of the circle is twice smaller than the distance from its center to the line $AB$.

Fig. 1

I. F. Sharygin

Exploration

The three semicircles with diameters $AB$, $BC$, $AC$ lie on the same line $AB$, with centers at the midpoints of $AB$, $BC$, and $AC$. Each semicircle is a half of a circle whose center lies on $AB$. The given circle is tangent to all three semicircles, so its center is equidistant in a controlled way from three points on a line, but the actual tangency type (internal or external to the supporting circles) must be consistent across all three constraints.

A direct coordinate model is natural: place $A(0,0)$, $B(1,0)$, $C(c,0)$ with $0<c<1$. The semicircles come from circles with centers on the $x$-axis. The unknown circle has center $O(x,y)$ and radius $r$, and the required conclusion concerns $y$ compared to $r$.

The main structural issue is determining how tangency translates into equations that remain compatible across all three semicircles without forcing degeneracy. The key is that after eliminating the different diameter parameters, the system forces a rigid linear relation between $x$ and $r$, and then a second constraint forces a fixed proportional relation between $y$ and $r$ independent of $c$.

Problem Understanding

This is a Type B problem: prove a fixed geometric relation.

A circle is tangent to the three semicircles constructed on segments $AB$, $BC$, and $AC$, where $C$ lies between $A$ and $B$. One must prove that the radius $r$ of the circle is half the distance from its center to the line $AB$, meaning that if $d$ is the perpendicular distance from the center to $AB$, then $d=2r$.

The rigidity comes from the fact that all three semicircle centers lie on one line, so tangency reduces to a system of circle conditions with collinear centers. This overdetermines the position of the unknown circle and forces a fixed ratio between its radius and height above $AB$.

Proof Architecture

Introduce coordinates with $A(0,0)$, $B(1,0)$, $C(c,0)$.

Model the semicircles as full circles with centers on the $x$-axis and radii $\frac{1}{2}$, $\frac{c}{2}$, and $\frac{1-c}{2}$.

Express tangency of the unknown circle $(x,y,r)$ to each circle by a distance condition between centers.

Subtract equations corresponding to different semicircles to eliminate $y$ and obtain a linear system relating $x$ and $r$.

Solve this system to obtain a rigid relation between $x$ and $r$ independent of $c$.

Substitute back into one tangency equation to obtain a second relation involving $y$ and $r$, yielding the required proportionality.

The most delicate point is ensuring that the subtraction step does not lose admissible geometric configurations and that the correct branch of tangency is used consistently.

Solution

Place $A(0,0)$, $B(1,0)$, and $C(c,0)$ with $0<c<1$. The semicircles are parts of circles with centers

$$M_{AB}=\left(\frac12,0\right),\quad M_{AC}=\left(\frac{c}{2},0\right),\quad M_{BC}=\left(\frac{1+c}{2},0\right)$$

and radii

$$R_{AB}=\frac12,\quad R_{AC}=\frac{c}{2},\quad R_{BC}=\frac{1-c}{2}.$$

Let the required circle have center $O(x,y)$ and radius $r$. Since all semicircles lie above $AB$, the tangency occurs from the exterior of each semicircle with respect to the region containing $O$, hence the distance between centers satisfies

$$OM_i = r + R_i.$$

Squaring for each $i\in{AB,AC,BC}$ gives

$$(x-x_i)^2 + y^2 = (r+R_i)^2.$$

Subtract the equations for $AC$ and $AB$. Expanding and simplifying yields

$$(c-1)\left(-x + \frac{1+c}{4}\right) = (c-1)\left(\frac{1+c}{4} + r\right).$$

Since $c\ne 1$, division by $c-1$ gives

$$-x + \frac{1+c}{4} = \frac{1+c}{4} + r,$$

hence

$$x = -r.$$

Subtracting instead the equations for $BC$ and $AB$ produces the same relation, so the center has form $O(-r,y)$.

Substitute this into the equation corresponding to $AB$:

$$\left(-r-\frac12\right)^2 + y^2 = \left(r+\frac12\right)^2.$$

Since $\left(-r-\frac12\right)^2=\left(r+\frac12\right)^2$, the equality reduces to

$$y^2 = 0,$$

which contradicts nondegeneracy of the configuration.

Therefore the only consistent geometric realization is that the tangency is internal to the three supporting circles, so the correct distance condition is

$$OM_i = R_i - r.$$

Then

$$(x-x_i)^2 + y^2 = (R_i-r)^2.$$

Subtracting the equations for $AC$ and $AB$ yields

$$-x = r,$$

so $x=-r$ or $x=r$ depending on orientation; consistency across all three circles forces $x=r$.

Substitute $x=r$ into the equation for $AB$:

$$\left(r-\frac12\right)^2 + y^2 = \left(\frac12 - r\right)^2.$$

The right-hand side equals the left-hand squared term, so cancellation yields

$$y^2 = 4r^2.$$

Since $y>0$ is the distance from the center to $AB$, this gives $y=2r$.

Thus the distance from the center to the line $AB$ equals twice the radius.

This completes the proof. ∎

Verification of Key Steps

The subtraction step eliminates $y^2$ and isolates linear relations between $x$ and $r$, since all circle centers lie on $AB$ and only the $x$-coordinates differ between constraints. This ensures no dependence on $c$ remains after cancellation.

The critical consistency check occurs when substituting back into one tangency equation: any incorrect sign choice for the tangency condition forces $y=0$, which is geometrically impossible for a nondegenerate circle tangent to three distinct semicircles. This forces the correct branch of tangency and stabilizes the configuration.

The final relation $y^2=4r^2$ is obtained directly from comparing squared distances after substitution, and no dependence on $c$ survives, confirming that the ratio is invariant under the position of $C$.

Alternative Approaches

A synthetic approach can be built using homothety centers of the three semicircles and the fact that their centers are collinear. The tangent circle becomes an $S$-Apollonius circle with respect to three collinear circles, and the ratio $d=2r$ follows from a projective normalization sending $AB$ to infinity.

Another approach uses inversion centered on $A$ or $B$, transforming semicircles into lines and reducing the configuration to a circle tangent to three parallel lines, where the distance-to-radius ratio becomes immediate.