Kvant Math Problem 1256
The circles are equal and tangent to each other.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 9m48s
Source on kvant.digital
Problem
Two equal circles touch each other. Construct a trapezoid such that each of the circles is tangent to three of its sides, and the centers of the circles lie on the diagonals of the trapezoid.
V. A. Senderov
Exploration
The circles are equal and tangent to each other. Let their centers be $O_1$ and $O_2$, and let the common radius be $r$. Since each circle is tangent to three sides of the trapezoid, the two circles must occupy symmetric positions inside the trapezoid.
A first guess is that each circle is tangent to one base and to the two nonparallel sides. If the circles touched different bases, the symmetry of the condition involving the diagonals would be difficult to realize. Thus it is natural to seek an isosceles trapezoid whose legs are tangent to both circles, while the upper base is tangent to one circle and the lower base to the other.
Take coordinates
$$O_1=(-r,0),\qquad O_2=(r,0).$$
The circles are
$$(x+r)^2+y^2=r^2,\qquad (x-r)^2+y^2=r^2.$$
Suppose the legs are the common external tangents to the two circles. Since the circles are equal and lie on the same horizontal line, these tangents are simply
$$y=\pm r.$$
If these were chosen as bases, every circle would touch only two sides, so this does not work.
The other pair of common tangents are the two oblique external tangents. For equal tangent circles, these pass through the midpoint of the segment joining the centers. Writing one of them as
$$y=mx+b,$$
the tangency condition to either circle gives
$$\frac{|b-mr|}{\sqrt{1+m^2}}=r.$$
Taking the tangent above the circles yields
$$b=r\bigl(\sqrt{1+m^2}+m\bigr).$$
Now choose horizontal bases tangent to the circles individually:
$$y=r,\qquad y=-r.$$
The oblique common tangents intersect these lines and form an isosceles trapezoid. The remaining condition is that the centers lie on the diagonals.
This is the crucial point. Computing the vertices and imposing that a diagonal pass through $O_1$ (and hence by symmetry the other through $O_2$) should determine $m$.
Problem Understanding
We are given two equal tangent circles. We must construct a trapezoid such that each circle is tangent to three sides of the trapezoid, and the center of each circle lies on a diagonal of the trapezoid.
This is a Type D problem. We must exhibit a construction and verify all required properties.
The core difficulty is finding the correct inclination of the nonparallel sides. Once the trapezoid is described, the tangency conditions are straightforward; the nontrivial requirement is that the centers lie on the diagonals.
The answer will be an isosceles trapezoid whose bases are tangent to the circles at their highest and lowest points, and whose legs are the two common external tangents to the circles making a specific angle determined by the diagonal condition.
Proof Architecture
First, place the circles in coordinates with centers $(-r,0)$ and $(r,0)$.
Second, describe all common external tangents to the circles and write them as
$$y=\pm(mx+b),$$
with
$$b=r(\sqrt{1+m^2}+m).$$
Third, intersect these tangents with the horizontal lines $y=\pm r$ to obtain an isosceles trapezoid.
Fourth, compute the vertices explicitly and derive the equation of a diagonal.
Fifth, impose that the diagonal pass through one center. This yields an equation for $m$.
Sixth, solve the equation and obtain the unique positive slope.
Finally, verify that each circle is tangent to exactly three sides and that the centers lie on the diagonals.
The most delicate lemma is the determination of $m$ from the diagonal condition.
Solution
Let the two equal tangent circles have radius $r$ and centers
$$O_1=(-r,0),\qquad O_2=(r,0).$$
The circles touch at the origin.
Take the horizontal lines
$$y=r,\qquad y=-r.$$
The line $y=r$ is tangent to the left circle, and the line $y=-r$ is tangent to the right circle.
Let one leg of the trapezoid be a common external tangent to the two circles. Write it as
$$y=mx+b,$$
with $m>0$.
Since the line is tangent to the right circle,
$$\frac{b-mr}{\sqrt{1+m^2}}=r,$$
hence
$$b=r\bigl(\sqrt{1+m^2}+m\bigr).$$
The second common external tangent is
$$y=-mx+b.$$
These two lines together with $y=\pm r$ form an isosceles trapezoid.
Let
$$A=(x_A,r),\qquad B=(x_B,r)$$
be the intersections of $y=r$ with $y=-mx+b$ and $y=mx+b$, respectively. Solving,
$$x_A=\frac{b-r}{m},\qquad x_B=\frac{r-b}{m}.$$
Similarly, let
$$D=(x_D,-r),\qquad C=(x_C,-r)$$
be the intersections of $y=-r$ with $y=-mx+b$ and $y=mx+b$. Then
$$x_D=\frac{b+r}{m},\qquad x_C=-,\frac{b+r}{m}.$$
Consider the diagonal $AC$.
Its slope equals
$$\frac{-r-r}{x_C-x_A} = \frac{-2r} {-\frac{b+r}{m}-\frac{b-r}{m}} = \frac{mr}{b}.$$
Hence the equation of $AC$ is
$$y-r=\frac{mr}{b} \left(x-\frac{b-r}{m}\right).$$
The condition that $O_1=(-r,0)$ lie on this diagonal gives
$$-r=\frac{m}{b} \left( -r-\frac{b-r}{m} \right).$$
Multiplying by $b$,
$$-b=-mr-b+r,$$
so
$$r=mr,$$
and therefore
$$m=1.$$
Thus the legs are
$$y=x+r(1+\sqrt2), \qquad y=-x+r(1+\sqrt2).$$
This determines the trapezoid uniquely up to similarity.
It remains to verify the required properties.
The line $y=r$ is tangent to the left circle. Since the distance from $O_1$ to each leg equals $r$, the left circle is tangent to both legs as well. Hence the left circle is tangent to three sides.
Similarly, the line $y=-r$ is tangent to the right circle, and the right circle is tangent to both legs. Hence the right circle is tangent to three sides.
By construction, $O_1$ lies on the diagonal $AC$. The trapezoid is symmetric about the $y$-axis, so the other diagonal $BD$ passes through the symmetric point $O_2$.
Consequently both centers lie on diagonals of the trapezoid.
The required construction is therefore:
Construct the two horizontal tangents $y=r$ and $y=-r$ through the highest point of one circle and the lowest point of the other. Construct the pair of common external tangents to the circles making angle $45^\circ$ with the horizontal. Their intersections with the horizontal tangents form the desired trapezoid.
$$\boxed{\text{The required trapezoid is the isosceles trapezoid whose legs are the common external tangents of slope }\pm1.}$$
Verification of Key Steps
The first delicate step is the formula for the common external tangent. For a line $y=mx+b$, the distance from $O_2=(r,0)$ equals
$$\frac{|b-mr|}{\sqrt{1+m^2}}.$$
Tangency requires this distance to be $r$, yielding
$$b=r(\sqrt{1+m^2}+m).$$
Using the opposite sign would place the line between the circles rather than outside them.
The second delicate step is the diagonal condition. Substituting $O_1=(-r,0)$ into the equation of $AC$ produces
$$-b=-mr-b+r.$$
No cancellation error is permissible here. The equation reduces immediately to
$$r=mr,$$
hence $m=1$. Any algebraic slip at this point gives an incorrect family of trapezoids.
The third delicate step is the tangency count. The upper base $y=r$ is tangent only to the left circle, while the lower base $y=-r$ is tangent only to the right circle. Each leg is a common tangent to both circles. Thus each circle touches exactly three sides, namely one base and the two legs.
Alternative Approaches
A synthetic solution avoids coordinates. Let the bases be the tangents at the highest point of one circle and the lowest point of the other. Let the legs be common external tangents. Denote by $M$ the midpoint of the segment joining the centers. Because the circles are equal, the trapezoid is symmetric about the perpendicular through $M$.
One then expresses the condition that a diagonal pass through a center using similar triangles formed by the diagonal and the sides of the trapezoid. The resulting proportionality forces the horizontal and vertical displacements along a leg to be equal, so the leg makes an angle of $45^\circ$ with the bases. This determines the trapezoid.
The coordinate method is preferable because the diagonal condition translates into a single linear equation whose solution is immediate.