Given non-collinear points $A$, $B$, and $C$, construct three circles that are pairwise tangent at these points. Is it always possible?

Question

Given three non-collinear points $A$, $B$, and $C$, construct three circles that are pairwise tangent at these points. Are there any cases where such circles do not exist?

I am not sure how to start the first part of this problem. For the second part, I found that all the angles must be acute, after a lot of angle bashing, but am still stuck on the first part.

Any help would be appreciated, thanks!

Answer 1

Hint: For a pair of tangents of a circle that are not parallel, their distance from the tangent point to the intersection point are equal.

By construction, let the three common tangents intersect at a point, say $O$, which satisfies $OA=OB=OC$.

For your second question, have you considered the circles may be tangent to another internally? I think the only case where such circles do not exist is when one of the radii is infinitely large.

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Find the circumcentre $O$ of $\triangle ABC$, which must exist. Each of the radii $OA$, $OB$, $OC$ acts as a common tangent. Construct perpendicular lines passing through $A$, $B$ and $C$ respectively, each perpendicular to its radius. Then the intersection for each pair of perpendicular lines is a centre of your required circles.

When is such arrangement impossible? When one of the angles of $\triangle ABC$ is $90^\circ$, the circumcentre is the mid-point on the hypotenuse, and the perpendicular lines originated from the endpoints of the hypotenuse are parallel. So the "circle" becomes the hypotenuse line which, depending on your definition, is not a circle.

If $\triangle ABC$ is obtuse, the circumcentre is outside the triangle, but the circles can still be tangent to each other:

Answer 2

Assume that $\Gamma_A,\Gamma_B,\Gamma_C$ are three circles with centers in $O_A,O_B,O_C$ and $\Gamma_A\cap\Gamma_B=I_C$ and so on. Since $AI_B=AI_C$, then $I_A,I_B,I_C$ are the projections of the incenter $I$ of $O_A O_B O_C$ on the sides of $O_A O_B O_C$. Since $II_A=II_B=II_C=r$, $I$ is the circumcenter of its pedal triangle.

So, in order to solve our problem, let $O$ be the circumcenter of $ABC$, $l_A$ be the line through $A$ that is orthogonal to $OA$ and so on. Then $l_A,l_B,l_C$ meet in the centers of three circles pairwise tangent in $A,B,C$, as wanted.

This is what we get when $\widehat{BAC}$ is obtuse: