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Take a quadrilateral $ABCD$ and consider a line parallel to each of its side such that: 1- the distance between the parallel line and the side is some fixed amount $x$ and 2- the parallel line are "outside" of the quadrilateral (this is a clumsy formulation when the quadrilateral is concave, but I think the idea is clear).

These lines define a new quadrilateral $A'B'C'D'$ where $A'$ is the intersection of the parallels to $AB$ and $AD$, and likewise for the others.

Question: Assume that for all $x$, $A'B'C'D'$ is similar $ABCD$. Is $ABCD$ a tangential quadrilateral?

Note that if $ABCD$ is a tangential quadrilateral, then $A'B'C'D'$ is similar to $ABCD$ for all $x$. Indeed, up to similarity, one can always assume a tangential quadrilateral has an incircle of radius one, and this circle touches the quadrilateral at a point which is at 0°. So, up to similarity, what defines a tangential quadrilateral is the location (say in degrees) of the other 3 points on the circle where the tangents touch. Upon preforming the transformation, the point of contact remain at the same angles (on a larger circle).

My question asks for the converse of this observation.

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A stronger result holds: if $A'B'C'D'$ is similar to $ABCD$ for some $x$, then $ABCD$ is a tangential quadrilateral.

Let's show first of all that lines $AA'$, $BB'$, $CC'$ and $DD'$ are concurrent. Let $O$ be the intersection point of lines $AA'$ and $BB'$: we have then $A'B'/AB=OB'/OB=1+BB'/OB$. Let $O'$ be the intersection point of lines $BB'$ and $CC'$: we have then $B'C'/BC=O'B'/O'B=1+BB'/O'B$. But by hypothesis $A'B'/AB=B'C'/BC$, hence $OB=O'B$ and $O=O'$. The same reasoning can be repeated for $CC'$ and $DD'$, hence all four lines concur at point $O$.

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If the perpendicular from $O$ to $AB$ intersects $AB$ at $H$ and $A'B'$ at $H'$ we also have $A'B'/AB=OH'/OH=1+HH'/OH$. And in the same way, if $OKK'$ is perpendicular to $BC$ we get $B'C'/BC=OK'/OK=1+KK'/OK$. But $A'B'/AB=B'C'/BC$ and $HH'=KK'=x$, thus $OH=OK$. The same reasoning can be repeated for sides $CD$ and $DA$: we then obtain that point $O$ has the same distance $r$ from the four sides of $ABCD$, that is the sides of $ABCD$ are tangent to the circle of center $O$ and radius $r$.

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  • $\begingroup$ Very nice use of Thalès' Intercept theorem! thanks! $\endgroup$ – ARG Aug 20 '18 at 13:14

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