Consider a circle with center $O$ and radius $R,$ and let $A$ and $B$ be two points in the plane of this circle. a.) Draw a chord $CD$ of the circle such that $CD$ is parallel to $AB,$ and the point of the intersection $P$ of the lines $AC$ and $BD$ lies on the circle. b.) Show that generally, one gets two possible points $P$ ($P_{1}$ and $P_{2}$) satisfying the condition of the above problem, and compute the distance between these two points, if the lengths $OA=a,$ $OB=b$ and $AB=d$ are given.
Problem
Source: IMO LongList 1959-1966 Problem 39
Tags: geometry, circumcircle, circle, Intersection, IMO Shortlist, IMO Longlist
grobber
02.09.2004 21:56
A remark (which solves a)): The homothety of center $P$ mapping $C\to A,\ D\to B$ turns $\mathcal C(O,R)$ into a circle tangent to $\mathcal C$ at $P$ and passing through $A,B$. Conversely, if one has a circle tangent to $\mathcal C$ and passing through $A,B$, then their point of tangency, $P$, satisfies the conditions. This means that the construction is equivalent to finding the two circles tangent (internally and externally) to $\mathcal C$ and passing through $A,B$. In order to do this, we may first construct the image $B'$ of $B$ through an inversion of pole $A$. We also construct the image $\mathcal C'$ of $\mathcal C$, and we construct the two tangents from $B'$ to $\mathcal C'$. When we perform the inversion again, the two lines are turned into the two circles we are looking for.
Sowmitra
28.04.2014 15:33
Solution to (a): Draw an arbitrary circle $\Gamma$ which intersects $\mathcal{C}(O,R)$ at $M$ and $N$. Let, $T=AB\cap MN$. Draw tangents $TP_1$ and $TP_2$ from $T$ on to the circle $\mathcal{C}(O,R)$. Then, $P_1$ and $P_2$ are the required points.(Because, the circumcircles of $\triangle ABP_1$ and $\triangle ABP_2$ are tangent to $\mathcal{C}(O,R)$. As a result, $P_1$ and $P_2$ are the centres of Homothety which take $AB$ to $CD$.) Stuck at (b)...any help...