Two parallel lines $r,s$ and two points $P \in r$ and $Q \in s$ are given in a plane. Consider all pairs of circles $(C_P, C_Q)$ in that plane such that $C_P$ touches $r$ at $P$ and $C_Q$ touches $s$ at $Q$ and which touch each other externally at some point $T$. Find the locus of $T$.
Problem
Source: itamo 2004, p2
Tags: geometry, geometric transformation, homothety, geometry proposed
Radar
27.05.2013 20:38
Consider negative homothety $h$ with the center in $T$, in which image of $C_P$ is $C_Q$. Obviously, this homothety really exists. Now because $r$ and $s$ are parallel and touch $C_P$ and $C_Q$, it's quite obvious that $r$ is image of $s$ in $h$. So $P$ is image of $Q$. Therefore $P$, $Q$ and $T$ lie on a single line. Therefore it's clear, that the locus is segment $PQ$ without $P$ and $Q$. (Obviously we can construct our circles for every such $T$.) Q.E.D.
MindFlyer
16.04.2014 20:35
Radar wrote: Consider negative homothety $h$ with the center in $T$, in which image of $C_P$ is $C_Q$. Obviously, this homothety really exists. Now because $r$ and $s$ are parallel and touch $C_P$ and $C_Q$, it's quite obvious that $r$ is image of $s$ in $h$. So $P$ is image of $Q$. Therefore $P$, $Q$ and $T$ lie on a single line. Therefore it's clear, that the locus is segment $PQ$ without $P$ and $Q$. (Obviously we can construct our circles for every such $T$.) Q.E.D. This solution gets 3 points. (Hint: did you consider that the circles may lie on either side of the lines?) It's funny how a solution that contains the word "obvious" a grand total of 3 times would get 3 points. (Hint: no, saying "obvious" 7 times won't grant you 7 points.)
yunxiu
18.04.2014 13:22
MindFlyer wrote: Radar wrote: Consider negative homothety $h$ with the center in $T$, in which image of $C_P$ is $C_Q$. Obviously, this homothety really exists. Now because $r$ and $s$ are parallel and touch $C_P$ and $C_Q$, it's quite obvious that $r$ is image of $s$ in $h$. So $P$ is image of $Q$. Therefore $P$, $Q$ and $T$ lie on a single line. Therefore it's clear, that the locus is segment $PQ$ without $P$ and $Q$. (Obviously we can construct our circles for every such $T$.) Q.E.D. This solution gets 3 points. (Hint: did you consider that the circles may lie on either side of the lines?) It's funny how a solution that contains the word "obvious" a grand total of 3 times would get 3 points. (Hint: no, saying "obvious" 7 times won't grant you 7 points.)
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fmasroor
21.04.2014 05:27
That diagram is probably the only reason why this problem was not the first problem of the competition.
MindFlyer
21.04.2014 05:31
fmasroor wrote: That diagram is probably the only reason why this problem was not the first problem of the competition. Maybe answer with a complete solution, instead of a totally irrelevant and arbitrary comment? Just saying.