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"Let $k$ and $l$ be two circles that intersect in two points $P$ and $Q$. Construct (with a straightedge and a compass) the line $m$ through $P$, not containing $Q$, with the property that if $m$ intersects $k$ in $B$ and $P$, and $m$ intersects $l$ in $C$ and $P$, then $\lvert PB \rvert = \lvert PC\rvert$. (Hint: Solve for $B$.)"

I first drew the lines $BQ$ and $CQ$. Then, assuming that we already have the desired line $BC$, I tried using the law of sines and cosines, hoping to get some identity I can easily use. However, I couldn't get much further than some complicated equations and some equations relating the ratio $$\frac{\lvert BQ \rvert}{\lvert CQ\rvert}$$ to ratio of some angles. I am not sure what the hint means, since the angle $PB'Q$ and the angle $PC'Q$ are constant for any choice of points $B'$ and $C'$ on circles $k$ and $l$ (respectively). I also realize that the angle $PBQ$ and $PCQ$, and hence the angle $BQC$, are all constant.

Could you provide a hint?

dantopa
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User New
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1 Answers1

7

The construction:

  • Take $l$, the smallest circle.

  • Take $L$, the center of $l$.

  • Draw a circle $a$ with center $P$ and radius $PL$.

  • Extend $PL$ to intersect $a$ in $L'$.

  • Draw a circle $l'$ with center $L'$ and radius $L'P$.

  • $l'$ will intersect $k$ in $B$.

  • Draw the line $BP$.

  • $BP$ will intersect $l$ in $C$.

By construction:

  • $BP$ and $PC$ are colinear,

  • $l$ and $l'$ are congruent and tangent in $P$,

  • $BP$ and $PC$ subtend the same angle, hence they are equal length,

Hence $B$ and $C$ are the requested points.

Brethlosze
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