Locus of Q as P Varies Along L

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Locus of Q as P Varies Along L

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Source: 1989 IrMO Paper 2 Problem 1

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#1 h Sep 29, 2017, 3:50 AM • 1 Y

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Suppose $L$ is a fixed line, and $A$ is a fixed point not on $L$ . Let $k$ be a fixed nonzero real number. For $P$ a point on $L$ , let $Q$ be a point on the line $AP$ with $|AP|\cdot |AQ|=k^2$ . Determine the locus of $Q$ as $P$ varies along the line $L$ .

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#2 h Sep 10, 2024, 6:11 AM

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inversion says it's a circle

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#3 h Sep 10, 2024, 2:13 PM

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WLOG, let the line $L$ coincide with the $x$ -axis such that $P(p, 0) \in L$ . Let $A(0, a)$ be a fixed point on the $y$ -axis with real $k \neq 0$ . If the line $AP$ be represented by the line $y = -(a/p)x + a$ , then let $Q(q, -(a/p)q + a)$ . If $|AP||AQ| = k^2$ , then:

$|AP||AQ| = \sqrt{a^2 + p^2} \cdot \sqrt{q^2 +(-aq/p + a)^2} = \sqrt{a^2 + p^2}\sqrt{q^2 + a^2 - 2(a^2q/p) + a^2q^2/p^2} = k^2$ ;

or $\sqrt{a^2+p^2}\sqrt{\frac{p^2(q^2+a^2) - 2a^2pq + a^2q^2}{p^2}} = k^2$ ;

or $\sqrt{(a^2+p^2)q^2 -2a^2pq + a^2p^2} = \frac{k^2p}{\sqrt{a^2+p^2}}$ ;

or $(a^2+p^2)q^2 -2a^2pq + a^2p^2 = \frac{k^4p^2}{a^2+p^2}$ ;

or $q^2 -\frac{2a^2p}{a^2+p^2}q = \frac{k^4p^2}{(a^2+p^2)^2} - \frac{a^2p^2}{a^2 + p^2}$ ;

or $(q - \frac{a^2p}{a^2+p^2})^2 - \frac{a^4p^2}{(a^2+p^2)^2} = \frac{k^4p^2}{(a^2+p^2)^2} - \frac{a^2p^2}{a^2+p^2}$ ;

or $(q - \frac{a^2p}{a^2+p^2})^2 = \frac{a^4p^2}{(a^2+p^2)^2} + \frac{k^4p^2}{(a^2+p^2)^2} - \frac{a^2p^2(a^2+p^2)}{(a^2+p^2)^2}$ ;

or $q = \frac{p}{a^2+p^2} \cdot (a^2 \pm \sqrt{k^4 - a^2p^2})$

This post has been edited 8 times. Last edited by Mathzeus1024, Sep 12, 2024, 10:03 AM
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