Locus of point in the line connecting foot of tangents

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Locus of point in the line connecting foot of tangents

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Source: 2012 Indonesia Round 2 TST 4 Problem 2

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#1 h Mar 18, 2012, 2:24 PM • 2 Y

Y by Adventure10, Mango247

Let $\omega$ be a circle with center $O$ , and let $l$ be a line not intersecting $\omega$ . $E$ is a point on $l$ such that $OE$ is perpendicular with $l$ . Let $M$ be an arbitrary point on $M$ different from $E$ . Let $A$ and $B$ be distinct points on the circle $\omega$ such that $MA$ and $MB$ are tangents to $\omega$ . Let $C$ and $D$ be the foot of perpendiculars from $E$ to $MA$ and $MB$ respectively. Let $F$ be the intersection of $CD$ and $OE$ . As $M$ moves, determine the locus of $F$ .

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#2 h Mar 19, 2012, 9:32 PM • 2 Y

Y by Adventure10, Mango247

I think this problem has been posted several times before.

Anyways, let $E'$ be the inverse of $E$ wrt $w$ . I claim $F$ is the midpoint of $EE'$ .

Clearly, since $M$ is in the polar of $E'$ then $E'$ is in the polar of $M$ , and therefore $A-E' - B$ .

It is easy to see that $M,E,B,O,A$ lie on a circle with diameter $MO$ . From this, it's very natural to think of droping a perpendicular from $E$ to $AB$ . Let $G$ be the foot of this perpendicular.

By simpson, we have $C-D-F-G$ . Also, using that $DEGB$ is cyclic we easily get $\angle FGE = \angle DBE = \angle MOE = \angle FEG$ the last because $MO \parallel EG$ . Hence $FG= FE$ and we get that $F$ is midpoint of $EE'$ since $EGE'$ is a right triangle.

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