Problem 4

Let

R

and

S

be different points on a circle

\Omega

such that

RS

is not a diameter. Let

\ell

be the tangent line to

\Omega

R

. Point

T

is such that

S

is the midpoint of

RT

. Point

J

is chosen on the shorter arc

RS

\Omega

so that the circumcircle

\Gamma

of triangle

JST

intersects

\ell

at two distinct points. Let

A

be the common point of

\Gamma

and

\ell

that is closer to

R

. Line

AJ

meets

\Omega

again at

K

. Prove that line

KT

is tangent to

\Gamma

Step 3 of 6: Apply Reim's theorem

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R,K,B\text{ are collinear}

Detailed analysis

Apply Reim's theorem to the two circles Omega and Gamma, which meet at S and J. Their chords RK and TA are parallel by the preceding step. With the parallelogram relation from the first step, Reim's theorem gives that R, K, and B are collinear. This is the standard corresponding-endpoints conclusion of Reim's theorem in this configuration.