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 5 of 6: Transfer the angle

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\angle KTA=\angle TKB=\angle TSB=\angle TSA

Detailed analysis

Because AT is parallel to RB and R, K, B are collinear, the first two angles in the displayed chain are equal. The cyclic quadrilateral TBKS gives the next equality, since both angles subtend the same chord TB. Finally A, S, B are collinear, so the last equality follows.