So this is a problem from my country's first test for choosing avid students for the summer program. In the following diagram which is not to scale, chord $AB$ is the diameter of the smaller circle and this circle passes through $O$, center of the bigger circle. We choose point $C$ on line $AB$ such that $\angle BCO=30^o$.
Then we draw a tangent line from $C$ that is tangent to the big circle at $D$. If the length of $DK$ is $3\sqrt 2$, What's the length of $CD$? I've tried Power of the point $C$ and doing some angle calculation with the given angle to find some equal chords, but it was to no avail.
Best Answer
Connect $O$ and $O'$. It is clear that $\angle OO'A = 90^o$.
Hence from the assumption $\angle BCO = 30^o$, it follows that $\angle KOO' = 60^o$.
But $KO' = OO'$, so the triangle $KOO'$ is a regular triangle and we have $\angle KO'C = 90^o - 60^o = 30^o = \angle KCO'$.
Therefore $KC = KO' = KO = OO' = \frac 1 2 CO$.
From this point, it might be helpful to do some length calculations.
Write $a = OO'$. We then have $CK = a$ and $CO = 2a$.
Therefore $CD^2 = CA \times CB = CK \times CO = 2a^2$, which shows $CD = \sqrt 2 a$.
But also $OD = OA = \sqrt 2 a$, and $CO = 2a$, so the triangle $OCD$ is a $45 - 45 -90$ degree triangle.
Since $K$ is the middle point of $CO$, it is clear that $CD = \sqrt2 DK$.