National Math Camp 2020 Exam 2 Problem 1

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FuadAlAlam
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National Math Camp 2020 Exam 2 Problem 1

Unread post by FuadAlAlam » Mon Dec 07, 2020 1:30 pm

Two circles $\Gamma_1$ and $\Gamma_2$ meet at $A$ and $B$. Let $r$ be a line through $B$ that meets $\Gamma_1$ at $C$ and $\Gamma_2$ at $D$, such that $B$ is between $C$ and $D$. Let $s$ be the line parallel to $AD$, which is tangent to $\Gamma_1$ in $E$ and has the minimal distance from $AD$. $EA$ meets $\Gamma_2$ in $F$, and let $t$ be the line through $F$ which is tangent to $\Gamma_2$. Prove that,
a) $t \parallel AC$.
b) $r,s$ and $t$ are concurrent.

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FuadAlAlam
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Re: National Math Camp 2020 Exam 2 Problem 1

Unread post by FuadAlAlam » Mon Dec 14, 2020 12:43 pm

For part (a), let $s$ and $t$ intersect at $T$. We have $$\measuredangle AFT = \measuredangle ABF = \measuredangle ABD+\measuredangle DBF = \measuredangle ABC+\measuredangle DAF = \measuredangle AEC+\measuredangle TEA = \measuredangle AEC+\measuredangle ECA = \measuredangle EAC = \measuredangle FAC$$
Thus, $t \parallel AC$, as desired.

For part (b), let $r$ and $s$ intersect at $T'$. We have $$ \measuredangle FBT' = \measuredangle FBD = \measuredangle FAD = \measuredangle FET'$$ ,which implies $E, B, F$ and $T'$ are concyclic.
We now have $$\measuredangle AFT' = \measuredangle EFT' = \measuredangle EBT' = \measuredangle EBD = \measuredangle EBA+\measuredangle ABD = \measuredangle ECA+\measuredangle ABC = \measuredangle ECA+\measuredangle AEC = \measuredangle EAC = \measuredangle FAC = \measuredangle AFT$$
The last equality follows from part (a).
Therefore, $FT' \equiv t$ and $T=T'$, as desired.

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