Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Find the answer

6. Given that $A B$ is the diameter of a semicircle $O$ with a radius of 4, $C$ is the midpoint of the semicircle arc, $E$ is the midpoint of $B C$ , a small circle is tangent to $B C$ at $E$ and tangent to $\text{B C}$ at $F$ , then the length of the tangent $A T$ drawn from $A$ to the small circle is:

Pick one

Solution

6. A.

Connect $A C$ , then $A C \perp B C$ , connect $A E$ intersecting the smaller circle at $D$ , connect $O F$ which must pass through $E$ , and connect $D F$ . In the right triangle $\triangle A C E$ ,
$\sqrt{A C^{2}+C E^{2}}=\sqrt{(2 \sqrt{2})^{2}+(\sqrt{2})^{2}}=\sqrt{10}.$
$A E=$
It is also easy to prove that right triangles $\triangle E D F \sim \triangle A C E$ , so $\frac{E D}{A C}=\frac{E F}{A E}$ . Therefore, $E D=\frac{A C(O F-O E)}{A E}=\frac{2 \sqrt{2}(2-\sqrt{2})}{\sqrt{10}}$ , hence $A D=A E+E D$ ,
which means $A T=\sqrt{A E \cdot A J}=\sqrt{5+4 \sqrt{2}}=2+\sqrt{2}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.