Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 3.5 AMC 10/12 Find the answer

A circle with center $O$ is tangent to the coordinate axes and to the hypotenuse of the $30^\circ$ - $60^\circ$ - $90^\circ$ triangle $ABC$ as shown, where $AB=1$ . To the nearest hundredth, what is the radius of the circle?

Pick one

Solution

Draw radii $OE$ and $OD$ to the axes, and label the point of tangency to triangle $ABC$ point $F$ . Let the radius of the circle $O$ be $r$ . Square $OEAD$ has side length $r$ .
Because $BD$ and $BF$ are tangents from a common point $B$ , $BD = BF$ .
$AD = AB + BD$
$r = 1 + BD$
$r = 1 + BF$
Similarly, $CF = CE$ , and we can write:
$AE = AC + CE$
$r = \sqrt{3} + CF$
Equating the radii lengths, we have $1 + BF = \sqrt{3} + CF$
This means $BF - CF = \sqrt{3} - 1$
$BF + CF = 2$ by the 30-60-90 triangle.
Therefore, $2BF = 2 + \sqrt{3} - 1$ , and we get $BF = \frac{1}{2} + \frac{\sqrt{3}}{2}$
The radius of the circle is $AD$ , which is $BF + 1 = \frac{3}{2} + \frac{\sqrt{3}}{2}$
Using decimal approximations, $r \approx 1.5 + \frac{1.73^+}{2} \approx 2.37$ , and the answer is $\boxed{D}$ .

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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.