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# Geometry

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Semicircles are constructed on AB, AC, and BC.  A circle is tangent to all three semicircles.  Find the radius of the circle.

Feb 27, 2024

#1
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Let the center of the small circle be O, and the point where circle O is tangent to the circle with diameter AC be D, and let the radius of circle O be r. Lastly, let the midpoints of AB and BC be points M and N, respectively.

Since B is the center of a semicircle, BD must also be the same length as AB (because they are the radius) = 2.

MO = 1 + r, and MB = 1. OMB forms a right triangle with angel OBM = 90 degrees, since O lies on the perpendicular bisector of AC.

This means that OB, by the pythagorean theorem, has length $$\sqrt{(1+r)^2-1}=\sqrt{r^2+2r}$$.

Additionally, DO = r, and lines on the same line as OB. Recalling that DB = 2, we can set an equation: DO + OB = 2:

$$r + \sqrt{r^2+2r}=2$$

$$\sqrt{r^2+2r}=2-r$$

$$r^2+2r=(2-r)^2=r^2-4r+4$$

$$2r = -4r+4$$

$$6r=4$$

$$r={2\over3}$$

Feb 27, 2024
#2
+1

Let the center of the smaller semi-circle be $$X$$ and the center of the small circle be $$O$$
$$\triangle BOX$$ is a right angle triangle with right angle at $$B$$. So:

$$BO^2+BX^2=OX^2\\ (2-r)^2+1^2=(1+r)^2\\ 6r=4\\ \boxed{r=\displaystyle\frac{2}{3}}$$

Feb 27, 2024