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

booboo44 Feb 7, 2024

#1**+2 **

Label the tangent point between Semicircle AB and the red circle as D, the tangent point between arc BC and the red circle as E, and the tangent point between the red circle and semicircle AC as F. Label the center of semicircle AB to C_{1 }the center of semicircle of BC to C_{2} and the radius of the red circle to C_{3}.

Additionally, set the radius of the red circle is x.

We know BF is 2, and so BC_{3} is 2-x.

Similarly, C_{1}C_{3} is 1+x.

Construct the right triangle, C_{1}C_{3}B.

Using the pythagorean theorem, we can set an equation:

\({1}^{2}+{(2-x)}^{2}={(1+x)}^{2}\). In the equation, the x's conveniently cancel out, and we get **radius = 2/3.**

hairyberry Feb 7, 2024

#1**+2 **

Best Answer

Label the tangent point between Semicircle AB and the red circle as D, the tangent point between arc BC and the red circle as E, and the tangent point between the red circle and semicircle AC as F. Label the center of semicircle AB to C_{1 }the center of semicircle of BC to C_{2} and the radius of the red circle to C_{3}.

Additionally, set the radius of the red circle is x.

We know BF is 2, and so BC_{3} is 2-x.

Similarly, C_{1}C_{3} is 1+x.

Construct the right triangle, C_{1}C_{3}B.

Using the pythagorean theorem, we can set an equation:

\({1}^{2}+{(2-x)}^{2}={(1+x)}^{2}\). In the equation, the x's conveniently cancel out, and we get **radius = 2/3.**

hairyberry Feb 7, 2024