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

RedDragonl Jun 7, 2024

#1**+1 **

Let's let 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 C1 the center of semicircle of BC to C2 and the radius of the red circle to C3.Also, let the radius be x.

We know from the problem that \(BF=2\) so that means \(BC_3 = 2-x\)

We also know that \(C_1C_3 = 1+x.\)

Now, draw \(C_1C_3B.\) Notice this is a right triangle.

Now, we can use the pythaogrean thereom to finish this poblem.

\({1}^{2}+{(2-x)}^{2}={(1+x)}^{2}\) The x in the equation is cancelled out to get the radius of 2/3.

2/3 is our answer.

Thanks! :)

NotThatSmart Jun 7, 2024

#1**+1 **

Best Answer

Let's let 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 C1 the center of semicircle of BC to C2 and the radius of the red circle to C3.Also, let the radius be x.

We know from the problem that \(BF=2\) so that means \(BC_3 = 2-x\)

We also know that \(C_1C_3 = 1+x.\)

Now, draw \(C_1C_3B.\) Notice this is a right triangle.

Now, we can use the pythaogrean thereom to finish this poblem.

\({1}^{2}+{(2-x)}^{2}={(1+x)}^{2}\) The x in the equation is cancelled out to get the radius of 2/3.

2/3 is our answer.

Thanks! :)

NotThatSmart Jun 7, 2024