Mind your decisions, I'm Presh Talwalkar.
In semicircle A,
construct semicircles B and C along the diameter of semicircle A.
Then construct circle D
which is internally tangent to semicircle A
and externally tangent to semicircles B and C.
If the radius of semicircle B is equal to 2
and the radius of semicircle C is equal to 1,
solve for the radius of circle D.
This problem appeared on the 2017 AMC 12a exam.
Pause the video, if you'd like to give this problem a try
and when you're ready, keep watching to learn how to solve this problem.
We begin with a preliminary result.
If two circles are tangent,
their centers and the tangent points are collinear.
To see why, draw a tangent line through Point C.
By definition, AC will meet this at a right angle
and also BC will meet this tangent line at a right angle.
Therefore, ACB is a straight line
and all points A, C, and B are along the same line.
This is true for externally tangent circles
and it's also true for internally tangent circles.
Returning to the original problem we'll calculate the radius of circle A.
Recall that circle B has a radius of 2 and circle C has a radius of 1
Therefore, circle A has a diameter equal to 6 and a radius that's equal to 3.
This means AB has a length of 1 and AC has a length of 2.
Now construct BD.
B, D and its tangent point will be collinear.
Therefore, B to its radius will be a distance of 2
and suppose circle D has a radius equal to r.
Similarly, construct DC.
This will be the radius of D plus the radius of C.
Now construct a line through A, D and the tangent point between A and D.
This entire length will be equal to the radius of A which is equal to 3.
Since circle D has a radius of r.
The distance between A and D will be (3 - r).
Now, let's just focus on the triangle.
Well clean this diagram up a little bit.
And now we'll be able to calculate some things from this diagram.
We'll use al-Kāshī's law of cosines on the entire triangle BDC
and we'll label angle C as theta.
Now by al-Kāshī's law of cosines,
the quantity (2 + r)^2 is equal to the quantity (r + 1)^2 plus 3^2 minus 2
multiplied by the quantity (r + 1) multiplied by 3 multiplied by the cosine of theta.
We can make a similar equation for triangle ADC.
Now we can consider just these two equations.
What we'll do is we'll multiply the second equation by 1.5.
This makes sure that the term with cosine of theta will be equal in both equations.
When we subtract the second equation from the first,
the cosine of data terms will cancel out.
We then get a single equation in one variable r.
We do some careful simplification
and we get some magical cancellations.
The result is that 14r is equal to 12.
And therefore, r is equal to 6/7
and that's our answer.
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