+101 Semicircles are constructed on AB, AC, and BC. A circle is tangent to all three semicircles. Find the radius of the circle. 
+1632 Law of Cosines Method: Trigonometry Rules!!!
Let the center of the semicircle be P, and the center of the small circle be Q. Additionally, let the midpoint of AB = M, and the midpoint of BC = N. I'm about to spit out constructions, so try to follow along:
We can construct a triangle MQP, with side lengths MQ = 2 + r, QP = 5 - r, and PM = 5 - 2 = 3.
Additionally, triangle NQP, with side lengths NQ = 3 + r, QP = 5 - r, and PN = 5 - 3 = 2.
Using the Law of Cosines (which I expect that you should know already) on NQP:
\((3 + r)^2 = (5 - r)^2+2^2-2(5-r)(2)\cos({QPN})=r^2+6r+9=r^2-10r+29-(20-4r)\cos({QPN})\)
\({{16r-20}\over{4r-20}}=\cos({QPN})={4r-5\over{r-5}}\)
Using the Law of Cosines on MQP: (and substitution)
\((2+r)^2=3^2+(5-r)^2-2(3)(5-r)\cos({QPM})=r^2+4r+4=r^2-10r+34+{(30-6r)(4r-5)\over{(r-5)}}\)
\(14r-30={-(6r-30)(4r-5)\over{(r-5)}}=-6(4r-5)=-24r+30\)
Note that cos(QPM) = -cos(QPN)
Solving this equation, you get 38r = 60, and r = 30/19.
+1632 Law of Cosines Method: Trigonometry Rules!!!
Let the center of the semicircle be P, and the center of the small circle be Q. Additionally, let the midpoint of AB = M, and the midpoint of BC = N. I'm about to spit out constructions, so try to follow along:
We can construct a triangle MQP, with side lengths MQ = 2 + r, QP = 5 - r, and PM = 5 - 2 = 3.
Additionally, triangle NQP, with side lengths NQ = 3 + r, QP = 5 - r, and PN = 5 - 3 = 2.
Using the Law of Cosines (which I expect that you should know already) on NQP:
\((3 + r)^2 = (5 - r)^2+2^2-2(5-r)(2)\cos({QPN})=r^2+6r+9=r^2-10r+29-(20-4r)\cos({QPN})\)
\({{16r-20}\over{4r-20}}=\cos({QPN})={4r-5\over{r-5}}\)
Using the Law of Cosines on MQP: (and substitution)
\((2+r)^2=3^2+(5-r)^2-2(3)(5-r)\cos({QPM})=r^2+4r+4=r^2-10r+34+{(30-6r)(4r-5)\over{(r-5)}}\)
\(14r-30={-(6r-30)(4r-5)\over{(r-5)}}=-6(4r-5)=-24r+30\)
Note that cos(QPM) = -cos(QPN)
Solving this equation, you get 38r = 60, and r = 30/19.
+195 Let O be the center of the circle, let X, Y, and Z be the points of tangency with AB, AC, and BC, respectively, and let D be the midpoint of BC. Let r be the radius of the circle.
Since BC is a diameter of the circle, DO=r. Therefore, by Pythagoras on triangle BOD, BD=BC2−DO2=36−r2.
Since △COD is isosceles with CD a leg, CO=DO2+CD2=r2+9. Similarly, AO=r2+4.
By Power of a Point, OX2=AO⋅AX, so r2=(r2+4)(r+DX), or r3+4r=r2+4r+rDX, which simplifies to rDX=0. Since DX=0, we have r=0, which is impossible.
Therefore, the only solution is r=2.
+1632 Hey boseo--- there are a couple of inaccuracies here... If O is the center of the (smaller) circle, and D by your construction is the midpoint of BC, BOD is not necessarily a right triangle and thus you cannot operate pythagorean theorem. Additionally, DO is clearly not r, but r + 3.
Additionally, COD is most definitely not isosceles, and there appears to be a random pythagorean again...
+1632 Let the midpoint of AB be M and the midpoint of BC be N, let the center of the smaller circle be O, and the point tangent to circle O with arc BC be point P:
Common fallacies:
NOM is a right triangle with angle NOM = 90 degrees.
P, O, and N are collinear.
These assumptions should not be made, especially when our diagram is not to scale- also be clear in your statements of your variables so that you also don't get lost in your work... One can quickly use the Law of Cosines to see that r cannot be 2... good attempt still Boseo!!
+399 Here is a different solution, which is a generalization of proyaop's method. Here is the graph below, labeled with some extra points and lines. P.S. - In a math competition, or any other place where you don't have good access to resources, I suggest not using this method, as Stewart's theorem is just a generalization of the double law of cosines, proyaop's method. Stewart's theorem is also notoriously hard to memorize. But, it does shorten the amount of work significantly.
Stewart's theorem is perfect for this problem:
First, we know \(\overline{C1C3} = 3, \overline{C2C3}=2\).
We use a mnemonic to memorize Stewart's theorem, man + dad = bmb + cnc, where these all recommend certain side lengths of the triangle. A man and his dad put a bomb in the sink, though this does not exactly make it easier to memorize, as we also have to memorize the configuration of the sides.
So onto the solution,
Substituting the values into the equation we get \(30+5{(5-x)}^{2}=3{(3+x)}^{2}+2{(2+x)}^{2}\).
When solving, we get a convient cancellation of the quadratic terms, and we are left with x = 120/76, or x = 30/19.
Note while this is still easier, I would mostly still recommend proyaop's method.
