Prove radius of fourth circle is equal to semi-perimeter of the right triangle formed by three circles
Question: There are 3 circles that touch. A right triangle forms the center of the circles. Find the radius of the fourth circle given that it is circumscribed around all three circles. Express using the semi-perimeter of the triangle.
I am very confused as to how to start the question, and any help would be greatly appreciated.
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$\begingroup$Let $\triangle ABC$ be the right triangle with $AB$ as the long edge, and let $O$ be the midpoint of $AB$, i.e., the circumcenter of $\triangle ABC$. Extend $CO$ to $D$ on the excircle of $\triangle ABC$, and then draw three points $E,F,G$, such that $DA, DC, DB$ touches the three circles at $E,F,G$ respectively. It suffices to prove that $DE = DF = DG = \frac{AB+AC+BC}{2}$.
This can be proved by noticing that $ACBD$ is a rectangle, and that $\frac{AB+AC+BC}{2}=r_A+r_B+r_C$. Then, notice that $$DE = DA+AE = BC + AE =r_A+r_B+r_C$$ $$DF = DC+CF = AB + CF = r_A+r_B+r_C$$ $$DG = DB+BG = AC+BG = r_A+r_B+r_C$$ and hence $D$ is the circumcenter of $\triangle EFG$ with radius $r_A+r_B+r_C=\frac{AB+AC+BC}{2}$. The remaining proof is to show that the circumcircle of $\triangle EFG$ is tangent to the three circles, which is true because $E,F,G$ are constructed by extending $DA, DC, DB$ to the farthest points on the circles with centers at $A,C,B$. This concludes our proof.
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