Problem 6

Consider an isosceles triangle. Let

R

be the radius of its circumcircle and

r

the radius of its inscribed circle. Prove that the distance

d

between the centers of these two circles is

\sqrt{R(R-2r)}

Step 6 of 6: Evaluate the diameter secant and conclude

Previous step Next step

2Rr=PI\cdot QI=(R+d)(R-d)=R^2-d^2

Detailed analysis

Let $d=OI$ . Since $PO=QO=R$ , the positions on the diameter line give $PI=PO+OI=R+d$ and $QI=QO-OI=R-d$ . Combining the preceding equalities gives $2Rr=R^2-d^2$ , so $d=\sqrt{R(R-2r)}$ .