IMO 1962 · Problem 6

Consider an isosceles triangle. Let \(r\) be the radius of its circumscribed circle and \(\rho\) the radius of its inscribed circle. Prove that the distance \(d\) between the centers of these two circles is \[ d=\sqrt{r(r-2\rho)}. \]

Official text of the problem. Official problem paper (PDF)

Solution

Let the triangle have vertices \[ A=(0,h),\qquad B=(-b,0),\qquad C=(b,0), \] where \(b,h>0\). Write \(AB=AC=s\), so that \[ s^2=b^2+h^2. \]

By symmetry, the incenter \(I\) lies on the altitude from \(A\). Its distance from \(BC\) is \(\rho\), hence \(I=(0,\rho)\). Joining \(I\) to the vertices divides the triangle into three triangles of height \(\rho\); comparing their total area with \(bh\) gives \[ bh=\rho(s+b). \] Consequently, \[ \rho=\frac{bh}{s+b} =\frac{b(s-b)}{h}, \] where the second equality uses \(h^2=(s-b)(s+b)\).

The circumcenter \(O\) lies on the perpendicular bisector of \(BC\), so write \(O=(0,t)\), allowing \(t\) to be negative. Since \(OA=OB\), \[ (h-t)^2=b^2+t^2, \] and therefore \[ t=\frac{h^2-b^2}{2h} =\frac{s^2-2b^2}{2h}. \] Moreover, \[ h-t=\frac{h^2+b^2}{2h}=\frac{s^2}{2h}>0, \] so the circumradius is \(r=OA=s^2/(2h)\).

Thus \[ d^2=(t-\rho)^2 =\left(\frac{s^2-2b^2-2b(s-b)}{2h}\right)^2 =\frac{s^2(s-2b)^2}{4h^2}. \] On the other hand, \[ r-2\rho =\frac{s^2-4b(s-b)}{2h} =\frac{(s-2b)^2}{2h}, \] whence \[ r(r-2\rho)=\frac{s^2(s-2b)^2}{4h^2}=d^2. \] Taking the nonnegative square root proves the required formula.

The calculation permits either sign of \(t\), so it includes obtuse triangles. It also includes the equilateral case \(s=2b\), when \(d=0\).

The solution was checked against John Scholes' IMO solution archive (Kalva).