Problem: A circle of radius r is inscribed in a right isosceles triangle, and a circle of radius R is circumscribed about the triangle. Then rRβ equals
Answer Choices:
A. 1+2β
B. 22+2ββ
C. 22ββ1β
D. 21+2ββ
E. 2(2β2β)
Solution:
In the adjoining figure. β³ABC18 a right isosceles triangle, with β BAC=90β and AB=AC, Inscribed in a circle with center O and radius R The line segment AO has length R and bisects line segment BC and β BAC. A circle with center Oβ² lying on AO and radius r is inscribed in β³ABC. The sides AB and AC are tangent to the inscribed circle with points of tangency T and Tβ², respectively. Since β³ATOβ² has angles 45ββ45ββ90β and Oβ²Tβr, AT=r and Oβ²A=r2β, then R=r+r2β and R/r=1+2β.