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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below to enlarge it.
Since a₁² = 2R²(1+cos2A) = 4R² cos²A, thus a₁ = 2R cosA. r = 4R sin(A/2) sin(B/2) sin(C/2)r₁ = 4R sin(A/2) cos(B/2) cos(C/2)r₁ − r = 4R sin(A/2) cos(B/2 + C/2) = 4R sin²(A/2) = 2R(1−cosA)a₁ + r₁ − r = 2Rr₁ + a₁ = 2R + r
Let IE cut circumcircle at P. By properties of incenter, P is circumcenter of BICE.Let A' be midpoint of BCra-r=IE*sin(<A/2)=2*IP*sin(<A/2)=2*BP*sin(<A/2)=2*A'P,(ra-r)/2=A'PIt is also known that a1/2=OA'Summing up previous equations gives(ra-r)/2+a1/2=A'P+OA'=R, or ra+a1-r=2R