Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Monday, December 1, 2014
Geometry Problem 1065: Triangle, Acute Angle, Orthocenter, Circumradius, Inradius, Exradius, Distance, Diameter
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Since a₁² = 2R²(1+cos2A) = 4R² cos²A, thus a₁ = 2R cosA.
ReplyDeleter = 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 = 2R
r₁ + a₁ = 2R + r
Let IE cut circumcircle at P. By properties of incenter, P is circumcenter of BICE.
ReplyDeleteLet A' be midpoint of BC
ra-r=IE*sin(<A/2)=2*IP*sin(<A/2)=2*BP*sin(<A/2)=2*A'P,
(ra-r)/2=A'P
It 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