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.

## Monday, December 1, 2014

### Geometry Problem 1065: Triangle, Acute Angle, Orthocenter, Circumradius, Inradius, Exradius, Distance, Diameter

Labels:
acute,
angle,
circumradius,
diameter,
distance,
exradius,
inradius,
orthocenter,
triangle

Subscribe to:
Post Comments (Atom)

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