Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Monday, November 17, 2014

### Geometry Problem 1061: Triangle, Inradius (r), Circumradius (R), Circumcircle, Angle Bisector, Distance from a point to a Line, Perpendicular

Labels:
angle bisector,
circumcircle,
circumradius,
distance,
inradius,
perpendicular,
triangle

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Join A₁B, A₁C. Then A₁B=A₁C=2Rsin(A/2).

ReplyDeleteBy Ptolemy Theorem on ABA₁C,

a×AA₁ = 2Rsin(A/2)×(b+c)

a×AA₁sin(A/2) = 2Rsin²(A/2)×(b+c)

a×a₁ = 2Rsin²(A/2)×(b+c)

sinA×a₁ = 2Rsin²(A/2)(sinB+sinC)

2sin(A/2)cos(A/2)×a₁ = 4Rsin²(A/2) cos(A/2)cos((B-C)/2)

a₁ = 2Rsin(A/2)cos((B-C)/2) = R(cosB+cosC)

Similarly,

b₁ = R(cosA+cosC)

c₁ = R(cosA+cosB)

a₁+b₁+c₁

= 2R (cosA + cosB + cosC)

= 2R + 8Rsin(A/2)sin(B/2)sin(C/2)

= 2R + 2r

= 2(R+r)

From property of incenter I, A1, B1, and C1 are circumcenters of BIC, AIC, and AIB respectively.

ReplyDeleteLet C' and A' be perpendiculars from C1 and A1 onto AB and BC respectively.

Due to first statement,

c1-r=IC1*sin(<C/2)=C1C'

a1-r=IA1*sin(<A/2)=A1A'

Let P be perpendicular from O to B1B2.

<OB1P=<C=<BOC', and OB1=OB=R, so triangles B1OP and OBC' are congruent. Therefore,

b1-OA'=B1P=OC',

b1=OA'+OC'

Summing up previous equations gives

(c1-r)+(a1-r)+(b1)=(C1C')+(A1A')+(OA'+OC')=2R.