See complete Problem 158

Relation between the Circumradius, Inradius and Exradii of a triangle. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Thursday, August 7, 2008

### Elearn Geometry Problem 158

Labels:
chord,
circumcircle,
circumradius,
diagonal,
diameter,
excircle,
exradius,
incircle,
inradius,
midpoint,
perpendicular,
tangent,
trapezoid,
triangle

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name G, I meet AC, P meet BC, T meet AB

ReplyDeletename K, E2 meet AC, H meet BC, M meet BA

name R, E3 meet CA, S, E1 meet AC

E is midpoint of AC, midpoint of RS ( RA = CS see P117)

draw from O perpendicular to r1 and r3

E1²=R²+2Rr1, E3²=R²+2Rr3 ( from P 157 )

E1²-(r1-OE)² =E3²-(r3-OE)² (tr perpen from O tor1,r3)

=>

OE = ( r1 + r3 - 2R )/2

=>

DE = R + OE = R + (r1+r3-2R)/2

DE = ( r1 + r3 )/2

2)

join B to F extend , it meet E2

( BIF, F is midpoint of arc AC, P154, FC=FA P156 )

BM = BH => BT+AT+AM=BP+PC+CH => AT+AM=PC+CH =>

AG+AK=CG+CK => AK+KG+AK=CG+CG+KG => 2AK=2CG

AK = CG

▲NIG~▲NEF => NG/r = NE/EF => NG/NE = r/EF (1)

▲ENF~▲GKE2 => NE/EF = NK/r2 => NK/NE = r2/E2

=> (NE+NG+NE)/NE =r2/E2 =>NG/NE=r2/E2-2 (2)

from (1) and (2) r/EF = (r2/E2)-2

EF = ( r2 - r )/2

3)

DE + EF = 2R =>

R = ( r1 + r2 + r3 - r )/4

( and now 159 is done )

http://ahmetelmas.wordpress.com/2010/05/15/geo-geo/

ReplyDeleteSolution to problem 158.

ReplyDelete1) Let be p the semiperimeter of ABC, G and H the tangent points of the circumferences E1 and E3 with AC, respectively.

The angle bisector BI passes through E2 and meets the circumcircle at F, which is midpoint of arc AFC. As ang(DBF) = ang(IBE) = 90º, then points E1, B, D and E3 are collinear. Since E is midpoint of AC, and AH = CG = p-b, then E is midpoint of GH. DE links the midpoints of sides E1E3 and GH of the trapezoid E1GHE3. From that we conclude that DE = (r1+r3)/2.

2) Let be J and K the tangent points of circles I and E2 with AC, respectively. We have

AK = CJ = p-a, so E is midpoint of JK. The line drown by F, parallel to AC, cuts KE2 at M, and the extension of IJ at L, so that FL = FM. Hence triangles FME2 and FLI are congruent, IL = ME2, r + EF = r2 – EF and EF = (r2-r)/2.

3) From items (1) and (2) we have DE + EF = (r+r2+r3-r)/2 = DF = 2R, thus

R = (r1+r2+r3-r)/4.