See complete Problem 155

Euler's Theorem: Distance from the Incenter to the Circumcenter. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Sunday, August 3, 2008

### Elearn Geometry Problem 155

Labels:
chords theorem,
circumcenter,
circumradius,
diameter,
distance,
Euler,
incenter,
inradius,
orthic triangle,
Problem 154,
Problem 155

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Let's take origin B,x-axis: BC

ReplyDeletethe coordinates of O and I are:

O(a/2;RcosA),I((p-b);r)

d²=(a/2-(p-b))²+(RcosA-r)²

=R²sin²A-a(p-b)+(p-b)²+R²cos²A-2rRcosA+r²

we know:

cos²A+sin²A=1;cosA=1-2sin²(A/2)

d²=R²-2rR+4rRsin²(A/2)+r²-a(p-b)+(p-b)²

but abc=4RS;S=pr;sin²(A/2)=(p-b)(p-c)/(bc)

then 4rRsin²(A/2)=a(p-b)(p-c)/p

because r²=(p-a)(p-b)(p-c)/p

r²-a(p-b)+(p-b)²=-a(p-b)(p-c)/p

then

d²=R²-2Rr

.-.

a more geometric proof

ReplyDeleteE touching point on AC

in right triangle AEI , AI=r/sin(A/2)

bisector AI meets the circumcircle at D,

draw diameter BF

ang(BFD)=ang(BAD)=A/2(inscribed angles)

in right triangle BFD ,BD=2Rsin(A/2)

in triangle IBD

ang(ACB)=ang(ADB)=C;(inscribed angles)

hence ang(BID)=ang(IBD)=(A+B)/2

IBD is isoscele,BD=ID

the circle power of point I with respect to the circumcercle is:

P(I)=-IA.ID= -IA.BD= -2Rr

P(I)= d²-R²=-2Rr

d²=R²-2Rr

.-.

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

ReplyDeleteSolution to Problem 155.

ReplyDeleteThe extension of OI meets the circumcircle at P (from I to O) and Q (from O to I).

Taking the circle power of point I with respect to the circumcircle we have BI.ID = PI.IQ. But BI.ID = 2Rr (as proved in problem 154), and PI = R+d and IQ = R-d. So 2Rr = (R+d)(R-d) then d2 = R2 – 2Rr.