Sunday, August 3, 2008

Elearn Geometry Problem 155

See complete Problem 155
Euler's Theorem: Distance from the Incenter to the Circumcenter. Level: High School, SAT Prep, College geometry

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

2. a more geometric proof
E touching point on AC
in right triangle AEI , AI=r/sin(A/2)
bisector AI meets the circumcircle at D,
draw diameter BF
in right triangle BFD ,BD=2Rsin(A/2)
in triangle IBD
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
.-.

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

4. Solution to Problem 155.
The 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.