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

Post your solutions or ideas in the comments.

4 comments:

  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
    .-.

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  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
    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
    .-.

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

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  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.

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