An Euler Circle ProofDate: 03/26/99 at 13:02:48 From: Bernadette Gural Subject: The Euler Circle I am having trouble proving the following result: The distance, d, between the circumcenter O and the incenter, I, of a triangle is given by the relation: d^2 = R(R-2r), where R is the circumradius and r is the inradius. I know that the nine points are consisted of the following: J, K, L - the Euler points D, E, F - the feet of ha, hb, hc of triangle ABC A', B', C' - the midpoints of triangle ABC I guess I am having trouble with the incenter (not sure what exactly an incenter is) and also with the inradius. Is there any help you can give me to prove the above result? I appreciate both your time, effort and patience in dealing with yet another stumped geometry student. Date: 03/26/99 at 14:41:52 From: Doctor Floor Subject: Re: The Euler Circle First: the question you pose is a theorem by Euler (1765, an earlier statement is by Chapple 1746), but it is not about the nine point circle. Second: the incenter is the point of intersection of the three (internal) angle bisectors in a triangle. It is also the center of the circle inscribed in the triangle (= incircle). That means that this circle is inside the triangle and is tangent to the three sides. Third: The proof. Let ABC be the triangle, O its circumcenter and I its incenter. Let R be the radius of the circumcircle and r of the incircle. Let E be the point where the incircle meets BC, and let D and D' be the midpoints of arcs AB over the circumcircle. Finally, let d = OI. Here is a picture of the situation: We observe that angle ACD = 0.5* angle C, since D bisects arc AB. As a consequence, I is on segment CD. It is also easy to see that DD' is a diameter of the circumcircle. Note that angle DBI = angle DBA + angle ABI = angle DCA + 0.5*angle B (cuts off same arc) = angle ICA + 0.5*angle B = 0.5*angle C + 0.5*angle B Note also that angle BDI = angle BDC = angle A (cuts off same arc) So angle BID = 180 - angle A - 0.5*angle C - 0.5*angle B = 0.5*angle C + 0.5*angle B So, triangle BID is isosceles, and DB = DI. It is not difficult to see that triangles D'BD and CEI are similar, since angle DD'B = angle DCB = angle ICE and angle D'BD = angle CEI = 90 degrees. We can use this to get the following ratio: DD':DB = IC:IE From which we can deduce: DD'*IE = IC*DB [DD' = 2R, IE = r and DB = DI] 2R*r = IC*ID But IC*ID is equal to the power of incenter I w.r.t. the circumcircle. As a consequence IC*ID = R^2 - OI^2 or: 2Rr = R^2 - d^2 d^2 = R^2 - 2Rr d^2 = R(R - 2r) And the statement is proven. - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/