Circumcenter on Ray (Posted on 2013-10-12)

	Submitted by Bractals
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Place a co-ordinate system on ΔABC such that A (0,2a), B (2b,0), and C (-2b,0), where a,b > 0. We need to show that the circumcenter of ΔBHO lies on line y = a(2b-x)/b with x ≤ 2b. Let A' and B' be the midpoints of sides BC and AC respectively. The orthocenter and circumcenter of ΔABC lie on the y-axis with H (0,2h) and O (0,2k), where h > 0 and k < a. Let M (m,h+k) be the circumcenter of ΔBHO [the y-coordinate is h+k since M lies on the perpendicular bisector of HO]. We need to show that h+k = a(2b-m)/b with m ≤ 2b. The lengths |MB|, |MH|, and |MO| as radii must be equal. Therefore,      (m-2b)2 + (h+k)2 = |MB|2 = |MH|2 = |MO|2 = m2 + (h-k)2                  or      m = (b2+hk)/b ΔBA'H ~ ΔAB'O ~ ΔAA'C ~ ΔAA'B. Therefore,      2h/2b = |A'H|/|A'B| = |A'B|/|A'A| = 2b/2a            or      h = b2/a B'O ⊥ AC implies      (a-2k)/(0-b) = slope(B'O) = -1/slope(AC) = -b/a         or       k = (a2-b2)/(2a) --------------------------------------------           0 < b(a2+b2)       ⇒ 0 < a2b+b3       ⇒ 3a2b < 4a2b+b3       ⇒ 3a2b-b3 < 4a2b       ⇒ b(3a2-b2) < 4a2b       ⇒ b(3a2-b2)/(2a2) < 2b       ⇒ m < 2b --------------------------------------------       h+k = [b2/a]+[(a2-b2)/(2a)]              = (a2+b2)/(2a)              = (4a2-3a2+b2)/(2a)              = a[2b-b(3a2-b2/(2a2)]/b              = a[2b-m]/b -------------------------------------------- QED

	Subject	Author	Date
	Solution	Harry	2013-10-16 12:07:50