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  "body": "\n## Definition\n\nA set $S$ of points of the Euclidean plane is an *orthocentric system* if both conditions hold.\n\n1. $S$ has exactly four points.\n2. Some three points of $S$ form a [[triangle|0JDB8YJBSR]] whose [[orthocentre|0QMYQ1X03H]] is the fourth point.\n\n## Remarks\n\nCondition 1 forces the triangle in condition 2 to have no right angle.\nFor a right triangle the orthocentre is the vertex of the right angle, so only three distinct points would remain.\n\nExample: let $A = (0, 2)$, $B = (-1, 0)$, $C = (1, 0)$ and $H = (0, \\frac{1}{2})$.\nThe altitude from $A$ is the line $x = 0$.\nThe altitude from $B$ is perpendicular to $C - A = (1, -2)$, so it is the line $(x + 1) - 2y = 0$, which meets $x = 0$ at $(0, \\frac{1}{2})$.\nSo $H$ is the orthocentre of $ABC$, and $\\set{A, B, C, H}$ is an orthocentric system.\n\nNon-example: $(0, 0)$, $(1, 0)$, $(0, 1)$ and $(1, 1)$.\nThe orthocentre of each three of them is a vertex of that triangle, since each such triangle has a right angle, so condition 2 fails.\n\nIf $\\set{A, B, C, H}$ is an orthocentric system, then each point is the orthocentre of the other three.\nIndeed, $H$ is the orthocentre of $ABC$ when $AH \\perp BC$, $BH \\perp CA$ and $CH \\perp AB$.\nThe same three relations say that $A$ is the orthocentre of $HBC$, and similarly for $B$ and $C$.\n",
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