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Thread: Geometric Construction #2

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    Geometric Construction #2

    Given three medians of a triangle ABC, construct the triangle.
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    Quote Originally Posted by MATNTRNG View Post
    Given three medians of a triangle ABC, construct the triangle.
    $\displaystyle \setlength{\unitlength}{1.3mm}
    \begin{picture}(50,45)
    \put(24.6,38){$\bullet$}
    \put(19.6,5.6){$\bullet$}
    \put(33.7,3.9){$\bullet$}
    \put(21.6,39){$A$}
    \put(0,6){$B$}
    \put(36,3.2){$C$}
    \put(24,16){$O$}
    \put(20,3){$D$}
    \put(36,22.5){$X$}
    \put(43,11){$Y$}
    \put(25.6,38.6){\line(3,-5){20}}
    \put(45,11){\line(-5,-3){20}}
    \thicklines
    \put(39,0){\line(-1,1){34}}
    \put(26,41){\line(-1,-6){7}}
    \put(42,29){\line(-5,-3){42}}
    \end{picture}$

    Given lines AO, BO, CO, you want to locate the positions of A, B, C on those lines so that they form the vertices of a triangle having those lines as medians. The size of the triangle is not determined by that condition, so you can fix A arbitrarily on the line AO. Construct the point D on the line AO, on the opposite side of O from A, so that $\displaystyle DO = \tfrac12AO$. Then D will lie on the side BC of the triangle.

    The centroid O of the triangle is the centre of mass of three equal masses at the vertices. So the vertices A and C must be equidistant from the median BO. Construct a line AXY through A, perpendicular to BO, meeting BO at X, and such that AX = XY. Then construct a line through Y perpendicular to AXY (and therefore parallel to BOX). This line meets CO at the vertex C of the triangle, and thus determines the position of C.

    Finally, you can construct B as the point where the line CD meets the median BOX.
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