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Math Help - 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.
    \setlength{\unitlength}{1.3mm}<br />
\begin{picture}(50,45)<br />
\put(24.6,38){$\bullet$}<br />
\put(19.6,5.6){$\bullet$}<br />
\put(33.7,3.9){$\bullet$}<br />
\put(21.6,39){$A$}<br />
\put(0,6){$B$}<br />
\put(36,3.2){$C$}<br />
\put(24,16){$O$}<br />
\put(20,3){$D$}<br />
\put(36,22.5){$X$}<br />
\put(43,11){$Y$}<br />
\put(25.6,38.6){\line(3,-5){20}}<br />
\put(45,11){\line(-5,-3){20}}<br />
\thicklines<br />
\put(39,0){\line(-1,1){34}}<br />
\put(26,41){\line(-1,-6){7}}<br />
\put(42,29){\line(-5,-3){42}}<br />
\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 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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