Dissection of Triangle into Rhombus

by Hubert Shutrick

The applet below illustrates Problem 8 from the 2010 All-Russian Olympiad:

In an acute triangle ABC, the median AM is longer than side AB. Prove that you can cut the triangle into three parts out of which you can construct a rhombus.

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

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Let M be the midpoint of BC and B_m the midpoint AC. Assume that BM < AB. Then there is point N on AB such that MN = AB. Cut over NM and B_mM and swivel the two pieces 180°, one around B_m, the other around M. N' and N'' are the corresponding images of N and M' is the image of M under the rotation around B_m.

The proof that the construction leads to a rhombus is rather straightforward: all four sides of the resulting quadrilateral N'N''MM' equal AB.

(There is another attempt at solving the problem.)

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