A piece of cardboard has the shape of a pentagon

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A piece of cardboard has the shape of a pentagon

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Source: Romanian JBMO TST, Day 4, Problem 17

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Valentin Vornicu

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Valentin Vornicu

#1 h Mar 18, 2006, 10:33 PM • 1 Y

Y by Adventure10

A piece of cardboard has the shape of a pentagon $ABCDE$ in which $BCDE$ is a square and $ABE$ is an isosceles triangle with a right angle at $A$ . Prove that the pentagon can be divided in two different ways in three parts that can be rearranged in order to recompose a right isosceles triangle.

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yetti

#2 h May 1, 2006, 4:06 AM • 2 Y

Y by Adventure10, Mango247

Let a = BC be the square side. Area $|ABCDE| = \frac{5a^2}{4}.$ Let x be the leg of the isosceles right triangle with the area $\frac{x^2}{2} = \frac 5 4 a^2.$ Then $x = a\ \sqrt{\frac 5 2.}$ Since $AC = AD = a\ \sqrt{\frac 1 4 + \frac 9 4} = a\ \sqrt{\frac 5 2}$ , one of these is the 1st cut, say AD. Rotate the triangle $\triangle ADE$ by $90^\circ$ around A, so that AE coincides with AB and D goes into $D' \in BE$ . Let M be the midpoint of BC. Since the triangles $\triangle MCD \cong \triangle MBD'$ are congruent, DM is the 2nd cut.

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