Sunday Times Teaser 2687 – Creased
by H Bradley and C Higgins
I took a rectangular piece of paper whose long side was three times its short side, both being whole numbers of centimetres.
I folded the rectangle along a diagonal and then folded the resulting shape along its line of symmetry. When unfolded, the resulting creases divided the rectangle into two equal triangles and two equal quadrilaterals.
The areas of these triangles and quadrilaterals were comprised of the same three digits but in a different order.
What was the area of each triangle?
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AR=2(AT+AQ)=3(a^2)
AT=(1/2)(a/2)((3a/2)+x) and AQ=(1/2)(a/2)(3a/2)+(a/2)(3a/2)-(1/2)(x)(a/2) where :
AR=area of the rectangular paper , AU=area of the triangle , AQ=area of the quadrilateral
a=shorter side of the rectangle and x=the difference between one side of the triangle and the half of the longer side of the rectangle
Hence we get .
a^2=(8AT-2ax)/3=(8AQ+2ax)/9 as well as AQ=3AT-ax and taking a=16,17,18… and x=1,2,3,… we can get to the solution by trial and error.