IMO 1985 SL 15

Problem

5a.(FRA 3) Let K and K′ be two squares in the same plane, their sides of equal length. Is it possible to decompose K into a finite number of tri- angles T1, T2, . . . , Tp with mutually disjoint interiors and find translations t1, t2, . . . , tp such that K′ = p2 i=1 ti(Ti)?

Solution

There is no loss of generality if we assume K = ABCD, K′ = AB′C′D′, and that K′ is obtained from K bya clockwise rotation around A by \varphi, 0 \leq\varphi \leq\pi/4. Let C′D′, B′C′, and the parallel to AB through D′ meet the line BC at E, F, and G respectively. Let us now choose points E′ \inAB′, G′ \inAB, C′′ \inAD′, and E′′ \inAD such that A B B′ C C′ C′′ D D′ D′′ E E′ E′′ F G G′ H H′ the triangles AE′G′ and AC′′E′′ are translates of the triangles D′EG and FC′E respectively. Since AE′ = D′E and AC′′ = FC′, we have C′′E′′ = C′E = B′E′ and C′′D′ = B′F, which imply that \triangleE′′C′′D′ is a

translate of \triangleE′B′F, and consequently E′′D′ = E′F and E′′D′ \parallelE′F. It follows that there exist points H \inCD, H′ \inBF, and D′′ \inE′G′ such that E′′D′HD is a translate of E′FH′D′′. The remaining parts of K and K′ are the rectangles D′GCH and D′′H′BG′ of equal area. We shall now show that two rectangles with parallel sides and equal ar- eas can be decomposed into translation invariant parts. Let the sides of the rectangles XY ZT and X′Y ′Z′T ′ (XY \parallelX′Y ′) satisfy X′Y ′ < XY , Y ′Z′ > Y Z, and X′Y ′ \cdot Y ′Z′ = XY \cdot Y Z. Suppose that 2X′Y ′ > XY (otherwise, we may cut offcongruent rectangles from both the original ones until we reduce them to the case of 2X′Y ′ > XY ). Let U \inXY and V \inZT be points such that Y U = TV = X′Y ′ and W \inXV be a point such that UW \parallelXT . Then translating \triangleXUW to a triangle V ZR and \triangleXV T to a triangle WRS results in a rectangle UY RS congruent to X′Y ′Z′T ′. Thus we have partitioned K and K′ into translation-invariant parts. Al- though not all the parts are triangles, we may simply triangulate them.