Problem:
A rectangle that is inscribed in a larger rectangle (with one vertex on each side) is called unstuck if it is possible to rotate (however slightly) the smaller rectangle about its center within the confines of the larger. Of all the rectangles that can be inscribed unstuck in a 6 by 8 rectangle, the smallest perimeter has the form N, for a positive integer N. Find N.
Solution:
Let ABCD be the outer rectangle, with AB=8. Let PQRS be an inscribed rectangle, with P,Q,R,S on AB,BC,CD,DA respectively. Note that angles QPB, RQC,SRD, and PSA are congruent; let θ denote their common measure. Let x=PQ and y=QR. Without loss of generality, we may assume that the figure has been labeled so that y<x and θ<45∘. (Why?) Then
BA=BP+PA=xcosθ+ysinθ=8BC=BQ+QC=xsinθ+ycosθ=6
and these equations can be solved simultaneously to give
x=cos2θ−sin2θ8cosθ−6sinθ and y=cos2θ−sin2θ6cosθ−8sinθ
Thus the perimeter of PQRS is
2x+2y=cosθ+sinθ28
Since cosθ+sinθ=2sin(θ+45∘) and 0<θ<45∘, it follows that the perimeter of PQRS is a decreasing function of θ. It will be shown below that BQ is an increasing function of θ, and that PQRS is stuck if and only if 3<BQ. Therefore, we seek the perimeter of the rectangle that results when Q is the midpoint of BC (and S is the midpoint of DA). Because Q and S are midpoints of opposite sides of ABCD, rectangle PQRS has area 24 and diagonal 8, yielding the equations
xy=24x2+y2=64
Combining these, we find that (x+y)2=64+2⋅24=112, so 2(x+y)=2112=448 is the smallest perimeter for an unstuck inscribed rectangle.
To complete the demonstration, let O be the intersection of AC and BD, and consider circles centered at O and intersecting all four sides of ABCD. These circles all have diameters between 8 and 10. Except for the extreme cases, each such circle intersects ABCD at eight points, P,P′,Q,Q′,R,R′,S,S′, given in cyclical order so that P and P′ are on AB,Q and Q′ are on BC, etc. Note that PQRS and PQ′RS′ are inscribed rectangles.
Rectangle PQRS is unstuck (Figure 1), because its vertices can be moved along the arcs PS′,QP′,RQ′, and SR′, which lie inside ABCD. Note that ∠QOP′=2θ. As the diameter decreases, both 2θ and BQ increase, because P′ and Q drift away from B. This shows that BQ is an increasing function of θ.
Rectangle PQ′RS′ is stuck (Figure 2) when the diameter exceeds 8 (and 3<BQ′). This is because arcs Q′Q and RR′ are outside ABCD, so Q′ and R are free to move only along arc Q′R, which is impossible.
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