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All-Soviet-Union Mathematical Olympiad

Geometry Difficulty 4.5 AIME Prove it Soviet Union

Problem:
A rectangular box has sides x<y<zx < y < z. Its perimeter is p=4(x+y+z)p = 4(x + y + z), its surface area is s=2(xy+yz+zx)s = 2(xy + yz + zx) and its main diagonal has length d=x2+y2+z2d = \sqrt{x^2 + y^2 + z^2}. Show that 3x<(p/4−d2−s/2)3x < (p/4 - \sqrt{d^2 - s/2}) and 3z>(p/4+d2−s/2)3z > (p/4 + \sqrt{d^2 - s/2}).

Solution

Solution:
We have 3(y−x)(z−x)>03(y - x)(z - x) > 0, so 3x2+3yz>3xy+3xz3x^2 + 3yz > 3xy + 3xz. Hence y2+z2+4x2+2yz−4xy−4xz>x2+y2+z2−xy−yz−xzy^2 + z^2 + 4x^2 + 2yz - 4xy - 4xz > x^2 + y^2 + z^2 - xy - yz - xz or (y+z−2x)2>(d2−s/2)(y + z - 2x)^2 > (d^2 - s/2). Hence (x+y+z)>3x+d2−s/2(x + y + z) > 3x + \sqrt{d^2 - s/2}. So 3x<p/4−d2−s/23x < p/4 - \sqrt{d^2 - s/2}.

Similarly, 3(z−x)(z−y)>03(z - x)(z - y) > 0, so x2+y2+4z2>x2+y2+z2+3zx+3zy−3xyx^2 + y^2 + 4z^2 > x^2 + y^2 + z^2 + 3zx + 3zy - 3xy, so (2z−x−y)2>x2+y2+z2−xy−yz−zx(2z - x - y)^2 > x^2 + y^2 + z^2 - xy - yz - zx or (3z−p/4)2>(d2−s/2)(3z - p/4)^2 > (d^2 - s/2). Hence 3z>p/4+d2−s/23z > p/4 + \sqrt{d^2 - s/2}.

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