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Additional info for 44th International Mathematical Olympiad: Short-listed problems and solutions
Thus c must have 2n + 1 digits. Let a = 10x + s, where x is an n-digit number (in case x = 0 we set n = 0). Then b = 10n s + x, c = 102n s2 + 2 · 10n sx + x2 , d = 10(c − 10m−1 f ) + f = 102n+1 s2 + 20 · 10n sx + 10x2 − 10m f + f, where m is the number of digits of c. , a = 6 · · · 6 3 or a = 4 · · · 4 2 or a = 2 · · · 2 1, n n n for n ≥ 0. The first two possibilities must be rejected for n ≥ 1, since a2 = d would have 2n + 2 digits, which means that c would have to have at least 2n + 2 digits, but we already know that c must have 2n + 1 digits.
O1 D φ1 θ4 θ1 A O4 φ4 C P φ2 θ3 θ2 O2 φ3 O3 B Put θ1 = ∠O4 O1 O2 , θ2 = ∠O1 O2 O3 , θ3 = ∠O2 O3 O4 , θ4 = ∠O3 O4 O1 and φ1 = ∠P O1 O4 , φ2 = ∠P O2 O3 , φ3 = ∠P O3 O2 , φ4 = ∠P O4 O1 . By the law of sines, we have O1 O2 : O1 O3 = sin φ3 : sin θ2 , O3 O4 : O1 O3 = sin φ1 : sin θ4 , O3 O4 : O2 O4 = sin φ2 : sin θ3 , O1 O2 : O2 O4 = sin φ4 : sin θ1 . Since the segment P A is the common chord of Γ1 and Γ2 , the segment P A is the altitude from P to O1 O2 . Similarly P B , P C , P D are the altitudes from P to O2 O3 , O3 O4 , O4 O1 , respectively.
D+e+f All we have to prove is that √ r 1 √ = 2 + 3. ≥ 2g 2− 3 49 Since r = 2g def d+e+f 1 1 1 + + +2 d e f d+e+f def =√ x+y+z + 2, xy + yz + zx where x = 1/d, y = 1/e, z = 1/f , it suffices to prove that (x + y + z)2 ≥ 3. xy + yz + zx This inequality is true because (x + y + z)2 − 3(xy + yz + zx) = 1 (x − y)2 + (y − z)2 + (z − x)2 ≥ 0. 2 Solution 2. We prove that t > s/2 in the same way as in Solution 1. Put g = t − s/2. Γd Γe D d e E (−e, 0) r/2 Γr/2 g Γg f Γf F (f, 0) Now set the coordinate system so that E(−e, 0), F (f, 0), and the y-coordinate of D is positive.
44th International Mathematical Olympiad: Short-listed problems and solutions