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Solution. Let’s euler 5 this problem. Let ρ be the circumradius of the triangle ABC. It’s easy to show that BC = 2ρ sin A and EF = 2ρ sin A cos A. Since DQ = 2ρ sin C cos B cos A, DR = 2ρ sin B cos C cos A, and ∠F DE = π − 2A, the Cosine Law gives us or where QR 2 = DQ 2 + DR 2 − 2DQ · DR cos(π − 2A) [ ] = 4ρ 2 cos 2 A (sin C cos B) 2 + (sin B cos C) 2 + 2 sin C cos B sin B cos C cos(2A) QR = 2ρ cos A √ f(A, B, C), f(A, B, C) = (sin C cos B) 2 + (sin B cos C) 2 + 2 sin C cos B sin B cos C cos(2A). So, what we need to attack is the following inequality: ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎝ ∑ 2ρ sin A⎠ ⎝ ∑ 2ρ cos A √ f(A, B, C) ⎠ ≥ ⎝ ∑ 2ρ sin A cos A⎠ or cyclic cyclic cyclic cyclic cyclic ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎝ ∑ sin A⎠ ⎝ ∑ cos A √ f(A, B, C) ⎠ ≥ ⎝ ∑ 2 sin A cos A⎠ . Our job is now to find a reasonable lower bound of √ f(A, B, C). Once again, we express f(A, B, C) as the sum of two squares. We observe that cyclic f(A, B, C) = (sin C cos B) 2 + (sin B cos C) 2 + 2 sin C cos B sin B cos C cos(2A) = (sin C cos B + sin B cos C) 2 + 2 sin C cos B sin B cos C [−1 + cos(2A)] = sin 2 (C + B) − 2 sin C cos B sin B cos C · 2 sin 2 A = sin 2 A [1 − 4 sin B sin C cos B cos C] . So, we shall express 1 − 4 sin B sin C cos B cos C as the sum of two squares. The trick is to replace 1 with ( sin 2 B + cos 2 B ) ( sin 2 C + cos 2 C ) . Indeed, we get 2 1 − 4 sin B sin C cos B cos C = ( sin 2 B + cos 2 B ) ( sin 2 C + cos 2 C ) − 4 sin B sin C cos B cos C = (sin B cos C − sin C cos B) 2 + (cos B cos C − sin B sin C) 2 = sin 2 (B − C) + cos 2 (B + C) = sin 2 (B − C) + cos 2 A. It therefore follows that f(A, B, C) = sin 2 A [ sin 2 (B − C) + cos 2 A ] ≥ sin 2 A cos 2 A so that ∑ cos A √ f(A, B, C) ≥ ∑ sin A cos 2 A. cyclic cyclic So, we can complete the proof if we establish that ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎝ ∑ sin A⎠ ⎝ ∑ sin A cos 2 A⎠ ≥ ⎝ ∑ 2 sin A cos A⎠ . cyclic cyclic cyclic Indeed, one sees that it’s a direct consequence of the Cauchy-Schwarz inequality (p + q + r)(x + y + z) ≥ ( √ px + √ qy + √ rz) 2 , where p, q, r, x, y and z are positive real numbers. 5 euler v. (in Mathematics) transform the problems in triangle geometry to trigonometric ones 8
Alternatively, one may obtain another lower bound of f(A, B, C): f(A, B, C) = (sin C cos B) 2 + (sin B cos C) 2 + 2 sin C cos B sin B cos C cos(2A) = (sin C cos B − sin B cos C) 2 + 2 sin C cos B sin B cos C [1 + cos(2A)] = sin 2 (B − C) + 2 sin(2B) · sin(2C) · 2 cos 2 A 2 2 ≥ cos 2 A sin(2B) sin(2C). Then, we can use this to offer a lower bound of the perimeter of triangle P QR: p(P QR) = ∑ 2ρ cos A √ f(A, B, C) ≥ ∑ 2ρ cos 2 A √ sin 2B sin 2C cyclic cyclic So, one may consider the following inequality: p(ABC) ∑ 2ρ cos 2 A √ sin 2B sin 2C ≥ p(DEF ) 2 or or cyclic cyclic cyclic ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎝2ρ ∑ sin A⎠ ⎝ ∑ 2ρ cos 2 A √ sin 2B sin 2C⎠ ≥ ⎝2ρ ∑ 2 sin A cos A⎠ . cyclic cyclic cyclic cyclic ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎝ ∑ sin A⎠ ⎝ ∑ cos 2 A √ sin 2B sin 2C⎠ ≥ ⎝ ∑ 2 sin A cos A⎠ . However, it turned out that this doesn’t hold. Try to disprove this! Problem 5. (IMO 2001/1) Let ABC be an acute-angled triangle with O as its circumcenter. Let P on line BC be the foot of the altitude from A. Assume that ∠BCA ≥ ∠ABC + 30 ◦ . Prove that ∠CAB + ∠COP < 90 ◦ . Proof. The angle inequality ∠CAB + ∠COP < 90 ◦ can be written as ∠COP < ∠P CO. This can be shown if we establish the length inequality OP > P C. Since the power of P with respect to the circumcircle of ABC is OP 2 = R 2 − BP · P C, where R is the circumradius of the triangle ABC, it becomes R 2 − BP · P C > P C 2 or R 2 > BC · P C. We euler this. It’s an easy job to get BC = 2R sin A and P C = 2R sin B cos C. Hence, we show the inequality R 2 > 2R sin A · 2R sin B cos C or sin A sin B cos C < 1 4 . Since sin A < 1, it suffices to show that sin A sin B cos C < 1 4 . Finally, we use the angle condition ∠C ≥ ∠B + 30◦ to obtain the trigonometric inequality sin B cos C = sin(B + C) − sin(C − B) 2 ≤ 1 − sin(C − B) 2 ≤ 1 − sin 30◦ 2 = 1 4 . We close this section with Barrows’ inequality stronger than Erdös-Mordell Theorem. following trigonometric inequality: We need the Proposition 1.2.1. Let x, y, z, θ 1 , θ 2 , θ 3 be real numbers with θ 1 + θ 2 + θ 3 = π. Then, x 2 + y 2 + z 2 ≥ 2(yz cos θ 1 + zx cos θ 2 + xy cos θ 3 ). Proof. Using θ 3 = π − (θ 1 + θ 2 ), it’s an easy job to check the following identity x 2 + y 2 + z 2 − 2(yz cos θ 1 + zx cos θ 2 + xy cos θ 3 ) = (z − (x cos θ 2 + y cos θ 1 )) 2 + (x sin θ 2 − y sin θ 1 ) 2 . 9
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Page 3 and 4: Chapter 1 Geometric Inequalities It
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Page 9: 1.2 Trigonometric Methods In this s
Page 13 and 14: 1.3 Applications of Complex Numbers
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Page 17 and 18: 2.2 Algebraic Substitutions We know
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Page 23 and 24: (IMO 2000/2) Let a, b, c be positiv
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Page 27 and 28: Third Solution. (by an IMO 2005 con
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Page 35 and 36: Second Solution. It’s equivalent
Page 37 and 38: Second Solution. After setting a =
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Page 41 and 42: (Iran 1998) Prove that, for all x,
Page 43 and 44: Let a, b > 0 and m, n ∈ N. Take x
Page 45 and 46: (3) ⇒ (2) : Let x 1 , · · · ,
Page 47 and 48: 4.2 Power Means Convexity is one of
Page 49 and 50: 4.3 Majorization Inequality We say
Page 51 and 52: Lemma 4.4.1. Let f : (a, b) −→
Page 53 and 54: M 6. (Korea 2001) Let x 1 , · ·
Page 55 and 56: M 24. (IMO Short List 1998) Let r 1
Page 57 and 58: M 41. (C1578, O. Johnson, C. S. Goo
Page 59 and 60: M 57. (China 2002) Given c ∈ ( 1
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P 12. Putnam 98B1 Find the minimum
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P 33. (MM Q197, Norman Schaumberger
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MEK Marcin E. Kuczma, Problem 1940,
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