Maths Olympiad Prep

Library / /4 of 133

Geometry Difficulty 4.5 AIME Prove it Saudi Arabia

Show that in any triangle ABCABC with A^=90∘\widehat{A}=90^{\circ} the following inequality holds:
(AB−AC)2(BC2+4AB⋅AC)2≤2BC6. (AB-AC)^{2}\left(BC^{2}+4 AB \cdot AC\right)^{2} \leq 2 BC^{6} .

Solutions — 2

Solution 1

The inequality is equivalent to
(ABBC−ACBC)2(1+4ABBC⋅ACBC)2≤2, \left(\frac{AB}{BC}-\frac{AC}{BC}\right)^{2}\left(1+4 \frac{AB}{BC} \cdot \frac{AC}{BC}\right)^{2} \leq 2,
hence
(cos⁡B−sin⁡B)2(1+4cos⁡Bsin⁡B)2≤2, (\cos B-\sin B)^{2}(1+4 \cos B \sin B)^{2} \leq 2,
that is
(1−sin⁡2B)(1+4sin⁡2B+4sin⁡22B)≤2. (1-\sin 2B)\left(1+4 \sin 2B+4 \sin ^{2} 2B\right) \leq 2 .
After easy computations we get the equivalent inequality
(1+sin⁡2B)(2sin⁡2B−1)2≥0. (1+\sin 2B)(2 \sin 2B-1)^{2} \geq 0 .
We have equality if and only if sin⁡2B=12\sin 2B=\frac{1}{2}, i.e. B^=π12\widehat{B}=\frac{\pi}{12}.

Solution 2

Let α=ABBC,β=ACBC\alpha=\frac{AB}{BC}, \beta=\frac{AC}{BC}. We have α2+β2=1\alpha^{2}+\beta^{2}=1, and the inequality is equivalent to:
(α−β)2(1+4αβ)2≤2, (\alpha-\beta)^{2}(1+4 \alpha \beta)^{2} \leq 2,
hence
(1−2αβ)(1+4αβ)2≤2. (1-2 \alpha \beta)(1+4 \alpha \beta)^{2} \leq 2 .
Put x=2αβx=2 \alpha \beta and obtain
(1−x)(1+2x)2≤2 or 4x2−3x+1≥0, (1-x)(1+2 x)^{2} \leq 2 \text{ or } 4 x^{2}-3 x+1 \geq 0,
which is equivalent to (2x−1)2(x+1)≥0(2 x-1)^{2}(x+1) \geq 0.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.