Maths Olympiad Prep

Library / /520 of 520

Algebra Difficulty 6.3 National olympiad Prove it

Lemma In △ABC\triangle A B C, prove that:
tan⁡2A2+tan⁡2B2+tan⁡2C2⩾2−8sin⁡A2⋅sin⁡B2⋅sin⁡C2, \begin{array}{l} \tan ^{2} \frac{A}{2}+\tan ^{2} \frac{B}{2}+\tan ^{2} \frac{C}{2} \\ \geqslant 2-8 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}, \end{array}
The equality holds if and only if △ABC\triangle A B C is an equilateral triangle.

Solution

Lemma Proof: Let i,j,ki, j, k be unit vectors in a plane, and j\boldsymbol{j} forms an angle of π−∠A\pi-\angle A with kk, kk forms an angle of π−∠B\pi-\angle B with ii, and ii forms an angle of π−∠C\pi-\angle C with jj. Then
(itan⁡A2+jtan⁡B2+ktan⁡C2)2⩾0. \left(i \tan \frac{A}{2}+j \tan \frac{B}{2}+k \tan \frac{C}{2}\right)^{2} \geqslant 0 .

Thus,
tan⁡2A2+tan⁡2B2+tan⁡2C2⩾2tan⁡A2⋅tan⁡B2⋅cos⁡C+2tan⁡B2⋅tan⁡C2⋅cos⁡A+2tan⁡C2⋅tan⁡A2⋅cos⁡B=2tan⁡A2⋅tan⁡B2(1−2sin⁡2C2)+2tan⁡B2⋅tan⁡C2(1−2sin⁡2A2)+2tan⁡C2⋅tan⁡A2(1−2sin⁡2B2)=2(tan⁡A2⋅tan⁡B2+tan⁡B2⋅tan⁡C2+tan⁡C2⋅tan⁡A2)−4sin⁡A2⋅sin⁡B2⋅sin⁡C2(sin⁡A2cos⁡B2⋅cos⁡C2+sin⁡B2cos⁡C2⋅cos⁡A2+sin⁡C2cos⁡A2⋅cos⁡B2)=2−4sin⁡A2⋅sin⁡B2⋅sin⁡C2.sin⁡A+sin⁡B+sin⁡C2cos⁡A2⋅cos⁡B2⋅cos⁡C2=2−8sin⁡A2⋅sin⁡B2⋅sin⁡C2. \tan ^{2} \frac{A}{2}+\tan ^{2} \frac{B}{2}+\tan ^{2} \frac{C}{2} \geqslant 2 \tan \frac{A}{2} \cdot \tan \frac{B}{2} \cdot \cos C + 2 \tan \frac{B}{2} \cdot \tan \frac{C}{2} \cdot \cos A + 2 \tan \frac{C}{2} \cdot \tan \frac{A}{2} \cdot \cos B =2 \tan \frac{A}{2} \cdot \tan \frac{B}{2}\left(1-2 \sin ^{2} \frac{C}{2}\right) + 2 \tan \frac{B}{2} \cdot \tan \frac{C}{2}\left(1-2 \sin ^{2} \frac{A}{2}\right) + 2 \tan \frac{C}{2} \cdot \tan \frac{A}{2}\left(1-2 \sin ^{2} \frac{B}{2}\right) =2\left(\tan \frac{A}{2} \cdot \tan \frac{B}{2}+\tan \frac{B}{2} \cdot \tan \frac{C}{2}+\tan \frac{C}{2} \cdot \tan \frac{A}{2}\right)- 4 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}\left(\frac{\sin \frac{A}{2}}{\cos \frac{B}{2} \cdot \cos \frac{C}{2}}+ \frac{\sin \frac{B}{2}}{\cos \frac{C}{2} \cdot \cos \frac{A}{2}}+\frac{\sin \frac{C}{2}}{\cos \frac{A}{2} \cdot \cos \frac{B}{2}}\right) =2-4 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}. \sin A+\sin B+\sin C 2 \cos \frac{A}{2} \cdot \cos \frac{B}{2} \cdot \cos \frac{C}{2} =2-8 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}.
Note that, in △ABC\triangle ABC, there is a well-known identity:
tan⁡A2⋅tan⁡B2+tan⁡B2⋅tan⁡C2+tan⁡C2⋅tan⁡A2=1. \tan \frac{A}{2} \cdot \tan \frac{B}{2}+\tan \frac{B}{2} \cdot \tan \frac{C}{2}+\tan \frac{C}{2} \cdot \tan \frac{A}{2}=1.
Thus, equation (1) holds.
With the above lemma, the solution to this problem is much easier. Below is the solution to this problem.

Solution: As in Approach 2, we can form an acute △ABC\triangle ABC with sides a,b,ca, b, c.
 Let x=cos⁡A,y=cos⁡B,z=cos⁡C, then f(x,y,z)=cos⁡2A1+cos⁡A+cos⁡2B1+cos⁡B+cos⁡2C1+cos⁡C=1−sin⁡2A2cos⁡2A2+1−sin⁡2B2cos⁡2B2+1−sin⁡2C2cos⁡2C2=1−4sin⁡2A2⋅cos⁡2A22cos⁡2A2+1−4sin⁡2B2⋅cos⁡2B22cos⁡2B2+1−4sin⁡2C2⋅cos⁡2C22cos⁡2C2=sin⁡2A2+cos⁡2A2−4sin⁡2A2⋅cos⁡2A22cos⁡2A2+sin⁡2B2+cos⁡2B2−4sin⁡2B2⋅cos⁡2B22cos⁡2B2+sin⁡2C2+cos⁡2C2−4sin⁡2C2⋅cos⁡2C22cos⁡2C2=32+12(tan⁡2A2+tan⁡2B2+tan⁡2C2)−2(sin⁡2A2+sin⁡2B2+sin⁡2C2)=32+12(tan⁡2A2+tan⁡2B2+tan⁡2C2) - 2(1−2sin⁡A2⋅sin⁡B2⋅sin⁡C2)⩾32+12(2−8sin⁡A2⋅sin⁡B2⋅sin⁡C2)−2(1−2sin⁡A2⋅sin⁡B2⋅sin⁡C2) (by the lemma) =12.  \begin{array}{l} \text { Let } x=\cos A, y=\cos B, z=\cos C, \text { then } \\ f(x, y, z) \\ =\frac{\cos ^{2} A}{1+\cos A}+\frac{\cos ^{2} B}{1+\cos B}+\frac{\cos ^{2} C}{1+\cos C} \\ =\frac{1-\sin ^{2} A}{2 \cos ^{2} \frac{A}{2}}+\frac{1-\sin ^{2} B}{2 \cos ^{2} \frac{B}{2}}+\frac{1-\sin ^{2} C}{2 \cos ^{2} \frac{C}{2}} \\ =\frac{1-4 \sin ^{2} \frac{A}{2} \cdot \cos ^{2} \frac{A}{2}}{2 \cos ^{2} \frac{A}{2}}+\frac{1-4 \sin ^{2} \frac{B}{2} \cdot \cos ^{2} \frac{B}{2}}{2 \cos ^{2} \frac{B}{2}}+ \\ \frac{1-4 \sin ^{2} \frac{C}{2} \cdot \cos ^{2} \frac{C}{2}}{2 \cos ^{2} \frac{C}{2}} \\ =\frac{\sin ^{2} \frac{A}{2}+\cos ^{2} \frac{A}{2}-4 \sin ^{2} \frac{A}{2} \cdot \cos ^{2} \frac{A}{2}}{2 \cos ^{2} \frac{A}{2}}+ \\ \frac{\sin ^{2} \frac{B}{2}+\cos ^{2} \frac{B}{2}-4 \sin ^{2} \frac{B}{2} \cdot \cos ^{2} \frac{B}{2}}{2 \cos ^{2} \frac{B}{2}}+ \\ \sin ^{2} \frac{C}{2}+\cos ^{2} \frac{C}{2}-4 \sin ^{2} \frac{C}{2} \cdot \cos ^{2} \frac{C}{2} \\ 2 \cos ^{2} \frac{C}{2} \\ =\frac{3}{2}+\frac{1}{2}\left(\tan ^{2} \frac{A}{2}+\tan ^{2} \frac{B}{2}+\tan ^{2} \frac{C}{2}\right)- \\ 2\left(\sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}\right) \\ =\frac{3}{2}+\frac{1}{2}\left(\tan ^{2} \frac{A}{2}+\tan ^{2} \frac{B}{2}+\tan ^{2} \frac{C}{2}\right) \text { - } \\ 2\left(1-2 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}\right) \\ \geqslant \frac{3}{2}+\frac{1}{2}\left(2-8 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}\right)- \\ 2\left(1-2 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}\right) \text { (by the lemma) } \\ =\frac{1}{2} \text {. } \\ \end{array}

Clearly, when ∠A=∠B=∠C=60∘\angle A=\angle B=\angle C=60^{\circ}, the equality holds, so the minimum value of f(x,y,z)f(x, y, z) is 12\frac{1}{2}.

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: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.