1981 USAMO Problems/Problem 3


Show that for any triangle, $\frac{3\sqrt{3}}{2}\ge \sin(3A) + \sin(3B) + \sin (3C) \ge -2$.

When does the equality hold?


Given three angles that add to 180°, one can construct a triangle from them. This is true even if the angles are negative; however, the resulting triangle could then be recognized as having only positive angles, and the interpretation with negative angles (and, likely, negative side lengths) might be considered perverse. In this case, the problem must be interpreted as ruling out such "perverse" triangles; otherwise, its angles could be -30°, -20°, and 230°, in which case sin(3x) becomes -1, -√3/2, and -1/2 respectively, which add to about -2.3, contradicting the problem statement.

The constraints on the angles of a non-perverse triangle are: 0 ≤ A,B,C ≤ 180°, and A+B+C = 180°. Any

In fact, at this point, we only care about 3A, 3B, and 3C. Let us call them x, y, and z. We have:

0 ≤ x,y,z ≤ 540°; x+y+z = 540°. Prove $\frac{3\sqrt{3}}{2}\ge \sin(x) + \sin(y) + \sin (z) \ge -2$.

Now. Without loss of generality, assume x ≤ y ≤ z. It follows that x ≤ 180° and z ≥ 180° [otherwise, x+y+z would be strictly greater than 180°, or strictly less than 180°, respectively].

Since sin(u) is nonnegative over the interval [0, 180°], and -1 ≤ sin u ≤ 1 for all real u, we can immediately use the x ≤ 180° result to show that sin x + sin y + sin z ≤ 0 + -1 + -1 = -2. This proves the lower bound on sin x + sin y + sin z; equality occurs when sin x = 0 and sin y = sin z = -1, and this is reachable only when y = z = 270° and x = 0°, which translates into A = 0° and B = C = 90°. (This might be considered a degenerate triangle and ruled out; in that case, the lower bound could be stated as a strict one.)

Now for the upper bound. It is true that sin(u) is non-positive over the interval [180°, 360°]. Also, 3√3/2 is greater than 2 (it is roughly 2.6): if you square it, you get 27/4, while if you square 2, you get 16/4. So... We know z ≥ 180°. If z ≤ 360°, then sin z ≤ 0, and then sin x + sin y + sin z ≤ 1 + 1 + 0 = 2 < 3√3/2. Therefore, we need merely handle the case where z ≥ 360°. sin(u + 360°) = sin(u), so we may as well subtract 360° from z. Then the problem effectively becomes:

"Given 0 ≤ x,y,z ≤ 180° and x+y+z = 180°, prove that sin x + sin y + sin z ≤ 3√3/2. Also find when you get equality."

This is probably a well-known theorem, but I shall address it here anyway.

Let us suppose that {x,y,z} is the selection of x,y,z given the above constraints with the largest value of sin x + sin y + sin z. It suffices to show that, for this selection, sin x + sin y + sin z ≤ 3√3/2. We shall see that, given these constraints, sin x + sin y + sin z increases when you move any two variables closer together, and it is maximized when x=y=z.

Lemma with some general expository use: A special case of a result called "Jensen's inequality". If f(x) is well-behaved and f'(x) is strictly decreasing over an open interval (a, b), then, for any {x,y} selected from that interval with x ≠ y, $2f(\frac{x + y}{2}) > f(x) + f(y).$

Proof of lemma: It is true that, for any well-behaved function f, f(x+d) = f(x) + ∫(f'(u), u from x to x+d), and likewise f(x-d) = f(x) - ∫(f'(u), u from x-d to x). Addressing the lemma, WLOG suppose x < y; let d = (y-x)/2. Since f'(u) is given to be strictly decreasing over the relevant interval, we have f'(u) > f'(x+d) for all u in (x, x+d); therefore f(x+d) = f(x) + ∫(f'(u), u from x to x+d) > f(x) + ∫(f'(x+d), u from x to x+d) = f(x) + d*f'(x+d). Likewise, f'(u) < f'(y-d) for all u in (y-d, y); therefore f(y-d) = f(y) - ∫(f'(u), u from y-d to y) > f(y) - ∫(f'(y-d), u from y-d to y) = f(y) - d*f'(y-d). Therefore, adding our inequalities together, $f(x+d) + f(y-d) > f(x) + d\cdot f'(x+d) + f(y) - d*f'(y-d)$. But $x+d = y-d = \frac{x+y}{2}$. In that case, the $d\cdot f'(x+d)$ and $-d\cdot f'(y-d)$ terms cancel, and we get our desired result: $2 f(\frac{x+y}{2}) > f(x) + f(y).$

So. sin(u) is well-behaved, and sin'(u) = cos(u), which is strictly decreasing over the open interval (0, 180°). Suppose that {x,y,z} are not all the same value. Then, without loss of generality, suppose x ≠ y. Let a = (x+y)/2. Then the selection {a,a,z} also satisfies the constraints 0 ≤ a,a,z ≤ 180° and a+a+z ≤ 180°. Furthermore, by the above lemma, 2*sin a > sin x + sin y, so this new selection {a,a,z} has a larger corresponding value sin a + sin a + sin z than the selection {x,y,z}. This contradicts our original assumption that {x,y,z} was chosen to have the largest value of sin x + sin y + sin z. Therefore, x=y=z.

Then, since x+y+z = 180°, we conclude x=y=z=60°. Then sin x + sin y + sin z = 3 * sin 60° = 3 * √3/2. So 3√3/2 is indeed the maximum value of sin x + sin y + sin z, which was to be proven. Translating back into the previous problem (0 ≤ x,y,z ≤ 540°; x+y+z = 540°), the maximum occurs when x=y=60° and z=420°; translating into the original problem, we have proven the upper bound, and equality with the upper bound occurs when A=B=20° and C = 140°.

See Also

1981 USAMO (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5
All USAMO Problems and Solutions

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