1981 USAMO Problems/Problem 3

Problem

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

When does the equality hold?

Solution

Given three angles that add to $180^\circ$ , one can construct a triangle from them. However, its angles must all be nonnegative; thus the constraints on the angles of a triangle are $0^\circ\le A,B,C\le180^\circ$ and $A+B+C=180^\circ$ .

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

$0 \le x,y,z \le 540^\circ, x+y+z = 540^\circ$

and we must prove that

$\frac{3\sqrt{3}}{2}\ge \sin(x) + \sin(y) + \sin (z) \ge -2$

Without loss of generality, assume $x \le y \le z$ . It follows that $x \le 180^\circ$ and $z \ge 180^\circ$ (otherwise, $x+y+z$ would be strictly greater than $180^\circ$ or strictly less than $180^\circ$ , respectively).

Since $\sin(u)$ is nonnegative over the interval $[0^\circ, 180^\circ]$ and $-1 \le \sin (u) \le 1$ for all real $u$ , we can immediately use the $x \le 180^\circ$ result to show that

$\sin( x) + \sin (y) + \sin (z) \le 0 + -1 + -1 = -2$

This proves the lower bound; equality occurs when $\sin(x) = 0$ and $\sin (y) = \sin (z) = -1$ , and this is reachable only when $y = z = 270^\circ$ and $x = 0^\circ$ , which translates into $A = 0^\circ$ and $B = C = 90^\circ$ .

Now for the upper bound. It is true that $\sin(u)$ is non-positive over the interval $[180^\circ, 360^\circ]$ . Also, $\frac{3\sqrt{3}}{2}>2$ (by a simple squaring argument). Then $z \ge 180^\circ$ . If $z \le 360^\circ$ , then $\sin (z) \le 0$ , and then $\sin (x) + \sin (y) + \sin (z) \le 1 + 1 + 0 = 2 < \frac{3\sqrt{3}}{2}$ Therefore, we need to handle the case where $z \ge 360^\circ$ . $\sin(u + 360^\circ) = \sin(u)$ , so we can subtract $360^\circ$ from $z$ due to periodicity. Then the constraints become $0^\circ \le x,y,z \le 180^\circ$ and $x+y+z = 180^\circ$ .

Let us suppose that $(x,y,z)$ is the selection of $x,y,z$ given the above constraints with the largest value of $\sin (x) + \sin (y) + \sin (z)$ . We shall see that, given these constraints, this value increases when the distance between two of these variables decreases, and it is maximized when $x=y=z$ .

$\textbf{Lemma:}$ 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 $x \ne y$ , then $2f\left(\frac{x + y}{2}\right) > f(x) + f(y).$

$\textbf{Proof of lemma:}$ It is true that, for any well-behaved function $f(x)$ , $f(x+d) = f(x) + \int_{x}^{x+d} f'(u)\,du$ , and likewise, $f(x-d) = f(x) - \int_{x-d}^x f'(u)\,du$ . Without loss of generality, suppose $x < y$ ; let $d = \frac{1}{2}(y-x)$ . 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 $f(x+d) = f(x) + \int_x^{x+d} f'(u)\,du > f(x) + \int_x^{x+d} f'(x+d)\,du = f(x) + d\cdot f'(x+d)$ Likewise, $f'(u) < f'(y-d)$ for all $u\in(y-d, y)$ ; therefore $f(y-d) = f(y) - \int_{y-d}^y f'(u)\,du > f(y) - \int_{y-d}^y f'(y-d)\,du = f(y) - d\cdot f'(y-d)$ Therefore, adding our inequalities together, $f(x+d) + f(y-d) > f(x) + d\cdot f'(x+d) + f(y) - d\cdot f'(y-d)$ 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: $2 f\left(\frac{x+y}{2}\right) > f(x) + f(y)$ .

Since $\sin(u)$ is well-behaved, and $\sin'(u) = \cos(u)$ , which is strictly decreasing over the open interval $(0^\circ, 180^\circ)$ , then we can apply our lemma. Suppose for the sake of contradiction that $x,y,z$ are not all the same value, and this produces a maximum. Then, without loss of generality, suppose $x \ne y$ . Let $a = \frac{x+y}{2}$ . Then the selection $(a,a,z)$ also satisfies the constraints $0^\circ \le a,a,z \le 180^\circ$ and $a+a+z \le 180^\circ$ . Furthermore, by the above lemma, $2\cdot\sin (a) > \sin (x) + \sin (y)$ , 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 maximum. Therefore, $x=y=z$ .

Then, since $x+y+z = 180^\circ$ , we conclude $x=y=z=60^\circ$ . Then $\sin (x) + \sin (y) + \sin (z) = 3 \cdot \sin (60^\circ) = 3 \cdot \frac{\sqrt{3}}{2}$ So $\frac{\sqrt{3}}{2}$ is indeed the maximum value of $\sin(x) + \sin (y) + \sin (z)$ , proving the bound. Translating back into the previous problem, the maximum occurs when $x=y=60^\circ$ and $z=420^\circ$ ; translating into the original problem, we have proven the upper bound, and equality with the upper bound occurs when $A=B=20^\circ$ and $C=140^\circ$ .

~ $\LaTeX$ by eevee9046

1981 USAMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
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1981 USAMO Problems/Problem 3

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