Difference between revisions of "1981 USAMO Problems/Problem 3"

Latest revision as of 00:05, 13 March 2025

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

@@ Line 5: / Line 5: @@
 ==Solution==
-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.
+Given three angles that add to <math>180^\circ</math>, one can construct a triangle from them. However, its angles must all be nonnegative; thus the constraints on the angles of a triangle are <math>0^\circ\le A,B,C\le180^\circ</math> and <math>A+B+C=180^\circ</math>.
-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 <math>3A, 3B,</math> and <math>3C</math>.  Let us call them <math>x, y,</math> and <math>z</math>.  We have:
-In fact, at this point, we only care about 3A, 3B, and 3C.  Let us call them x, y, and z.  We have:
+<cmath>0 \le x,y,z \le 540^\circ, x+y+z = 540^\circ</cmath>
-≤ x,y,z ≤ 540°; x+y+z = 540°.  Prove <math>\frac{3\sqrt{3}}{2}\ge \sin(x) + \sin(y) + \sin (z) \ge -2</math>.
+and we must prove that
-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].
+<cmath>\frac{3\sqrt{3}}{2}\ge \sin(x) + \sin(y) + \sin (z) \ge -2</cmath>
-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.)
+Without loss of generality, assume <math>x \le y \le z</math>.  It follows that <math>x \le 180^\circ</math> and <math>z \ge 180^\circ</math> (otherwise, <math>x+y+z</math> would be strictly greater than <math>180^\circ</math> or strictly less than <math>180^\circ</math>, respectively).
-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:
+Since <math>\sin(u)</math> is nonnegative over the interval <math>[0^\circ, 180^\circ]</math> and <math>-1 \le \sin (u) \le 1</math> for all real <math>u</math>, we can immediately use the <math>x \le 180^\circ</math> result to show that
-"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."
+<cmath>\sin( x) + \sin (y) + \sin (z) \le 0 + -1 + -1 = -2</cmath>
-This is probably a well-known theorem, but I shall address it here anyway.
+This proves the lower bound; equality occurs when <math>\sin(x) = 0</math> and <math>\sin (y) = \sin (z) = -1</math>, and this is reachable only when <math>y = z = 270^\circ</math> and <math>x = 0^\circ</math>, which translates into <math>A = 0^\circ</math> and <math>B = C = 90^\circ</math>.
-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.
+Now for the upper bound. It is true that <math>\sin(u)</math> is non-positive over the interval <math>[180^\circ, 360^\circ]</math>.  Also, <math>\frac{3\sqrt{3}}{2}>2</math> (by a simple squaring argument). Then <math>z \ge 180^\circ</math>.  If <math>z \le 360^\circ</math>, then <math>\sin (z) \le 0</math>, and then
+<cmath>\sin (x) + \sin (y) + \sin (z) \le 1 + 1 + 0 = 2 < \frac{3\sqrt{3}}{2}</cmath>
+Therefore, we need to handle the case where <math>z \ge 360^\circ</math>. <math>\sin(u + 360^\circ) = \sin(u)</math>, so we can subtract <math>360^\circ</math> from <math>z</math> due to periodicity. Then the constraints become <math>0^\circ \le x,y,z \le 180^\circ</math> and <math>x+y+z = 180^\circ</math>.
-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, <math>2f(\frac{x + y}{2}) > f(x) + f(y).</math>
+Let us suppose that <math>(x,y,z)</math> is the selection of <math>x,y,z</math> given the above constraints with the largest value of <math>\sin (x) + \sin (y) + \sin (z)</math>. We shall see that, given these constraints, this value increases when the distance between two of these variables decreases, and it is maximized when <math>x=y=z</math>.
-: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, <math>f(x+d) + f(y-d) > f(x) + d\cdot f'(x+d) + f(y) - d*f'(y-d)</math>.  But <math>x+d = y-d = \frac{x+y}{2}</math>.  In that case, the <math>d\cdot f'(x+d)</math> and <math>-d\cdot f'(y-d)</math> terms cancel, and we get our desired result: <math>2 f(\frac{x+y}{2}) > f(x) + f(y).</math>
+<math>\textbf{Lemma:}</math> If <math>f(x)</math> is well-behaved and <math>f'(x)</math> is strictly decreasing over an open interval <math>(a, b)</math>, then, for any <math>x,y</math> selected from that interval with <math>x \ne y</math>, then <math>2f\left(\frac{x + y}{2}\right) > f(x) + f(y).</math>
-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.
+<math>\textbf{Proof of lemma:}</math> It is true that, for any well-behaved function <math>f(x)</math>, <math>f(x+d) = f(x) + \int_{x}^{x+d} f'(u)\,du</math>, and likewise, <math>f(x-d) = f(x) - \int_{x-d}^x f'(u)\,du</math>. Without loss of generality, suppose <math>x < y</math>; let <math>d = \frac{1}{2}(y-x)</math>.  Since <math>f'(u)</math> is given to be strictly decreasing over the relevant interval, we have <math>f'(u) > f'(x+d)</math> for all <math>u\in(x, x+d)</math>; therefore
+<cmath>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)</cmath>
+Likewise, <math>f'(u) < f'(y-d)</math> for all <math>u\in(y-d, y)</math>; therefore
+<cmath>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)</cmath>
+Therefore, adding our inequalities together,
+<cmath>f(x+d) + f(y-d) > f(x) + d\cdot f'(x+d) + f(y) - d\cdot f'(y-d)</cmath>
+But <math>x+d = y-d = \frac{x+y}{2}</math>. In that case, the <math>d\cdot f'(x+d)</math> and <math>-d\cdot f'(y-d)</math> terms cancel, and we get our desired result: <math>2 f\left(\frac{x+y}{2}\right) > f(x) + f(y)</math>.
-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°.
+Since <math>\sin(u)</math> is well-behaved, and <math>\sin'(u) = \cos(u)</math>, which is strictly decreasing over the open interval <math>(0^\circ, 180^\circ)</math>, then we can apply our lemma. Suppose for the sake of contradiction that <math>x,y,z</math> are not all the same value, and this produces a maximum. Then, without loss of generality, suppose <math>x \ne y</math>. Let <math>a = \frac{x+y}{2}</math>. Then the selection <math>(a,a,z)</math> also satisfies the constraints <math>0^\circ \le a,a,z \le 180^\circ</math> and <math>a+a+z \le 180^\circ</math>. Furthermore, by the above lemma, <math>2\cdot\sin (a) > \sin (x) + \sin (y)</math>, so this new selection <math>(a,a,z)</math> has a larger corresponding value <math>\sin (a) + \sin (a) + \sin (z)</math> than the selection <math>(x,y,z)</math>. This contradicts our original assumption that <math>(x,y,z)</math> was chosen to have the maximum. Therefore, <math>x=y=z</math>.
+Then, since <math>x+y+z = 180^\circ</math>, we conclude <math>x=y=z=60^\circ</math>. Then
+<cmath>\sin (x) + \sin (y) + \sin (z) = 3 \cdot \sin (60^\circ) = 3 \cdot \frac{\sqrt{3}}{2}</cmath>
+So <math>\frac{\sqrt{3}}{2}</math> is indeed the maximum value of <math>\sin(x) + \sin (y) + \sin (z)</math>, proving the bound. Translating back into the previous problem, the maximum occurs when <math>x=y=60^\circ</math> and <math>z=420^\circ</math>; translating into the original problem, we have proven the upper bound, and equality with the upper bound occurs when <math>A=B=20^\circ</math> and <math>C=140^\circ</math>.
+~ <math>\LaTeX</math> by [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9046]
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Difference between revisions of "1981 USAMO Problems/Problem 3"

Latest revision as of 00:05, 13 March 2025

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