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

(Created page with "==Problem== Show that for any triangle, <math>\frac{3\sqrt{3}}{2}\ge \sin(3A) + \sin(3B) + \sin (3C) \ge -2</math>. When does the equality hold? ==Solution== {{solution}} ==S...")
 
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==Solution==
 
==Solution==
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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>.
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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:
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<cmath>0 \le x,y,z \le 540^\circ, x+y+z = 540^\circ</cmath>
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and we must prove that
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<cmath>\frac{3\sqrt{3}}{2}\ge \sin(x) + \sin(y) + \sin (z) \ge -2</cmath>
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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).
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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
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<cmath>\sin( x) + \sin (y) + \sin (z) \le 0 + -1 + -1 = -2</cmath>
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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>.
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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
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<cmath>\sin (x) + \sin (y) + \sin (z) \le 1 + 1 + 0 = 2 < \frac{3\sqrt{3}}{2}</cmath>
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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>.
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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>.
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<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>
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<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
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<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>
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Likewise, <math>f'(u) < f'(y-d)</math> for all <math>u\in(y-d, y)</math>; therefore
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<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>
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Therefore, adding our inequalities together,
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<cmath>f(x+d) + f(y-d) > f(x) + d\cdot f'(x+d) + f(y) - d\cdot f'(y-d)</cmath>
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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>.
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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>.
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Then, since <math>x+y+z = 180^\circ</math>, we conclude <math>x=y=z=60^\circ</math>. Then
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<cmath>\sin (x) + \sin (y) + \sin (z) = 3 \cdot \sin (60^\circ) = 3 \cdot \frac{\sqrt{3}}{2}</cmath>
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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>.
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~ <math>\LaTeX</math> by [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9046]
  
 
==See Also==
 
==See Also==
 
{{USAMO box|year=1981|num-b=2|num-a=4}}
 
{{USAMO box|year=1981|num-b=2|num-a=4}}
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{{MAA Notice}}
  
 
[[Category:Olympiad Trigonometry Problems]]
 
[[Category:Olympiad Trigonometry Problems]]

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

See Also

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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