If $A,B,C$ are the angles of a triangle, prove that \[-2 \le \sin{3A}+\sin{3B}+\sin{3C} \le \frac{3\sqrt{3}}{2}\] and determine when equality holds.
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Tags: AMC, USA(J)MO, USAMO, trigonometry, inequalities, geometry unsolved, geometry
26.07.2011 23:25
hello, if $A=B=20^{\circ},C=140^{\circ}$ (or permutations) we get $\sin(60^{\circ})+\sin(60^{\circ})+\sin(520^{\circ})=\frac{3\sqrt{3}}{2}$. If $A=0^{\circ},B=C=90^{\circ}$ (or permutations) we get $\sin(0^{\circ})+\sin(270^{\circ})+\sin(270^{\circ})=-2$. $\sin(3x)$ is positive for $x<60^{\circ}$ or $x>120^{\circ}$ thus on angle must be greater than $120^{\circ}$, now we assume $A\le B<60^{\circ}$ and $C>120^{\circ}$, let $C^{\star}=C-120^{\circ}$ and we have $3A+3B+3C^{\star}=180^{\circ}$. Further we have that $\sin(x)$ is concave for $0\le x\le 180^{\circ}$ thus $-\sin(x)$ is convex and we get by Jensen $-\sin(x)-\sin(y)-\sin(z)\geq3\left(-\sin(\frac{x+y+z}{3})\right)=-3\sin(60^{\circ})$ thus we get $\sin(x)+\sin(y)+\sin(z)\le 3\sin(60^{\circ})=\frac{3\sqrt{3}}{2}$ therefore $\sin(3A)+\sin(3B)+\sin(3C)\le \frac{3\sqrt{3}}{2}$. Sonnhard.
16.11.2020 01:01
I guess to prove the lower bound (since above merely showed that -2 is obtainable but didn't show that we can't go any lower than -2): From above, $\sin(3x)$ only when negative for $60^{\circ} < x < 120^{\circ}$. Since $A + B + C = 180^{\circ}$, at most $2$ of $\sin(3A)$, $\sin(3B)$, $\sin(3C)$ may be negative. The sin function obtains its minimum at $-1$, so the bare minimum of $\sin(3A)+\sin(3B)+\sin(3C)$ is $(-1) + (-1) + 0 = 2$. If we assume $\sin(3B)$ and $\sin(3C)$ are the negative values, we have $B + C > 120^{\circ}$, so $\sin(3A) = 0$ only happens when $A = 0^{\circ}$. Then, $B = C = 90^{\circ}$ lets the two other sin values be $-1$. We can permute $A, B, C$ freely. Yes, I know 9 year bump, but it doesn't actually hurt anyone.
17.11.2020 06:06
$3/2\,\sqrt {3}-\sin \left( 2\,A \right) -\sin \left( 2\,B \right) - \sin \left( 2\,C \right) =1/6\,{\frac {\sqrt {3} \left( -4\,\sqrt {3}s r+9\,{R}^{2} \right) }{{R}^{2}}} $ So ,is easy
17.11.2020 06:08
$\sin \left( A \right) +\sin \left( B \right) +\sin \left( C \right) - \sin \left( 2\,A \right) -\sin \left( 2\,B \right) -\sin \left( 2\,C \right) ={\frac { \left( R-2\,r \right) s}{{R}^{2}}} $
17.11.2020 06:13
We have $\sum{\sin{A}\cos^4{\frac{B-C}{2}}}-\sin \left( 2\,A \right) -\sin \left( 2\,B \right) -\sin \left( 2\,C \right) =1/16\,{ \frac { \left( R-2\,r \right) s \left( \left( R-2\,r \right) \left( 2\,r+4\,R \right) +4\,{R}^{2}+4\,Rr+3\,{r}^{2}-{s}^{2} \right) }{{R}^{ 4}}}\geq 0$
17.11.2020 06:20
sorry,is $\sum{3A}$
17.11.2020 06:58
Mrdavid445 wrote: If $A,B,C$ are the angles of a triangle, prove that \[-2 \le \sin{3A}+\sin{3B}+\sin{3C} \le \frac{3\sqrt{3}}{2}\]and determine when equality holds. stronger is: In triangle,prove that \[\frac{3}{2}\,\sqrt {3-24\,\cos \left( A \right) \cos \left( B \right) \cos \left( C \right) }\geq \sin \left( 3\,A \right) +\sin \left( 3\,B \right) +\sin \left( 3\,C \right) \]
17.11.2020 07:02
xzlbq wrote: Mrdavid445 wrote: If $A,B,C$ are the angles of a triangle, prove that \[-2 \le \sin{3A}+\sin{3B}+\sin{3C} \le \frac{3\sqrt{3}}{2}\]and determine when equality holds. stronger is: In triangle,prove that \[\frac{3}{2}\,\sqrt {3-24\,\cos \left( A \right) \cos \left( B \right) \cos \left( C \right) }\geq \sin \left( 3\,A \right) +\sin \left( 3\,B \right) +\sin \left( 3\,C \right) \] <=> \[f=-4\,{s}^{6}+24\, \left( R+r \right) ^{2}{s}^{4}+ \left( -144\,{R}^{3}r -216\,{R}^{2}{r}^{2}-90\,{R}^{4}-144\,R{r}^{3}-36\,{r}^{4} \right) {s} ^{2}+27\,{R}^{4} \left( 8\,Rr+2\,{r}^{2}+9\,{R}^{2} \right) \geq 0\]But \[f=3\,{\it u1}\,{\it u2}\,r{R}^{2}+{\it u1}\,{\it u3}\, \left( 272\,{r} ^{3}+152\,R{r}^{2}+83\,r{R}^{2}+8\,{\it u1}\,{\it u3} \right) +4\,{ \it u2}\,{\it u3}\,{s}^{2}+{\it u1}\, \left( 11\,{R}^{5}+56\,{R}^{3}{r }^{2}+58\,{R}^{4}r \right) +{\it u3}\, \left( 26\,{R}^{4}+448\,{r}^{4} \right) \]as \[\left\{ {\it u1}=R-2\,r,{\it u2}={s}^{2}-16\,Rr+5\,{r}^{2},{\it u3}=4 \,{R}^{2}+4\,Rr+3\,{r}^{2}-{s}^{2} \right\} \]
17.11.2020 12:53
Mrdavid445 wrote: If $A,B,C$ are the angles of a triangle, prove that \[-2 \le \sin{3A}+\sin{3B}+\sin{3C} \le \frac{3\sqrt{3}}{2}\]and determine when equality holds. In triangle,have \[\sin \left( 3\,A \right) +\sin \left( 3\,B \right) +\sin \left( 3\,C \right) +2\geq 54\,{\frac {\sqrt {3}{r}^{2}}{s \left( {\it h_a}+{\it h_b}+{ \it h_c} \right) }}\]<=> \[-\sqrt {3}{s}^{6}+\sqrt {3} \left( 2\,Rr+3\,{R}^{2}+2\,{r}^{2} \right) {s}^{4}+2\,{R}^{3}{s}^{3}\sqrt {3}+3\,\sqrt {3} \left( 4\,R+r \right) r \left( R+r \right) ^{2}{s}^{2}+2\,\sqrt {3} \left( 4\,R+r \right) r{R}^{3}s-324\,{R}^{4}{r}^{2}\geq 0\]<=> ${{\it u2}}^{2} \left( -{\frac {14103}{2}}+{\frac {16287}{4}}\,\sqrt {3 } \right) {r}^{2}+36\,{\it u10}\,{\it u2}\,\sqrt {3}r+{{\it u7}}^{2} \left( 510-144\,\sqrt {3} \right) {r}^{4}+{\it u10}\, \left( -846+ 1116\,\sqrt {3} \right) {r}^{3}+{\it u1}\, \left( 19242-10980\,\sqrt { 3} \right) {r}^{5}+{\it u1}\,{\it u7}\, \left( \left( {\frac {46875}{ 8}}-{\frac {26639}{8}}\,\sqrt {3} \right) {r}^{4}+ \left( -{\frac {45} {8}}+{\frac {97}{16}}\,\sqrt {3} \right) {s}^{3}r \right) +{\it u9}\,{ \it u7}\, \left( \left( {\frac {963}{8}}-{\frac {211}{8}}\,\sqrt {3} \right) {r}^{2}+\sqrt {3}Rs+ \left( {\frac {45}{8}}-{\frac {33}{16}} \,\sqrt {3} \right) sr \right) +5\,{\it u7}\,{\it u3}\,\sqrt {3}Rrs+{ \it u1}\,{\it u9}\, \left( \left( 28017-16175\,\sqrt {3} \right) {r}^ {2}+ \left( -9+8\,\sqrt {3} \right) sr \right) +{\it u2}\,{\it u7}\, \left( \left( -{\frac {2673}{16}}+{\frac {2839}{16}}\,\sqrt {3} \right) {r}^{3}+ \left( {\frac {675}{8}}-{\frac {1253}{64}}\,\sqrt {3 } \right) {r}^{2}s \right) +{\it u2}\,{\it u3}\, \left( \left( -{ \frac {113715}{16}}+{\frac {65663}{16}}\,\sqrt {3} \right) {r}^{2}+ \sqrt {3}{\it u2} \right) +{\it u1}\,{\it u10}\, \left( -81+320\, \sqrt {3} \right) {r}^{2}+{\it u7}\, \left( \left( -{\frac {549}{16}} +{\frac {1565}{64}}\,\sqrt {3} \right) {s}^{3}{r}^{2}+ \left( {\frac { 34215}{16}}-{\frac {60687}{64}}\,\sqrt {3} \right) s{r}^{4} \right) \geq 0 $ <=> $ 0.977877\,{{\it u2}}^{2}{r}^{2}+ 62.35382909\,{\it u10}\,{\it u2}\,r+ 260.5846836\,{{\it u7}}^{2}{r}^{4}+ 1086.968702\,{\it u10}\,{r}^{3}+ 224.08213\,{\it u1}\,{r}^{5}+{\it u1}\,{\it u7}\, \left( 91.862316\, {r}^{4}+ 4.875558020\,{s}^{3}r \right) +{\it u9}\,{\it u7}\, \left( 74.69215994\,{r}^{2}+ 1.732050808\,Rs+ 2.052645208\,sr \right) + 8.660254040\,{\it u7}\,{\it u3}\,Rrs+{\it u1}\,{\it u9}\, \left( 1.07818\,{r}^{2}+ 4.85640646\,sr \right) +{\it u2}\,{\it u7}\, \left( 140.2682652\,{r}^{3}+ 50.46469277\,{r}^{2}s \right) +{\it u2} \,{\it u3}\, \left( 1.040763\,{r}^{2}+ 1.732050808\,{\it u2} \right) + 473.2562586\,{\it u1}\,{\it u10}\,{r}^{2}+{\it u7}\, \left( 8.04155491\,{s}^{3}{r}^{2}+ 496.047385\,s{r}^{4} \right) \geq 0. $ as $[{\it u1}=R-2\,r,{\it u10}= \left( 4\,{R}^{2}+4\,Rr+3\,{r}^{2}-{s}^{2} \right) \left( R-r \right) -{r}^{2} \left( R-2\,r \right) ,{\it u2}= {s}^{2}-16\,Rr+5\,{r}^{2},{\it u3}=4\,{R}^{2}+4\,Rr+3\,{r}^{2}-{s}^{2} ,{\it u7}=2\,R-4\,r-s+3\,\sqrt {3}r,{\it u9}= \left( {s}^{2}-16\,Rr+5 \,{r}^{2} \right) \left( R-r \right) -{r}^{2} \left( R-2\,r \right) ] $