Example - 14

Let \(S\) be an interior point of triangle \(ABC.\) Show that at least one of \(\begin{align}\angle SAB,\ \angle SBC,\ \text{and}\ \angle SCA\end{align}\) is less than or equal to 30º.

Solution: The given conditions in the problem motivate us to consider the Brocard point \(P\) of triangle \(\begin{align}ABC\ \text{with}\ \alpha =\angle PAB=\angle PBC=\angle PCA\end{align}\). Because S (see figure) lies inside or on the boundary of at least one of the triangles \(PAB, PBC,\) and \(PCA,\) at least one of \(\begin{align}\angle SAB,\ \angle SBC,\ \text{and}\ \angle SCA\end{align}\) is less than or equal to \(\alpha \). It suffices to show that \(\begin{align}\alpha \le 30{}^\text{o}\end{align}\); that is, \(\begin{align}\sin \alpha \le \frac{1}{2}\ \text{or}\ {{\csc }^{2}}\alpha \ge 4\end{align}\), by considering the range of \(\alpha \).

We have shown that

\[{{\csc }^{2}}\alpha ={{\csc }^{2}}A+{{\csc }^{2}}B+{{\csc }^{2}}C\]

and using the Cauchy-Schwarz inequality, we have

\[\begin{align}\frac{9}{4}{\csc ^2}\alpha  \ge \left( {{{\sin }^2}A + {{\sin }^2}B + {{\sin }^2}C} \right)\left( {{{\csc }^2}A + {{\csc }^2}B + {{\csc }^2}C} \right) \ge 9,\end{align}\]

implying that \(\begin{align}{\csc ^2}\alpha  \ge 4\end{align}\), as desired.

 

Example - 15

Let \(x, y, z\) be positive real numbers.

(a) Prove that

\(\begin{align}    \frac{x}{{\sqrt {1 + {x^2}} }} + \frac{y}{{\sqrt {1 + {y^2}} }} + \frac{z}{{\sqrt {1 + {z^2}} }} \le \frac{{3\sqrt 3 }}{2}\\    {\text{if}}\;x + y + z = xyz; \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad    \end{align}\)

(b) Prove that

\(\begin{align}   \frac{x}{{1 - {x^2}}} + \frac{y}{{1 - {y^2}}} + \frac{z}{{1 - {z^2}}} \ge \frac{{3\sqrt 3 }}{2}\\
{\text{if}}\;0 < x,\;y,\;z < 1\;{\rm{and}}\;xy + yz + zx = 1. \qquad \qquad \qquad \qquad \qquad    \end{align}\)

Solution: Both problems can be solved by trigonometric substitutions.

(a) Let there be there is an acute triangle \(ABC\) with

\(\tan A = x,\;\tan B = y\), and \(\tan C = z\). Note that

\[\begin{align}\frac{{\tan A}}{{\sqrt {1 + {{\tan }^2}A} }} = \frac{{\tan A}}{{\sec A}} = \sin A.\end{align}\]

The desired inequality becomes

\[\begin{align}\sin A + \sin B + \sin C \le \frac{{3\sqrt 3 }}{2}\end{align}\]

which we’ve covered earlier.

(b) From the given condition, we can assume that there is an acute triangle \(ABC\) such that

\[\begin{align}\tan \frac{A}{2} = x, \qquad \tan \frac{B}{2} = y, \qquad \tan \frac{C}{2} = z.\end{align}\]

By the double-angle formulas, it suffices to prove that

\[\begin{align}\tan A + \tan B + \tan C \ge 3\sqrt 3 \end{align}\]

which is another inequality we’ve encountered already.

Example - 16

Compute the sums

\[\left( \begin{matrix}     n  \\      1  \\    \end{matrix} \right)\sin a+\left( \begin{matrix}      n  \\       2  \\     \end{matrix} \right)\sin 2a+....+\left( \begin{matrix}        n  \\        n  \\    \end{matrix} \right)\sin na\]

and

\[\left( \begin{matrix}      n  \\       1  \\     \end{matrix} \right)\cos a+\left( \begin{matrix}       n  \\       2  \\   \end{matrix} \right)\cos 2a+....+\left( \begin{matrix}       n  \\       n  \\     \end{matrix} \right)\cos na\]

Solution: Let Sn and Tn denote the first and second sums, respectively. Set the complex number \(z=\cos a+i\sin a\). Then, by de Moivre’s formula, we have \(\begin{align}{{z}^{n}}=\cos na+i\sin na\end{align}\). By the binomial theorem, we obtain.

\[\begin{align}      1+{{T}_{n}}+i\,{{S}_{n}}& =1+\left( \begin{matrix}      n  \\       1  \\    \end{matrix} \right)(\cos a+i\ \sin a)+\left( \begin{matrix}      n  \\       2  \\    \end{matrix} \right)(\cos 2a+i\ \sin 2a) \\     & \quad \;\;\;\; +...+\left( \begin{matrix}       n  \\       n  \\    \end{matrix} \right)(\cos na+i\ \sin na) \\     & =\left( \begin{matrix}        n  \\       0  \\     \end{matrix} \right){{z}^{0}}+\left( \begin{matrix}        n  \\      1  \\    \end{matrix} \right)z+\left( \begin{matrix}        n  \\       2  \\     \end{matrix} \right)    {{z}^{2}}+....+\left( \begin{matrix}        n  \\       n  \\     \end{matrix} \right){{z}^{n}} \\      & ={{(1+z)}^{n}} \\     \end{align}\]

Because

\[\begin{align}   1 + z = 1 + \cos a + i\;\sin a & = 2{\cos ^2}\frac{a}{2} + 2i\sin \frac{a}{2}\cos \frac{a}{2}\\
& = 2\cos \frac{a}{2}\left( {\cos \frac{a}{2} + i\;\sin \frac{a}{2}} \right),    \end{align}\]

if follows that

\[\begin{align}{(1 + z)^n} = {2^n}\cos \frac{a}{2}\left( {\cos \frac{{na}}{2} + i\;\sin \frac{{na}}{2}} \right),\end{align}\]

again by de Moivre’s formula. Therefore,

\[\begin{align}(1 + {T_n}) + i\,{S_n} = \left( {{2^n}{{\cos }^n}\frac{a}{2}\cos \frac{{na}}{2}} \right) + i\,\left( {{2^n}{{\cos }^n}\frac{a}{2}\sin \frac{{na}}{2}} \right),\end{align}\]

and so

\[\begin{align}{S_n} = {2^n}{\cos ^n}\frac{a}{2}\sin \frac{{na}}{2}\;{\rm{and}}\;{T_n} =  - 1 + {2^n}{\cos ^n}\frac{a}{2}\cos \frac{{na}}{2}.\end{align}\]

 

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