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Example - 17

Let \(ABC\) be a triangle such that

\[\begin{align}{\left( {\cot \frac{A}{2}} \right)^2} + {\left( {2\cot \frac{B}{2}} \right)^2} + {\left( {3\cot \frac{C}{2}} \right)^2} = {\left( {\frac{{6s}}{{7r}}} \right)^2}\end{align}\]

where \(s\) and \(r\) denote its semiperimeter and its inradius, respectively. Prove that triangle \(ABC\) is similar to a triangle \(T\) whose side lengths are all positive integers with no common divisor and determine these integers.

Soluton: Define

\[\begin{align}u = \cot \frac{A}{2},\; & v = \cot \frac{B}{2},\; & w = \cot \frac{C}{2}.\end{align}\]

As shown in Figure, denote the incenter of triangle \(ABC\) by \(I,\) and let \(D, E,\) and \(F\) be the points of tangency of the incircle with sides \(BC, CA, \)and \(AB,\) respectively. Then \(\left| EI \right|\text{ }=r\), and by the standard formula, \(\left| AE \right|\text{ }=s-a.\)

We have

\[\begin{align}u = \cot \frac{A}{2} = \frac{{|AE|}}{{|EI|}} = \frac{{s - a}}{r},\end{align}\]

and similarly \(\begin{align}v = \frac{{s - b}}{r},\;w = \frac{{s - c}}{r}\end{align}\). Because

\[\begin{align}\frac{s}{r} = \frac{{(s - a) + (s - b) + (s - c)}}{r} = u + v + w,\end{align}\]

we can rewrite the given relation as

\[\begin{align}49\left[ {{u^2} + 4{v^2} + 9{w^2}} \right] = 36{(u + v + w)^2}\end{align}\]

Expanding the last equality and canceling like terms, we obtain

\[\begin{align}13{u^2} + 160{v^2} + 405{w^2} - 72(uv + vw + wv) = 0,\end{align}\]

or

\[\begin{align}{(3u - 12\upsilon )^2} + {(4\upsilon  - 9\omega )^2} + {(18\omega  - 2u)^2} = 0\end{align}\]

Therefore, \(\begin{align}u:v:w = 1:\frac{1}{4}:\frac{1}{9}\end{align}\). This can also be realized by recognizing that the given relation corresponds to equality in Cauchy-Schwarz inequality.

\[\begin{align}({6^2} + {3^2} + {2^2})\left[ {{u^2} + {{(2v)}^2} + {{(3w)}^2}} \right] \ge {(6 \cdot u + 3 \cdot 2v + 2 \cdot 3w)^2}\end{align}\]

After multiplying by \(r,\) we see that

\[\begin{align}    \frac{{s - a}}{{36}} & = \frac{{s - b}}{9} = \frac{{s - c}}{4} = \frac{{2s - b - c}}{{9 + 4}} = \frac{{2s - c - a}}{{4 + 36}} = \frac{{2s - a - b}}{{36 + 9}}\\    & = \frac{a}{{13}} = \frac{b}{{40}} = \frac{c}{{45}};    \end{align}\]

that is, triangle \(ABC\) is similar to a triangle with side lengths 13, 40, 45.

 

Example - 18

Prove that the average of the numbers

\[\begin{align}2\sin 2{}^\text{o},\ \ 4\sin 4{}^\text{o},\ \ 6\sin 6{}^\text{o},\ ...,\ \ 180\sin 180{}^\text{o}\end{align}\]

is \(\begin{align}\cot {{1}^{{}^\circ }}\end{align}\).

Solution: We need to prove that

\[\begin{align}2\sin 2{}^\text{o}+4\sin 4{}^\text{o}+.....+178\sin 178{}^\text{o}=90\cot 1{}^\text{o},\end{align}\]

which is equivalent to

\[\begin{align}2\sin 2{}^\text{o}\cdot \sin 1{}^\text{o}+2(2\sin 4{}^\text{o}\cdot \sin 1{}^\text{o})+....+89(2\sin 178{}^\text{o}\cdot \sin 1{}^\text{o})=90\cos 1{}^\text{o}\end{align}\]

Note that

\[\begin{align}2\sin 2k{}^\text{o}\sin 1{}^\text{o}=\cos (2k-1){}^\text{o}-\cos (2k+1){}^\text{o}.\end{align}\]

We have

\[\begin{align} 2\sin 2{}^\text{o}\cdot &\sin 1{}^\text{o}+2(2\sin 4{}^\text{o}\cdot \sin 1{}^\text{o})+.....+89(2\sin 178{}^\text{o}\cdot \sin 1{}^\text{o}) \\    & =(\cos 1{}^\text{o}-\cos 3{}^\text{o})+2(\cos 3{}^\text{o}-\cos 5{}^\text{o})+.....+89(\cos 177{}^\text{o}-\cos 179{}^\text{o}) \\    & =\cos 1{}^\text{o}+\cos 3{}^\text{o}+....+\cos 177{}^\text{o}-89\cos 179{}^\text{o} \\    & =\cos 1{}^\text{o}+(\cos 3{}^\text{o}+\cos 177{}^\text{o})+.....+(\cos 89{}^\text{o}+\cos 91{}^\text{o})+\ 89\cos 1{}^\text{o} \\  & =\cos 1{}^\text{o}+89\cos 1{}^\text{o}=90\cos 1{}^\text{o}, \end{align}\] as desired.

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