Example - 11

A circle of radius 1 is randomly placed in a 15-by-36 rectangle \(ABCD\) to that the circle lies completely within the rectangle. Given that the probability that the circle will not touch diagonal \(AC\) is \(\begin{align}\frac{m}{n}\end{align}\), where \(m\) and \(n\) are relatively prime positive integers, find \(m + n.\)

Solution: In order for the circle to lie completely within the rectangle, the center of the circle must lie in a rectangle that is (15 – 2) × (36 – 2), or 13 × 34. The requested probability is equal to the probability that the distance from the circle’s center to the diagonal \(AC\) is greater than 1, which equals the probability that the distance from a randomly selected point in the 13 × 34 rectangle to each side of triangles \(ABC\) and \(CDA\) is greater than 1. Let | AB | = 36 and | BC | = 15 (and so | AC | = 39). Draw three segments that are 1 unit away from each side of triangle \(ABC\) and whose endpoints are on the sides. Let \(E, F,\) and \(G\) be the three points of intersection nearest to \(A, B,\) and \(C,\) respectively, of the three segments. Because the corresponding sides of triangle \(ABC\) and \(EFG\) are parallel, the two triangles are similar to each other. The desired probability is equal to

\[\begin{align}\frac{{2\left[ {EFG} \right]}}{{13 \cdot 34}} = {\left( {\frac{{\left| {EF} \right|}}{{\left| {AB} \right|}}} \right)^2} \cdot \frac{{2\left[ {ABC} \right]}}{{13 \cdot 34}} = {\left( {\frac{{\left| {EF} \right|}}{{\left| {AB} \right|}}} \right)^2} \cdot \frac{{15 \cdot 36}}{{13 \cdot 34}} = {\left( {\frac{{\left| {EF} \right|}}{{\left| {AB} \right|}}} \right)^2} \cdot \frac{{270}}{{221}}.\end{align}\]

Because \(E\) is equidistant from sides \(AB\)  and \(AC, E\) lies on the bisector of \(CAB.\) Similarly, \(F\) and \(G\) lie on the bisectors of \(\begin{align}\angle ABC\;{\rm{and}}\;\angle BCA\end{align}\), respectively. Hence lines \(AE, BF,\) and \(CG\) meet \(I,\) the incenter of triangle \(ABC.\)

Let \(\begin{align}{E_1}\;{\rm{and}}\;{F_1}\end{align}\) be the feet of the perpendiculars from \(E\) and \(F\) to segment \(AB,\) respectively. Then \(\begin{align}|EF| = |{E_1}{F_1}|\end{align}\). It is not difficult to see that \(\begin{align}|B{F_1}| = |F{F_1}| = |E{E_1}| = 1\end{align}\). Set \(\begin{align}\theta  = \angle EAB\end{align}\). Then \(\begin{align}\angle CAB = 2\theta ,\;\sin 2\theta  = \frac{5}{{12}},\;\cos 2\theta  = \frac{{12}}{{13}},\;{\rm{and}}\;2\theta  = \frac{5}{{12}}\end{align}\). By either the double-angle formulas or the half-angle formulas.

\[\begin{align}\tan 2\theta  = \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\;\;{\rm{or}}\;\;\tan \,\theta  = \frac{{1 - \cos 2\theta }}{{\sin 2\theta }},\end{align}\]

and we obtain \(\begin{align}\tan \theta  = \frac{1}{5}\end{align}\). It follows that \(\begin{align}\left| {\frac{{E{E_1}}}{{A{E_1}}}} \right| = \tan \theta  = \frac{1}{5},\;{\rm{or}}\;|A{E_1}| = 5\end{align}\). Consequently, \(\begin{align}|EF|\; = \;|{E_1}\;{F_1}| = 30\end{align}\). Hence \(\begin{align}\frac{m}{n} = {\left( {\frac{{30}}{{36}}} \right)^2} \cdot \frac{{270}}{{221}} = \frac{{375}}{{442}}\;{\rm{and}}\;m + n = 817.\end{align}\)


Example - 12


\[\cos a\cos 2a\cos 3a.....\cos 999a\],

where \(\begin{align}a = \frac{{2\pi }}{{1999}}\end{align}\).

Solution: Let P denote the desired product, and let

\[Q = \sin a\sin 2a\sin 3a.....\sin 999a.\]


\[\begin{align}   {2^{999}}PQ & = (2\sin a\cos a)(2\sin 2a\cos 2a)....(2\sin 999a\cos 999a)\\   & = \sin 2a\sin 4a....\sin 1998a\\   & = (\sin 2a\sin 4a...\sin 998a)[ - \sin (2\pi  - 1000a)]\\    & \cdot \;[ - \sin (2\pi  - 1002a)]...[ - \sin (2\pi  - 1998a)]\\    & = \sin 2a\sin 4a...\sin 998a\sin 999a\sin 997a...\sin a = Q   \end{align}\]

It is easy to see that \(\begin{align}Q \ne 0\end{align}\). Hence the desired product is \(\begin{align}P = \frac{1}{{{2^{999}}}}.\end{align}\)


Example - 13

Let \(ABC\) be a triangle with \(\begin{align}\angle BAC=40{}^\text{o}\ \text{and}\ \angle ABC=60{}^\text{o}\end{align}\). Let \(D\) and \(E\) be the points lying on the sides \(AC\) and \(AB,\) respectively, such that \(\begin{align}\angle CBD=40{}^\text{o}\ \text{and}\ \angle BCE=70{}^\text{o}\end{align}\). Segments \(BD\) and \(CE\) meet at \(F.\) Show that \(\begin{align}AF\bot BC\end{align}\).


Note that \(\begin{align}\angle ABD=20{}^\text{o},\ \angle BCA=80{}^\text{o},\ \text{and}\ \angle ACE=10{}^\text{o}.\end{align}\) Let \(G\) be the foot of the altitude from \(A\) to \(BC.\) Then \(\begin{align}\angle BAG=90{}^\text{o}-\angle ABC=30{}^\text{o}\ \text{and}\ \angle CAG=90{}^\text{o}-\angle BCA=10{}^\text{o}\end{align}\). Now,

\[\begin{align}      \frac{\sin \angle BAG\sin \angle ACE\sin \angle CBD}{\sin \angle CAG\sin \angle BCE\sin \angle ABD} & =\frac{\sin 30{}^\text{o}\sin 10{}^\text{o}\sin 40{}^\text{o}}{\sin 10{}^\text{o}\sin 70{}^\text{o}\sin 20{}^\text{o}} \\     & =\frac{\frac{1}{2}(\sin 10{}^\text{o})(2\sin 20{}^\text{o}\cos 20{}^\text{o})}{\sin 10{}^\text{o}\cos 20{}^\text{o}\sin 20{}^\text{o}} \\     & =1. \\    \end{align}\]

Then by the trigonometric form of Ceva’s theorem, lines \(AB, BD\) and \(CE\) are concurrent. Therefore, \(F\) lies on segment \(AG,\) and so line \(AF\) is perpendicular to the line \(BC,\) as desired.


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