Properties of Triangles 1 Question 19
19. Let $A _1, A _2, \ldots, A _n$ be the vertices of an $n$-sided regular polygon such that $\frac{1}{A _1 A _2}=\frac{1}{A _1 A _3}+\frac{1}{A _1 A _4}$. Find the value of $n$.
$(1994,4$ M)
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Answer:
Correct Answer: 19. 2
的 $\angle A=75$
Solution:
- Let $O$ be the centre and $r$ be the radius of the circle passing through the vertices $A _1, A _2, \ldots, A _n$.
Then,
$$ \begin{aligned} \angle A _1 O A _2 & =\frac{2 \pi}{n} \\ O A _1 & =O A _2=r \end{aligned} $$
also
Again, by cos formula, we know that,
$$ \cos \frac{2 \pi}{n}=\frac{O A _1^{2}+O A _2^{2}-A _1 A _2^{2}}{2\left(O A _1\right)\left(O A _2\right)} $$
$$ \begin{aligned} & \Rightarrow \quad \cos \frac{2 \pi}{n}=\frac{r^{2}+r^{2}-A _1 A _2^{2}}{2(r)(r)} \\ & \Rightarrow \quad 2 r^{2} \cos \frac{2 \pi}{n}=2 r^{2}-A _1 A _2^{2} \\ & \Rightarrow \quad A _1 A _2^{2}=2 r^{2}-2 r^{2} \cos \frac{2 \pi}{n} \\ & \Rightarrow \quad A _1 A _2^{2}=2 r^{2} 1-\cos \frac{2 \pi}{n} \\ & \Rightarrow \quad A _1 A _2^{2}=2 r^{2} \cdot 2 \sin ^{2} \frac{\pi}{n} \\ & \Rightarrow \quad A _1 A _2^{2}=4 r^{2} \sin ^{2} \frac{\pi}{n} \\ & \Rightarrow \quad A _1 A _2=2 r \sin \frac{\pi}{n} \\ & \text { Similarly, } \quad A _1 A _3=2 r \sin \frac{2 \pi}{n} \\ & \text { and } \quad A _1 A _4=2 r \sin \frac{3 \pi}{n} \\ & \text { Since, } \quad \frac{1}{A _1 A _2}=\frac{1}{A _1 A _3}+\frac{1}{A _1 A _4} \\ & \text { [given] } \\ & \Rightarrow \quad \frac{1}{2 r \sin (\pi / n)}=\frac{1}{2 r \sin (2 \pi / n)}+\frac{1}{2 r \sin (3 \pi / n)} \\ & \Rightarrow \quad \frac{1}{\sin (\pi / n)}=\frac{1}{\sin (2 \pi / n)}+\frac{1}{\sin (3 \pi / n)} \\ & \Rightarrow \quad \frac{1}{\sin (\pi / n)}=\frac{\sin \frac{3 \pi}{n}+\sin \frac{2 \pi}{n}}{\sin (2 \pi / n) \sin (3 \pi / n)} \\ & \Rightarrow \quad \sin \frac{2 \pi}{n} \cdot \sin \frac{3 \pi}{n}=\sin \frac{\pi}{n} \sin \frac{3 \pi}{n} \\ & +\sin \frac{\pi}{n} \cdot \sin \frac{2 \pi}{n} \\ & \Rightarrow \sin \frac{2 \pi}{n} \sin \frac{3 \pi}{n}-\sin \frac{\pi}{n}=\sin \frac{\pi}{n} \cdot \sin \frac{3 \pi}{n} \\ & \Rightarrow \sin \frac{2 \pi}{n} \quad 2 \cos \frac{3 \pi+\pi}{2 n} \sin \frac{3 \pi-\pi}{2 n} \\ & =\sin \frac{\pi}{n} \cdot \sin \frac{3 \pi}{n} \\ & \Rightarrow 2 \sin \frac{2 \pi}{n} \cdot \cos \frac{2 \pi}{n} \cdot \sin \frac{\pi}{n}=\sin \frac{\pi}{n} \sin \frac{3 \pi}{n} \\ & \Rightarrow \quad 2 \sin \frac{2 \pi}{n} \cos \frac{2 \pi}{n}=\sin \frac{3 \pi}{n} \\ & \Rightarrow \quad \sin \frac{4 \pi}{n}=\sin \frac{3 \pi}{n} \\ & \Rightarrow \quad \frac{4 \pi}{n}=\pi-\frac{3 \pi}{n} \\ & \Rightarrow \quad \frac{7 \pi}{n}=\pi \\ & \Rightarrow \quad n=7 \end{aligned} $$
