In the figure below, $MON$ is a straight line. $OP$ is a random ray with endpoint $O$. The ray

Question

In the figure below, $MON$ is a straight line. $OP$ is a random ray with endpoint $O$. The rays $OQ_1$ and $OR_1$ are the bisectors of the angle $MOP$ and angle $NOP$, respectively. Find the size of angle $Q_1OR_1$. By extending the figure in (i), the rays $OQ_2$ and $OR_2$ are the bisectors of angle $Q_1OP$ and angle $R_1OP$, respectively. Find the size of angle $Q_2OR_2$ and state a pattern or assumption you find about angle $Q_2OR_2$ and angle $Q_nOR_n$ where $n > 2$. By extending the figure in (ii), the rays $OQ_3$ and $OR_3$ are the bisectors of angle $Q_2OP$ and angle $R_2OP$, respectively. Find the size of angle $Q_3OR_3$ and test whether your assumption also holds for angle $Q_3OR_3$. Based on the pattern you find, complete the following table, where the 4th and 5th patterns are given. Pattern Number, $n$Size of angle $Q_nOR_n$ 1 2 3 4$11.25^\circ = 180^\circ \times (\frac{1}{2})^4$ 5$5.625^\circ = 180^\circ \times (\frac{1}{2})^5$ ...... Based on your previous answers, find the general term of size of $Q_nOR_n$ in terms of $n$, where $OQ_n$, $OR_n$ are the bisectors of angle $Q_{n-1}OP$ and angle $R_{n-1}OP$ respectively and $n$ is an integer that is greater than 3.

Accepted Answer

(i) $90^\circ$ (ii) $45^\circ$, with $\angle Q_nOR_n=\dfrac12\angle Q_{n-1}OR_{n-1}$ (iii) $22.5^\circ$, and the recurrence holds (iv) $90^\circ=180^\circ\left(\dfrac12\right)^1$, $45^\circ=180^\circ\left(\dfrac12\right)^2$, $22.5^\circ=180^\circ\left(\dfrac12\right)^3$ (v) $\angle Q_nOR_n=180^\circ\left(\dfrac12\right)^n$

Pattern Number, \(n\)	Size of angle \(Q_nOR_n\)
1
2
3
4	\(11.25^\circ = 180^\circ \times (\frac{1}{2})^4\)
5	\(5.625^\circ = 180^\circ \times (\frac{1}{2})^5\)
...	...

Angles

Angles

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