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Let $C$ be the smallest acute angle such that $\tan{4C}=4$. Then the above figure is constructed bu following the points

  1. Now let $OA$, $OB$ be two perpendicular radii of a circle. Make $OI$ one-fourth of $OB$ and the angle $OIE$ (with $E$ in $OA$) one-fourth of the angle $OIA$.
  2. Find on $AO$ produced a point $F$ such that $EIF$ = $\frac{\pi}{4}$.
  3. Let the circle on $AF$ as diameter tut $OB$ in $K$, and let the circle whose centre is $E$ and radius $EK$ cut $OA$ in $N_3$ and $N_5$ ($N_3$ on $OA$, $N_5$ on $AO$ produced).
  4. Draw $N_3P_3$, $N_5P_5$ perpendicular to $OA$ to tut the circumference of the original circle in $P_3$ and $P_5$.

Hardy and Wright describe the above construction and then state

$$2\cos{AOP_3}.2\cos{AOP_5} = -4 \frac{ON_3 - ON_5}{OA^2} ...(1) $$ $$ = -4 \frac{OK^2}{OA^2} ...(2) $$ $$ = -4\frac{OF}{OA} ...(3) $$

My doubt is how do we get $(3)$ from $(2)$ ?

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2 Answers 2

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We have FKA is a right triangle. Then by Geometric mean theorem \begin{equation} OK^2 = OF\times OA. \end{equation}

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As $FA$ is a diameter of circle $\text{circ}(FKA)$ and $N_3N_5$ is a diameter of circle $\text{circ}(N_3KN_5)$. Therefore $$OK^2=OF \cdot OA = ON_3 \cdot ON_5$$ because both triangles $FKA$ and $N_3KN_5$ are right angled triangles as triangles inscribed in circles $\text{circ}(FKA)$ and $\text{circ}(N_3KN_5)$ respectively and their edges $FA$ and $N_3N_5$ are diameters of their respective circumcircles $\text{circ}(FKA)$ and $\text{circ}(N_3KN_5)$. Then the proper chain of equalities is (you have a typo or something in your post): \begin{align}2\cos{(\angle AOP_3)}\cdot 2\cos{(\angle AOP_5)} &= -4 \frac{ON_3}{OA} \cdot \frac{ON_5}{OA} = -4 \frac{ON_3 \cdot ON_5}{OA^2} \\ &=-4\frac{OK^2}{OA^2}\\ &= -4\frac{OF \cdot OA}{OA^2}\\ &= -4\frac{OF}{OA} \end{align}

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