Engineering Geometry #2

Suppose that $O_{1}X\perp AB$ ,

Since $AO_{1}=O_{1}B$

then $AX=XB=20$

Also, $AO_{1}X=XO_{1}B$

Then, let $\angle AO_{1}X=\angle XO_{1}B=\theta$

$sin\theta=\cfrac{20}{25}=\cfrac{4}{5}$

$cos2\theta=1-2(sin\theta)^{2}=1-2(\cfrac{4}{5})^{2}=\cfrac{-7}{25}<0$

$90^{\circ}<2\theta<180^{\circ}$

$2\theta=180^{\circ}-\cos^{-1}\cfrac{7}{25}$

Let $\angle AO_{1}R=\alpha$

$\sin\alpha=\cfrac{7}{25}$

$0^{\circ}<\alpha<90^{\circ}$

$\alpha=\sin^{-1}\cfrac{7}{25}$

---

Let $\angle BO_{1}Q=\beta$

$\alpha+2\theta+\beta=180^{\circ}$

$\sin^{-1}\cfrac{7}{25}+180^{\circ}-\cos^{-1}\cfrac{7}{25}+\beta=180^{\circ}$

$\beta=\cos^{-1}\cfrac{7}{25}-\sin^{-1}\cfrac{7}{25}$

$\beta=(\cos^{-1}\cfrac{7}{25}+\sin^{-1}\cfrac{7}{25})-2\sin^{-1}\cfrac{7}{25}$

$\beta=90^{\circ}-2\sin^{-1}\cfrac{7}{25}$

$\beta=90^{\circ}-2\alpha$

$\cos\beta=\cos(90^{\circ}-2\alpha)$

$\cos\beta=\sin2\alpha$

$\cos\beta=2\sin\alpha\cos\alpha$

$\cos\beta=2(\cfrac{7}{25})(\cfrac{24}{25})$

$\cos\beta=\cfrac{336}{625}$

---

$\cos\beta=\cfrac{O_{1}Q}{25}$

$\cfrac{O_{1}Q}{25}=\cfrac{336}{625}$

$O_{1}Q=\cfrac{336}{25}$

$O_{1}O_{2}=2\times\cfrac{336}{25}=\cfrac{672}{25}$

So, $a+b=672+25=697$

The mirror line through B is the perpendicular bisector of O1O2. So that O1O2 = 2 PQ. PQ = 25 sin O1BQ, < O1BQ = <PO1B, <PO1B= <AO1B – cos -1(7/25) Angle AO1B can be calculated by cos rule, cos AO1B = (〖25〗^2+〖25〗^2-〖40〗^2)/2(25)(25) =-7/25. So < O1BQ = cos -1(- 7/25) - cos -1(7/25) = 32.52. O1O2 = 2 (25) sin32.52 = 26.88 = 672/25. So a+b=697