Tangent with Arbelos\Large{\purple{Tangent\space with\space Arbelos}}

Here is an Arbelos with a tangent drawn to it. I didn't uploaded this because it was larger than 10MB10MB.

Now what is the Area and what is the Maximum possible Area of the shaded region if the Diameter of the bigger semicircle is fixed?\boxed{\red{Now\space what\space is\space the\space Area\space and\space what\space is\space the\space Maximum\space possible\space Area\space of\space the\space shaded\space region\space if\space the\space Diameter\space of\space the\space bigger\space semicircle\space is\space fixed?}}

I found the following result :

If RR is the radius of the bigger semi-circle and r1,r2r_1,r_2 are the radii of the smaller semi-circles then, Area of the shaded region=R22cos⁡−1(1−8CR4)−2CR4−4CR2\purple{Area\space of\space the\space shaded\space region}=\dfrac{R^{2}}{2}\cos^{-1}\left(1-\dfrac{8C}{R^{4}}\right)-\dfrac{2\sqrt{C}\sqrt{R^{4}-4C}}{R^{2}} where C=(R2−r1r2)r1r2where\space C=\left(R^{2}-r_1r_2\right)r_1r_2

#Geometry

Note by Zakir Husain
1 month, 3 weeks ago

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Comments

Neat problem. How did you arrive at the area of the shaded region? Area of the sector - triangle = Segment?

Mahdi Raza - 1 month, 3 weeks ago

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The whole Process:

I took the whole case in a coordinate plane with the center of the large semi-circle as the origin and it's diameter as x-axis.

Then the equation of the tangent became : y=x(r2−r1)+r12+t222r1r2y=\dfrac{x(r_2-r_1)+r_1^2+t_2^2}{2\sqrt{r_1r_2}}

Then I found the length of PQ‾=4R2−r1r2r1r2R\overline{PQ}=\dfrac{4\sqrt{R^{2}-r_1r_2}\sqrt{r_1r_2}}{R}

It is then easier to find the area of the shaded region as you now know the length of the chord...

Zakir Husain - 1 month, 3 weeks ago

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Oh wow! That’s well made

Mahdi Raza - 1 month, 3 weeks ago

Using just "elementary" geometry: Let r1{{r}_{1}} be the radius of the left semicircle and r2{{r}_{2}} the radius of the right one. WLOG we take r1≥r2{{r}_{1}}\ge {{r}_{2}}. Denote by d=ODd=OD the apothem of the cord FEFE. Denote by KK, OO and LL the centers of the three semicircles as seen in the figure. Let GG, HH be the points of tangency of the cord with the semicircles. Let LNLN and OPOP be perpendicular to KGKG. Then we have,

R=r1+r2⇒KO=AO−AK=R−r1=r2\begin{aligned} & R={{r}_{1}}+{{r}_{2}} \\ & \Rightarrow KO=AO-AK=R-{{r}_{1}}={{r}_{2}} \\ \end{aligned} Since right triangles PKOPKO and NKLNKL are similar, PKKO=NKKL⇒r1−dr2=r1−r2r1+r2⇒r1−dr2=r1−r2R\frac{PK}{KO}=\frac{NK}{KL}\Rightarrow \frac{{{r}_{1}}-d}{{{r}_{2}}}=\frac{{{r}_{1}}-{{r}_{2}}}{{{r}_{1}}+{{r}_{2}}}\Rightarrow \frac{{{r}_{1}}-d}{{{r}_{2}}}=\frac{{{r}_{1}}-{{r}_{2}}}{R} which solves to

d=1R(2r12−2Rr1+R2)d=\dfrac{1}{R}\left( 2{{r}_{1}}^{2}-2R{{r}_{1}}+{{R}^{2}} \right) Now, the area of the circular segment gets maximised when its cord is maximised, which in turn occurs when the apothem dd is minimised. The latter happens when r1{{r}_{1}} equals the xx-coordinate of the vertex of the parabola y=1R(2x2−2Rx+R2)y=\dfrac{1}{R}\left( 2{{x}^{2}}-2Rx+{{R}^{2}} \right), i.e. when r1=R2{{r}_{1}}=\dfrac{R}{2}.

In this case, d=R2d=\dfrac{R}{2} which is the apothem of the inscribed equilateral triangle, thus ∠FOE=2π3\angle FOE=\dfrac{2\pi }{3} and the maximal area of the circular segment is

A=122π3R2−12R2sin⁡2π3=(π3−34)R2A=\dfrac{1}{2}\dfrac{2\pi }{3}{{R}^{2}}-\dfrac{1}{2}{{R}^{2}}\sin \dfrac{2\pi }{3}=\boxed{\left( \dfrac{\pi }{3}-\dfrac{\sqrt{3}}{4} \right){{R}^{2}}}

    \ \ \

In general, the area of the circular segment can be expressed as

A=R22cos⁡−1(2d2R2−1)−dR2−d2A=\dfrac{{{R}^{2}}}{2}{{\cos }^{-1}}\left( 2\dfrac{{{d}^{2}}}{{{R}^{2}}}-1 \right)-d\sqrt{{{R}^{2}}-{{d}^{2}}} where d=1R(2r12−2Rr1+R2)=1R(r12+r22)d=\frac{1}{R}\left( 2{{r}_{1}}^{2}-2R{{r}_{1}}+{{R}^{2}} \right)=\frac{1}{R}\left( {{r}_{1}}^{2}+{{r}_{2}}^{2} \right)

Thanos Petropoulos - 1 month, 3 weeks ago

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Very neat solution, I didn't even thought of Apothem!

Zakir Husain - 1 month, 3 weeks ago

I found a mistake in my previous formula so fixed it, now it's absolutely correct

Zakir Husain - 1 month, 3 weeks ago

We can substitute r1=x,r2=R−xr_1=x,r_2=R-x

Then, Area=R22cos⁡−1(1−8CR4)−2CR4−4CR2=R22cos⁡−1(2d2R2−1)−dR2−d2Area=\dfrac{R^{2}}{2}\cos^{-1}\left(1-\dfrac{8C}{R^{4}}\right)-\dfrac{2\sqrt{C}\sqrt{R^{4}-4C}}{R^{2}}=\dfrac{R^{2}}{2}\cos^{-1}\left(\dfrac{2d^{2}}{R^{2}}-1\right)-d\sqrt{R^{2}-d^{2}} where C=(R2−x(R−x))x(R−x),d=2x2−2Rx+R2Rwhere\space C=\left(R^{2}-x\left(R-x\right)\right)x\left(R-x\right),d=\dfrac{2x^{2}-2Rx+R^{2}}{R} The Graph of the equation looks like this

The maxima of the graph is x=R2x=\dfrac{R}{2} (solved by @Thanos Petropoulos ) ⇒A=R2(π3−34)\Rightarrow \boxed{A=R^{2}\left(\dfrac{\pi}{3}-\frac{\sqrt{3}}{4}\right)}

Zakir Husain - 1 month, 3 weeks ago
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