0

enter image description here

I got this question from mind your decision channel in youtube , and there the tutor has provided an elegant solution with euclidean geometry.

I am trying to work it out with coordinate geometry .
The figure is fairly symmetric and we need ratio of radius.

Taking the vertex of quadrat as origin, we write equation of bigger circle as $$x^2+y^2=R^2$$

Taking the center of the smaller semicircle as (a,a), the equation of the circle touching both the axes is $$x^2+y^2-2ax-2ay+a^2=0. $$

Now ,distance between center should be equal to $a\sqrt2=R-a-$sagitta of the larger circle
or $$a\sqrt2=R-a-((R-\sqrt{R^2-a^2})$$ but this is not fetching the correct result. can you point out my mistake ? Thanks.

Quanto
  • 97,352
George carlin
  • 558
  • Also, the distance between centres cannot be equal to $R-a$. This should be obvious from the diagram, and intuitively since you would be measuring the distance from the small circle's centre to the large circle's edge, which is not necessarily (indeed, shouldn't expect to be) half the large circle's radius. – Nij Sep 22 '19 at 09:58

2 Answers2

2

I don't see why $a\sqrt{2}$, the distance from the quarter-circle's right-angle to the red semicircle's base's centre, should be equal to $R-a$, i.e. the quarter-circle with the semicircle's radius subtracted. That's equivalent to claiming the larger circle's radius joining the two centres has an excess length $a$ beyond the semicircle's base, which doesn't look true.

What you should do instead is note that the shapes' equations imply $x+y=\frac{a^2+R^2}{2a}$ is the equation of the semicircle's base, and the endpoints satisfy $xy=\frac{(x+y)^2-x^2-y^2}{2}=\frac{(a^2-R^2)^2}{8a^2}$. Thus $x,\,y$ are, in some order, the roots of $t^2-\frac{a^2+R^2}{2a}t+\frac{(a^2-R^2)^2}{8a^2}=0$ (changing the order reflects one endpoint in $y=x$ to give the other). These roots are $$t_\pm:=\frac{a^2+R^2\pm\sqrt{6a^2R^2-a^4-R^4}}{4a}.$$The squared distance between $(t_+,\,t_-)$ and $(t_-,\,t_+)$ is $$4a^2=2(t_+-t_-)^2=\frac{6a^2R^2-a^4-R^4}{2a^2}.$$This rearranges to $0=(3a^2-R^2)^2$, i.e. $a=R/\sqrt{3}$.

Now we've proven that, let's answer the original question: $$\frac{\frac{\pi}{2}\left(\frac{R}{\sqrt{3}}\right)^2}{\frac{\pi}{4}R^2}=\frac{2}{3}.$$

J.G.
  • 115,835
  • how do you get the equation of the base of semicircle $a^2$+$r^2$/2a part ? – George carlin Sep 22 '19 at 10:17
  • 1
    @Georgecarlin That equation is deducible from the circles' equations, so is true at each endpoint of the base. Therefore, it's true everywhere on that base too, because it's the equation of a line. – J.G. Sep 22 '19 at 10:23
  • i have modified my question. can you point out the mistake in my approach ? – George carlin Sep 22 '19 at 11:34
  • the base of semicircle is basically equation of common chord obtained by subtracting the two circles, i got that part, and other equation is elimination of circle and common chord . thanks for your cool answer ! – George carlin Sep 22 '19 at 11:54
2

Observe that the triangle formed by the centers of the two circles and one of the diameter end points of the red circle is a right triangle. Thus,

$$R^2 = a^2+(\sqrt{2}a)^2= 3a^2$$

There is an error in your setup. The correct relationship should be $\sqrt{2} a = R - \text{sagitta}$.

Quanto
  • 97,352