Circle The Ratio

Consider 2 overlapping congruent circles, the area formed by the overlap area shown in red is equal to the remaining area shown in blue. If O is the centre of the circle, and A, O, B and C are colinear points

Then find $\frac{AB}{AC}$

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Let $D$ and $F$ be the points of intersection of the circle, let $E$ be the intersection of $AB$ and $DF$ , let $r$ be the radii of each circle, and let $\theta = \angle DOF$ .

Then the area of sector $ODAF$ is $A_{ODAF} = \frac{1}{2}r^2 \theta$ and the area of $\triangle ODF$ is $A_{\triangle ODF} = \frac{1}{2}r^2 \sin \theta$ , so the difference between them is $\frac{1}{2}r^2(\theta - \sin \theta)$ , which makes the area of the red region twice that much or $A_{\text{red}} = r^2(\theta - \sin \theta)$ . Since the red region and the blue region make a circle and the red region and the blue region are the same, the area of the red region is also the area of half the circle or $A_{\text{red}} = \frac{1}{2}\pi r^2$ . Therefore, $A_{\text{red}} = r^2(\theta - \sin \theta) = \frac{1}{2}\pi r^2$ or $2\theta - 2\sin \theta = \pi$ which solves numerically to $\theta \approx 2.3098815$ .

By trigonometry on $\triangle OED$ , $OE = r \cos \frac{1}{2}\theta$ , which means $AB = 2r - 2OE = 2r - 2r \cos \frac{1}{2}\theta$ . Since $AC = 2r$ , $\frac{AB}{AC} = \frac{2r - 2r \cos \frac{1}{2}\theta}{2r} = 1 - \cos \frac{1}{2}\theta$ . With $\theta \approx 2.3098815$ , this makes the ratio $\frac{AB}{AC} \approx \boxed{0.596027265}$ .

David Vreken - 2 years, 3 months ago

This would be a good problem for the geometry or calculus section

Steven Chase - 2 years, 3 months ago

It's very elegant indeed, and I'm surprised I never saw it before, so simple to formulate it is. But I'm not seeing any analytical or exact solution to it. The best I can get so far is an equation mixing angles and cosines, which I'm not seeing how to solve unless through numeric methods. I'll think more on it, later.

João Pedro Afonso - 2 years, 3 months ago

Hi sir how you r editing the fig which software

Uj Al Solution - 5 months, 3 weeks ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$