#6**+33 **

**Here's a puzzle for when the site activity is low!**

**In the diagram below there are two quarter circles of radius 1 and 2 intersecting as shown. What is the area of their intersection (i.e. the shaded region)? **

**\(circle_1: x^2 + y^2 = 4 \qquad r_1 = 2\\ circle_2: (x-2)^2 + (y-2)^2 = 1 \qquad r_2 = 1\\\)**

**intersections \(S_1 \text{ and } S_2\):**

\(2 x_{s}^2 - 2a \cdot x_s + a^2 - r_1^2 = 0 \qquad a = \frac{3r_1^2-r_2^2}{2r_1} = \frac{11}{4}\\ \vec{S_1} =\dbinom{ \frac18(11-\sqrt{7}) }{ \frac18(11+\sqrt{7}) }\\ \vec{S_2} =\dbinom{ \frac18(11+\sqrt{7}) }{ \frac18(11-\sqrt{7}) }\)

**sectors angels (cosinus-rule) \(\alpha_1 \text{ and } \alpha_2\):**

\( \cos{(\alpha_1)} = \frac{57}{64} \qquad \alpha_1 = 27.0481105464^{\circ}\\ \cos{(\alpha_2)} = \frac{9}{16} \qquad \alpha_2 = 55.7711336722^{\circ}\\\)

**Areas sectors \(A_{s_1} \text{ and } A_{s_2}\):**

\( A_{s_1} = 4\pi \frac{27.0481105464^{\circ}}{360^{\circ}}\\ A_{s_2} = \pi \frac{55.7711336722^{\circ}}{360^{\circ}}\\\)

**Areas triangles \(A_{t_1} \text{ and } A_{t_2}\):**

\( A_{t_1} = \frac{1}{2} \cdot \left| \vec{S_2}\times \vec{S_1} \right| = \frac{11}{32}\sqrt{7}\\ A_{t_2} = \frac{1}{2} \cdot \left| \left[\vec{S_1}-\binom{2}{2}\right] \times \left[\vec{S_2}-\binom{2}{2}\right] \right|= \frac{5}{32}\sqrt{7} \)

\( \text{The area of their intersection }\quad A = A_{s_1}-A_{t_1}+A_{s_2}-A_{t_2}\)

\(A=\frac{\pi}{360}(4\cdot 27.0481105464^{\circ}+55.7711336722^{\circ})-\frac{ \sqrt{7} }{2}\\ A= 1.43085212603-1.32287565553\\ \mathbf{A=0.10797647050}\)

.heureka Oct 28, 2015

#2**+7 **

I could have solved this with some messy Algebra, but....I decided to cheat....LOL!!!

Placing the center of the larger circle at the origin, we have these two equations :

x^2 + y^2 = 4 and (x - 2)^2 + (y - 2)^2 = 1

Here's a diagram :

The area of triangle ABC = 2*sin(27.048) = about .909 units^2

And the area of the sector ABC = about 944 units^2

Similarly, the area of triangle DAB = (1/2)sin(55.771) = about .413 units^2

And the area of sector DAB = about .4867 units^2

So...the approximate total area of the area of intersection = [ .944 - .909] + [ .4867 - .413] = about .1087 units^2

OOPS....you are correct Alan....I have provided an edit......duh!!!....I think I mis-placed my decimal point.....Is my answer now correct???

CPhill Oct 28, 2015

#4**0 **

Yes I know that Alan - That is why I wrote that it was not correct.

I suppose i should jut have deleted it. I will delete it now.

I am working on a solution that uses co-ordinate geometry. I am sure I can do it that way but it is quite painful especially if I want to keep the answer exact. ://

There you go Chris, Yours is wrong too. LOL

**But your diagram is fantastic.**

I am only joking Chris, You answer is great too :))

Thanks for this distraction Alan. I have enjoyed it.

Melody Oct 28, 2015

#5**0 **

I find it fascinating that the question is much more complicated than it appears to be at first sight.

Alan Oct 28, 2015

#6**+33 **

Best Answer

**Here's a puzzle for when the site activity is low!**

**In the diagram below there are two quarter circles of radius 1 and 2 intersecting as shown. What is the area of their intersection (i.e. the shaded region)? **

**\(circle_1: x^2 + y^2 = 4 \qquad r_1 = 2\\ circle_2: (x-2)^2 + (y-2)^2 = 1 \qquad r_2 = 1\\\)**

**intersections \(S_1 \text{ and } S_2\):**

\(2 x_{s}^2 - 2a \cdot x_s + a^2 - r_1^2 = 0 \qquad a = \frac{3r_1^2-r_2^2}{2r_1} = \frac{11}{4}\\ \vec{S_1} =\dbinom{ \frac18(11-\sqrt{7}) }{ \frac18(11+\sqrt{7}) }\\ \vec{S_2} =\dbinom{ \frac18(11+\sqrt{7}) }{ \frac18(11-\sqrt{7}) }\)

**sectors angels (cosinus-rule) \(\alpha_1 \text{ and } \alpha_2\):**

\( \cos{(\alpha_1)} = \frac{57}{64} \qquad \alpha_1 = 27.0481105464^{\circ}\\ \cos{(\alpha_2)} = \frac{9}{16} \qquad \alpha_2 = 55.7711336722^{\circ}\\\)

**Areas sectors \(A_{s_1} \text{ and } A_{s_2}\):**

\( A_{s_1} = 4\pi \frac{27.0481105464^{\circ}}{360^{\circ}}\\ A_{s_2} = \pi \frac{55.7711336722^{\circ}}{360^{\circ}}\\\)

**Areas triangles \(A_{t_1} \text{ and } A_{t_2}\):**

\( A_{t_1} = \frac{1}{2} \cdot \left| \vec{S_2}\times \vec{S_1} \right| = \frac{11}{32}\sqrt{7}\\ A_{t_2} = \frac{1}{2} \cdot \left| \left[\vec{S_1}-\binom{2}{2}\right] \times \left[\vec{S_2}-\binom{2}{2}\right] \right|= \frac{5}{32}\sqrt{7} \)

\( \text{The area of their intersection }\quad A = A_{s_1}-A_{t_1}+A_{s_2}-A_{t_2}\)

\(A=\frac{\pi}{360}(4\cdot 27.0481105464^{\circ}+55.7711336722^{\circ})-\frac{ \sqrt{7} }{2}\\ A= 1.43085212603-1.32287565553\\ \mathbf{A=0.10797647050}\)

heureka Oct 28, 2015

#7**0 **

I just had my decimal point in the wrong place....Alan said mine is [fairly] correct....

Soooooo.....

A BIG GOLD STAR FOR HEUREKA AND ME....!!!!

And for Melody????....A BAG OF SWITCHES....!!!!

CPhill Oct 28, 2015

#8**0 **

Area of the square : 2 x 2 = 4

Area of Big arc = 1/4 pi 2^2 = pi

Area of small arc = 1/4 pi 1^2 = 1/4 pi

Area OUTSIDE of Small arc = 4 - 1/4 pi

Area OUTSIDE of Big arc = 4 - pi

Add the two outside areas and subtract the square area

4-1/4pi + 4 - pi -4 = 4 - 5/4 pi = ,073 Square units

Guest Oct 28, 2015