## Excel in math and science

### Master concepts by solving fun, challenging problems.

## It's hard to learn from lectures and videos

### Learn more effectively through short, conceptual quizzes.

## Our wiki is made for math and science

###
Master advanced concepts through explanations,

examples, and problems from the community.

## Used and loved by 4 million people

###
Learn from a vibrant community of students and enthusiasts,

including olympiad champions, researchers, and professionals.

## Comments

Sort by:

TopNewestLet side length be 1.

Area of an overlap = \(\frac{\pi}{2}\) - 1.

Area wanted = \(\frac{\pi}{4}\) - \((\frac{\pi}{2} - 1) + \) 2 \(\Delta\) = 1 - \(\frac{\pi}{4}\) + 2 \(\Delta\).

We are in a challenge of finding 2 \(\Delta\).

Perhaps we ought to allocate coordinates to find critical interception points via geometry to solve.

\(x^2\) + \(y^2\) = \(\frac14\) intercepts with \((x - \frac12)^2\) + \((y + \frac12)^2\) = 1 for coordinates of positive x and positive y.

(x, y) of (\(\frac{\sqrt{7}-1}{8}\), \(\frac{\sqrt{7}+1}{8}\)) is a critical interception point. 4 of them are just a matter of plus and minus, however I think we should find for \(\frac12 \Delta\) at first. (\(\frac{\sqrt{2}}{4}\), \(\frac{\sqrt{2}}{4}\)) is another critical point. Area of new triangle found by joining three points which includes (\(\frac12\),\(\frac12\)), two segments from two different curves are to be considered for their areas.

I think the problem is solved provided areas of the two different segments centered at (0, 0) to minus and (\(\frac12\),\(-\frac12\)) to plus respectively are calculated. Angles formed for each sector is critical but they can always been found with all known coordinates.

\(\theta_m\) = \(tan^{-1}\) \(\frac{m_1 - m_2}{1 + m_1 m_2}\) = \(tan^{-1}\) \(\frac{\sqrt{7}}{7}\) \(\Rightarrow sin \theta_m = \frac{\sqrt2}{4}\).

\(\delta_m = \frac12(\frac12)^2 (\theta_m-sin\theta_m)\) = \(\frac18(tan^{-1}\frac{\sqrt7}{7}-\frac{\sqrt2}{4})\) = (9.7671666417925542359581190853027+)e-4

\(tan (\frac{\pi}{2} - \theta_p)\) = \(\displaystyle |\frac{\frac{\sqrt7+1}{8}+\frac12}{\frac{\sqrt7-1}{8}-\frac12}|\) = \(|-\frac{16+5\sqrt7}{9}|\) and \(\theta_p = \frac{\pi}{2} - tan^{-1}|\frac{16+5\sqrt7}{9}|\Rightarrow cos\theta_p = \frac{5+\sqrt7}{8}and \) \(sin\theta_p = \frac{5-\sqrt7}{8}\)

\(\delta_p = \frac12(1)^2 (\theta_p-sin\theta_p)\) = \(\frac12(\frac{\pi}{2} - tan^{-1}\frac{16+5\sqrt7}{9}-\frac{5-\sqrt7}{8})\) = 0.0022110611028744654823035761735159+

\(\delta_t = \frac{1}{16}(1-\frac{\sqrt2}{2})\) = 0.018305826175840779724947227368447+, by joining (\(\frac12\),\(\frac12\)), (\(\frac{\sqrt{7}-1}{8}\), \(\frac{\sqrt{7}+1}{8}\)) and (\(\frac{\sqrt{2}}{4}\), \(\frac{\sqrt{2}}{4}\)).

\(\delta_t\) + \(\delta_p\) - \(\delta_m\) = \(\frac12\Delta\)

2 \(\Delta\) = 4 (\(\delta_t\) + \(\delta_p\) - \(\delta_m\))

Area wanted = 1 - \(\frac{\pi}{4}\) + 4 (\(\delta_t\) + \(\delta_p\) - \(\delta_m\)) = 0.292762519060696+.

If side length of square is S, then the area is (0.292762519060696+) \(S^2\)

Do you think (29.2762519060696+)% of area is there?

Rectified area wanted: 1 - \(\frac{\pi}{4}\) + \(\frac14(1-\frac{\sqrt2}{2})\) + \(2\)(\(\)\(\sin^{-1}\frac {5-\sqrt7}{8} - \frac {5-\sqrt7}{8})\) - \(\frac12\)(\(\)\(\sin^{-1}\frac {\sqrt2}{4} - \frac {\sqrt2}{4})\)

\(=\frac{\sqrt7-\pi}{4}\) + \(2\)(\(\)\(\sin^{-1}\frac {5-\sqrt7}{8})\) - \(\frac12 sin^{-1}\frac {\sqrt2}{4}\)

Area = [\(\displaystyle\frac{\sqrt7-\pi}{4}\) + \(\frac12 (\displaystyle\cos^{-1}\frac {47}{128})\) - \(\frac14\)\(\displaystyle\cos^{-1}\frac {3}{4}\)] \(S^2\) = (0.29276251906069564951895912071385+) \(S^2\)

// May be many other forms. // Example:

[\(\displaystyle\frac{\sqrt7-\pi}{4}\) + \(\frac14 (\displaystyle\cos^{-1}\frac {-5983}{8192})\) - \(\frac14\)\(\displaystyle\cos^{-1}\frac {3}{4}\)] \(S^2\) for \(\frac14\) to be factored out.

Area wanted = \(\displaystyle \frac 14 (cos^{-1}\frac{-5983}{8192} - cos^{-1}\frac34 + \sqrt7 - \pi) S^2\)

Luis Ortiz introduced an elegant solution here within an hour ago.

\(SQRT(7)/4+ACOS(-3/4)/4-ACOS(9/16)\) is another form by the author. The second method introduced should be a most preferred method. Only trigonometry is required rather than geometry! I recommend the method.

http://mathworld.wolfram.com/Lune.html – Lu Chee Ket · 1 year, 5 months ago

Log in to reply

– Jay Rathod · 1 year, 4 months ago

Lol thats a bad method! this question is just a matter of 2 minutes!Log in to reply

– Lu Chee Ket · 1 year, 4 months ago

With known constant, this requires seconds rather than minutes.Log in to reply

– Jay Rathod · 1 year, 4 months ago

but this requires nothing more thn primary mathematics !Log in to reply

– Lu Chee Ket · 1 year, 4 months ago

Please write your solution to this question.Log in to reply

A = 2 [2Δ + a² * arccos( (b²-a²-c²) / (2ac) ) - b² * arccos( (b²+c²-a²) / (2bc) )]

Introduced by Luis Ortiz is correct. a = \(\frac12\), b = 1 and c = \(\frac{\sqrt2}{2}\) – Lu Chee Ket · 1 year, 5 months ago

Log in to reply

– Shyam R · 1 year, 5 months ago

How could, Area of an overlap = pi/2 - 1.Log in to reply

Two of them each with center at diagonal corners added up to \(\frac{\pi}{2}\).

But such area equals to area of 1 sq. unit with a redundant of shape of a lens.

Wanting to find one time of area of the lens only, we minus away 1: \(\frac{\pi}{2} - 1\). – Lu Chee Ket · 1 year, 5 months ago

Log in to reply

Consider the square is of length r. So the radius of circle is r/2. If we subtract the area of circle(πr^2/4) from the area of the square(r^2) we get the area of 4 corners i.e. [r^2(1-π/4)]. Now the radii of the bigger arc is r(=length of side of square). So its area is πr^2/4. If we subtract the areas of two corners from this area ,we will get the area of white patch and one of the black patch inside the circle i.e. [3πr^2/8-r^2/2]. If we subtract this from the area of circle , we get the area of one black patch which is r^2/2-πr^2/8. If we double that we will get area of shaded part. So the area of shaded part is r^2 -πr^2/4. – Rushi Panmand · 1 year, 5 months ago

Log in to reply

– Jay Rathod · 1 year, 4 months ago

Now thats the way to do this!! Perfect!!Log in to reply

Let side length be 10.

Pic

\(cos\frac{θ_{1}}{2}=\frac{（5\sqrt{2}）^{2}+10^{2}-5^{2}}{2×5\sqrt{2}×10}=\frac{5}{4\sqrt{2}}\)

\(cosθ_{1}=2(cos\frac{θ_{1}}{2})^2-1=\frac{9}{16}\)

\(cos\frac{θ_{2}}{2}=-cos∠AOQ=\frac{5^2+（5\sqrt{2}）^{2}-10^{2}}{2×5×5\sqrt{2}}=\frac{1}{2\sqrt{2}}\)

\(cosθ_{2}=-\frac{3}{4}\)

\(sinθ_{1}=\frac{5\sqrt{7}}{16}\)

\(sinθ_{2}=\frac{\sqrt{7}}{4}\)

\(θ_{1}=arccos(\frac{9}{16}) \)

\(θ_{2}=arccos(-\frac{3}{4})\)

\(S_1+S_2=\frac{1}{2}×5^2×[arccos(-\frac{3}{4})-\frac{\sqrt{7}}{4}]\)

\(S_2=\frac{1}{2}×10^2×[arccos(\frac{9}{16})-\frac{5\sqrt{7}}{16}]\)

\(2S_1=25\sqrt{7}+25arccos(-\frac{3}{4})-100arccos（\frac{9}{16}）≈29.2763\) – Z Xy · 1 year, 5 months ago

Log in to reply

– Lu Chee Ket · 1 year, 5 months ago

Nice! \(\pi\) is inclusive in inverse trigonometric functions and not appeared. This method should gain the first place on this page.Log in to reply

– Z Xy · 1 year, 5 months ago

Thanks😊Log in to reply

– Lu Chee Ket · 1 year, 5 months ago

All right.Log in to reply

assume diagonals as x and y axes so centre is origin.....eq of smaller circle x^2+y^2=(a^2)/4 eq of larger circle x^2+(y+a/root2)^2=a^2.find pt of intersection x=0.467a or -0.467a Integrate diffrence between y of both eqs for limits x=0 to x=0.467a and multiply it by 4.....hope this helps... – Rushikesh Joshi · 1 year, 5 months ago

Log in to reply

Let side=a Total area=a

a Area excluding the left top corner(shaded and unshaded part=0.25api Therefore area of shaded part={(piaa/2)-(0.25api)}2 – Shubhayu Das · 1 year, 5 months agoLog in to reply

An amazing question – Tushar Kumar · 1 year, 5 months ago

Log in to reply

Hi – Aditya Singh · 1 year, 5 months ago

Log in to reply

Let side of square be a then area of square will be a^2 and area circle will be pia^2/4 now subtract the area of circle from the area of square.next add the area of two quadrents and subtract the area of circle now subtract the area obtained from the first difference now the area obtained from the last difference is subtracted from the area of the circlenow we getvthe required area – Himanahu Singh · 1 year, 5 months ago

Log in to reply

– Rishabh Dhawan · 1 year, 5 months ago

No it won't giveLog in to reply

If side of square = a Area of shaded region = (3

a^2)((4-pi)/8) – Jay Rathod · 1 year, 4 months agoLog in to reply

How the hell do you work this out ?!??!?!?!? – Madison Stevens · 1 year, 4 months ago

Log in to reply

Thats a lot of work haha. Can you just grid the whole square with metric units and then do simple fractions? – Samath Oung · 1 year, 4 months ago

Log in to reply

I'm not going into details... That's pedantic Instead I wish to say "Square, quarter circle, circle"!!! – HaaraaD B J · 1 year, 4 months ago

Log in to reply

Let R be the radius of the Quadrant circle,

Let the Side of the square be '2a'.

and, let radius of the circle that touching the all four sides of the Square = r.

Then , we have,

R = 2a

r = a.

Area of the Square = (2a)^2 = 4a^2

Area of Circle = πr^2 = πa^2.

Area of a Quadrant = 1/4(πR^2) , where R =2a

Therefore,

Required area of the shaded region = Area of Square + Area of Circle ― 2(area of each quadrant)

=4a^2 + πa^2 ― 2[1/4(πR^2)]

= 4a^2 + πa^2 ― 2(1/4(π(4a^2))) , [ since R = 2a ]

= 4a^2 + πa^2 ― 2πa^2

= 4a^2 ― πa^2

= [4 ― π]a^2

= [4 ― π]r^2 sq units [ as, a = r ] – Avinash Kumar · 1 year, 5 months ago

Log in to reply

1/4th of the square – Aditya Singh · 1 year, 5 months ago

Log in to reply

If square length is x Then the area of the shaded is (5πx^2)/8 – محمد النموس · 1 year, 5 months ago

Log in to reply

Idk – Lori Oconnell · 1 year, 5 months ago

Log in to reply

Basically you have to times the sub divided pattern by the integral Base and then divide it by four because a square has four sides – Caitlin Harwood · 1 year, 5 months ago

Log in to reply

Let's assume the square has side length (s). Then, we can say that the radii of both fourth-of-circle, that form the symmetrical lens (eye shape) with their intersection in the middle of the figure, is (s).

The black, small circle has a radius of s/2 and the distance between the center of the black circle and the center of either fourth-of-circle is (s)*(sq.root 2)/2. (Isosceles triangle)

Knowing those 3 things (radii and distance between centers of circles) we can use the formula for finding the area of the lunes (black areas). Since the lunes are the same, we can find the area of a single lune and multiply by 2.

Area of lune =0.146381259542932(s)^2 ................. Total area=0.292762519(s)^2

Or roughly 30% of the area of the square.

Formulas: Δ = (1/4)√( (a+b+c) * (b+c-a) * (c+a-b) * (a+b-c))

A = 2Δ + a² * arcsec( (2ac)/ (b²-a²-c²) ) - b² * arcsec( (2bc) / (b²+c²-a²) )

Where a = radius of small circle, b = radius of big circle, c = distance between centers of circles. – Luis Ortiz · 1 year, 5 months ago

Log in to reply

I found the area of another overlaps since long ago. Just add another lens onto another diagonal, what is the curved-sides' square area in the middle with side length of 1 unit for the square enclosing it? This question has nothing to do with different circles but only one radius. A challenge to you, how are you going to sort for this? – Lu Chee Ket · 1 year, 5 months ago

Log in to reply

– Luis Ortiz · 1 year, 5 months ago

Is it 34%?Log in to reply

Finally got this from Started Problems. You can try the question! – Lu Chee Ket · 1 year, 5 months ago

Log in to reply

– Lu Chee Ket · 1 year, 5 months ago

Less than 34%. I tried to find a question I found in brilliant for you concerning the question being modified a little bit. Unfortunately the system seems to have removed an activity that I wrote something on the page. I try to scroll the whole list but I couldn't get it. They provided with a clear diagram. I think you can score from it when you have solved the problem.Log in to reply

A = 2Δ + a² * arccos( (b²-a²-c²) / (2ac) ) - b² * arccos( (b²+c²-a²) / (2bc) )

I tried with the formula you introduced. Not giving the same answer. – Lu Chee Ket · 1 year, 5 months ago

Log in to reply

– Luis Ortiz · 1 year, 5 months ago

Sorry, I think now I wrote it correctly. I heavily believe there is a way to simplify those formulae for this particular problem, I might post them later here. It might help that the source of the formulas is wolfram mathworld, you can find them at mathworld.wolfram.com/Lune.html.Log in to reply

Times two again 0.292762519060696+ is obtainable. All right!

Congratulation Luis Ortiz! I have never known that you can express in this elegant way. Thanks! – Lu Chee Ket · 1 year, 5 months ago

Log in to reply

– Luis Ortiz · 1 year, 5 months ago

No, thank you. I love to see the way you did it simply because its complexity is really beautiful to me. It actually inspires to get better.Log in to reply

– Lu Chee Ket · 1 year, 5 months ago

The way elaborated is not using special tool and therefore complexity appears. However, I think the one you introduced is important. However, to solve 4 \(\times\) quarter circles centered onto every corner for area at the middle must apply the way I elaborated, I think. Try to get another constant of it!Log in to reply

Let us consider the picture at the following link: https://www.dropbox.com/s/t70y79qb8d4e19t/aaa.jpg?dl=0

Referring to the previous figure, I will indicate areas by capital letters of the points placed on their perimeter.

Be x=NGEM y=HNG z= HIN j= INML the areas in figure.

x, y, z, j can be obtained as the solution of the following system of 4 equations.

where the areas GFE, HGFEDMN, GFE, HILDMN, ACLMEGNI can be simply obtained as

Solving the system, x is the area needed. – Andrea La Monaca · 1 year, 4 months ago

Log in to reply

Let side length be 20.

Set up coordinate system.

Pic

Circle M：\((x-10)^2+(y-10)^2=10^2\)

Circle N：\(x^2+(y-20)^2=20^2\)

Solve x1,x2

\(x_1=\frac{25}{2}-\frac{5\sqrt{7}}{2}\)

\(x_2=\frac{25}{2}+\frac{5\sqrt{7}}{2}\)

LM:Lower half part of Circle M:\(y=10-\sqrt{10^2-(x-10)^2}\)

LN:Lower half part of Circle N:\(y=20-\sqrt{20^2-x^2}\)

\(S_1=\int_{x_1}^{x_2}(LN-LM)dt=\int_{x_1}^{x_2}10-\sqrt{20^2-t^2}+\sqrt{10^2-(t-10)^2}dt\)

\(S_2=S_3=\int_{x_2}^{20}(10-LM)dt=\int_{x_2}^{20}\sqrt{10^2-(t-10)^2}dt\)

\(2S=2(S_1+S_2+S_3)\)

```

## Use matlab

clear

syms x y;

[X ,Y]=solve('(x-10)^2+(y-10)^2=100','x^2+(y-20)^2=400');

x1=X(2);x2=X(1);

S1=int(10-(20^2-x^2)^0.5+(10^2-(x-10)^2)^0.5, x1, x2);

S2=int((10^2-(x-10)^2)^0.5,x2,20);

S=S1+S2+S2+S1+S2+S2

S =

100

7^(1/2)-5/2(50-25/27^(1/2))^(1/2)-5/2(50-25/27^(1/2))^(1/2)7^(1/2)-100asin(-1/4-1/47^(1/2))+400asin(-5/8-1/87^(1/2))-25/2(200-125/27^(1/2))^(1/2)-5/2(200-125/27^(1/2))^(1/2)7^(1/2)-5/2(50+25/27^(1/2))^(1/2)+5/2(50+25/27^(1/2))^(1/2)7^(1/2)-100asin(1/4-1/47^(1/2))+400asin(5/8-1/87^(1/2))+25/2(200+125/27^(1/2))^(1/2)-5/2(200+125/27^(1/2))^(1/2)7^(1/2)+100pi-200asin(1/4+1/47^(1/2))?100

7^(1/2)-5/2(50-25/27^(1/2))^(1/2)-5/2(50-25/27^(1/2))^(1/2)7^(1/2)-100asin(-1/4-1/47^(1/2))+400asin(-5/8-1/87^(1/2))-25/2(200-125/27^(1/2))^(1/2)-5/2(200-125/27^(1/2))^(1/2)7^(1/2)-5/2(50+25/27^(1/2))^(1/2)+5/2(50+25/27^(1/2))^(1/2)7^(1/2)-100asin(1/4-1/47^(1/2))+400asin(5/8-1/87^(1/2))+25/2(200+125/27^(1/2))^(1/2)-5/2(200+125/27^(1/2))^(1/2)7^(1/2)+100pi-200asin(1/4+1/47^(1/2))ans =

117.1050

``` – Z Xy · 1 year, 5 months ago

Log in to reply