# Elementary geometry

1. Apr 7, 2016

### davon806

1. The problem statement, all variables and given/known data
In the attached diagram,the edge of the square is a. find the area of the shaded region

2. Relevant equations
area of circle = πr^2 ,area of square,triangles
(Please avoid using integration/radian angle/tangent...Since this problem is found in a maths exercise suitable for a 12-year-old..)Thanks

3. The attempt at a solution
The unshaded corners of the square is easy.The "diagonal leaf" is easy as well.But I have been struggling to find the small portion(marked as A in my diagram) Looking for some insight.Thanks!
Btw,the book mentioned that it might be helpful to draw the same leaf on the left diagonal

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2. Apr 7, 2016

### blue_leaf77

You can construct a system of equation with four unknowns and four equations, the unknowns are $A,B,C,$ and $D$. $A$ is the area of one of the shaded region. $B$ is the portion of the circle inside the diagonal leaf. $2C$ is the remainder of the area inside the diagonal leaf (those two curved triangles at the square's corners). $D$ the small area which you denoted as A. You want to find $2A$.

3. Apr 7, 2016

### davon806

I tried using simulataneous eq. to solve but I couldn't solve the variables I want(i.e. 2A you have mentioned).I ended up with obtaining the same equation after solving some of the variables.
So I look for some alternative method,though I have used trigonometry....But I am highly unsure on the answer I have calculated.I hope someone can comment on those steps,especially MEN~AEF(I simply think they look similar,I don't know whether it is true.)

Thanks

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4. Apr 8, 2016

### haruspex

Have you actually solved it that way? I have to say, I'm struggling to find a fourth equation. There are three known areas, the circle, the quadrant and the square, and each yields one equation.

5. Apr 8, 2016

### blue_leaf77

Yes I have done that but I fail to notice that upon elliminating some unknowns, I got two identical equations, like the OP said. I must apologize that my method doesn't work without finding another equation.

6. Apr 8, 2016

### haruspex

Ok, thanks for confirming.
I can find the angles, e.g. the angle the lower left intersection point and the bottom left corner subtend at the bottom right corner. But that's well beyond a 12-year old's standard, and still a long way to go to find the area.

7. Apr 8, 2016

### blue_leaf77

Yeah, the use of some trigonometrics can relieve the problem up to some extent. But I also believe that the use of some identities on a circle can also help solve it.

8. Apr 8, 2016

### LCKurtz

I think this is an interesting and nontrivial problem. @davon806 I am curious where you got this problem. Using advanced tools unavailable to a 12 year old, I have gotten the area of one of the shaded regions, (taking $a = 1$, for general a just multiply by $a^2$):

This expression certainly hides the $\pi$'s. I was thinking I might be able to reverse engineer from the answer to see what geometry to use. I tell you what -- those $\sqrt 7$'s don't ring any bells for me. I am curious if anyone has made progress on an "elementary" solution.

9. Apr 10, 2016

### haruspex

I get the same result - at least, when I evaluate my expression and yours they seem to be numerically equal. I got it into the form $\frac{\sqrt 7}8-\frac 14 \arcsin(\frac{23}{128}\sqrt{14})$, but that's scarcely any nicer.

10. Apr 12, 2016

### LCKurtz

I finally took time to read the images. It is definitely not true that MEN~AEF. The curved boundaries of AEF have equal radii and those of MEN do not. Again @davon806 please tell us where you found this problem.

11. Apr 12, 2016

### davon806

My friend sent me a picture and he told me it was found from some sort of exercise suitable for a junior form student in China(about 12-13 years old).I don't know about the Chinese curriculum but I guess if there is an elementary solution,a lot of lines and variables are needed to be drawn on the diagram before we can finally reach the answer.Btw,can you show me your method?I failed to solve it :d Thanks

12. Apr 12, 2016

### LCKurtz

I used integral calculus to find the answer, definitely not an "elementary geometrical" argument, and not an argument accessible without calculus.

13. Apr 12, 2016

### haruspex

My method was purely geometric and trigonometric, but quite complicated.

14. Apr 12, 2016

### LCKurtz

The shaded region is called a Lune. I have learned that there is quite a bit of info on the internet about lunes. It turns out there are 5 quadrable lunes (those for which you can construct a square with the same area). I don't know much except what I have recently read about them. I don't think the present example is one of the quadrable lunes, but I don't know whether that has any bearing on whether there is a purely geometric/pythagorean theorem solution or not. I'm thinking that solutions like ours with arcsines in them which hide the $\pi$'s don't qualify. But, for all I know, maybe there isn't a simpler answer.

15. Apr 12, 2016

### haruspex

Here's an outline of what I did:
Label the centre of the unit square O and the corners ABCD anticlockwise from lower left.
E is the midpoint of AD, F is the point inside the quadrilateral ABOE where the arcs intersect. So BF is length 1 and OF is length 1/2.
The sines of the angles EOF and ABF can be found through a couple of simultaneous equations.
That allows the area ABFOE to be found. The areas of the sectors OEF and BAF are subtracted to find the area of the fillet AFE (following the arcs).
Once this is found, it is reasonably straightforward to determine all the areas in the figure.

[Edit: of course, it would have been a little simpler to find angles DBF and DOF (α and β respectively in the notation at the link below) and get the lune area from those sector areas and that of triangle OFD. It is easy to show that sin(α) = (√14)/8.]

I chased up your reference to quadrable lunes. According to http://www4.wittenberg.edu/academics/mathcomp/bjsdir/TheFiveLunes120408.pdf,there [Broken] are only five, and none of them look to match the case in this problem. On the other hand, as you say, it does not have to be a quadrable lune for there to be an easy solution.

Last edited by a moderator: May 7, 2017
16. Apr 12, 2016

### haruspex

Is there any chance the picture got changed somewhere along the line? At first glance, it looks rather like Figure 3 at the link I posted just above.

17. Apr 12, 2016

### LCKurtz

Actually, I don't know, and I do wonder, whether that statement is true. It it's false then, of course, there would be no easy solution to the OP's lune problem.

18. Apr 12, 2016

### haruspex

Consider a degenerate case: the 'lune' is a complete circle. It is not quadrable, but the solution is easy. I feel sure there would be non-degenerate cases with the same feature. If the solution involves pi (irreducibly) then it is not quadrable.
Edit: here's one... two identical circles, each with its centre on the circumference of the other.

19. Apr 15, 2016

### LCKurtz

Done editing, I think the typos are gone.

Not sure anyone still cares, but here's an approach that I think, plausibly, a bright kid with a little trig might possibly come up with. Lets start with a picture:

The area of the large lens CDFC is the area of the sector BDFC minus the area of triangle BDC. Similarly, the area of the small lens CDGC is the area of sector ADGC minus the triangle ADC. So$$A_{small~lens} = \frac 1 2 b^2(2\beta - \sin(2\beta))$$ $$A_{large ~lens} = \frac 1 2 a^2(2\alpha - \sin(2\alpha))$$The area of the upper left blue lune is the difference:$$A_{lune} = a^2\alpha -b^2\beta + \left(\frac 1 2 b^2\sin(2\beta) -\frac 1 2 a^2\sin(2\alpha)\right)$$Notice that the quantity in the large parentheses is the difference in areas of the triangles ACD and BCD, which is the area of the quadrilateral ADBC, which is twice the area of triangle ABC which has sides a,b, and c. I will call the area of this triangle $A_t$. So we have$$A_{lune} = a^2\alpha - b^2\beta + 2A_t$$Now the only reason I think a geometry student might do this is that the area $A_t$ of triangle ABC can be calculated directly from the three sides a,b, and c using Heron's formula, though I would have to look up that formula. And $\beta$ and $\pi -\alpha$ can be calculated from the law of cosines from triangle ABC:$$\cos\beta = \frac{b^2+c^2-a^2}{2ac}$$ and $$\cos(\pi - \alpha) = -cos(\alpha) = \frac{a^2+c^2-b^2}{2ac}$$So $\alpha$ and $\beta$ can both be expressed in terms of arccosines and everything can be expressed in terms of a,b, and c.

Last edited: Apr 16, 2016
20. Apr 15, 2016

### haruspex

I think that's the same as the Edit paragraph in my post #15.