Thursday, July 21, 2011

Solution for the area, Problem for the cost of pencils

I'm sorry for not posting in awhile. I kind of got a little lazy. Thanks everyone for trying out the problem and posting in the comments!

Problem:A square of side length 1 and a circle of radius (√3)/3 share the same center. What is the area inside the circle, but outside the square?

π is the pi symbol
You can easily find the distance from the center of the square to the corner of the square to be 1/√(2). 1/√(2) is greater than 1/√(3) therefore the circle does not extend out of the corner of the square. However the length from the center of the square to the side of the square is 1/2. Thus, we get something like this:
(We're solving for the shaded regions)

The method I'm going to use is find the area of one of the shaded regions (top) and multiply that by 4 to get the total area of all 4 regions. We can plot this onto a graph with the center of the circle and square as (0,0). For the circle we have an equation of x^2 + y^2 = 1/3. For the line of the square at the top we have y = 1/2. We can now find the points of intersection of the square and the circle by substituting 1/2 for y.  

x^2 + y^2 = 1/3
x^2 + (1/2)^2 = 1/3
x^2 + 1/4 = 1/3
x^2 = 1/4 - 1/3
x^2 = 1/12
x = ±1/√(12)

We can draw a triangle from the center of the square to the intersection points like this:
The height of the triangle is 1/2 (because it is half the length of the square) and the length of the base of the triangle is 1/√3. Also, the other two sides of the triangle are 1/√3 because it is the radius of the circle. Using simple trigonometry, you can find the angles of the intersection points to be 60°. That means the angle at the center is also 60° because 180-60-60=60. Now, we can take the area of the part of the circle and subtract the area of the triangle to find the shaded region.


π/3 is the area of the entire circle. However, we only want the area of the 60° slice of the circle. Therefore we multiply the total area by 1/6 because 60° is 1/6 of 360°.

π/3 * 1/6 = π/18

Now we have to subtract the area of the triangle from π/18.

1/√3 * (1/2)/2

Thus, the area of one of the shaded regions is π/18 - 1/4√3. We must now multiply it by 4 to find the total area of all the shaded regions.

4(π/18) - 4(1/4√3)

2π/9 - 1/√3

Therefore the answer is 2π/9 - 1/√3

A majority of the 30 students in Ms. Demeanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than 1. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the 
pencils was $17.71. What was the cost of a pencil in cents?

Good luck! and remember, calculators are not allowed.


  1. Great math problem, although i don't think I can solve it hehe +1 follower

  2. Damn math! hahaha, Anyways i'll jolt my brain for this problem as I am going back to school soon.

  3. you're awesome! I'll try this one out but by just looking, I can say I won't make it

  4. yay, I'm a mathematics student on Gdańsk University of Technology. I guess it's going to become one of my favourite blogs! Keep it up!

  5. Dude, I suck at math. But I need to brush up on my skill. I will have to come here to get my daily dose of smarts.

  6. holy crap I couldn't do it. followed+ though, maybe ill try easier ones

  7. 19 Cents? Subscribing to a math blog over the summer, starting to have second thoughts.

    Keep us updated on what the solution really is, Im pretty sure it's $0.19 if every student bought 3 pencils.

  8. I feel stupid...did I really forget this much since school? Oh, well, I think I'm going to follow and try to get some of my lost intelligence back :P

  9. I really tried hard but my result(83)sadly is not right
    I used 1771=30*(x+1)*(x+2)
    I think this would have beentoo easy:D waiting to see the solution

  10. wow...feels like I'm back in school +1 follower

  11. NO CALCULATORS!? What, it wasn't hard enough before?

  12. Whaaaaaaaaaaat. I should have paid more attention in school.

  13. i cant do this. maybe if i read your blog everyday i will begin to grasp the maths.

  14. Oh boy, trying this now. Here we go..

  15. Ah, math. I may not enjoy this blog the most during the summer, but as soon as school starts back up here I'll be more than glad that I'm following you now.

  16. okay, I'm glad someone finally explained this. I had nothing but crappy math teachers through high school, who couldn't teach to save their lives.

  17. oh thinkink, havent done it in a while, feels weird ;)

  18. Sadly I dare not even attempt this! My maths skills have been terrible and I fear they always will be! Science is my area!

  19. nice problem. Don't think i'd be able to do it. still in mental shock from my as maths exams :P