School-Safe Puzzle Games

Rightagon Inc.- a math/geometry puzzle

rightagon-geometry-math-puzzle

Now, not everyone enjoys these math problems,  so please be sure to check out all the other puzzle types we’ve just put up tonight.

Created by Michaelc!

Rightagon Inc. has a lot in downtown Metropolis in which they wish to build upon. The area that they are allowed to build a structure on is a 10,000 ft (squared) right triangle with 2 equal sides.

If they want to build an Octagon building with all sides equal, what is the maximum area they can build?

Can you solve by Friday?

On a side note, it looks like I’ve gotten into the habbit the past year of posting just really hard math puzzles and brain teasers; however, from here on out, will try to put up a nicer mixture of hard and medium ones.

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21 Comments to “Rightagon Inc.- a math/geometry puzzle”


  1. .mau. | Profile

    I got 5685+ sq ft.
    To be more precise, the ratio between the area of the Octagon and the area of the real estate is


    8 (1+sqrt(2)) / (17 + 12 sqrt(2))


    but I may be wrong.


  2. mrman | Profile

    Whoops I wrote down the wrong number. I meant to submit 5685.425 sqft. Is there no way to edit my previous submission?


  3. Shawn | PUZZLE GRANDMASTER | Profile

    I’m coming up with 5,685.43 ft2.


    Hard to explain without a diagram, but I’ll try.


    I drew the right triangle with one upright leg, another leg at 90 degrees from the bottom and heading right, and the hypotenuse connecting them. I then dropped a perpendicular from the right angle to the middle of the hypotenuse. The angles in the 10,000ft2 triangle are perfect to situate the octagon, with a flat side resting on each of the legs of the large triangle. When you draw the octagon, you can see that the original perpendicular perfectly bisects the octagon.


    The sides of the original triangle measure 100*sqrt(2), so both the hypotenuse andits perpendicular must each measure 100.


    Looking at the intersection of the perpendicular and the hypotenuse, I dropped ANOTHER hypotenuse for the new right triangle that had formed. The length of this line must be 100*sqrt(2)/2, and now it is easy to find the length of each octagon side. Setting “x” equal to the length of a side, I looked at each of the line segments within the octagon as I traveled down the hypotenuse, and found the following equation:


    0.5*x*cos45 + x*cos45 + x = 100*sqrt(2)/2


    Solving for x = 34.3146 feet


    Formula for an octagon is 2 * (1 + sqrt(2) * (side)^2 = 5,685.43 ft2


  4. LadyInsomnia | Profile

    The largest regular octagon that could be built in that triangle has an area approximately 4142 sq ft [to be precise 10,000*(sqrt2 - 1)].


  5. LadyInsomnia | Profile

    http://ladyinsomnia.googlepages.com/rightagon.JPG


    As an explanation:
    From properties of isosceles right triangles we know 10000 = 1/2 L^2, so L = 100*sqrt2. Using right-triangle properties again, H = 200, then the perpendicular bisector (of H) called h = 100, and finally the perpendicular bisectors (of L) called b = 50*sqrt2. We have a square of side length b = 50*sqrt2 in which to inscribe our regular octagon. It will have area A = S^2 – a^2 where {S = side length of square = b} and {a = side length of octagon = S/(1+sqrt2)} — these are properties of a regular octagon inscribed maximally in a square. So with substitution of b = 50*sqrt2, we have A = b^2 – (b/(1+sqrt2))^2 = 10000*(sqrt2 – 1)


  6. bilbao | Profile

    I have tried to solve this puzzle graphically. I have drawn the right triangle with an octagon inside. I have divided the right triangle with a grid, thus forming little squares and triangles. If every little square is area 1 and every little triangle is area 1/2, I get an area of 7 inside the octagon and 5 1/2 outside the octagon. So area of the octagon would be 5.600 sq.feet inside the 10.000 sq.feet right triangle.


  7. ruddwd | Profile

    ~5700 square feet
    CAD makes things a little easier.
    But here’s the math I started with.
    A right triangle of area 10,000 ft^2
    AREA = 1/2 b X h
    b = h
    AREA = 1/2 b^2
    10,000 = 1/2 b^2
    20,000 = b^2
    b = 141.4213
    The hypotenus (sp?) is c^2 = a^2 + b^2
    a = b
    c^2 = 2b^2
    c^2 = 40,000
    c = 200


    Now you need to find out the maximum size of octagon that can fit in that space. this is where I was lost, so I turned to my friend CAD.


  8. bizarette18 | PUZZLE MASTER | Profile

    5685.4 sq ft?


  9. ignaciordc | Profile

    This problem is a pure geometric problem
    The area is a right rectangle, hence its size is equal to S=100*sqr(2)
    The octagon we are seeking is one that has 3 of its sides touching the internal sides of that right triangle.
    If we consider the right angle in the origin of coordinates, and the hypotenuse in the positive quadrant, then those sides will be 1,4 and 7, 1 being the bottom horizontal side of the octagon. If the “center” of the octagon is in point (x,y), then x=y; but at the same time, that value is the distance from the center to the hypotenuse, whose equation is x+y-100*sqr(2)=0 -> x = (2*x – 100*sr(2)) / sqr(2)
    Solving that, we find that the center of the octagon is at X=Y=100/(1+sqr(2)), and that is the height of all 8 triangles of the octagon.


    We can calculate now the proportions between base and height for any right triangle to find out that the maximal area is
    40000 * sqrt[sqrt(2)*(2-sqrt(2)] / (3+2*sqrt(2))
    which is close to 6,246,486 square feet


  10. ignaciordc | Profile

    Ooops!, that is 6,246.486 square feet, a little over 62% of the land


  11. MFox | Profile

    I believe the area of the largest octagon they could fit into that triangle (zoning laws and rounding error notwithstanding) is approximately 5685.3704 feet squared.


    I drew an equal-sided unit octagon, and then fit a triangle around it. I also extended the sides of the octagons to make squares. Looking at the drawing, I came up with the following relationships:
    [C is the length of a side of the octagon.]
    [When you extend two sides of the octagon to meet in a right angle, A is the length of that extended segment. Or, in other words, if you make an isosceles right triangle with C as the base, A is the length of one of the sides.]
    [B is the length of either side of the plot of land, and D is the hypotenuse of the plot.]


    B = 3A + 2C
    D = 3C + 4A


    I calculated that the sides of our particular plot of land would have to be ~141.42 feet long, and the hypotenuse of the same plot would be 200 feet exactly. So, solving algebraically, C must be 34.32 feet, and A must be 24.26 feet.


    With these numbers in hand, I worked out that the area of the octagon must be 5685.3704 square feet.
    The area of the big square that contains the octagon is (C + 2A)^2, so 82.84′^2 = 6862.4656′sq.
    But the area of the 4 corner triangles must be removed, which adds up to (A^2) * 2, or 1177.0952′sq.


    6862.4656 – 1177.0952 = 5685.3704


  12. mimi112 | Profile

    I think the area is 4852.813741 ft(squared)


  13. MFox | Profile

    I posted an image to help interpret all the verbiage. It can be found at http://www.tulane.edu/~venom/i.....ilding.jpg


  14. TheMadEgyptian | Profile

    I get 5685.4 square feet, based on a regular octagon of side 34.3145 ft and an equation for area I pulled off of Wikipedia (Geometry class was a long time ago). I calculated the side of the octagon as Sqrt(20000)/(2 + 3/2 * Sqrt(2)) taking advantage of multiple congruent 45/45/90 triangles. Any more specific explanations would require a diagram, which I’m just too lazy to post.


  15. Mashplum | PUZZLE MASTER | Profile

    I get about 5,685 feet squared. Or 10,100*(40*(2^0.5)-56) to be exact. The octagon should touch all 3 sides of the triangle.


  16. RK | Founder | Profile

    .mau., mrman, Shawn, Bizarette18, Mashplum, TheMadEgyptian (#1 Score on Swizzlepop!), MFox, got this one.


    The correct answer to MichaelC’s challenge is 5685.4, nice explanations can be found by Mfox and Shawn above.


    Ruddwd, Bilbao, Ladyinsomnia and ignaciordc, looks like you were all on the right track


  17. michaelc | Profile

    Great job guys!


    This should have been a great trigonomic work out for most folks, it was for me.


    Ok everyone, let’s jazzercise now! :)


  18. ignaciordc | Profile

    Second oops!
    First steps of my solution were correct, since indeed 100/(1+sqr(2))= h = 41.4213.. is the height of the 8 isosceles triangles. However, I messed up the final part. The base of those triangles should be b=2*h/(1+sqr(2)) = 34.314… as you all pointed out, and the octagonal area = 4*h*b = 40000*2/(1+sqr(2))^2,
    which is 5685.37… again reached by many.
    I liked very much the approach MFox has taken, simple and effective.


  19. .mau. | Profile

    @michaelc: no trigonometry, just rectangles and triangles :-)


  20. brianu | Profile

    5685.4249492377 sq. ft.


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