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Maximum area of triangle inside a square

  1. Mar 16, 2017 #1
    1. The problem statement, all variables and given/known data
    ##ABCD## is a square piece of paper with sides of length 1 m. A quarter-circle is drawn from
    ##B## to ##D## with center ##A##. The piece of paper is folded along ##EF##, with ##E## on ##AB## and ##F## on ##AD##,
    so that ##A## falls on the quarter-circle. Determine the maximum and minimum areas that the
    ##\triangle AEF## can have.

    2. Relevant equations
    Area of triangle

    3. The attempt at a solution
    Now when the paper is folded along ##EF##, the segment connecting from ##A## to the point on half circle is perpendicular to ##EF## and the perpendicular from ##A## on ##EF## has length of ##1/2##. Let ##G## be the foot of perpendicular from ##A## on ##EF##. So let's draw a half circle of radius ##1/2## with center at ##A##. Then ##EF## is the tangent to this half circle. Now the area of ##\triangle AEF## can be found in two ways. ##\mbox{Area } = \frac{1}{2}(AE)(AF) = \frac{1}{2}(EF)(AG)##. But since ##AG=1/2##, we get ##\mbox{Area } = \frac{1}{2}(AE)(AF) = \frac{1}{4}(EF)##. This leads to ##EF = 2(AE)(AF)##. Now let ##\theta = \angle AEF##. Then ##\tan(\theta) = \frac{1/2}{EG}## and ##\angle AFE = 90 -\theta## , so ##\tan(90 - \theta) = \frac{1/2}{GF}##. Hence we have ##EG = \frac{\cot(\theta)}{2}## and ##GF = \frac{\tan(\theta)}{2}##. So we have ##EF = (1/2)(\tan(\theta) + \cot(\theta) )##. So the area of ##\triangle AEF## would be ##\mbox{Area } = (1/4) EF = (1/8) (\tan(\theta) + \cot(\theta) )##. This can be simplified as ##\mbox{Area } = \frac{1}{4\sin(2\theta)}##. Once we have figured out the formula for the area, we can now think about the possible values of ##\theta##. Now at the most extreme, point ##F## can slide along the segment ##AD## and be at point ##D##. So ##AF = 1##. Since ##EF = 2(AE)(AF)##, this means ## EF = 2(AE)##. So ##\sin(\theta) = \frac{AE}{EF} = 1/2##. This means that ##\angle AEF = \theta = 30^{\circ}## and ##\angle AFE = 60^{\circ}##. This shows that the acute angles of ##\triangle AEF## vary between ##30^{\circ}## and ##60^{\circ}##. Now the area is ##\frac{1}{4\sin(2\theta)}##. This will be minimum when ##\sin(2\theta)## becomes maximum. So minimum area will be when ##\theta=45^{\circ}## and maximum area will be when ##\theta=30^{\circ}##. With this is mind, we finally get that $$\frac{1}{4}\leqslant \mbox{Area} \leqslant \frac{1}{2\sqrt{3}}$$ Does this look correct ?
     
  2. jcsd
  3. Mar 17, 2017 #2

    haruspex

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    Looks ok to me. I got the same answer independently.
     
  4. Mar 17, 2017 #3
    What method did you use ?
     
  5. Mar 17, 2017 #4

    haruspex

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    It was substantially the same. I read the problem statement then solved it before reading your solution.
     
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