Maximum area of triangle inside a square

[issacnewton]

Homework Statement

$ABCD$ 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.

Homework Equations

Area of triangle

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 ?

 

[haruspex]

Homework Statement

$ABCD$ 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.

Homework Equations

Area of triangle

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 ?

Looks ok to me. I got the same answer independently.

 

[issacnewton]

Looks ok to me. I got the same answer independently.

What method did you use ?

 

[haruspex]

What method did you use ?

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