Area after rotation

My son's proposed solution is correct, but inelegant.

The green area consists of a large right isosceles triangle (short edges of length = 1), and a small right isosceles triangle (short edges of length = √2-1).

Thus, the green area = 2 -√2, and p^2=2.

$\text{red} = \text{brown} - \text{blue} = \sqrt 2 - 1.$ $\text{yellow} = \tfrac12\cdot \text{blue} \cdot \text{red} = \tfrac12(\sqrt 2 - 1).$ $\text{green} = \text{unit square} - 2\cdot \text{yellow} = 1 - (\sqrt 2 - 1) = \boxed{2 - \sqrt 2}.$

The line of symmetry cuts the kite into a pair of right triangles with angles of 22.5, 67.5, and 90 degrees.

By definition, this triangle has a short:long:hypotenuse ratio of 1 : 1+√2 : √(4+2√2).

Since this is a unit square,

• the "short" length = $\frac{1}{1+√2}$
• the white area = $\frac{1}{1+√2}$
• the green area = 1 - $\frac{1}{1+√2}$
• the green area can also be written as 2-√2 (i.e. multiply numerator and denominator by 1-√2)
• Solve $p^2 -p = 2 -√2$

$p^2 = 2$

Rather than bothering to actually solve the problem, we can use logic to eliminate answer choices. $\frac12$ is clearly not correct (just look at the diagram...). Right triangles with a 45 degree angle have the hypotenuse as $\sqrt2$ times each leg, so $\sqrt2$ must be involved in the answer, and there is no way $\sqrt3$ can be part of the answer. Therefore, $2-\sqrt2$ is the only sensible answer.

With 45 degrees, anything root-three is out (30 or 60 degrees would be another story). If the side of the rotated square extended to the opposite corner, the shaded region would be 1/2; it doesn't, so that answer is also out. So three out of four answers are eliminated.

I know this explanation doesn't prove anything, but it is a good multiple choice testing strategy.

Divide into two smaller problems: The large green triangle which consists of half the area of the entire square: 1^2/2 The small triangle whose sides are: Sqrt(2)-1 --> area=(sqrt(2)-1)^2/2 Add these areas to get: 1/2 + (sqrt(2))^2/2 -2*sqrt(2)/2 + 1/2 = 1/2 + 1/2 + 1 - sqrt(2) = 2 - sqrt(2)

$\begin{aligned} A &= \dfrac{x^{2}}{2} - \dfrac{1^{2}}{2} + 1^{2} \\ &= \dfrac{\left ( \sqrt{2}-1 \right )^{2}}{2} + \dfrac{1}{2} \\ &= \dfrac{2-2\sqrt{2}+1+1}{2} \\ &= \dfrac{4-2\sqrt{2}}{2} = \boxed{2 - \sqrt{2}}\end{aligned}$

We can use the fact that the shape has reflection symmetry with respect to line through points $A$ and $E$ to get: From here we get the following: $1+x= \sqrt 2,\\ A_{\text{yellow}}+A_{\text{green}} = 1,\\ A_{\text{yellow}} = x,$ and this gives us $A_{\text{green}} = 1 - x = 1 - (\sqrt 2 - 1)= \boxed{\hphantom{\,}2 - \sqrt 2\hphantom{\,}}.$

The reflection argument. Any rotation is composition of two reflections. To get rotation by $\alpha$ centered at $A$ choose any two lines through $A$ with angle between them $\alpha/2$ . Composition of reflections with respect to those lines gives rotation by $\alpha$ . Switching the order of applying the reflections switches between clockwise and anticlockwise rotation.

In our example, instead of rotating square $ABCD$ by $45^\circ$ clockwise, we can first reflect it with respect to line $AC$ followed by reflection with respect to line $AE$ . Since square $ABCD$ is invariant under reflection with respect to line $AC$ , we can get square $AB'C'D'$ just by reflecting it with respect to line $AE$ (up to relabeling of vertices).

I deduced a general formula for the green area $A$ of a square of longitude $L$ and which has been rotated an angle of $\theta$ (where $0º\le\theta < 90º$ ) by doing the difference of the total area and the intersection area between both squares, which can be divided in a rectangle triangle and a trapezium.

I’m not going to write here the steps of the demonstration, but applying the previous idea and simplifying the expression it’s not complicated to obtain it. This general formula is: $A=L^2\left(1-\frac{1-\sin{(\theta)}}{\cos{(\theta)}}\right)$ In this particular case, with $L=1$ and $\theta=45º$ , the green area is: $A=1^2\left(1-\frac{1-\sin{(45º)}}{\cos{(45º)}}\right)=1-\frac{1-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}=\boxed{2-\sqrt{2}}$

$A_{\triangle{AEF}} = \dfrac{1}{2}$

$A_{\triangle{FLD}} = \dfrac{1}{2}(\sqrt{2} - 1)^2$

$A_{AELD} = A_{\triangle{AEF}} - A_{\triangle{FLD}} = \sqrt{2} - 1$

$\therefore$ The desired green area above is $A = 1 - A_{AELD} = 2 - \sqrt{2}$ .

Well if you don't want to mess with calculation just eliminate the options

Clearly area is bigger than 0.5

Next we see that the area is kind of unsymmetrical,so it will not be a/b form

Also ( root 3 )deals with 30 and 60 degrees so we get our answer

Yes I know that's not at all a solution

tan(67.5°) = a / b = 1 / b

2.4142 ≈ 1 / b

b ≈ 0.4142

The area of one blue triangle: 1 * 0.4142 / 2

The are of 2 blue triangles: 0.4142

The green area: 1 - 0.4142 ≈ 0.5858 ≈ $2 - \sqrt{2}$

If instead of 45° the square were to be rotated counterclockwise about A by a general angle θ (0<θ<90°) we can show that the resulting green area = tan(θ) - sec(θ) + 1 = 2 -√2 at θ = 45°.