Geometry Challenge – 2025/06/07

Solution: Edge $AD$ and $CB$ of unit square $ABCD$ are folded along diagonal $AC$, forming parallelogram $AECF$. Paralleogram $AECF$ is folded along $GH$, forming trapezoid $AIJB$.

Let $x=BE$, Because $B’E=BE$, and $\triangle{AB’E}$ is an isosceles right triangle, and $AE=AB-BE=1-x$, by Pythagorean Theorem, we have $$x^2+x^2=(1-x)^2$$

Solving the above equation and ignoring the invalid $x$ value, we have $x=\sqrt{2}-1$. Therefore

$$Area(\triangle{EBC})=\dfrac{1}{2}\cdot BE\cdot CB=\dfrac{1}{2}\cdot (\sqrt{2}-1)\cdot 1=\dfrac{\sqrt{2}-1}{2}$$

$$Area(\triangle{BHJ})=\dfrac{1}{4}Area(\triangle{EBC})=\dfrac{1}{4}\cdot\dfrac{\sqrt{2}-1}{2}=\dfrac{\sqrt{2}-1}{8}$$

$$Area(AIJB)=Area(AGHB)-Area(\triangle{AGI})-Area(\triangle{BHJ})$$

$$=\dfrac{1}{2}-2\cdot Area(\triangle{BHJ})=\dfrac{1}{2}-2\cdot\dfrac{\sqrt{2}-1}{8}=\boxed{\dfrac{3-\sqrt{2}}{4}}$$