# Solution to the Triangle in Two Tilted Squares Puzzle +-- {.image} [[TriangleinTwoTiltedSquares.png:pic]] > Two squares. What’s the shaded area? =-- ## Solution by [[Transformations]] +-- {.image} [[TriangleinTwoTiltedSquaresLabelled.png:pic]] =-- The vertex $E$ is free to move on the line $A D$. The vertex $G$ is the image of $E$ after a combined transformation of a scaling by scale factor $\sqrt{2}$ and a rotation clockwise by $45^\circ$. It therefore also moves on a line. When $E$ is at $A$, $G$ is at $D$. When $E$ is at $D$, the square $B F G E$ is tilted at $45^\circ$. So $G$ moves along a line that is at $45^\circ$ to the horizontal passing through $D$. This line is parallel to $A C$, which we can view as the "base" of the triangle $A C G$, and so the area of the triangle is independent of where the point $G$ is on this line. In particular, if it is at $D$ then the area of the triangle is evidently half of the area of the square, and thus is $72$. ## Solution by [[Invariance Principle]] +-- {.image} [[TriangleinTwoTiltedSquaresExtremes.png:pic]] =-- There are two special cases of this problem: when the vertex $E$ is at $A$ and when it is at $D$. If it is at $A$, $G$ is at $D$ and the area is half of that of the square, whence $72$. If it is at $D$ then the height of the triangle above $A C$ is half of $D B$ and so again the area of the triangle is half of that of the smaller square.