# Solution to the Two Overlapping Triangles (II) Puzzle +-- {.image} [[TwoOverlappingTrianglesII.png:pic]] > The largest of these two equilateral triangles has area $8$. What’s the blue area? =-- ## Solution by Calculating Lengths +-- {.image} [[TwoOverlappingTrianglesIILabelled.png:pic]] =-- The blue area is a [[triangle]], so to calculate its area we can start with picking a side to be the "base". The side $B C$ coincides with one of the sides of the equilateral triangle whose area is given, so that seems a reasonable side to pick. The perpendicular height of the triangle from this base is the length $E F$. The triangle $E F C$ is congruent to triangle $C G D$, meaning that length $E F = C G$. This is the height of the equilateral triangle above $A B$. As the triangle $A B C$ is equilateral, $A B = B C$ and so the two triangles have the same base and perpendicular height, and thus the same area. ## Solution by [[Symmetry]] There is a strategy for solving this which exploits [[symmetry]]. To use this, draw a copy of the larger triangle rotated by $60^\circ$ clockwise. +-- {.image} [[TwoOverlappingTrianglesIIRotated.png:pic]] =-- The side $D C$ of the smaller triangle becomes the side $E C$, as the smaller triangle is also equilateral. The point $E$ lies on the side $A F$ of the new triangle which is parallel to the side $B C$ of the original triangle. This then establishes the perpendicular height above $B C$ of the blue area as being the height of the larger triangle above the same side, and therefore that the blue area has the same area as the larger triangle. ## Solution by the [[Invariance Principle]] Both the weak and strong [[Invariance Principles]] can be used to solve this problem. The position of $D$ along the side $A B$ is not fixed and so could be anywhere along that edge. For the strong principle, we can look at either of the extremes. If $D$ coincides with $B$ then the blue area is exactly the same as the larger triangle and so has area $8$. If $D$ coincides with $A$ then the "smaller" triangle is a copy of the original triangle and the blue area then consists of two halves of that triangle, and so has area $8$. The weak principle says that the area is independent of the position of $D$ and so the height of the blue area, which is the perpendicular distance of $E$ above $B C$, must be constant. This means that as $D$ moves on $A B$, the point $E$ moves on a straight line parallel to $B C$ that passes through $A$. Drawing in this line leads to the symmetry solution above.