# Solution to the [[Three Squares in a Circle]] Puzzle +-- {.image} [[ThreeSquaresinaCircle.png:pic]] > Three squares in a circle ... If the blue triangle has area $5$, what's the area of the red triangle? =-- ## Solution by the [[Intersecting Chords Theorem]], Lengths in a [[Square]], and [[Area of a Triangle]] +-- {.image} [[ThreeSquaresinaCircleAnnotated.png:pic]] =-- With the points as labelled in the diagram, $A P D$ and $B P E$ are chords of the circle that intersect at $P$. Therefore, by the [[intersecting chords theorem]], the lengths of the segments satisfy: $$ A P \cdot P D = B P \cdot P E $$ Now, $B P$ is the diagonal of the [[square]] with $A P$ as side length, so $B P$ is $\sqrt{2} \cdot A P$. This means that $P D$ must be $\sqrt{2} \cdot P E$. Since $P E$ is the diagonal of the square with $P F$ as side, and $P C$ is the diagonal with $P D$ as side: $$ P C = \sqrt{2} \cdot P D = \sqrt{2} \cdot \sqrt{2} \cdot P E = \sqrt{2} \cdot \sqrt{2} \cdot \sqrt{2} \cdot P F $$ Lastly, triangle $B P C$ is [[right-angled triangle|right-angled]] since $B P$ and $P C$ are diagonals of aligned squares. Therefore, the area of triangle $B P C$ is: $$ \frac{1}{2} B P \cdot P C = \frac{1}{2} \sqrt{2} \cdot A P \cdot 2 \sqrt{2} \cdot P F = 4 \times \frac{1}{2} A P \cdot P F = 4 \times 5 = 20 $$