Notes
three squares in a circle solution

Solution to the Three Squares in a Circle Puzzle

Three Squares in a Circle

Three squares in a circle … If the blue triangle has area 55, what’s the area of the red triangle?

Solution by the Intersecting Chords Theorem, Lengths in a Square, and Area of a Triangle

Three squares in a circle annotated

With the points as labelled in the diagram, APDA P D and BPEB P E are chords of the circle that intersect at PP. Therefore, by the intersecting chords theorem, the lengths of the segments satisfy:

APPD=BPPE A P \cdot P D = B P \cdot P E

Now, BPB P is the diagonal of the square with APA P as side length, so BPB P is 2AP\sqrt{2} \cdot A P. This means that PDP D must be 2PE\sqrt{2} \cdot P E. Since PEP E is the diagonal of the square with PFP F as side, and PCP C is the diagonal with PDP D as side:

PC=2PD=22PE=222PF 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 BPCB P C is right-angled since BPB P and PCP C are diagonals of aligned squares.

Therefore, the area of triangle BPCB P C is:

12BPPC=122AP22PF=4×12APPF=4×5=20 \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