Painting - Squared Circle

This oil painting on pressed wood, #52 in the series, shows an original construction of Crockett Johnson. He executed this work in 1968, three years after he began creating mathematical paintings. It is evident that the artist was very proud of this construction because he drew four paintings dealing with the problem of squaring the circle. The construction was part of Crockett Johnson's first original mathematical work, published in The Mathematical Gazette in early 1970. A diagram relating to the painting was published there.

To "square a circle," mathematically speaking, is to construct a square whose area is equal to that of a given circle using only a straightedge (an unmarked ruler) and a compass. It is an ancient problem dating from the time of Euclid and is one of three problems that eluded Greek geometers and continued to elude mathematicians for 2,000 years. In 1880, the German mathematician Ferdinand von Lindermann showed that squaring a circle in this way is impossible - pi is a transcendental number. Because this proof is complicated and difficult to understand, the problem of squaring a circle continues to attract amateur mathematicians like Crockett Johnson. Although he ultimately understood that the circle cannot be squared with a straightedge and compass, he managed to construct an approximate squaring.

Crockett Johnson began his construction with a circle of radius one. In this circle he inscribed a square. Therefore, in the figure, AO=OB=1 and OC=BC=√(2) / 2. AC=AO+OC=1 + √(2) / 2 and AB=√(AC² + BC²) which equals the square root of the quantity (2+√(2)). Crockett Johnson let N be the midpoint of OT and constructed KN parallel to AC. K is thus the midpoint of AB, and KN=AO - (AC)/2=1/2 - √(2) / 4. Next, he let P be the midpoint of OG, and he drew KP, which intersects AO at X. Crockett Johnson then computed NP=NO+OP=(√(2))/4+(1/2). Triangle POX is similar to triangle PNK, so XO/OP=KN/NP. From this equality it follows that XO=(3-2√(2))/2.

Also, AX=AO-XO=(2√(2)-1)/2 and XC=XO+OC=(3-√(2))/2. Crockett Johnson continued his approximation by constructing XY parallel to AB. It is evident that triangle XYC is similar to triangle ABC, and so XY/XC=AB/AC. This implies that XY=[√((2+√(2)) × (8-5√(2))]/2. Finally he constructed XZ=XY and computed AZ=AX+XZ=[2√(2)-1+(√(2+√(2)) × (8-5√(2))]/2 which approximately equals 1.7724386. Crockett Johnson knew that the square root of pi approximately equals 1.772454, and thus AZ is approximately equal to √(Π) - 0.000019. Knowing this value, he constructed a square with each side equal to AZ. The area of this square is (AZ)² = 3.1415258. This differs from the area of the circle by less than 0.0001. Thus, Crockett Johnson approximately squared the circle.

The painting is signed: CJ68. It is marked on the back: SQUARED CIRCLE* (/) Crockett Johnson 1968 (/) FLAT OIL ON PRESSED WOOD) (/) MATHEMATICALLY (/) DEMONSTRATED (/) TO √π + 0.000000001. It has a white wooden frame. Compare to painting #91 (1979.1093.60).

References: Crockett Johnson, “On the Mathematics of Geometry in My Abstract Paintings,” Leonardo 5 (1972): p. 98.

C. Johnson, “A Geometrical look at √π," Mathematical Gazette, 54 (1970): p. 59–60. the figure is from p. 59.

Currently not on view