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Épreuve de section européenne

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Baccalauréat, toutes séries Session de juin 2008

Épreuve de section européenne

Squaring the circle

bO

bP bRbH bT

b

Q

bS

b l

M

bN

bK

bL b

C

bD

LetP QRbe a circle with centreO, of which a diameter isP R. BisectP OatHand letT be the point of trisection of OR nearer R. Draw T Q perpendicular toP R and place the chord RS=T Q.

Join P S, and draw OM and T N parallel to RS, with M and N on RS. Place a chord P K=P M, and draw the tangentP L=M N.

Join RL, RK and KL. Place C on RK such that RC = RH. Draw CD parallel to KL, meetingRLat D.

Then the square onRDwill be equal to the circleP QR approximately.

For

RS2 = 5 26d2 where dis the diameter of the circle. Therefore

P S2= 31 26d2.

ButP Land P K are equal to M N and P M respectively. Therefore P K2 = 31

144d2 and P L2= 31 324d2. Hence

RK2=P R2−P K2 = 113

144d2 and RL2 =P R2+P L2= 355 324d2. But

RK

RL = RC RD = 3

2 r113

355

and RC = 3

4d. Therefore

RD= d 2

r355 113 =r√

π, very nearly.

Note.-If the area of the circle be 140,000 square miles, then RD is greater than the true length by about an inch.

FromJournal of the Indian Mathematical Society, 1913, by Srinivasa Ramanujan

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Questions

1. Explain each step of the proof.

2. Is this construction exact ? If not so, what information is given in the text about its preci- sion ?

2007-04 – Squaring the circle

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