The great Persian poet, Omar Khayyam (circa 1050-1130), found a geometric soluti
ID: 3004104 • Letter: T
Question
The great Persian poet, Omar Khayyam (circa 1050-1130), found a geometric solution of the cubic equation x^3 + a^2x = b by using a pair of intersecting conic sections. In modem notation, he first constructed the parabola x^2 = ay. Then he drew a semicircle with diameter AC = b/a^2 on the x-axis, and let P be the point of intersection of the semicircle with the parabola (see the figure). A perpendicular is dropped from P to the x-axis to produce a point Q. Complete the details in the following proof that the x-coordinate of P, that is, the length of segment AQ, is the root of the given cubic. AQ^2 = a(PQ). Triangles AQP and PQC are similar, so that AQ/PQ = PQ/QC' or PQ^2 =AQ (b/a^2 - AQ). Substitution gives AQ^3 + a^2AQ = b.Explanation / Answer
a) Let AQ be x.
P lies on the parabola x2=ay. So the coordinate of P can be expressed as (x,x2/a). Thus length of PQ becomes x2/a.
Thus PQ=AQ2/a. Or, AQ2=a(PQ)
b) In triangles AQP and PQC, angle (APC) = 90° (as APCQ is a semi circle). Let angle (PCQ) be z°. So angle (PAC) becomes (90-z)°. Also PQ is a perpendicular from P to AC. Thus angle (PQC) becomes 90°. So in triangle PQC, angle PCQ is z°, angle PQC is 90°. Thus angle QPC becomes (90-z)°.
So angle (APQ)= angle (APC) - angle (QPC) = 90-(90-z)= z°
In triangle APQ, angle (APQ)=z°, angle (PQA)=90°. Thus angle (PAQ)= 180-90-z= (90-z)°
Thus in triangle AQP and PQC,
(I) angle (PAQ)= angle (QPC)
(II) angle (APQ) = angle (PCQ)
(III) angle (AQP) = angle (PQC)
Thus these triangles are similar. Thus AQ/PQ = PQ/QC
QC= AC - AQ= b/a2-AQ
Thus PQ2= AQ*QC = AQ(b/a2-AQ)
Now PQ=AQ2/a, so PQ2= AQ4/a2
But PQ2= AQ(b/a2-AQ)=AQ4/a2
Or,(b/a2-AQ) = AQ3/a2
Multiplying both sides by a2 we get,
b-a2AQ=AQ3
Thus AQ3 + a2AQ = b ( Proved)
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