Intersection of semicircle and parabola (Omar Khayyam)

Question

(Source: The History of Mathematics 7th Edition, David Burton)

I can't even get through question a.. Could someone give a hint? The only thing I can think of is the Pythagorean Theorem, but it turns out that that is not going to help me..

Answer 1

The parabola is given as $x^2 = ay$, so $y = \frac{1}{a}x^2$.

Now, $P$ is on the parabola, so its $y$-value, $PQ$, is such that $PQ = \frac{1}{a}(AQ)^2$; that is, $a(PQ) = (AQ)^2$.

Answer 2

Beautiful!

It's hard to resist the classics, whether they be Euclid, Archimedes, or, in the present case, Omar Khayyam; so here's my two ruppee's worth:

Fleshing out the details:

(a.) We have

$(AQ)^2 = a(PQ) \tag{1}$

by "construction"; that is, (1) is hypothesized as a given, since the curve $AP$ represents the parabola $x^2 = ay$; though it should by all rights be recalled that it is impossible to draw such a curve with a straightedge and compass;

(b.) Why is $\triangle AQP \sim \triangle PQC$? This is Euclid, pure and simple: since, again by "construction", PQ is perpendicular to AC (and this one can be done with compass and ruler), we have

$\angle AQP = \angle CQP = \dfrac{\pi}{2}; \tag{2}$

that is, they are both right angles. Furthermore, $\angle APC$ is also a right angle, since it subtends an arc of measure $\pi$; this arc is not shown in the figure, but it is the "lower" semicircular arc of the of circle diameter $AC$, the "upper" semicircular arc $APC$ of which is shown; the arc subtended by $\angle APC$ is in fact the reflection of arc $APC$ across the line $AC$. In affirming that $\angle APC = \pi/2$, we are invoking the classical Euclidean result that an inscribed angle is equal to half of the arc it subtends; see https://en.m.wikipedia.org/wiki/Inscribed_angle.

Since $\angle APC$ is a right angle, it follows that

$\angle APQ + \angle CPQ = \dfrac{\pi}{2}; \tag{3}$

by (2), we see that

$\angle APQ + \angle QAP = \dfrac{\pi}{2}, \tag{4}$

and

$\angle CPQ + \angle QCP = \dfrac{\pi}{2}; \tag{5}$

comparing (3), (4), (5) we conclude that

$\angle CPQ = \angle QAP; \tag{6}$

$\angle APQ = \angle QCP; \tag{7}$

(2), (6), and (7) show that $\triangle APQ \sim \triangle PQC$, since they share all the same angles. From this it immediately follows that

$\dfrac{AQ}{PQ} = \dfrac{PQ}{QC}, \tag{8}$

whence

$(PQ)^2 = (AQ)(QC), \tag{9}$

and, since

$QC = AC - AQ, \tag{10}$

we have

$(PQ)^2 = (AQ)(AC - AQ); \tag{11}$

finally, using

$AC = \dfrac{b}{a^2}, \tag{12}$

$(PQ)^2 = (AQ)(\dfrac{b}{a^2} - AQ), \tag{13}$

completing the demonstration of part (b);

(c.) by (1),

$a^2(PQ)^2 = (AQ)^4; \tag{14}$

multiplying (13) through by $a^2$:

$a^2(PQ)^2 = a^2(AQ)(\dfrac{b}{a^2} - AQ) = (AQ)(b - a^2(AQ)); \tag{15}$

substituting (14) into (15),

$(AQ)^4 = (AQ)(b - a^2(AQ)) \tag{16}$

or

$(AQ)^4 + a^2(AQ)^2 = b(AQ); \tag{17}$

and if we now grant that $AQ \ne 0$, we may cancel a factor of $AQ$ from each side of (17), yielding

$(AQ)^3 + a^2(AQ) = b, \tag{18}$

as per request. QED!

So wrote Omar:

"By the help of God and with His precious assistance, I say that Algebra is a scientific art. The objects with which it deals are absolute numbers and measurable quantities which, though themselves unknown, are related to "things" which are known, whereby the determination of the unknown quantities is possible."

"Whoever thinks algebra is a trick in obtaining unknowns has thought it in vain. No attention should be paid to the fact that algebra and geometry are different in appearance. Algebras (jabbre and maqabeleh) are geometric facts which are proved by propositions five and six of Book two of Elements."

"Ah, Love! could you and I with Him conspire/ To grasp this sorry Scheme of Things entire,/ Would not we shatter it to bits — and then/ Re-mould it nearer to the Heart's Desire!"

"Whereat some one of the loquacious Lot — I think a Sufi pipkin-waxing hot — "All this of Pot and Potter — Tell me then, Who is the Potter, pray, and who the Pot?"

from https://en.m.wikiquote.org/wiki/Omar_Khayy%C3%A1m