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In the Wikipedia article on lever mechanics from the Archimedes codex where

"The first proposition states:

The area of the triangle ABC is exactly three times the area bounded by the parabola and the secant line AB."

enter image description here

The wikipedia proof ends prematurely in my view:

"In other words, it suffices to show that $EF:GD = EH :JD$. But that is a routine consequence of the equation of the parabola. Q.E.D."

I cannot see this is obvious - in fact it seems counterintuitive since the parabola equation is quadratic while the other dimensions are linear.

Please could you see if there is a explanation for the Q.E.D bit.

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  • $\begingroup$ You can use analytic geometry (boring but straightforward). Or you can see here (p. 21) how Archimedes himself proved that. $\endgroup$ – Aretino Jul 7 '17 at 17:43
  • $\begingroup$ It says on p24, "From the geometry of Figure 4.3, Archimedes proved the following result in The Method:4" In reference 4,p16 it then refers to conic sections. So this seems a paper chase rather than an explanation. Routine and straight forward these proofs may be after being understood. Beforehand they are confusing to me. $\endgroup$ – rupert Jul 7 '17 at 21:34
  • $\begingroup$ I've tried to fill in the argument as Archimedes would make it. The steps may be "routine", but you are right: there are more of them than wikipedia might lead you to believe. $\endgroup$ – Edward Porcella Jul 10 '17 at 20:07
  • $\begingroup$ The statement that it's a routine consequence does not mean that it's an obvious consequence, but only that it can be routinely verified with a bit of elbow grease. The fact that $31\times 43 = 1333$ may not be obvious, but a fourth-grader knows how to show it. However, the method by which Archimedes showed it would probably look foreign to us now. $\endgroup$ – Michael Hardy Sep 27 '17 at 20:43
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An analytic proof. Let $y=ax^2$ be the equation of a generic parabola, and $A=(x_1,ax_1^2)$, $B=(x_2,ax_2^2)$ any two points on it. The equation of tangent $BC$ is then $y-ax_2^2=2ax_2(x-x_2)$ and $C=(x_1,2ax_1x_2-ax_2^2)$. Point $D$ is the midpoint of $AC$, thus: $D=(x_1,ax_1x_2+a(x_1^2-x_2^2)/2)$.

Let now $x$ be the abscissa of $E$. We have immediately: \begin{align} & E=(x, ax(x_1+x_2)-ax_1x_2),\quad F=(x, ax^2),\quad H=(x, 2axx_2-ax_2^2), \\[8pt] & G=\left(x, {1\over2}ax(x_1+3x_2)-{1\over2}ax_2(x_1+x_2)\right). \end{align}

We get then: $$ {EF\over EH}={ax^2-ax(x_1+x_2)+ax_1x_2\over2axx_2-ax_2^2-ax(x_1+x_2)+ax_1x_2} ={a(x-x_2)(x-x_1)\over a(x-x_2)(x_2-x_1)}={x-x_1\over x_2-x_1} ={GD\over BD}, $$ but this is the same as $EF:GD=EH:JD$, because $JD=BD$.

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    $\begingroup$ much appreciated. $\endgroup$ – rupert Jul 8 '17 at 21:54
  • $\begingroup$ $+1,$ but I wonder why you include the constant $a$ instead of just letting $a=1.$ There is no parabola whose equation is not $y=x^2$ if you locate and scale the two axes suitably. $\qquad$ $\endgroup$ – Michael Hardy Oct 1 '17 at 14:30
  • $\begingroup$ @MichaelHardy You are right, but leaving the $a$ there costs very little in terms of algebra and can be more convincing if the OP is not an "expert" about parabolas. In other words: explaining why I can set $a=1$ can be more complicated than leaving $a$ in place. $\endgroup$ – Aretino Oct 1 '17 at 14:47
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parabolic segmentThe proportion Archimedes needs here rests on $Props. 4$ and $5$ of his Quadrature of the Parabola. In the posted figure, bisect $AB$ at $K$ and draw $KP$ parallel to $AC$, cutting the parabola at $M$, and from $F$ draw $FL$ parallel to $AB$, meeting $JB$ at $N$. $Prop. 4$ proves that $$\frac{BK}{KE}=\frac{EG}{FG}$$For since $MK$ is parallel to the axis, and $AK$=$KB$, then $BK$ is an ordinate to diameter $MK$ and $$\frac{MK}{ML}=\frac{BK^2}{FL^2}$$ This is the fundamental proportion of the parabola (see Apollonius, Conics, I, 20). Thus by Archimedes' $Prop. 2$, $MK$=$MP$. Therefore, $M$ lies on $JB$, so that by similar triangles $$\frac{MK}{ML}=\frac{BM}{MN}$$ and $$\frac{BK}{FL}=\frac{BK}{EK}=\frac{BM}{MG}$$ Therefore by substitution$$\frac{BM}{NM}=\frac{BM^2}{MG^2}$$ Hence $BM, MG, MN$ are in continued proportion:$$\frac{BM}{MG}=\frac{MG}{MN}=\frac{BM+MG}{MG+MN}=\frac{BG}{NG}$$ But by similar triangles$$\frac{BG}{NG}=\frac{EG}{FG}$$and$$\frac{BM}{MG}=\frac{BK}{KE}$$Therefore, $$\frac{BK}{KE}=\frac{EG}{FG}$$$QED$ for $Prop. 4$. From this, and the fact that $AD=DC$ and $EG=GH$, Archimedes proves $Prop. 5$: $$\frac{BE}{AE}=\frac{HF}{EF}$$Since from $Prop. 4$$$\frac{BK}{BK-KE}=\frac{EG}{EG-FG}$$then$$\frac{BK}{AE}=\frac{EG}{EF}$$whence $$\frac{2BK}{AE}=\frac{2EG}{EF}$$so that$$\frac{AB}{AE}=\frac{EH}{EF}$$hence$$\frac{AB-AE}{AE}=\frac{EH-EF}{EF}$$or$$\frac{BE}{AE}=\frac{HF}{EF}$$which proves $Prop. 5$. But by similarity$$\frac{BE}{AE}=\frac{BG}{DG}$$Therefore$$\frac{HF}{EF}=\frac{BG}{DG}$$and so$$\frac{HF+EF}{EF}=\frac{BG+DG}{DG}$$or$$\frac{EH}{EF}=\frac{DB}{DG}$$Alternating this gives the needed proportion$$\frac{EF}{DG}=\frac{EH}{JD}$$since $JD=DB$.

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