Euclid, Catoptrics (Mirrors) 30 (an unsound demonstration), with scholia©
translated by Henry Mendell (Cal. State U., L.A.)

Return to Vignettes of Ancient Mathematics
Return to Catoptrics, introduction
Go to prop. 29

Prop. 30: From concave mirrors being positioned towards the sun a fire will ignite.

(general diagram )

(diagram 1) Let there be a concave mirror, ABG, and sun, EZ, and center of the mirror, Q, (diagram 2) and, having been joined from some point, D, to center Q, let DQ be extended to B, (diagram 3) and let ray DG fall on and be reflected at K. In fact, it will be reflected above center Q. (diagram 4) For the angle at the circular-arc, P, is smaller than that at the circular-arc of a remainder, that by BGD.schol 56 (diagram 5) And let circumference AB be equal to BG, and from D let some other ray fall on it, DA. And so, it is obvious that ray AD, being reflected, falls onto K due to circular-arc AB being equal to BG. Similarly, it will be shown that all those falling from D to the mirror, taking out equal angles will fall together to the same point on BQ above Q.schol 57

Scholion (Heiberg 56): (diagram 6) For if we join G to Q, the angle by HGQ is equal to that by QGB. For they are angles of semicircles. Accordingly, that by HGD is smaller than that by QGB. Much more than that by DPB.

Note: the scholiast must mean that the angles by HGQ and QGB are horn angles, that is formed by the radius and the circular-arc of the circle (or semicircle). Therefore, the other angles are horn angles as well. This is also clear from the fact that lines BGH are not constructed and that the claims would be false.

But why is the reflected not joined to the center? Since, in fact, the sight-lines are reflected in equal angles, but the angle at P is to be smaller than that by QGB, it is thus necessary that to make the reflection equal to the angle at P taken away from the larger, that by QGB, be somewhere higher, as to K.

Scholion (Heiberg 57): (diagram 7) And so, it is obvious that if we join from Q to G and to A, it will be clear thus. Since two, KQG, are equal to two, KQA, and an angle to an angle, since to the circular-arcs, all will be equal to all. Thus, an angle, that by KAQ, is equal to an angle, that by KGQ. Again, since that by AQD is equal to that by GQD, due to the stated equal angles from four right-angles leave the remaining two equal, two, GQD, are equal to two, AQD, and an angle to an angle. Therefore, the angle by QAD is equal to that by QGD. And so, since a whole, that by QAB, is equal to that by QGB, since semiciricles fit onto one another, from which that by KAQ is equal to that by KGQ, therefore, a remainder, that by KAB, is equal to that by KGB. But that by KGB is equal that by DPH. Therefore, that by DGH is equal to that by KAB. But that by DGH is equal to that by DAR. Therefore, that by KAB is also equal to that by DAR.


top

(general diagram )

(diagram 1) Let there again be a concave mirror, ABG, and sun, DEZ, and from some point, E, through center Q, let there be EQB, (diagram 2) and from other points through D, Z, let there be DQG, ZQA. (diagram 3) Accordingly, we have shown earlier that the rays from A will fall together to themselves due to angles P, R being equal. For they are diameters. (diagram 4) But those from Z due to angles K, L, (diagram 5) and those from L to DG, due to angles N, X being equal. And it is clear that all these are reflected to themselves. (diagram 6 = gen. diag.) For they make semicircles from the center, and the angles of semicircles are equal. Therefore, the reflections come about through equal angles. And so, they are reflected to themselves. Therefore all will fall together from all points to the [rays] through the centerschol 58 and at the center. And so, when these rays become hot about the center fire gathers. Thus, flax placed here

Scholion (Heiberg 58): "To those through the center," that is, all at one and another point of BQ, but from-each-side, just as GKA (in the first diagram?).


top