angle that lies in one of the lines AF, AG coincides with the leg of another triangle that lies in the other of these two lines; but this exception is only apparent, for the value of n in these two triangles is the same as that of n in the two triangles A B E, ABE', .. when the hypothenuse of one triangle coincides with a leg of another triangle, the value of n, that corresponds to A in one of the triangles, is the same as in the other triangle. "Cor. We can now easily find the value of m that corresponds to the vertical angle of an isosceles triangle whose base-angles are represented by n (or to which n corresponds). BI a BI "For let AI be drawn from the base-angle A of the isosceles triangle ABC, perpendicular to the opposite side BC, meeting it in I, then from what has been shown we get CI= n b, or (since b 2na) CI=2n2 a, .. BI-a-CI=a(1-2 n2), and (since by (1) of prop. 2, m = 1-2 n2) we get m= which can also be easily obtained from other considerations. And since n corresponds equally to the base-angles of all the isosceles triangles represented by ABC, A B'C', and since m-1-2 n2, it follows that all isosceles triangles whose base-angles are equal will have equal values of m corresponding to their vertical angles. "Prop. 4. If there are two right-angled triangles, such that a leg of the one divided by its hypothenuse gives the same quotient as a leg of the other divided by its hypothenuse, then shall the angle included by the leg and hypothenuse of the one triangle be equal to the angle included by the leg and hypothenuse of the other triangle. “Let ABC, DFE be two triangles right-angled at C and E, n; then shall the angle B A C equal the angle such that AC AB DE DF FD E. For on the longer leg D E of the one take D C equal to A C, and through C" draw CG perpendicular to DE, meeting D F in G, then shall the triangles ABC, D G C be identical; for since the right triangles DFE, DG C' have the angle D common, we B. I., .. 4, DB' = A C, we get D G = - AB. If we take C'B' = CB, and draw a right line from D to B', the triangles A B C, D B'C' are identical, Sim., p. A B, and of course D B' D G, which cannot be unless B' falls on G; for of the base-angles D B'G, D G B, one is acute and the other obtuse, and the same holds true whether B' is within or without the triangle DFE; hence we cannot have DB' = D G unless B' coincides with G, Sim., B. I., p. 19. Hence, since the triangles A B C, D B'C' are identical, and that B' falls on G, the triangles ABC, D G C are identical, and the angles BA C, FDE are equal, as required. m= BI AB m= BI B'K B'K "Cor. If we draw A K at right angles to B'C', meeting it in K (see fig., prop. 3), then since the base-angles of the isosceles triangles represented by ABC, A B'C' are equal, we have, by cor. to prop. 3, and hence the vertical angles of the isosceles triangles are equal; and the same holds true whether the perpendiculars AI, AK fall within the triangles (as in the figure) or without them; for when the perpendiculars fall without the triangles, the equality shows that the supplements of the vertical angles of the triangles are equal, and consequently the vertical angles are equal; and it is evident, by cor. to prop. 3, that the perpendiculars will both at the same time fall within or without the triangles; the case when 0, or BI=0, B'K=0, is too evident to require any expla nation, for the vertical angles are evidently right. Hence the angles A BE, AB'E', the halves of the vertical angles of the isosceles triangles, are equal; hence it follows that all those right-angled triangles which have an acute angle common, or equal, have their other acute angles also equal. Hence (see fig. 3) from the right triangles ABE, ABE, having their angles B, B' equal, we get that is, if we have two (or more) right triangles which have an acute angle common or equal, then if we divide the leg of any one of them which is opposite to the (equal) angle by the leg adjacent to the angle, the quotient will equal the corresponding quotient obtained in like manner from any other (one) of the triangles; and the converse of what is here affirmed is also easily proved to be true in a manner very analogous to that given in proving the proposition above. BE = AB B'E and since AE "Prop. 5. The sum of the acute angles of a right triangle equals a right angle, and the sum of the squares of the legs equals the square of the hypothenuse. "Let ACB be the triangle, having the angle C right; from C draw CD, at right angles to the hypothenuse, meeting it at D; hence cor., prop. 4, since the right triangles ACD, ACB have the angle A common, their other acute angles A CD and B are equal; also since the (right) triangles B CD, C A D B ABC have a common angle, B, their other acute angles BC D and A are equal. Hence the sum of the angles A and B is equal to the sum of the angles BCD and ACD, which compose the right angle A CB, and of course the sum of the angles A and B is equal to a right angle; and consequently the sum of all the angles of the triangle ACB is equal to two right angles. Again, the right triangles ACB, ACD having the common angle A, by prop. 3, give the equality or A C2 AB. AD; and in the same way we get from the triangles ACB, BCD, BC2=AB. BD; and consequently A C2 + B C2 = A B2, as required. AC AB AD "Cor. Since the right triangles A CD, BCD have the angles CAD, BCD equal, they (by the cor. to prop. 4) give the equality or CD2 = AD. BD. CD AD BD CD' "Prop. 6. The sum of the angles of any triangle is equal to two right angles. "Let A CB denote any triangle; and suppose that the angle Cis not less than either of the other angles of the triangle, and that the perpendicular CD is drawn from C to the opposite side AB; then, Sim., p. 17, B. I., CD will fall within the triangle A CB. Hence, since the triangles A CD, BC D B are right-angled at D, by the last prop. the sum of the acute angles A and ACD of the first of these triangles is equal to a right angle; and in the same way the sum of the acute angles B and B CD of the second triangle is equal to a right angle; but the sum of the acute angles of these triangles equals the sum of the angles of the triangle A CB; consequently, the sum of the angles of the triangle A CB is equal to two right angles, as required. "In conclusion, we will remark that the relation of what has been done to the doctrine of similar triangles and the science of trigonometry is too evident to require any comment." Two hundred and ninety-fifth Meeting. May 4, 1847. - MONTHLY MEETING. The VICE-PRESIDENT in the chair. Professor Peirce announced that he had continued and nearly completed his researches into the irregularities of motion exhibited by Uranus, and was more strongly than ever of the opinion that they were not to be attributed to the influence of the newly discovered planet Neptune. He had obtained several possible solutions of the problem, which are different from those of Leverrier and Adams, and which are published in a communication to the Boston Courier, dated April 29, 1847, and which he now proposes to lay before the Academy. "The problem of the perturbations of Uranus admits of three solutions, which are decidedly different from each other, and from those of Leverrier and Adams, and equally complete with theirs. The present place of the theoretical planet, which might have caused the observed irregularities in the motions of Uranus, would, in two of them, be about one hundred and twenty degrees from that of Neptune, the one being behind, and the other before, this planet. If the above geometers had fallen upon either of these solutions instead of that which was obtained, Neptune would not have been discovered in consequence of geometrical prediction. The following are the approximate elements for the three solutions at the epoch of Jan. 1, 1847. 0.0001187 In each of them (the mass of the sun being unity) The mass is "The period of sidereal revolution is double that of Uranus. It will be observed that the mean distance in all these cases is the same with that of Neptune, and that, in the first of them, the present direction The first of these solutions is corrected from the one which was published in is not more than seven degrees from it; and in another solution which I have obtained, the present direction is almost identical with Neptune's. But the coincidence fails in a most important point; for, whereas Walker and Adams both demonstrate, from incontrovertible data and a simple but indisputable argument, that the new planet cannot be more than ninety degrees from its perihelion, either of these two latter geometrical planets would now be in aphelion and at much too great a distance from the sun. "All my attempts to reconcile the observed motions of Neptune with the assumption that it is the principal source of the unexplained irregularities in the motions of Uranus, have been frustrated. Whatever orbit is attributed to this planet in my analysis, whether Walker's, or Valtz's, or Encke's, or Adams's, or any other which I can suppose, and which is not unquestionably irreconcilable with observation, and whatever may be supposed to be its mass, I cannot materially diminish the amount of residual perturbation, but leave it full as great as it was previously to Galle's discovery. Notwithstanding my repeated examinations, it would be presumptuous in me to claim for my investigations a freedom from error which the greatest geometers have not escaped, especially in the face of the vastly improbable conclusion to which my analysis tends; namely, that the influence of the new planet is wholly different from that demanded by the problem whose solution led to its discovery. It may, however, be asked whether the attraction of Uranus might not be exhibited in the motions of Neptune, in such a way as to modify the orbit deduced from observation, and thus reconcile it with theory; but this question cannot be answered without further investigation." Professor Peirce stated that the above solutions were not to be regarded as actual solutions, but merely as theoretical and possible; that is, if a planet had moved in either of the above orbits, the perturbations which it would have produced in Uranus would have been precisely those which have been manifested. But the influence of the planet Neptune has been wholly disregarded in obtaining these solutions, precisely because the nature of that influence must remain unknown, a previous communication to the Boston Courier, and which was vitiated by an oversight in the date for which the computations had been made. |
