CHAPTER III. RECENT DEVELOPMENTS OF POINT O THEOREM. SECTION I. THE BROCARD POINTS AND CIRCLE OF A TRIANGLE. 27. Brocard Points Q, Q.-In Art. 20 if the inscribed triangle PQR is similar to ABC and PA, Q=B, R=C, then BOC=A+P = 2A,similarly COA = 2B and AOB = 2C; therefore is the centre of the circum-circle. Secondly, let PB, QC and R=A. Then similarly and BOC=A+PA+B=π-C; = COA B+Q=B+C=π-A, T-C. Thirdly, let P=C, Q=A_and_ R= B. It follows as in the last case that BOC=π-В, COA = π- -C and AOB -A. = Thus we see that a triangle PQR similar to a given. one may be inscribed in the latter in three different ways; and that the point O in each case may be found as in the general method by describing segments of circles on two of the sides containing given angles. In the second and third positions the points of intersection of the circles are usually denoted by the letters and . They are termed the Brocard Points of the triangle ABC, and are distinguished as Positive (N) and Negative ('). 28. Brocard Angle (w).-Since BOC is the supplement of C, QBC+QCB-C or QBC=QCA. For a similar reason QCA = ΩΑΒ, hence QBC QCA =QAB=w = = w (say). The angle w is called the Brocard Angle of the triangle ABC. We may remark that the angle subtended at by the base c is the supplement of B, the angle at the right extremity of AB, and at equal to the supplement of A, the angle at the other extremity of AB. The same relations hold for the sides a and b; hence the names Positive and Negative Brocard points. The value of w as a function of the sides or angles is thus found. Let x, y, z denote the lengths of AN, BO and CO respectively. Then in the triangle BNC It is proved in like manner for 'that 'CB-QAC='BA, and that the value of these angles is also given by (1). .(1) or cot wcot A+ cot B+cot C.........(2) 2. The distances of 2 from the sides of ABC are 2R sin2w, 2R sin, 2R sin2; and of N', 2R sin, 2R sin2, 2R sin2. α [For let the distances of a=y sin w= α The ratios of the distances are evidently as follows: and and also a: B: y=c2a: a2b : b2c, a': ẞ': y' = ab2 : bc2 : ca2, aa'=BB'=yy'=4R2 sin*w.] C * Or Trilinear Co-ordinates of the points with respect to the triangle, which is also called the Triangle of Reference. 3. AD is the bisector of the angle A of a triangle ABC, and w1, w2 the Brocard angles of the triangles ABD and ACD respectively; prove that cot w1+cot w2=2 cosec A+cot A + cot w, with similar expressions for the triangles formed by the bisectors of the angles B and C. 4. If W1 and W2 denote the Brocard angles of the triangles CAD and BAD, where AD is the median to the side BC, with similar expressions for the medians BE and CF. 5. Hence prove that cot w1+cot w2+cot w;=cot w2+cot w1+cot wε 2(a2+b2+c2) A and cot wi = 6. If ABC is divided as in the previous exercises by the symmedians, prove that Σ(b2+c2) (cot w1 — cot w2) = 0. 7. and ' are Brocard points of their pedal triangles PQR and PQR. (Euc. III. 21.) 8. The triangles PQR and P'Q'R' are equal in area. similarly therefore QPQ: NBC=NP2 : NB2=sin2; QQR: NCA=NRP : NAB=sin2; PQR=P'QR' ABC. sin2w.] = 9. The Brocard points are equidistant from the circum-centre. [By Ex. 8 and Art 23, Ex. 1.] 10. If A', B', C' be the points of intersection of the pairs of lines y, z: z, x': x, y', prove that the six points A', B', C', O, N, N' lie on a circle. [For the triangles BCA', CAB' and ABC" are isosceles and similar, their base angles each being equal to w, hence OA′, OB′, OC' are the bisectors of their vertical angles. In the quadrilateral ΟΩΩ' Α' we have ON=ON' and OA' the bisector of the angle NA'N'; therefore O is a point on the circum-circle of NA'N', and the quadrilateral is therefore cyclic. Similarly B' and C' are on the circum-circle of the triangle ONN'.] DEF. This is called the Brocard Circle, and A'B'C' the First Brocard Triangle of ABC. 11. To find the distance of the Brocard points from the circum centre (09-ON′ = 8). [By Art. 23, Ex. 1, 2PQR (R2-62)sinA sinB sinC, but (Ex. 8) hence PQR=ABC sin2w=2R2sinA sinB sinC sin2w, R2-82-4R2sin2w or 8=R1-4sin w.] 12. The angle subtended at the circum-centre by 'N=2w. (By Ex. 10 and Euc. III. 22.) 13. To find the distance ′ between the Brocard points. [Since ONN' is an isosceles triangle, NN'--20N sin w=2R sin o√1-4 sin'w, by Ex. 11.] 14. The diameter of the Brocard circle is equal to R sec w 1-4 sin2w. [For it equals S/sin 2w; therefore, etc.] 15. The altitudes of the similar isosceles triangles BCA', CAB', ABC" are equal to the distances of the symmedian point (K) from the sides. therefore, etc., by Art. 28, (1).] 16. The circle on OK as diameter is the Brocard circle. [For AA' is parallel and 04′ perpendicular to BC, hence OK subtends a right angle at '; similarly for the points B and C'"'; therefore, etc.] 17. Brocard's first triangle is Inversely Similar to ABC; ie, by rotation in the plane of the paper their sides cannot be brought into a position of parallelism with each other. [For BC" subtends equal angles at A and A, but KB and KC are respectively parallel to CA and AB, and therefore contain an angle ; similarly the angles B and C are equal to B and C.] 18. Having given the base e and Brocard angle e of a triangle ABC, find the locus of the vertex (Neuberg). [Let p be the median CZ and & the angle between it and PZ. om 12 cotu=(y2+H+ Y. CR and 2+H=12+2p2, we have |