BC, CA, and AB are given lines. Hence any vertex (C) is found by describing a segment of a circle upon PQ containing an angle equal to C, and with O as centre and OC as radius describing circle. Where these circles meet

is the required position of C.

Again in the triangle BOC when BC is a maximum. OC is a maximum, and is therefore a diameter of the circle OPQC. Then OPC is a right angle. Hence the maximum triangle of given species escribed to a given one is that whose sides are perpendicular to OP, OQ, OR.

COR. If the sides of the given escribed triangle be λ, μ, and v, and a, B, y the distances of O from P, Q, R, λa+uß+vy a minimum.

=

Hence required to find a point, given multiples of whose distances from three fixed points is a minimum when any two of the multiples are together greater than the third.

EXAMPLES.

1. If d denote the distance of the point O from the circumcentre H of the triangle ABC; prove that twice the area of the minimum triangle PQR is (R2d2) sin A sin B sin C.

[For join 40 and produce it to meet the circum-circle again in C';

join BC'.

Now since we have

but

LR=AOB-C=AOB-C' OBC' (Euc. I. 32),

=

2PQR=RP. RQ sin R=RP. RQ sin OBC".......(1) RP OB sin B and RQ=OA sin A.

Substituting these values in (1) and putting

OB sin OBC' OC' sin C',

=

2PQR AO. BO sin A sin B sin OBC'

=AO.OC' sin A sin B sin C

=(R2d2) sin A sin B sin C.]

NOTE.-If the point O is on the circum-circle Rd and the area of the triangle vanishes, hence if from any point on the circum-circle of a triangle perpendiculars be let fall upon the sides their feet lie in a line. This is termed a Simson Line of the triangle, and the collinearity of the points admits of an easy direct proof.

2. If the pedal triangle PQR of a point O is constant in area the locus of the point is a circle.

[Concentric with the circum-circle by the equation of Ex. 1.] 2a. The theorem holds generally for a polygon.

3. Having given of a triangle the base c, and ab sin (C-a) where a is a given angle, find the locus of the vertex.

[In Ex. 1 we have

2PQR AO. BO sin A sin B sin (AOB-C)

∞ AO. BO sin (AOB-C),

and the locus of O is in that case a circle. Hence in the triangle AOB we have the data in question; therefore the locus of the vertex is a circle concentric with H.]

4. To inscribe a quadrilateral of given species PQRS in a given quadrilateral ABCD.

Find the point 0, of the triangle PQR of given species inscribed in a given one, viz., that formed by three of the sides, AB, BC, CD of

the quadrilateral. Similarly find O2 of the triangle PQS inscribed in a given one. Now by Art. 19, since the species of each of the triangles O,PQ and O,PQ is given, we have LO,PO2=O2PQ ~ O1PQ= a known quantity; therefore the point P is determined.

5. To escribe a quadrilateral ABCD of given species to a given one PQRS.

[Take any quadrilateral abcd of the same species as ABCD. Inscribe in it by Ex. 4 a quadrilateral pqrs of the species PQRS. It is obvious that SPA=spa, since the figures are similar, hence the problem reduces to drawing lines in known directions through P, Q, R, S.

Otherwise thus:

Upon a pair of opposite sides PQ and RS describe segments of circles containing angles equal to B and D respectively. Find a point M such that the arcs PM and QM subtend angles equal to ABD and CBD respectively. Similarly find N such that CDN and ADN may be equal to the known segments of the angle C. Join MN; where it meets the circles in B and D are two of the required vertices of the quadrilateral ABCD.]

6. To escribe a square ABCD to a quadrilateral PQRS

=

[By Ex. 5 or simply thus :-Join PR and let fall a perpendicular from Qupon it. Make QS' PR. SS' is a side of the required square. This construction depends upon the property that any two rectangular lines terminated by the opposite sides of a square are equal to one another (Mathesis).

7. From any point P on the base of a triangle perpendiculars PX and PY are drawn to the sides, find the locus of the middle point N of XY.

[Bisect CP in M, join MX, MY and MN. It is easy to see that MXY is an isosceles triangle of given species, each of its base angles being the complement of C; and since its vertices X, M, Y move on fixed lines, any point N invariably connected with it describes a line. By taking P to coincide alternately with A and B the locus is seen to be the line joining the middle points of the perpendiculars from the extremities of the base of the triangle ABC.]

8. The sides of the pedal triangle PQR are in the ratios a. AO:b. BO: c. CO.

[For QR AO sin Ax a. AO, etc.]

9. Extension of Ptolemy's Theorem.-If the three pairs of opposite connectors of four points be denoted by a, c; b, d; &, &' to prove the relation

828'2=a2c2+b2d2-2abcd cos (0+0'),

where 0+ is the sum of a pair of opposite angles of the quadrilateral.

[Let A, B, C, O be the four points. From any one of them let fall perpendiculars OP, OQ, OR on the sides of the triangle ABC formed by the remaining three; then since

PQ2 QR2+RP2-2QR. RP cos R,

=

substituting for PQ, QR, RP the values in Ex. 8, and reducing, the above equation follows at once (M'Cay).]

9a. What does this theorem reduce to for the quadrilateral ABCP in the figure of Ex. 7? Deduce the relation of Art. 3, Ex. 5, as a further particular case.

10. A variable circle passes through the vertex of an angle and a second fixed point; find the locus of the intersection of tangents at the extremities of its chord of intersection.

11. If a, ẞ, y denote the distances of any point 0 from the sides of a triangle; to prove that

where S and S' are the rectangles under the segments of a variable chord through O* of the circum-circles of ABC and of the pedal triangle of the point 0 (M'Vicker).

[In Ex. 1 let K be the point where RO meets the circum-circle of PQR; then y=S'|OK=S' sinP/B sinOQK.

But sinOQK=sin(4+P)=sinBOC; .. By=S' sinP/sin BOC. Also a=OB. OC sin BOC/a, therefore aẞy=S'. OB. OC sinP/a. Again OB=RP/sinB, etc. therefore by substitution

12. In the particular cases when O coincides with the in- or excentres of the triangle ABC, the formula in Ex. 11 reduces to 82=R2-2Rr or d12=R2+2Rr, etc.

24. Theorem.—When three points P, Q, R are taken collinearly on the sides of a triangle, the circles circumscribing the four triangles QRA, RPB, PQC, ABC meet in a point.

This theorem may be easily proved directly, but it is obviously a particular case of Art. 19, for the circles QRA, RPB, PQC meet in a point 0 (Art. 19) such that COA = Q-B, which in this case is 180-B; therefore, etc. Euc. III. 22.

The transversal PQR to the sides of ABC is the limiting case of a triangle inscribed in ABC, the angles at P and

*The constant rectangle under the segments of a variable chord of a circle passing through a fixed point has been termed by Steiner the Power of the Point with respect to the circle.

D