a2a + ß3b + y2c = a (PO2 + AO2 – 2 PO. AO. cos POA) + c (PO2 + CO2 – 2PO.CO. cos POC) = (a+b+c) PO + a. 40+b. BO2 + c. CO -2PO(a.AO.cos POA+b.BO.cos POB+c. CO. cos POC) COS A B. C (6 B (coscose-cos sin (+0) + cos sin (0-2)} 2 (a + b + c) PO2 + abc 2 Next let P be without the plane ABC; draw PQ perpendicular to the plane, and join QA, QB, QC', QO, then PA2 = QP2 + QA2, PO2 = QP2 + QO°; and a3a+ß3b+y3c = a(QP2+QA2)+b(QP2+QB2)+c(QP2+QC2) = (a + b + c) QP2 + a. QA2 + b. QB + c. QC" (a + b + c) QP2 + (a + b + c) QO2 + abc (by the first case) = (a + b + c) PO2 + abc, and is therefore invariable, since P and O are two fixed points. ST JOHN'S COLLEGE. DEC. 1831. (No. II.) 1. Ir four magnitudes of the same kind are proportionals, the greatest and least of them together are greater than the other two together. 2. If two straight lines meeting one another be parallel to two others that meet one another, and are not in the same plane with the first two; the first two and the other two shall contain equal angles. 3. The sum of the perpendiculars from any point in the base of an isosceles triangle is equal to a line of fixed length. 4. To find a point in the side or side produced of any parallelogram, such that the angle it makes with the line joining the point and one extremity of the opposite side, may be bisected by the line joining it with the other extremity. 5. The lines which bisect the vertical angles of all triangles on the same base and with the same vertical angle, all intersect in one point. 6. If a semicircle be described on the hypothenuse AB of a right-angled triangle ABC, and from the centre E the radius ED be drawn at right angles to AB, shew that the difference of the segments on the two sides equal twice the sector CED. 7. The locus of the centres of the circles which are inscribed in all right-angled triangles on the same hypothenuse is the quadrant described on the hypothenuse. 8. Of all the angles which a straight line makes with any straight lines drawn in a given plane to meet it, the least is that which measures the inclination of the line to the plane. 9. Find the sine of the inclination to each other of two straight lines whose equations are given. 10. Find the length of the perpendicular from the origin of co-ordinates upon the line whose equation is a (x − a) + b (y — b) = 0, and the part of the line intercepted between the co-ordinate axes. 11. The equation to a circle is y2+ x2 = a (y + x); what is the equation to that diameter which passes through the origin of co-ordinates? 12. A side of a triangle being assumed as the axis of x, the equations to the equations to the other sides are y = ax + b, and y = d'x; determine the sides and angles of the triangle. 13. If through any point of a quadrant whose radius is R, two circles be drawn touching the bounding radii of the quadrant, and r, be the radii of these circles, shew that rr'′ = R2. SOLUTIONS TO (No. II.) 1. EUCLID, Prop. 25. Book v. 2. Euclid, Prop. 10. Book xI. 3. Let D (fig. 12) be a point in the base AB of the isosceles triangle ABC; draw DE, DF perpendicular to AC, BC respectively; then DE + DF = AD. sin A + DB . sin B = (AD + DB) sin A AB. sin A equal the perpendicular from B upon the side AC, and is therefore constant wherever D be taken in AC. 4. Let ABCD (fig. 13) be the parallelogram; with centre B and radius BA describe a circle cutting DC in the points E, F; join AF, BF; then BF = BA and ▲ BFA = L BAF = L AFD, Similarly if or ▲ DFB is bisected by the straight line AF. AE, BE be joined, ▲ DEB is bisected by AE. Also if with centre A and radius AB a circle be described cutting DC in the points G, H, the angles CGA, CHA will be bisected by the straight lines BG, BH respectively. 5. Let ACB (fig. 14) be one of the triangles; about it describe the circle ACBD; then since the vertical angle is constant, the vertices of all the triangles will be in the circumference ACB. Bisect the arc AB in D, and join CD; then since arc AD = arc DB, ▲ ACD = L BCD, and CD bisects the angle ACB; hence the line which bisects the angle ACB will always pass through the fixed point D. = 6. Let ABC (fig. 15) be the right-angled triangle, ADCB the semicircle described upon the hypothenuse AB; then since AE EB, A AEC = A BEC; and segment on AC segment on BC = (segment on AC + ▲ AEC) – (segment on BC + ▲ BEC) = sector AEC = sector AEC - sector BEC sector BEC = (sector AED + sector DEC) - (sector BED – sector DEC) sector DEC) = 2 sector DEC since quadrant AED = quadrant BED. 7. Let 0 (fig. 16) be the centre of the circle inscribed in the AACB; join OA, OB; then ▲ AOB =T · (OAB + OBA) and is constant. Hence the locus of the point O is a segment of a circle containing an angle therefore the angle subtended by AOB at the centre E = π ~ C. therefore AOB is the quadrant described upon the hypothenuse AB. 8. Let P (fig. 17) be any point in the line AP which meets the plane in the point 4; draw PM perpendicular to the plane; join AM, and through A draw any other line AQ in the plane; draw PQ perpendicular to AQ, and join QM; then if 0, 0', be the inclinations of AP to AM, AQ respectively, but PQ2 = PM2 + MQ since PMQ is a right angle, or PQ is greater than PM; therefore sin is less than sin' and is less than '. Now 0 measures the inclination of the line AP to the plane, which is therefore less than the inclination of the line AP to any other line AQ drawn to meet it in that plane. 9. Let y = ax + b, y a' x + b' be the equations to the two straight lines; then if 0, 0' be the angles which they respectively make with the axis of x, |