THURSDAY, Jan. 5. 9 to 12.

SENIOR MODERATOR. Roman numbers.
JUNIOR MODERATOR. Arabic numbers.

i. THREE concentric circles are drawn in the same plane. Draw a straight line, such that one of its segments between the inner and outer circumference may be bisected at one of the points in which the line meets the middle circumference.

=

Let O be the common centre. Take any point P on the circumference of the middle circle; join OP and produce to Q making PQ OP. With centre Q and radius equal to that of the smallest circle describe a circle, and let one of the points in which it meets the outermost circle be R. Again, with centre Q and radius equal to that of the largest circle describe a circle, and let one of the points in which it meets the innermost circle be S. Then if R and S be taken properly RP and PS shall be in one straight line which line will also satisfy the required condition. Fig. 13.

Join

OR, QS, OS, QR.

Then ORQS is a parallelogram, because its opposite sides are equal, and from this, together with the property that diagonals of a parallelogram bisect each other, the truth of the proposition is obvious.

ii. If a quadrilateral circumscribes an ellipse, prove that either pair of opposite sides subtends supplementary angles at either focus.

Let ABCD be the quadrilateral; P, Q, R, T the points of contact of the respective sides, and S one of one foci. Fig. 14.

Join S with the angles of the quadrilateral and the points of contact.

=

By a property of the ellipse ASP TSA, applying this to the eight angles at S taken in pairs, we get

and since all the angles at S are together equal to four right angles the truth of the proposition is evident.

iii. If a polygon of a given number of sides circumscribes an ellipse, prove that, when its area is a minimum, any side is parallel to the line joining the points of contact of the two adjacent sides.

The polygon of minimum area and given number of sides circumscribing a circle is the regular polygon, and any side is therefore parallel to the line joining the points of contact of the two adjacent sides.

Hence by projecting this circle into an ellipse the truth of the proposition is obvious.

Or we may prove it thus: Fig. 15.

Let EA, AB, BF be three consecutive sides of the polygon. Then if the area is a minimum a small displacement given to AB while EA and DF remain fixed cannot alter the area of the polygon. Let AB be displaced to A'B'. The point of intersection of these lines will ultimately coincide with the point of contact P, and we have

▲APA' =▲BPB' ultimately;

.. AP. A'P sin APA' = BP. B'P sin BPB',

AP.A'P=BP.B'P;

.. AP= BP,

since A'P and B'P are ultimately equal to AP and BP respectively.

Hence the diameter which bisects chords parallel to EF meets the ellipse in P, and therefore the tangent at P is parallel to EF.

4. If the tangent at any point P of an hyperbola cut an asymptote in T, and if SP cut the same asymptote in Q, then SQ QT. See Fig. 16.

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If from any point T, two tangents are drawn to a conic, they subtend equal angles at either focus. One of these tangents in this problem becomes an asymptote, the other is TP. Therefore if SR be drawn parallel to QT, the angles TSQ, TSR are equal, and therefore the angles QST, QTS are equal, or SQ=QT.

5. Prove that the sum of the products of the first n natural numbers taken two and two together is

(n-1)n (n+1) (3n+2)

24

Since

(a+b+c+...)2= a2 + b2 + c2 + ... + 2 (ab + bc + ...),

which reduces to _(n − 1) n. (n + 1) (3n + 2)

24

6. The centres of the escribed circles of a triangle must lie without the circumscribing circle, and cannot be equidistant from it unless the triangle be equilateral.

It may be proved, as in Todhunter's Trigonometry (Art. 253), if Q be the centre of the circumscribing circle, P of any one of the escribed circles, and R, r, their radii, that PQ2= R2+2Rr,; whence it follows that PQ must be greater than R, and the three distances cannot be equal, unless the radii of the escribed circles are equal. The formulæ for these A A where a, b, c are the

A

radii are respectively S-a' S-b' S-c'

sides and ▲ the area of the triangle. Whence it follows that a, b, c are all equal.

This may also be proved independently of the proposition quoted from Todhunter.

vii. If perpendiculars be drawn from the angles of an equilateral triangle upon any tangent to the inscribed circle, prove that the sum of the reciprocals of those perpendiculars which fall upon the same side of the tangent is equal to the reciprocal of that perpendicular which falls upon the opposite

side.

Let ABC be the triangle, O the centre of the inscribed circle, and P be the point of contact of the tangent in question so that OP makes the angle 0 with AO, (0 being the centre of the circle). Fig. 17.

Then the inclinations of OB and OC to OP produced are

Hence remembering that in this case the radius of the circumscribing circle = 2 the radius of the inscribed = 2r, suppose perpendicular from A= 2r cos 0-r=r (2 cos 0-1),

0)+r=r{20
(2

·0)+1}

B = 2r cos (0) +r=r {2008 (−)+1}

C = 2r cos

3

COS

3

viii. Four equal particles are mutually repulsive, the law of force being that of the inverse distance. If they be joined

together by four strings of given length so as to form a quadrilateral, prove that, when there is equilibrium, the four particles lie in a circle. Fig. 18.

When there is equilibrium, the action of C on A : action of B on A

:: sin DAB sin DAC,

also action of D on B: action of A on B

:: sin ABC sin DBC;

:

.. action of Con A: action of D on B

:: sin DAB. sin DBC: sin DAC. sin ABC;

.. DB: AC :: sin DAB sin DBC: sin DAC sin ABC.

(It is to be observed that the action between A and B is the difference between the repulsive force and the tension, it therefore follows no law);

:: sin DAB. sin DBC: sin DAC sin ABC, sin ACB sin ADB:: sin DBC: sin DAC;

BO.sin DBC AO sin DAC

sin DBC sin DAC;

9. A heavy rod is placed in any manner resting on two points A and B of a rough horizontal curve, and a string attached to the middle point C of the chord is pulled in any direction so that the rod is on the point of motion. Prove that the locus of the intersection of the string with the directions of the frictions at the points of support is an arc of a circle and a part of a straight line.

Find also how the force must be applied that its intersections with the frictions may trace out the remainder of the circle. See Fig. 19.