90. To find the angular radius of the small circle described so as to touch one side of a given triangle, and the other sides produced.

B

C

Let ABC be the triangle; and suppose we require the radius of the small circle which touches BC, and AB and AC produced. Produce AB and AC to meet at A'; then we require the radius of the small circle inscribed in A'BC, and the sides of A'BC are a, π-b, π-c respectively. Hence if r ̧ be the required radius, and s denote as usual 1⁄2 (a+b+c), we have from Art. 89,

A

tan r
= tan

sin s...

2

.(1).

From this result we may derive other equivalent forms as in the preceding article; or we may make use of those forms immediately, observing that the angles of the triangle A'BC are A, -B, C respectively. Hence s being (a+b+c) and S being (A+B+C) we shall obtain

These results may also be found independently by bisecting two of the angles of the triangle ABC, so as to determine the pole of the small circle, and proceeding as in Art. 89.

91. A circle which touches one side of a triangle and the other sides produced is called an escribed circle; thus there are three escribed circles belonging to a given triangle. We may denote the radii of the escribed circles which touch CA and AB respectively by r, and r,, and values of tan r, and tan r, may be found from what has been already given with respect to tan r, by appropriate changes in the letters which denote the sides and angles.

In the preceding article a triangle A'BC was formed by producing AB and AC to meet again at A'; similarly another triangle may be formed by producing BC and BA to meet again, and another by producing CA and CB to meet again. The original triangle ABC and the three formed from it have been called associated triangles, ABC being the fundamental triangle. Thus the inscribed and escribed circles of a given triangle are the same as the circles inscribed in the system of associated triangles of which the given triangle is the fundamental triangle.

92. To find the angular radius of the small circle described about a given triangle.

Let ABC be the given triangle; bisect the sides CB, CA in D and E respectively, and draw from D and E arcs perpendicular to CB and CA respectively, and let P be the intersection of these

arcs.

Then P will be the pole of the small circle described about ABC. For draw PA, PB, PC; then from the right-angled triangles PCD and PBD it follows that PBPC; and from the right-angled triangles PCE and PAE it follows that PA = PC; hence PA= PB = PC. Also the angle PAB = the angle PBA, the angle PBC = the angle PCB, and the angle PCA = the angle PAC; therefore PCB + A = ¿ (A + B + C), and PCB =S-A.

The value of tan R may be expressed in various forms; thus

a

if we substitute for tan from Art. 49, we obtain

2

cos S

cos (S – A) cos (S – B) cos (S – C)

cos (S – A) = cos {≥ (B + C') – ↓ A}

= cos (B + C) cos § 4 + sin ≥ (B + C) sin & A

Substitute in the last expression the value of sin A from

It may be shewn, by common trigonometrical formulæ, that

4 sina sin b sin c=sin (s-a) + sin (s- b) + sin (sc) - sin 8; hence we have from (4)

tan R=

1

2n

{sin (s — a) + sin (s — ¿) + sin (§ — c) – si . . . ...........

93. To find the angular radii of the small circles described round the triangles associated with a given fundamental triangle.

Let R, denote the radius of the circle described round the triangle formed by producing AB and AC to meet again at A'; similarly let R, and R, denote the radii of the circles described round the other two triangles which are similarly formed. Then we may deduce expressions for tan R1, tan R2, and tan R, from those found in Art. 92 for tan R. The sides of the triangle ABC are a, b, π c, and its angles are A, π-B, π- С; hence if s = 1⁄2 (a + b + c) and S Art. 92

=

(A + B + C) we shall obtain from

1 sin A sin

=

=

b sin c

2 sina cosb cos c

√{sin s sin (s — a) sin (s — b) sin (s — c)}

sin 8 — sin (8 − a) + sin (8 − 6) + sin (s – c)} ...

Similarly we may find expressions for tan R, and tan R.

2

94. Many examples may be proposed involving properties of the circles inscribed in and described about the associated triangles. We will give one that will be of use hereafter.

therefore

2 (1

4n2 = 1 - cosa - cos3 b-cos c + 2 cos a cos b cos c;

(sin a + sin b + sin c)2 - 4n2

= 2 ( 1 + sin a sin b + sin b sin c + sin c sin a

sc).
(8-0)};

and by squaring both members of this equation the required result will be obtained. For it may be shewn by reduction that sin3 s + sin2 (s — a) + sin2 (s — 6) + sin2 (s — c) = 2 − 2 cos a cos b cos c,

b) + sin s sin (sc)

+ sin (s − a) sin (s — b) + sin (s — b) sin (s — c) + sin (s — c) sin (s — a)

=

- sin a sin b + sin b sin c + sin c sin a.

Similarly we may prove that

95. In the figure to Art. 89, suppose DP produced through P to a point A' such that DA' is a quadrant, then A′ is a pole of

BC, and PA'

r; similarly, suppose EP produced through P

to a point B such that EB is a quadrant, and FP produced through P to a point C' such that FC is a quadrant. Then

A'B'C' is the polar triangle of ABC, and PA′ = PB′ = PC′ =

Thus P is the pole of the small circle described round the polar triangle, and the angular radius of the small circle described round the polar triangle is the complement of the angular radius of the

T. S. T.

F