Circumscribed and Inscribed Circles

Excerpts

• Circumscribed and Inscribed Circles

  Let $ABC$ be the triangle; bisect the angles $A$ and $B$ by arcs meeting at $P$; from $P$ draw $PD$, $PE$, $PF$ perpendicular to the sides. Then it may be shewn that $PD$, $PE$, $PF$ are all equal; also that $AE = AF$, $BF = BD$, $CD = CE$.

• Circumscribed and Inscribed Circles

  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.

• Circumscribed and Inscribed Circles

  Then $P$ will be the pole of the small circle described about $ABC$.

• Circumscribed and Inscribed Circles

  Many examples may be proposed involving properties of the circles inscribed in and described about the associated triangles.

• Circumscribed and Inscribed Circles

  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 small circle inscribed in the primitive triangle.

• Circumscribed and Inscribed Circles

  Shew that in an equilateral triangle $\tan R = 2\tan r$.

Equations

• Circumscribed and Inscribed Circles

  \tan r = \tan\dfrac{A}{2} \sin (s-a)

  The tangent of the angular radius of the small circle inscribed in a spherical triangle equals tan(A/2) times sin(s−a).

• Circumscribed and Inscribed Circles

  \tan \dfrac{A}{2} = \Surd {\frac{\sin (s - b) \sin (s - c)}{\sin s\, \sin (s - a)} }

  The tangent of half an angle of a spherical triangle is the square root of sin(s−b)sin(s−c) divided by sin s sin(s−a), quoted from Article 45.

• Circumscribed and Inscribed Circles

  \tan r = \Surd{\left\{ \dfrac{\sin (s - a) \sin (s - b) \sin (s - c)}{\sin s} \right\}} = \dfrac{n}{\sin s}

  The tangent of the inscribed circle's angular radius equals the square root of sin(s−a)sin(s−b)sin(s−c)/sin s, which equals n/sin s.

• Circumscribed and Inscribed Circles

  \tan r = \dfrac{\sin\tfrac{1}{2}B \sin \tfrac{1}{2}C}{\cos \tfrac{1}{2}A} \sin a

  The tangent of the inscribed circle's angular radius equals sin(B/2)sin(C/2)/cos(A/2) times sin a.

• Circumscribed and Inscribed Circles

  \tan r = \dfrac{\surd\{-\cos S \cos (S - A) \cos (S - B) \cos (S - C)\}}{2 \cos \tfrac{1}{2}A \cos \tfrac{1}{2}B \cos \tfrac{1}{2}C}

  The tangent of the inscribed circle's angular radius equals the square root of −cos S cos(S−A)cos(S−B)cos(S−C), divided by 2 cos(A/2)cos(B/2)cos(C/2); the book sets this equal to N over the same denominator.

• Circumscribed and Inscribed Circles

  4 \cos\tfrac{1}{2}A \cos\tfrac{1}{2}B \cos\tfrac{1}{2}C = \cos S + \cos (S - A) + \cos (S - B) + \cos (S - C)

  Four times the product of the cosines of half the angles equals the sum of cos S and the three cos(S−A), cos(S−B), cos(S−C).

• Circumscribed and Inscribed Circles

  \cot r = \frac{1}{2N} \bigl\{\cos S + \cos (S - A) + \cos (S - B) + \cos (S - C)\bigr\}

  The cotangent of the inscribed circle's angular radius equals the sum cos S + cos(S−A) + cos(S−B) + cos(S−C) divided by 2N.

• Circumscribed and Inscribed Circles

  \tan r_1 = \tan\dfrac{A}{2}\sin s

  For the small circle touching BC and the other two sides produced, tan r_1 equals tan(A/2) times sin s.

• Circumscribed and Inscribed Circles

  \tan r_1 = \Surd{\left\{\dfrac{\sin s \sin(s-b)\sin(s-c)} {\sin(s-a)}\right\}} = \dfrac{n}{\sin(s-a)}

  For the escribed circle touching BC, tan r_1 equals the square root of sin s sin(s−b)sin(s−c)/sin(s−a), which equals n/sin(s−a).

• Circumscribed and Inscribed Circles

  \tan r_1 = \dfrac{\cos\tfrac{1}{2}B \cos\tfrac{1}{2}C} {\cos\tfrac{1}{2}A} \sin a

  For the escribed circle touching BC, tan r_1 equals cos(B/2)cos(C/2)/cos(A/2) times sin a.

• Circumscribed and Inscribed Circles

  \cot r_1 = \dfrac{1}{2N} \{-c\cos S - \cos(S-A) + \cos(S-B) + \cos(S-C) \}

  The cotangent of the escribed circle's angular radius r_1 equals [−c cos S − cos(S−A) + cos(S−B) + cos(S−C)] divided by 2N, as printed; the leading term reads '−c cos S', which looks like a misprint for −cos S and is FLAGGED, not corrected, pending check against the source.

• Circumscribed and Inscribed Circles

  \tan R = \dfrac{\tan\tfrac{1}{2}a}{\cos(S-A)}

  The tangent of the angular radius R of the small circle described about the triangle equals tan(a/2) divided by cos(S−A).

• Circumscribed and Inscribed Circles

  \tan R = \Surd{\left\{\dfrac{-\cos S}{\cos(S-A)\cos(S-B)\cos(S-C)}\right\}} = \dfrac{\cos S}{N}

  The tangent of the circumscribed small circle's angular radius R equals the square root of −cos S divided by cos(S−A)cos(S−B)cos(S−C), which equals cos S / N.

• Circumscribed and Inscribed Circles

  \tan R = \dfrac{\sin\tfrac{1}{2}a } {\sin A\cos\tfrac{1}{2}b\cos\tfrac{1}{2}c }

  The tangent of the circumscribed small circle's angular radius R equals sin(a/2) divided by sin A cos(b/2) cos(c/2).

• Circumscribed and Inscribed Circles

  \tan R = \dfrac{2\sin\tfrac{1}{2}a \sin\tfrac{1}{2}b \sin\tfrac{1}{2}c} {\surd{\left\{\sin s\sin(s-a) \sin(s-b) \sin(s-c) \right\}}}

  The tangent of the circumscribed small circle's angular radius R equals 2 sin(a/2)sin(b/2)sin(c/2) divided by the square root of sin s sin(s−a)sin(s−b)sin(s−c), which is 2 sin(a/2)sin(b/2)sin(c/2)/n.

• Circumscribed and Inscribed Circles

  4\sin\tfrac{1}{2}a\sin\tfrac{1}{2}b\sin\tfrac{1}{2}c = \sin(s-a) + \sin(s-b) + \sin(s-c)-\sin s

  Four times the product of the sines of half the sides equals sin(s−a)+sin(s−b)+sin(s−c)−sin s.

• Circumscribed and Inscribed Circles

  \tan R=\dfrac{1}{2n}\{ \sin(s-a)+\sin(s-b)+\sin(s-c)-\sin s \}

  The tangent of the circumscribed small circle's angular radius R equals [sin(s−a)+sin(s−b)+sin(s−c)−sin s] divided by 2n.

• Circumscribed and Inscribed Circles

  \tan R_1 = \frac{\tan\frac{1}{2}a}{-\cos S}

  For the circle described about the associated triangle A'BC, tan R_1 equals tan(a/2) divided by −cos S.

• Circumscribed and Inscribed Circles

  \tan R_1 = \Surd{\left\{ \frac{\cos(S-A)}{-\cos S\cos(S-B)\cos(S-C)}\right\}} = \frac{\cos(S-A)}{N}

  For the circle described about the associated triangle A'BC, tan R_1 equals the square root of cos(S−A) divided by −cos S cos(S−B)cos(S−C), which equals cos(S−A)/N.

• Circumscribed and Inscribed Circles

  \tan R_1 = \frac{\sin\frac{1}{2}a} {\sin A\sin\frac{1}{2}b\sin\frac{1}{2}c}

  For the circle described about the associated triangle A'BC, tan R_1 equals sin(a/2) divided by sin A sin(b/2) sin(c/2).

• Circumscribed and Inscribed Circles

  \tan R_1 = \frac{2\sin\frac{1}{2}a\cos\frac{1}{2}b\cos\frac{1}{2}c} {\surd{\{\sin s\sin(s-a)\sin(s-b)\sin(s-c)\}}}

  For the circle described about the associated triangle A'BC, tan R_1 equals 2 sin(a/2)cos(b/2)cos(c/2) divided by the square root of sin s sin(s−a)sin(s−b)sin(s−c).

• Circumscribed and Inscribed Circles

  \tan R_1 =\frac{1}{2n} \{\sin s - \sin(s-a) + \sin(s-b) + \sin(s-c) \}

  For the circle described about the associated triangle A'BC, tan R_1 equals [sin s − sin(s−a) + sin(s−b) + sin(s−c)] divided by 2n.

• Circumscribed and Inscribed Circles

  (\cot r + \tan R)^2=\dfrac{1}{4n^2}(\sin a+\sin b+\sin c)^2 -1

  The square of cot r plus tan R equals (sin a + sin b + sin c)² divided by 4n², minus 1 (Article 94 result).

• Circumscribed and Inscribed Circles

  (\cot r_1-\tan R)^2=\dfrac{1}{4n^2}(\sin b+\sin c-\sin a)^2 -1

  The square of cot r_1 minus tan R equals (sin b + sin c − sin a)² divided by 4n², minus 1 (Article 94, stated as 'similarly').

• Circumscribed and Inscribed Circles

  4n^2=1-\cos^2 a-\cos^2 b-\cos^2 c+2\cos a\cos b\cos c

  Four times n squared equals 1 minus the squares of the cosines of the sides plus twice the product of the cosines of the sides.

• Circumscribed and Inscribed Circles

  \cot r + \tan R = \dfrac{1}{2n}\Bigl\{\sin s + \sin(s-a)+\sin(s-b)+\sin (s-c)\Bigr\}

  The sum of cot r and tan R equals [sin s + sin(s−a) + sin(s−b) + sin(s−c)] divided by 2n.

• Circumscribed and Inscribed Circles

  \sin^2 s + \sin^2 (s-a) + \sin^2 (s-b) + \sin^2 (s-c) = 2-2 \cos a \cos b \cos c

  The sum of the squared sines of s and of s−a, s−b, s−c equals 2 minus 2 cos a cos b cos c.

• Circumscribed and Inscribed Circles

  PA' = PB' = PC' = \dfrac{\pi}{2}-r

  The point P is at angular distance π/2 − r from each vertex A', B', C' of the polar triangle, so the circle about the polar triangle has angular radius complementary to r.

Problems