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Concept 04 / 11Trigonometry

Circles Connected with a Triangle: Circumradius, Inradius and Exradii

Formulas for R, r and the exradii r1, r2, r3 in terms of sides, angles and area, r = 4R sin(A/2) sin(B/2) sin(C/2), distances such as OI^2 = R^2 - 2Rr, relations among r, r1, r2, r3, lengths of tangents from vertices, and inequalities like R >= 2r.

Figure 1
Circumcircle: the circle through the three vertices, centred at the circumcentre OO (intersection of perpendicular bisectors), radius RR: R=a2sin⁡A=abc4ΔR = \frac{a}{2\sin A} = \frac{abc}{4\Delta}. For a right triangle, RR is half the hypotenuse. Incircle: the circle touching all three sides, centred at the incentre II (intersection of internal angle bisectors), radius rr: r=Δs=(s−a)tan⁡A2=4Rsin⁡A2sin⁡B2sin⁡C2r = \frac{\Delta}{s} = (s - a)\tan\frac{A}{2} = 4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}. The first formula follows by splitting the triangle into three triangles with apex II, each with height rr: Δ=12r(a+b+c)=rs\Delta = \frac{1}{2}r(a + b + c) = rs. The tangent lengths from the vertices to the incircle are s−as - a, s−bs - b, s−cs - c (from AA, BB, CC respectively). For a right triangle with legs a,ba, b and hypotenuse cc: r=a+b−c2r = \frac{a + b - c}{2}. Excircles: the circle opposite AA touches side aa and the extensions of bb and cc; its centre I1I_{1} is the intersection of the internal bisector of AA and the external bisectors of BB and CC. Its radius is r1=Δs−a=stan⁡A2=4Rsin⁡A2cos⁡B2cos⁡C2r_{1} = \frac{\Delta}{s - a} = s\tan\frac{A}{2} = 4R\sin\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}, and similarly r2=Δs−br_{2} = \frac{\Delta}{s - b}, r3=Δs−cr_{3} = \frac{\Delta}{s - c}. The tangent lengths from AA to the excircle opposite AA equal ss. Relations. 1r1+1r2+1r3=1r\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}} = \frac{1}{r} (add s−aΔ\frac{s - a}{\Delta} and so on). r1+r2+r3−r=4Rr_{1} + r_{2} + r_{3} - r = 4R. r1r2+r2r3+r3r1=s2r_{1}r_{2} + r_{2}r_{3} + r_{3}r_{1} = s^{2}. rr1r2r3=Δ2rr_{1}r_{2}r_{3} = \Delta^{2}. cos⁡A+cos⁡B+cos⁡C=1+rR\cos A + \cos B + \cos C = 1 + \frac{r}{R}. Distances between centres. Euler's formula: OI2=R2−2RrOI^{2} = R^{2} - 2Rr. Since OI2≥0OI^{2} \ge 0, this gives R≥2rR \ge 2r, with equality iff the triangle is equilateral. OI12=R2+2Rr1OI_{1}^{2} = R^{2} + 2Rr_{1}. AI=rsin⁡(A/2)AI = \frac{r}{\sin(A/2)} and AI1=r1sin⁡(A/2)AI_{1} = \frac{r_{1}}{\sin(A/2)}. Distance from OO to side aa: Rcos⁡AR\cos A; from the orthocentre HH to vertex AA: 2Rcos⁡A2R\cos A; and OH2=9R2−(a2+b2+c2)OH^{2} = 9R^{2} - (a^{2} + b^{2} + c^{2}). Equilateral triangle of side aa: R=a3R = \frac{a}{\sqrt{3}}, r=a23r = \frac{a}{2\sqrt{3}}, r1=r2=r3=3a2r_{1} = r_{2} = r_{3} = \frac{\sqrt{3}a}{2}, and all centres coincide (except the excentres). Key Takeaways: - R=abc4Δ=a2sin⁡AR = \frac{abc}{4\Delta} = \frac{a}{2\sin A}; r=Δsr = \frac{\Delta}{s}; r1=Δs−ar_{1} = \frac{\Delta}{s - a}. - r=4R∏sin⁡A2r = 4R\prod\sin\frac{A}{2}; r1=4Rsin⁡A2cos⁡B2cos⁡C2r_{1} = 4R\sin\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}. - Tangent lengths to the incircle: s−a,s−b,s−cs - a, s - b, s - c. - 1r=∑1r1\frac{1}{r} = \sum\frac{1}{r_{1}}; r1+r2+r3−r=4Rr_{1} + r_{2} + r_{3} - r = 4R; rr1r2r3=Δ2rr_{1}r_{2}r_{3} = \Delta^{2}. - OI2=R2−2RrOI^{2} = R^{2} - 2Rr, so R≥2rR \ge 2r. - Right triangle: R=c2R = \frac{c}{2}, r=a+b−c2r = \frac{a + b - c}{2}. - Pitfall: confusing s−as - a (tangent length from AA) with the side aa in the inradius formulas. Brain-Teaser 1: For the triangle with sides 13,14,1513, 14, 15, find RR, rr and r1r_{1} (opposite the side 1313). Solution: s=21s = 21, Δ=21×8×7×6=7056=84\Delta = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84. R=13×14×154×84=2730336=658R = \frac{13 \times 14 \times 15}{4 \times 84} = \frac{2730}{336} = \frac{65}{8}. r=8421=4r = \frac{84}{21} = 4. r1=8421−13=212r_{1} = \frac{84}{21 - 13} = \frac{21}{2}. Check Euler: OI2=422564−2×658×4=4225−416064=6564>0OI^{2} = \frac{4225}{64} - 2 \times \frac{65}{8} \times 4 = \frac{4225 - 4160}{64} = \frac{65}{64} > 0. Consistent. Brain-Teaser 2: Show that r1+r2+r3−r=4Rr_{1} + r_{2} + r_{3} - r = 4R. Solution: r1+r2=Δ(1s−a+1s−b)=Δc(s−a)(s−b)r_{1} + r_{2} = \Delta\left(\frac{1}{s - a} + \frac{1}{s - b}\right) = \frac{\Delta c}{(s - a)(s - b)} and r3−r=Δ(1s−c−1s)=Δcs(s−c)r_{3} - r = \Delta\left(\frac{1}{s - c} - \frac{1}{s}\right) = \frac{\Delta c}{s(s - c)}. Adding: Δcs(s−c)+(s−a)(s−b)s(s−a)(s−b)(s−c)=Δcs(s−c)+(s−a)(s−b)Δ2\Delta c\frac{s(s - c) + (s - a)(s - b)}{s(s - a)(s - b)(s - c)} = \Delta c\frac{s(s - c) + (s - a)(s - b)}{\Delta^{2}}. The numerator s(s−c)+(s−a)(s−b)=2s2−s(a+b+c)+ab=abs(s - c) + (s - a)(s - b) = 2s^{2} - s(a + b + c) + ab = ab. So the sum is abcΔ=4R\frac{abc}{\Delta} = 4R. Pairing the terms so that each pair produces the factor cc makes the algebra collapse. Brain-Teaser 3: If in a triangle r1=2r2=3r3r_{1} = 2r_{2} = 3r_{3}, find the ratio a:b:ca : b : c. Solution: r1:r2:r3=1s−a:1s−b:1s−c=6:3:2r_{1} : r_{2} : r_{3} = \frac{1}{s - a} : \frac{1}{s - b} : \frac{1}{s - c} = 6 : 3 : 2 (since r1=6kr_{1} = 6k, r2=3kr_{2} = 3k, r3=2kr_{3} = 2k). So s−a:s−b:s−c=16:13:12=1:2:3s - a : s - b : s - c = \frac{1}{6} : \frac{1}{3} : \frac{1}{2} = 1 : 2 : 3. Let s−a=ts - a = t, s−b=2ts - b = 2t, s−c=3ts - c = 3t; adding, 3s−2s=s=6t3s - 2s = s = 6t. Then a=5ta = 5t, b=4tb = 4t, c=3tc = 3t, so a:b:c=5:4:3a : b : c = 5 : 4 : 3, a right triangle. Working with s−as - a, s−bs - b, s−cs - c as the unknowns linearises the problem.

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Circles Connected with a Triangle: Circumradius, Inradius and Exradii — DPP

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