Olympiad Geometry III: Circumcircle, Power of a Point, Incircle

Study Sheet

Olympiad Geometry III: Circumcircle, Power of a Point, Incircle

Arc midpoints, the incenter–excenter lemma, power of a point, Euler's OI², bisector lengths

Circles Doing the Work

Tip
Power of a point, three ways

For a point PP and a circle with center OO, radius rr: every line through PP meeting the circle at X,YX, Y has PX⋅PY=∣PO2−r2∣PX\cdot PY = |PO^2 - r^2| — equal to the tangent length squared when PP is outside.

The proof is one similarity: for two chords XYXY, X′Y′X'Y' through PP, the inscribed angles make PXX′∼PY′YPXX' \sim PY'Y, so PX⋅PY=PX′⋅PY′PX\cdot PY = PX'\cdot PY'. Use it (i) to compute a length from a secant and a tangent, (ii) to prove four points concyclic (the converse holds: PA⋅PB=PC⋅PDPA\cdot PB = PC\cdot PD with the right configuration forces ABCDABCD cyclic), and (iii) through the radical axis — the locus of equal power with respect to two circles is a line, and three radical axes concur.

Tip
The arc midpoint and the incenter–excenter lemma

If the bisector from AA meets the circumcircle again at MM, then MB=MC=MI=MIAMB = MC = MI = MI_A: MM is the center of the circle through BB, CC, the incenter II, and the AA-excenter IAI_A.

Angle chase: ∠MBI=∠MBC+∠CBI=A2+B2\angle MBI = \angle MBC + \angle CBI = \tfrac A2 + \tfrac B2 and ∠MIB=∠IAB+∠IBA=A2+B2\angle MIB = \angle IAB + \angle IBA = \tfrac A2 + \tfrac B2 (exterior angle of AIBAIB), so MB=MIMB = MI. Combined with the power of DD (where the bisector meets BCBC), DB⋅DC=DA⋅DMDB\cdot DC = DA\cdot DM, and the bisector-length formula AD2=bc−DB⋅DCAD^2 = bc - DB\cdot DC, every length on the bisector is computable — including AM=bcADAM = \tfrac{bc}{AD} from ABD∼AMCABD \sim AMC. Euler's OI2=R(R−2r)OI^2 = R(R - 2r) is the power of II: R2−OI2=AI⋅IM=AI⋅MB=rsin⁡(A/2)⋅2Rsin⁡A2=2RrR^2 - OI^2 = AI\cdot IM = AI\cdot MB = \tfrac{r}{\sin(A/2)}\cdot 2R\sin\tfrac A2 = 2Rr.

Tip
Incircle lengths

Tangent lengths s−a,s−b,s−cs - a, s - b, s - c; r=Ksr = \tfrac{K}{s}; AI=rsin⁡(A/2)AI = \tfrac{r}{\sin(A/2)}, so AI2=(s−a)2+r2AI^2 = (s-a)^2 + r^2; the AA-excircle has radius ra=Ks−ar_a = \tfrac{K}{s - a} and tangent length ss from AA.

Every incircle problem reduces to right triangles formed by a radius and a tangent. The 1313-1414-1515 triangle is the standard test case: s=21s = 21, K=84K = 84, r=4r = 4, R=658R = \tfrac{65}{8}, AI2=72+42=65AI^2 = 7^2 + 4^2 = 65, OI2=6564OI^2 = \tfrac{65}{64} — keep these numbers in your head to sanity-check formulas under pressure.

Side note
Homothety with circles

Two circles are related by two homotheties whose centers are the internal and external centers of similitude — the intersections of the common tangents. For tangent circles the tangency point is one of them, so a line through the tangency point meets the circles at points P,QP, Q with TQTP=r2r1\tfrac{TQ}{TP} = \tfrac{r_2}{r_1}. This is the tool for "the line TPTP passes through the arc midpoint" problems.

Try it
Try it: Euler in numbers

A triangle has R=5R = 5 and r=2r = 2. Compute OIOI.

OI2=R(R−2r)=5⋅1=5OI^2 = R(R - 2r) = 5\cdot1 = 5, so OI=5OI = \sqrt5. Note R≥2rR \ge 2r always (Euler's inequality) since OI2≥0OI^2 \ge 0.