Olympiad Geometry II: Circles, Tangents & Cyclic Quadrilaterals

The three pairs of equal tangents from $A$, $B$, $C$ give a linear system whose solution is $s - a$, $s - b$, $s - c$ — in the $13$-$14$-$15$ triangle, $BD = 21 - 15 = 6$ and $DC = 8$. Combined with $r = \tfrac{\text{Area}}{s}$ ($= 4$ there) and $R = \tfrac{abc}{4\cdot\text{Area}}$ ($= \tfrac{65}{8}$), every incircle and circumcircle length is available before any angle is chased. Common tangent lengths between two circles are right triangles: external $\sqrt{d^2 - (r_1 - r_2)^2}$, internal $\sqrt{d^2 - (r_1 + r_2)^2}$.

Ptolemy is the workhorse: it computes the diagonal of an inscribed quadrilateral from its sides, and its equality case (Ptolemy's INEQUALITY holds for all quadrilaterals, with equality iff cyclic) is a cyclicity criterion in disguise. Brahmagupta is Heron with a fourth factor; sides $5, 6, 7, 8$ enclose $4\sqrt{105}$. The cyclic configuration maximizes area among quadrilaterals with given sides — Bretschneider's formula shows why.

The British flag theorem makes $PC^2 - PD^2 = PB^2 - PA^2$ regardless of where $P$ is; Stewart computes any cevian from the sides; Apollonius is Stewart's special case. All three are Pythagoras or the law of cosines applied twice, and all three turn "find the length" problems into arithmetic — worth memorizing precisely because they are so easy to rederive under pressure that people never do.

A rectangle $ABCD$ with sides $a, b$ is cyclic (opposite angles $90^\circ$). Ptolemy: $AC\cdot BD = AB\cdot CD + AD\cdot BC$ becomes $d\cdot d = a\cdot a + b\cdot b$, i.e. $d^2 = a^2 + b^2$. Pythagoras is Ptolemy for rectangles.

Choose $E$ on diagonal $AC$ with $\angle ABE = \angle DBC$. Then triangles $ABE$ and $DBC$ are similar (two angles: the constructed one, and $\angle BAE = \angle BDC$ as inscribed angles on arc $BC$), giving $\tfrac{AE}{DC} = \tfrac{AB}{DB}$, i.e. $AE\cdot DB = AB\cdot DC$. Similarly triangles $EBC$ and $ABD$ are similar, giving $EC\cdot DB = BC\cdot AD$. Add: $(AE + EC)\cdot DB = AB\cdot DC + BC\cdot AD$, and $AE + EC = AC$. $\blacksquare$ Significance: the same construction with $E$ off the diagonal proves Ptolemy's inequality for non-cyclic quadrilaterals — the equality case is exactly cyclicity.

Let the tangent lengths from $A$, $B$, $C$ be $x$, $y$, $z$ (equal tangents from each vertex). Then $y + z = a$, $z + x = b$, $x + y = c$; adding, $x + y + z = s$, so $y = s - (z + x) = s - b$. $\blacksquare$ The identical bookkeeping with an EXCIRCLE gives tangent lengths $s$ and $s - c$ etc., which is why $s - a$ appears everywhere in triangle geometry.

Step 1 — the angle bisector theorem: $\tfrac{BD}{DC} = \tfrac{AB}{AC} = \tfrac79$, so $BD = \tfrac{7}{16}\cdot8 = 3.5$, $DC = 4.5$.

Step 2 — Stewart with $a = 8$, $m = 3.5$, $n = 4.5$, $b = 9$, $c = 7$: $81\cdot3.5 + 49\cdot4.5 = 8(d^2 + 3.5\cdot4.5)$, i.e. $283.5 + 220.5 = 8d^2 + 126$, so $8d^2 = 378$, $d^2 = 47.25$, $d = \tfrac{3\sqrt{21}}{2}$.

Step 3 — cross-check with the bisector-length formula $AD^2 = bc - mn = 63 - 15.75 = 47.25$ ✓. Two formulas, one number — always run the second when the first is arithmetic-heavy.

Step 1 — the perpendicular bisector of $AB$ is the locus of points equidistant from $A$ and $B$; likewise for $BC$.

Step 2 — their intersection $O$ (they are not parallel since $AB$ and $BC$ are not) satisfies $OA = OB$ and $OB = OC$, hence $OA = OC$.

Step 3 — so $O$ lies on the perpendicular bisector of $AC$ as well: all three concur, and the circle centered at $O$ through $A$ passes through $B$ and $C$. The proof is one sentence once "perpendicular bisector = equidistance locus" is internalized; the same locus argument gives the incenter (angle bisectors = equidistance from lines).