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

The proof is one similarity: for two chords $XY$, $X'Y'$ through $P$, the inscribed angles make $PXX' \sim PY'Y$, so $PX\cdot PY = PX'\cdot PY'$. Use it (i) to compute a length from a secant and a tangent, (ii) to prove four points concyclic (the converse holds: $PA\cdot PB = PC\cdot PD$ with the right configuration forces $ABCD$ 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.

Angle chase: $\angle MBI = \angle MBC + \angle CBI = \tfrac A2 + \tfrac B2$ and $\angle MIB = \angle IAB + \angle IBA = \tfrac A2 + \tfrac B2$ (exterior angle of $AIB$), so $MB = MI$. Combined with the power of $D$ (where the bisector meets $BC$), $DB\cdot DC = DA\cdot DM$, and the bisector-length formula $AD^2 = bc - DB\cdot DC$, every length on the bisector is computable — including $AM = \tfrac{bc}{AD}$ from $ABD \sim AMC$. Euler's $OI^2 = R(R - 2r)$ is the power of $I$: $R^2 - OI^2 = AI\cdot IM = AI\cdot MB = \tfrac{r}{\sin(A/2)}\cdot 2R\sin\tfrac A2 = 2Rr$.

Every incircle problem reduces to right triangles formed by a radius and a tangent. The $13$-$14$-$15$ triangle is the standard test case: $s = 21$, $K = 84$, $r = 4$, $R = \tfrac{65}{8}$, $AI^2 = 7^2 + 4^2 = 65$, $OI^2 = \tfrac{65}{64}$ — keep these numbers in your head to sanity-check formulas under pressure.

Everything comes from one similarity. Since $OP\cdot OP^* = OQ\cdot OQ^*$, we have $\tfrac{OP}{OQ} = \tfrac{OQ^*}{OP^*}$, and the triangles $OPQ$ and $OQ^*P^*$ share the angle at $O$, so they are similar (SAS). Reading off the third side gives $\tfrac{P^*Q^*}{PQ} = \tfrac{OQ^*}{OP} = \tfrac{k^2}{OP\cdot OQ}$.

For a line $\ell$ not through $O$, let $F$ be the foot of the perpendicular from $O$. For any $P$ on $\ell$ the same similarity gives $\angle OP^*F^* = \angle OFP = 90^\circ$, so $P^*$ lies on the circle with diameter $OF^*$ — a circle through $O$. Inversion is its own inverse, so that circle comes back to $\ell$. A circle not through $O$ is handled the same way with the two points where the diameter through $O$ meets it. The payoff is strategic: choose the center at the point where the most circles meet, and those circles straighten into lines.

Step 1 — invert about $A$ with radius $1$. By the distance formula, $B^*C^* = \dfrac{BC}{AB\cdot AC}$, $\;C^*D^* = \dfrac{CD}{AC\cdot AD}$, $\;B^*D^* = \dfrac{BD}{AB\cdot AD}$.

Step 2 — the triangle inequality $B^*D^* \le B^*C^* + C^*D^*$ becomes, after multiplying by $AB\cdot AC\cdot AD$,

Step 3 — equality holds iff $C^*$ lies on segment $B^*D^*$. Three points $B^*, C^*, D^*$ are collinear iff $B, C, D$ lie on a line or circle through $A$ (the image of a line not through $A$ is a circle through $A$); and $C^*$ between $B^*$ and $D^*$ means $C$ is on the arc $BD$ not containing $A$. That is the order $A, B, C, D$ on a circle (or on a line, when $B, C, D$ are collinear with $A$). $\blacksquare$

How to find it. Products of two lengths, one point in every product: invert at that point and the products become single distances, where the triangle inequality lives.

Step 1 — fix $n$ and invert about $A$ with $k^2$ equal to the power of $A$ with respect to $\mathcal C_n$. Every line through $A$ meets $\mathcal C_n$ at $X, Y$ with $AX\cdot AY = k^2$, so the inversion swaps $X$ and $Y$: $\mathcal C_n$ is mapped to itself.

Step 2 — the semicircles on $AB$ and $AC$ pass through $A$ and have their centers on line $AC$, so their images are lines perpendicular to $AC$: two parallel lines, bounding a strip of some width $w$.

Step 3 — the circles $\mathcal C_0, \mathcal C_1, \ldots, \mathcal C_n$ are tangent to both semicircles, and inversion preserves tangency, so their images are circles tangent to both parallel lines: each has diameter $w$, and consecutive images touch. They form a vertical stack of equal circles. $\mathcal C_0$ is symmetric in line $AC$, and inversion about $A$ commutes with that reflection, so the image of $\mathcal C_0$ has its center on $AC$; the image of $\mathcal C_n$ then has its center at height $n w$.

Step 4 — the image of $\mathcal C_n$ is $\mathcal C_n$ itself, so $d_n = w$ and its center is at height $n w = n d_n$. $\blacksquare$ With $AB = 4$ and $AC = 6$ the first three chain circles have diameters $\tfrac{12}{7}$, $\tfrac65$, $\tfrac45$ and centers at heights $\tfrac{12}{7}$, $\tfrac{12}{5}$, $\tfrac{12}{5}$ — exactly $1$, $2$, $3$ diameters.

How to find it. A chain of circles all tangent to two circles that meet at a point: invert at that point and the two circles become parallel lines, where a chain of tangent circles is trivial. The clever part is choosing $k$ so the circle you care about does not move.

Step 1 — invert about $T_1$, the tangency point of $\omega_1$ and $\omega_2$. Both circles pass through $T_1$ and are tangent there, so they become two parallel lines $\ell_1, \ell_2$ (tangent curves stay tangent, and two lines tangent at the point at infinity are parallel). The exteriors of $\omega_1$ and $\omega_2$ go to the open strip between the lines.

Step 2 — $\omega_3$ and $\omega_4$ avoid $T_1$ and lie outside $\omega_1$ and $\omega_2$ (by the tangencies and the disjointness hypotheses), so their images $\gamma_3, \gamma_4$ are circles inside the strip, $\gamma_3$ tangent to $\ell_2$ at $T_2^*$, $\gamma_4$ tangent to $\ell_1$ at $T_4^*$, and $\gamma_3, \gamma_4$ externally tangent at $T_3^*$.

Step 3 — the homothety with center $T_3^*$ and negative ratio $-\tfrac{\rho_4}{\rho_3}$ sends $\gamma_3$ to $\gamma_4$. It sends $\ell_2$, the tangent to $\gamma_3$ at $T_2^*$, to a tangent of $\gamma_4$ parallel to $\ell_2$; since the ratio is negative, $\gamma_4$ lies on the side of the image line facing $\ell_2$, and the only such tangent is $\ell_1$. So $T_2^* \mapsto T_4^*$, and $T_2^*, T_3^*, T_4^*$ are collinear.

Step 4 — that line does not pass through $T_1$ (it crosses the strip, while $T_1$ is the center of inversion, sent to infinity), so it inverts back to a circle through $T_1$ containing $T_2, T_3, T_4$. $\blacksquare$

How to find it. Four tangency points, four circles: inverting at one tangency point turns two circles into lines at once. Then the problem is a statement about two circles in a strip, which a homothety settles.

Put $O$ at the origin, so $|A| = |B| = |C| = R$. The point $H = A + B + C$ satisfies $(H - A)\cdot(B - C) = (B + C)\cdot(B - C) = |B|^2 - |C|^2 = 0$, so it is on the altitude from $A$, and by symmetry on the others: (3). The reflection of $H$ in $M = \tfrac{B + C}{2}$ is $B + C - H = -A$, the antipode of $A$: (2), and $AH = |B + C| = 2|M| = 2\,OM$. For (4), $N = \tfrac H2$ is at distance $\left|\tfrac{A + B + C}{2} - \tfrac{B + C}{2}\right| = \tfrac R2$ from $M$, and at distance $\tfrac{|A|}{2}$ from the midpoint $\tfrac{A + H}{2}$ of $AH$; those two midpoints are ends of a diameter of the circle, and the foot $D$ of the altitude sees that diameter at a right angle, so $D$ is on it too. The homothety $X \mapsto \tfrac{H + X}{2}$ moves $|X| = R$ to $|\tfrac{H + X}{2} - \tfrac H2| = \tfrac R2$. For (1), the reflection $H_a$ of $H$ in $BC$ has $\angle BH_aC = \angle BHC = 180^\circ - A$, which puts $H_a$ on $\Gamma$.

Step 1 — the altitude from $A$ is $x = 5$. The altitude from $B$ is perpendicular to $AC$, whose direction is $(9, -12)$, so it is $9x - 12y = 0$, i.e. $y = \tfrac34x$. They meet at $H = \left(5, \tfrac{15}{4}\right)$.

Step 2 — $O$ is on $x = 7$ with $OB = OA$: $49 + y^2 = 4 + (12 - y)^2$ gives $y = \tfrac{33}{8}$, so $O = \left(7, \tfrac{33}{8}\right)$ and $R^2 = 49 + \tfrac{1089}{64} = \tfrac{4225}{64}$, $R = \tfrac{65}{8}$.

Step 3 — checks. With $O$ not at the origin, fact 3 reads $H = A + B + C - 2O = (19, 12) - \left(14, \tfrac{33}{4}\right) = \left(5, \tfrac{15}{4}\right)$ ✓. Fact 2: $AH = 12 - \tfrac{15}{4} = \tfrac{33}{4} = 2\cdot OM$ with $M = (7, 0)$ ✓. Fact 1: the reflection $\left(5, -\tfrac{15}{4}\right)$ has $(5 - 7)^2 + \left(-\tfrac{15}{4} - \tfrac{33}{8}\right)^2 = 4 + \tfrac{3969}{64} = \tfrac{4225}{64} = R^2$ ✓.

Step 4 — $N = \tfrac{O + H}{2} = \left(6, \tfrac{63}{16}\right)$ and the nine-point radius is $\tfrac{65}{16}$; indeed $NM^2 = 1 + \tfrac{3969}{256} = \tfrac{4225}{256}$ ✓. The centroid $G = \left(\tfrac{19}{3}, 4\right)$ sits a third of the way from $O$ to $H$.

How to find it. Compute two altitudes, not three; use the third fact as the check. A computed configuration is the fastest way to believe a synthetic statement before proving it.

Step 1 — by fact 2, the reflection of $H$ in $M$ is the antipode $A'$ of $A$. So $M$ is the midpoint of $HA'$, and $H$, $M$, $A'$ are collinear with $M$ between $H$ and $A'$.

Step 2 — $H$ is inside $\Gamma$ (the triangle is acute), so line $HA'$ meets $\Gamma$ at $A'$ and at a second point on the other side of $H$; that second point is on ray $MH$ beyond $H$, so it is $X$.

Step 3 — $AA'$ is a diameter, so $\angle AXA' = 90^\circ$. Since $X, H, A'$ are collinear, $\angle AXH = 90^\circ$. $\blacksquare$ The circle on diameter $AH$ also passes through the feet of the altitudes from $B$ and $C$, so $X$ is the second intersection of $\Gamma$ with the circle through $A$ and those two feet — a point that appears in many olympiad problems.

How to find it. A midpoint of a side and the orthocenter in one statement: reflect $H$ in the midpoint and the antipode appears.

Step 1 — let $H_a, H_b$ be the reflections of $H$ in $BC$ and $CA$; by fact 1 they lie on $\Gamma$. Reflection in $BC$ sends line $H_aP$ to line $HP_a$, and reverses directed angles with $BC$: $\angle(BC, HP_a) = -\angle(BC, H_aP)$.

Step 2 — since $AH_a \perp BC$ and $H_a, C$ both see chord $AP$ of $\Gamma$, $\angle(BC, H_aP) = \angle(BC, H_aA) + \angle(H_aA, H_aP) = 90^\circ + \angle(CA, CP)$. So $\angle(BC, HP_a) = 90^\circ - \angle(CA, CP)$ (mod $180^\circ$). In the same way, with $BH_b \perp CA$ and chord $BP$: $\angle(CA, HP_b) = 90^\circ - \angle(CB, CP)$.

Step 3 — add up:

So $HP_a \parallel HP_b$: the points $H, P_a, P_b$ are collinear, and likewise $P_c$ is on this line. (If $P$ happens to coincide with $H_a$ or $H_b$, use the two other sides.)

Step 4 — the homothety with center $P$ and ratio $\tfrac12$ sends $P_a, P_b, P_c$ to the feet of the perpendiculars from $P$ to the sides, and sends $H$ to the midpoint of $PH$. So the feet are collinear (the Simson line) and their line passes through the midpoint of $PH$. $\blacksquare$ That midpoint lies on the nine-point circle, by the homothety in fact 4.

How to find it. The Simson line is a line of feet; doubling every foot turns it into a line of reflections, and reflections of $P$ talk to reflections of $H$, which live on $\Gamma$.