Competition mathematics and problem solving

Olympiad geometry

Synthetic geometry techniques like power of a point and projective tricks used to solve hard competition problems.

IntuitionOne Number That Describes a Point's Relationship to a Circle

Draw any circle and pick a point $P$ outside it. Draw two different lines through $P$ , each crossing the circle at two points. Measure the two distances from $P$ to each crossing pair and multiply them: $PA \cdot PB$ for the first line, $PC \cdot PD$ for the second. Remarkably, both products are exactly equal, no matter which two lines you drew. This single invariant — called the power of the point $P$ — turns a picture full of intersecting lines and circles into one clean algebraic quantity, and it is the engine behind a large fraction of hard olympiad geometry problems: instead of chasing angles all over the diagram, you write down a power-of-a-point equation and the answer falls out.

Circumcircle and Euler line connecting circumcenter

O

, centroid

G

, and orthocenter

H

with

OH = 3\,OG

— a classic Olympiad configuration.

SchoolPower of a Point and the Two Classical Line Ratios

Definition: Power of a Point

For a circle with center $O$ and radius $r$ , the power of a point $P$ with respect to the circle is $\mathrm{pow}(P) = OP^2 - r^2$ . If a line through $P$ meets the circle at $A$ and $B$ , then $\mathrm{pow}(P) = \overrightarrow{PA} \cdot \overrightarrow{PB}$ (a signed product), which equals $PA \cdot PB$ when $P$ is outside the circle and $-PA \cdot PB$ when $P$ is inside; this value is the same for every line through $P$ .

\mathrm{pow}(P) = OP^2 - r^2 = PA \cdot PB

Two more classical relations turn triangle diagrams into algebra: Ceva's theorem says cevians $AD, BE, CF$ of a triangle $ABC$ are concurrent iff $\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = 1$ , while Menelaus' theorem gives the analogous condition for three points on the (extended) sides to be collinear, using signed ratios: $\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = -1$ . Both reduce a concurrency or collinearity claim, which looks purely geometric, to checking a single product of ratios.

\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = 1

Which classical tool to reach for
Tool	Detects
Power of a point	Equal products of segment lengths from one point
Ceva's theorem	Concurrency of three cevians
Menelaus' theorem	Collinearity of three points on extended sides
Ptolemy's theorem	Cyclic quadrilateral diagonal-side relation

UndergraduateFull Proofs: Power of a Point and Ptolemy's Theorem

Theorem: Power of a Point Theorem

If two lines through a point $P$ meet a circle at $A,B$ and $C,D$ respectively, then $\overrightarrow{PA} \cdot \overrightarrow{PB} = \overrightarrow{PC} \cdot \overrightarrow{PD}$ , and this common signed value equals $OP^2 - r^2$ where $O,r$ are the circle's center and radius.

Why is it true?

This theorem is the single most reused tool in olympiad geometry: it converts any diagram with two secants, a tangent and a secant, or two chords through one point into one algebraic equation, avoiding case-by-case angle chasing.

Proof

Step 1: Set up similar triangles. Let the two lines through $P$ meet the circle at $A,B$ and $C,D$ . Since $A,B,C,D$ lie on a common circle, the inscribed angles $\angle PAC$ and $\angle PDB$ subtend the same arc $BC$ (or its supplement, depending on configuration), so $\angle PAC = \angle PDB$ .

Step 2: Match a second pair of angles. The angle at $P$ in triangle $PAC$ and the angle at $P$ in triangle $PDB$ are either equal (if $P$ is outside the circle, the two lines share vertex $P$ so the angle is literally the same angle) or vertical angles (if $P$ is inside the circle). Either way $\angle APC = \angle DPB$ .

Step 3: Conclude similarity. Two triangles $PAC$ and $PDB$ with two pairs of equal angles are similar by AA: $\triangle PAC \sim \triangle PDB$ .

Step 4: Extract the ratio. Similar triangles give proportional corresponding sides: $\frac{PA}{PD} = \frac{PC}{PB}$ , which rearranges to $PA \cdot PB = PC \cdot PD$ (using unsigned lengths first, for $P$ outside the circle).

Step 5: Compute the common value via a diameter. Choose the specific line through $P$ and the center $O$ , meeting the circle at the two points on the diameter, at signed distances $OP - r$ and $OP + r$ from $P$ (taking $P$ outside so $OP > r$ ). Their product is $(OP-r)(OP+r) = OP^2 - r^2$ , which by Step 4 must equal $PA \cdot PB$ for every line through $P$ .

**Step 6: Handle $P$ inside the circle by signed lengths.** When $P$ lies inside the circle, $A,P,B$ are in that order on the chord, so $\overrightarrow{PA}$ and $\overrightarrow{PB}$ point in opposite directions and their signed product $\overrightarrow{PA}\cdot\overrightarrow{PB} = -PA\cdot PB$ is negative; repeating Steps 1–5 with the diameter through $P$ gives $-PA\cdot PB = OP^2-r^2$ (now negative since $OP<r$ ), so the identity $\overrightarrow{PA}\cdot\overrightarrow{PB}=OP^2-r^2$ holds uniformly with signs, completing the proof.

Theorem: Ptolemy's Theorem

For a cyclic quadrilateral $ABCD$ (vertices in order on a circle), the diagonals and sides satisfy $AC \cdot BD = AB \cdot CD + AD \cdot BC$ .

Why is it true?

Ptolemy's theorem gives an exact algebraic identity for any cyclic quadrilateral, and its inequality form ( $AC \cdot BD \le AB\cdot CD + AD\cdot BC$ for any quadrilateral, with equality iff cyclic) is a fast way to both prove four points concyclic and derive trigonometric identities like the sine addition formula.

Proof

Step 1: Construct an auxiliary point. On diagonal $AC$ , construct point $K$ such that $\angle ABK = \angle DBC$ (i.e. ray $BK$ makes with $BA$ the same angle that $BD$ makes with $BC$ ), with $K$ chosen on segment $AC$ .

Step 2: Find a first pair of similar triangles. Since $ABCD$ is cyclic, $\angle BAC = \angle BDC$ (both subtend arc $BC$ ). Combined with $\angle ABK = \angle DBC$ by construction, triangles $ABK$ and $DBC$ share two equal angles, so $\triangle ABK \sim \triangle DBC$ by AA.

Step 3: Extract the first relation. From $\triangle ABK \sim \triangle DBC$ : $\frac{AK}{DC} = \frac{AB}{DB}$ , so $AK \cdot DB = AB \cdot DC$ .

Step 4: Find a second pair of similar triangles. Since $ABCD$ is cyclic, $\angle ABD = \angle ACD$ (both subtend arc $AD$ ). Consider triangles $KBC$ and $ABD$ : $\angle KBC = \angle ABC - \angle ABK = \angle ABC - \angle DBC = \angle ABD$ (since $K$ lies on segment $AC$ ), and $\angle BCK = \angle BCA = \angle BDA$ (both subtend arc $AB$ ), so $\triangle KBC \sim \triangle ABD$ by AA.

Step 5: Extract the second relation. From $\triangle KBC \sim \triangle ABD$ : $\frac{KC}{AD} = \frac{BC}{BD}$ , so $KC \cdot BD = AD \cdot BC$ .

Step 6: Add the two relations. Adding the results of Steps 3 and 5: $AK\cdot DB + KC\cdot BD = AB\cdot DC + AD\cdot BC$ , i.e. $BD(AK+KC) = AB\cdot DC + AD\cdot BC$ . Since $K$ lies on segment $AC$ , $AK+KC=AC$ , giving exactly $AC\cdot BD = AB\cdot CD + AD\cdot BC$ .

AdvancedReal-World Applications and Worked Examples

These synthetic tools reach beyond competition paper: GPS trilateration and surveying use power-of-a-point-style circle intersection equations to resolve a receiver's position from distance measurements to multiple satellites treated as circles/spheres of known radius. In computer graphics and robotics, Ptolemy's inequality underlies fast tests for whether a point lies inside a circle (a primitive used constantly in Delaunay triangulation for mesh generation and motion planning), since the inequality direction flips exactly at the boundary of the circumcircle.

Example: Locating a Receiver by Circle Intersection

Two ground beacons at points $S_1, S_2$ broadcast signals that let a receiver $P$ compute its distances $d_1 = PS_1$ and $d_2 = PS_2$ . If a fixed reference circle passes through both beacons, express $\mathrm{pow}(P)$ using $d_1, d_2$ along the two beacon lines, and explain why a third beacon is needed to pin down $P$ uniquely.

Solution

Step 1: If the line $PS_1$ extended meets the reference circle again at a second point $S_1'$ , then by the power of a point $\mathrm{pow}(P) = \overrightarrow{PS_1}\cdot\overrightarrow{PS_1'} = d_1 \cdot PS_1'$ ; similarly along the other line $\mathrm{pow}(P) = d_2 \cdot PS_2'$ using the second intersection $S_2'$ .

Step 2: Since $\mathrm{pow}(P)$ is a single number depending only on $P$ and the fixed circle (not on which line through $P$ was used), these two expressions must be equal: $d_1 \cdot PS_1' = d_2 \cdot PS_2'$ , one algebraic constraint linking $P$ 's two measured distances.

Step 3: This single equation restricts $P$ to lie on a curve (in general a circle or line, by the same power-of-a-point argument run in reverse — the locus of points with a fixed power is itself a circle concentric with the reference circle), not to a single point: one distance measurement to a beacon pair only constrains $P$ to a curve, so a single pair of beacons cannot pin down a unique location.

Step 4: A third beacon $S_3$ not on the line through $S_1,S_2$ contributes an independent circle-power equation, and generically two such curves (each a specific circle) intersect in at most two points, so a third measurement is what collapses the remaining ambiguity to (at most) a unique receiver position — exactly why real positioning systems (GPS, indoor beacon trilateration) require three or more reference points.

Example: Testing Concyclicity with Ptolemy's Inequality

A mesh-generation algorithm needs to decide whether a query point $D$ lies inside the circumcircle of a triangle $ABC$ (the Delaunay condition). Using Ptolemy's inequality $AC\cdot BD \le AB\cdot CD + AD\cdot BC$ (equality iff $ABCD$ cyclic), explain how the sign of $AB\cdot CD + AD\cdot BC - AC\cdot BD$ tells you whether $D$ is inside, on, or outside the circumcircle.

Solution

Step 1: Fix triangle $ABC$ and its circumcircle. Ptolemy's inequality holds for any four points, with equality precisely when the four points are concyclic in the order $A,B,C,D$ : $AC\cdot BD = AB\cdot CD + AD\cdot BC \iff D$ lies exactly on the circumcircle of $ABC$ .

Step 2: Define $f(D) = AB\cdot CD + AD\cdot BC - AC\cdot BD$ . By Step 1, $f(D) = 0$ exactly when $D$ is on the circumcircle.

Step 3: As $D$ moves continuously from far outside the circle to the center, $f(D)$ varies continuously (it is built from continuous distance functions), and general position arguments (checking one interior point, e.g. the circumcenter, and one exterior point, e.g. a point at infinity along a ray) show $f(D) > 0$ throughout the open interior (Ptolemy's inequality is strict there) and $f(D) < 0$ throughout the exterior.

Step 4: Therefore the sign of $f(D) = AB\cdot CD + AD\cdot BC - AC\cdot BD$ directly classifies $D$ : positive means $D$ is strictly inside the circumcircle (satisfies the Delaunay in-circle test), zero means $D$ is exactly on it, and negative means $D$ is strictly outside — turning a geometric containment question into a single arithmetic sign check, exactly the primitive used millions of times per mesh in Delaunay triangulation software.

Common mistake. A common mistake with the power of a point is forgetting the sign: when

P

is inside the circle,

\mathrm{pow}(P) = OP^2-r^2

is negative, and the correct relation is

PA \cdot PB = r^2 - OP^2

(both measured as positive lengths) or equivalently

\overrightarrow{PA}\cdot\overrightarrow{PB} = OP^2-r^2

with a genuine sign flip — plugging in unsigned lengths blindly at an interior point gives the wrong equation. With Ceva and Menelaus, the frequent error is mixing up when ratios should be signed: Ceva's product equals

+1

for genuinely concurrent cevians (unsigned lengths, or signed with an even number of external points), while Menelaus' collinearity condition needs the product to be

-1

using signed ratios — swapping the two signs is the single most common wrong answer on transversal problems.

Research frontier as of 2026

Olympiad geometry itself became a research testbed for AI reasoning: in 2024, Google DeepMind's AlphaGeometry system solved 25 of 30 IMO geometry problems from the past 25 years at a level matching a human gold medalist, by combining a neural language model with a symbolic deduction engine that chains exactly the classical tools discussed here (power of a point, similar triangles, cyclic quadrilaterals) together with auxiliary-point construction; the successor AlphaGeometry2 (2024) further improved coverage to about 84% of a larger historical IMO geometry problem set. Beyond AI, active research directions include projective and inversive generalizations of Ceva/Menelaus (e.g. for conics rather than lines), and computational formalization of olympiad-style synthetic proofs in proof assistants like Lean, an ongoing effort to make classical Euclidean arguments machine-checkable.

A point $P$ is outside a circle of radius $r=5$ with center $O$ , and $OP=13$ . A secant from $P$ meets the circle at $A$ and $B$ with $PA=4$ . What is $PB$ ?

In triangle $ABC$ , cevians $AD, BE, CF$ satisfy $\frac{BD}{DC}=2$ , $\frac{CE}{EA}=3$ . For the cevians to be concurrent by Ceva's theorem, what must $\frac{AF}{FB}$ equal?

Square $ABCD$ has side length $1$ and is inscribed in a circle. Using Ptolemy's theorem on cyclic quadrilateral $ABCD$ , compute the diagonal length $AC$ .

Which 2024 AI system solved 25 of 30 historical IMO geometry problems near gold-medalist level by combining a neural model with a symbolic deduction engine using power-of-a-point-style synthetic tools?

References

Trieu Trinh, Yuhuai Wu, Quoc Le, He He, Thang Luong (2024). AlphaGeometry: An Olympiad-level AI system for geometry
Evan Chen (2016). Euclidean Geometry in Mathematical Olympiads
H.S.M. Coxeter, S.L. Greitzer (1967). Geometry Revisited