Course Theory
Book 2. Olympiad Geometry Methods
Book 2. Olympiad Geometry Methods
- 1. Advanced Angle Chasing
- 2. Power of a Point
- 3. Radical Axis
- 4. Homothety and Spiral Similarity
- 5. Inversion I: First Contact
- 6. Ceva and Menelaus
- 7. Area Method II
- 8. Complete Quadrilaterals and Miquel Points
- 9. Geometry with Coordinates and Vectors
- 10. Mixed Problems II
Chapter
Advanced Angle Chasing
Key Idea
In Book 1 we often wrote equalities of ordinary angles. At the intermediate olympiad level it is more convenient to work with oriented angles modulo \(180^\circ\). This notation avoids separate casework about which side of a line the points lie on and makes proofs shorter.
The main pattern of the module is: if points \(A,B,C,D\) lie on one circle, then \(\angle ABC\equiv\angle ADC\pmod{180^\circ}\), because both angles stand on chord \(AC\). Conversely, if such an angle equality holds and the configuration is non-degenerate, it often proves cyclicity.
Basic Facts
An oriented angle \(\angle(\ell_1,\ell_2)\) is the angle of rotation from line \(\ell_1\) to line \(\ell_2\), considered modulo \(180^\circ\). Therefore parallel lines have oriented angle zero.
For four non-degenerate points: \(A,B,C,D\) lie on one circle if and only if \(\angle ABC\equiv\angle ADC\pmod{180^\circ}\). This is the useful cyclicity criterion.
The tangent-chord theorem in oriented form says: the angle between the tangent to a circle at \(A\) and chord \(AB\) equals the angle standing on chord \(AB\): \(\angle(t,AB)\equiv\angle ACB\pmod{180^\circ}\).
The angle between two circles at their common point is the angle between the tangents to the circles at that point. It is often found using the tangent-chord theorem.
When to Use This Method
Use oriented angles when a problem has several circles, points may lie on different sides of lines, exterior angles appear, tangents are involved, or cyclicity must be proved without long casework.
If you need to prove parallelism, it is enough to get \(\angle(\ell_1,\ell_2)\equiv0\pmod{180^\circ}\). If you need to prove tangency, it is enough to show that the angle between the proposed tangent and a chord equals the angle in the opposite arc.
How to Recognise the Method
Signals include: four points almost form a circle; angles of the form \(\angle ABC\) and \(\angle ADC\) appear; a tangent touches a circumcircle; two circles intersect; one must prove that a line is tangent or that two lines are parallel.
A good move is to choose one chord and replace all angles standing on it. When a tangent appears, immediately ask: with which chord does it form a useful angle?
Typical Mistakes
Do not mix ordinary non-oriented angles and oriented angles in one chain without explanation. If you write \(\equiv\pmod{180^\circ}\), you may move through exterior angles, but you cannot suddenly conclude equality of lengths.
Do not apply the cyclicity criterion if points coincide or if three of the needed points are collinear. In tangent problems, track which circle the tangent belongs to and which chord is being used.
Mini-Checklist
1. Is there a chord seen by two angles? 2. Can the angles be written modulo \(180^\circ\)? 3. Do we need to prove a circle or use an existing one? 4. Is there a tangent and a suitable chord? 5. Do we need to prove parallelism via zero oriented angle? 6. If two circles intersect, can the angle between them be replaced by angles on the common chord?
Example 1. A Cyclic Quadrilateral in New Notation
This example shows why oriented angles are introduced.
Problem. Points \(A,B,C,D\) lie on one circle. Prove that \(\angle ABC\equiv\angle ADC\pmod{180^\circ}\).
Both angles stand on chord \(AC\). If points \(B\) and \(D\) lie on the same side of \(AC\), the angles are equal as ordinary angles. If they lie on opposite sides, the angles are supplementary. In both cases the oriented notation gives \(\angle ABC\equiv\angle ADC\pmod{180^\circ}\).
Comment. One notation replaces two positional cases.
Example 2. The Reverse Move: Proving a Circle
Equality of oriented angles is often a cyclicity criterion.
Problem. For four distinct points \(A,B,C,D\), with no three of them collinear, it is known that \(\angle ABC\equiv\angle ADC\pmod{180^\circ}\). Prove that points \(A,B,C,D\) lie on one circle.
Consider the circle through \(A,B,C\). A point \(D'\) on this circle with the same angular view of chord \(AC\) satisfies \(\angle AD'C\equiv\angle ABC\pmod{180^\circ}\). By the condition, point \(D\) gives the same angle. The locus of points from which segment \(AC\) is seen under a fixed oriented angle is an arc of a circle through \(A\) and \(C\). Therefore \(D\) lies on the same circle.
Example 3. Tangent and Chord
A tangent becomes an angle in the opposite arc.
Problem. In triangle \(ABC\), line \(t\) is tangent to the circumcircle at \(A\). Prove that \(\angle(t,AB)\equiv\angle ACB\pmod{180^\circ}\).
Draw radius \(OA\). It is perpendicular to the tangent. The central angle \(\angle AOB\) is twice the inscribed angle \(\angle ACB\). Therefore the angle between the tangent and chord \(AB\) equals half of the corresponding central angle, that is \(\angle ACB\). In oriented notation this gives \(\angle(t,AB)\equiv\angle ACB\pmod{180^\circ}\).
Example 4. How to Recognise a Tangent
If a line forms the correct angle with a chord, it is a tangent.
Problem. Points \(A,B,C\) lie on a circle. Line \(l\) passes through \(A\), and \(\angle(l,AB)\equiv\angle ACB\pmod{180^\circ}\). Prove that \(l\) is tangent to the circle at \(A\).
The tangent \(t\) to the circle at \(A\) satisfies \(\angle(t,AB)\equiv\angle ACB\pmod{180^\circ}\). By the condition, the same is true for line \(l\). Through point \(A\) there is a unique line forming the given oriented angle with \(AB\), so \(l=t\). Hence \(l\) is the tangent.
Example 5. Angle Between Two Circles
Two circles are conveniently compared through their tangents at a common point.
Problem. Circles \(\omega_1\) and \(\omega_2\) intersect at \(A\) and \(B\). Point \(C\) lies on \(\omega_1\), and point \(D\) lies on \(\omega_2\). Prove that the angle between the circles at \(A\) equals \(\angle ACB-\angle ADB\) in oriented notation.
Let \(t_1\) and \(t_2\) be the tangents to \(\omega_1\) and \(\omega_2\) at \(A\). By the tangent-chord theorem, \(\angle(t_1,AB)\equiv\angle ACB\), and \(\angle(t_2,AB)\equiv\angle ADB\). Therefore \(\angle(t_1,t_2)\equiv\angle ACB-\angle ADB\pmod{180^\circ}\).
Example 6. Antiparallel Lines in a Triangle
A cyclic quadruple inside an angle gives similar but not parallel angles.
Problem. In triangle \(ABC\), points \(D\) and \(E\) lie on \(AB\) and \(AC\). If \(B,C,D,E\) lie on one circle, prove that \(\triangle ADE\sim\triangle ACB\).
Since \(B,C,D,E\) are cyclic, \(\angle BDE\equiv\angle BCE\) and \(\angle BED\equiv\angle BCD\). But \(BD\) lies on \(BA\), and \(CE\) lies on \(CA\). Hence \(\angle ADE\equiv\angle ACB\) and \(\angle AED\equiv\angle ABC\). Therefore \(\triangle ADE\sim\triangle ACB\).
Example 7. Two Tangents to a Circumcircle
A classical intermediate-level move: the smaller angle between tangents is expressed through a central angle.
Problem. Tangents to the circumcircle of triangle \(ABC\), where \(\angle BAC<90^\circ\), are drawn at \(B\) and \(C\), meeting at \(T\). Prove that the smaller angle \(\angle BTC=180^\circ-2\angle BAC\).
Let \(O\) be the centre of the circumcircle. Radii \(OB\) and \(OC\) are perpendicular to tangents \(TB\) and \(TC\). Since \(\angle BAC<90^\circ\), the smaller central angle \(\angle BOC\), standing on arc \(BC\), equals \(2\angle BAC\). The smaller angle between the tangents is supplementary to it, so \(\angle BTC=180^\circ-2\angle BAC\).
Comment. The smaller angle between the tangents is specified deliberately; in oriented notation one can also use the exterior angle, but it must be named separately.
Example 8. Reim's Theorem as Angle Chasing
This is a more mature technique: two circles and two secants give parallelism.
Problem. Two circles intersect at \(A\) and \(B\). A line through \(A\) meets the first circle again at \(C\) and the second again at \(D\). A line through \(B\) meets the first circle again at \(E\) and the second again at \(F\). Prove that \(CE\parallel DF\).
Since \(A,B,C,E\) lie on the first circle, \(\angle(CE,EB)\equiv\angle(CA,AB)\). Since \(A,B,D,F\) lie on the second circle, \(\angle(DF,FB)\equiv\angle(DA,AB)\). But \(C,A,D\) are collinear, and \(E,B,F\) are collinear. Thus lines \(CE\) and \(DF\) form equal oriented angles with the same line, so \(CE\parallel DF\).
Chapter
Power of a Point
Key Idea
Power of a point turns a circle into an equality of products of segments. If two secants from point \(P\) meet a circle at \(A,B\) and \(C,D\), then \(PA\cdot PB=PC\cdot PD\). If \(PT\) is a tangent from \(P\), then \(PT^2=PA\cdot PB\).
In an olympiad problem this often means: do not try to find angles or lengths separately. See two secants, one tangent and one secant, or two chords, and write the correct product.
Basic Facts
Secant-secant: if point \(P\) lies outside a circle, and secants through \(P\) meet the circle in the order \(P,A,B\) and \(P,C,D\), then \(PA\cdot PB=PC\cdot PD\).
Tangent-secant: if \(PT\) is a tangent and a secant through \(P\) meets the circle at \(A,B\), then \(PT^2=PA\cdot PB\).
Intersecting chords: if chords \(AB\) and \(CD\) meet at point \(X\) inside the circle, then \(XA\cdot XB=XC\cdot XD\).
Converse criterion: if points \(A,B\) lie on one line with \(P\), points \(C,D\) lie on another, and \(PA\cdot PB=PC\cdot PD\), then points \(A,B,C,D\) lie on one circle in a non-degenerate configuration.
When to Use This Method
Look for power of a point when there is a circle and a point from which two lines go to the circle. Strong signals are: a tangent, two secants, intersecting chords, a product of segments, a square of a length, or a ratio like \(PA:PB\).
The method is useful for finding a length, proving equality of products, proving cyclicity, obtaining a ratio, and preparing for the radical axis.
How to Recognise the Method
If the statement has a point outside a circle and two lines through it, almost always try to write \(PA\cdot PB=PC\cdot PD\). If there is a tangent, immediately look for \(PT^2=PA\cdot PB\). If two chords meet inside the circle, use the products of the four small segments.
If the circle is not given explicitly but an equality of products appears, try to prove that four points lie on one circle. This is sometimes faster than angle chasing.
Typical Mistakes
Do not confuse the near and far secant segments: the formula uses the distances from the external point to both intersection points with the circle, not only the part inside the circle.
In the tangent formula \(PT^2=PA\cdot PB\), point \(T\) is the point of tangency to this particular circle. For points inside a circle and for side extensions, directed segments may be needed; in this module the position is stated explicitly to avoid sign issues.
Mini-Checklist
1. Where is point \(P\)? 2. Which two lines through it meet the circle? 3. Which points are near and far? 4. Is there a tangent? 5. Are there intersecting chords inside the circle? 6. Can cyclicity be proved by the converse criterion? 7. Do we first need to extend a side to a second intersection point?
Example 1. Two Secants From One Point
Basic technique: correctly choose the near and far points on each secant.
Problem. From point \(P\) outside a circle, two secants are drawn. The first meets the circle at \(A,B\), the second at \(C,D\), with \(P,A,B\) and \(P,C,D\) in that order. If \(PA=6\), \(PB=15\), \(PC=9\), find \(PD\).
By the secant theorem, \(PA\cdot PB=PC\cdot PD\). Hence \(6\cdot15=9\cdot PD\), so \(PD=10\).
Example 2. Tangent and Secant
The square of the tangent equals the product of the whole secant and its external part.
Problem. From point \(P\), tangent \(PT\) and secant \(PAB\) are drawn, where \(A\) is the nearer point of the circle. If \(PA=4\), \(PB=25\), find \(PT\).
By the tangent-secant theorem, \(PT^2=PA\cdot PB=4\cdot25=100\). Therefore \(PT=10\).
Example 3. Intersecting Chords
If the point is inside the circle, we use products of the parts of chords.
Problem. Chords \(AB\) and \(CD\) meet at point \(X\). It is known that \(XA=3\), \(XB=12\), \(XC=4\). Find \(XD\).
For intersecting chords, \(XA\cdot XB=XC\cdot XD\). Hence \(3\cdot12=4\cdot XD\), and \(XD=9\).
Example 4. Proof Through Similarity
The secant theorem is not magic: it follows from similarity of triangles.
Problem. From point \(P\) outside a circle, secants \(PAB\) and \(PCD\) are drawn. Prove that \(PA\cdot PB=PC\cdot PD\).
Join \(A\) to \(C\), and \(B\) to \(D\). Consider triangles \(PAC\) and \(PDB\). The angle at \(P\) is common, because \(P,A,B\) and \(P,C,D\) are collinear. Also, \(\angle PAC=\angle BAC\), \(\angle PDB=\angle CDB\), and angles \(\angle BAC\) and \(\angle CDB\) are equal as inscribed angles standing on chord \(BC\). Therefore \(\triangle PAC\sim\triangle PDB\). From similarity, \(\frac{PA}{PD}=\frac{PC}{PB}\), hence \(PA\cdot PB=PC\cdot PD\).
Example 5. Converse Circle Criterion
An equality of products may create a circle instead of using one.
Problem. Points \(A,B\) lie on one ray starting at \(P\), points \(C,D\) lie on another ray, and \(PA\cdot PB=PC\cdot PD\). Prove that \(A,B,C,D\) lie on one circle.
Draw the circle through \(A,B,C\). Let its second intersection with ray \(PC\) be \(D'\). By the secant theorem, \(PA\cdot PB=PC\cdot PD'\). By the condition, \(PA\cdot PB=PC\cdot PD\). Since \(PC>0\), we get \(PD'=PD\), hence \(D'=D\). Therefore \(A,B,C,D\) are cyclic.
Example 6. A Small Circle in a Triangle
Power of a point often gives a ratio on sides of a triangle.
Problem. In triangle \(ABC\), points \(D\) and \(E\) lie on \(AB\) and \(AC\). If \(B,C,D,E\) lie on one circle, prove that \(AD\cdot AB=AE\cdot AC\).
Consider point \(A\) with respect to circle \((BCDE)\). Lines \(AB\) and \(AC\) are secants: the first meets the circle at \(D,B\), the second at \(E,C\). Therefore \(AD\cdot AB=AE\cdot AC\).
Example 7. Tangent to a Circumcircle
Here power of a point combines with the previous module's tangent-chord result.
Problem. The tangent to the circumcircle of triangle \(ABC\) at \(A\) meets line \(BC\) at point \(T\), with \(T\) outside segment \(BC\). Prove that \(TB:TC=AB^2:AC^2\).
By the tangent-secant theorem, \(TA^2=TB\cdot TC\). Also, by the tangent-chord theorem, \(\triangle TAB\sim\triangle TCA\). From similarity, \(\frac{TB}{TA}=\frac{AB}{AC}\) and \(\frac{TA}{TC}=\frac{AB}{AC}\). Multiplying, we get \(\frac{TB}{TC}=\frac{AB^2}{AC^2}\).
Example 8. Equal Powers
This prepares for the radical axis: equal powers mean equal products of secants.
Problem. Two circles intersect at \(A\) and \(B\). Point \(P\) lies on line \(AB\) outside both circles. Prove that the powers of point \(P\) with respect to these circles are equal.
Line \(PAB\) is a secant for both circles, and the intersection points with each circle are the same points \(A\) and \(B\). Therefore the power of \(P\) with respect to the first circle is \(PA\cdot PB\), and with respect to the second circle it is also \(PA\cdot PB\). Hence the powers are equal.
Chapter
Radical Axis
Key Idea
The radical axis of two circles is the set of points having equal powers with respect to these circles. If point \(X\) lies on the radical axis of circles \(\omega_1\) and \(\omega_2\), then \(\operatorname{Pow}_{\omega_1}(X)=\operatorname{Pow}_{\omega_2}(X)\).
In olympiad problems, the radical axis is most often used to prove collinearity: to prove that several points lie on one line, it is enough to show that each of them has equal powers with respect to two fixed circles.
Basic Facts
If two circles intersect at points \(A\) and \(B\), then their radical axis is line \(AB\). Indeed, for every point \(X\) on \(AB\), the powers with respect to both circles are equal to \(XA\cdot XB\).
If the circles are tangent at point \(A\), then their radical axis is the common tangent at \(A\). For every point \(X\) on this tangent, the power to each circle equals \(XA^2\).
If the circles have centres \(O_1,O_2\) and radii \(r_1,r_2\), then equality of powers is written as \(XO_1^2-r_1^2=XO_2^2-r_2^2\). Therefore the radical axis is perpendicular to line \(O_1O_2\).
For three circles, the radical axes of the pairwise pairs are either parallel or meet at one point. This point is called the radical center.
When to Use This Method
Use the radical axis when a problem has two or three circles, a common chord, equal tangents, equal products of secants, or a request to prove that three points are collinear.
The method is especially useful when angle chasing becomes long but products of segments are already visible: \(XA\cdot XB\), \(XT^2\), equal tangents, or intersections of common chords.
How to Recognise the Method
If two circles intersect, immediately mark their common chord: it is the radical axis. If the circles do not intersect, look for two points with equal powers, for example points from which tangent lengths to the two circles are equal.
If there are three circles, try to find two radical axes. Their intersection automatically lies on the third radical axis.
Typical Mistakes
Do not confuse the radical axis with the line of centres: in fact, the radical axis is perpendicular to the line of centres. Do not say “common chord” if the circles do not intersect; in that case the radical axis still exists, but it must be found through equal powers.
Do not use the radical center without checking: one must show that the point has equal powers with respect to at least two pairs of circles. Also track signs if the point lies inside a circle; in this module the configurations state the needed products explicitly.
Mini-Checklist
1. Which two circles are being compared? 2. Do they have a common chord? 3. Can equal powers be written through secants or tangents? 4. Do we need to prove collinearity? 5. Is there a third circle and a radical center? 6. Where is the line of centres, and is the radical axis perpendicular to it? 7. Is the radical axis hidden in an equality of products?
Example 1. Common Chord
The first and most important way to see a radical axis is to find the common chord of two circles.
Problem. Circles \(\omega_1\) and \(\omega_2\) intersect at points \(A\) and \(B\). Prove that line \(AB\) is their radical axis.
Take any point \(X\) on line \(AB\). For the first circle, secant \(XAB\) gives power \(XA\cdot XB\). For the second circle, the same secant has the same intersection points \(A\) and \(B\), so the power is also \(XA\cdot XB\). Thus the powers are equal for all points \(X\) on \(AB\), and \(AB\) is the radical axis.
Example 2. Tangent Circles
If circles are tangent, the radical axis does not disappear: it becomes the common tangent.
Problem. Two circles are tangent at point \(A\). Prove that their common tangent at \(A\) is the radical axis.
Let \(X\) be a point on the common tangent. With respect to the first circle, \(XA\) is a tangent, so the power equals \(XA^2\). With respect to the second circle, the same line is also tangent at \(A\), so the power is again \(XA^2\). Therefore all points of the common tangent have equal powers.
Example 3. The Radical Axis is Perpendicular to the Line of Centres
This explains the shape of the radical axis even when the circles do not intersect.
Problem. Circles have centres \(O_1,O_2\) and radii \(r_1,r_2\). Prove that their radical axis is perpendicular to \(O_1O_2\).
For a point \(X\) on the radical axis, \(XO_1^2-r_1^2=XO_2^2-r_2^2\), that is \(XO_1^2-XO_2^2=r_1^2-r_2^2\). The locus of points with a constant difference of squares of distances to two fixed points \(O_1,O_2\) is a line perpendicular to \(O_1O_2\). Hence the radical axis is perpendicular to the line of centres.
Example 4. Radical Center
Three circles do not give three random lines, but one intersection point of radical axes.
Problem. The radical axes of circles \(\omega_1,\omega_2\) and \(\omega_2,\omega_3\) meet at point \(R\). Prove that \(R\) lies on the radical axis of \(\omega_1\) and \(\omega_3\).
Since \(R\) lies on the radical axis of \(\omega_1\) and \(\omega_2\), we have \(\operatorname{Pow}_{\omega_1}(R)=\operatorname{Pow}_{\omega_2}(R)\). Since \(R\) lies on the radical axis of \(\omega_2\) and \(\omega_3\), we have \(\operatorname{Pow}_{\omega_2}(R)=\operatorname{Pow}_{\omega_3}(R)\). Therefore \(\operatorname{Pow}_{\omega_1}(R)=\operatorname{Pow}_{\omega_3}(R)\), so \(R\) lies on the third radical axis.
Example 5. Collinearity Through Equal Tangents
One of the main applications of the method is to prove that a point lies on a known line.
Problem. Two circles intersect at points \(A\) and \(B\). Point \(P\) lies outside both circles. Tangents \(PT_1\) and \(PT_2\) are drawn from \(P\) to them, and \(PT_1=PT_2\). Prove that \(P,A,B\) are collinear.
The power of point \(P\) with respect to the first circle is \(PT_1^2\), and with respect to the second circle is \(PT_2^2\). By the condition these are equal, so \(P\) lies on the radical axis of the two circles. Since the circles intersect at \(A\) and \(B\), their radical axis is line \(AB\). Therefore \(P,A,B\) are collinear.
Example 6. Common Chords of Three Circles
The radical center often appears as the intersection point of common chords.
Problem. Three circles \(\omega_1,\omega_2,\omega_3\) intersect pairwise. The common chord of \(\omega_1\) and \(\omega_2\) meets the common chord of \(\omega_2\) and \(\omega_3\) at point \(R\). Prove that \(R\) lies on the common chord of \(\omega_1\) and \(\omega_3\).
The common chord of two circles is their radical axis. Therefore \(R\) lies on the radical axis of \(\omega_1,\omega_2\) and on the radical axis of \(\omega_2,\omega_3\). By the radical center theorem, \(R\) also lies on the radical axis of \(\omega_1,\omega_3\). Since these circles intersect, their radical axis is their common chord.
Example 7. A Hidden Radical Axis
Sometimes the radical axis is not given as a common chord; it must be recognised from equal products.
Problem. For two circles \(\omega_1\) and \(\omega_2\), point \(P\) has secants \(PAB\) to \(\omega_1\) and \(PCD\) to \(\omega_2\). It is known that \(PA\cdot PB=PC\cdot PD\). Prove that \(P\) lies on the radical axis of these circles.
The product \(PA\cdot PB\) is the power of point \(P\) with respect to \(\omega_1\), while \(PC\cdot PD\) is the power with respect to \(\omega_2\). These products are equal, so the powers are equal. Therefore \(P\) lies on the radical axis.
Example 8. Orthogonal Circles and the Radical Axis
This is a stronger idea: the centre of a circle orthogonal to two given circles lies on their radical axis.
Problem. Circle \(\gamma\) with centre \(X\) and radius \(\rho\) is orthogonal to circles \(\omega_1(O_1,r_1)\) and \(\omega_2(O_2,r_2)\). Prove that \(X\) lies on the radical axis of \(\omega_1\) and \(\omega_2\).
Orthogonality of circles \(\gamma\) and \(\omega_1\) gives \(XO_1^2=\rho^2+r_1^2\), hence \(XO_1^2-r_1^2= ho^2\). Similarly, \(XO_2^2-r_2^2= ho^2\). Therefore the powers of point \(X\) with respect to \(\omega_1\) and \(\omega_2\) are equal, and \(X\) lies on their radical axis.
Chapter
Homothety and Spiral Similarity
Key Idea
A homothety preserves the shape of a figure and sends each point to a point on the same line through the centre. Therefore it immediately gives parallel sides, equal angles, and ratios of lengths.
A spiral similarity does the same with a rotation: one segment is sent to another after a rotation and a dilation. If two pairs of segments are seen from one point under equal angles and with the same ratio of lengths, that point is often the centre of a spiral similarity.
Basic Facts
If a homothety with centre \(O\) sends \(A\) to \(C\) and \(B\) to \(D\), then \(O,A,C\) are collinear, \(O,B,D\) are collinear, and \(AB \parallel CD\). Moreover, \(\frac{OA}{OC}=\frac{OB}{OD}=\frac{AB}{CD}\), with corresponding lengths taken correctly.
The converse is often used in problems: if \(A,C,O\) are collinear, \(B,D,O\) are collinear, and \(\frac{OA}{OC}=\frac{OB}{OD}\), then \(AB \parallel CD\), and triangles \(OAB\) and \(OCD\) are similar.
A point \(P\) is the centre of a spiral similarity sending segment \(AC\) to segment \(BD\) if \(\triangle PAC \sim \triangle PBD\) with correspondences \(PA \leftrightarrow PB\), \(PC \leftrightarrow PD\), \(AC \leftrightarrow BD\).
When to Use This Method
Homothety is useful when the problem contains parallel segments, several points on two rays from one point, midpoints of sides, trapezoids, tangent circles, or circles with common tangents.
Spiral similarity is useful when two pairs of segments appear, when similar triangles are arranged around one point, when two circles intersect, or when one has to prove equality of angles between different segments.
How to Recognise the Method
Look for the centre: the intersection of the lines joining corresponding vertices. For a homothety, corresponding sides should be parallel. For a spiral similarity, instead of parallelism there is usually equality of angles and the same ratio of two pairs of distances.
A useful signal is that the solution wants to prove \(AB \parallel CD\), \(\frac{OA}{OC}=\frac{OB}{OD}\), \(\angle APC=\angle BPD\), or \(\triangle PAC \sim \triangle PBD\).
Typical Mistakes
Do not confuse the centre of homothety with a midpoint: the centre may lie outside the figure. Do not write a length ratio before checking the correspondence of points. In spiral similarity, the order of vertices matters: a wrong correspondence may give plausible numbers but wrong angles.
Another common mistake is trying to prove a spiral similarity from only one equal angle. Usually one needs either similar triangles, or an equal angle together with a ratio of corresponding sides.
Mini-Checklist
1. Find a possible centre: the intersection of lines through corresponding points.
2. Check collinearity for homothety or equality of angles for spiral similarity.
3. Write the correct ratio of corresponding sides.
4. Translate the claim into similarity of triangles.
5. After proving similarity, return to the target: parallelism, collinearity, a ratio, or equality of angles.
Example 1. Homothety from Parallel Segments
This example shows the basic link between parallelism and a centre of homothety.
Problem. Lines \(AC\) and \(BD\) meet at \(O\), and \(AB \parallel CD\). Prove that \(\frac{OA}{OC}=\frac{OB}{OD}=\frac{AB}{CD}\).
Since \(AB \parallel CD\), we have \(\angle OAB=\angle OCD\) and \(\angle OBA=\angle ODC\). Hence \(\triangle OAB \sim \triangle OCD\). From similarity, \(\frac{OA}{OC}=\frac{OB}{OD}=\frac{AB}{CD}\). Thus \(O\) is the centre of the homothety sending \(AB\) to \(CD\).
Comment. In problems with trapezoids and parallel segments, this is one of the fastest moves.
Example 2. A Midline as a Homothety
Here homothety explains a familiar fact without a separate angle computation.
Problem. In triangle \(ABC\), points \(D\) and \(E\) are the midpoints of \(AB\) and \(AC\). Prove that \(DE \parallel BC\) and \(DE=\frac{1}{2}BC\).
The homothety with centre \(A\) and ratio \(\frac{1}{2}\) sends \(B\) to \(D\) and \(C\) to \(E\). Therefore the image of segment \(BC\) is segment \(DE\). Corresponding segments under a homothety are parallel, and lengths are multiplied by the ratio. Hence \(DE \parallel BC\) and \(DE=\frac{1}{2}BC\).
Example 3. Centre of Homothety of Two Circles
This example teaches students to look for the centre on the line of centres, not necessarily at a tangency point.
Problem. Circles with centres \(O_1\) and \(O_2\) and radii \(r_1\) and \(r_2\) have an external centre of homothety \(H\). Prove that \(H,O_1,O_2\) are collinear and \(\frac{HO_1}{HO_2}=\frac{r_1}{r_2}\).
The homothety sends the first circle to the second one, so it sends the centre of the first circle to the centre of the second: \(O_1 \mapsto O_2\). Under a homothety, a point, its image, and the centre of homothety are collinear. Therefore \(H,O_1,O_2\) are collinear. The ratio of the homothety equals the ratio of the radii, so \(\frac{HO_1}{HO_2}=\frac{r_1}{r_2}\).
Example 4. Parallelism from a Ratio
This is the reverse move: instead of using given parallelism, we prove it through similarity.
Problem. Points \(A,C,O\) lie on one line, points \(B,D,O\) lie on another line, and \(\frac{OA}{OC}=\frac{OB}{OD}\). Prove that \(AB \parallel CD\).
Triangles \(OAB\) and \(OCD\) have a common or vertical angle at \(O\), and the sides adjacent to this angle are proportional. Therefore \(\triangle OAB \sim \triangle OCD\). From similarity, \(\angle OAB=\angle OCD\). These are corresponding angles for lines \(AB\) and \(CD\), hence \(AB \parallel CD\).
Example 5. Spiral Similarity via Similar Triangles
This is the basic criterion: first prove similarity, then identify the centre.
Problem. For a point \(P\), \(\triangle PAC \sim \triangle PBD\). Prove that \(P\) is the centre of a spiral similarity sending \(AC\) to \(BD\).
From similarity, \(\frac{PA}{PB}=\frac{PC}{PD}=\frac{AC}{BD}\), and the corresponding angles are equal. Thus ray \(PA\) turns into ray \(PB\), and ray \(PC\) turns into ray \(PD\) by the same angle; at the same time, all corresponding distances are multiplied by one ratio. Hence a rotation about \(P\) followed by a dilation sends \(A\) to \(B\), \(C\) to \(D\), and segment \(AC\) to segment \(BD\).
Example 6. Hidden Similarity from a Spiral Centre
The problem shows how a spiral centre produces a new useful triangle similarity.
Problem. Let \(P\) be the centre of a spiral similarity sending \(A\) to \(B\) and \(C\) to \(D\). Prove that \(\triangle PAC \sim \triangle PBD\).
By the definition of a spiral similarity, there is one rotation angle, so \(\angle APB=\angle CPD\), and one dilation ratio, so \(\frac{PA}{PB}=\frac{PC}{PD}\). Then \(\angle APC=\angle BPD\), because the same rotation part is removed from both angles. Thus two sides around an equal included angle are proportional, and \(\triangle PAC \sim \triangle PBD\).
Comment. After this step, the needed ratios \(AC:BD\) and equal angles often appear immediately.
Example 7. A Trapezoid and a Centre of Homothety
Homothety helps work with the diagonals of a trapezoid without long computations.
Problem. In trapezoid \(ABCD\), bases \(AD\) and \(BC\) are parallel, and the diagonals meet at \(O\). Prove that \(\frac{AO}{OC}=\frac{DO}{OB}=\frac{AD}{BC}\).
Consider triangles \(AOD\) and \(COB\). Angles \(\angle AOD\) and \(\angle COB\) are vertical. Since \(AD \parallel BC\), we also have \(\angle ADO=\angle CBO\) and \(\angle DAO=\angle BCO\). Therefore \(\triangle AOD \sim \triangle COB\). Hence \(\frac{AO}{OC}=\frac{DO}{OB}=\frac{AD}{BC}\).
Example 8. Miquel Point Preview
The full theory of the Miquel point is not needed here; the goal is to see why intersections of circles often lead to spiral similarity.
Problem. Points \(B,A,D\) are collinear, and points \(C,A,E\) are collinear. Circles \((ABC)\) and \((ADE)\) meet again at \(M\). Prove that \(\angle BMC=\angle DME\).
Since \(M\) lies on circle \((ABC)\), \(\angle BMC=\angle BAC\). Since \(M\) lies on circle \((ADE)\), \(\angle DME=\angle DAE\). But rays \(AB\) and \(AD\) lie on one line, and rays \(AC\) and \(AE\) lie on another line, so \(\angle BAC=\angle DAE\). Therefore \(\angle BMC=\angle DME\).
Comment. This configuration is often the first step toward a spiral similarity sending one segment to another.
Chapter
Inversion I: First Contact
Key Idea
An inversion with centre \(O\) and radius \(R\) sends a point \(P\ne O\) to the point \(P'\) on ray \(OP\), where \(OP \cdot OP'=R^2\). Nearby points move far away, while distant points move closer to the centre.
The main olympiad use of inversion is that a circle through the centre becomes a line, while a line not passing through the centre becomes a circle through the centre. This often simplifies tangencies, angles, and configurations with several circles.
Basic Facts
If \(P'\) is the image of \(P\), then \(O,P,P'\) are collinear and \(OP \cdot OP'=R^2\). Points on the circle of inversion \(OP=R\) remain fixed.
A line through \(O\) maps to itself. A line not passing through \(O\) maps to a circle through \(O\). A circle through \(O\) maps to a line not passing through \(O\).
Inversion preserves angles between curves at points different from the centre. Therefore tangency maps to tangency if the tangency point is not \(O\).
When to Use This Method
Inversion is worth trying when a problem contains many circles through one point, several tangencies, products such as \(OA \cdot OB\), or when it would be useful to replace a circle through a chosen point by a line.
At the first level, it is especially useful to choose the centre at a common point of circles or at a point from which tangents are drawn. The radius is often chosen so that two important points swap places.
How to Recognise the Method
Look for a common point of circles, a tangency point, an expression \(OA \cdot OB\), and pairs of objects: a line and a circle, two circles through one point, or tangents from one point.
If after choosing the centre several circles pass through \(O\), inversion often turns them into lines. Then a complicated cyclic picture becomes a problem about lines and angles.
Typical Mistakes
The centre of inversion \(O\) has no finite image. Not every circle maps to a line: only a circle passing through the centre of inversion does. Not every line maps to a circle: a line through the centre remains a line.
Remember that \(P'\) lies on ray \(OP\), not merely on line \(OP\). In angle problems, inversion preserves the size of an angle, but the drawing may look reversed.
Mini-Checklist
1. Choose the centre \(O\): a common point of circles, a tangency point, or a point from which tangents are drawn.
2. Choose the radius: it is often convenient to take \(R^2=OA \cdot OB\), so that \(A\) and \(B\) swap places.
3. Determine the images of lines and circles.
4. Translate the goal into the image configuration: collinearity, cyclicity, tangency, or an angle.
5. Solve the simplified problem and return to the original one.
Example 1. An Inverse Point
The first skill is to find the distance to the image quickly.
Problem. An inversion has centre \(O\) and radius \(6\). If \(OP=4\), find \(OP'\).
By definition, \(OP \cdot OP'=R^2\). Therefore \(4 \cdot OP'=36\), so \(OP'=9\). Point \(P'\) lies on ray \(OP\).
Example 2. Fixed Points
This example shows why the circle of inversion is fixed pointwise.
Problem. Prove that a point \(A\) remains fixed if and only if \(OA=R\).
If \(A'=A\), then \(OA^2=OA \cdot OA'=R^2\), so \(OA=R\). Conversely, if \(OA=R\), then \(OA \cdot OA'=R^2=OA^2\), hence \(OA'=OA\), and on ray \(OA\) we get \(A'=A\).
Example 3. A Concentric Circle
The simplest image of a circle occurs when its centre coincides with the centre of inversion.
Problem. An inversion has radius \(10\). What is the image of the circle with centre \(O\) and radius \(5\)?
For every point \(P\) on the circle, \(OP=5\). Then \(5 \cdot OP'=100\), so \(OP'=20\). The image is the circle with centre \(O\) and radius \(20\).
Example 4. Image of a Line
This is the main technical fact of the first encounter with inversion.
Problem. Line \(l\) does not pass through \(O\). Let \(H\) be the foot of the perpendicular from \(O\) to \(l\), and let \(H'\) be the image of \(H\). Prove that the image of \(l\) lies on the circle with diameter \(OH'\).
Take \(P\in l\), and let \(P'\) be its image. We have \(OP \cdot OP'=OH \cdot OH'=R^2\), so \(\frac{OP}{OH'}=\frac{OH}{OP'}\). The angle at \(O\) is common, hence \(\triangle OPH \sim \triangle OH'P'\). Since \(\angle OHP=90^\circ\), we get \(\angle OP'H'=90^\circ\). Therefore \(P'\) lies on the circle with diameter \(OH'\).
Example 5. A Circle Through the Centre
This is the reverse move to the previous example.
Problem. A circle \(\omega\) passes through \(O\). Prove that its image is a line not passing through \(O\).
Inversion is its own inverse. By the previous example, a line not passing through \(O\) maps to a circle through \(O\). Therefore the inverse image of such a circle must be a line not passing through \(O\).
Example 6. Three Points Become Collinear
A circle through the centre is often replaced by one line.
Problem. Points \(A,B,C\) lie on a circle passing through \(O\). Prove that their images \(A',B',C'\) are collinear.
The image of a circle through the centre of inversion is a line. Therefore all images of points of this circle, except the centre itself, lie on one line. Hence \(A',B',C'\) are collinear.
Example 7. Tangency Is Preserved
Inversion is useful in tangency problems because it preserves angles.
Problem. Line \(l\) is tangent to circle \(\omega\) at \(T\ne O\). Prove that their images are also tangent at \(T'\).
Tangency means that the angle between the line and the circle is \(0^\circ\). Inversion preserves angles at points different from the centre, so the angle between the images at \(T'\) is also \(0^\circ\). Thus the images are tangent.
Example 8. Choosing the Radius
The radius is often chosen so that important points swap places.
Problem. On a ray from \(O\), points \(A\) and \(B\) satisfy \(OA=3\), \(OB=12\). Find the radius of the inversion sending \(A\) to \(B\).
We need \(OA \cdot OB=R^2\). Therefore \(R^2=3\cdot 12=36\), so \(R=6\).
Chapter
Ceva and Menelaus
Key Idea
Ceva's and Menelaus' theorems translate geometric statements about three lines or three points into a product of ratios on the sides of a triangle.
Ceva answers the question: when do three cevians pass through one point? Menelaus answers the question: when do three points on the sides or extensions of the sides lie on one line?
Basic Facts
Let in triangle \(ABC\), point \(D\) lie on \(BC\), point \(E\) on \(CA\), and point \(F\) on \(AB\). Then lines \(AD\), \(BE\), \(CF\) are concurrent if and only if
\[\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1.\]
If points \(D,E,F\) lie on lines \(BC,CA,AB\), then they are collinear if and only if, for directed ratios,
\[\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=-1.\]
In training problems with ordinary positive lengths, a common Menelaus version is: if exactly one of the three points lies on an extension of a side, the product of the corresponding positive ratios is \(1\).
When to Use This Method
Use Ceva when you need to prove that three lines pass through one point, or find the ratio for which this is possible. Use Menelaus when you need to prove that three points are collinear, or find where a line intersects a side of a triangle.
Both theorems are especially useful when the problem already gives ratios on the sides of a triangle, or when those ratios can be obtained from similarity, areas, angle bisectors, or parallel lines.
How to Recognise the Method
Signals for Ceva: "Prove that the lines meet at one point", "three cevians", "the intersection point of two lines lies on the third".
Signals for Menelaus: "Prove that the points are collinear", "a line intersects the sides of a triangle", "a point lies on an extension of a side".
Typical Mistakes
The most common mistake is confusing Ceva and Menelaus. Ceva is about concurrence of lines; Menelaus is about collinearity of points.
The second mistake is writing ratios in inconsistent order. If you start with \(\frac{BD}{DC}\), continue cyclically: \(\frac{CE}{EA}\), \(\frac{AF}{FB}\). In problems with points on extensions, remember directed ratios or explicitly switch to the positive-length version.
Mini-Checklist
1. Draw the triangle to which you will apply the theorem.
2. Mark the three points on the sides or extensions of the sides.
3. Decide whether you need concurrence of lines or collinearity of points.
4. Write the product of ratios in one cyclic order.
5. Find the unknown ratio or check equality to \(1\) for Ceva and \(-1\) for directed Menelaus.
Example 1. Finding a Ratio by Ceva
A basic example of direct substitution into Ceva's formula.
Problem. In triangle \(ABC\), points \(D,E,F\) lie on \(BC,CA,AB\). It is known that \(BD:DC=2:3\), \(CE:EA=3:4\). Find \(AF:FB\) if \(AD\), \(BE\), \(CF\) are concurrent.
By Ceva's theorem, \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1\). Therefore \(\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{AF}{FB}=1\). We get \(\frac{1}{2}\cdot\frac{AF}{FB}=1\), hence \(AF:FB=2:1\).
Example 2. Medians Meet at One Point
Ceva quickly proves the concurrence of medians.
Problem. Prove that the medians of a triangle are concurrent.
Let \(D,E,F\) be the midpoints of \(BC,CA,AB\). Then \(\frac{BD}{DC}=\frac{CE}{EA}=\frac{AF}{FB}=1\). The product is \(1\), so by Ceva's theorem the lines \(AD\), \(BE\), \(CF\), that is, the medians, are concurrent.
Example 3. Menelaus with One External Point
Here it is important to see that the statement is about collinearity of points, not concurrence of lines.
Problem. In triangle \(ABC\), points \(D\in BC\), \(E\in CA\), and point \(F\) lies on the extension of \(AB\) beyond \(B\). Let \(D,E,F\) be collinear, \(BD:DC=2:5\), \(CE:EA=5:3\). Find \(AF:FB\).
For positive lengths with one external point, we use Menelaus in the form \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1\). Hence \(\frac{2}{5}\cdot\frac{5}{3}\cdot\frac{AF}{FB}=1\), so \(\frac{2}{3}\cdot\frac{AF}{FB}=1\). Therefore \(AF:FB=3:2\).
Example 4. Checking Collinearity
Menelaus often proves that three points lie on one line.
Problem. In triangle \(ABC\), points \(D\in BC\), \(E\in CA\), and \(F\) lies on the extension of \(AB\) beyond \(B\). Given \(BD:DC=2:3\), \(CE:EA=3:4\), \(AF:FB=2:1\). Prove that \(D,E,F\) are collinear.
Compute the product of positive ratios: \(\frac{2}{3}\cdot\frac{3}{4}\cdot 2=1\). Since exactly one point lies on an extension of a side, by the positive form of Menelaus' theorem, points \(D,E,F\) are collinear.
Example 5. Do Not Confuse Ceva and Menelaus
The same product of ratios may correspond to different geometric goals.
Problem. In triangle \(ABC\), points \(D,E,F\) lie on sides \(BC,CA,AB\), and \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1\). What follows from Ceva's theorem?
If the points lie on the sides of the triangle and the product is \(1\), then by Ceva's theorem the lines \(AD\), \(BE\), \(CF\) are concurrent. This is not a statement about the collinearity of points \(D,E,F\).
Example 6. Angle Bisectors via Ceva
Ceva connects with the familiar angle bisector theorem.
Problem. Prove that the internal angle bisectors of a triangle are concurrent.
Let the angle bisector from \(A\) meet \(BC\) at \(D\), from \(B\) meet \(CA\) at \(E\), and from \(C\) meet \(AB\) at \(F\). By the angle bisector theorem, \(\frac{BD}{DC}=\frac{AB}{AC}\), \(\frac{CE}{EA}=\frac{BC}{BA}\), \(\frac{AF}{FB}=\frac{CA}{CB}\). The product is \(1\), hence by Ceva the angle bisectors are concurrent.
Example 7. Two Cevians Determine the Third
If two lines already meet, Ceva helps find where the third one must meet the side.
Problem. In triangle \(ABC\), points \(D\in BC\), \(E\in CA\), \(BD:DC=3:2\), \(CE:EA=5:6\). Lines \(AD\) and \(BE\) meet at \(P\), and \(CP\) meets \(AB\) at \(F\). Find \(AF:FB\).
Lines \(AD\), \(BE\), \(CF\) are concurrent at \(P\). By Ceva, \(\frac{3}{2}\cdot\frac{5}{6}\cdot\frac{AF}{FB}=1\). The first two factors give \(\frac{5}{4}\), so \(\frac{AF}{FB}=\frac{4}{5}\). Therefore \(AF:FB=4:5\).
Example 8. Ceva and Menelaus Together
In stronger problems, the same pair of points may be used by two different theorems.
Problem. In triangle \(ABC\), points \(D\in BC\), \(E\in CA\). Lines \(AD\) and \(BE\) meet at \(P\), \(CP\) meets \(AB\) at \(F\), and line \(DE\) meets the extension of \(AB\) at \(X\). Prove that \(\frac{AF}{FB}=\frac{AX}{XB}\) for ordinary lengths.
By Ceva for concurrent \(AD,BE,CF\): \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1\). By Menelaus for collinear \(D,E,X\): \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AX}{XB}=1\). The first two factors on the left are the same, hence \(\frac{AF}{FB}=\frac{AX}{XB}\).
Chapter
Area Method II
Key Idea
The area method turns ratios of segments into ratios of areas. If two triangles have a common altitude, their areas are in the ratio of their bases. If they have a common base, their areas are in the ratio of their altitudes.
In Book 2, the area method becomes a tool for working with cevians: through the areas of the small triangles around an interior point, one can quickly find ratios on the sides and prove concurrence.
Basic Facts
If \(D\in BC\), then \(\frac{[ABD]}{[ADC]}=\frac{BD}{DC}\), because triangles \(ABD\) and \(ADC\) have a common altitude from \(A\).
Let point \(P\) lie inside triangle \(ABC\). Denote \(S_A=[PBC]\), \(S_B=[PCA]\), \(S_C=[PAB]\). If \(AP\), \(BP\), \(CP\) meet \(BC\), \(CA\), \(AB\) at \(D,E,F\), then
\[\frac{BD}{DC}=\frac{S_C}{S_B},\qquad \frac{CE}{EA}=\frac{S_A}{S_C},\qquad \frac{AF}{FB}=\frac{S_B}{S_A}.\]
Also, if \(P\in AD\), where \(D\in BC\), then \(\frac{[PBC]}{[ABC]}=\frac{PD}{AD}\). This is the first step toward mass points: areas show how a point divides a cevian.
When to Use This Method
The area method is useful when a problem contains ratios on sides, cevians, medians, intersection points inside a triangle, or when one needs to prove equality of ratios without trigonometry.
It is especially powerful after Ceva and Menelaus: sometimes instead of searching for similarity, it is enough to express three ratios through three small areas.
How to Recognise the Method
Look for triangles with a common altitude or a common base. If a point lies inside a triangle and cevians are drawn through it, you can almost always denote three areas \(S_A\), \(S_B\), \(S_C\).
Signals include: "Find the ratio", "Prove equality of ratios", "intersection point of cevians", "median", and "without trigonometry".
Typical Mistakes
Do not compare areas of triangles unless they have a common altitude or a common base. First explicitly name which altitude is common.
Do not confuse the areas around point \(P\): \(S_A=[PBC]\) lies opposite vertex \(A\), not near it. In cevian formulas, the order of vertices matters.
Mini-Checklist
1. Find pairs of triangles with a common altitude or a common base.
2. If there is an interior point \(P\), denote \(S_A=[PBC]\), \(S_B=[PCA]\), \(S_C=[PAB]\).
3. Translate side ratios into area ratios.
4. Use Ceva if you need to prove concurrence.
5. For a ratio on a cevian, compare the area of the triangle with base on the side and the area of the whole triangle.
Example 1. Common Base or Common Altitude
The first example fixes the main translation: side segments become areas.
Problem. In triangle \(ABC\), point \(D\in BC\), \(BD:DC=3:5\). Find \([ABD]:[ADC]\).
Triangles \(ABD\) and \(ADC\) have a common altitude from \(A\) to line \(BC\). Therefore their areas are in the ratio of their bases: \([ABD]:[ADC]=BD:DC=3:5\).
Example 2. From Area to Segment
The method also works in the reverse direction.
Problem. In triangle \(ABC\), point \(D\in BC\), and \([ABD]:[ADC]=7:4\). Find \(BD:DC\).
Triangles \(ABD\) and \(ADC\) again have a common altitude from \(A\). Hence the ratio of areas equals the ratio of bases, so \(BD:DC=7:4\).
Example 3. An Interior Point and a Cevian
This shows how an area around point \(P\) gives a ratio on a side.
Problem. Point \(P\) lies inside triangle \(ABC\). Line \(AP\) meets \(BC\) at \(D\). Prove that \(\frac{BD}{DC}=\frac{[ABP]}{[ACP]}\).
Triangles \(ABP\) and \(ACP\) have common base \(AP\). Their altitudes to \(AP\), drawn from \(B\) and \(C\), are in the ratio \(BD:DC\), because points \(B,D,C\) are collinear. Therefore \(\frac{[ABP]}{[ACP]}=\frac{BD}{DC}\).
Example 4. Three Small Areas
From the three areas around a point, one can read the three ratios on the sides immediately.
Problem. Inside triangle \(ABC\), point \(P\) is given. Let \([PBC]=8\), \([PCA]=12\), \([PAB]=18\). Lines \(AP\), \(BP\), \(CP\) meet \(BC\), \(CA\), \(AB\) at \(D,E,F\). Find \(BD:DC\), \(CE:EA\), \(AF:FB\).
Use the formulas: \(\frac{BD}{DC}=\frac{[PAB]}{[PCA]}=\frac{18}{12}=3:2\). Next, \(\frac{CE}{EA}=\frac{[PBC]}{[PAB]}=\frac{8}{18}=4:9\). Finally, \(\frac{AF}{FB}=\frac{[PCA]}{[PBC]}=\frac{12}{8}=3:2\).
Example 5. Ceva from Areas
Areas explain why the product in Ceva equals \(1\).
Problem. Point \(P\) lies inside triangle \(ABC\), and \(AP\), \(BP\), \(CP\) meet the sides at \(D,E,F\). Prove that \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1\).
Denote \(S_A=[PBC]\), \(S_B=[PCA]\), \(S_C=[PAB]\). Then \(\frac{BD}{DC}=\frac{S_C}{S_B}\), \(\frac{CE}{EA}=\frac{S_A}{S_C}\), \(\frac{AF}{FB}=\frac{S_B}{S_A}\). The product is \(1\).
Example 6. Ratio on a Cevian
Here the first meaning of mass points appears: a point divides a cevian according to areas.
Problem. In triangle \(ABC\), point \(D\in BC\), \(P\in AD\). It is known that \([PBC]:[ABC]=2:7\). Find \(AP:PD\).
Triangles \(PBC\) and \(ABC\) have common base \(BC\), so their areas are in the ratio of the altitudes to \(BC\). On segment \(AD\), the altitude from \(P\) to \(BC\) equals \(\frac{PD}{AD}\) times the altitude from \(A\). Hence \(\frac{PD}{AD}=\frac{2}{7}\). Therefore \(AP:PD=(AD-PD):PD=5:2\).
Example 7. Finding Areas Around a Point
Two cevians determine the ratios of the three small areas.
Problem. In triangle \(ABC\), points \(D\in BC\), \(E\in CA\), \(BD:DC=2:3\), \(CE:EA=4:5\). Lines \(AD\) and \(BE\) meet at \(P\). Find \([PAB]:[PBC]:[PCA]\).
Let \(x=[PAB]\), \(y=[PBC]\), \(z=[PCA]\). From \(BD:DC=2:3\), we get \(x:z=2:3\). From \(CE:EA=4:5\), we get \(y:x=4:5\). Take \(x=10\); then \(z=15\), \(y=8\). Therefore \([PAB]:[PBC]:[PCA]=10:8:15\).
Example 8. Without Trigonometry
Sometimes areas replace heavier computations.
Problem. Point \(P\) lies inside triangle \(ABC\), and \([PAB]=[PAC]\). Line \(AP\) meets \(BC\) at \(D\). Prove that \(D\) is the midpoint of \(BC\).
By the cevian formula, \(\frac{BD}{DC}=\frac{[PAB]}{[PAC]}\). The areas are equal by the condition, so \(\frac{BD}{DC}=1\). Hence \(BD=DC\), that is, \(D\) is the midpoint of \(BC\).
Chapter
Complete Quadrilaterals and Miquel Points
Key Idea
A complete quadrilateral is formed by four lines. Every two lines meet, so six points appear. The four triples of lines form four triangles, and their circumcircles often pass through one common point, the Miquel point.
The Miquel point turns long angle chasing into one strong observation: if two circles from the complete configuration meet again at a point \(M\), then \(M\) lies on the other two circles as well.
Basic Facts
For four lines \(l_1,l_2,l_3,l_4\), denote \(A=l_1\cap l_2\), \(B=l_2\cap l_3\), \(C=l_3\cap l_4\), \(D=l_4\cap l_1\), \(E=l_1\cap l_3\), \(F=l_2\cap l_4\). Then the four circles
\[(ABE),\quad (ADF),\quad (CDE),\quad (BCF)\]
pass through one common point \(M\). This point is called the Miquel point of the complete quadrilateral.
The main working cyclicity criterion: points \(X,Y,Z,T\) lie on one circle if \(\angle XZY=\angle XTY\), or if a pair of opposite angles sums to \(180^\circ\).
When to Use This Method
Look for the Miquel point when a problem contains four lines, many intersections, several circles through triples of points, or asks you to prove that several circles have one common point.
The method is especially useful for proving concyclicity: instead of constructing a new circle from scratch, it is often enough to recognise a complete quadrilateral and name its Miquel point.
How to Recognise the Method
Signals include: four lines, six intersection points, circles on triangles, and phrases such as "Prove that several circles pass through one point" or "Prove that four points are concyclic".
A useful move is to label the four lines first, then write down the six intersection points. After that, it becomes clear which four circles belong to the complete quadrilateral.
Typical Mistakes
Do not confuse a complete quadrilateral with an ordinary quadrilateral. In a complete quadrilateral, all six intersection points of the four lines matter.
Do not call a point the Miquel point before checking which four circles are involved. Each circle must pass through three vertices formed by three of the four lines.
Mini-Checklist
1. Find the four lines of the configuration.
2. Label the six intersection points.
3. List the four triangles formed by triples of lines.
4. Take the second intersection of two circles and check by angles that it lies on the third and fourth.
5. To prove cyclicity, use equality of inscribed angles.
Example 1. A Cyclicity Criterion
Before the Miquel point, students need a confident angle criterion for cyclicity.
Problem. Prove that if \(\angle AXB=\angle AYB\), then points \(A,B,X,Y\) lie on one circle.
The equal angles \(\angle AXB\) and \(\angle AYB\) subtend the same segment \(AB\). By the converse of the inscribed angle criterion, points \(X\) and \(Y\) lie on one circle with \(A\) and \(B\).
Example 2. A Complete Quadrilateral from Four Lines
This example teaches students to label the six points correctly.
Problem. Four lines \(l_1,l_2,l_3,l_4\) are given. Label the six intersection points and list the four circles of the complete quadrilateral.
Let \(A=l_1\cap l_2\), \(B=l_2\cap l_3\), \(C=l_3\cap l_4\), \(D=l_4\cap l_1\), \(E=l_1\cap l_3\), \(F=l_2\cap l_4\). The triples of lines form triangles \(ABE\), \(BCF\), \(CDE\), \(ADF\). The corresponding circles are \((ABE)\), \((BCF)\), \((CDE)\), \((ADF)\).
Example 3. Two Circles Determine the Candidate
The Miquel point is conveniently constructed as the second intersection of two circles.
Problem. In the notation of the previous example, circles \((ABE)\) and \((ADF)\) meet at \(A\) and \(M\). Prove that \(\angle BME=\angle BAE\) and \(\angle DMF=\angle DAF\).
Since \(A,B,E,M\) are concyclic, angles \(\angle BME\) and \(\angle BAE\) subtend chord \(BE\), so they are equal. Similarly, from cyclicity of \(A,D,F,M\), we get \(\angle DMF=\angle DAF\).
Example 4. First Step Toward Miquel's Theorem
We show how two circles give the third one.
Problem. In the complete quadrilateral formed by lines \(l_1,l_2,l_3,l_4\), point \(M\) lies on circles \((ABE)\) and \((ADF)\). Prove that \(B,C,F,M\) are concyclic.
It is enough to prove \(\angle BMF=\angle BCF\). Split the angle: \(\angle BMF=\angle BMA+\angle AMF\). Since \(A,B,E,M\) are cyclic, \(\angle BMA=\angle BEA\), the angle between \(l_3\) and \(l_1\). Since \(A,D,F,M\) are cyclic, \(\angle AMF=\angle ADF\), the angle between \(l_1\) and \(l_4\). Their sum is the angle between \(l_3\) and \(l_4\), which is \(\angle BCF\). Hence \(B,C,F,M\) are concyclic.
Example 5. Full Miquel Theorem
Now we complete the proof for all four circles.
Problem. Prove that circles \((ABE)\), \((ADF)\), \((BCF)\), \((CDE)\) have one common point.
Let \(M\) be the second intersection of circles \((ABE)\) and \((ADF)\). By the previous example, \(M\in (BCF)\). By an analogous angle computation, \(\angle CME=\angle CDE\), so \(C,D,E,M\) are cyclic, that is, \(M\in (CDE)\). Therefore all four circles pass through \(M\).
Example 6. An Ordinary Quadrilateral as a Complete One
An ordinary quadrilateral also has the complete quadrilateral of its side lines.
Problem. In quadrilateral \(ABCD\), lines \(AB\) and \(CD\) meet at \(E\), and lines \(AD\) and \(BC\) meet at \(F\). Which four circles pass through the Miquel point of these four lines?
The four lines are \(AB\), \(BC\), \(CD\), \(DA\). The triples of neighbouring lines give circles \((ABF)\), \((BCE)\), \((CDF)\), \((DAE)\). By Miquel's theorem, they pass through one point.
Example 7. Equal Angles from the Miquel Point
After finding the Miquel point, one can read new angles.
Problem. Let \(M\) be the Miquel point of a complete quadrilateral, and \(M\in (BCF)\). Prove that \(\angle BMF=\angle BCF\).
Points \(B,C,F,M\) lie on one circle. Angles \(\angle BMF\) and \(\angle BCF\) subtend the same chord \(BF\), so they are equal.
Example 8. Spiral Similarity at the Miquel Point
The Miquel point is often also a centre of spiral similarity.
Problem. If \(M\) lies on circles \((ABE)\) and \((ADF)\), prove that \(\angle BME=\angle DMF\), with the angles between \(l_1,l_2\) taken consistently.
From \(A,B,E,M\) cyclic, we get \(\angle BME=\angle BAE\). From \(A,D,F,M\) cyclic, we get \(\angle DMF=\angle DAF\). But \(\angle BAE\) and \(\angle DAF\) are the same angle between lines \(l_2\) and \(l_1\). Therefore \(\angle BME=\angle DMF\).
Chapter
Geometry with Coordinates and Vectors
Key Idea
Coordinates and vectors in olympiad geometry are not meant for blind computation. They are a language choice. With good axes, parallelism, perpendicularity, lengths, midpoints, and circles become short calculations.
The main principle is: first understand the geometry of the configuration, then choose the origin, axes, and scale so that as many points as possible have simple coordinates.
Basic Facts
If \(A(x_1,y_1)\), \(B(x_2,y_2)\), then \(\overrightarrow{AB}=(x_2-x_1,y_2-y_1)\), and \(AB^2=(x_2-x_1)^2+(y_2-y_1)^2\).
The dot product is \((u_1,u_2)\cdot(v_1,v_2)=u_1v_1+u_2v_2\). Two vectors are perpendicular if and only if their dot product is \(0\).
The midpoint of \(AB\) has coordinates \(\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\). A point dividing \(AB\) in ratio \(m:n\) is often easiest to write using a parameter.
A circle has equation \(x^2+y^2+ux+vy+w=0\). A line has equation \(ax+by+c=0\). For a circle through the origin, usually \(w=0\), which greatly shortens the computation.
When to Use This Method
The coordinate method is especially useful when a problem contains right angles, parallel lines, midpoints, ratios on segments, a circle with a convenient diameter, or a need to prove that a point lies on a line or circle.
Vectors are useful for proving parallelism, midpoint statements, concurrence of medians, and problems where one wants to avoid heavy angle chasing.
How to Recognise the Method
If one can put a side on the \(Ox\)-axis, another side on the \(Oy\)-axis, the centre of a circle at the origin, or a midpoint at the origin, coordinates may simplify the problem strongly.
Phrases such as “Prove perpendicularity”, “Find a length”, “The point lies on a circle”, and “Find a ratio” often point toward the dot product, the distance formula, or the equation of a circle.
Typical Mistakes
The first mistake is choosing coordinates that are too general, for instance three arbitrary points instead of \(A(0,0)\), \(B(1,0)\), \(C(u,v)\). The second is proving equality of lengths by using lengths themselves, when comparing squares is enough.
The third mistake is forgetting that a coordinate solution must justify the coordinate choice. If the choice does not simplify the problem, the method loses its value.
Mini-Checklist
Before calculating, ask: where should the origin be? Which axes create zero coordinates? Can lengths be replaced by squared lengths? Is there perpendicularity that can be checked by the dot product? Can a circle be written as \(x^2+y^2+ux+vy+w=0\)?
Example 1. A Good Choice of Axes
This example shows how an axis of symmetry turns a proof into one line.
Problem. In an isosceles triangle \(CA=CB\), prove that the median from \(C\) to the base \(AB\) is an altitude.
Set \(A(-1,0)\), \(B(1,0)\), \(C(0,h)\). The midpoint of \(AB\) is \(M(0,0)\). Then \(CM\) is vertical and \(AB\) is horizontal, so \(CM\perp AB\).
Comment. The coordinates were chosen so that the symmetry is immediately visible.
Example 2. The Dot Product
Perpendicularity is often easier to prove by a zero dot product than by angles.
Problem. Let \(A(0,0)\), \(B(a,b)\), \(C(-b,a)\). Prove that \(AB\perp AC\) and \(AB=AC\).
We have \(\overrightarrow{AB}=(a,b)\), \(\overrightarrow{AC}=(-b,a)\). Then \(\overrightarrow{AB}\cdot\overrightarrow{AC}=a(-b)+ba=0\), hence \(AB\perp AC\). Also, \(AB^2=a^2+b^2\) and \(AC^2=(-b)^2+a^2=a^2+b^2\), so \(AB=AC\).
Example 3. A Circle Through the Origin
If one point of the circle is the origin, the equation becomes shorter.
Problem. Find the circle through \(A(0,0)\), \(B(a,0)\), \(C(0,b)\), where \(a,b\ne 0\).
Let the circle have equation \(x^2+y^2+ux+vy+w=0\). From \(A(0,0)\), we get \(w=0\). From \(B(a,0)\): \(a^2+ua=0\), so \(u=-a\). From \(C(0,b)\): \(b^2+vb=0\), so \(v=-b\). Hence the circle is \(x^2+y^2-ax-by=0\).
Example 4. The Midpoint of the Hypotenuse
A classical statement becomes a direct check of squared distances.
Problem. In the right triangle \(A(0,0)\), \(B(m,0)\), \(C(0,n)\), prove that the midpoint of the hypotenuse is equidistant from all three vertices.
The midpoint of \(BC\) is \(M\left(\frac m2,\frac n2\right)\). Then \(MA^2=\frac{m^2+n^2}{4}\), \(MB^2=\left(\frac m2\right)^2+\left(\frac n2\right)^2\), and \(MC^2=\left(\frac m2\right)^2+\left(\frac n2\right)^2\). All three squares are equal, so \(MA=MB=MC\).
Example 5. A Ratio on a Side
Coordinates are convenient for points defined by segment ratios.
Problem. In triangle \(A(0,0)\), \(B(6,0)\), \(C(0,6)\), let \(P(2,1)\). Lines \(AP\), \(BP\), \(CP\) meet the opposite sides at \(D,E,F\). Find the ratios \(BD:DC\), \(CE:EA\), \(AF:FB\).
Line \(AP\) has equation \(y=\frac{x}{2}\), while \(BC\) has equation \(x+y=6\). Thus \(D(4,2)\), so \(BD:DC=1:2\). Line \(BP\) gives \(E(0,\frac32)\), hence \(CE:EA=3:1\). Line \(CP\) meets \(AB\) at \(F(\frac{12}{5},0)\), hence \(AF:FB=2:3\).
Example 6. A Vector Proof of the Midline
Vectors work especially well with midpoints.
Problem. In triangle \(ABC\), points \(M,N\) are the midpoints of \(AB\) and \(AC\). Prove that \(MN\parallel BC\) and \(MN=\frac12BC\).
Denote the position vectors of \(A,B,C\) by \(a,b,c\). Then \(m=\frac{a+b}{2}\), \(n=\frac{a+c}{2}\). Therefore \(\overrightarrow{MN}=n-m=\frac{c-b}{2}=\frac12\overrightarrow{BC}\). Hence \(MN\parallel BC\) and \(MN=\frac12BC\).
Example 7. Lying on a Circle
Sometimes it is enough to substitute coordinates into the equation of a circle.
Problem. Prove that points \(A(-3,0)\), \(B(3,0)\), \(C(2,2)\), \(D(-2,2)\) lie on one circle.
Check the circle \(x^2+\left(y+\frac14\right)^2=\frac{145}{16}\). For \(A\) and \(B\), we get \(9+\frac{1}{16}=\frac{145}{16}\). For \(C\) and \(D\), we get \(4+\left(\frac94\right)^2=4+\frac{81}{16}=\frac{145}{16}\). Thus all four points lie on one circle.
Example 8. Coordinates as a Strong Method
A general result can be proved by computation if the coordinates are chosen carefully.
Problem. In triangle \(A(0,0)\), \(B(1,0)\), \(C(u,v)\), \(v\ne 0\), prove that the orthocenter has coordinates \(H\left(u,\frac{u(1-u)}{v}\right)\).
The altitude from \(C\) is perpendicular to \(AB\), so it has equation \(x=u\). Line \(AC\) has direction vector \((u,v)\), so the altitude from \(B\) has direction vector \((v,-u)\). Its equation is \(y=-\frac{u}{v}(x-1)\). At \(x=u\), we get \(y=-\frac{u}{v}(u-1)=\frac{u(1-u)}{v}\). Hence \(H\left(u,\frac{u(1-u)}{v}\right)\).
Chapter
Mixed Problems II
Key Idea
In a mixed problem, the method is not announced in advance. First one has to see the structure: circles, tangents, ratios, midpoints, parallel lines, and intersections of lines. A good solution begins not with computation, but with choosing the tool.
The same fact can often be proved in several ways. For example, concyclicity may follow from angles, power of a point, a circle equation, or a Miquel point. The goal of this module is to learn how to choose the shortest route.
Basic Facts
For circles, use inscribed angles, the tangent-chord theorem, power of a point \(PA\cdot PB=PC\cdot PD\), and equality of powers on the radical axis.
For ratios on sides, use Ceva and Menelaus, areas with a common height, similarity, and homothety. For configurations of four lines, a Miquel point often appears.
If the configuration contains right angles, midpoints, a circle equation, or too many lengths, coordinates and vectors can replace a long synthetic solution.
When to Use This Method
Try power of a point when two secants, or a tangent and a secant, come from one point. The radical axis is useful when one needs to prove collinearity of points related to several circles.
Ceva points to concurrence, Menelaus to collinearity. Areas help when there is a common height or a ratio on a side. The Miquel point appears in configurations of four lines and three or four circles.
How to Recognise the Method
Ask: what must be proved? Concyclicity usually calls for angles or power of a point. Collinearity involving circles often calls for the radical axis. Concurrence of cevians calls for Ceva or areas.
If the problem contains an “intersection of tangents”, look for power of a point or homothety. If there are “four lines”, check for a complete quadrilateral and a Miquel point. If the synthetic picture is overloaded, choose coordinates.
Typical Mistakes
The first mistake is starting with a favourite method instead of the clues in the problem. The second is applying Ceva when the goal is collinearity, or Menelaus when the goal is concurrence.
The third mistake is forgetting directed lengths in problems with exterior points. The fourth is proving concyclicity by angles when a one-line power-of-a-point argument already solves the problem.
Mini-Checklist
Before solving, mark the goal: concyclicity, collinearity, concurrence, a ratio, or a length. Then find the main object: a circle, a point, a line, a ratio, a tangent, or a complete quadrilateral. After that, choose the tool and only then begin computations or angle chasing.
Example 1. Recognising Power of a Point
If a tangent and a secant come from one point, angle chasing is usually unnecessary.
Problem. From point \(P\), a tangent \(PT\) and a secant \(PAB\) are drawn to a circle. Given \(PA=4\), \(PT=6\), find \(PB\).
By power of a point, \(PT^2=PA\cdot PB\). Hence \(36=4\cdot PB\), so \(PB=9\).
Comment. The main clue is a tangent and a secant from the same point.
Example 2. Concyclicity by Right Angles
Sometimes the circle should be seen as a circle with a diameter.
Problem. In triangle \(ABC\), points \(D\) and \(E\) are the feet of the altitudes from \(B\) and \(C\). Prove that \(B,C,D,E\) lie on one circle.
We have \(\angle BDC=90^\circ\) and \(\angle BEC=90^\circ\). Thus points \(D\) and \(E\) lie on the circle with diameter \(BC\). Therefore \(B,C,D,E\) are concyclic.
Example 3. Radical Axis Instead of Computation
Collinearity in a problem with several circles often points to radical axes.
Problem. Circles \(\omega_1\) and \(\omega_2\) meet at \(A,B\). Point \(P\) lies on line \(AB\). A line through \(P\) meets \(\omega_1\) at \(X,Y\), and another line through \(P\) meets \(\omega_2\) at \(U,V\). Prove that \(PX\cdot PY=PU\cdot PV\).
Line \(AB\) is the radical axis of circles \(\omega_1\) and \(\omega_2\). Hence the powers of point \(P\) with respect to the two circles are equal. Therefore \(PX\cdot PY=PU\cdot PV\).
Example 4. Ceva or Menelaus
If one needs to prove that three lines meet, first check Ceva.
Problem. In triangle \(ABC\), points \(D,E,F\) lie on \(BC,CA,AB\), respectively. Suppose \(BD:DC=2:3\), \(CE:EA=3:4\), \(AF:FB=2:1\). Prove that \(AD,BE,CF\) are concurrent.
By Ceva's theorem it is enough to check the product: \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=\frac{2}{3}\cdot\frac{3}{4}\cdot 2=1\). Hence the lines are concurrent.
Example 5. Areas Instead of Similarity
A ratio on a side is often equal to a ratio of areas with a common height.
Problem. In triangle \(ABC\), point \(D\) lies on \(BC\), and \(BD:DC=3:5\). Prove that \([ABD]:[ACD]=3:5\).
Triangles \(ABD\) and \(ACD\) have the same altitude from \(A\) to line \(BC\). Therefore their areas are proportional to their bases: \([ABD]:[ACD]=BD:DC=3:5\).
Example 6. Recognising a Miquel Point
Four lines and four circles are almost always a signal for a Miquel point.
Problem. In triangle \(ABC\), points \(D\in AB\), \(E\in AC\), and lines \(BE\) and \(CD\) meet at \(P\). Prove that circles \((ABE)\), \((ACD)\), \((BDP)\), \((CEP)\) have one common point.
Consider the four lines \(AB,AC,BE,CD\). Their triples form the circles \((ABE)\), \((ACD)\), \((BDP)\), \((CEP)\). By Miquel's theorem for a complete quadrilateral, these circles pass through one point.
Example 7. Coordinates as a Rescue
If a ratio comes from a projection, coordinates give a short solution.
Problem. In triangle \(A(0,0)\), \(B(6,0)\), \(C(0,8)\), point \(P\) is the foot of the perpendicular from \(A\) to \(BC\). Find \(BP:PC\).
Write \(P=B+t(C-B)=(6-6t,8t)\). The vector \(BC=(-6,8)\), and the condition \(AP\perp BC\) gives \((6-6t,8t)\cdot(-6,8)=0\). Thus \(-36+100t=0\), so \(t=\frac{9}{25}\). Therefore \(BP:PC=t:(1-t)=9:16\).
Example 8. Tangent and Symmedian
A strong problem is often solved by mixing tangents, angles, and the sine rule.
Problem. In triangle \(ABC\), the tangent to the circumcircle at \(A\) meets line \(BC\) at \(T\). Prove that \(TB:TC=AB^2:AC^2\).
By the tangent-chord theorem, \(\angle TAB=\angle ACB\), and \(\angle TAC=\angle ABC\). Applying the sine rule in triangles \(TAB\) and \(TAC\), we get \(\frac{TB}{TC}=\frac{\sin^2\angle ACB}{\sin^2\angle ABC}\). By the sine rule in \(ABC\), this equals \(\frac{AB^2}{AC^2}\).