Course Theory
Book 3. Advanced Olympiad Geometry
Book 3. Advanced Olympiad Geometry
- 1. Inversion II
- 2. Projective Geometry I
- 3. Poles and Polars
- 4. Simson Line and Pedal Geometry
- 5. Brocard, Napoleon, and Special Points
- 6. Trigonometric Geometry
Chapter
Inversion II
Key Idea
Inversion replaces a complicated configuration of circles by a simpler one: circles through the center of inversion become lines, lines not through the center become circles through the center, while tangencies and angles are preserved. The main question in this module is not how to compute, but which center and radius to choose.
Basic Facts
If an inversion has center \(O\) and radius \(R\), then point \(A\) maps to point \(A^*\) on ray \(OA\), with \(OA\cdot OA^*=R^2\). A line through \(O\) maps to itself. A line not through \(O\) maps to a circle through \(O\). A circle through \(O\) maps to a line not through \(O\). A circle orthogonal to the circle of inversion maps to itself.
Inversion preserves angles between curves, where the angle is understood as the angle between tangents. Thus tangency, orthogonality of circles, and many angle conditions survive the transformation.
When To Use This Method
The method is especially strong when the problem contains many circles, tangencies, common points, chains of tangent circles, or several circles through one fixed point. A good signal is a point through which several circles pass: inversion centered there turns them into lines.
How To Recognise The Method
Look for a point that is a tangency point, a common point of several circles, a vertex of an angle, or a center of a pencil. If choosing this point makes several circles become lines, the problem often becomes much simpler.
Typical Mistakes
Do not choose the radius randomly: the power should fix important points or interchange useful ones. Remember that the center of inversion has no image. Do not transfer lengths directly: inversion preserves angles, but distances change nonlinearly.
Mini-Checklist
- Which point should be the center of inversion?
- Which circles will become lines?
- Which power fixes or swaps the important points?
- What happens to tangency?
- How do we translate the result back?
Example 1. Inverse Points and Similarity
This example shows why inversion naturally creates similar triangles.
Problem. Under an inversion centered at \(O\), points \(A\) and \(B\) map to \(A^*\) and \(B^*\). Prove that \(\triangle OAB\sim\triangle OB^*A^*\).
By definition, \(OA\cdot OA^*=OB\cdot OB^*=R^2\). Hence \(\frac{OA}{OB}=\frac{OB^*}{OA^*}\), and the angle at \(O\) is common: \(\angle AOB=\angle B^*OA^*\). Therefore the triangles are similar.
Comment: this basic mechanism later replaces long angle computations.
Example 2. Image of a Line
We learn how a line becomes a circle.
Problem. A line \(l\) does not pass through the inversion center \(O\). Let \(C\) be the foot of the perpendicular from \(O\) to \(l\), and let \(C^*\) be the image of \(C\). Prove that the image of \(l\) is the circle with diameter \(OC^*\).
For any point \(M\in l\), its image \(M^*\) lies on ray \(OM\). From the similarity of \(\triangle OCM\) and \(\triangle OM^*C^*\), we get \(\angle OM^*C^*=90^\circ\). Hence \(M^*\) lies on the circle with diameter \(OC^*\). The reverse inclusion follows by the same argument backwards.
Example 3. A Circle Through the Center Becomes a Line
This is the most frequent move: choose the inversion center at a common point of circles.
Problem. A circle \(\omega\) passes through \(O\). Prove that its image under inversion centered at \(O\) is a line.
Let \(A\) be the second intersection of \(\omega\) with the line joining \(O\) to the center of \(\omega\). For any point \(M\in\omega\), \(\angle OMA=90^\circ\). After inversion this becomes the condition that \(M^*\) lies on the line perpendicular to \(OA\) through \(A^*\).
Example 4. An Orthogonal Circle Is Fixed
An important way to choose the radius of inversion.
Problem. A circle \(\gamma\) is orthogonal to the circle of inversion centered at \(O\) with radius \(R\). Prove that \(\gamma\) maps to itself.
If \(X\in\gamma\), line \(OX\) meets \(\gamma\) again at \(Y\). Orthogonality implies that the power of \(O\) with respect to \(\gamma\) is \(R^2\), so \(OX\cdot OY=R^2\). Therefore \(Y=X^*\). Hence the image of every point of \(\gamma\) lies again on \(\gamma\).
Example 5. Tangency Is Preserved
Tangency often becomes parallelism or tangency of lines.
Problem. Two circles are tangent at point \(T\), not equal to the inversion center. Prove that their images are also tangent.
Inversion preserves the angle between curves. For tangent circles, the angle between the tangents at \(T\) is \(0^\circ\). After inversion, the angle between the images is still \(0^\circ\), so the images have a common tangent at \(T^*\), hence are tangent.
Example 6. Two Circles Through One Point
Here we see how inversion turns circles into lines.
Problem. Circles \(\omega_1\) and \(\omega_2\) pass through \(A\) and meet again at \(B\). What happens to them under inversion centered at \(A\)?
Each circle passes through the inversion center, so each maps to a line. Both image lines pass through \(B^*\), because \(B\) belongs to both original circles. Thus a configuration of two circles becomes two lines meeting at \(B^*\).
Example 7. Choosing the Center at a Tangency Point
This is one of the main moves in difficult problems.
Problem. Two circles are tangent at \(A\). Prove that after inversion centered at \(A\), they become two parallel lines.
Both circles pass through the inversion center, so their images are lines. Since the original circles are tangent at \(A\), the angle between them is \(0^\circ\). Inversion preserves angle, so the angle between the image lines is \(0^\circ\). Two distinct lines with zero angle are parallel.
Example 8. A Chain of Circles
The final idea of the module: inversion can turn a chain of tangencies into equal circles between concentric circles.
Problem. Two disjoint circles can be sent by an inversion and a homothety to concentric circles. Explain why a chain of circles tangent to both becomes easier to study after this transformation.
Tangencies and angles are preserved. If the two given circles become concentric, then every circle tangent to both has radius equal to half the difference of the concentric radii. Thus all circles in the chain become equal. Many statements then reduce to rotations about the common center.
Chapter
Projective Geometry I
Key Idea
Projective geometry lets us change the picture while preserving lines, intersections, tangencies, and collinearity. A complicated circle or conic may become a simpler circle, and an inconvenient line may become the line at infinity. Then parallelism, degenerate cases, and Pascal or Desargues type arguments often become visible.
The main idea of this module is: if a problem is only about points, lines, intersections, tangents, and points lying on one conic, look for a projective transformation that sends the configuration to a simpler model.
Basic Facts
Central projection sends lines to lines and preserves incidence: if a point lay on a line, its image lies on the image of that line.
The cross-ratio of four points on a line is
\[ (A B C D)=\frac{AC}{BC}:\frac{AD}{BD}. \]
Central projection, and more generally any projective transformation of a line, preserves cross-ratio.
A projective transformation of a line is uniquely determined by the images of three distinct points. Hence if such a transformation has three distinct fixed points, it is the identity.
Desargues' theorem connects perspective triangles: if the lines joining corresponding vertices are concurrent, then the intersections of corresponding sides are collinear. The converse is also true.
Pascal's theorem says that if six points lie on one conic, then the three intersections of opposite sides of the corresponding hexagon are collinear. Degenerate forms of Pascal produce useful tangent statements.
When to Use This Method
The projective method is especially useful when a problem contains a conic, many intersections of extended sides, tangents, a hexagon on a circle, perspective triangles, or the phrase “prove that three points are collinear”.
It is also worth trying when metric information seems distracting: lengths and angles look secondary, while intersections and incidence with lines or a circle matter most.
How to Recognise the Method
Look for three signs: two configurations differing only by perspective; four points on a line where cross-ratio can be used; six points on a circle or conic where opposite sides give three intersections.
If two lines meet in an inconvenient distant point, try sending their intersection to infinity. If a conic is awkward, try replacing it by a circle, because incidence properties are preserved.
Typical Mistakes
Do not transfer lengths, midpoints, perpendicularity, or equality of angles through a projective transformation unless this is separately justified. Projective geometry preserves straightness, intersections, tangency, cross-ratio, and conic incidence, but not ordinary metric data.
A second common mistake is applying Pascal with the wrong order of vertices. In a hexagon the opposite side pairs are first with fourth, second with fifth, and third with sixth.
A third mistake is forgetting degenerate vertices. If two neighbouring vertices of the hexagon coincide, the corresponding side becomes the tangent to the conic.
Mini-Checklist
Before solving, ask: which properties in the problem are genuinely projective? Can one line be sent to infinity? Are there four collinear points with a useful cross-ratio? Is there a hexagon on a circle or conic? Can the statement be proved by showing that a projectivity has three fixed points?
Example 1. Cross-Ratio as a Projection Invariant
This example shows why projection does not destroy the main numerical relation on a line.
Problem. Four rays from a point \(O\) meet a line \(l\) at \(A,B,C,D\), and a line \(m\) at \(A_1,B_1,C_1,D_1\). Prove that \((A B C D)=(A_1 B_1 C_1 D_1)\).
The projection with center \(O\) sends \(A,B,C,D\) to \(A_1,B_1,C_1,D_1\). For four rays through one point, the cross-ratio can be expressed using sines of the angles between the rays. If the rays are cut by any line not passing through \(O\), the same ratios are expressed through segments on that line.
Thus the cross-ratio belongs to the pencil of four rays rather than to the particular line \(l\). Therefore it is unchanged when we pass from \(l\) to \(m\).
Comment. This is the basic mechanism of most problems in this module: projection changes lengths, but preserves cross-ratio.
Example 2. Three Fixed Points
This example teaches how to prove that a projective transformation is the identity without computation.
Problem. A projective transformation of a line \(l\) fixes three distinct points \(A,B,C\). Prove that it fixes every point of the line.
Let \(X\) be any point of the line, and let \(X'\) be its image. Since the transformation is projective, it preserves cross-ratio:
\[ (A B C X)=(A B C X'). \]
For fixed distinct points \(A,B,C\), the value \((A B C X)\) uniquely determines \(X\). Hence \(X'=X\). Therefore every point is fixed.
Comment. This is especially powerful when a complicated composition of projections becomes a self-map of one line.
Example 3. Desargues Through a Convenient Perspective
Here a projective transformation replaces a central perspective by a simpler parallel picture.
Problem. Triangles \(ABC\) and \(A_1B_1C_1\) are such that the lines \(AA_1\), \(BB_1\), \(CC_1\) pass through one point. Prove that the points \(AB\cap A_1B_1\), \(BC\cap B_1C_1\), \(CA\cap C_1A_1\) are collinear.
Apply a projective transformation sending the center of perspective to a point at infinity. Then \(AA_1\), \(BB_1\), \(CC_1\) become parallel.
In the new affine picture, the vertices of the second triangle are obtained from the vertices of the first by shifts in one direction, possibly with different coefficients. The affine form of Desargues' theorem gives that the intersections of corresponding sides are collinear.
Collinearity is preserved by the inverse projective transformation, so the original statement follows.
Example 4. Pascal with Parallel Sides
This example shows how a point at infinity turns collinearity into parallelism.
Problem. Points \(A,B,C,D,E,F\) lie on one circle, and \(AB\parallel DE\). Let \(Q=BC\cap EF\) and \(R=CD\cap FA\). Prove that \(QR\parallel AB\).
By Pascal's theorem for the hexagon \(ABCDEF\), the points \(P=AB\cap DE\), \(Q=BC\cap EF\), and \(R=CD\cap FA\) are collinear.
Since \(AB\parallel DE\), the point \(P\) is the point at infinity in the direction of \(AB\). Therefore the line \(QR\), which passes through \(P\), has the same direction. Hence \(QR\parallel AB\).
Example 5. Degenerate Pascal and Tangents
Here neighbouring vertices of the hexagon merge, and a side becomes a tangent.
Problem. Points \(A,B,C,D\) lie on one circle. The tangents at \(A\) and \(C\) meet at \(X\). Let \(Y=AB\cap CD\) and \(Z=BC\cap AD\). Prove that \(X,Y,Z\) are collinear.
Apply Pascal's theorem to the degenerate hexagon \(A,A,B,C,C,D\). The side \(AA\) means the tangent at \(A\), and the side \(CC\) means the tangent at \(C\).
The three pairs of opposite sides give the points \(X\), \(Y\), and \(Z\). Hence Pascal's theorem gives their collinearity.
Comment. This is one of the most common ways to obtain “polar” lines without developing the full theory of poles and polars.
Example 6. Projecting a Circle to a Line and Back
This example explains why compositions of projections often give projective transformations of a line.
Problem. A circle \(\omega\), a line \(l\), and two points \(M,N\) on the circle are given, with \(M,N\notin l\). For \(X\in l\), define \(X'\in l\) as follows: the line \(MX\) meets the circle again at \(Y\), and the line \(NY\) meets \(l\) at \(X'\). Prove that \(X\mapsto X'\) is a projective transformation of \(l\).
The first step \(X\mapsto Y\) is central projection from the line \(l\) to the circle with center \(M\). The second step \(Y\mapsto X'\) is central projection from the circle to \(l\) with center \(N\).
If the circle is identified with a line by any fixed projection, both steps become projective maps between lines. A composition of projective maps is projective.
Example 7. Sending a Line to Infinity
This example shows how to choose a convenient model instead of doing direct angle computations.
Problem. In an incidence proof about a conic, a line \(s\) appears which is not tangent to the conic. Explain why one may choose a projective model in which the image of \(s\) is the line at infinity.
Projective transformations allow us to send a chosen ordinary line to the line at infinity, provided we work only with incidence and do not require lengths or angles to be preserved. The image of a conic is again a conic.
After this, all lines that used to meet on \(s\) become parallel. If the final statement is a collinearity, concurrence, or tangency statement, it may be proved in the new model and then transformed back.
Example 8. Pappus as a Limiting Case of Pascal
This example connects two classical projective theorems.
Problem. Points \(A,B,C\) lie on one line, and \(A_1,B_1,C_1\) lie on another. Let \(P=AB_1\cap A_1B\), \(Q=AC_1\cap A_1C\), and \(R=BC_1\cap B_1C\). Prove that \(P,Q,R\) are collinear.
This is Pappus' theorem. It may be viewed as a degenerate case of Pascal's theorem: the conic splits into two lines, one containing \(A,B,C\), the other containing \(A_1,B_1,C_1\).
Applying the Pascal scheme to six points on this degenerate conic gives the collinearity of \(P,Q,R\).
Chapter
Poles and Polars
Key Idea
Poles and polars turn tangent problems into incidence problems about points and lines. Instead of chasing angles between tangents and chords every time, we replace a point \(P\) by its polar \(p\), and a line \(p\) by its pole \(P\). Then many collinearities and concurrences follow from one principle: if a point lies on the polar of another point, the converse is also true.
Basic Facts
Let \(\omega\) be a circle with center \(O\) and radius \(R\). If a point \(P\) lies outside the circle and \(PA\), \(PB\) are tangents, then the line \(AB\) is called the polar of \(P\).
For any point \(P\ne O\), the polar may be defined as follows: if \(H=OP\cap p\), then
\[ OP\cdot OH=R^2,\qquad p\perp OP. \]
With respect to the unit circle \(x^2+y^2=1\), the polar of \(P(p,q)\) is
\[ px+qy=1. \]
La Hire's theorem: if a point \(Q\) lies on the polar of \(P\), then \(P\) lies on the polar of \(Q\).
If a secant through \(P\) meets the circle at \(A\) and \(B\), then the intersection of the tangents at \(A\) and \(B\) lies on the polar of \(P\). For a complete quadrangle inscribed in a circle, the diagonal triangle is self-polar.
When to Use This Method
Try this method when a problem contains tangents to one circle, intersections of tangents, a complete quadrangle on a circle, a tangential polygon, or a request to prove that a tangent intersection is collinear with two chord intersections.
How to Recognise the Method
Look for a “hidden polar”: a line joining two contact points; a point which is the intersection of two tangents; two secants passing through one point; four circle points \(A,B,C,D\) together with intersections \(AB\cap CD\), \(AC\cap BD\), \(AD\cap BC\).
Typical Mistakes
Do not confuse the polar of a point with an arbitrary chord through that point. If \(P\) is outside the circle, the polar is the chord of contact, not a secant through \(P\). If \(P\) is inside the circle, no tangents exist, but the polar still exists by \(OP\cdot OH=R^2\).
A second mistake is forgetting which circle defines the polar. A problem may contain two circles, and the statement is valid only with respect to the chosen one.
A third mistake is transferring metric properties through polarity. Polarity preserves incidence in the dual sense, but not lengths or angles.
Mini-Checklist
Before solving, mark the circle with respect to which polarity is used. Find intersections of tangents. Find possible contact chords. Check whether the target line is the polar of a known point. If you need concurrence, try replacing it by the dual collinearity of poles.
Example 1. The Chord of Contact as a Polar
This example introduces the main formula for the polar.
Problem. From a point \(P\), tangents \(PA\) and \(PB\) are drawn to a circle with center \(O\) and radius \(R\). Let \(H=AB\cap OP\). Prove that \(AB\perp OP\) and \(OP\cdot OH=R^2\).
The triangles \(OAP\) and \(OBP\) are right triangles, have the common hypotenuse \(OP\), and have equal legs \(OA=OB\). Hence they are congruent, and \(A\) and \(B\) are symmetric with respect to \(OP\). Therefore \(AB\perp OP\).
In the right triangle \(OAP\), the altitude \(AH\) to the hypotenuse gives \(OA^2=OH\cdot OP\). Since \(OA=R\), we get \(OP\cdot OH=R^2\).
Example 2. Coordinate Formula
The formula helps check La Hire and simple polar statements quickly.
Problem. For the unit circle \(x^2+y^2=1\), find the polar of the point \(P(p,q)\).
If \(X(x,y)\) lies on the polar of \(P\), then the vector \(OX\) has a constant projection on the direction \(OP\). The condition \(OP\cdot OH=1\) gives the scalar equation
\[ px+qy=1. \]
This line is perpendicular to \(OP\) and meets \(OP\) at a point \(H\) such that \(OP\cdot OH=1\). Hence it is the polar of \(P\).
Example 3. La Hire's Theorem
This is the main symmetric principle of the module.
Problem. Prove that if \(Q\) lies on the polar of \(P\), then \(P\) lies on the polar of \(Q\).
For the unit circle, let \(P=(p,q)\), \(Q=(u,v)\). The condition \(Q\in p\) means \(pu+qv=1\).
The polar of \(Q\) has equation \(ux+vy=1\). Substituting the coordinates of \(P\), we get \(up+vq=1\). Hence \(P\) lies on the polar of \(Q\).
Example 4. Pole of a Line
Here the polar is used backwards: from a line to a point.
Problem. A line \(l\) does not pass through the center \(O\) of a circle of radius \(R\). Let \(H\) be the foot of the perpendicular from \(O\) to \(l\), and let \(P\) lie on the ray \(OH\) with \(OP\cdot OH=R^2\). Prove that \(P\) is the pole of \(l\).
By the definition of the polar, the intersection of the polar of \(P\) with \(OP\) must be a point \(H'\) such that \(OP\cdot OH'=R^2\), and the polar must be perpendicular to \(OP\). But \(l\perp OP\) and \(OP\cdot OH=R^2\). Hence \(l\) is the polar of \(P\).
Example 5. A Secant Through a Point and Tangent Intersection
This is the main working fact for hidden polar problems.
Problem. A secant through \(P\) meets a circle at \(A\) and \(B\). The tangents at \(A\) and \(B\) meet at \(T\). Prove that \(T\) lies on the polar of \(P\).
The point \(T\) has polar \(AB\), because \(TA\) and \(TB\) are tangents. The point \(P\) lies on \(AB\), that is, on the polar of \(T\). By La Hire's theorem, \(T\) lies on the polar of \(P\).
Example 6. Self-Polar Diagonal Triangle
This is a standard way to see polars in a complete quadrangle.
Problem. Points \(A,B,C,D\) lie on a circle. Let \(P=AB\cap CD\), \(Q=AC\cap BD\), and \(R=AD\cap BC\). Prove that the polar of \(P\) is the line \(QR\).
Take the secant \(PAB\). The intersection of the tangents at \(A\) and \(B\) lies on the polar of \(P\). The same is true for the secant \(PCD\). The line through these two tangent intersections is the polar of \(P\).
By a degenerate form of Pascal for the four points \(A,B,C,D\), this line passes through \(Q\) and \(R\). Hence the polar of \(P\) is \(QR\).
Example 7. Collinearity Through One Polar
This shows how tangents immediately produce the needed line.
Problem. Points \(A,B,C,D\) lie on a circle. The tangents at \(A\) and \(D\) meet at \(S\). Let \(P=AB\cap CD\) and \(Q=AC\cap BD\). Prove that \(P,Q,S\) are collinear.
Let \(R=AD\cap BC\). For the complete quadrangle \(ABCD\), the polar of \(R\) is the line \(PQ\). But \(S\) lies on the polar of \(R\), since \(R,A,D\) are collinear and \(S\) is the intersection of the tangents at \(A\) and \(D\). Therefore \(S\in PQ\), so \(P,Q,S\) are collinear.
Example 8. Brianchon as Dual Pascal
This example prepares strong problems with tangential polygons.
Problem. A hexagon \(ABCDEF\) is circumscribed about a circle. Prove that \(AD\), \(BE\), and \(CF\) are concurrent.
Consider the six points of tangency with the circle. Pascal's theorem for these six points, after passing to poles and polars, becomes the dual statement: the three lines joining opposite vertices of the tangential hexagon pass through one point.
These lines are \(AD\), \(BE\), and \(CF\). Hence they are concurrent.
Chapter
Simson Line and Pedal Geometry
Key Idea
Pedal geometry studies the feet of perpendiculars from a point to the sides of a triangle. If the point lies on the circumcircle of the triangle, its pedal triangle degenerates into a line. This line is called the Simson line, or the Wallace-Simson line.
In olympiad problems, the Simson line often appears as a hidden collinearity of three projections. The method is to prove cyclicity of small right-angle quadrilaterals and then compare directed angles.
Basic Facts
Let \(ABC\) be a triangle, \(P\) a point in the plane, and \(A_1,B_1,C_1\) the projections of \(P\) onto the lines \(BC,CA,AB\). The triangle \(A_1B_1C_1\) is called the pedal triangle of \(P\).
The points \(P,A_1,C,B_1\) lie on one circle with diameter \(PC\). Similarly, \(P,B_1,A,C_1\) lie on a circle with diameter \(PA\), and \(P,C_1,B,A_1\) lie on a circle with diameter \(PB\).
Wallace-Simson theorem: the points \(A_1,B_1,C_1\) are collinear if and only if \(P\) lies on the circumcircle of \(ABC\).
If \(H\) is the orthocenter of \(ABC\), and \(P\) lies on the circumcircle, then the Simson line of \(P\) passes through the midpoint of \(PH\).
The Simson lines of antipodal points of the circumcircle are perpendicular, and their intersection lies on the nine-point circle.
When to Use This Method
Use the method when a problem contains three perpendicular feet from one point to the sides of a triangle, asks for collinearity of projections, or involves a point on the circumcircle together with the orthocenter and the nine-point circle.
How to Recognise the Method
Look for right angles \(PA_1\perp BC\), \(PB_1\perp CA\), \(PC_1\perp AB\). If two such right angles subtend the same segment, a circle with diameter \(PC\), \(PA\), or \(PB\) appears. If three points must be proved collinear, check whether the original point \(P\) lies on the circumcircle.
Typical Mistakes
Remember that projections are taken onto the side lines, not only onto the segments. For a point \(P\) on an arc, one of the projections often falls on an extension of a side.
A second mistake is proving collinearity without directed angles. In Simson configurations ordinary angles easily change orientation, so directed angles modulo \(180^\circ\) are safer.
A third mistake is confusing the pedal triangle of an arbitrary point with the Simson line. The pedal triangle degenerates into a line only when \(P\) lies on the circumcircle.
Mini-Checklist
Mark the three projections \(A_1,B_1,C_1\). Find cyclic quadrilaterals with diameters \(PA,PB,PC\). Check whether \(P\) lies on the circumcircle. For orthocenter problems, find the midpoint of \(PH\). For two opposite points, check perpendicularity of their Simson lines.
Example 1. First Circles of the Pedal Triangle
This example gives the building block for almost every proof in the module.
Problem. From a point \(P\), perpendiculars are dropped to the lines \(BC\) and \(CA\), with feet \(A_1\) and \(B_1\). Prove that \(P,A_1,C,B_1\) lie on one circle.
Since \(PA_1\perp BC\), and \(A_1\in BC\), we have \(\angle PA_1C=90^\circ\). Similarly, \(\angle PB_1C=90^\circ\). Thus \(A_1\) and \(B_1\) lie on the circle with diameter \(PC\). Therefore \(P,A_1,C,B_1\) are cyclic.
Example 2. Wallace-Simson Theorem
The main theorem: a point on the circumcircle produces a line from three projections.
Problem. A point \(P\) lies on the circumcircle of triangle \(ABC\). Let \(A_1,B_1,C_1\) be its projections onto \(BC,CA,AB\). Prove that \(A_1,B_1,C_1\) are collinear.
The right angles give cyclic quadrilaterals \(P,A_1,C,B_1\) and \(P,B_1,A,C_1\). Hence the directed angles between \(B_1A_1\), \(B_1C_1\), and the sides of the triangle can be expressed through \(\angle PCA\) and \(\angle PAB\).
Since \(A,B,C,P\) lie on one circle, the corresponding inscribed angles are equal. Hence \(\angle A_1B_1C=\angle C_1B_1A\), so the rays \(B_1A_1\) and \(B_1C_1\) are opposite. Therefore \(A_1,B_1,C_1\) are collinear.
Example 3. Converse of the Simson Theorem
Collinearity of pedal points recognises membership in the circumcircle.
Problem. For a point \(P\), the projections onto \(BC,CA,AB\) of triangle \(ABC\) are collinear. Prove that \(P\) lies on the circumcircle of \(ABC\).
Repeat the angle chain from the direct theorem backwards. The cyclicity of \(P,A_1,C,B_1\) and \(P,B_1,A,C_1\) expresses the angles at \(B_1\) through \(\angle PCA\) and \(\angle PAB\). The collinearity of \(A_1,B_1,C_1\) means that these directed angles are equal.
Therefore \(\angle PCA=\angle PBA\) in the directed sense. This is exactly the criterion that \(A,B,C,P\) lie on one circle.
Example 4. The Pedal Triangle Degenerates Only on the Circumcircle
This is the working formulation of the Simson theorem.
Problem. Prove that the pedal triangle of a point \(P\) with respect to \(ABC\) degenerates if and only if \(P\) lies on the circumcircle of \(ABC\).
If \(P\) lies on the circumcircle, then by the Wallace-Simson theorem the three projections are collinear, so the pedal triangle degenerates.
Conversely, if the pedal triangle is degenerate, its three vertices are collinear. By the converse Simson theorem, \(P\) lies on the circumcircle.
Example 5. Direction of the Simson Line
The Simson line changes direction half as fast as the point moves around the circle.
Problem. A point \(P\) moves on the circumcircle of \(ABC\). Explain why if the radius \(OP\) rotates by \(2\varphi\), the Simson line rotates by \(\varphi\).
The direction of the Simson line can be expressed through inscribed angles subtending arcs with endpoint \(P\). When \(P\) moves along an arc of angular measure \(2\varphi\), each corresponding inscribed angle changes by \(\varphi\).
Since the direction of the Simson line is determined by such an inscribed angle, it rotates by half of the angular displacement of \(P\).
Example 6. The Midpoint of \(PH\)
The connection with the orthocenter turns the Simson line into a tool for nine-point circle problems.
Problem. Let \(H\) be the orthocenter of \(ABC\), and let \(P\) lie on the circumcircle. Prove that the Simson line of \(P\) passes through the midpoint of \(PH\).
Denote the projections of \(P\) by \(A_1,B_1,C_1\), and let \(M\) be the midpoint of \(PH\). From the parallelisms \(BH\perp AC\), \(CH\perp AB\), and the definitions of \(B_1,C_1\), the projections of \(M\) in directions parallel to \(AB\) and \(AC\) lie on the line \(B_1C_1\).
More precisely, the homothety with center \(P\) and ratio \(\frac12\) sends segments related to the altitudes from \(H\) to the segments ending at the perpendicular feet from \(P\). Hence \(M\in B_1C_1\). Since \(A_1,B_1,C_1\) are collinear, \(M\) lies on the Simson line.
Example 7. Antipodal Points
Two opposite points of the circumcircle give perpendicular Simson lines.
Problem. Points \(P\) and \(Q\) are antipodal on the circumcircle of \(ABC\). Prove that their Simson lines are perpendicular.
Passing from \(P\) to the antipodal point \(Q\) rotates the radius \(OP\) by \(180^\circ\). By the direction fact for the Simson line, its direction rotates by half of this angle, that is by \(90^\circ\). Hence the Simson lines of \(P\) and \(Q\) are perpendicular.
Example 8. Generalised Simson Line
If perpendiculars are replaced by equal oblique angles, collinearity is preserved after rotation.
Problem. A point \(P\) lies on the circumcircle of \(ABC\). Through \(P\), draw three lines meeting \(BC,CA,AB\) respectively at the same directed angle \(\alpha\). Prove that the three intersection points are collinear.
For \(\alpha=90^\circ\), this is the usual Simson theorem. For general \(\alpha\), rotate each of the three perpendicular projections around \(P\) by the angle \(90^\circ-\alpha\). Since the angle of rotation is the same, the correspondence between the ordinary feet and the new points is given by one spiral homothety centered at \(P\).
A spiral homothety sends a line to a line. Therefore the image of the usual Simson line is again a line, and the three new points lie on it.
Chapter
Brocard, Napoleon, and Special Points
Key Idea
Special points of a triangle are not just a list of names; they usually appear as centers of transformations: homotheties, rotations, isogonal conjugation, symmetries, and projections. In an olympiad problem, the useful skill is not to recognise a name, but to understand which configuration forces concurrence or concyclicity.
This module uses three groups of ideas: the Euler line and the nine-point circle, Napoleon/Fermat constructions with equilateral triangles, and symmedians, the Lemoine point, and Brocard points.
Basic Facts
If \(O\), \(G\), \(H\) are the circumcenter, centroid, and orthocenter of triangle \(ABC\), then they are collinear and \(OG:GH=1:2\). The midpoint of \(OH\) is the center of the nine-point circle.
A symmedian is the isogonal image of a median. If \(AS\) is the \(A\)-symmedian, then \(\frac{BS}{CS}=\frac{AB^2}{AC^2}\). The three symmedians meet at the Lemoine point \(K\).
If external equilateral triangles are constructed on the sides of a triangle, their centers form an equilateral triangle. The Fermat point is connected with \(120^\circ\) angles and arises from the same \(60^\circ\) rotations.
The first Brocard point \(P\) is defined by \(\angle ABP=\angle BCP=\angle CAP\). The second Brocard point is obtained cyclically in the opposite direction; these two points are isogonal conjugates.
When to Use This Method
Use it when the problem contains centroids, orthocenters, midpoints of altitudes, the nine-point circle, equilateral triangles on sides, \(60^\circ\) or \(120^\circ\) angles, tangents to the circumcircle, symmedians, antiparallels, or repeated cyclic angles.
Very often this replaces a long angle chase by a short transformation: a homothety with ratio \(\frac12\), a \(60^\circ\) rotation, an isogonal reflection, or a passage to area ratios.
How to Recognise the Method
Look for “non-accidental” centers: a point on the Euler line, the midpoint of \(OH\), an intersection of lines symmetric about angle bisectors, a center of spiral similarity, a point with three \(120^\circ\) angles, or a point from which the sides are seen under equal cyclic angles.
If squared side lengths appear in ratios, a symmedian or the Lemoine point is probably nearby. If three equilateral triangles appear, test a \(60^\circ\) rotation. If cyclic equal angles repeat around the three sides, this is often a trace of a Brocard point.
Typical Mistakes
Do not replace proof by naming a point: even if a point looks like a Fermat or Lemoine point, its defining property must be proved. A common mistake in Napoleon configurations is forgetting orientation. Another common mistake is confusing a median and a symmedian: a symmedian divides the opposite side in the ratio of squares of adjacent sides, not the sides themselves.
In Brocard problems, non-oriented angles easily create false equalities. It is safer to use directed angles modulo \(180^\circ\).
Mini-Checklist
1. Is there a homothety with ratio \(\frac12\)? 2. Is there a \(60^\circ\) or \(120^\circ\) rotation? 3. Can a median be replaced by its isogonal image, giving a symmedian? 4. Do ratios \(AB^2:AC^2\) appear? 5. Can the special point be proved through its defining property rather than through its name?
Example 1. The Euler Line by Vectors
This example shows why the Euler line is not accidental.
Problem. Let \(O\), \(G\), \(H\) be the circumcenter, centroid, and orthocenter of triangle \(ABC\). Prove that \(O,G,H\) are collinear and \(OG:GH=1:2\).
Take \(O\) as the origin and denote the position vectors of the vertices by \(\vec a,\vec b,\vec c\). For the point \(H\) with position vector \(\vec h=\vec a+\vec b+\vec c\), we have \(AH\perp BC\), since \((\vec h-\vec a)\cdot(\vec b-\vec c)=(\vec b+\vec c)\cdot(\vec b-\vec c)=|\vec b|^2-|\vec c|^2=0\). Similarly, the other two altitudes pass through this point, so it is the orthocenter.
The centroid has position vector \(\vec g=\frac{\vec a+\vec b+\vec c}{3}=\frac{\vec h}{3}\). Hence \(O,G,H\) are collinear and \(OG:GH=1:2\).
Comment. This vector trick is worth remembering: from the circumcenter, the orthocenter is the sum of the position vectors of the vertices.
Example 2. The Nine-Point Circle as an Image of the Circumcircle
The main tool here is a homothety with ratio \(\frac12\).
Problem. Prove that the midpoints of \(AH\), \(BH\), \(CH\) lie on the circle centered at the midpoint of \(OH\) with radius \(\frac R2\), where \(R\) is the circumradius of \(ABC\).
Let \(N\) be the midpoint of \(OH\). The homothety centered at \(H\) with ratio \(\frac12\) sends \(A,B,C\) to the midpoints of \(AH,BH,CH\). It also sends the circumcircle of \(ABC\), centered at \(O\) with radius \(R\), to the circle centered at \(N\) with radius \(\frac R2\).
Thus the three indicated midpoints lie on this circle. In the full statement, the same idea together with right angles shows that the side midpoints and altitude feet lie on it as well.
Comment. The nine-point circle is often easiest to build as the image of the circumcircle.
Example 3. Napoleon's Theorem
A \(60^\circ\) rotation turns a complicated picture into a symmetric one.
Problem. External equilateral triangles are constructed on the sides of \(ABC\). Prove that their centers form an equilateral triangle.
Let \(X,Y,Z\) be the centers of the equilateral triangles constructed on \(BC,CA,AB\). A \(60^\circ\) rotation sending one side of an equilateral triangle to the other sends one segment between centers to the next such segment.
The rotation relation gives \(XY=YZ=ZX\), and the angle between consecutive segments is \(60^\circ\). Hence \(XYZ\) is equilateral.
Comment. In Napoleon configurations, the orientation of the constructed equilateral triangles must be fixed consistently.
Example 4. The Fermat Point via an Equilateral Triangle
The Fermat point is recognised by three \(120^\circ\) angles.
Problem. Assume all angles of \(ABC\) are less than \(120^\circ\). Construct an external equilateral triangle \(BCX\). If \(AX\) meets the analogous line from another vertex at \(T\), prove that \(\angle ATB=\angle BTC=\angle CTA=120^\circ\).
Consider the \(60^\circ\) rotation sending \(B\) to \(C\) about \(X\). It sends one line joining a vertex of the original triangle to the vertex of an equilateral construction to the corresponding line of the next construction.
Therefore their intersection sees two sides under angle \(120^\circ\). Repeating cyclically gives \(\angle ATB=\angle BTC=\angle CTA=120^\circ\).
Comment. If one angle of the original triangle is at least \(120^\circ\), the role of the Fermat point changes: the optimal point is that vertex.
Example 5. Symmedian and the Squared Ratio
This is the basic technique behind the Lemoine point.
Problem. Let \(AM\) be a median of \(ABC\), and let \(AS\) be its isogonal image. Prove that \(\frac{BS}{CS}=\frac{AB^2}{AC^2}\).
For an arbitrary cevian \(AX\), the sine rule in triangles \(ABX\) and \(ACX\) gives \(\frac{BX}{CX}=\frac{AB\sin\angle BAX}{AC\sin\angle XAC}\).
For the median \(AM\), the ratio is \(1\), so \(\frac{\sin\angle BAM}{\sin\angle MAC}=\frac{AC}{AB}\). Since \(AS\) is isogonal to \(AM\), \(\angle BAS=\angle MAC\), \(\angle SAC=\angle BAM\). Therefore \(\frac{BS}{CS}=\frac{AB\sin\angle MAC}{AC\sin\angle BAM}=\frac{AB^2}{AC^2}\).
Comment. This is why the Lemoine point has barycentric coordinates \((a^2:b^2:c^2)\).
Example 6. Tangents and a Symmedian
Tangents to the circumcircle often hide a symmedian.
Problem. The tangents to the circumcircle of \(ABC\) at \(B\) and \(C\) meet at \(P\). Prove that \(AP\) is the \(A\)-symmedian.
By the tangent-chord theorem, \(\angle PBA=\angle ACB\) and \(\angle PCA=\angle ABC\). Thus the line \(AP\) is the isogonal image of the median direction that gives equal division on \(BC\).
Equivalently, let \(AP\cap BC=S\). Similar triangles obtained from the tangent angles give \(\frac{BS}{CS}=\frac{AB^2}{AC^2}\). By the criterion from the previous example, \(AS\) is a symmedian.
Comment. This configuration is often the fastest way to see the Lemoine point.
Example 7. Brocard Points as an Isogonal Pair
This example introduces Brocard points without overloading the full theory.
Problem. Let \(P\) be an interior point of \(ABC\) such that \(\angle ABP=\angle BCP=\angle CAP\). Show that its isogonal conjugate satisfies \(\angle BAQ=\angle ACQ=\angle CBQ\).
Reflect the lines \(AP,BP,CP\) in the corresponding angle bisectors. The three reflected lines meet at a point \(Q\), because isogonal conjugation preserves the trigonometric Ceva condition.
The angle equalities for \(P\), after reflection, become \(\angle BAQ=\angle ACQ=\angle CBQ\). Hence \(Q\) is the second Brocard point.
Comment. In Brocard problems, it is useful to keep isogonal conjugation nearby.
Example 8. The Brocard Circle as a Preview
The final example shows how Lemoine and Brocard geometry join one picture.
Problem. Let \(O\) be the circumcenter, \(K\) the Lemoine point, and \(P,Q\) the two Brocard points. Explain why it is natural to expect \(P\) and \(Q\) to lie on the circle with diameter \(OK\).
The Lemoine point controls symmedians and squared side ratios, while Brocard points arise as centers of spiral similarities between figures built on the sides. The circle with diameter \(OK\) appears as the similarity circle of the three sides of the triangle.
In the full proof, one takes three similar figures built on \(BC,CA,AB\). The corresponding lines through a Brocard point meet at a point of the similarity circle. Since this circle has diameter \(OK\), both Brocard points lie on it.
Comment. This is a preview fact: it is useful as a map of the topic even before the full theory of the similarity circle is developed.
Chapter
Trigonometric Geometry
Key Idea
Trigonometric geometry is useful when ordinary similarity and angle chasing do not provide enough length information. Sines convert angles into segment ratios, while cosines turn a complicated configuration into a single equation.
The main principle is: if a problem contains cevians, points on sides, segment ratios, or concurrence, first try to express the needed ratio through sines of adjacent angles.
Basic Facts
Sine rule: \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\). Cosine rule: \(a^2=b^2+c^2-2bc\cos A\).
For a cevian \(AD\) in triangle \(ABC\): \(\frac{BD}{DC}=\frac{AB\sin\angle BAD}{AC\sin\angle CAD}\).
Trigonometric Ceva: lines \(AA_1\), \(BB_1\), \(CC_1\) are concurrent if and only if \[ \frac{\sin\angle BAA_1}{\sin\angle CAA_1}\cdot \frac{\sin\angle CBB_1}{\sin\angle ABB_1}\cdot \frac{\sin\angle ACC_1}{\sin\angle BCC_1}=1. \]
Trigonometric Menelaus is the analogous condition for collinearity of three points on sides or extensions; signs are handled by directed segments, while in training problems it is often enough to verify the product of sine ratios.
When to Use This Method
Use trigonometry when there is concurrence of three lines, collinearity of three side points, angles such as \(10^\circ,20^\circ,30^\circ,40^\circ\), cevians with prescribed angles, isogonal lines, symmedians, circle radii, or expressions involving \(R,r,p\).
How to Recognise the Method
If you need to prove three lines are concurrent, try trig Ceva. If you need to prove three points are collinear, try trig Menelaus. If a ratio on a side is given, try expressing it through areas or through the sines of two angles at a vertex.
Typical Mistakes
Do not mix ordinary and directed angles without checking signs. Do not cancel \(\sin x\) with \(x\): they are different quantities. In trig Ceva, the order of angles matters; permuting numerators often changes the problem.
Mini-Checklist
1. Which three ratios must be multiplied? 2. Are all angles written at the correct vertices? 3. Are directed segments needed? 4. Can a side be replaced by \(2R\sin A\)? 5. Is there a simple sine identity that finishes the problem?
Example 1. A Cevian and a Sine Ratio
This is the main local tool of the module.
Problem. In triangle \(ABC\), cevian \(AD\) meets \(BC\) at \(D\). Prove that \(\frac{BD}{DC}=\frac{AB\sin\angle BAD}{AC\sin\angle CAD}\).
Triangles \(ABD\) and \(ACD\) have altitudes to the same line \(BC\), so \(\frac{BD}{DC}=\frac{S_{ABD}}{S_{ACD}}\).
Also \(S_{ABD}=\frac12 AB\cdot AD\sin\angle BAD\), and \(S_{ACD}=\frac12 AC\cdot AD\sin\angle CAD\). Dividing gives the required ratio.
Comment. Almost all of trig Ceva is built from this formula.
Example 2. Trigonometric Ceva
This example shows how local ratios turn into concurrence.
Problem. Prove trig Ceva for points \(A_1\in BC\), \(B_1\in CA\), \(C_1\in AB\).
By the previous example, \(\frac{BA_1}{CA_1}=\frac{AB\sin\angle BAA_1}{AC\sin\angle CAA_1}\). Write the analogous ratios for the other two sides.
Ordinary Ceva requires the product \(\frac{BA_1}{CA_1}\cdot\frac{CB_1}{AB_1}\cdot\frac{AC_1}{BC_1}\) to be \(1\). When multiplying, the side factors \(AB,BC,CA\) cancel, leaving exactly the trigonometric condition.
Comment. The converse is proved the same way through ordinary Ceva.
Example 3. Isogonal Lines and Product of Ratios
Here trigonometry explains symmedians with little extra construction.
Problem. Lines \(AX\) and \(AY\) are isogonal in angle \(A\) and meet \(BC\) at \(X,Y\). Prove that \(\frac{BX}{CX}\cdot\frac{BY}{CY}=\frac{AB^2}{AC^2}\).
By the cevian formula, \(\frac{BX}{CX}=\frac{AB\sin\angle BAX}{AC\sin\angle XAC}\). For \(Y\), \(\frac{BY}{CY}=\frac{AB\sin\angle BAY}{AC\sin\angle YAC}\).
Since \(AX\) and \(AY\) are isogonal, \(\angle BAX=\angle YAC\), \(\angle XAC=\angle BAY\). Multiplying cancels the sine factors and leaves \(\frac{AB^2}{AC^2}\).
Comment. If \(X\) is the midpoint of \(BC\), then \(Y\) lies on the symmedian.
Example 4. Trigonometric Menelaus
This is the collinear analogue of trig Ceva.
Problem. Points \(A_1\in BC\), \(B_1\in CA\), \(C_1\in AB\) are collinear. State the trigonometric Menelaus condition.
Ordinary Menelaus with directed segments gives \(\frac{BA_1}{CA_1}\cdot\frac{CB_1}{AB_1}\cdot\frac{AC_1}{BC_1}=-1\).
Each ratio is expressed through the sines of the angles made by the transversal with the two sides of the triangle. After side factors cancel, we get a product of three sine ratios; with directed angles it equals \(-1\), while in non-directed form one checks equality of absolute values.
Comment. In problems with extensions, the sign is often the main trap.
Example 5. A Symmedian via Trig Ceva
A symmedian is a typical object best recognised through sines.
Problem. Prove that if \(AS\) is the \(A\)-symmedian, then \(\frac{BS}{CS}=\frac{AB^2}{AC^2}\).
Let \(AM\) be the median isogonal to \(AS\). For the median, \(\frac{BM}{CM}=1\), so \(\frac{AB\sin\angle BAM}{AC\sin\angle MAC}=1\).
Since \(AS\) is isogonal to \(AM\), \(\angle BAS=\angle MAC\), \(\angle SAC=\angle BAM\). Substituting into the cevian formula gives \(\frac{BS}{CS}=\frac{AB^2}{AC^2}\).
Comment. This connects the present module to the Lemoine point from the previous one.
Example 6. Kiepert Cevians
This example shows why similar triangles on the sides give one point.
Problem. On sides \(BC,CA,AB\), external similar triangles with apex angle \(\varphi\) are constructed. Prove that the lines from \(A,B,C\) to the corresponding outer vertices are concurrent.
Let the corresponding vertex on side \(BC\) be \(A_1\). Then angles \(\angle BAA_1\) and \(\angle CAA_1\) are expressed through \(B,C,\varphi\). The two other cevians are analogous.
Substitution into trig Ceva gives a product of the form \(\frac{\sin(B+\varphi)}{\sin(C+\varphi)}\cdot\frac{\sin(C+\varphi)}{\sin(A+\varphi)}\cdot\frac{\sin(A+\varphi)}{\sin(B+\varphi)}=1\). Hence the three lines are concurrent.
Comment. This is one entrance to Kiepert geometry.
Example 7. Special Angles and a Sine Identity
Sometimes the whole problem reduces to one clean product of sines.
Problem. Prove that if three cevians in a triangle make angle pairs \(10^\circ,70^\circ\), \(30^\circ,20^\circ\), \(40^\circ,10^\circ\) at the vertices, then the cevians are concurrent.
By trig Ceva it is enough to check \(\frac{\sin10^\circ}{\sin70^\circ}\cdot\frac{\sin30^\circ}{\sin20^\circ}\cdot\frac{\sin40^\circ}{\sin10^\circ}=1\).
After cancelling \(\sin10^\circ\), it remains to prove \(\sin30^\circ\sin40^\circ=\sin20^\circ\sin70^\circ\). This is true because \(\sin30^\circ=\frac12\), and \(\sin70^\circ=\cos20^\circ\), so the right side is \(\sin20^\circ\cos20^\circ=\frac12\sin40^\circ\).
Comment. Such problems are excellent practice for recognising trig Ceva.
Example 8. Cosines in a Parallelism Condition
Cosines help when geometry must become one numerical condition.
Problem. Let \(H\) be the orthocenter and \(O\) the circumcenter. Explain why \(OH\parallel BC\) can be reduced to \(\tan B\tan C=3\).
We know \(AH=2R\cos A\), and the altitude \(AA_h=2R\sin B\sin C\). If the Euler line is parallel to \(BC\), the centroid divides the median in ratio \(2:1\), and projecting the condition onto the altitude from \(A\) gives \(AH:AA_h=2:3\).
Thus \(\frac{2R\cos A}{2R\sin B\sin C}=\frac23\), so \(3\cos A=2\sin B\sin C\). Substituting \(\cos A=\sin B\sin C-\cos B\cos C\), we obtain \(\sin B\sin C=3\cos B\cos C\), or \(\tan B\tan C=3\).
Comment. This example shows how trigonometry works with the Euler line.