Olympiad Geometry: Circles, Ceva and Computation — IOQM, RMO and INMO

Weightage: Geometry gives one problem in most RMO and INMO papers and a few in the IOQM. A diagram and a short list of theorems solve most of them, and the typical failure is not the theorem but the diagram: a figure that is too special hides the true relationships.

1. Draw first, and draw generally

Make a large, accurate figure with a scalene triangle, not an isosceles or right one. A special figure creates false coincidences. Mark every equal angle and equal length as you find it, and list what you know before deciding what to prove.

2. Angle chasing and cyclic quadrilaterals

The foundation is the inscribed angle theorem: an angle at the circumference is half the central angle on the same arc, and angles on the same arc are equal.

A quadrilateral $ABCD$ is cyclic if and only if any of these holds:

• Opposite angles sum to $180^{\circ }$.
• An exterior angle equals the interior opposite angle.
• $\angle ABD=\angle ACD$ (equal angles subtended by the same side).

The tangent-chord angle equals the inscribed angle on the chord's other side. When a problem says four points are concyclic, prove it by one of these angle conditions, and when you are given a circle, use it to move angles around.

3. Similar triangles and lengths

Triangles are similar by AA, SAS (with proportional sides) or SSS. Similar triangles turn an angle equality into a length ratio.

The angle bisector theorem: if $AD$ bisects angle $A$ in triangle $ABC$ with $D$ on $BC$, then $BD/DC=AB/AC$.

Stewart's theorem: for a cevian $AD$ to side $BC$ with $BD=m$, $DC=n$, $BC=a$, $AD=d$,

$b^{2}m+c^{2}n=a(d^2+mn).$

Ptolemy's theorem: for a cyclic quadrilateral, $AC\cdot BD=AB\cdot CD+AD\cdot BC$. For a unit square this gives $\sqrt{2}\cdot \sqrt{2}=1+1$.

4. Power of a point and radical axes

For a point $P$ and a circle, power of a point says that for any line through $P$ meeting the circle at $A$ and $B$, the product $PA\cdot PB$ is constant. If $PT$ is a tangent, $P{T}^2=PA\cdot PB$.

Worked example. From an external point $P$, a secant meets a circle at $A$ and $B$ with $PA=4$ and $AB=5$. Then $PB=9$ and the tangent length is $PT=\sqrt{4\times 9}=6$.

The radical axis of two circles is the locus of points with equal power to both, and it is a straight line perpendicular to the line of centres. The three radical axes of three circles meet at one point (the radical centre) or are parallel. Use it to show that three lines are concurrent.

5. Ceva, Menelaus and areas

Ceva's theorem. Cevians $AD,BE,CF$ of triangle $ABC$ are concurrent if and only if

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

Medians, angle bisectors and altitudes of an acute triangle all satisfy it. The trigonometric form uses the sines of the divided angles.

Menelaus' theorem. Points $D,E,F$ on the lines $BC,CA,AB$ are collinear if and only if the same product equals $-1$ in signed lengths, or $1$ in unsigned lengths when an odd number of the points (one or three) lie on the extensions of the sides.

Area ratios. Triangles with the same height have areas in the ratio of their bases, and triangles with the same base have areas in the ratio of their heights. Many length ratios are easiest to find through areas.

6. Formulas for a triangle

With sides $a,b,c$, semi-perimeter $s$, area $K$:

• Heron: $K=\sqrt{s(s-a)(s-b)(s-c)}$.
• Circumradius: $R=\frac{abc}{4K}$, and $a=2R\sin A$.
• Inradius: $r=K/s$.
• Law of cosines: $a^2=b^2+c^2-2bc\cos A$.

Worked example. For the 13-14-15 triangle, $s=21$ and $K=\sqrt{21\cdot 8\cdot 7\cdot 6}=84$. So $r=84/21=4$ and $R=\frac{13\cdot 14\cdot 15}{4\cdot 84}=\frac{65}{8}$.

Other useful facts: the centroid divides each median $2:1$, the orthocentre, centroid and circumcentre are collinear (the Euler line) with $OG:GH=1:2$, and the midpoints of the sides, the feet of the altitudes and the midpoints of the segments from the orthocentre to the vertices lie on the nine-point circle, of radius $R/2$.

7. Transformations

A homothety (a scaling about a point) maps a figure to a similar one and preserves angles. It explains why the centroid, the circumcentre and the orthocentre line up, and why a tangent circle and a larger circle share a centre of similarity. A reflection or rotation can turn a configuration into a simpler one. Inversion about a circle sends lines and circles to lines and circles, and it can straighten a tangency problem.

8. When to compute

If synthetic methods stall, compute.

• Coordinates: place a vertex at the origin and one side on an axis. Good for perpendicularity and collinearity.
• Trigonometry: use the law of sines to express every length by angles.
• Complex numbers: for configurations on a circle, place it as the unit circle.

A computation is a proof as long as every step is justified, and it is a safe choice when you cannot find a clever solution.

9. Geometry in the IOQM

Answers are integers, so lengths and areas often come out as integers or are asked in a scaled form. After finding the structure with a figure, compute the value with Heron, similar triangles or power of a point, and check that the figure is consistent with the answer.

Common traps

• A special diagram that gives a false coincidence.
• Assuming a point lies on a line because it looks that way.
• Using Ceva with unsigned lengths for external points.
• Forgetting that cyclic needs the converse, since the forward theorem does not prove concyclicity.
• Ptolemy for a non-cyclic quadrilateral. It then gives only an inequality.

Memory aids

• "Same arc, same angle": inscribed angles.
• "Product of segments": power of a point.
• "Ratio product equals one": Ceva.

Summary

Geometry is solved by a faithful diagram, angle chasing around circles, similar triangles and a few named theorems: power of a point, Ceva, Menelaus, Ptolemy and Stewart. Triangle formulas link sides, area and radii.

When the synthetic approach fails, coordinates, trigonometry or complex numbers give a reliable computation.

Exam protocol

• Draw a scalene, general figure before reasoning.
• State each theorem with its hypothesis.
• Prove concyclicity or collinearity with the converse.
• Choose a computation if the synthetic idea does not come within ten minutes.