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Course · Geometry

Plane geometry III

The national rounds: sustained angle chases, transformations, and the incircle's whole family.

The hardest band of the corpus, drawn from the Serbian drzavno and republicko rounds and the Slovenian drzavno competition for first-year students, category A. It opens with the national-round configuration problems - long angle chases, the auxiliary constructions that unlock them, and collinearity as the conclusion - then makes transformations a first-class tool: reflections that straighten paths, half-turns that complete parallelograms, rotations that transplant segments, and homothety that scales whole circles. The course closes on the incircle's full family: tangent-length bookkeeping, Pitot's theorem, excircles, the touch chord and the arc-midpoint lemma. Two lessons carry the new theory; every other concept enters through a warm-up problem.

3 chapters · 2 lessons · 33 problems

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National-round configurations

The step up to the national rounds is not new machinery but longer arguments. The opening third chases angles through circles that first have to be produced; the middle third finds or builds the triangle the figure is missing, from a straddling congruent pair to an invented rectangle and an equilateral copy; the closing third ends the way hard geometry ends, with three points forced onto one line. Every tool comes from the earlier courses - what is new is the discipline of holding a ten-step configuration together.

Circle angle chasesAuxiliary constructionsCollinearity and concurrency

  1. Warm-upLet \(ABC\) be a triangle and let \(M\), \(N\), \(P\) be points on its sides \(AB\), \(BC\), \(AC\) respectively, chosen so that \(AMNP\) is a parallelogram. Let \(k_1\) be the circle circumscribed about …Next upRepublic
  2. PracticeLet \(ABCD\) be a parallelogram whose interior angle at \(A\) is acute, and let \(E\) be a point of the plane such that \(EA \perp AB\) and \(EC \perp CB\). Prove that \[ \angle AED = \angle CEB . \]Republic
  3. PracticeLet \(ABC\) be an acute triangle and let \(D\) be the foot of the altitude from \(A\), so that \(D\) lies on the side \(BC\). A point \(P\) is chosen on the segment \(AD\) in such a way that \[ \angle PBA = \angle PCA . \] …National
  4. ConsolidateIn a triangle \(ABC\) we have \(\angle ABC = 45^\circ\) and \(\angle CAB = 15^\circ\). Let \(M\) be the point of the ray \(BC\) for which \[ \overrightarrow{BM} = 3 \cdot \overrightarrow{BC} . \] Determine …National
  5. Warm-upA square \(ABCD\) is given, together with points \(E\) and \(F\) lying outside the square such that the triangles \(BEC\) and \(CFD\) are equilateral. Prove that the triangle \(AEF\) is equilateral as …National
  6. PracticeA triangle \(ABC\) has a point \(D\) on side \(AB\) and a point \(E\) on side \(AC\) such that \[ |AE| = |ED| = |DB| \qquad\text{and}\qquad |AD| = |DC| = |CB| . \] Determine the angles of triangle \(ABC\). …National
  7. PracticeLet \(ABC\) be an isosceles triangle with \(AB = AC\). Let \(D\) be the point of the side \(AC\) for which \(CD = 2 \, AD\), and let \(P\) be a point of the segment \(BD\) with \(\angle APC = 90^\circ\). …Republic
  8. ConsolidateIn a triangle \(ABC\) the angle at \(B\) equals \(80^\circ\). Three further points are marked: the point \(D\) on the side \(BC\) with \(AB = AD = CD\); the point \(F\) on the side \(AB\) with \(AF = BD\); …National
  9. Warm-upIn a rectangle \(ABCD\) with \(|AB| > |BC|\), the perpendicular bisector of the diagonal \(AC\) meets the side \(CD\) at the point \(E\). The circle with center \(E\) and radius \(|AE|\) meets the side …National
  10. PracticeLet \(D\) be the midpoint of side \(AB\) of an acute triangle \(ABC\). Points \(A'\) and \(B'\) are chosen on the segments \(AC\) and \(BC\) so that the triangles \(ADA'\) and \(DBB'\) are both isosceles …National
  11. ChallengeLet \(ABC\) be a triangle, let \(D\) be the foot of the altitude from \(A\) to the line \(BC\), and let \(\omega\) be the circle whose diameter is the segment \(AD\). Denote by \(G\) the second common …National

Transformations

The national-round instinct: stop chasing points and move the whole configuration. A homothety blows the circumcircle up from a chosen centre, a mirror folds a path straight, and a half-turn about a midpoint turns an orthocentre into an antipode - one map, and every length and angle comes along. Eleven problems from the Serbian and Slovenian national ladders in which naming the right transformation replaces a page of case analysis.

HomothetyReflectionsRotations and central symmetry

  1. LessonTransformations
  2. Warm-upA point \(D\) is chosen on the side \(BC\) of a triangle \(ABC\). Points \(E\) and \(F\), both different from \(D\), are chosen on the line \(BC\) so that \[ BE = BD \qquad \text{and} \qquad CF = CD . \] …Regional
  3. PracticeLet \(k > 0\). On the sides \(A_1B_1\), \(B_1C_1\) and \(C_1A_1\) of a triangle \(A_1B_1C_1\), points \(C_2\), \(A_2\) and \(B_2\) are chosen, respectively, so that \[ \frac{A_1C_2}{C_2B_1} = \frac{B_1A_2}{A_2C_1} = \frac{C_1B_2}{B_2A_1} = k . \] …National
  4. ConsolidateLet \(M\) be the midpoint of the base \(AB\) of a trapezoid \(ABCD\), so that \(AB \parallel CD\). A point \(E\) is taken in the interior of the segment \(AC\) in such a way that the lines \(BC\) and \(ME\) …National
  5. Warm-upTwo vertical poles stand on level ground, at a distance of \(9\) m from each other; one pole is \(11\) m high and the other is \(15\) m high. A rope of length \(15\) m is fastened to the top of one pole …Republic
  6. PracticeA circle \(k\) has radius \(31\,\mathrm{mm}\), and \(\ell\) is a broken line of length \(61\,\mathrm{mm}\) whose two endpoints both lie on \(k\). Prove that there is a line \(p\) passing through the centre …Republic
  7. ConsolidateThe lines containing the two diagonals of a quadrilateral \(ABCD\) meet at an angle of \(60^{\circ}\). Each vertex of the quadrilateral is reflected in the line containing the diagonal that joins its two …National
  8. Warm-upA parallelogram \(ABCD\) satisfies \(|AB| = |BD|\). Let \(K\) be the point of line \(AB\), different from \(A\), with \(|KD| = |AD|\). Let \(M\) be the image of \(C\) under the half-turn about \(K\) (so …National
  9. PracticeLet \(H\) and \(O\) be the orthocentre and the circumcentre of a triangle \(ABC\) with \(AB \neq AC\). The lines \(AH\) and \(AO\) meet the circumcircle of \(ABC\) a second time at \(M\) and \(N\) respectively. …National
  10. PracticeLet \(ABC\) be an isosceles triangle with \(AB = AC\). A point \(P\) is taken inside the triangle so that \[ \angle BPC = 90^\circ + \tfrac{1}{2}\angle BAC , \] and a point \(Q\) is taken so that \(\angle BPQ = \angle PQA = 90^\circ\). …National
  11. ConsolidateLet \(ABC\) be a triangle and let \(k\) be its circumcircle, with centre \(O\). Construct a point \(D\) on \(k\) such that the centroids of the triangles \(ABC\) and \(ABD\) are collinear with the point …National
  12. ChallengeCircumscribe about a given triangle \(ABC\) an equilateral triangle \(PQR\) whose side is as long as possible. (Here \(\triangle PQR\) is called circumscribed about \(\triangle ABC\) when \(A \in QR\), …National

The incircle and its relatives

The closing chapter of the course belongs to the circle that touches all three sides. Naming the tangent lengths s - a, s - b, s - c turns tangency into arithmetic, Pitot's theorem does the same for quadrilaterals, and the excircles extend the bookkeeping past the vertices with the semiperimeter itself. On top of the lengths sit two configurations the national rounds return to again and again: the touch chord with its known direction, and the arc midpoint that holds B, C and the incentre on one circle.

Tangent lengths and pitotIncentre configurationsExcircles and touch chords

  1. LessonThe incircle and its relatives
  2. Warm-upA circle can be inscribed in the trapezoid \(ABCD\), whose parallel sides are \(AB\) and \(CD\). Prove that the circle having \(BC\) as a diameter and the circle having \(AD\) as a diameter touch each …National
  3. PracticeA disc \(K\) of radius \(R\) is cut into three circular sectors in such a way that the areas of the two smaller sectors add up to the area of the largest sector, while the difference of the areas of the …National
  4. ConsolidateThree points \(A\), \(B\), \(C\), not lying on one line, are given. Construct a point \(D\) for which the quadrilateral \(ABCD\) is at the same time cyclic and tangential, that is, admits both a circumscribed …National
  5. Warm-upLet \(ABC\) be a triangle. Points \(D\) and \(E\) lie on the rays \(CA\) and \(CB\) respectively, but not on the sides of the triangle \(ABC\), and are chosen so that \[ |AD| = |BE| = |AB| . \] Let \(G\) …National
  6. PracticeLet \(ABC\) be a triangle with \(\angle CAB = 60^\circ\). Denote by \(O\) and \(I\) the centres of the circle circumscribed about it and of the circle inscribed in it, respectively, and let \(A'\) be the …National
  7. ConsolidateLet \(I\) be the incenter of a triangle \(ABC\), and suppose that \[ |CA| + |AI| = |BC| . \] Determine the ratio of the sizes of the angles \(\angle BAC\) and \(\angle CBA\).National
  8. Warm-upA triangle \(ABC\) is given. Consider all lines which cut the side \(AC\) at a point \(M\) and the side \(BC\) at a point \(N\) in such a way that \(MN = AM + BN\). Prove that there is a circle \(k\) which …Republic
  9. PracticeThe circle inscribed in a triangle \(ABC\) touches the sides \(AB\) and \(AC\) at the points \(M\) and \(N\) respectively. Let \(P\) be the point in which the bisector of the angle \(ABC\) meets the line …Republic
  10. PracticeLet \(I\) be the centre of the inscribed circle of a triangle \(ABC\), and let \(A_1\), \(B_1\), \(C_1\) be the feet of the perpendiculars dropped from \(I\) to the sides \(BC\), \(AC\) and \(AB\). The …National
  11. ConsolidateLet \(ABC\) be a triangle. Prove that the following three lines all pass through one point: the bisector of the angle at \(A\); the line through the midpoints of the sides \(CA\) and \(CB\); and the line …National
  12. ChallengeLet \(ABC\) be a right triangle with hypotenuse \(AB\), and let \(D\) be the midpoint of \(AB\). Let \(k\) be the circumcircle of the triangle \(BCD\), and let \(E\) be an arbitrary point of the shorter …National

Problems from the Serbian drzavno, republicko and okruzno rounds, high school grade I, category A, 1996-2026, organized by the Mathematical Society of Serbia (DMS), and from the Slovenian drzavno competition (1. letnik, kategorija A, 2002-2023) organized by DMFA Slovenije. Statements and solutions are re-expressed in English; the mathematical content follows the official papers.