Publications - Geometry

These are articles and problems I have published during the (roughly) 5 years I have spent doing elementary geometry (2002-2007). This list is not likely to grow much in the future, since I have almost completely left elementary geometry now.

Electronically avaliable publications:

• Cosmin Pohoata and Darij Grinberg, Problem F07-1, The Harvard College Mathematics Review, Vol. 1, No. 2, Fall 2007. (Link to the problem statement. The deadline for submitting solutions is over.)
  Consider ABC an arbitrary triangle and P a point in its plane. Let D, E, and F be three points on the lines through P perpendicular to the lines BC, CA, and AB, respectively. Prove that if triangle DEF is equilateral, and if the point P lies on the Euler line of triangle ABC, then the center of triangle DEF also lies on the Euler line of triangle ABC.

• Cosmin Pohoata and Darij Grinberg, Problem O65, Mathematical Reflections 5/2007. (Link to the problem statement. The deadline for submitting solutions is over.)
  Let ABC be a triangle, and let D, E, and F be the tangency points of its incircle with BC, CA, and AB, respectively. Let I be the center of this incircle. Let X₁ and X₂ be the intersections of line EF with the circumcircle of triangle ABC. Similarly, define Y₁ and Y₂ as well as Z₁ and Z₂. Prove that the radical center of the circles DX₁X₂, EY₁Y₂, and FZ₁Z₂ lies on the line OI, where O is the circumcenter of triangle ABC.

• Darij Grinberg, Problem S64, Mathematical Reflections 5/2007. (Link to the problem statement. The deadline for submitting solutions is over.)
  Proposed solution as a PDF file.
  Let ABC be a triangle with centroid G, and let g be a line through G. Line g intersects BC at a point X. The parallels to lines BG and CG through A intersect line g at points X_b and X_c, respectively. Prove that 1 / GX + 1 / GX_b + 1 / GX_c = 0, where the segments are directed.

• Cezar Lupu and Darij Grinberg, Problem O49, Mathematical Reflections 3/2007. (Link to the solution. For the problem see 3/2007.)
  Proposed solution as a PDF file.
  Let A₁, B₁, C₁ be points on the sides BC, CA, AB of a triangle ABC. The lines AA₁, BB₁, CC₁ intersect the circumcircle of triangle ABC at the points A₂, B₂, C₂, respectively (apart from A, B, C). Prove that AA₁ / A₁A₂ + BB₁ / B₁B₂ + CC₁ / C₁C₂ >= 3s² / (r (4R+r)), where s, r, R are the semiperimeter, inradius, and circumradius of triangle ABC, respectively.

• Darij Grinberg, The Neuberg-Mineur circle, Mathematical Reflections 3/2007.
  PDF file. See http://reflections.awesomemath.org/2007_3/NeubergMineur.pdf for the same file on the Mathematical Reflections server.
  
  In a Mathésis article from 1931, V. Thébault and A. Mineur established a remarkable property of quadrilaterals:
  Let ABCD be a quadrilateral, and let X, Y, Z, W be the points on the lines AB, BC, CD, DA which divide the sides AB, BC, CD, DA externally in the ratios of the squares of the adjacent sides, i. e. which satisfy
  AX / XB = - DA² / BC²; BY / YC = - AB² / CD²; CZ / ZD = - BC² / DA²; DW / WA = - CD² / AB², where the segments are directed.
  Then, the points X, Y, Z, W lie on one circle.
  In the above note, I show a proof of this fact (not having access to the original paper, I cannot decide whether this proof is new) as well as an additional result: If ABCD is a cyclic quadrilateral, then the circle through the points X, Y, Z, W degenerates into a line. This additional result is proven in two different ways, one of them yielding a characterization of this line as a radical axis.
  
  Thanks to the help of Francisco Bellot Rosado, I have obtained the Mathésis article by V. Thébault and A. Mineur referenced as [6] in my article. It provides an algebraic proof of the concyclicity of X, Y, Z, W.

• Darij Grinberg, Problem O25, Mathematical Reflections 6/2006. (Link to the solution. For the problem see 5/2006.)
  Proposed solution as a PDF file.
  For any triangle ABC, prove that cos A/2 cot A/2 + cos B/2 cot B/2 + cos C/2 cot C/2 >= sqrt(3)/2 (cot A/2 + cot B/2 + cot C/2).

• Darij Grinberg, From Baltic Way to Feuerbach - A Geometrical Excursion, Mathematical Reflections 2/2006.
  Zipped PDF file. See From Baltic Way to Feuerbach - A Geometrical Excursion for the published version; however, it has some graphical errors.
  
  We start our journey with a geometry problem from the Baltic Way Mathematical Contest 1995:
  Let ABC be a triangle, and B' the midpoint of its side CA. Denote by H_b the foot of the B-altitude of triangle ABC, and by P and Q the orthogonal projections of the points A and C on the bisector of angle ABC. Then, the points H_b, B', P, Q lie on one circle.
  
  By studying further properties of the points P and Q and of the circle (i. e., the center of the circle lies on the nine-point circle of triangle ABC), we arrive at some more results. A greater step is done by identifying the incenter of triangle PH_bQ as the point where the incircle of triangle ABC touches CA. Using these observations, we establish a relationship between the configuration and the famous Feuerbach theorem, stating that the incircle of a triangle touches its nine-point circle. We succeed to prove this Feuerbach theorem in a new way. Further investigations, partly based on results obtained before, yield new proofs of two known characterizations of the Feuerbach point (the point of tangency of the incircle and the nine-point circle of triangle ABC) as Anti-Steiner point.

• Darij Grinberg and Alexei Myakishev, A Generalization of the Kiepert Hyperbola, Forum Geometricorum 4 (2004) pages 253-260.
  
  Consider an arbitrary point P in the plane of triangle ABC with cevian triangle A₁B₁C₁. Erecting similar isosceles triangles on the segments BA₁, CA₁, CB₁, AB₁, AC₁, BC₁, we get six apices. If the apices of the two isosceles triangles with bases BA₁ and CA₁ are connected by a line, and the two similar lines for B₁ and C₁ are drawn, then these three lines form a new triangle, which is perspective to triangle ABC. For fixed P and varying base angle of the isosceles triangles, the perspector draws a hyperbola. Some properties of this hyperbola are studied in the paper.

• Atul Dixit and Darij Grinberg, Orthopoles and the Pappus theorem, Forum Geometricorum 4 (2004) pages 53-59.
  
  If the vertices of a triangle are projected onto a given line, the perpendiculars from the projections to the corresponding sidelines of the triangle intersect at one point, the orthopole of the line with respect to the triangle. We prove several theorems on orthopoles using the Pappus theorem, a fundamental result of projective geometry.

• Darij Grinberg, On the Kosnita Point and the Reflection Triangle, Forum Geometricorum 3 (2003) pages 105-111.
  
  The Kosnita point of a triangle is the isogonal conjugate of the nine-point center. We prove a few results relating the reflections of the vertices of a triangle in their opposite sides to triangle centers associated with the Kosnita point.
  
  See also a geometry-college posting related to the paper.

• Darij Grinberg and Paul Yiu, The Apollonius Circle as a Tucker Circle, Forum Geometricorum 2 (2002) pages 175-182.
  
  We give a simple construction of the circular hull of the excircles of a triangle as a Tucker circle.
  
  See also the Hyacinthos message related to the paper.

Non-electronic publications include:

• Darij Grinberg, Mircea Lascu, Marius Pachitariu, Marian Tetiva, Din Nou Despre Inegalitati Geometrice, Gazeta Matematica - Seria B, 6/2006 pages 285-292. (in Romanian)
  This note is devoted to the inequality 8R² + 4r² >= a² + b² + c², where R is the circumradius and r is the inradius of an acute-angled triangle ABC. This inequality is also known in its equivalent form (1 - cos A) (1 - cos B) (1 - cos C) >= cos A cos B cos C (again, only for acute-angled triangles). Five different proofs of this inequality are given, as well as a sharper version and a geometrical interpretation.

• Darij Grinberg, Cezar Lupu, Problem 896, College Mathematics Journal, 2009 no. 2 (March 2009).
  Let a, b, c be positive reals such that a + b + c = 3. Show that 1/a + 1/b + 1/c >= 1 + 2 sqrt((a^2+b^2+c^2)/(3abc)).
  The deadline for submitting solutions is June 15, 2009.

• Floor van Lamoen and Darij Grinberg, Problem 11025, American Mathematical Monthly 110 (2003) page 543.

Publications - Geometry

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Darij Grinberg