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the future, since I have almost completely left elementary geometry now; only occasionally I publish some of my older unpublished work.

Electronically avaliable publications:

• Darij Grinberg, Ehrmann's Third Lemoine Circle, Journal of Classical Geometry 1 (2012) pages 40-52.
Local copy of the PDF. Also, last draft version and its TeX source.

The symmedian point of a triangle is known to give rise to two circles, obtained by drawing parallels (for the first circle) and antiparallels (for the second one) to the sides of the triangle through the symmedian point. In this note we will explore a third circle with a similar construction - discovered by Jean-Pierre Ehrmann in Hyacinthos message #6098. It is obtained by drawing circles through the symmedian point and two vertices of the triangle, and intersecting these circles with the triangle's sides. We prove the existence of this circle and identify its center and radius.

• 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. Erect similar isosceles triangles on the segments BA1, CA1, CB1, AB1, AC1, BC1, and consider the apices of these triangles. 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, which is called 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 giving a new twist 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.

• Cosmin Pohoata and Darij Grinberg, Problem F07-1, The Harvard College Mathematics Review, Vol. 1, No. 2, Fall 2007. (Link to the problem statement. For the solution see Vol. 2, No. 1, Spring 2008.)
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.)
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.)
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 Xb and Xc, respectively. Prove that 1 / GX + 1 / GXb + 1 / GXc = 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  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 Hb 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 Hb, 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 PHbQ 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.

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)).

Publications - Geometry

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Darij Grinberg