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Hadamard: Elementary Geometry HADAMARD: ELEMENTARY GEOMETRY SOLUTIONS AND NOTES TO SUPPLEMENTARY PROBLEMS Mark Saul Center for Mathematical Talent Courant Institute of Mathematical Sciences New York University 1 2 FOREWORD This addendum to the Reader's Companion completes the solutions and an- notations to the problems in Hadamard's elementary geometry text (Lessons in Geometry. I. Plane Geometry, Jacques Hadamard, Amer. Math. Soc. (2008)). Because of space constraints, they were not included in the printed volume con- taining solutions to most of Hadamard's problems (Hadamard's Plane Geometry: A Reader's Companion, Mark Saul, Amer. Math. Soc. (2010)). Hadamard revised his geometry textbook a dozen times during his long life, and kept adding interesting problems to the miscellany he had provided in the original edition. So this collection of problems is the result of Hadamard's lifetime love of synthetic Euclidean geometry. The textbook itself is written in an abbreviated, almost telegraphic style. The problems offered for each chapter invite the reader to unfold the compressed exposition, discovering its implications through hands-on experience with geometry. These problems have a somewhat different feel to them. Their results sometimes take us far from the main stream of the exposition in the text. Often, they require a deep, almost virtuosic control of the techniques and theorems offered in the text. Many of them generalize or build on problems mentioned earlier. While some of these problems fall more easily to advanced (i.e. analytic) methods, we have here provided solutions for the most part within the bounds of Hadamard's text and his synthetic approach. As with most large efforts, much credit must be given to others. Behzad Mehrdad patiently reviewed each piece of the manuscript, offering valuable insights, correcting errors, and in some cases providing improved solutions to the problem. Alexei Kopylov likewise suggested significant corrections to and improvements on the original manuscript. I am also indebted to the Education Development Center (EDC) and Al Cuoco in particular for significant and sustained support in this translation project. But the lion's share of the work was actually done by others. More than is the case with the problems in the textbook, these solutions are often based on those of D. I. Perepelkin and his colleagues, who prepared the Russian edition of Hadamard's book (Gosudarstvenoye Uchebno-Pedagogicheskoye Izdatelstvo Minis- terstva Prosveshcheniya RSFSR Moscow, 1957). Perepelkin's solutions are written for a professional mathematical audience, and even where his ideas are central to the solution, the exposition has been reworked significantly. Of course, I take responsibility for any errors that have crept in during this process. The work on Hadamard's Elementary Geometry was initiated under National Science Foundation grant ESI 0242476-03 to the Educational Development Center. Additional support was provided by the John Templeton Foundation through a grant to the Mathematical Sciences Research Institute, and the Alfred P. Sloan Foundation through a grant to the Courant Institute of Mathematical Sciences at New York University. Mark Saul New York City August 2011 3 PROBLEMS, SOLUTIONS, AND NOTES Exercises 349, 350, 353, 354,384, 386, 387, 393, 394, 400, 404, 413 are taken from the General Competition of Lyc´eesand Colleges. Exercises 365, 374, 397, 406, 409, 412, 421 were taken from the contest of the Assembly of the Mathematical Sciences. We have not felt obligated to give these problems in the form in which they were originally proposed; we have, in particular, made certain changes in their formulation to correspond to other exercises given in the rest of this work. Problem 343. If A; B; C; D are four points on a circle (in that order), and if a; b; c; d are the midpoints of arcs AB; BC; CD; DA, show that lines ac; bd are perpendicular. 1 1 1 1 Solution. We have aB= 2 AB; Bb= 2 BC; cD= 2 CD; Dd= 2 DA : 1 ◦ Thus ab + cd= 2 AB + BC + CD + DA = 180 , and since the angle be- tween lines ac and bd is measured by half this arc, it must be a right angle. Figure t344a Problem 344. We take points D; E; F on sides BC; CA; AB of a triangle, and construct circles AEF , BF D, and CDE. Prove that : 1o. These three circles are concurrent at a point O; 2o. If an arbitrary point P in the plane is joined to A; B; C, then the new points a; b; c where P A; P B; P C intersect these circles belong to a circle passing through O and P . Solution. (1◦) Suppose circles AEF and BF D intersect at point O (fig. t344a). We must show that O is on the circle through C; E, and D as well. We will do this by showing that quadrilateral CEOD is cyclic, using the criterion of 80. 4 Since quadrilateral AEOF is cyclic, we know (79) that EOF\ = 180◦ − CAB\. Since quadrilateral BF OD is cyclic, we have F\ OD = 180◦ − ABC\. By adding the angles about point O, we see that EOD\ = 360◦ − (EOF\ + F\ OD) = 360◦ − (180◦ − CAB\ + 180◦ − ABC\) = CAB\ + CBA\ = 180◦ − ACB\ (this last because the three angles involved are the angles of triangle ABC). Thus the opposite angles of CEOD are supplementary, and point O must lie on the circle through C; E, and D. Note. This is a relatively easy problem. If students need a hint, it is usually enough to suggest using cyclic quadrilaterals and the criteria of 79 and 80. Some- times students have a slight problem imagining that O is the intersection of only two of the circles, and will make some assumption in their argument equivalent to the assumption that O is on the third circle as well. This error is worth examining in detail in the classroom. In constructing a diagram with dynamic software, students may discover that the proof needs modification in certain cases. For some positions of D; E, and F , point O will end up outside the triangle, and the quadrilaterals referred to above won't lead to a solution. But other quadrilaterals will, and students can look at various such cases to convince themselves that the argument does not essentially change. Certain angles that are equal in some cases are supplementary in others, and vice versa. In more advanced work, greater unity is achieved if we consider angles as ori- ented, defining equality of angles slightly differently. See the note below. (2◦) The solution to this exercise is again straightforward, but the relationship of the elements in the figure offer many different cases for exploration. We give just one here. We will show that the quadrilateral with vertices P; a; O; b (in some order) is cyclic. The argument can be repeated for the quadrilateral with vertices P; b; c; O, which proves the theorem. In figure t344b, P[ aO is supplementary to AaO[, which in turn is supplementary to AF[ O (because these are opposite angles in cyclic quadrilateral AaOF ). Hence P[ aO = AF[ O. Similarly, using cyclic quadrilateral BbOF , we see that BbO[ is supplementary to both P[ bO and BF\ O, so these last two angles are equal. But clearly AF[ O and BF\ O are supplementary, so P[ aO and P[ bO are also supplementary, showing that P aOb is cyclic. To show that P cOb is cyclic, we note that P[ bO supplements BbO[, which is equal to BDO\ (they both intercept arc BF O). And BDO\ supplements CDO\, so P[ bO = CDO\. Finally, from cyclic quadrilateral CDOc, we see that CDO\ supplements CcO[ = P[ cO, so P[ bO also supplements P[ cO. This last statement shows that quadrilateral P cOb is cyclic, and the argument is completed as indicated above. Note. In other cases, we can make use of angles P[ aO; AaO;[ OF\ B, and so on. In each case, certain angles are equal, and others are supplementary. In advanced work, we consider oriented angles, and define `equality' so that `equal' oriented angles are either equal or supplementary, if orientation is not considered. 5 Figure t344b The solution we have presented here can motivate this step, which students and teachers can take using other references. See, for example, Roger A. Johnson, Advanced Euclidean Geometry, New York: Dover Books, 2007 (reprint of 1929 edition). See also exercise 214. Problem 345. With each side of a cyclic quadrilateral ABCD as a chord, we draw an arbitrary circular segment. The four new points A0;B0 ;C0;D0 where each of these four circles S1;S2;S3;S4 intersects the next are also the vertices of a cyclic quadrilateral. Solution. Since quadrilateral ABB0A0 (figure t345) is cyclic, we know that A\0AB+BB\0A0 = 180◦. Since quadrilateral BCC0B0 is cyclic, we know that BCC\0+ BB\0C0 = 180◦. Adding, we find that A\0AB + BB\0A0 + BCC\0 + BB\0C0 = 360◦ = BB\0A0 +A\0B0C0 +BB\0C0; or A\0AB +BCC\0 = A\0B0C0. Similarly, using other cyclic quadrilaterals, we can prove that DAA\0 + DCC\0 = A\0D0C0. Adding again, we find A\0AB + BCC\0 + DAA\0 + DCC\0 = A\0B0C0 + A\0D0C0. But the first sum is equal to DAB\ + DCB\ = 180◦, because these are the opposite angles of cyclic quadrilateral ABCD. Hence D\0A0B0 + D\0C0B0 = 180◦, so that A0B0C0D0 is cyclic. Note. The result is slightly more general than is indicated by the figure. The four points can be in any order on the circle, and the result will hold. That is, the given cyclic quadrilateral can have `sides' which intersect (we don't usually consider these true quadrilaterals), and the result will hold. 6 Figure t345 Students can be challenged to show that this is true.
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