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Also some readers wrote in after the initial publication of the problem, pointing out that the original result could not be true, and suggesting possible corrections. Thank you. The following note describes a few uses of a relatively less known result in plane geometry, the Generalized Ptolemy Theorem GPT, for short , also known as Casey's Theorem see Johnson. Featured will be two proofs of the problem proposed by India for the 33rd IMO in Moscow [ ; 86].
Circles 1 and 2 are externally tangent at a point I , and both are enclosed by and tangent to a third circle. Then I is the incentre of triangle ABC. Figure 1. Proof 1. Notation: Let tij refer to the length of the external common tangent to circles i and j thus the two circles lie on the same side of the tangent.
We use the GPT, which we state in the following manner. For a proof of the GPT and its converse, please refer to . Figure 2. This proves the result. Proof 2. Let M be the midpoint of S2, and consider all possible circles. Figure 3. Proof of Lemma. Proof of Theorem 1. See gure 4.
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Using the above lemma,. Therefore M has equal powers with respect to 1 and 2 and lies on their radical axis, namely AI. It follows that I is the incentre of 4ABC. For non-believers, here are two more illustrations of the power and economy of the GPT. See Figure 5. But this is just the radius of the incircle of 4ABC. The next illustration concerns one of the most celebrated discoveries in elementary geometry made during the last two centuries. Feuerbach's Theorem The incircle and nine-point circle of a triangle are tangent to one another.
Let a, b, c be the sides of 4ABC , and let s be its semi-perimeter. We now consider the 4-tuple of circles D; E; F;.
37: No 8 December / Décembre 2011
Here is what we nd:. But this is immediate! Since the circle passing through D, E , F is the nine-point circle of the triangle, it follows that and the nine-point circle are tangent to one another. Then the centres of 1 and 2 are collinear with the incentre of 4ABC. I have, however, not been able to nd such a proof, and I leave the problem to the interested reader. We note in passing that Thebault's theorem provides yet another proof of Theorem 1. References:  R. Johnson, Advanced Euclidean Geometry, Dover, Acknowledgements: I thank the referee for making several valuable comments that helped tidy up the presentation of the paper.
My thanks go to Professor T. Lewis, the University of Alberta for forwarding me a copy. Students have 90 minutes to complete their contest. It is mostly written by students in Grade XII, but has often been won by students in earlier grades. A circle and a parabola are drawn on a piece of paper. The number of regions they divide the paper into is at most A. By what percentage has your height increased during the last two years? The largest even integer you use is A.
A rectangle contains three circles as in the diagram, all tangent to the rectangle and to each other. If the height of the rectangle is 4, then the width of the rectangle 4 p2 is p p 4 A. Mary Lou works a full day and gets her usual pay. Her total pay that day is equivalent to 12 hours at her usual hourly salary. The number of hours that she usually works each day is A. A fair coin is tossed 10; times. The probability p of obtaining at least three heads in a row satis es A. In the plane, the angles of a regular polygon with n sides add up to less than n2 degrees.
The smallest possible value of n satis es: A. The value of P 6 is A. Which of the following conditions does not guarantee that the convex quadrilateral ABCD is a parallelogram? Intermediate Mathematical Challenge, written February 2, Here are answers. A That completes this month's Skoliad Corner. I need materials of a suitable level to build up a bank of contests.
Please send me suitable materials as well as comments, criticisms, and suggestions. I would like to have some feedback too about how your students do with these materials. Because we are now publishing eight numbers of the Corner rather than ten, I am giving two Olympiad Sets for your pleasure. Besides, here in Canada it is winter, and quite cold, so having a stock of problems to contemplate in a warm spot is a good idea.
Many thanks to him for gathering a wide sample of contests. We begin with the Telecom Australian Mathematical Olympiad. For each positive integer n, let. Note: If x is a real number, then [x] is the largest integer not exceeding x. The vertices of triangle ABC in the x,y plane have integer coordinates, and its sides do not contain any other points having integer coordinates.
The interior of ABC contains only one point, G, that has integer coordinates. Prove that G is the centroid of ABC. Prove that A or B is not periodic. Let d n be the largest odd number which divides a given number n. In a contest, x students took part, and y problems were posed. For every problem, the number of students who solved it was the same. For each pair of students, just three problems were solved by both of them. Determine all possible pairs x; y. Five radii of a sphere are given so that no three of them are in a common plane.
Among the 32 possible choices of an end point from each segment, nd out the number of choices for which the 5 points are in a hemisphere. Last issue we gave a set of six Klamkin Quickies. Many thanks go to Murray Klamkin, the University of Alberta, for sending them to me. That they are equalpis an identity of Ramanujan. For other related radical identities of Ramanujan, see Susan Landau, How to tangle with a nested radical, Math. Intelligencer, 16 , pp. Express 1,1! Prove that if the line joining the incentre to the centroid of a triangle is parallel to one of the sides of the triangle, then the sides are in arithmetic progression and, conversely, if the sides of a triangle are in arithmetic progression then the line joining the incentre to the centroid is parallel to one of the sides of the triangle.
Let A, B, C denote vectors to the respective vertices A, B , C of the triangle from a point outside the plane of the triangle. To nd the remaining solution s , we multiply the given equation by the least common denominator to give. Next we turn to the readers' solutions of problems from earlier numbers of the Corner. Smeenk, Zaltbommel, the Netherlands and Chris Wildhagen, Rotterdam, the Netherlands, for sending in nice solutions to some of the problems of the Canadian and U.
Mathematical Olympiads. Next, correspondence from Murray Klamkin, the University of Alberta,. ABCD is a quadrilateral inscribed in a circle of radius r. Give a reason for your answer. Klamkin's Comment. A simpler solution than that published, plus a generalization, is given in Crux, , p. This problem appeared as problem in The Math. Magazine, April Now we turn to readers' solutions of problems proposed to the Jury, but not used, at the 34th International Mathematical Olympiad at Istanbul.
Solution by D. Smeenk, Zaltbommel, The Netherlands. Similarly, B. Inequalities, 2. Show that I is the midpoint of DE. Then F lies on the production of AI. Let G be the foot of the perpendicular from I 0 to AB. So I 0 coincides with I. Klamkin, The University of Alberta. A natural number n is said to have the property P if, whenver n divides an , 1 for some integer a, n2 also necessarily divides an , 1.
Solution by E. Suppose 2p j a2p , 1. See e. Rosen, 3rd Edition. For suppose that 4 j a4 , 1. Smeenk, Zaltbommel, the Netherlands. See Figure 1. Assume OQ? See Figure 2. We want to show that OQ? This direction could be shortened by using the law of sines in triangles BQF and CQF , but the given solution is, I think, more elementary, and therefore more elegant. That completes the material we have available for this number. The Olympiad Season is fast approaching. Please collect your contests and send them to me. Also send me your nice solutions to problems posed in the Corner. The second puzzle, attributed there to Mike Dawes of the University of Western Ontario, asked: Suppose that you take a physical model of this logo, and manipulate it to turn the in nity sign into a circle.
What does the circle turn into? It turns out that the same question occurred to several other people at the same time, including some members of the Netherlands team at IMO Less than half an hour prior to the start of the competition, the IMO logo inspired one of the contestants RvL to pose the following problem: Consider the logo as a knot composed of a blue string forming the in nity-sign and a red string forming the zero.
Can the knot be rearranged so that the blue string forms the zero and the red string forms the in nity-sign without using scissors? The contestant solved the problem shortly after the competition and, upon returning home, wrote a program together with one of the other members of his team DG. Ronald van Luijk rmluijk cs. See, for example, L. Tucker, Preparing for a New Calculus, edited by Anita Solow, Reviewed by Jack Macki, University of Alberta. Anyone who attended the January joint meetings in San Francisco could not help but be struck by the contrast between the evident malaise on the research side of the meeting and the enthusiasm, energy and high quality of the discourse on the educational side.
They describe the past and present of reform in undergraduate calculus, and evaluate the situation in a thorough and realistic manner. The authors describe the debate between Anthony Ralston and Ron Douglas which led to the famous Tulane workshop in with 25 participants, four of whom were research mathematicians.
They follow the entire development of the reform movement, carefully describing the key players, and the roles of NRC and NSF. They present all kinds of surprising at least to this reviewer information along the way did you know that in the IEEE, deeply dissatis ed with existing texts, published a calculus book? In the nine years since that seminal Tulane conference, reform has clearly moved into a central position in the mathematical life of North America.
Very determined department chairs at Stony Brook and Michigan convinced their departments to adopt reform texts. The cake is calculus. The main part of this book has only 44 pages of text, and then concludes with 50 pages of important and highly readable appendices: data on enrolments, copies of surveys, text-by-text description of reform materials presently available, and a very detailed list of NSF-funded projects.
If you have colleagues who pooh-pooh the reform movement and who doesn't? Preparing for a New Calculus is the proceedings of a conference held at the University of Illinois in April, Eighty individuals attended, 40 of them associated with projects at colleges or universities, 25 of them associated with projects at the high school level, seven from community college projects, and eight from organizations like the MAA, NSF, NCTM, etc. Part I contains seven background papers, which were provided to participants before the conference.
Teaching reform courses is very hard work.
At the U. Calculus starts in term two! And don't expect all students to love the changes students who are successful in standard courses often react negatively to being saddled with lab partners and being evaluated in unfamiliar ways. In fact, several authors emphasize that student questionnaires cannot be the sole means of teaching evaluation one needs to talk to students who have gone down the road a way in order to get a fair picture.
The second part of Preparing for a New Calculus reports on the workshops, one each on the topics of content, teaching strategies, and institutional context. Each report has a long and useful list of suggestions and observations. The third section consists of ve contributed papers of very high quality on topics ranging from calculator courses, to the gateway exam at Michigan. The nal piece is a thoughtful article by Peter Renz of Academic Press on Publishers, Innovation and Technology, which should be required reading for all university mathematics professors.
Many of the articles in this book have extensive lists of very timely and useful references.
These may be typewritten or neatly hand-written, and should be mailed to the Editor-in-Chief, to arrive no later that 1 October Graphics les should be in epic format, or plain postscript. ABCD is a square with incircle ,. Find all polynomials f such that f p is a prime for every prime p. A triangle has sides a; b; c and area F. Let P be an interior point of the side AC.
The primitive Pythagorean triangle with sides and and hypotenuse has area , which is an 8-digit number of the form abcdabcd. Find another primitive Pythagorean triangle whose area is of this form. Little Sam is a unique child and his math marks show it. What's more, the average of these scores is the same as the average if each score is reversed so 94 becomes 49, for example , and this average is an integer none of whose digits is equal to any of the digits in the scores. What is Sam's average? It is known e. Rosen's Elementary Number Theory and its Applications, Third Edition that every natural number greater than 6 is the sum of two relatively prime integers, each greater than 1.
Find all natural numbers which can be expressed as the sum of three pairwise relatively prime integers, each greater than 1. Proposed by Catherine Shevlin, Wallsend, England. Prove that AD? Johnson, Advanced Euclidean Geometry, p. Let p be the probability that the sum of the chosen numbers is divisible by 5. Paul, Minnesota, USA. Wilf, When are subset sums equidistributed modulo m? That condition is:. While this does not directly answer problem since it gives no information about the speci c remainder 0 mod 5 , the paper does discuss many intriguingopen questions related to the mod m distribution of subset sums.
Thus it strikes me that, because some of your readers were successful at generalizing the problem as stated, they would be interested in this reference. The problem arose from lottery considerations.
When are the tickets in a lottery equidistributed with respect to the mod m value of their sums? Find angle A. Solution by the proposer. There was one incorrect solution. Recall that stands for the repeating decimal : : : , for example, and that the period of a repeating decimal is the number of digits in the repeating part.
Solution by Christopher J. In order to be a multiple of it must have 3k nines for some integer k. When this number is divided by , the quotient has k ones interspersed with pairs of zeros: : : : In order that this quotient be divisible by 99 as well, k must be divisible by 9 and must be even using the well known rules for divisibility by 9 and That the number has the full period of 54 results from the fact that 19 and are both coprime to 99 and ensuring no fortuitous cancellations. Cross notes that the product in b equals. This formula, which can be more simply written as x; z [a; b] where x; z is the greatest common divisor of x and z , gives precisely 54 in our case.
Unfortunately, the reference given for Oliver's result is page of Math. Monthly, Vol. Monthly, which only started publishing in Can any reader supply us with more information? A reference will be given when the solution is published. Case 1. Each element of fn , 1; n , 2; : : : ; 3; 2g may be in either the rst or second sequence, but not both. Choosing which elements are in these sequences determines the positions of all of fn , 1; n , 2; : : : ; 2; 1g uniquely.
Thus there are n2n,2 permutations with minimal sum 2n , 2. Thus the term xi contributes 2xi to the sum. Suppose that j elements have no contribution to the sum. In each pair, one number is greater and one is less than the other. Thus there are n,2 j numbers xi which contribute 2xi to the sum, and n,2 j numbers xi which contribute ,2xi to the sum. If we can arrange the numbers xi which contribute 2xi to the sum to be the largest n,j numbers, and the numbers xi which contribute ,2xi to the sum to be 2 the smallest n,2 j numbers, then a maximum value is clearly attained with j as small as possible.
This is indeed possible. So there are 2 k! We can do this in two directions, clockwise or counterclockwise. The vertex of the equilateral triangle on C is rotated to the vertex on the original diameter. We see that there are two such equilateral triangles four if the extension of the diameter is allowed. To construct the pair of rotated circles, nd their centres as the third vertices of the equilateral triangles that have PO as base, where O is the centre C ; their radius is half the length of the given diameter.
One incorrect solution was received. Most of the solvers used an argument similar to the featured solution. Hence exactly one of x and y is negative. Ten incorrect or incomplete solutions were also received. Is this a record? This was shown by Flanigan and Mane. Find the locus of P so that. Solution by P. Penning, Delft, the Netherlands. Thus the locus consists of the three medians of the triangle excluding the end points. So we need to nd the locus of P such that. Now, this is exactly when P lies on a median of triangle ABC excluding the end points.
Find the smallest integer of the form.
B denotes the six-digitinteger formed by placing A and B side by side. Solution by M. Parmenter, Memorial University of Newfoundland, St. I claim that the answer to the problem is G Observe that. Some solvers made use of computers to search for all possible solutions. One incorrect solution was submitted.
Janous suggested an extension to:. Multiply the second, the third, the fourth and the fth fractions on the left by a , ab , abc and abcd respectively. Then the desired inequality is equivalent to: 5! Then the. Klamkin, University of Alberta, Edmonton, Alberta. Generalize this result to an n-dimensional simplex. Let O be the centre of a circumhypersphere, let R be the radius, and let A1 , A2 , : : : ; An be the vertices. Then n X. Quadrilateral ABCD is inscribed in a circle ,, and has an incircle as well.
EF is a diameter of , with EF? Note: It is necessary that A and E lie on the same side of the line BD, otherwise the result is false. Thus C F. These solvers assumed implicitly that A and E lay on the same side of the line BD. Let g t be the sum of the digits in the binary representation of t.
Kuczma notes that the problem can be solved without any computation as follows: Look at Pascal's triangle, rows 0 through 2 , modulo 2, and visualize it geometrically as an equilateral triangle with black and white spots zeros and ones. It has three geometric symmetry axes, and each one of these symmetries preserves the spot design.
This fact can be considered as known. The base row is all ones. These observations provide a scheme for an induction proof of the claim for rows 0 through 2 , 1. The symmetry with respect to one of the oblique axes is precisely the contents of the problem, expressed in algebraic terms. Comment by the Editor-in-Chief Choosing a solution to print is not an easy task. Each collection of submitted solutions is assigned to a member of the Editorial Board, and these editors work independently of one another. The choice of which solution to highlight is left entirely to that editor.
Sometimes it happens, as in this issue, that several solutions are chosen from the same solver. I would like to assure all subscribers that every submission is very important to CRUX, and that we encourage everyone to submit solutions, as well as proposals for problems, articles for publication, and contributions to the other corners.
Every subscriber is very important to us and we really value all contributions. And while I am on the subject of contributions, please continue to send in proposals for problems. We publish per year, and we do not have too many in reserve at this time. Without your contributions, there would be no CRUX. All three authors wrote that contest. Afterwards, they felt that the problem would have been more meaningful had they been asked to cut non-isosceles triangles into isosceles ones. We say that a triangle is n-dissectible if it can be dissected into n isosceles triangles where n is a positive integer.
The isosceles triangles themselves are the only ones that are 1-dissectible, and of course 1-critical. However, there are two exceptions. The solution to the contest problem, which motivated this study, shows that almost all 1-dissectible triangles are not 2-dissectible.
We will point out later that some 2-dissectible ones are not 3-dissectible. On the other hand, it is easy to see that all 1-dissectible triangles are 3-dissectible. Figure 1 illustrates the three cases where the vertical angle is acute, right, and obtuse respectively. Clearly, such a triangle can only be cut into two triangles by drawing a line from a vertex to the opposite side, as illustrated in Figure 2. There are three ways in which CAD may become an isosceles triangle.
Case 2. Case 3. This class consists of all right triangles. We should point out that while our three classes of 2-dissectible triangles are exhaustive, they are not mutually exclusive. We now consider 3-dissectible triangles. Suppose one of the cuts does not pass through any vertices. Then it divides the triangle into a triangle and a quadrilateral. The latter must then be cut into two triangles, and this cut must pass through a vertex.
Hence at least one cut passes through a vertex. The only possible con guration is the one illustrated in Figure 1 a. It follows that the three arms have equal length and this point is the circumcentre of the original triangle. Since it is an interior point, the triangle is acute. Thus all acute triangles are 3-dissectible. In all other cases, one of the cuts go from a vertex to the opposite side, dividing the triangle into an isosceles one and a 2-dissectible one. There are quite a number of cases, but the argument is essentially an elaboration of that used to determine all 2-dissectible triangles.
We leave the details to the reader, and will just summarize our ndings in the following statement. A triangle is 3-dissectible if and only if it satis es at least one of the following conditions: 1. It is an isosceles triangle. It is an acute triangle. It is a right triangle. Each can then be cut into two isosceles triangles by cutting along the median from the right angle to the hypotenuse, as illustrated in Figure 3. Consider any triangle. Divide it by a line through a vertex into an isosceles triangle and another one. By the induction hypothesis, the second can be dissected into n triangles.
Announcement For the information of readers, we are saddened to note the deaths of two mathematicians who are well-known to readers of CRUX. Leroy F. Meyers died last November. He was a long time contributor to CRUX. He is well-known for his work on Geometric Inequalities, in particular, being co-author of Geometric Inequalities and Recent Advances in Geometric Inequalities.
This issue we feature the Eleventh W. Blundon Contest, written February 23, Find the two integers. A geometric sequence is a sequence of numbers in which each term, after the rst, can be obtained from the previous term, by multiplying by the same xed constant, called the common ratio. A square is inscribed in an equilateral triangle. Find the ratio of the area of the square to the area of the triangle.
ABCD is a square. Three parallel lines l1 , l2 and l3 pass through A, B and C respectively. The distance between l1 and l2 is 5 and the distance between l2 and l3 is 7. Find the area of ABCD. The sum of the lengths of the three sides of a right triangle is The sum of the squares of the lengths of the three sides is Find the area of the triangle. A palindrome is a word or number that reads the same backwards and forwards. For example, is a palindromic number.
How many palindromic numbers are there between 1 and 99; inclusive? Find the area of the parallelogram. Two numbers are such that the sum of their cubes is 5 and the sum of their squares is 3. Find the sum of the two numbers. This month we give the solutions. How well did you do? D The parabola divides the plane into two regions. The circle intersects the parabola in at most four points, so that it is divided by the parabola into at most four arcs. Each arc carves an existing region into two.
C Suppose the height was two years ago. Then it was. C All factors of are distinct and odd, with the largest one being Hence the last even number used is If the height of the rectangle is 4, then the width of the rectangle is 4 p p p 4 2 2 2 A. A Let O be the centre of the large circle, P that of one of the small circles, and Q the point of tangency of the small circles. E Suppose Mary Lou usually works x hours per day but y on that day.
D Partition the tosses into consecutive groups of three, discarding the last one. If we never get 3 heads in a row, none of the groups can consist of 3 heads. The probability of this is 78 , which is clearly less than C Since 0; 0 , 1; 2 and 2; 4 are collinear, a circle passes through at most two of them. Since 2; 0 , 3; 1 , 3; 3 and 4; 3 are not concyclic, a circle passes through at most three of them.
The circle with centre 1; 2 and passing through 0; 0 also passes through 2; 0 , 3; 1 , 3; 3 and 2; 4. Note that one factor is odd and the other even, and that the rst is smaller than the second. Then ABCD is not a parallelogram. It is easy to see that the other three conditions do guarantee parallelograms. That completes the Skoliad Corner for this issue. Send me contest materials, as well as your comments, suggestions, and desires for future directions for the Skoliad Corner.
A professor at the University of Waterloo and founding member of the Canadian Mathematics Competition, he has dedicated his career to encouraging excellence in students.