{"id":259782,"date":"2026-07-13T17:13:27","date_gmt":"2026-07-13T14:13:27","guid":{"rendered":"https:\/\/1kitap1.com\/en\/engaging-young-student-math-vol-1-geretschlager-robert\/"},"modified":"2026-07-13T17:13:27","modified_gmt":"2026-07-13T14:13:27","slug":"engaging-young-student-math-vol-1-geretschlager-robert","status":"publish","type":"post","link":"https:\/\/1kitap1.com\/en\/engaging-young-student-math-vol-1-geretschlager-robert\/","title":{"rendered":"Engaging Young Student Math Vol 1 – Geretschlager Robert"},"content":{"rendered":"
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Adding the relationships 9p2 = 2(a2 +c2)\u2212a2, 9q2 = 2(a2 +c2)\u2212b2, and 9r2 = 2(a2 +b2)\u2212c2 gives us 3(p2 + q2 + r2) = a2 + b2 + c2, and substituting in the \ufb01rst relationship then yields 9p2 = 2(3p2 +3q2 +3r2 \u2212a2)\u2212a2, and therefore 3a2 = 6(q2 + r2) \u22123p2. From this, we obtain a2 = 2(q2 + r2) \u2212p2 = 2(202 + 152) \u2212172 = 961 or a = \u221a 961 = 31.<\/p>\n

Problem 2.10. In a semicircle with diameter AB = 2R, the two chords AP and BQ are each tangent to two small circles of equal size as shown in the \ufb01gure. If AP = BQ = 10, determine the length of AB = 2R. Answer. 2R \u224811.18. This problem was proposed by Isomura Chikayuki. Solution by H. Fukagawa. Let H denote the midpoint of AB and \u03b8 as shown in the \ufb01gure, we have cos \u03b8 = BQ BT with AB = 2R, BQ = 10, and BT = \u0005 4 = \u221a 2 R.<\/p>\n

Since BH = R holds, we therefore obtain = \u221a = 10 \u221a = 5 \u221a 5 \u224811.18. Problem 2.11. At a party, n guests and m attendants participate and drink many cups of sake. Each guest drinks sake from a 350-ml mug and each attendant drinks from a 250-ml mug. Altogether, they empty one entire 3.5 liter barrel of sake. The total amount of sake consumed by the guests is equal to the total amount consumed by the attendants. Determine the number of guests and attendants at the party.<\/p>\n

Answer. n = 5 and m = 7. This problem was proposed by Isomura Chikayuki. Solution by H. Fukagawa. We are given 350 n = 250 m, or 7 n = 5 m, and 350 n + 250 m = 3500. We therefore obtain m = 7, n = 5. Sangaku: Traditional Japanese Mathematics Problem 2.12. We are given a trapezoid ABC D with BC||AD, and AD\u22a5DC. Three roads AE, FG, and HC are constructed such that the triangles \u0004DAE, \u0004E FG and \u0004GHC are similar. Furthermore, we are given BC = b = 10 m, AD = a = 80 m and C D = h = 105 m. Deter- mine the ratio k of the lengths of the legs of the similar triangles.<\/p>\n

Answer. The ratio is equal to k = 3 4. This problem was proposed by Ishizuka Masakatsu. Solution by H. Fukagawa.<\/p>\n

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New York City College of Technology – City University of New York There are countless applications that would be considered problem solving in mathematics and beyond. One could even argue that most of mathematics in one way or another involves solving problems. However, this series is intended to be of interest to the general audience with the sole purpose of demonstrating the power and beauty of mathematics through clever problem-solving experiences.<\/p>\n

Each of the books will be aimed at the general audience, which implies that the writing level will be such that it will not engulfed in technical language \u2014 rather the language will be simple everyday language so that the focus can remain on the content and not be distracted by unnecessarily sophiscated language. Again, the primary purpose of this series is to approach the topic of mathematics problem- solving in a most appealing and attractive way in order to win more of the general public to appreciate his most important subject rather than to fear it.<\/p>\n

At the same time we expect that professionals in the scientific community will also find these books attractive, as they will provide many entertaining surprises for the unsuspecting reader. Published Vol. 13 Engaging Young Students in Mathematics through Competitions \u2014 World Perspectives and Practices: Volume I \u2014 Competition-ready Mathematics; Entertaining and Informative Problems from the WFNMC8 Congress in Semriach\/Austria 2018 edited by Robert Geretschl\u00e4ger Vol. 12 The Psychology of Problem Solving: The Background to Successful Mathematics Thinking by Alfred S. Posamentier, Gary Kose, Danielle Sauro Virgadamo and Kathleen Keefe-Cooperman Vol. 11 Solving Problems in Our Spatial World by Guenter Maresch and Alfred S.<\/p>\n

Posamentier For the complete list of volumes in this series, please visit www.worldscientific.com\/series\/psmb Published by World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224 USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE Library of Congress Control Number: 2019030293 British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library.<\/p>\n

Problem Solving in Mathematics and Beyond \u2014 Vol. 13 ENGAGING YOUNG STUDENTS IN MATHEMATICS THROUGH COMPETITIONS \u2014 WORLD PERSPECTIVES AND PRACTICES Volume I \u2014 Competition-ready Mathematics; Entertaining and Informative Problems from the WFNMC8 Congress in Semriach\/Austria 2018 Copyright \u00a9 2020 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the publisher.<\/p>\n

For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission to photocopy is not required from the publisher.<\/p>\n<\/blockquote>\n

This is a short excerpt from the opening of “” by Unknown, quoted for review and introduction purposes. All rights belong to the copyright holders.<\/em><\/p>\n

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