{"id":265614,"date":"2026-07-16T14:18:43","date_gmt":"2026-07-16T11:18:43","guid":{"rendered":"https:\/\/1kitap1.com\/en\/learning-geometry-through-problem-solving-boris-pritsker\/"},"modified":"2026-07-16T14:18:43","modified_gmt":"2026-07-16T11:18:43","slug":"learning-geometry-through-problem-solving-boris-pritsker","status":"publish","type":"post","link":"https:\/\/1kitap1.com\/en\/learning-geometry-through-problem-solving-boris-pritsker\/","title":{"rendered":"Learning Geometry Through Problem Solving &#8211; Boris Pritsker"},"content":{"rendered":"<figure style=\"text-align:center;margin:0 auto 1.5em;\"><img decoding=\"async\" src=\"https:\/\/1kitap1.com\/en\/wp-content\/uploads\/2026\/07\/8b21c6484c6562ee.jpg\" alt=\" - Unknown book cover\" style=\"max-width:300px;width:100%;height:auto;box-shadow:0 4px 12px rgba(0,0,0,.25);border-radius:4px;\"\/><\/figure>\n<p>Applying the Pythagorean Theorem, = + , we obtain the following quadratic equation to solve for x ( ) = + \u2212 ( ) . Squaring and simplifying we arrive at the equation 2 \u2212 \u2212 = . Solving this quadratic equation, we have x = \u00b1 = \u00b1 2 2 3 and rejecting the negative root 1 \u2212 (it does not satisfy the conditions of the problem, the length can\u2019t be a negative number), we are left with x = + , i.e., AO = + .<\/p>\n<p>Therefore, AC = = + 3, and respec- tively, MC = \u2212 = + \u2212 = 3. Section C Problem 4.31.* Proof. Direct statement. Let ABCD be a convex quadrilateral in which the sums of the squares of the opposite sides are equal to each other, + = + . The goal is to prove that AC \u22a5 . Assume that AC is not perpendicular to BD and draw BM \u22a5 and \u22a5 . As it is shown in figure above, in that assumption, AK < and CM < .<\/p>\n<p>168 Learning Geometry Through Problem Solving Applying the Pythagorean Theorem to right triangles AMB, CKD, CMB, and AKD gives = = , = + , = + . Adding the first two equalities and the last two equalities yields + = + + + and + = + + . Now, recalling that by conditions of the problem, AB + = + , we obtain AM + + + = + + + .<\/p>\n<p>This is simplified to + = + \b (*) As we assumed above, AK < and CM < K, leading to + < + , which contradicts equality (*). It implies that our assumption was wrong, and points K and M must coincide with O, the point of intersection of the diagonals. Respectively, AC \u22a5 , as it was desired to be proved. One of the nice alternative ways to solving many geometric prob- lems is applying Euclidean vector techniques. Proof of the direct statement utilizing vector algebra techniques was provided in B.<\/p>\n<blockquote>\n<p>New York City College of Technology &#8211; 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<p>Each of the books will be aimed at the general audience, which implies that the writing level will be such that it will not be 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 this most important subject rather than to fear it.<\/p>\n<p>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. 42 Learning Geometry Through Problem Solving by Boris Pritsker Vol. 39 Fun Math: Problem Solving Beyond the Classroom by Alfred S Posamentier Vol. 38 Mathematics: The Quest for Truth and Beauty by James D Stein Vol. 37 Mathematics in Architecture, Art, Nature, and Beyond by Alfred S Posamentier and G\u00fcnter J Maresch Vol. 36 Geometric Gems: An Appreciation for Geometric Curiosities Volume III: The Wonders of Circles by Alfred S Posamentier and Robert Geretschl\u00e4ger For the complete list of volumes in this series, please visit www.worldscientific.com\/series\/psmb Boris Pritsker Problem Solving in Mathematics and Beyond Volume 42 NEW JERSEY \u2022 LONDON \u2022 SINGAPORE \u2022 BEIJING \u2022 SHANGHAI \u2022 TAIPEI \u2022 CHENNAI World Scientific Learning Geometry Through Problem Solving 2 2 Published by World Scientific Publishing Co.<\/p>\n<p>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: 2025032291 British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library. Problem Solving in Mathematics and Beyond \u2014 Vol. 42 LEARNING GEOMETRY THROUGH PROBLEM SOLVING Copyright \u00a9 2026 by Boris Pritsker All rights reserved. ISBN 978-981-98-1670-5 (hardcover) ISBN 978-981-98-1764-1 (paperback) ISBN 978-981-98-1671-2 (ebook for institutions) ISBN 978-981-98-1672-9 (ebook for individuals) For any available supplementary material, please visit https:\/\/www.worldscientific.com\/worldscibooks\/10.1142\/14406#t=suppl Desk Editors: Soundararajan Raghuraman\/Rok Ting Tan Typeset by Stallion Press Email: enquiries@stallionpress.com Printed in Singapore In loving memory of Genadi Aleksandrovich Mikhalin, a prominent Soviet-Ukrainian mathematician, my first college calculus professor, and scientific advisor.<\/p>\n<p>With admiration and deepest respect. This page intentionally left blank vii Preface The knowledge of which geometry aims is the knowledge of the eternal.<\/p>\n<\/blockquote>\n<p><em>This is a short excerpt from the opening of &ldquo;&rdquo; by Unknown, quoted for review and introduction purposes. All rights belong to the copyright holders.<\/em><\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_85 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\" style=\"cursor:inherit\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 ' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/1kitap1.com\/en\/learning-geometry-through-problem-solving-boris-pritsker\/#Book_Information\" >Book Information<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/1kitap1.com\/en\/learning-geometry-through-problem-solving-boris-pritsker\/#Reading_Word_Statistics\" >Reading &amp; Word Statistics<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/1kitap1.com\/en\/learning-geometry-through-problem-solving-boris-pritsker\/#Most_Frequent_Words\" >Most Frequent Words<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/1kitap1.com\/en\/learning-geometry-through-problem-solving-boris-pritsker\/#PDF_Download\" >PDF Download<\/a><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"Book_Information\"><\/span>Book Information<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li><strong>Unique ID:<\/strong> 8b21c6484c6562ee<\/li>\n<li><strong>File Extension:<\/strong> .pdf<\/li>\n<li><strong>File Size:<\/strong> 43,892,212 bytes (41.859 MB)<\/li>\n<li><strong>Title:<\/strong> &#8211;<\/li>\n<li><strong>Author:<\/strong> Unknown<\/li>\n<li><strong>ISBN:<\/strong> 9789819816705, 9789819817641, 9789819816712, 9789819816729<\/li>\n<li><strong>Pages:<\/strong> 409<\/li>\n<li><strong>Language:<\/strong> English (en)<\/li>\n<\/ul>\n<h2><span class=\"ez-toc-section\" id=\"Reading_Word_Statistics\"><\/span>Reading &amp; Word Statistics<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li><strong>Estimated Reading Time:<\/strong> 470.69 minutes<\/li>\n<li><strong>Total Words:<\/strong> 94,137<\/li>\n<li><strong>Total Characters:<\/strong> 441,140<\/li>\n<li><strong>Average Words per Page:<\/strong> 230.16<\/li>\n<li><strong>Average Characters per Page:<\/strong> 1078.58<\/li>\n<\/ul>\n<h2><span class=\"ez-toc-section\" id=\"Most_Frequent_Words\"><\/span>Most Frequent Words<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>triangle (1053), problem (881), abc (490), angle (441), sin (390), circle (379), point (376), triangles (359), right (354), angles (344), sides (301), area (297), given (281), draw (275), equal (272), two (268), solving (266), length (244), find (237), abcd (234), geometry (231), prove (231), therefore (231), learning (200), cos (196), theorem (183), inscribed (180), points (178), solutions (177), applying (162), isosceles (151), parallelogram (148), respectively (147), answers (144), side (144), quadrilateral (142), now (141), trapezoid (137), also (132), center (131), method (129), since (129), see (128), intersection (126), areas (124), acb (124), one (122), sum (120), get (119), respective (119), line (116), similar (111), altitude (111), congruent (105), note (105), figure (105), desired (104), implies (101), let\u2019s (101), thus (100), square (98), follows (97), diagonals (97), parallel (97), three (96), lengths (96), radius (95), opposite (93), straight (92), bisector (91), denote (87), equilateral (86), construction (85), diameter (85), perpendicular (83), intersecting (82), pythagorean (81), consider (80), abd (79), equality (75), half (75), rectangle (74), common (72), drawn (72), observe (72), aob (72), arrive (71), vertices (71), well (70), following (69), solution (69), substituting (69), gives (69), first (68), property (68), problems (67), circles (67), dropped (67), vertex (66), segment (65).<\/p>\n<h2><span class=\"ez-toc-section\" id=\"PDF_Download\"><\/span>PDF Download<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p style=\"text-align:center;\"><a href=\"https:\/\/1kitap1.com\/en\/wp-content\/uploads\/2026\/07\/learning-geometry-through-problem-solving-boris-pritsker.pdf\" download rel=\"nofollow\" style=\"display:inline-block;background:#2271b1;color:#ffffff;padding:14px 36px;border-radius:6px;text-decoration:none;font-weight:bold;font-size:1.05em;\">&#11015;&#65039; PDF Download<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Applying the Pythagorean Theorem, = + , we obtain the following quadratic equation to solve for x ( ) = + \u2212 ( ) . Squaring and simplifying we arrive at the equation 2 \u2212 \u2212 = . Solving this quadratic equation, we have x = \u00b1 = \u00b1 2 2 3 and rejecting the [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":265612,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[8],"tags":[],"class_list":["post-265614","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-english"],"blocksy_meta":[],"_links":{"self":[{"href":"https:\/\/1kitap1.com\/en\/wp-json\/wp\/v2\/posts\/265614","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/1kitap1.com\/en\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/1kitap1.com\/en\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/1kitap1.com\/en\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/1kitap1.com\/en\/wp-json\/wp\/v2\/comments?post=265614"}],"version-history":[{"count":0,"href":"https:\/\/1kitap1.com\/en\/wp-json\/wp\/v2\/posts\/265614\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/1kitap1.com\/en\/wp-json\/wp\/v2\/media\/265612"}],"wp:attachment":[{"href":"https:\/\/1kitap1.com\/en\/wp-json\/wp\/v2\/media?parent=265614"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/1kitap1.com\/en\/wp-json\/wp\/v2\/categories?post=265614"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/1kitap1.com\/en\/wp-json\/wp\/v2\/tags?post=265614"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}