Shortcuts Sitemap Contact Newsletter Store Books Features Gallery E-cards Games  ••• Related links Puzzles workshops for schools & museums. Editorial content and syndication puzzles for the media, editors & publishers. Abracada-Brain Teasers and Puzzles! Numbers, just numbers... Have a Math question? Ask Dr. Math! ••• ••• Logic smile! Life is full of misery, loneliness, and suffering - and it's all over much too soon!      - Woody Allen One martini is all right, two is two many, three is not enough...      - James Thurber You can't trust water: Even a straight stick turns crooked in it!      - W.C. Fields ••• # Previous Puzzles of the Month + Solutions

May-June 2009, Puzzle nr 121 Back to Puzzle-of-the-Month page | Home Puzzle # 121 Difficulty level:   , general math knowledge. Circle vs Square    The diameter a of the large semicircle below is 10 cm long. Knowing that one of the vertices of the square meets the circumference of the small circle at point P (see diagram above), try to guess the area of the blue square WITHOUT using trigonometry! Keywords: sangaku, tangent property, bisector theorem. Related puzzles: - Perpendicular or not? - Ratio of similar triangles. Italiano - Area incognita Il diametro a del semicerchio qui sopra vale 10 cm; sapendo che un vertice del quadrato (punto P) appartiene anche alla circonferenza piccola, qual è l'area del quadrato? È vietato utilizzare la trigonometria per risolvere questo problema! Parole chiave: sangaku. Suggerisci un'altra soluzione Chiudi Français - La surface X Le diamètre a du demi-cercle ci-dessus vaut 10 cm; sachant qu'un des sommets du carré (point P) appartient aussi à la circonférence du petit cercle, peux-tu nous dire quelle est l'aire du carré bleu? Il est interdit d'utiliser la trigonométrie pour résoudre ce problème! Mots clés: sangaku. Propose une autre solution Fermer Source of the puzzle: BrainTrainer, issue #23. © G. Sarcone. You cannot reproduce any part of this page without prior written permission. There are 2 possible solutions to this 'sangaku' problem: SOLUTION 1 A) Build a large square with base QR (see illustration opposite) and prolong the diagonal d to the vertex S, as illustrated. According to the drawing: q = a/2 - R d = QP = a/2 + x x = d - a/2 QR = RS = a QR ⊥ ON QP ⊥ OP ∠NQO = ∠OQP B) According to the ‘angle bisector theorem’: if a ray or segment bisects an angle of a triangle, then it divides the two segments on either side proportionally. In other terms, given in the triangle QRS the bisectrix QM, then: MS / MR = QS / QR that is (a-m)/m = a 2/a We obtain, by resolving the equation: a/(1 + 2) = m m = 10/2.414 = 4.14 cm C) The right triangles QNO and QRM are similar, then: (a/2 + x) / R = d / R = a / m = 10 / 4.14 d = R(2.42) D) Thank to the ‘tangent property’, we find that: a/2 + x = d Substituting x with d - a/2 in the following Pythagorean equation: (a/2 - R)2 = x2 + R2 = (d - a/2)2 + R2 We can also substitute d with R(2.42) (a/2 - R)2 = (d - a/2)2 + R2 = [R(2.42) - a/2]2 + R2 and we will obtain: (5 - R)2 = [R(2.42) - 5]2 + R2 25 - 10R + R2 = (5.86R2 - 24.2R + 25) + R2 R = 14.2/5.86 = 2.423 E) The area of the blue square is therefore: d2/2 = [R(2.42)]2/2 = [2.423 x 2.42]2/2 = 17.2 cm2 SOLUTION 2 F) According to the drawing: q = a/2 - R ; d = QP = a/2 + x ; x = d - a/2 QP, the diagonal d of the blue square, is tangent to point P. OP is perpendicular to QP, then: the angle QPS = the angle OPS = 45 degrees. G) The triangle OMP is isosceles, then the height PS of the blue square equals: ON + PM = R + R/ 2 = R(1 + 2)/ 2 = R(1.71) and the diagonal QP equals: R(1 + 2) = R(2.42) H) Thank to the ‘tangent property’ (as argument D further above), we find that: a/2 + x = d (the diagonal QP) Substituting x with d - a/2 in the following Pythagorean equation: (a/2 - R)2 = x2 + R2 = (d - a/2)2 + R2 We can also substitute d with R(2.42) (a/2 - R)2 = (d - a/2)2 + R2 = [R(2.42) - a/2]2 + R2 and we will obtain: (5 - R)2 = [R(2.42) - 5]2 + R2 25 - 10R + R2 = (5.86R2 - 24.2R + 25) + R2 R = 14.2/5.86 = 2.423 I) The area of the blue square is therefore: PS2 = R(1.71)2 = (2.423 x 1.71)2 = 17.2 cm2 The Winners of the Puzzle of the Month are: Nheo Le, USA Mauro Zoffoli, Italy Hasan Kurt, the Netherlands Amaresh G. S., India Congratulations! © 2006 G. Sarcone, www.archimedes-lab.org You can re-use content from Archimedes’ Lab on the ONLY condition that you provide credit to the authors (© G. Sarcone and/or M.-J. Waeber) and a link back to our site. You CANNOT reproduce the content of this page for commercial purposes. Reactions from our visitors --- Posted: Fri, 28 May 2010 18:12:51 by Philippe Chevanne Dear Gianni, I just discovered your web site from a friend. Great site, and congratulations! Lurking a few problems there, I got at problem #76 (may-June 2009) I think the solution given is false, as it is not stated that the diagonal QP should be tangent to the small yellow circle. This unstated condition is necessary to solve the puzzle, and is used in the given solutions. Discarding this condition, the area of the square can take ANY value. For instance see attached picture. To construct the figure from *any* point P, that is here any point P on a 45° diagonal line, is an "Appolonius problem" in the case Point(P) - Line(diameter) - Circle(large circle). That is to construct a circle (yellow) touching the three objects considered as three "circles": P = circle with null radius, Line = circle with infinite radius, and given (half)circle. This can be solved in several classical ways. (for instance at my own web site http://mathafou.free.fr/pbg_en/jsp136e.html for english version) The result is that for any point P on the 45° line, there is a corresponding square of area = 2QP2 = any value, and a yellow circle through P and tangent to diameter and large half circle. (In fact two such circles, the other being "on the left" to P) The condition "QP tangent to yellow circle" is equivallent to "yellow circle minimal", restricting the yellow circle to be centered in the right quarter circle, otherwise the minimum is 0, with P=Q. So that the missing condition "when yellow circle is minimal" should be added to the statement. Best Regards. You're encouraged to expand and/or improve this article. Send your comments, feedback or suggestions to Gianni A. Sarcone. Thanks!

 More Math Facts behind the puzzle Angle bisector and proportions If a segment bisects an angle of a triangle then it divides the two segments on either side proportionally: CA / CP = BA / PB and CA x PB = CP x BA Angle bisector theorem on Wikipedia.

 Previous puzzles of the month...   Back to Puzzle-of-the-Month page | Home   Send a comment Recommend this page Share it on FaceBook Rate it on StumbleUpon
Archimedes' Laboratory™ | How to contact us | Come contattarci | Comment nous contacter    