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September-October 2008, Puzzle nr 119 Back to Puzzle-of-the-Month page | Home Puzzle # 119 Difficulty level:   , general math knowledge. Perpendicular or not?   The large circle (O1, B) above circumscribes 2 smaller circles (with centers O2 and O3, respectively) and an isosceles triangle, whose base stands on the diameter of the large circle. As shown in the drawing, each geometric shape touches the 3 other ones. Points A, O1, O2 and B are aligned. Demonstrate that the segment O3A is perpendicular to the diameter AB. Keywords: incircles, isosceles, perpendicular, secant-tangent. Related puzzles: - Ratio of similar triangles. - Soccer ball problem. Italiano - Ad angolo retto? Il grande cerchio (O1, B) circoscrive 2 cerchi minori (con i rispettivi centri O2 e O3) e un triangolo isoscele, la cui base si confonde col diametro del grande cerchio. Ogni figura geometrica tocca ognuna delle altre tre. I punti A, O1, O2 e B sono allineati. Dimostrare che il segmento O3A è perpendicolare al diametro AB. Parole chiave: cerchio inscritto, isoscelo, perpendicolare. Suggerisci un'altra soluzione Chiudi Français - A angle droit? Le grand cercle (O1, B) circonscrit 2 autres cercles plus petits (ayant pour centres O2 et O3, respectivement) ainsi qu’un triangle isocèle, dont la base se trouve sur le diamètre du grand cercle. Chaque figure géométrique touche les 3 autres. Les points A, O1, O2 et B sont alignés. Démontrer que le segment O3A est bien perpendiculaire au diamètre AB. Mots clés: cercle inscrit, isocèle, perpendiculaire. Propose une autre solution Fermer Source of the puzzle: Sangaku tablet of Gunma prefecture (群馬県; Gunma-ken), 1803. ©G. Sarcone. Some interesting considerations • In fig. 1, we can see that triangles A'BC' and ABC are similar. • In fig. 2, we can see that AC' and AE' are tangent to the circle with center O3. The segment O3A represents then the symmetry axis along which the right triangle AO3D is reflected on AE'. What we have to prove If we prove that angle γ (gamma, see fig. 2) of the right triangle AO3D is complementary to angle α (alpha) of the isosceles triangle AC'A', then we prove that points E and E' are identical, and consecutively the segment O3A is perpendicular to the diameter AB.  a) It is given: radius O1B = R (see fig. 3) radius O2B = r radius O3D = p segment O3A = t β = 90° - α b) According to the drawing: A’C’ = AC’ MA = MA' = R - r h ⊥ AA' p ⊥ AC’ O1O3 = R - p O1A = 2r - R O1M = O2B = r O2O3 = r + p c) According to the 'Secant-Tangent' theorem (see further below): t2 = p(p + 2r) d) According to the Pythagorean theorem: t2 + O1A2 = O1O32 → t2 + (2r - R)2 = (R - p)2 e) Solving the equations in c) and d) together, we obtain: t = 2r [2R(R - r)] / (R + r) and p = 2r(R - r) / (R + r) f) According to the Pythagorean theorem: 1) h = (O1C'2 - O1M2) = (O1B2 - O1M2) = (R2 - r2) and 2) AC’2 = h2 + MA2 = (R2 - r2) + (R - r)2 → AC’ = [2R(R + r)] g) By solving the following equations: 1) (p / t)2 = p2 / t2 = [2r(R - r)]2/ 4r2[2R(R - r)] = = [2r(R - r)]2 / 4rR[2r(R - r)] = (R - r) / 2R and 2) (MA / AC')2 = MA2 / AC'2 = (R - r)2 / 2R(R + r) = (R - r) / 2R We proved that p / t = MA / AC', then the triangles AO3D and AC'M are similar, hence angle DÂO3 = angle β = 90° - α. In conclusion, angle γ (or DÂO3) being complementary to angle α, the segment O3A is perpendicular to the diameter AB. (You will find here a more geometric way to solve this puzzle) The Winners of the Puzzle of the Month are: No winners... Yes, the puzzle was a little bit harder. Sorry! © 2005 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. 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 Secant-tangent theorem: fig. 1) c2 = a(a + b) and fig. 2) c2 = a(a + 2r) The angle, formed between a tangent line and a chord, is congruent to the inscribed angle on the other side of the chord and subtended by the chord (angle α' = angle α, see fig. 3). Then, with similar triangles: c / (a + b) = a / c

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