Figures Circumscribing Circles Tom M
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USA Mathematical Talent Search Solutions to Problem 2/3/16
USA Mathematical Talent Search Solutions to Problem 2/3/16 www.usamts.org 2/3/16. Find three isosceles triangles, no two of which are congruent, with integer sides, such that each triangle’s area is numerically equal to 6 times its perimeter. Credit This is a slight modification of a problem provided by Suresh T. Thakar of India. The original problem asked for five isosceles triangles with integer sides such that the area is numerically 12 times the perimeter. Comments Many students simply set up an equation using Heron’s formula and then turned to a calculator or a computer for a solution. Below are presented more elegant solutions. Zachary Abel shows how to reduce this problem to finding Pythagorean triples which have 12 among the side lengths. Adam Hesterberg gives us a solution using Heron’s Create PDF with GO2PDFformula. for free, if Finally,you wish to remove Kristin this line, click Cordwell here to buy Virtual shows PDF Printer how to take an intelligent trial-and-error approach to construct the solutions. Solutions edited by Richard Rusczyk. Solution 1 by: Zachary Abel (11/TX) In 4ABC with AB = AC, we use the common notations r = inradius, s = semiperimeter, p = perimeter, K = area, a = BC, and b = AC. The diagram shows triangle ABC with its incircle centered at I and tangent to BC and AC at M and N respectively. A x h N 12 I a/2 12 B M a/2 C The area of the triangle is given by rs = K = 6p = 12s, which implies r = 12. -
Relationships Between Six Excircles
Sangaku Journal of Mathematics (SJM) c SJM ISSN 2534-9562 Volume 3 (2019), pp.73-90 Received 20 August 2019. Published on-line 30 September 2019 web: http://www.sangaku-journal.eu/ c The Author(s) This article is published with open access1. Relationships Between Six Excircles Stanley Rabinowitz 545 Elm St Unit 1, Milford, New Hampshire 03055, USA e-mail: [email protected] web: http://www.StanleyRabinowitz.com/ Abstract. If P is a point inside 4ABC, then the cevians through P divide 4ABC into smaller triangles of various sizes. We give theorems about the rela- tionship between the radii of certain excircles of some of these triangles. Keywords. Euclidean geometry, triangle geometry, excircles, exradii, cevians. Mathematics Subject Classification (2010). 51M04. 1. Introduction Let P be any point inside a triangle ABC. The cevians through P divide 4ABC into six smaller triangles. In a previous paper [5], we found relationships between the radii of the circles inscribed in these triangles. For example, if P is at the orthocenter H, as shown in Figure 1, then we found that r1r3r5 = r2r4r6, where the ri are radii of the incircles as shown in the figure. arXiv:1910.00418v1 [math.HO] 28 Sep 2019 Figure 1. r1r3r5 = r2r4r6 In this paper, we will find similar results using excircles instead of incircles. When the cevians through a point P interior to a triangle ABC are drawn, many smaller 1This article is distributed under the terms of the Creative Commons Attribution License which permits any use, distribution, and reproduction in any medium, provided the original author(s) and the source are credited. -
Point of Concurrency the Three Perpendicular Bisectors of a Triangle Intersect at a Single Point
3.1 Special Segments and Centers of Classifications of Triangles: Triangles By Side: 1. Equilateral: A triangle with three I CAN... congruent sides. Define and recognize perpendicular bisectors, angle bisectors, medians, 2. Isosceles: A triangle with at least and altitudes. two congruent sides. 3. Scalene: A triangle with three sides Define and recognize points of having different lengths. (no sides are concurrency. congruent) Jul 249:36 AM Jul 249:36 AM Classifications of Triangles: Special Segments and Centers in Triangles By angle A Perpendicular Bisector is a segment or line 1. Acute: A triangle with three acute that passes through the midpoint of a side and angles. is perpendicular to that side. 2. Obtuse: A triangle with one obtuse angle. 3. Right: A triangle with one right angle 4. Equiangular: A triangle with three congruent angles Jul 249:36 AM Jul 249:36 AM Point of Concurrency The three perpendicular bisectors of a triangle intersect at a single point. Two lines intersect at a point. The point of When three or more lines intersect at the concurrency of the same point, it is called a "Point of perpendicular Concurrency." bisectors is called the circumcenter. Jul 249:36 AM Jul 249:36 AM 1 Circumcenter Properties An angle bisector is a segment that divides 1. The circumcenter is an angle into two congruent angles. the center of the circumscribed circle. BD is an angle bisector. 2. The circumcenter is equidistant to each of the triangles vertices. m∠ABD= m∠DBC Jul 249:36 AM Jul 249:36 AM The three angle bisectors of a triangle Incenter properties intersect at a single point. -
Areas of Polygons and Circles
Chapter 8 Areas of Polygons and Circles Copyright © Cengage Learning. All rights reserved. Perimeter and Area of 8.2 Polygons Copyright © Cengage Learning. All rights reserved. Perimeter and Area of Polygons Definition The perimeter of a polygon is the sum of the lengths of all sides of the polygon. Table 8.1 summarizes perimeter formulas for types of triangles. 3 Perimeter and Area of Polygons Table 8.2 summarizes formulas for the perimeters of selected types of quadrilaterals. However, it is more important to understand the concept of perimeter than to memorize formulas. 4 Example 1 Find the perimeter of ABC shown in Figure 8.17 if: a) AB = 5 in., AC = 6 in., and BC = 7 in. b) AD = 8 cm, BC = 6 cm, and Solution: a) PABC = AB + AC + BC Figure 8.17 = 5 + 6 + 7 = 18 in. 5 Example 1 – Solution cont’d b) With , ABC is isosceles. Then is the bisector of If BC = 6, it follows that DC = 3. Using the Pythagorean Theorem, we have 2 2 2 (AD) + (DC) = (AC) 2 2 2 8 + 3 = (AC) 6 Example 1 – Solution cont’d 64 + 9 = (AC)2 AC = Now Note: Because x + x = 2x, we have 7 HERON’S FORMULA 8 Heron’s Formula If the lengths of the sides of a triangle are known, the formula generally used to calculate the area is Heron’s Formula. One of the numbers found in this formula is the semiperimeter of a triangle, which is defined as one-half the perimeter. For the triangle that has sides of lengths a, b, and c, the semiperimeter is s = (a + b + c). -
Stml043-Endmatter.Pdf
http://dx.doi.org/10.1090/stml/043 STUDENT MATHEMATICAL LIBRARY Volume 43 Elementary Geometry Ilka Agricola Thomas Friedric h Translated by Philip G . Spain #AM^S^fcj S AMERICAN MATHEMATICA L SOCIET Y Providence, Rhode Islan d Editorial Boar d Gerald B . Follan d Bra d G . Osgoo d Robin Forma n (Chair ) Michae l Starbir d 2000 Mathematics Subject Classification. Primar y 51M04 , 51M09 , 51M15 . Originally publishe d i n Germa n b y Friedr . Viewe g & Sohn Verlag , 65189 Wiesbaden, Germany , a s "Ilk a Agricol a un d Thoma s Friedrich : Elementargeometrie. 1 . Auflag e (1s t edition)" . © Friedr . Viewe g & Sohn Verlag/GW V Fachverlag e GmbH , Wiesbaden , 2005 Translated b y Phili p G . Spai n For additiona l informatio n an d update s o n thi s book , visi t www.ams.org/bookpages/stml-43 Library o f Congress Cataloging-in-Publicatio n Dat a Agricola, Ilka , 1973- [Elementargeometrie. English ] Elementary geometr y / Ilk a Agricola , Thoma s Friedrich . p. cm . — (Student mathematica l librar y ; v. 43) Includes bibliographica l reference s an d index. ISBN-13 : 978-0-8218-4347- 5 (alk . paper ) ISBN-10 : 0-8218-4347- 8 (alk . paper ) 1. Geometry. I . Friedrich, Thomas , 1949 - II . Title. QA453.A37 200 7 516—dc22 200706084 4 Copying an d reprinting. Individua l reader s o f this publication , an d nonprofi t libraries actin g fo r them, ar e permitted t o mak e fai r us e of the material, suc h a s to copy a chapter fo r us e in teaching o r research. -
Searching for the Center
Searching For The Center Brief Overview: This is a three-lesson unit that discovers and applies points of concurrency of a triangle. The lessons are labs used to introduce the topics of incenter, circumcenter, centroid, circumscribed circles, and inscribed circles. The lesson is intended to provide practice and verification that the incenter must be constructed in order to find a point equidistant from the sides of any triangle, a circumcenter must be constructed in order to find a point equidistant from the vertices of a triangle, and a centroid must be constructed in order to distribute mass evenly. The labs provide a way to link this knowledge so that the students will be able to recall this information a month from now, 3 months from now, and so on. An application is included in each of the three labs in order to demonstrate why, in a real life situation, a person would want to create an incenter, a circumcenter and a centroid. NCTM Content Standard/National Science Education Standard: • Analyze characteristics and properties of two- and three-dimensional geometric shapes and develop mathematical arguments about geometric relationships. • Use visualization, spatial reasoning, and geometric modeling to solve problems. Grade/Level: These lessons were created as a linking/remembering device, especially for a co-taught classroom, but can be adapted or used for a regular ed, or even honors level in 9th through 12th Grade. With more modification, these lessons might be appropriate for middle school use as well. Duration/Length: Lesson #1 45 minutes Lesson #2 30 minutes Lesson #3 30 minutes Student Outcomes: Students will: • Define and differentiate between perpendicular bisector, angle bisector, segment, triangle, circle, radius, point, inscribed circle, circumscribed circle, incenter, circumcenter, and centroid. -
G.CO.C.10: Centroid, Orthocenter, Incenter and Circumcenter
Regents Exam Questions Name: ________________________ G.CO.C.10: Centroid, Orthocenter, Incenter and Circumcenter www.jmap.org G.CO.C.10: Centroid, Orthocenter, Incenter and Circumcenter 1 Which geometric principle is used in the 3 In the diagram below, point B is the incenter of construction shown below? FEC, and EBR, CBD, and FB are drawn. If m∠FEC = 84 and m∠ECF = 28, determine and state m∠BRC . 1) The intersection of the angle bisectors of a 4 In the diagram below of isosceles triangle ABC, triangle is the center of the inscribed circle. AB ≅ CB and angle bisectors AD, BF, and CE are 2) The intersection of the angle bisectors of a drawn and intersect at X. triangle is the center of the circumscribed circle. 3) The intersection of the perpendicular bisectors of the sides of a triangle is the center of the inscribed circle. 4) The intersection of the perpendicular bisectors of the sides of a triangle is the center of the circumscribed circle. 2 In the diagram below of ABC, CD is the bisector of ∠BCA, AE is the bisector of ∠CAB, and BG is drawn. If m∠BAC = 50°, find m∠AXC . Which statement must be true? 1) DG = EG 2) AG = BG 3) ∠AEB ≅ ∠AEC 4) ∠DBG ≅ ∠EBG 1 Regents Exam Questions Name: ________________________ G.CO.C.10: Centroid, Orthocenter, Incenter and Circumcenter www.jmap.org 5 The diagram below shows the construction of the 9 Triangle ABC is graphed on the set of axes below. center of the circle circumscribed about ABC. What are the coordinates of the point of intersection of the medians of ABC? 1) (−1,2) 2) (−3,2) 3) (0,2) This construction represents how to find the intersection of 4) (1,2) 1) the angle bisectors of ABC 10 The vertices of the triangle in the diagram below 2) the medians to the sides of ABC are A(7,9), B(3,3), and C(11,3). -
Half-Anticevian Triangle of the Incenter
International Journal of Computer Discovered Mathematics (IJCDM) ISSN 2367-7775 c IJCDM Volume 2, 2017, pp.55-71. Received 20 February 2017. Published on-line 28 February 2017 web: http://www.journal-1.eu/ c The Author(s) This article is published with open access1. Computer Discovered Mathematics: Half-Anticevian Triangle of the Incenter Sava Grozdeva, Hiroshi Okumurab and Deko Dekovc 2 a VUZF University of Finance, Business and Entrepreneurship, Gusla Street 1, 1618 Sofia, Bulgaria e-mail: [email protected] b Department of Mathematics, Yamato University, Osaka, Japan e-mail: [email protected] cZahari Knjazheski 81, 6000 Stara Zagora, Bulgaria e-mail: [email protected] web: http://www.ddekov.eu/ Abstract. By using the computer program ”Discoverer” we study the Half- Anticevian triangle of the Incenter. Keywords. Half-Anticevian triangle, triangle geometry, notable point, computer discovered mathematics, Euclidean geometry, “Discoverer”. Mathematics Subject Classification (2010). 51-04, 68T01, 68T99. 1. Introduction Let P be an arbitrary point in the plane of triangle ABC. Denote by P aP bP c the Anticevian triangle of point P . Let Ma be the midpoint of segment AP a and define Mb and Mc similarly. We call triangle MaMbMc the Half-Anticevian Triangle of Point P . Figure 1 illustrates the definition. In figure 1, P is an arbitrary point, P aP bP c is the Anticevian triangle, and MaMbMc is the Half-Anticevian Triangle of P . Note that the Half-Anticevian Triangle of the Centroid is the Medial Triangle. We encourage the students, teachers and researchers to investigate the Half- Anticevian triangles of other notable points, e.g. -
Cyclic Quadrilaterals
GI_PAGES19-42 3/13/03 7:02 PM Page 1 Cyclic Quadrilaterals Definition: Cyclic quadrilateral—a quadrilateral inscribed in a circle (Figure 1). Construct and Investigate: 1. Construct a circle on the Voyage™ 200 with Cabri screen, and label its center O. Using the Polygon tool, construct quadrilateral ABCD where A, B, C, and D are on circle O. By the definition given Figure 1 above, ABCD is a cyclic quadrilateral (Figure 1). Cyclic quadrilaterals have many interesting and surprising properties. Use the Voyage 200 with Cabri tools to investigate the properties of cyclic quadrilateral ABCD. See whether you can discover several relationships that appear to be true regardless of the size of the circle or the location of A, B, C, and D on the circle. 2. Measure the lengths of the sides and diagonals of quadrilateral ABCD. See whether you can discover a relationship that is always true of these six measurements for all cyclic quadrilaterals. This relationship has been known for 1800 years and is called Ptolemy’s Theorem after Alexandrian mathematician Claudius Ptolemaeus (A.D. 85 to 165). 3. Determine which quadrilaterals from the quadrilateral hierarchy can be cyclic quadrilaterals (Figure 2). 4. Over 1300 years ago, the Hindu mathematician Brahmagupta discovered that the area of a cyclic Figure 2 quadrilateral can be determined by the formula: A = (s – a)(s – b)(s – c)(s – d) where a, b, c, and d are the lengths of the sides of the a + b + c + d quadrilateral and s is the semiperimeter given by s = 2 . Using cyclic quadrilaterals, verify these relationships. -
Barycentric Coordinates in Olympiad Geometry
Barycentric Coordinates in Olympiad Geometry Max Schindler∗ Evan Cheny July 13, 2012 I suppose it is tempting, if the only tool you have is a hammer, to treat everything as if it were a nail. Abstract In this paper we present a powerful computational approach to large class of olympiad geometry problems{ barycentric coordinates. We then extend this method using some of the techniques from vector computations to greatly extend the scope of this technique. Special thanks to Amir Hossein and the other olympiad moderators for helping to get this article featured: I certainly did not have such ambitious goals in mind when I first wrote this! ∗Mewto55555, Missouri. I can be contacted at igoroogenfl[email protected]. yv Enhance, SFBA. I can be reached at [email protected]. 1 Contents Title Page 1 1 Preliminaries 4 1.1 Advantages of barycentric coordinates . .4 1.2 Notations and Conventions . .5 1.3 How to Use this Article . .5 2 The Basics 6 2.1 The Coordinates . .6 2.2 Lines . .6 2.2.1 The Equation of a Line . .6 2.2.2 Ceva and Menelaus . .7 2.3 Special points in barycentric coordinates . .7 3 Standard Strategies 9 3.1 EFFT: Perpendicular Lines . .9 3.2 Distance Formula . 11 3.3 Circles . 11 3.3.1 Equation of the Circle . 11 4 Trickier Tactics 12 4.1 Areas and Lines . 12 4.2 Non-normalized Coordinates . 13 4.3 O, H, and Strong EFFT . 13 4.4 Conway's Formula . 14 4.5 A Few Final Lemmas . 15 5 Example Problems 16 5.1 USAMO 2001/2 . -
Heron's Formula for Triangular Area Heron of Alexandria
Heron’s Formula for Triangular Area by Christy Williams, Crystal Holcomb, and Kayla Gifford Heron of Alexandria n Physicist, mathematician, and engineer n Taught at the museum in Alexandria n Interests were more practical (mechanics, engineering, measurement) than theoretical n He is placed somewhere around 75 A.D. (±150) 1 Heron’s Works n Automata n Catoptrica n Mechanica n Belopoecia n Dioptra n Geometrica n Metrica n Stereometrica n Pneumatica n Mensurae n Cheirobalistra The Aeolipile Heron’s Aeolipile was the first recorded steam engine. It was taken as being a toy but could have possibly caused an industrial revolution 2000 years before the original. 2 Metrica n Mathematicians knew of its existence for years but no traces of it existed n In 1894 mathematical historian Paul Tannery found a fragment of it in a 13th century Parisian manuscript n In 1896 R. Schöne found the complete manuscript in Constantinople. n Proposition I.8 of Metrica gives the proof of his formula for the area of a triangle How is Heron’s formula helpful? How would you find the area of the given triangle using the most common area formula? 1 A = 2 bh 25 17 Since no height is given, it becomes quite difficult… 26 3 Heron’s Formula Heron’s formula allows us to find the area of a triangle when only the lengths of the three sides are given. His formula states: K = s(s - a)(s - b)(s - c) Where a, b, and c, are the lengths of the sides and s is the semiperimeter of the triangle. -
Conic Construction of a Triangle from Its Incenter, Nine-Point Center, and a Vertex
Conic Construction of a Triangle From Its Incenter, Nine-point Center, and a Vertex Paul Yiu Abstract. We construct a triangle given its incenter, nine-point center and a vertex by locating the circumcenter as an intersection of two rectangular hyper- bolas. Some special configurations leading to solutions constructible with ruler and compass are studied. The related problem of construction of a triangle given its circumcenter, incenter, and one vertex is revisited, and it is established that such a triangle exists if and only if the incenter lies inside the cardioid relative to the circumcircle. 1. Introduction In this note we solve the construction problem of a triangle ABC given its in- center I, its nine-point center N, and one vertex A. This is Problem 35 in Harold Connelly’s list of construction problems [1]. We shall show that the triangle is in general not constructible with ruler and compass, but can be easily obtained by in- tersecting conics. Some special configurations of I, N, A from which the triangle can be constructible with ruler and compass are studied in details in 2, 6 – 8. §§ 2. The case of collinear I, N, A In this case, the triangle ABC is isosceles, with AB = AC. The incircle and the nine-point circle are tangent to each other internally at the midpoint of the base BC. We set up a rectangular coordinate system with N at the origin, I at ( d, 0), and A at (a, 0) for a > 0, and seek the inradius r. The circumradius is − A r R = 2(d + r).