For any arbitrary triangle ABC, let R denote its

circumradius and r its inradius (Figure 1). It was the

Swiss-German mathematician Leonhard Euler who

first observed that regardless of the shape of the triangle, the

following inequality is invariably true

Shirali, Shailesh

English

ePrints@APU

58 Azim Premji University At Right Angles, March 2020ClassRoomSHAILESH SHIRALIEuler’s Inequality for the Circumradius and Inradius of a TriangleFor any arbitrary triangle ABC, let R denote itscircumradius and r its inradius (Figure 1). It was theSwiss-German mathematician Leonhard Euler whofirst observed that regardless of the shape of the triangle, thefollowing inequality is invariably true:R ≥ 2r,equality precisely when the triangle is equilateral. AB CIOr RFigure 1.1Keywords: Euler, inequality, circumcircle, circumradius, incircle, inradius59Azim Premji University At Right Angles, March 2020The statement being so simple, one naturally longs for an equally simple proof of the result.Unfortunately, this is not readily forthcoming. (There are elegant and short proofs, but not simple proofs!)In this article, we present two very different proofs.Notation. We use standard symbols for the various elements of the triangle: A,B,C for the three angles;a, b, c for the three sides (named appropriately); s for the semi-perimeter; R for the circumradius; r for theinradius; and Δ for the area of the triangle.First proof. We first present a proof which has a strong component of algebra. The following formulas areall very well-known:2R =asin A=bsin B=csin C,Δ =12bc sin A =12ca sin B =12ab sin C,Δ = rs,Δ =√s(s − a)(s − b)(s − c).Combining the results in the first two lines, we obtain the additional resultΔ =abc4R.In addition, we shall need the most basic result in the theory of inequalities, namely, the arithmeticmean-geometric mean inequality. This is the statement that if x, y are any two non-negative real numbers,thenx + y2≥ √xy,with equality precisely when x = y. This gives rise to the following nice result. Let x, y, z be any threenon-negative real numbers. Then we have:x + y ≥ 2√xy,y + z ≥ 2√yz,z + x ≥ 2√zx.Hence by multiplication of the respective sides we obtain:(x + y)(y + z)(z + x) ≥ 8xyz. (1)Moreover, equality will hold in (1) precisely when x = y = z. Note that this is an interesting result in itsown right.Next we obtain the lengths of some segments associated with the incircle of a triangle (see Figure 2). LetD,E, F denote the points of contact of the incircle with the sides of the triangle, and let x, y, z denote thelengths of the segments as indicated.It is easy to obtain x, y, z in terms of a, b, c. We have:y + z = a,z + x = b,x + y = c.60 Azim Premji University At Right Angles, March 2020 AB CIDEFxyy zzxx = AE = AFy = BD = BFz = CD = CEFigure 2.By addition we get 2(x + y + z) = a + b + c = 2s, hence x + y + z = s. Therefore:x = s − a, y = s − b, z = s − c. (2)We now use these expressions for x, y, z in the inequality (1) (note that x, y, z are all positive; this followsfrom the triangle inequality, b + c > a, which yields s − a > 0; similarly for s − b and s − c). We obtain:abc ≥ 8(s − a)(s − b)(s − c). (3)This too is an interesting result in its own right; it holds for any triangle. Moreover, equality holdsprecisely when a = b = c, i.e., when the triangle is equilateral.We are now in a position to obtain Euler’s inequality. From the formulas stated earlier for the area of atriangle, we have:Δ =abc4R,∴ abc = 4RΔ,and:Δ =√s(s − a)(s − b)(s − c),∴ (s − a)(s − b)(s − c) =Δ2s.Therefore we have:4RΔ ≥ 8Δ2s,∴ R ≥ 2Δs.Since Δ = rs, the last line yields the desired result:R ≥ 2r, (4)with equality precisely when the triangle is equilateral.61Azim Premji University At Right Angles, March 2020Second proof. In contrast to the above, we now present a proof which is highly geometric. This is Euler’soriginal proof (1765). He obtains the inequality as an easy consequence of an important geometric result.Theorem (Euler). The distance d between the circumcentre and the incentre of a triangle is related to itscircumradius R and its inradius r by the following relation:d 2 = R(R − 2r).Remark. Before proceeding with the proof of the theorem, we note that the result instantly provides aproof of Euler’s inequality; for, we must have d 2 ≥ 0, and this yields R ≥ 2r.Proof of Euler’s theorem. In Figure 3, let AI extended meet the circumcircle at L; let LO extended meetthe circumcircle at M; let segment IO extended in both directions meet the circumcircle at P and Q; andfinally, let F be the foot of the perpendicular from I to AB.Consider △AFI and △MBL. They are similar to each other, for AFI and MBL are right angles, andFAI = BML (“angles in the same segment”). Hence:FIBL=AIML, i.e.,rBL=AI2R, (5)which yields 2Rr = AI · BL. Next, we claim that BL = IL, i.e., that △LBI is isosceles. This follows from asimple computation of angles. For we have,LBI = LBC + IBC = LAC + IBC =A2+B2,andLIB = LAB + ABI =A2+B2.It follows that BL = IL and so:2Rr = AI · IL. (6) AB CIOFPQLMFigure 3.62 Azim Premji University At Right Angles, March 2020Finally, from the intersecting chords theorem it follows thatAI · IL = PI · IQ.Since PI = OP − OI = R − d and IQ = QO + OI = R + d, we obtain:2Rr = (R − d)(R + d) = R2 − d 2,i.e.,d 2 = R2 − 2Rr = R(R − 2r). (7)Euler’s inequality now follows.For more proofs, the three references listed below may be consulted.References1. Wikipedia, “Euler’s theorem in geometry” (from Wikipedia, the free encyclopedia),https://en.wikipedia.org/wiki/Euler’s_theorem_in_geometry2. Wikipedia, “Incircle and excircles of a triangle” (from Wikipedia, the free encyclopedia),https://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle3. Samer Seraj, “A Short Proof of Euler’s Inequality”, Resonance – Journal of Science Education, Volume 20, Issue 1, January 2015 pp75-75, https://www.ias.ac.in/article/fulltext/reso/020/01/0075-0075SHAILESH SHIRALI is the Director of Sahyadri School (KFI), Pune, and heads the Community Mathematics Centre based in Rishi Valley School (AP) and Sahyadri School KFI. He has been closely involved with the Math Olympiad movement in India. He is the author of many mathematics books for high school students, and serves as Chief Editor for At Right Angles. He may be contacted at shailesh.shirali@gmail.com.1. Why was the fraction apprehensive about marrying the decimal?  Because he would have to convert…2. Why do plants hate math?  Because it gives them square roots…3. Why did the student get upset when his teacher called him average?  Well, it was a pretty mean thing to say!4. Why was the math book depressed?  Poor thing, it had too many problems.5. Why is the obtuse triangle always so frustrated?  Because it is never right.6. Why can you never trust a math teacher holding graphing paper?  He must be plotting something.Math Jokes and PunsContributed by Harin Hattangady, Azim Premji University, M.A. Education, batch of 2011-1363Azim Premji University At Right Angles, March 2020Introducing RobocompassA nifty tool for Geometrical ConstructionHow Geometrical Constructions are taught in SchoolsEuclid’s Elements – one of the most influential mathematical textbooks ever to have been written – is primarily a compendium of geometrical constructions created using straightedge and compass. But are we sure that these two geometrical tools which lie at the heart of such foundational ideas are being used effectively in the classrooms? The current practice is actually to use large wooden geometrical instruments in the classrooms, as the size of the real physical compass (in the ‘geometry box’) is not large enough to use conveniently as a demonstration tool.From a student’s perspective, pencil smudges, torn papers, hurt fingers due to sharp edges, changing measurements as they move the instruments are some of the inconveniences they need to deal with when working with a physical geometry box. Spatial reasoning is one of the skills all students should inculcate, but are the challenges of using these physical instruments holding students back?TechSpaceKeywords: geometry, construction, space, tools, softwareM SRINIVASAN

Euler’s inequality for the circumradius and inradius of a triangle

http://publications.azimpremjifoundation.org/2318/1/15_Eulers_inequality_for_the_circumradius_and_inradius_of_a_triangle.pdf

Euler’s inequality for the circumradius and inradius of a triangle

Abstract

Similar works

Full text

Available Versions

ePrints@APU