Bulging Triangles: Generalization of Reuleaux Triangles

Norihiro Someyama∗

∗Shin-yo-ji Buddhist Temple, Tokyo, Japan [email protected] ORCID iD: https://orcid.org/0000-0001-7579-5352

Abstract We introduce a bulging triangle like the generalization of the Reuleaux triangle. We may be able to propose various ways to bulge a triangle, but this paper presents the way so that its vertices are the same as them of the original triangle. We ﬁnd some properties and theorems of our bulging triangles. In particular, we investigate, via calculus, whether basic facts such as triangle inequalities and Pythagorean theorem hold for bulging triangles.

Keywords: Reuleaux triangle, Covexity, Triangle inequality, Pythagorean theorem. 2010 Mathematics Subject Classiﬁcation: 51M04, 51M05, 51N20.

1 Introduction • the “triangle inequality” holds, i.e., the sum of the lengths of any two “round sides” is There is the Reuleaux triangle (also called the greater than the length of the remaining one spherical triangle) as one of famous and notable “round side”, figures in geometry. It is named after F. Reuleaux (1829-1905), a German mechanical engineer. He • the three “round sides” are equal and the considered a figure in which an equilateral triangle common length is πa/3 ; 1.047a if the length is bulged in a natural way so that it has a constant of one side of 4ABC is a, √ √ width. To be more precise, it is defined as follows: 2 2 • its area is (π − 3)a /2 ; 1.628 × ( 3a /4) Let 4ABC be an equilateral triangle consisting (i.e., about 1.628 times the original area) if of vertices A, B and C. It namely holds that the length of one side of 4ABC is a.

|AB| = |BC| = |CA| Moreover, Reuleaux triangles are widely applied in various fields and there is no end to the list. where AB denotes the side whose end points are A See, e.g., [2] for the benefits and applications of and B, |AB| denotes its length, and the other edges Reuleaux triangles. It is also well known that the are similar. We have a bulging equilateral triangle Reuleaux triangle is one of solutions for the Kakeya arXiv:2108.07725v1 [math.GM] 11 Aug 2021 like Figure 1 by drawing arcs with center C and problem (see, e.g., [1] for more information). radius BC, with center A and radius CA and with Our aim in this paper is to introduce the “gen- center B and radius AB. eralized” Reuleaux triangle with different lengths The most notable property of the Reuleaux tri- on the three “round sides” and study the proper- angle is that it can rotate completely while being ties. within the square and touching all four sides of the square. This property has been applied to archi- 2 Main Topics tectural and engineering technologies. Any Reuleaux triangle has some trivial mathe- 2.1 Definition of Bulging Triangles matical properties, e.g., Let 4ABC be a triangle. For convenience, think • it is (strictly) convex, of this as an acute triangle. We would like to make

1 a figure that 4ABC is bulged by the following 2.2 Basic Properties of Bulging Trian- way. Such a figure is called the bulging triangles gle (stemmed) from 4ABC in this paper. Proposition 2.1. The AB-center P of the bulging Consider the perpendicular bisector ` of side ( AB, and find out the intersection point P of an- triangle 4ABC is on the longer of the remaining other side and `. Then, remark that P lies on the two sides BC, CA. In other words, P is the in- longer (BC in the case of Figure 2) of the remain- ternally dividing point of either BC or CA. The ing two sides BC and CA. This fact is almost clear, BC-center and CA-center are similar. but we will prove it later (Proposition 2.1). Deter- Proof. Assume ∠ABC < ∠BAC and denote by P mine intersection points Q and R in the same way the point such that |AP| = |BP|. Then, for sides BC and CA. Here each P, Q, R is, of course, uniquely determined. Note that ∠BAP = ∠ABC < ∠BAC and so P is on BC. Since P is the AB-center of |AP| = |BP|, ( 4ABC, this completes the proof. |BQ| = |CQ|, The bulging triangle that dents at a vertex is |CR| = |AR|. not ideal. We need to check the following fact.

We obtain the “round side” of 4ABC by draw- ( Proposition 2.2. Let 4ABC be any bulging tri- ing the circular arc whose center and radius are P angle from 4ABC. and |AP| (= |BP|), respectively. The other “round ( sides” are similarly obtained. From the above, we 1) If 4ABC is acute, then 4ABC is convex;

have been able to deﬁne the “bulging triangle” ( (Figure 3) that is naturally extended the Reuleaux 2) If 4ABC is right, then 4ABC is convex; triangle to the general case. We can treat vertices ( 3) If 4ABC is obtuse, then 4ABC is concave. of the original triangle as vertices of the bulging triangle. Proof. Without loss of generality, we can visualize ( In what follows, we set the following promises: 4ABC from 4ABC with ∠BCA = ∠R. Note that

( the BC-center and the CA-center match and the • We write 4 ABC for the bulging triangle common point is the middle point of AB. In other whose vertices are A, B and C. words, BC+f CAg (i.e., the arc formed by connecting ( ( • Its “round sides” are called edges of 4ABC BCf and CA)g coincides with semi-circular arc BA and are denoted by AB,g BCf and CA.g Their containing point C. By virtue of the above, the lengths are denoted by |ABg|, |BCf | and |CAg|, proof is easily obtained. respectively. In response to the result of Proposition 2.2, we • Three points P, Q and R defined above are exclusively consider bulging triangles from acute or called centers associated with edges of right-angled triangles in this paper. ( 4ABC. We often abbreviate such P as the For convenience, we put the following term. AB-center and the others are similar. Definition 2.1. Let 4ABC be a right-angled triangle with BCA = R. The bulging triangle from • The radius (i.e., AP, BP, and so on) of the ∠ ∠ ( sector consisting of A, B and P is called the this 4ABC is denoted by 4RBCA. The others are AB-radius. The others are similar. similar. Such a bulging triangle is called the right- angled bulging triangle in this paper. In partic- ( • The “length of the radius” is simply abbrevi- ular, we call 4RBCA the isosceles right-angled ated as “the radius” throughout this paper. bulging triangle if |BC| = |CA|. Remark 2.1. If 4ABC is an equilateral triangle, We study the edges of a bulging triangle from the AB-center is C, the BC-center is A and the the famous right triangle. In particular, the result CA-center is B. Hence, the bulging triangle defined for an isosceles right-angled bulging triangle can be above is certainly an extension of the Reuleaux tri- shown to be exactly the same as the conventional angle. one.

2 ( Proposition 2.3. For 4RBCA with |BC| = a, it Proof. Denote by P and Q the AB-center and CA- ( follows that center of 4 ABC, respectively (see Figure 5). It is obviously that QP k BC. Considering the per- 1) one has pendicular bisector L, passing through point A, of √ BC, it follows that Q is the point of symmetry of |ABg| : |BCf | : |CAg| = 2 : 1 : 1 P with respect to L. B is also, of course, the point if ∠ABC = π/4, of symmetry of C with respect to L. Hence, the claim clearly holds. This completes the proof. 2) one has √ √ Proposition 2.5. Suppose that 4ABC obeys |AB| : |BC| : |CA| = 4 : 3 : 2 3 g f g π 0 < ∠ABC < ∠BCA ≤ . (2.4) if ∠ABC = π/3. 2 ( Proof. We leave the proof of 1) to the reader and Then, 4ABC from such 4ABC satisﬁes that prove only 2). |CA| < |AB|. (2.5) Since |BC| = a and ∠ABC = π/3, we have g g √ π |AB| = 2a, |CA| = 3a, BAC = . Proof. Put ∠ 6 a := |BC|, b := |CA|, c := |AB| (2.6) Denoting by P the√ AB-center, P lies on CA, the AB-radius is 2a/ 3 and ∠APB = 2π/3. Hence we and have √ α := BAC, β := ABC, γ := BCA. (2.7) 4 3 ∠ ∠ ∠ |ABg| = πa. (2.1) 9 We prove tha claim by dividing into two patterns.

(See Figure 4.) Denoting by Q the BC-center, Q i) The case of α ≤ β: lies on AB, the BC-radius is a and ∠BPC = π/3. Hence we have By noting that β < γ, we have 1 b |BCf | = πa. (2.2) |CAg| = (π − 2α) 3 2 cos α Denoting by R the CA-center, R = Q, the CA- and radius is a and ∠CPA = 2π/3. Hence we have c |ABg| = (π − 2α). 2 2 cos α |CAg| = πa. (2.3) 3 From β < γ again, we obtain b < c. This (2.1)-(2.3) imply the claim, so the proof is com- implies (2.5). plete. ii) The case of β ≤ α: We investigate the relationship between the By noting that β < γ, we have edge of the bulging triangle and the angle of the b original triangle. Take note that we can meet ex- |CAg| = (π − 2α) actly the same results for a usual triangle. 2 cos α

( and Proposition 2.4. Let 4ABC be an isosceles (but c not necessarily right-angled) bulging triangle. That |ABg| = (π − 2β). is, suppose 2 cos β From β < γ again, we obtain b < c. In order ABC = BCA, i.e., |AB| = |CA|. ∠ ∠ to gain (2.5), we should prove Then, one has π − 2α π − 2β < . (2.8) |ABg| = |CAg|. cos α cos β

3 ( Consider the function with respect to α ∈ Theorem 2.1. For any 4ABC, one has (0, π/2), |ABg| < |BCf | + |CAg|. (2.9) F (α) := (π − 2β) cos α − (π − 2α) cos β, Proof. The policy of this proof is similar to Propo- by ﬁxing β. Since 0 < β ≤ α < π/2 and sition 2.5. Put (2.6) and (2.7) again.

F 0(α) = −(π − 2β) sin α + 2 cos β, i) The case of α ≤ β: By noting that β < γ, we have we have a |BCf | = (π − 2β), F 00(α) = −(π − 2β) cos α < 0. 2 cos β

We can thus concludes that F is monotone b |CAg| = (π − 2α) decreasing. 2 cos α lim F (α) = 0 and α↑π/2 c |ABg| = (π − 2α). regardless of the value of β, and so F (α) > 0 2 cos α on (0, π/2). Therefore, we have shown (2.8) The triangle inequality derives b+c > a, and and can obtain (2.5). so Hence this completes the proof. b + c |CAg| + |ABg| = (π − 2α) 2 cos α Regarding the converse of Proposition 2.5, it a(π − 2α) > . seems that yet another suﬃcient condition is nec- 2 cos α essary. This point is diﬀerent from the property of a usual triangle. In order to gain (2.9), we should prove

( π − 2α π − 2β Proposition 2.6. Let 4 ABC be a bulging tri- > . (2.10) cos α cos β angle that ∠BAC is the minimum internal angle of 4ABC. Then, |ABg| > |CAg| implies ∠BCA > Consider the function with respect to α ∈ ∠ABC. (0, π/2),

Proof. We adopt the notations, (2.6) and (2.7), F (α) := −(π − 2β) cos α + (π − 2α) cos β, again. We should prove that β < γ if α < min{β, γ} and |CAg| < |ABg|. Since α < β and by ﬁxing β. Since 0 < α ≤ β < π/2 and α < γ by the assumption, we have F 0(α) = (π − 2β) sin α − 2 cos β, c |ABg| = (π − 2α), 2 cos α we have b |CAg| = (π − 2α). 00 2 cos α F (α) = (π − 2β) cos α > 0.

Another assumption, |ABg| > |CAg|, then implies We can thus concludes that F is monotone c > b, and so we concludes γ > β. Hence this increasing. We now put completes the proof. G(β) := lim F (α) = π cos β + 2β − π. α↓0 2.3 Theorems and the Proofs Then it is easy to see that G(β) > 0 on Recall that the triangle inequality always holds for (0, π/2), so we leave this conﬁrmation to the any general triangle (see [7] for the proofs). Let us reader. It namely follows that F (α) > 0 on ﬁrst show that a property like the triangle inequal- (0, π/2). Therefore, we have shown (2.10) ity holds for a bulging triangle. and can obtain (2.9).

4 ii) The case of β ≤ α: By virtue of (2.11)-(2.13), we have

The same discussion is repeated, so we leave a2 + b2 2 b |ABg|2 = Arctan2 (2.14) it to the reader. b a Hence this completes the proof. and 2 2 Let us next verify that “Pythagorean Theo- |BCf | + |CAg| rem” holds only for a right-angled bulging triangle π2 b b =(a2 + b2) − πArctan + 2Arctan2 . whose two edges are equal. 4 a a

( (2.15) Theorem 2.2. For 4RBCA, it follows that We here put t := Arctan(b/a) and ( 2 2 2 |ABg| ≥ |BCf | + |CAg| if ∠ABC ∈ [π/4, θ0], (|BCf |2 + |CAg|2) − |ABg|2 2 2 2 F (t) := |ABg| < |BCf | + |CAg| if ∠ABC ∈ (θ0, π/2), a2 + b2 b2 − a2 π2 where = t2 − πt + . b2 4 π|CA| Note that its domain is the interval [π/4, π/2) from θ0 := . 2(|BC| + |CA|) the assumption “a ≤ b”. The axis of the parabola F which is convex downward is The equality holds if ABC = θ or |BC| = |CA|, ∠ 0 πb2 i.e., ABC = BAC = π/4. t = α := ∠ ∠ 2(b2 − a2) Proof. Put a := |BC|, b := |CA|, c := |AB|. With- and out loss of generality, we can set that a ≤ b. b2 a2 π We prove the claim by algebraic geometry. Set 2 2 − 1 = 2 2 > 0, i.e., < α. A(a, b), B(0, 0) and C(a, 0). Consider the AB- b − a b − a 2 perpendicular bisector ÀB. This is given by Moreover, we have π2(b2 − a2) a a b F (π/4) = ≥ 0 y = − x − + . 2 b 2 2 16b and By finding the intersection point of ` and line AB 2 2 2 π a x = a, we have the AB-center P(a, −a /2b + b/2). lim F (t) = − < 0. t↑π/2 4b2 Thus, the AB-radius rAB is given by We find, by the intermediate value theorem, that a2 b a2 + b2 r = |AP| = b − − + = . F has a unique zero point AB 2b 2 2b πb θ0 = , Denote by M the midpoint of segment AB and put 2(a + b) θ := APM = BPM (see Figure 6). It follows ∠ ∠ and we concludes that F ≥ 0 on [π/4, θ0] and tan θ = b/a, since obviously ABC = θ. Hence, ∠ F < 0 on (θ0, π/2). Therefore, the claim has been a2 + b2 b proved and we have found that “Pythagorean the- |ABg| = Arctan . (2.11) orem” does not generally hold. b a We find the condition for the equal sign to hold. Similarly, we obtain It is obvious that the equality holds if θ = θ0 by virtue of the above discussion. If a = b, we have p π b |BCf | = a2 + b2 − Arctan (2.12) θ = π/4 and Arctan(b/a) = 1. Hence, from (2.14) 2 a and (2.15), “Pythagorean theorem” holds: and π2a2 |ABg|2 = |BCf |2 + |CAg|2 = . p b 4 |CAg| = a2 + b2Arctan . (2.13) a This completes the proof.

5 Remark 2.2. We check that the “triangle inequal- Then, it is suﬃcient to prove that the CA-radius is ity” holds for right-angled bulging triangle via the shorter than the radius of the circumscribed circle, proof of Theorem 2.2. Since i.e., a2 + b2 b |ABg| = Arctan |AP| < |XP|. (2.18) b a and It has already been obtained that

π p 2 2 |BCf | + |CAg| = a2 + b2, a + b 2 |AP| = (2.19) 2b we need to prove √ in the proof of Theorem 2.2. Since a2 + b2 b π √ Arctan < (2.16) c a2 + b2 b a 2 |XM| = |AM| = = 2 2 so as to see and √ |ABg| < |BCf | + |CAg|. a a2 + b2 |MP| = We obtain π/4 ≤ Arctan(b/a) < π/2 because of 2b the assumption “a ≤ b”. The left-hand side of from (2.17), we have (2.16) can be rewritten as √ √ (a + b) a2 + b2 a2 + b2 b Arctan(b/a) |XP| = |XM| + |MP| = . (2.20) Arctan = , 2b b a sin(Arctan(b/a)) We should thus check so it is suﬃcient to verify that √ (a + b) a2 + b2 a2 + b2 x > (2.21) f(x) = , x ∈ [π/4, π/2), 2b 2b sin x is strictly monotone increasing and f(x) < π/2. It to prove (2.18), according to (2.19) and (2.20). We is easy to see that fact, so we leave it to the reader. gain that We have (2.16) as a result. √ (a + b) a2 + b2 a2 + b2 We ﬁnally investigate the relationship between − √ 2b 2b the bulging triangle and the circle. It is well known a2 + b2 p = {(a + b) − a2 + b2}, that any triangle is inscribed in a circle. We here 2b try a similar argument for right-angled bulging triangles. The method of the proof of Theorem 2.2 is but the triangle inequality for 4ABC implies that useful for this. p 2 2 ( (a + b) − a + b = (a + b) − c > 0. Theorem 2.3. Any 4RBCA is inscribed in a circle whose center is the midpoint M of A and B and So (2.21), i.e., (2.18) has obtained. This completes whose radius is |AM|. the proof.

Proof. We saw that BCf + CAg is semi-circular arc ( Remark 2.3. The best feature of the above proof BA containing C in the proof of Proposition 2.2. is to show (2.21), i.e., Moreover, the center and radius of that semi- p circular arc are M and |AM|, respectively. Denote a + b > a2 + b2. by C the circle whose center is M and whose radius is |AM|. Consider the line PM (i.e., perpendicular This was a paraphrase of the triangle inequality. bisector ` of AB) and the intersection point X of ` The triangle inequality seems to have already been and C (see Figure 7). Recall considered trivial in ancient Greek times, but it is quite interesting that such a trivial property is the M(a/2, b/2), P(a, (−a2 + b2)/2b). (2.17) essence of Theorem 2.3.

6 3 Conclusion We would like to make research on them a future subject. We have defined the generalized Reuleaux triangles by one of many ways. Beside this way, there is for instance a way to construct a generalization References of the Reuleaux triangle 4ABC as the closed and [1] Besicovitch, A.S.: The Kakeya Problem, The convex curve consisting of three arcs centered on American Mathematical Monthly, Vol. 70, No. 7 each A, B, C. That can be seen a little in [6]. As (1963), pp. 697-706. we can see from this, “generalization” has many [2] Barrallo, J., Gonzalez-Quintial,´ F. and implications. Sanchez-Beitia,´ S.: An Introduction to the We have proved that “triangle inequalities” al- Vesica Piscis, the Reuleaux Triangle and Related ways hold but “Pythagorean theorems” do not gen- Geometric Constructions in Modern Architecture, erally hold for bulging triangles defined in this pa- Nexus Network Journal (2015) 17, pp. 671-684. per. Also, right-angled bulging triangles have good [3] Dillon, M.I.: Geometry Through History, qualities (the latter claim of Theorem 2.2 and Re- Springer (2018). mark 2.2). In that sense, our definition of the gen- [4] Fitzpatrick, R. (Edited): Euclid’s Elements of eralization of Reuleaux triangles would be moder- Geometry (from Euclidis Elementa, edidit et La- ately appropriate. tine interpretatus est I.L. Heiberg, in aedibus B.G. We may however be able to define new bulging Teubneri), ISBN 978-0-6151-7984-1 (Revised and triangles satisfying the above two properties. In corrected - 2008). addition, there are a number of topics left by this [5] Kodaira, K.: Invitation to Geometry, Iwanami paper: Shoten (in Japanese, 2019).

• We can easily obtain the area of a Reuleaux [6] Ridley, J.N.: A Generalization of Reuleaux Tri- triangle (see Introduction of this paper), but angle, it seems diﬃcult to ﬁnd that of a general http://frink.machighway.com/∼dynamicm/constant- bulging triangle. diameter-curves.pdf, 2 pages. • It is unclear if any (not necessarily right- [7] Someyama, N. and Borongan, M.L.A.: New angled) bulging triangle is inscribed in a cir- Proofs of Triangle Inequalities, Journal of Ra- cle. manujan Society of Mathematics and Mathemati- cal Sciences, Vol. 7, No. 1 (Dedicated to Prof. A.K. • The application of bulging triangles is our Agarwal’s 70th Birth Anniversary, 2019), pp. 85- concern. etc. 92.

7 Figure 1: Reuleaux Triangle

Figure 2: AB-center P

Figure 3: Bulging Triangle

Figure 4: Right-Angled Bulging Triangle

8 Figure 5: Isosceles Right-Angled Bulging Triangle

Figure 6: Coordinate Representation of Isosceles Right-Angled Bulging Triangle

Figure 7: Circumscribed Circle of Right-Angled Bulging Triangle