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The Vertex-Midpoint-Centroid Triangles

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The Vertex-Midpoint-Centroid Triangles Forum Geometricorum b Volume 4 (2004) 97–109. bbb FORUM GEOM ISSN 1534-1178 The Vertex-Midpoint-Centroid Triangles Zvonko Cerinˇ Abstract. This paper explores six triangles that have a vertex, a midpoint of a side, and the centroid of the base triangle ABC as vertices. They have many in- teresting properties and here we study how they monitor the shape of ABC. Our results show that certain geometric properties of these six triangles are equivalent to ABC being either equilateral or isosceles. Let A, B, C be midpoints of the sides BC, CA, AB of the triangle ABC and − let G be its centroid (i.e., the intersection of medians AA, BB , CC ). Let Ga , + − + − + Ga , Gb , Gb , Gc , Gc be triangles BGA , CGA , CGB , AGB , AGC , BGC (see Figure 1). C − + Gb Ga B A G + − Gb Ga − + Gc Gc ABC Figure 1. Six vertex–midpoint–centroid triangles of ABC. This set of six triangles associated to the triangle ABC is a special case of the cevasix configuration (see [5] and [7]) when the chosen point is the centroid G.It has the following peculiar property (see [1]). Theorem 1. The triangle ABC is equilateral if and only if any three of the trian- = { − + − + − +} gles from the set σG Ga ,Ga ,Gb ,Gb ,Gc ,Gc have the same either perime- ter or inradius. In this paper we wish to show several similar results. The idea is to replace perimeter and inradius with other geometric notions (like k-perimeter and Brocard angle) and to use various central points (like the circumcenter and the orthocenter – see [4]) of these six triangles. Publication Date: July 14, 2004. Communicating Editor: Paul Yiu. 98 Z. Cerinˇ Let a, b, c be lengths of sides of the base triangle ABC. For a real number k, k k k the sum pk = pk(ABC)=a + b + c is called the k-perimeter of ABC.Of course, the 1-perimeter p1(ABC) is just the perimeter p(ABC). The above theo- rem suggests the following problem. Problem. Find the set Ω of all real numbers k such that the following is true: The triangle ABC is equilateral if and only if any three of the triangles from σG have the same k-perimeter. Our first goal is to show that the set Ω contains some values of k besides the value k =1. We start with k =2and k =4. Theorem 2. The triangle ABC is equilateral if and only if any three of the trian- gles in σG have the same either 2-perimeter or 4-perimeter. Proof for k =2. We shall position the triangle ABC in the following fashion with respect to the rectangular coordinate system in order to simplify our calculations. The vertex A is the origin with coordinates (0, 0), the vertex B is on the x-axis and 2 ( ( + ) 0) rg(f −1) 2rfg has coordinates r f g , , and the vertex C has coordinates fg−1 , fg−1 . The three parameters r, f, and g are the inradius and the cotangents of half of angles at vertices A and B. Without loss of generality, we can assume that both f and g are larger than 1 (i.e., that angles A and B are acute). Nice features of this placement are that many important points of the triangle have rational functions in f, g, and r as coordinates and that we can easily switch from f, g, and r to side lengths a, b, and c and back with substitutions ( 2+1) ( 2+1) = rf g = rg f = ( + ) a fg−1 ,b fg−1 ,cr f g , 2 2 2 2 = (b+c) −a = (a+c) −b = 2∆ f 4∆ ,g 4∆ ,r a+b+c , 1 where the area ∆ is 4 (a + b + c)(b + c − a)(a − b + c)(a + b − c). There are 20 ways in which we can choose 3 triangles from the set σG. The following three cases are important because all other cases are similar to one of these. ( − + −) 2 ( −) ( +) Case 1: Ga ,Ga ,Gb . When we compute the -perimeters p2 Ga , p2 Ga , ( −) and p2 Gb and convert to lengths of sides we get ( − )( + ) ( −) − ( +)= c b c b p2 Ga p2 Ga 3 , 2 2 2 − − a b c p2(G ) − p2(G )= − + . a b 6 2 3 Both of these differences are by assumption zero. From the first we get b = c and (a−c)(a+c) when we substitute this into the second the conclusion is 6 =0. Hence, b = c = a so that ABC is equilateral. ( − + +) Case 2: Ga ,Ga ,Gb . Now we have The vertex-midpoint-centroid triangles 99 ( − )( + ) ( −) − ( +)= c b c b p2 Ga p2 Ga 3 , − + (a − b)(a + b) p2(G ) − p2(G )= , a b 2 which makes the conclusion easy. ( − − −) Case 3: Ga ,Gb ,Gc . This time we have 2 2 2 − − a b c p2(G ) − p2(G )= − + , a b 6 2 3 2 2 2 ( −) − ( −)=a − b − c p2 Ga p2 Gc 2 3 6 . The only solution of this linear system in a2 and b2 is a2 = c2 and b2 = c2. Thus the triangle ABC is equilateral because the lengths of sides are positive. Recall that the Brocard angle ω of the triangle ABC satisfies the relation ( ) cot = p2 ABC ω 4∆ . Since all triangles in σG have the same area, from Theorem 2 we get the following corollary. Corollary 3. The triangle ABC is equilateral if and only if any three of the trian- gles in σG have the same Brocard angle. √ On the other hand, when we put k = −2 then for a = −5+3 3 and b = c =1 − + − (−2) we find that the triangles Ga , Ga , and Gb have the same -perimeter while ABC is not equilateral. In other words the value −2 is not in Ω. The following result answers the final question in [1]. It shows that some pairs of triangles from the set σG could be used to detect if ABC is isosceles. Let τ de- ( − +)( − +) ( − +) ( + −) note the set whose elements are pairs Ga ,Ga Ga ,Gb , Ga ,Gc , Ga ,Gb , ( + −) ( − +) ( − +) ( + −) ( − +) Ga ,Gc , Gb ,Gb , Gb ,Gc , Gb ,Gc , Gc ,Gc . Theorem 4. The triangle ABC is isosceles if and only if triangles from some element of τ have the same perimeter. Proof. This time there are only two representative cases. Case 1: (G−,G+). By assumption, a a √ √ 2 2 − 2 +2 2 2 2 +2 2 − 2 ( −) − ( +)= a b c − a b c =0 p Ga p Ga 3 3 . When we move the second term to the right then take the square of both sides and (c−b)(c+b) move everything back to the left we obtain 3 =0. Hence, b = c and ABC is isosceles. + Case 2: (G−,G ). This time our assumption is a b √ √ a − b 2a2 − b2 +2c2 2c2 +2b2 − a2 p(G−) − p(G+)= + − =0. a b 2 6 6 100 Z. Cerinˇ When we move the third term to the right then take the square of both sides and move the right hand side back to the left and bring the only term with the square root to the right we obtain √ 2 2 − 3 + 2 ( − ) 2 2 − 2 +2 2 a ab b = b a a b c 6 6 . In order to eliminate the square root, we take the square of both sides and move the (a−b)2(a−b−c)(a−b+c) right hand side to the left to get 18 =0. Hence, a = b and the triangle ABC is again isosceles. Remark. The above theorem is true also when the perimeter is replaced with the 2-perimeter and the 4-perimeter. It is not true for k = −2 but it holds for any k =0 ( − +) ( − +) ( − +) when only pairs Ga ,Ga , Gb ,Gb , Gc ,Gc are considered. We continue with results that use various central points (see [4], [5, 6]) (like the centroid, the circumcenter, the orthocenter, the center of the nine-point circle, the symmedian or the Grebe-Lemoine point, and the Longchamps point) of the trian- gles from the set σG and try to detect when ABC is either equilateral or isosceles. Recall that triangles ABC and XY Z are homologic provided lines AX, BY , and CZ are concurrent. The point in which they concur is their homology center and the line containing intersections of pairs of lines (BC,Y Z), (CA,ZX), and (AB, XY ) is their homology axis. Instead of homologic, homology center, and homology axis many authors use the terms perspective, perspector, and perspec- trix. The triangles ABC and XY Z are orthologic when the perpendiculars at ver- tices of ABC onto the corresponding sides of XY Z are concurrent. The point of concurrence is [ABC, XY Z]. It is well-known that the relation of orthology for triangles is reflexive and symmetric. Hence, the perpendiculars at vertices of XY Z onto corresponding sides of ABC are concurrent at a point [XY Z,ABC]. By replacing in the above definition perpendiculars with parallels we get the analogous notion of paralogic triangles and two centers of paralogy ABC, XY Z and XY Z,ABC. The triangle ABC is paralogic to its first Brocard triangle AbBbCb which has the orthogonal projections of the symmedian point K onto the perpendicular bisectors of sides as vertices (see [2] and [3]). Theorem 5. The centroids G − , G + , G − , G + , G − , G + of the triangles Ga Ga Gb Gb Gc Gc from √σG lie on the image of the Steiner ellipse of ABC under the homothety 7 h(G, 6 ).
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