97 Dissecting Triangles into Isosceles Triangles Daniel Robbins student, Ecole Secondaire Beaumont, Beaumont Sudhakar Sivapalan student, Harry Ainlay Composite High School, Edmonton Matthew Wong student, University of Alberta, Edmonton Problem 2 of Part II of the 1993-1994 Alberta High School Mathematics Competition see CRUX [20:65] goes as follows: An isosceles triangle is called an amoeba if it can be divided into two isosceles triangles by a straight cut. How many di erent that is, not similar amoebas are there? All three authorswrote that contest. Afterwards, they felt that the problem would have been more meaningful had they been asked to cut non-isosceles triangles into isosceles ones. We say that a triangle is n-dissectible if it can be dissected into n isosce- les triangles where n is a positive integer. Sinceweare primarily interested in the minimum value of n, we also say that a triangle is n-critical if it is n-dissectible but not m-dissectible for any m<n. The isosceles triangles themselves are the only ones that are 1-dissectible, and of course 1-critical. Note that, in the second de nition, we should not replace \not m- dissectible for any m<n"by \not n 1-dissectible". It may appear that if a triangle is n-dissectible, then it must also be m-dissectible for all m>n. However, thereare two exceptions. The solution to the contest problem, which motivated this study, shows that almost all 1-dissectible triangles are not 2-dissectible. We will point out later that some 2-dissectible ones are not 3-dissectible. On the other hand, it is easyto see that all 1-dissectible triangles are 3-dissectible. Figure 1 illustrates the three cases where the vertical angle is acute, right, and obtuse respectively. A A @ A H @ H S H Z A @ H S H Z A A Z @ H S a b c Figure 1. 98 We now nd all 2-dissectible triangles. Clearly, such a triangle can only be cut into two triangles bydrawing a line fromavertexto the opposite side, as illustrated in Figure2. A S S S S S S S D 2 2 S B C Figure 2. Note that at least one of \AD B and \AD C is non-acute. We may assume that \AD B 90 .Inorder for BAD to be an isosceles triangle, we must have \BAD = \AB D . Denote their common value by . By the Exterior Angle Theorem, \AD C =2 . Thereare three ways in which CAD may become an isosceles triangle. Case 1. Here \AC D = \AD C =2 , as illustrated in Figure 2. Then \CAD = 180 4 >0 . This class consists of all triangles in which two of the angles are in the ratio 1:2, where the smaller angle satis es 0 < <45 .Of these, only the 36 ; 72 ; 72 triangle is 1-dissectible, but it turns out that every triangle hereis3-dissectible. Case 2. Here, \CAD = \AD C =2 . Then \CAB =3 and \AC D = 180 4 >0. This class consists of all triangles in which two of the angles are in the ratio 1:3, where the smaller angle satis es 0 < <45 . Of 540 540 180 ; ; triangles are 1- these, only the 36 ; 36 ; 108 and the 7 7 7 dissectible. It turns out that those triangles for which 30 < <45 , with a few exceptions, are not 3-dissectible. Case 3. Here, \AC D = \CAD. Then their common value is 90 so that \CAB =90 . This class consists of all right triangles. Of these, only the 45 ; 45 ; 90 triangle is 1-dissectible, and it turns out that every triangle hereis3-dissectible. We should point out that while our three classes of 2-dissectible tri- angles areexhaustive, theyare not mutually exclusive. For instance, the 30 ; 60 ; 90 triangle appears in all three classes, with the same dissection. The 20 ; 40 ; 120 triangle is the only one with two di erent dissections. We now consider 3-dissectible triangles. Suppose one of the cuts does not pass through any vertices. Then it divides the triangle into a triangle and a quadrilateral. The latter must then be cut into two triangles, and this cut must pass through a vertex. Hence at least one cut passes through a vertex. 99 Suppose no cut goes fromavertexto the opposite side. The only pos- sible con guration is the one illustrated in Figure 1a. Since the three angles at this point sum to 360 , at least two of them must be obtuse. It follows that the three arms have equal length and this point is the circumcentreof the original triangle. Since itisaninterior point, the triangle is acute. Thus all acute triangles are 3-dissectible. In all other cases, one of the cutsgofromavertexto the opposite side, dividing the triangle into an isosceles one and a 2-dissectible one. Thereare quite a number of cases, but the argument is essentially an elaboration of that used to determine all 2-dissectible triangles. We leave the details to the reader, and will just summarize our ndings in the following statement. Theorem. A triangle is 3-dissectible if and only if it satis es at least one of the following conditions: 1. It is an isosceles triangle. 2. It is an acute triangle. 3. It is a right triangle. 4. It has a 45 angle. 5. It has one of the following forms: a , 90 2 , 90 + , 0 < < 45 ; 3 b , 90 , 90 + , 0 < < 60 ; 2 2 c , 360 7 , 6 180 , 30 < <45 ; 3 d 2 , 90 , 90 , 0 < < 60 ; 2 2 e 3 , 90 2 , 90 , 0 < <45 ; f 180 2 , 180 , 3 180 , 60 < <90 ; g 180 4 , 180 3 , 7 180 , 30 < <45 . 6. Two of its angles are in the ratio p:q , with the smaller angle strictly between 0 and r , for the following values of p, q and r : p 1 1 1 1 1 1 2 3 3 q 2 3 4 5 6 7 3 4 5 r 60 30 22.5 30 22.5 22.5 60 67.5 67.5 100 The fact that all right triangles are 2-dissectible is important because every triangle can be divided into two right triangles by cutting along the altitude toits longest side. Each can then be cut into two isosceles triangles by cutting along the median from the right angle to the hypotenuse, as il- lustrated in Figure 3. It follows that all triangles are 4-dissectible, and that therearenon-critical triangles for n 5. S S S S H H S H H S H H S H H S Figure 3. We can proveby mathematical induction on n that all triangles are n- dissectible for all n 4. We have established this for n =4. Assume that all triangles are n-dissectible for some n 4. Consider any triangle. Divide it by a line through a vertex into an isosceles triangle and another one. By the induction hypothesis, the second can be dissected into n triangles. Hence the original triangle is n +1-dissectible. Announcement For the information of readers, we are saddened to note the deaths of two mathematicians who are well-known toreadersofCRUX. Leroy F. Meyers died last November. He was a long time contribu- tor to CRUX. See the February 1996 issue of the Notices of the American Mathematical Society. D.S. Mitrinovic died last March. He is well-known for his work on Ge- ometric Inequalities, in particular, being co-author of Geometric Inequalities and Recent Advances in Geometric Inequalities. See the July 1995 issue of the Notices of the American Mathematical Society. 101 THE SKOLIAD CORNER No. 13 R.E. Woodrow This issue we feature the Eleventh W.J. Blundon Contest, written February 23, 1994. This contest is sponsored by the Department of Math- ematics and Statistics of Memorial University of Newfoundland and is one of the \Provincial" contests that receives support from the Canadian Math- ematical Society THE ELEVENTH W.J. BLUNDON CONTEST February 23, 1994 1. a The lesser of two consecutive integers equals 5 more than three times the larger integer. Find the two integers. b If 4 x 6 and 2 y 3, nd the minimum values of x y x + y . 2. A geometric sequence is a sequence of numbers in which each term, after the rst, can be obtained from the previous term, by multiplying by the same xed constant, called the common ratio. If the second term of a geometric sequenceis12 and the fth term is 81=2, nd the rst term and the common ratio. 3. A square is inscribed in an equilateral triangle. Find the ratio of the area of the squareto the area of the triangle. 4. AB C D is a square. Three parallel lines l , l and l pass through 1 2 3 A, B and C respectively. The distance between l and l is 5 and the distance 1 2 between l and l is 7. Find the area of AB C D . 2 3 5. The sum of the lengths of the three sides of a right triangle is 18. The sum of the squares of the lengths of the three sides is 128. Find the area of the triangle. 6. A palindrome is a word or number that reads the same backwards and forwards. For example, 1991 is a palindromic number. How many palin- dromic numbersare there between 1 and 99; 999 inclusive? 2 2 2 7. Agraph of x 2xy + y x + y =12and y y 6= 0will produce four lines whose pointsofintersection are the vertices of a parallelogram.