Recreational Mathematics Paul Yiu Department of Mathematics Florida Atlantic University Summer 2003 Version 030131
Contents 1 Cross Number Puzzles 1 2 Sketchpad 7 3 Counting 9 4 Number Trivia 11
Chapter 1 Geometry Recreations 1.1 Routh s theorem If the sides BC, CA, AB of a triangle are divided by points X, Y, Z such that BX : XC = p :1, CY : YA= q :1, AZ : ZB = r :1, the area of the triangle bounded by the lines AX, BY, CZ is (pqr 1) 2 (pq + q +1)(qr + r +1)(rp + p +1). If p = q = r =2, for example, this area is 1 7. For what integers p, q, r is this a unit fraction? 1.2 Heron s formula The area of a triangle of given sides a, b, c is = s(s a)(s b)(s c), (1.1) where s = 1 2 (a + b + c). (1) Heron triangles can be formed by glueing two integer right triangles along a common side. But not every Heron triangle arises in this way. For example, the triangle (25,34,39) is one. Put the vertices of this triangle at points where the coordinates are integers. (2) There is an infinity of super-heron triangles: Heron triangles whose sides are three consecutive integers. If (a 1,a,a+1)is a Heron triangle, the middle side a is a term of the sequence (a n ), where a n+1 = a 2 n 2. Find the first term of this sequence, and a few more terms. Verify that each one of these does give a super Heron triangle.
2 Geometry Recreations 1.3 Proofs without words 1.4 Fun with the Geometer s Sketchpad 1.4.1 Packings of circles O b Oa a Or b r A l B 1. O a O b = a + b. 2. AB =2 ab. 3. l = a + b. 4. 1 r = 1 a + 1 b. If the middle circle has radius a and the other two have radii b, the radius of the left circle is c =2 ab + b. The width of the rectangle is 2(c + a) =2( a + b) 2. But this is also ( c + b) 2. From ( c + b) 2 =2(c + a). Eliminating a, wehave (b 2 c 2 ) 2 =16b 3 c. This diagram is not constructible with ruler and compass. How about c =( a + b) 2, ( c + b) 2 = c + a + (c + a) 2 a 2? 2c =( a + b) 2 ; ( c + b) 2 =( c + a) 2 +2a. 2 c( b a)=3a b; 4c( b a) 2 =(3a b) 2 ; 2(b a) 2 =(3a b) 2 ; 2b 2 4ab +2a 2 9a 2 +6ab b 2 =0; b 2 +2ab 7a 2 =0 (b + a) 2 =8a 2 ; b =(2 2 1)a.
1.4 Fun with the Geometer s Sketchpad 3 c Oc Oa a O b b b O b c Oc Oa O a a a Change the middle circle into radius d. Wehave2c =( a + b) 2. b +2 bc = a +2 d( a + c) From this, b( b +2 c) a d = 2( a + c) b( b + 2( a + b)) a = 2 a + 2( a + b) = ( 2+1)b + 2 ab a 2(( 2+1) a + b) = ( a + b)(( 2+1) b a) 2(( 2+1) a + b) Also, (b + d) 2 =(2 ad + a b) 2 +(2 ab + a d) 2. d = a( a + b)( a +( 2 1) b) = b +2 ab a a( a + b)( a +( 2 1) b) (( 2+1) a + b)( b ( 2 1) a).
4 Geometry Recreations Comparsion gives (( 2+1) b a) 2 = a( a +( 2 1) b) b ( 2 1) a. (1 + 2)b (4 2) ab a =0. If we take a =1, then 1+2 2+ 11 2 2 b = 2+. 2 a =1; b =1.88457; c =2.81508; d =1.05126. b O b c Oa O d Oc a d Japanese 5.6.7 c = 9 4 2 8 AB; a = 3 2 2 2 AB; c =(2 2 1) b. From these, it is easy to see that b = 1 8AB. This is the key to a simple construction. b a c Japanese 5.2.2 Since the inradius of the right triangle is 1 4 of a leg, the right triangle is a (3,4,5) triangle. The base angle of the isosceles triangle is 2arctan 3 4 = arccos 7 25. If we take the base to be 70, then the slant sides are 125 and the vertex is 120 above the base. The bottom incenter has coordinates (7,14).
1.4 Fun with the Geometer s Sketchpad 5 Japanese 5.1.1 2 ab + 3b = l, 2a +3c = 3l, 3b + 3c +2 bc = 2l. From (1), 4ab =(l 3b) 2. From (2), 4ab =2 3lb 3bc. Therefore, l 2 2 3lb +3b 2 =2 3lb 3bc. I had no patience for the elimination, and applied Mathematica and obtain a =0.202454, b =0.295107, c =0.442381. (3,4,5) triangle.
6 Geometry Recreations a +2 ab + 3b = l, a +2 ca +2c = 3l, 3b +2 bc + 3c = 2l. Write a = x 2, b = y 2, c = z 2. These equations become Completing the squares: x 2 +2xy + 3y 2 = 1, y 2 + 2 3 yz + y 2 = 2 3, 2z 2 +2zx + x 2 = 3. Too difficult. 3+ 3 4 (y z) 2 + 3+ 3 (y + z) 2 =1. 4
1.4 Fun with the Geometer s Sketchpad 7 1.4.2 Malfatti squares Is there an easy construction of the Malfatti squares? In JTG, p.132, the sides of the squares are given as 4 a 2 + b 2 + c 2 +4 m a, etc., where m a, m b, m c are the lengths of the medians.
8 Geometry Recreations 1.5 Bankoff s Christmas tree problem An isosceles right triangle is inscribed in a semicircle and the radius bisecting the other semicircle is drawn. Circles are inscribed in the triangle and in the two quadrants as the figure. Prove that these three smaller circles are equal.
1.5 Bankoff s Christmas tree problem 9 JRM 6.4, Problem 306 Does there exist a scalene triangle with m a = t b = h c? It does not look likely from a sketch, which suggests that if this condition holds, the triangle must be equilateral. I think this has to do with a comparison of the angles.
10 Geometry Recreations
Chapter 2 Cross Number Puzzles 5 A B C 4 3 D 2 1 E 0 0 1 2 3 4 5 6 7 In the following clues, each of the letters R through Y stands for a positive integer (which may have many digits). Each blank square isto contain a single digit. Thus, for example, A across will be an 7-digit number. Clues Across A. R!+S D. X Y + R E. W W T Down A. T S B. X + X + X + X
12 Cross Number Puzzles C. X X Y D. Y
13 5 4 3 A B C D E F G H 2 I J 1 0 K 0 1 2 3 4 5 Clues Across A. The square of H down. D. E across plus a factor of H down. E. The sum of all the digits in the puzzle. G. The sum of the digits in C down. I. The sum of the digits in K across. K. The square of C down. Down A. The root mean square of C down and H down. B. A multiple of G across. C. A prime. F. H down less one of its factors. H. A palindrome. J. The sum of the digits in A across.
14 Cross Number Puzzles 5 4 A B C 3 2 D E F 1 G 0 0 1 2 3 4 5 6 7 8 In the following clues, each of the letters N through W stands for a positive integer (which may have many digits). Each blank square is to contain a single digit. Thus, for example, A across will be an 8-digit number. Clues Across A. P! Q D. N R (N S 1) T F. SP R G. S! NW +1 Down A. W N B. NT C. N P + Q N D. W + T
15 Fill in the accompanying square with distinct 2-, 3-, and 4-digit numbers which are perfect squares, none of which begins with 0. 6 5 4 3 2 1 0 0 1 2 3 4 5 6
16 Cross Number Puzzles
Chapter 3 Sketchpad (Leon Bankoff, JRM, Problem 289). Show that the 90 angle of a right triangle is bisected by the line joining it to the center of the square on the hypotenuse. (L. Bankoff, JRM, Problem 290). What is the radius of the largest semicircle that touches all four sides of a unit square? (JRM, Problem 284). Determine the largest area that can be enclosed by a plane quadrilateral with sides 3, 7, 9, and 11 units in order. (JRM, Problem 464). Four points on a plane are at distances 1, 2, 3, 4 from a fixed point on the same plane. Find the largest possible area of the quadrilateral formed by the four points. (JRM, Problem 466). Given a triangle ABC, construct a square with two sides meeting at A and with the other two sides containing B and C respectively.
18 Sketchpad
Chapter 4 Counting JRM 7.1, Problem 315. How many integer sided triangles of perimeter N are there? How many of these are isosceles?
20 Counting
Chapter 5 Number Trivia sin 2 (6 2 ) =cos 2 ((2 6) 2 ) =0.3454 cos 2 (6 2 ) =sin 2 ((2 6) 2 ) =0.6545 1. What is the largest prime what square has no duplicate digit? 1 If p 2 has no duplicate digits, p 2 < 10 10 and p<10 5. There are π(10 5 ) below 10 5. We start from the top. 2. Twigg: 2 There is only one 5-digit number whose last three digits are alike and whose square has no duplicate digits. Thébault: 3 Determine the largest and smallest perfect squares which can be written with the 10 digits used one each in both cases. 4 3. JRM 2543: 5 Pandigital society number. 4. Knuth: 6 A positive integer is sorted if its digits in its decimal notation are nondecreasing from left to right. (i) If n =3 h 6 k 7, then n 2 is sorted. For example, 3 3 6 4 7 2 =1 3 3 3 4 2 6 3 8 4 9. What is the square of 3 h 6 k 7 1? (ii) Which positive integers n are such that n and n 2 are both sorted? See also Blecksmith and Nicol, Monotonic numbers, Math. Mag., 66 (1993) 257 262. 1 C. W. Trigg, E 385.396.S404. 2 Twigg, E389.397.S405. 3 E404.401.S408. 4 32043 and 99066. 5 JRM, 30 (1999 2000) 227. 6 MG1234.861.S871.
22 Number Trivia 5. Twigg, Permutation-twin square integers, JRM, 21 (1989) 170 173. N 2 and (N +1) 2 have the same digit set; they have the same digital root. The digital roots of successive integers form a periodic sequence 1, 4, 9, 7, 7, 4, 1, 9,... 13 2 = 169, 14 2 = 196; 157 2 = 24649, 158 2 = 24964; 913 2 = 833569, 159 2 = 835396; 4513 2 = 20367169, 4514 2 = 20376196. How about not requiring equality of multiplicities? 6. Charles Twigg, JRM, 24 (1992) 5: What three-digit squares have the following characteristics? (a) are palindromes. (b) are permutations of consecutive digits. (c) form reversal pairs. (d) are three permutations of the same digit set. (e) three of its permutations are prime. (f) the sum of the digits is 19. (g) is also a cube. (h) the central digit is perfect. (i) are composed of even digits. (j) the central digit is a nonzero cube. 7. JRM 2049, 25 (1993) 156 157. (a) Use each of the digits 1 through 9 once to form prime numbers such that the sum of the primes is as small as possible. What is this sum? (b) Use each of the digits 0 through 9 once to form prime numbers such that the sum of the primes is as small as possible. What is this sum? Answers: (a) are palindromes. 121, 484, 676 (b) are permutations of consecutive digits. 324, 576 (c) form reversal pairs. 144 and 441; 169 and 961 (d) are three permutations of the same digit set. 169, 196, 961 (e) three of its permutations are prime. 163, 613, 631 of 361 (f) the sum of the digits is 19. 289, 676, 784
23 (g) is also a cube. 729 (h) the central digit is perfect. 169, 361, 961 (i) are composed of even digits. 400, 484, (j) the central digit is a nonzero cube. 289, 484, 784