1/1/21. Fill in the circles in the picture t right with the digits 1-8, one digit in ech circle with no digit repeted, so tht no two circles tht re connected by line segment contin consecutive digits. In how mny wys cn this be done? We observe first tht the lower center circle (shded in the digrm below to the right) is connected to ll but one of the other circles. If we plce ny digit other thn 1 or 8 in this circle we will not be ble to complete the grid, since the digits 2 7 ech hve two other digits tht re not llowed to be djcent to it. Furthermore, if the lower center circle contins 1, then the top circle must contin 2, becuse 2 cnnot be plced djcent to 1. We notice tht if we re given solution to this problem, then 2 replcing every number x in the grph by 9 x gives us different solution. Furthermore this opertion gives 1-1 correspondence between the solutions with 1 in the lower center circle nd the solutions with 8 in the lower center circle. Thus, we count the number of solutions where 1 is in this circle, nd the totl will be twice this count. 1 The remining six empty circles form hexgon. We must plce the digit 3 in one of the bottom 3 circles (since it cnnot be djcent to 2). Once we hve chosen where to plce 3, we re left filling the hexgon with the digits 4 through 8. At this point the loctions of 1 nd 2 re irrelevnt, so we cn count ech of these three possibilities for the loction of 3 by counting three times the number of possibilites when 3 is plced t the bottom of hexgon nd the remining digits re plced elsewhere. We now see tht the nswer to the originl problem is 2 3 = 6 times the number of wys to fill six-circle loop with the digits 3 to 8 such tht 3 is on the bottom nd no two consecutive digits re djcent. Next we plce the digit 4. There re two cses: Cse 1: 4 is opposite from 3. From here we hve two choices for the loction of 5 (either of the circles djcent to 3), nd tht determines where 7 is plced uniquely, s in the figure to the right. At this point we cn plce 6 nd 8 in either of other two nodes, giving totl of 2 2 = 4 possibilities for this cse. 7 4 5 3
Cse 2: 4 is two positions wy from 3. There re two choices for the loction of 4, nd these re symmetric, so we ssume 4 is on the right-hnd side of the digrm (nd we remember tht we will get fctor of two from this symmetry). Now we hve two nonsymmetric choices for plcing 5. If we plce 5 djcent to 3 s in the left imge below, there is only one wy to complete the digrm, s shown. If we plce 5 distnce two from 3 (s in the picture to the right below), then the loction of 6 is now determined (s shown) nd 7 nd 8 my be plced freely. This gives two more options. So there re totl of 2 (1+2) = 6 possibilities in this cse. 6 8 4 5 4 5 7 6 3 3 Summing the two cses, this gives totl of 4 + 6 = 10 possibilities once 3 is plced. Reclling the fctor of 6 from erlier, we conclude tht the totl number of possibilities is 6 10 = 60.
2/1/21. The ordered pir of four-digit numbers (2025, 3136) hs the property tht ech number in the pir is perfect squre nd ech digit of the second number is 1 more thn the corresponding digit of the first number. Find, with proof, ll ordered pirs of five-digit numbers nd ordered pirs of six-digit numbers with the sme property: ech number in the pir is perfect squre nd ech digit of the second number is 1 more thn the corresponding digit of the first number. We observe first tht if the digits of n 2 re one less thn the digits of m 2, then m 2 n 2 = 11111 or m 2 n 2 = 111111 (depending on the lengths of m 2 nd n 2 ). Thus we strt by looking for solutions to these two equtions nd then test ll of our solutions. If (n 2, m 2 ) is n ordered pir of 5-digit numbers stisfying the desired property, then we must hve 11111 = m 2 n 2 = (m n)(m + n) The number 11111 hs only two fctoriztions into product of two fctors: 11111 = 41 271 nd 11111 = 1 11111. Checking the two fctoriztions we hve (n, m) = (115, 156) (n, m) = (5550, 5551) The second pir will fil since n > 1000 implies tht n 2 must hve t lest 7 digits. We check the first: (n 2, m 2 ) = (13225, 24336). This pir works. So the only 5-digit pir is (13225, 24336). Next we look for ordered pirs (n 2, m 2 ) of six-digit numbers with the desired property. We must hve 111111 = m 2 n 2 = (m n)(m + n). Agin we fctor: 111111 = 3 7 11 13 37. Thus, 111111 hs 2 5 = 32 divisors, so there re 16 wys to fctor 111111 into the product of pir of positive integers. But we must hve 100000 n 2 m 2 999999, so this severely restricts the possibilities: tking the squre root, we must hve 317 n < m 999. This mens tht the lrger fctor m+n must be t lest 635 but t most 1997. This restricts the choice of (m n, m + n) to the following pirs: These correspond to (n, m) equling one of {(143, 777), (111, 1001), (91, 1221), (77, 1443)}. {(317, 460), (445, 556), (565, 656), (683, 760)}.
We check these by computing (n 2, m 2 ): {(100489, 211600), (198025, 309136), (319225, 430336), (466489, 577600)}. None of these pirs stisfies the desired property. Our nswer is: There is one pir of 5-digit numbers which stisfies the property, (13225, 24336). There re no pirs of 6-digit numbers which stisfy the property.
3/1/21. A squre of side length 5 is inscribed in squre of side length 7. If we construct grid of 1 1 squres for both squres, s shown to the right, then we find tht the two grids hve 8 lttice points in common. If we do the sme construction by inscribing squre of side length 1489 in squre of side length 2009, nd construct grid of 1 1 squres in ech lrge squre, then how mny lttice points will the two grids of 1 1 squres hve in common? Let the two squres be ABCD nd P QRS, nd let x = AP = BQ = CR = DS. Then P B = QC = RD = SA = 2009 x. Then by the Pythgoren Theorem on tringle AP S, x 2 + (2009 x) 2 = 1489 2, A 2009 x which hs solutions x = 689 nd x = 1320. The lengths 689 nd 1320 produce equivlent digrms, so ssume x = 689. D R C For brevity, let = 689, b = 1320, nd c = 1489; note tht c is prime. Consider the digrm on the coordinte plne with D s the origin, so tht C = (2009, 0) nd A = (0, 2009). Then, by thinking of strting t point S nd moving in units in the directions of SR nd SP, we see tht points on the lttice of P QRS re points of the form ( ) ( b (0, ) + u c, + v c c, b ) c where 0 u c nd 0 v c. This is point on both lttices if nd only if the coordintes re integers; tht is, if ub + v u + vb nd c c re both integers. So our gol is to count ll (u, v) with 0 u c nd 0 v c such tht ub + v nd u + vb re integer multiples of c; tht is, S x P 1489 ub + v 0 (mod c), (1) u + vb 0 (mod c). (2) Clerly if (u, v) {(0, 0), (0, c), (c, 0), (c, c)} then this condition is stisfied these correspond to the four corners of the smller squre. We clim tht if 0 < u < c, then there exists unique v with 0 < v < c stisfying (1) nd (2). In prticulr, since c is prime, eqution (1) cn be rewritten s v ub (mod c), B Q
where division by is well-defined modulo c. (Indeed, with = 689, c = 1489 we cn check tht 1 389 (mod c), since (389)(689) = 268021 = (180)(1489) + 1.) So there is unique v with 0 < v < c stisfying (1). To verify tht this lso stisfies (2), note tht since 2 + b 2 = c 2, we hve b 2 2 (mod c), nd thus ( u + vb u + ub ) ) ) b ( + b2 u ( + 2 u 0 (mod c). Thus, ech 0 < u < c gives exctly one ordered pir (u, v) tht corresponds to lttice point of both grids. This gives c 1 such points, so together with the four corners of P QRS, we get totl of c + 3 points. Therefore, since c = 1489, the nswer is 1492.
4/1/21. Let ABCDEF be convex hexgon, such tht F A = AB, BC = CD, DE = EF, nd F AB = 2 EAC. Suppose tht the re of ABC is 25, the re of CDE is 10, the re of EF A is 25, nd the re of ACE is x. Find, with proof, ll possible vlues of x. Since F AB = 2 EAC, nd the hexgon is convex, we hve EAC = F AB EAC = F AE + BAC. Let X be the reflection of B cross AC. Then CX = CB = CD. Also, AX = AB = AF, nd XAE = EAC XAC = ( F AE + BAC) BAC = F AE. Since AX = AF nd XAE = F AE, we conclude tht X is lso the reflection of F cross AE. Hence EX = EF = ED. Since CX = CD nd EX = ED, we hve two possibilities: either X = D, or X is the reflection of D cross CE. Both cses re shown below. B A F B A F X C E C D D Let [P QR] denote the re of tringle P QR. In the first cse, E [ACE] = [ADC] + [ADE] [CDE] = [ABC] + [AF E] [CDE] = 25 + 25 10 = 40. In the second cse, [ACE] = [AXC] + [AXE] + [CXE] = [ABC] + [AF E] + [CDE] = 25 + 25 + 10 = 60. Hence, the re of tringle ACE must be either 40 or 60. However we hve not yet shown tht we cn chieve these vlues. It remins to construct convex hexgons with the initil conditions such tht tringle ACE hs these res.
For the first cse, let ACE be n equilterl tringle with re 40, nd let M be the midpoint of CE. Let CM =, so AM = 3. Let D be the point on the perpendiculr bisector of CE, on the opposite side of CE from A, such tht DM = 3. 4 Since tringle CDE shres bse with tringle ACE but its ltitude to tht bse is 1/4 the ltitude of D ACE, its re is 1/4 tht of ACE; thus, [CDE] = (40) = 10. Reflect D cross AC to get B, nd reflect D cross AE to get F. We hve [ABC] = [ADC] = [AMC] + [DMC] = 1(40) + 1 (10) = 25, nd similrly [AF E] = 25. 2 2 Also, tn MCD = ( 3/4) = 3 < 3 = tn 30, so MCD < 30. Thus, BCD = 4 3 2 ACD = 2( ACM + MCD) < 2(60 + 30 ) = 180. Similrly, F ED < 180, nd F AB = 2 EAC = 2(60 ) < 180. The remining ngles re obviously less thn 180, so hexgon ABCDEF is convex. We hve thus constructed the first cse. 1 4 B C A M E F For the second cse, let ACE be n equilterl tringle with re 60, nd let M be the midpoint of CE. Let X be the point on segment AM such tht XM = 1AM. Then [CXE] = 1[ACE] = 1 (60) = 10. 6 6 6 B Thus, [AXC] = 1([ACE] [CXE]) = 1 (60 10) = 25, 2 2 nd similrly [AXE] = 25. Let B, D, nd F be the reflections of X cross AC, CE, nd EA, respectively. Then [ABC] = 25, [CDE] = 10, nd [EF A] = 25. Also, F AB = 2 EAC = 2(60 ) < 180, nd similrly BCD < 180 nd DEF < 180. The remin- C ing ngles re obviously less thn 180, so hexgon ABCDEF is convex. Thus, we hve constructed the second cse. Therefore, we conclude tht 40 nd 60 re the two possible res of tringle ACE. A X D E F
5/1/21. The cubic eqution x 3 +2x 1 = 0 hs exctly one rel root r. Note tht 0.4 < r < 0.5. () Find, with proof, n incresing sequence of positive integers 1 < 2 < 3 < such tht 1 2 = r 1 + r 2 + r 3 +. (b) Prove tht the sequence tht you found in prt () is the unique incresing sequence with the bove property. () Since r is root of x 3 + 2x 1, we know tht 1 r 3 = 2r, so 1 2 = r 1 r = r(1 + 3 r3 + r 6 + r 9 + ) = r + r 4 + r 7 + r 10 + Furthermore, since r < 1, this sequence converges. Therefore n = 3n 2 solves the eqution bove. (b) Suppose tht 1 < 2 < 3 < nd b 1 < b 2 < b 3 < re distinct sequences of positive integers such tht Eliminting duplicte terms, we hve r 1 + r 2 + r 3 + = r b 1 + r b 2 + r b 3 + = 1 2. r s 1 + r s 2 + = r t 1 + r t 2 + = > 0, where s 1 < s 2 <... nd t 1 < t 2 <... re positive integers, nd ll the s s nd t s re distinct. (Note tht either series bove might in fct be finite, but this does not ffect the vlidity of the rgument to follow.) Assume with loss of generlity tht s 1 < t 1. Then r s 1 r s 1 + r s 2 + = r t 1 + r t 2 +..., so tht, fter dividing by r s 1, we hve 1 r u 1 + r u 2 +..., where u i = t i s 1 > 0. Thus, since 0 < r < 1 2, we hve 1 1 r contrdiction. < 2, nd hence 1 r u 1 + r u 2 + r + r 2 + r 3 + = 1 1 r 1 < 1,
Credits: Problem 2/1/21 is bsed on problem ppering in the Februry, 1997, issue of Crux Mthemticorum with Mthemticl Myhem. Problem 4/1/21 ws proposed by Gku Liu. All other problems nd solutions by USAMTS stff. c 2009 Art of Problem Solving Foundtion