18th Bay Area Mathematical Olympiad. Problems and Solutions. February 23, 2016

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1 18th Bay Area Mathematical Olympiad February 3, 016 Problems ad Solutios BAMO-8 ad BAMO-1 are each 5-questio essay-proof exams, for middle- ad high-school studets, respectively. The problems i each exam are i roughly icreasig order of difficulty, labeled A through E i BAMO-8 ad 1 through 5 i BAMO-1, ad the two exams overlap with three problems. Hece problem C o BAMO-8 is problem #1 o BAMO-1, problem D o BAMO-8 is # i BAMO-1, ad problem E i BAMO-8 is #3 i BAMO-1. The solutios below are sometimes just sketches. There are may other alterative solutios. We ivite you to thik about alteratives ad geeralizatios! A The diagram below is a example of a rectagle tiled by squares: Each square has bee labeled with its side legth. The squares fill the rectagle without overlappig. I a similar way, a rectagle ca be tiled by ie squares whose side legths are, 5, 7, 9, 16, 5, 8, 33, ad 36. Sketch a possible arragemet of those squares. They must fill the rectagle without overlappig. Label each square i your sketch by its side legth, as i the picture above. Solutio: To tile a rectagle, the areas of the squares must add to match the area of the rectagle. The total area of the 9 squares is: = 409. Whe we factor 409 we obtai 409 = To fit the largest square, the rectagle has to be at least 36 uits wide ad high, ad the oly way to do that with these three factors is a rectagle of size It s easy to get started by oticig that = 69 ad with so few squares the two largest squares must lie o oe edge of the rectagle of legth 69. From there it becomes easy to place the other oes, ad the followig figure illustrates a possible solutio (the ulabeled square has side legth of ). There are actually 3 additioal solutios that are obtaied from this oe by either a rotatio of 180 or takig a mirror image across either the horizotal or vertical axis.

2 BAMO 016 Problems ad Solutios March 17, B A weird calculator has a umerical display ad oly two buttos, D ad D. The first butto doubles the displayed umber ad the adds 1. The secod butto doubles the displayed umber ad the subtracts 1. For example, if the display is showig 5, the pressig the sequece D D D D will result i a display of 87. (a) Suppose the iitial displayed umber is 1. Give a sequece of exactly eight butto presses that will result i a display of 313. (b) Suppose the iitial displayed umber is 1, ad we the perform exactly eight butto presses. What are all the umbers that ca possibly result? Prove your aswer by showig that all of these umbers ca be produced, ad that o other umbers ca be produced. Solutio: (a) The sequece (uique) is D D D D D D D D, producig itermediate values 3,5,9,19,39,79,157,313. (b) The umbers produced will be the odd umbers from 1 to 511, iclusive. Clearly the oly umbers producible are odd, ad certaily the miimum ad maximum values will be 1 ad 511, respectively, sice that is what is produced after 7 presses of D ad 7 presses of D, respectively. We still eed to prove that ay odd umber betwee 1 ad 511 ca produced. There are may possible methods. Here is oe. Observe that if two umbers are differet ad we press the same butto, we have to get two differet outputs. Likewise, if we press differet buttos with the same startig value, we get differet outputs. However, it is possible to start with two differet umbers a,b ad get the same output, as log as we apply

3 BAMO 016 Problems ad Solutios March 17, D to the larger oe ad D to the smaller oe. For example, suppose that b > a. The we could have b 1 = a + 1, which is equivalet to b = a + 1. However, this will ever happe, sice the umbers that we have to work with are always odd! I other words, we start with 1, ad after oe butto press, we have two possible outputs, amely 1 ad 3. Applyig D to these two will yield two distict odd results, ad applyig D to them will also yield two distict odd results, ad there ca be o overlap! So after two butto presses, we will have 4 distict odd umbers, startig with 1 ad edig with 7. Clearly this patter persists to 7 presses. At each stage, we get twice as may distict odd umbers, ad after 7 presses, we will have 8 = 56 differet odd umbers, amely the set from 1 to 511. C/1 The distict prime factors of a iteger are its prime factors listed without repetitio. For example, the distict prime factors of 40 are ad 5. Let A = k ad B = k A, where k is a iteger (k > 1). Show that, for every choice of k, (a) A ad B have the same set of distict prime factors. (b) A + 1 ad B + 1 have the same set of distict prime factors. Solutio: (a) Sice B is give as a multiple of A, every prime that divides A also divides B. Coversely, suppose p is a prime that divides B. Sice B = k A, either p divides k or p divides A. If p divides k, the p =. But the p divides A ayway, because A = ( k 1 1). This shows that every prime that divides B also divides A. Sice every prime that divides A also divides B, ad vice versa, A ad B have the same set of distict prime factors. (b) Observe that A + 1 = k 1 ad B + 1 = k ( k ) + 1 = k k + 1 = ( k 1) = (A + 1). Sice B + 1 = (A + 1), every prime that divides A + 1 divides B + 1 ad vice versa. Therefore, A + 1 ad B + 1 have the same set of distict prime factors. D/ I a acute triagle ABC let K, L, ad M be the midpoits of sides AB, BC, ad CA, respectively. From each of K, L, ad M drop two perpediculars to the other two sides of the triagle; e.g., drop perpediculars from K to sides BC ad CA, etc. The resultig 6 perpediculars itersect at poits Q, S, ad T as i the figure to form a hexago KQLSMT iside triagle ABC. Prove that the area of the hexago KQLSMT is half of the area of the origial triagle ABC. B K A Q T L M S C

4 BAMO 016 Problems ad Solutios March 17, Solutio: Costruct segmets KL, LM ad MK. Next, costruct the three altitudes of triagle KLM that meet i its orthoceter, O. Sice KL coects the two midpoits of AB ad BC we kow that KL AB it is easy to see that MT KO LQ sice those lies are perpedicular to a pair of parallel lies. Similarly, KT MO LS ad KQ LO MS. Note that triagle KLM divides the origial triagle ABC ito four cogruet parts, all similar to the origial triagle (ad thus all acute), so the perpediculars we dropped from K, L, ad M are just altitudes of the smaller triagles. Sice all the smaller triagels are acute, all the altitudes will meet iside the respective triagles. Because of all the parallel lies, we kow that OLSM, OMT K ad OKQL are all parallelograms havig diagoals ML, KM, ad LK, respectively. The diagoal of a parallelogram divides it ito two cogruet triagles, so triagle LSM is cogruet to triagle MOL, triagle MT K is cogruet to triagle KOM, ad triagle KQL is cogruet to triagle LOK. If we cosider the hexago KQLSMT to be composed of triagle KLM ad the three outer pieces, we ca see that triagle KLM is composed of three smaller triagles that are cogruet to the correspodig outer pieces, so the area of the hexago is twice the area of triagle KLM. But triagle KLM coects the midpoits of the edges of triagle ABC so each of its sides is half the legth of the correspodig side of triagle ABC, so triagle KLM is similar to triagle ABC, but with 1/4 the area. We previously showed that the area of KQLSMT is twice the area of triagle KLM, so the area of the hexago is 1/ the area of triagle ABC. K B A Q T O L M S C E/3 For > 1, cosider a chessboard ad place pieces at the ceters of differet squares. (a) With chess pieces o the board, show that there are 4 pieces amog them that form the vertices of a parallelogram. (b) Show that there is a way to place ( 1) chess pieces so that o 4 of them form the vertices of a parallelogram. Solutio: (a) Sice there ca be at most pieces that are leftmost i their rows (some rows may be empty), there are at least pieces that are ot the leftmost i their row. Record the distaces (the umber of squares) betwee the leftmost piece ad the other pieces o the same row. There are the at least distaces recorded. Sice the distaces rage from 1 to 1, by the Pigeohole Priciple, at least two of these distaces are the same. This implies that there are at least two rows each cotaiig two pieces that are the same distace apart. These four pieces

5 BAMO 016 Problems ad Solutios March 17, yield a parallelogram. (b) If ( 1) pieces are placed, for example, o the squares of the first colum ad the first row, the there is o parallelogram. 4 Fid a positive iteger N ad a 1,a,...,a N, where a k = 1 or a k = 1 for each k = 1,,...,N, such that a a 3 + a a N N 3 = , or show that this is impossible. Solutio: It is possible, as log as the sum S desired is a multiple of 48, with N = S/6, which i this case is , ad the a k repeats the 8-term patter 1,1,1, 1,1, 1, 1,1. Use the observatio that if f (x) is a degree-k polyomial, the for ay costat h, the differece f (x +h) f (x) will be a degree-(k 1) polyomial. If we iterate this process three times, we ca fid a way to maipulate cosecutive cubes to always get a costat. More precisely, let c 0,c 1,c,c 3,c 4,c 5,c 6,c 7... be cosecutive cubes. I other words, c m = (m + u) 3 for some fixed startig iteger u. The the differeces c 1 c 0, c 3 c, c 5 c 4, c 7 c 6,... will be quadratic fuctios; i.e., if we defie a m := c m+1 c m, the a m is a quadratic fuctio of m (depedig o the parameter u, as well), ad the differeces are a 0,a,a 4,a 6,... Cotiuig, we see that the differeces a a 0, a 6 a 4,... will be a liear sequece; i.e., b m := a m+ a m is a liear fuctio of m (with parameter u), ad our differeces are b 0,b 4,... Fially, the sequece b 4 b 0, b 1 b 8,... is costat, o matter what the parameter u equals! We have b 4 b 0 = a 6 a 4 (a a 0 ) = a 6 a 4 a + a 0 = c 7 c 6 c 5 + c 4 c 3 + c + c 1 c 0, ad sice this is costat, we ca compute it usig ay value of u. Takig u = 3, the costat must equal ( 1) 3 + ( ) 3 ( 3) 3 = 48. I other words, if we defie s u := u 3 + (u + 1) 3 + (u + ) 3 (u + 3) 3 + (u + 4) 3 (u + 5) 3 (u + 6) 3 + (u + 7) 3, the s u = 48 for all values of u. Sice = , we ca easily write as a sum of elemet sum/differeces of = cosecutive cubes:

6 BAMO 016 Problems ad Solutios March 17, = k=0 s 8k+1 = NOTE: About half of the correct solutios were purely computatioal, makig use of the fact that = (( + 1)) /4 ad puttig = 95 gets you to a umber that is rather close to It is the possible to work out with great difficulty, by had values that work. Clearly this method is ot geeralizable, ulike the multiple-of-48 method above. 5 The corers of a fixed covex (but ot ecessarily regular) -go are labeled with distict letters. If a observer stads at a poit i the plae of the polygo, but outside the polygo, they see the letters i some order from left to right, ad they spell a word (that is, a strig of letters; it does t eed to be a word i ay laguage). For example, i the diagram below (where = 4), a observer at poit X would read BAMO, while a observer at poit Y would read MOAB. B A O Y M X Determie, as a formula i terms of, the maximum umber of distict -letter words which may be read i this maer from a sigle -go. Do ot cout words i which some letter is missig because it is directly behid aother letter from the viewer s positio. Solutio: Let us call our origial poits V 1,V,...,V. If A,B are two poits, the viewers o oe side of lie AB see A to the left of B, ad viewers o the other side see B to the left of A. Therefore, if we draw the ( ) lies determied by pairs {V i,v j } (1 i < j ), the differet views of V 1,V,...,V (from outside their covex hull) are i oe-to-oe correspodece with the regios formed outside the covex hull by these ( ) lies. These regios are what we will ow cout. Our strategy is to cout all regios, the subtract those regios that are iside the covex hull of V 1,V,...,V. We begi by statig ad provig a geeral lemma: Lemma. Let K be a covex regio of the plae. Let r lies pass through the iterior of K, makig a total of m itersectios i the iterior of K. Suppose further that o three of the lies meet at oe of these itersectios. The K is divided by the lies ito 1 + r + m regios.

7 BAMO 016 Problems ad Solutios March 17, Proof of Lemma. Number the lies l 1,l,...,l r. We imagie that we draw the lies oe at a time, i this order. Suppose m i is the umber of poits i the iterior of K where l i itersects l 1,l,...,l i 1. The at the stage whe we draw l i, it passes through m i + 1 existig regios, dividig each of them ito two regios ad thereby icreasig the umber of regios by m i + 1. Sice there is iitially oe regio (K itself), the fial umber of regios after all r lies are draw is 1 + (m 1 + 1) + (m + 1) + (m 3 + 1) + + (m r + 1). But observe that m 1 + m + + m r = m, sice each itersectio poit i the iterior of K is couted exactly oce o each side of the equatio. Thus the fial umber of regios is 1 + r(1) + (m 1 + m + + m r ) = 1 + r + m. By the lemma, if ( ) lies are i geeral positio (o two parallel, o three cocurret), the they divide the plae ito 1 + ( (( ) )) + regios. However, our ( ) lies are t i geeral positio; 1 lies meet at each of our origial poits V. Each V i is surrouded by ( 1) regios, but if we udged each lie by a tiy amout so as to separate all their pairwise itersectios, the these ( 1) regios would become ( ) 1 regios (by the lemma). Accoutig for this, the umber of regios formed by our ( ) lies is 1 + ( (( ) )) [ ( + + ( 1) ( ) )] 1. Fially, we subtract the regios iside the covex hull of V 1,V,...,V. The umber of lies passig through the covex hull (i.e. diagoals, ot sides) is ( ). Every set of four poits {V i,v j,v k,v m } (1 i < j < k < m ) determies a uique itersectio iside the covex hull, so there are ( ( 4) such itersectios. Thus by the lemma, the covex hull is cut ito 1 + ( ) + ) 4 regios. Subtractig this from our total cout of regios i the plae, we coclude that the umber of regios outside the covex hull (which is our fial aswer) is 1 + ( (( ) )) [ ( + + ( 1) ( ) )] 1 [ 1 + ( ( ) + 4)]. This may be simplified to 1 1 ( 1)( ). ALTERNATIVELY, there is aother expressio that this is equal to, amely ( ) ( ) +. 4 This more elegat expressio was derived by two studets, but oly oe gave a equally elegat combiatorial solutio that made it easy to see why this couts the umber of regios. For this work, Swapil Garg wo the brilliacy award for 016. We leave it to the reader to discover his argumet.