Russian Math Circle Problems

Transcription

1 Russia Math Circle Problems April 1, 2011 Istructios: Work as may problems as you ca. Eve if you ca t solve a problem, try to lear as much as you ca about it. Please write a complete solutio to each problem you solve, as if you were eterig it ito a math cotest ad had o ability to explai it to the grader. This will help you make sure that you ve thought of all the possibilities. 1. Suppose S is a set ad is a biary operatio o S. (This meas that for ay two elemets x ad y i S, the the elemet x y is also i S.) The oly thigs we kow about are these (i particular, is ot ecessarily associative): Show that x y = y x for all x, y S. x x = x for all x S (x y) z = (y z) x for all x, y, z S 2. The umber 26! = eds with a log strig of zeroes. Let N be the umber that remais after all those zeros are removed from the ed. Fid the largest value of k such that 12 k divides N. 3. Suppose two circles with equal radius r are iscribed as illustrated i the right triagle below havig side legths of 3, 4 ad 5. Fid r. 4. Fid the umber of ways to choose 1005 umbers from the set {1, 2, 3,..., 2009, 2010} i such a way that the sum of ay two chose umbers is either 2010 or Coi turig problems. I every problem below there are cois placed o a table with heads showig. Each move requires you to tur over k differet cois. The goal is to fiish with all showig tails. For each situatio below, show how to accomplish the goal or prove that it is impossible. (a) = 7 ad k = 2. (b) = 5 ad k = 3. (c) = 6 ad k = 4. (d) What ca you say about geeral ad k?

2 Russia Math Circle Problems April 1, 2011 Istructios: Work as may problems as you ca. Eve if you ca t solve a problem, try to lear as much as you ca about it. Please write a complete solutio to each problem you solve, as if you were eterig it ito a math cotest ad had o ability to explai it to the grader. This will help you make sure that you ve thought of all the possibilities. 1. Compute: (111, 111, 111, 111)(1, 000, 000, 000, 005) Let be a positive iteger. (a) If is a perfect square, show that + 1 is the sum of two perfect squares. (b) If is a perfect square, show that + 1 is the sum of three perfect squares. 3. I the followig figure, let the circle k cetered at O be perpedicular to the lie l at B with AOB beig a diameter of k. Choose a poit M o k ad exted AM to itersect l at C. Costruct a lie taget to k at M ad let that lie itersect l at D. Show that BD = DC. A k O M l B D C 4. Fid a set of positive itegers a 1, a 2, a 3,...a such that a 1 + a 2 + a a = 1000 ad such that the product a 1 a 2 a 3 a is as large as possible. 5. Suppose we begi with a 7 7 square grid of poits as i the figure below. (a) How may squares ca be formed whose vertices are grid poits? Remember to iclude squares like the oe i the figure. (b) What if we begi with a grid?

3 Russia Math Circle Solutios April 1, Suppose S is a set ad is a biary operatio o S. (This meas that for ay two elemets x ad y i S, the the elemet x y is also i S.) The oly thigs we kow about are these (i particular, is ot ecessarily associative): Show that x y = y x for all x, y S. Solutio: From the first rule, we have: x x = x for all x S (x y) z = (y z) x for all x, y, z S x y = (x y) (x y). We will ow apply the rules above repeatedly to covert the term o the right gradually to (y x): (x y) = (x y) (x y) = (y (x y)) x = ((x y) x) y = ((y x) x) y = ((x x) y) y = (x y) y = (y y) x = y x. 2. The umber 26! = eds with a log strig of zeroes. Let N be the umber that remais after all those zeros are removed from the ed. Fid the largest value of k such that 12 k divides N. Solutio: I 26! there are six factors of 5, twety three factors of 2 ad te factors of 3. There will be six zeros at the ed of 26! ad removig them will leave sevetee factors of 2 ad still te factors of 3. Each factor of 12 i the remaiig umber requires two factors of 2 ad oe of 3, so we are limited by the sevetee factors of 2. Thus the umber will be divisible by 12 8 ad ot 12 9, so k = Suppose two circles with equal radius r are iscribed as illustrated i the right triagle below havig side legths of 3, 4 ad 5. Fid r.

4 Solutio: Drop perpediculars to the sides of the triagle, coect the ceters of the iscribed circles, costruct lies through the ceters of the circles that meet to form a triagle ad label the poits as show below: C E D H F J I A G B Let CD = h 1, DA = h 2, BG = l 1, GA = l 2. Obviously, CD = CE ad BG = BF. We also ca see that HJ = h 2 r, that IJ = l 2 r, that HI = 2r ad that HIJ CBA. From the diagram ad the similarity of the large ad small triagle, we have the followig equatios: 5 = l 1 + h 1 + 2r (1) 3 = h 1 + h 2 (2) 4 = l 1 + l 2 (3) 4 = l 2 r 3 h 2 r (4) 5 2r = 3 h 2 r (5) We ow have five equatios ad five ukows, ad with a bit of relatively easy algebra, we ca calculate the value of r. From Equatio 5 we obtai h 2 = (11/5)r. From Equatio 3 we fid l 2 = 4 l 1 ad we ca substitute for l 2 ad h 2 i Equatio 4 to obtai: 4( 11 r r) 5 = 3(4 l1 r) 39 5 r = 12 3l 1 From Equatio 2 ad the value of h 2 i terms of r, we have: l 1 = 4 13 r. (6) 5 h 1 = 3 h 2 = 3 11 r. (7) 5 Substitutig the values of h 1 ad l 1 from Equatios 6 ad 7 ito Equatio 1, we obtai: 5 = r r + 2r 2 = 14 5 r 5 = r. 7

5 4. Fid the umber of ways to choose 1005 umbers from the set {1, 2, 3,..., 2009, 2010} i such a way that the sum of ay two chose umbers is either 2010 or Solutio 1: Let s form 1005 pairs: {1, 2010}, {2, 2009}, {3, 2008},..., {1005, 1006}. Clearly, a set of chose umbers must cotai exactly oe umber from each of these pairs. Let S be a chose set, ad suppose that x 2009 is a elemet of S. The 2010 x is ot chose, ad hece the secod umber i the pair {2010 x, y} must be chose. But it s obvious that y = x +1. Therefore, x +1 is also i S. Let be the smallest umber i S. If 1005, repeatig the above argumet we ca see that S must cotai, + 1,...,1005, ad the larger of the two elemets of each of the pairs {1, 2010}, {2, 2009},..., { 1, 2012 }, so that: S = {, + 1, + 2, , 2010, 2009,..., 2012 }. Thus we have exactly 1005 such sets for = 1, 2,..., The oly other possible solutio is: Hece the aswer is Solutio 2: S = {1006, 1007,..., 2010}. We ll illustrate the solutio with a smaller example, but oe with the same characteristics. Suppose the set just cotaied the umbers {1, 2, 3,..., 8} ad we wat to fid subsets of 4 of them such that o pair i the subset adds to either 8 or 9. I the diagram below all the umbers are listed, ad two umbers are coected by a lie segmet if the two caot appear i the same set. For example, 6 is coected to both 2 ad 3 sice ad yield 8 or The umbers 8 ad 4 are special i that they are the oly umbers that elimiate oly oe other possibility. There are 7 lie segmets i the drawig ad if we do ot choose either 4 or 8, our 4 chose umbers each will have 2 lies comig out which will require 8 total lies ad there are oly 7 available. Thus we are required to choose either 4 or 8 i our set. If we choose, say, 4 (choosig 8 is symmetric), the 5 caot be i the set, ad we might as well erase the lie from 3 to 5, so we have a zig-zag patter that is idetical, but two umbers shorter, from which we eed to choose 3 umbers. The same argumet applies: we eed to choose oe from oe of the two remaiig eds, et cetera. We ca see that ay fial set must cotai k (possibly zero) umbers startig from the 8 ad cotiuig right alog the lower lie ad (4 k) umbers begiig with 4 ad cotiuig left alog the upper lie. For this situatio, there are solutios for k = 0, 1, 2, 3, 4, so there will be exactly 5 subsets: {8, 7, 6, 5}, {8, 7, 6, 4}, {8, 7, 3, 4}, {8, 2, 3, 4}, {1, 2, 3,4}. The same argumet ca be used for the origial problem, ad there are 1006 subsets satisfyig the coditios. 6 5

6 5. Coi turig problems. I every problem below there are cois placed o a table with heads showig. Each move requires you to tur over k differet cois. The goal is to fiish with all showig tails. For each situatio below, show how to accomplish the goal or prove that it is impossible. (a) = 7 ad k = 2. (b) = 5 ad k = 3. (c) = 6 ad k = 4. (d) What ca you say about geeral ad k? Solutio: This is a ope-eded problem, so this solutio is ot complete, but here are some ideas: (a) This is impossible, sice turig two cois will leave the umber of heads odd. The goal is zero heads which is ot odd. This is a example of a ivariat: The umber of heads remais odd o matter which two cois are tured. (b) The ca be doe. Tur 1, 2, 3, the 3, 4, 5, the 1, 2, 4. The et result leaves all the cois heads except we have tured over coi 5. Thus we have a scheme that will tur over exactly oe coi. By reumberig the cois, we ca repeat the process five times ad tur all of them. This is clearly ot the most efficiet way to do it, but it does provide a proof that the result is possible. (c) Tur 1, 2, 3, 4, the 2, 3, 4, 5, ad fially 2, 3, 4, 6. (d) This is the ope-eded part. Here are some strategies to cosider: (a) Parity, as used i part (a). (b) Give ay sequece of moves, the et result will be that some of the cois are tured ad some will remai the same. If you ca fid a sequece that leaves all the cois except oe the same, you ca evetually tur all of them, as used i part (b). (c) The previous rule shows that if you ca fid a way to tur exactly oe coi for some pair (, k), the you ca solve the situatio for ay (m, k), where m. For example, with = 4 ad k = 3, the sequece 1, 2, 3, the 2, 3, 4, the 1, 2, 4 turs oly coi 2, so by choosig differet sets of 4 cois to apply this method, every set of cois with 3 ca be tured by some sequece of 3-at-a-time turs. Here is a complete solutio, thaks to Julia Ziegler Hut: We proceed uder the atural assumptios that k, Z + ad k, as otherwise the problem would make little sese. If k =, the it is clearly possible to complete our objective i a sigle move. Now, suppose that k, so that k + 1. Suppose that we label two of our cois a ad b. The pick k 1 other cois (which we ca do sice k + 1 ), ad label them x 1,...,x k 1. The perform two moves: first flip the cois labeled a, x 1,..., x k 1, the flip the cois labeled b, x 1,...,x k 1. The et result of performig the two moves is that a ad b each got flipped, ad all of the other cois remaied the same, so we ca, i two moves, flip a arbitrary pair of cois. This suggests that we must take the parity of ad k ito accout. I our first case, we suppose that is eve. The we ca split the cois ito 2 pairs of cois, the flip each pair of cois idividually by the process described above. I the secod case, is odd ad k is eve. Here, we caot achieve our objective (all cois flipped to tails), sice each time we flip a coi, the umber of tails chages from eve to odd or odd to eve, so that flippig k cois (a eve umber) preserves the parity of the umber of tails, so that there is always a eve umber of tails. But is odd, so our goal is to ed up with a odd umber of tails, which is impossible.

7 Our third case is where ad k are both odd. Here, we ca first flip k cois, edig up with a eve umber of cois left to flip, the splittig these ito pairs ad flippig each pair idividually. This solves the origial problem, as stated. We ca actually figure out the miimum umber of moves required to achieve our goal with oly a little difficulty, so we shall do so. First, a lemma similar to the oe ivolvig flippig ay give pair of cois: if, after two moves, exactly q cois have bee flipped, the q = 2r ad k +r for some r k, ad ay 2r cois with r k, k +r ca be flipped i exactly two moves. For the first part, suppose that the cois flipped by each move had exactly s cois i commo. The, lettig r = k s, we have that those s cois remaied the same, the other r flipped by each move chaged, ad the cois ot affected by either move remai the same. Thus there were exactly 2r cois flipped, there were s + 2r = k + r cois that were flipped by oe of the moves so that k + r, ad s 0 so r k. For the secod part, suppose we are give 2r cois which are to be flipped, with r k, k + r. The split the 2r cois ito two sets of r cois, ad pick some other k r cois (which ca be doe sice r k ad k + r ). The, for our first move, we flip the first set of r cois ad the k r cois, ad for our secod, we flip the secod set of r cois ad the k r cois. These are both valid moves, ad they have the et result of flippig the give 2r cois. Thus the lemma is proved. Now, to compute the miimum umber of moves, we ufortuately must break ito a umber of cases (fortuately, most of them are fairly simple). Case 1: If is odd ad k is eve, ad i this case we proved above that we ca t achieve our goal, so the miimum umber of moves is. Case 2: If = k, here the miimum umber of moves is 1. Case 3: If k < < 2k, we will split this case ito two sub-cases. We clearly ca t do this i oe move, ad if we could do it two moves, the we would have = 2r, r k, k + r, but the = 2r k + r, so r = k ad = 2k, a cotradictio. Case 3a: If ad k have the same parity, we ca do it i three moves, sice we ca first flip some k cois, be left with k cois to flip, which is a eve umber with k 2 k ad k + k 2, so we ca fiish i aother two moves. Case 3b: If, istead, k is odd ad is eve, the the same parity argumet used to show that odd, k eve shows that it ca t be doe i a odd umber (such as three) of moves (the parity of the umber of tails chages every move, so after a odd umber of moves it would be odd, but is eve). Thus, we ca cosider a solutio as a sequece of Moves, each of which cosists of two ormal moves. Each Move flips ay umber of the form 2r with r k cois, so the smallest umber of Moves is achieved by doig as may Moves that flip 2( k) cois as possible, the perhaps oe more Move to flip the remaiig cois. 2( k) This umber is give by miimum umber of moves is 2, where x is the uique iteger such that x 1 < x x, so the. 2( k) Case 4: If 2k, ad i this situatio we must cosider three sub-cases, depedig o the parities of ad k. First, though, i ay case, the miimum umber of moves is at least k, sice each move ca flip at most k cois from heads to tails. Case 4a: If ad k are both eve, the we flip k heads to tails each move util there are at most 2k heads left, at which poit, by the lemma (sice the umber of heads remaiig is eve ad 2k k + r), we ca fiish i two more moves, for a total of 2k k + 2 = k moves.

8 Case 4b: If is eve ad k is odd, the the umber of moves must be eve (see above), so we split ay solutio ito two-move Moves, ad try to solve for the miimum umber of Moves. A Move ca flip ay eve umber 2k of cois (sice k + r 2k for all r k), so we miimize the umber of Moves by doig 2k 1 Moves that flip cois, ad oe more Move that flips somewhere betwee 2 ad 2k cois, for a total of 2k Moves, or 2 2k moves. Case 4c: If ad k are both odd, we do a sigle move, so that we eed oly flip k more cois, ad the, by idetical logic to the previous case, this takes a additioal 2 k 2k moves, for a total of 2 k 2k + 1 moves. The results ca be summed up i the followig table: eve odd 1 if = k k eve 3 if k < < 2k k if 2k odd 1 if = k 4 if k < < 2k 2 2k if 2k 1 if = k 3 if k < < 2k 2 k 2k + 1 if 2k

9 Russia Math Circle Solutios April 1, Compute: (111, 111, 111, 111)(1, 000, 000, 000, 005)+ 1. Solutio: Let = 1, 000, 000, 000, 000. The we ca re-write the expressio above as follows: ( 1) ( + 5) + 1 = 1 ( 1)( + 5) Let be a positive iteger. = = 1 3 ( + 2) 2 = ( + 2)/3 = 333, 333, 333, 334. (a) If is a perfect square, show that + 1 is the sum of two perfect squares. (b) If is a perfect square, show that + 1 is the sum of three perfect squares. Solutio: (a) Sice 2+1 is odd, it must be the square of a odd umber, say (2k+1) 2 = 2+1. The 4k 2 +4k+1 = 2 + 1, or 2k 2 + 2k =. Thus + 1 = (k 2 ) + (k 2 + 2k + 1), or + 1 = (k 2 ) + (k + 1) 2 : the sum of two squares. (b) Sice 3+1 = x 2 is a perfect square, the x ±1(mod 3), so x = 3m±1. Thus 3+1 = (3m±1) 2 = 9m 2 ± 6m + 1. We have = 3m 2 ± 2m, so + 1 = 3m 2 ± 2m + 1 = m 2 + m 2 + (m 2 ± 2m + 1) = m 2 + m 2 + (m ± 1) I the followig figure, let the circle k cetered at O be perpedicular to the lie l at B with AOB beig a diameter of k. Choose a poit M o k ad exted AM to itersect l at C. Costruct a lie taget to k at M ad let that lie itersect l at D. Show that BD = DC. A k O M l B D C Solutio: Costruct the lie BM. Sice the lies DB ad DM are taget to the circle from D, we have DM = BD. This makes BDM isosceles, so DBM = DMB. Sice AMB is iscribed i a semicircle, it is a right agle, so BMC is a right triagle. Thus DBM + MCD = 90 ad

10 DMB + DMC = 90, so a little algebra gives us MCD = DMC. So CDM is also isosceles, ad DM = DC. Sice DM = DC ad DM = BD we have BD = DC. 4. Fid a set of positive itegers a 1, a 2, a 3,...a such that a 1 + a 2 + a a = 1000 ad such that the product a 1 a 2 a 3 a is as large as possible. Solutio: Sice we are lookig for a large product, it is poitless to have ay of the a i = 1. We ca reduce by 1 ad add that value of 1 to ay of the other umbers to leave the sum the same ad icrease the value of the product, so we kow that for all i, we have a i 2. Similarly, if ay of the a i 5 we ca obtai a larger product by breakig it as close to i half as possible, thus i the maximum product, for all i we have 2 a i 4. Ay solutio cotaiig a 4 is equivalet to a solutio with two 2 s, so ay optimal solutio ca be made to cotai oly 2 s ad 3 s. If there are three 2 s i a product, we ca do better if we replace them with two 3 s, sice < 3 3. Thus we seek a solutio with as may 3 s as possible. We could have up to 333 of them, but that leaves a extra 1, so the best solutio is either 332 copies of 3 ad either oe 4 or two 2 s. 5. Suppose we begi with a 7 7 square grid of poits as i the figure below. (a) How may squares ca be formed whose vertices are grid poits? Remember to iclude squares like the oe i the figure. (b) What if we begi with a grid? Solutio: For every k k square block of vertices, there are exactly k 1 squares (tilted ad ot) that will fit exactly i that square ad i o smaller square. See the figure below. I a grid, we eed to fid out how may k k square grids fit iside ad are aliged with the edges of the larger grid. These are determied by the positio of the upper-left corer of the smaller squares, ad it s easy to see that there are ( k + 1) 2 upper-left corers. To fit ay squares iside, k has to be at least two. Thus, the total umber of squares that ca be draw, tilted or ot, is give by: 1 ( 1) ( 2) ( 3) ( 1) 1 2.

11 This formula ca be prove to be equal to 2 ( 2 1)/12. For = 7, this yields: = 49 48/12 = 196.