19th Bay Area Mathematical Olympiad. Problems and Solutions. February 28, 2017

Transcription

1 th Bay Area Mathematical Olymiad February, 07 Problems and Solutions BAMO- and BAMO- are each 5-question essay-roof exams, for middle- and high-school students, resectively. The roblems in each exam are in roughly increasing order of difficulty, labeled A through E in BAMO- and through 5 in BAMO-, and the two exams overla with three roblems. Hence roblem C on BAMO- is roblem # on BAMO-, roblem D on BAMO- is # in BAMO-, and roblem E in BAMO- is # in BAMO-. The solutions below are sometimes just sketches. There are many other alternative solutions. We invite you to think about alternatives and generalizations! A Consider the multilication table below. The numbers in the first column multilied by the numbers in the first row give the remaining numbers in the table. For examle, the in the first column times the in the first row give the (= ) in the cell that is in the rd row and th column. We create a ath from the uer-left square to the lower-right square by always moving one cell either to the right or down. For examle, here is one such ossible ath, with all the numbers along the ath circled: If we add u the circled numbers in the examle above (including the start and end squares), we get. Considering all such ossible aths: (a) What is the smallest sum we can ossibly get when we add u the numbers along such a ath? Prove your answer is correct. (b) What is the largest sum we can ossibly get when we add u the numbers along such a ath? Prove your answer is correct.

2 BAMO 0 Problems and Solutions March, 07 B Two three-dimensional objects are said to have the same coloring if you can orient one object (by moving or turning it) so that it is indistinguishable from the other. For examle, suose we have two unit cubes sitting on a table, and the faces of one cube are all black excet for the to face which is red, and the the faces of the other cube are all black excet for the bottom face, which is colored red. Then these two cubes have the same coloring. In how many different ways can you color the edges of a regular tetrahedron, coloring two edges red, two edges black, and two edges green? (A regular tetrahedron has four faces that are each equilateral triangles. The figure below deicts one coloring of a tetrahedron, using thick, thin, and dashed lines to indicate three colors.) C/ Find all natural numbers n such that when we multily all divisors of n, we will obtain 0. Prove that your number(s) n works and that there are no other such numbers. (Note: A natural number n is a ositive integer; i.e., n is among the counting numbers,,,... Adivisor of n is a natural number that divides n without any remainder. For examle, 5 is a divisor of 0 because 0 5 = ; but 5 is not a divisor of 7 because 7 5 = with remainder. In this roblem we consider only ositive integer numbers n and ositive integer divisors of n. Thus, for examle, if we multily all divisors of we will obtain.) D/ The area of square ABCD is cm. Point E is inside the square, at the same distances from oints D and C, and such that \DEC = 50. What is the erimeter of ABE equal to? Prove your answer is correct. Consider the n n multilication table below. The numbers in the first column multilied by the numbers in the first row give the remaining numbers in the table. n n n n n n n We create a ath from the uer-left square to the lower-right square by always moving one cell either to the right or down. For examle, in the case n = 5, here is one such ossible ath, with all the numbers along the ath circled:

3 BAMO 0 Problems and Solutions March, If we add u the circled numbers in the examle above (including the start and end squares), we get. Considering all such ossible aths on the n n grid: (a) What is the smallest sum we can ossibly get when we add u the numbers along such a ath? Exress your answer in terms of n, and rove that it is correct. (b) What is the largest sum we can ossibly get when we add u the numbers along such a ath? Exress your answer in terms of n, and rove that it is correct. E/ Consider a convex n-gon A A A n. (Note: In a convex olygon, all interior angles are less than 0.) Let h be a ositive number. Using the sides of the olygon as bases, we draw n rectangles, each of height h, so that each rectangle is either entirely inside the n-gon or artially overlas the inside of the n-gon. As an examle, the left figure below shows a entagon with a correct configuration of rectangles, while the right figure shows an incorrect configuration of rectangles (since some of the rectangles do not overla with the entagon): A A A 5 A 5 A A A A A (a) Correct A (b) Incorrect Prove that it is always ossible to choose the number h so that the rectangles comletely cover the interior of the n-gon and the total area of the rectangles is no more than twice the area of the n-gon. 5 Call a number T ersistent if the following holds: Whenever a,b,c,d are real numbers different from 0 and such that a + b + c + d = T and we also have a + b + c + d = T, a + b + c + d = T.

4 BAMO 0 Problems and Solutions March, 07 Solutions (a) If T is ersistent, rove that T must be equal to. (b) Prove that is ersistent. Note: alternatively, we can just ask Show that there exists a unique ersistent number, and determine its value. A The minimum is and the maximum is 50. To see this more easily, tilt the grid 5 degrees: Now every ath must include exactly one number from each row. The smallest and largest numbers in each row are resectively at the edge and in the middle, so the smallest and largest totals are achieved by the aths below: These totals are and 50. B There are nine non-equivalent colorings where two edges are red, two are black, and two are green. In the illustrations below, we will indicate the three colors by drawing edges thick, thin, or dashed. We will artition the colorings into three cases, determined by the number of airs of edges that are colored the same: three, one, or zero (two is imossible). For each case, without loss of generality, we fix the osition of the two thick edges and carefully examine all ossible colorings, and determine which are equivalent. Each air of oosite edges is colored the same. Fixing the two thick edges, there are two ossible configurations shown below.

5 BAMO 0 Problems and Solutions March, 07 5 These two colorings are non-equivalent. To see why, note that the only rotations available are -fold (0-degree) rotations about axes joining midoints of oosite sides, or -fold (0-degree) rotations about axes joining a vertex with the center of the oosite face. The -fold rotations leave both ictures alone (none of them turns one icture into the other), and the -fold rotations do not kee the thick lines in lace. Exactly one air of oosite sides is the same color. For examle, suose that the two oosite sides are colored with thick lines. Keeing this air in lace, there are only two configurations. Note that these two colorings are in fact equivalent, since you can turn one into the other by erforming a -fold rotation about the axis joining the midoint of the two thick edges. Since there are three choices of colors we can use for the air, this case has a total of non-equivalent colorings. No airs of oosite edges have the same color. Again, we will fix two thick edges and carefully count the ossibilities. These are the only ossible choices, once we fix the thick edges, and they are all nonequivalent, since no rotation will kee the two thick edges in lace. Furthermore, the first two tetrahedra each have a thick-thick-thin face, but the first one has a face with edges colored thick-thin-dashed (going clockwise), whereas the second has a thick-thin-dashed face (going counterclockwise). For the next two tetrahedra, both have thick-thick-dashed faces, but the first has a thick-dashed-thin clockwise face, but the last has a thick-dashedthin counterclockwise face. In sum, there are + + = non-equivalent colorings. Brilliancy Prize Alternative Solution: Suraj Mathashery of Kennedy Middle School wrote a nice solution that was the only aer which correctly alied the count all the colorings as a string and divide out by the symmetries method. Many eole correctly counted (exercise) that there are 0 ways to arrange the letters RRGGBB as a string, and then divided this number because of rotations to get an answer. The roblem is that there is no single number one can choose! Here is a sketch of Suraj s method. You are urged to fill in the gas. There are 0 colorings if we don t worry about any rotations. These fall into two categories: The colorings where all three airs of oosite edges are colored the same way, and those colorings where this is not the case.

6 BAMO 0 Problems and Solutions March, 07 The first category has just colorings; the second has 0 =. The first category has just three rotations that change it; while the second category has. Hence the number of non-equivalent colorings is / + / = + 7 =. C/ Solution : Since the rime factorizaton of 0 = 5, the rime divisors of our natural number n are exactly and 5; i.e., n = a 5 b for some integer exonents a and b. When a = b =, we get the number n = 5 = 00, which actually works: the roduct of its divisors is: = ( 00) ( 50) ( 5) 0 = 00 0 = = 0. In fact, n = 00 is the only number that works. Indeed, if in n = a 5 b we have, say, b, we will end u with too many 5s in the roduct of all divisors! To see this, note that we will have at least the divisors 5, 5, 5, 5, 5, and 5. This yields at least = fives in 0, but we have only such 5s, a contradiction. Similarly, in a, we will end u with at least twos in 0, which is also a contraduction. We conclude that a ale and b ale. On the other hand, if, say b =, we will get too few fives in the roduct! Indeed, n = a 5 with a ale, so that the 5s in the roduct will articiate in at most the following divisors: 5, 5, 5; i.e., only at most fives, but we need fives in 0, a contradiction! Similarly, we will get too few s in the roduct if a =. We conclude that a = b = and the only number that works is n = 5 = 00. Solution : If we arrange all divisors of n in increasing order, d < d < d <...<d k, and multily the first and the last divisor, the second and the last but one divisor, etc., we will always obtain n (why?). This means that if we multily the roduct P = d d d d k by itself, we will obtain: P =(d d k )(d d k ) (d k d )=n n n {z } = n k, k where k is the number of divisors of n. We are given that the roduct P = 0, so we conclude that (0 ) = n k, i.e., n k = 0. From here it is easy to see that n must be a ower of 0; i.e., n = 0 c for some natural c such that (0 c ) k = 0. Thus, 0 ck = 0, ck =, and c is a divisor of. One may think that we need now check all cases for c, but there are faster ways to roceed. You may want to check that c = works (n = 00 works as in Solution ), argue that 0 and higher owers of 0 will yield larger than 0 roducts P and that lower owers of 0 (i.e., 0 ) will yield smaller than 0 roducts P. Alternatively, note that n = 0 c = c 5 c means that n will have k =(c + ) (c + ) divisors of n (why?), and hence ck = becomes c(c + )(c + )=, which has only one natural solution of c = (why?). Summarizing, n = 0 c = 0 = 00 is the only ossible solution. D/ Solution : Since the area of square ABCD is = (cm ), then the side of square ABCD is cm. We claim that ABE is equilateral. To rove this, we make a reverse construction, starting from an equilateral ABE 0, building u to square ABCD, and eventually showing that oints E and E 0 coincide. Thus,

7 BAMO 0 Problems and Solutions March, 07 7 Ste. Let E 0 be a oint inside square ABCD such that ABE 0 is equilateral. Hence, in articular, \ABE 0 = 0. Ste. Then \E 0 BC = 0 0 = 0. Ste. Since BE 0 = AB = BC (ABE 0 is equilateral and ABCD is a square), we conclude that CE 0 B is isosceles with BE 0 = BC and \E 0 BC = 0. Ste. This in turn imlies that both base angles of CE 0 B are (0 0 )=75. In articular, \BCE 0 = 75.(Note: Since \BCE 0 < 0, oint E 0 is between arallel lines AB and CD, and hence E 0 is indeed inside square ABCD, and not above CD or outside ABCD.) Ste 5. But then \DCE 0 = 0 75 = 5. Ste. Analogously (or by symmetry) we can show that \CDE 0 = 5. Ste 7. This means that DCE 0 is isosceles with DE 0 = CE 0 and base \DE 0 C = 50. Ste. Thus, E 0 is at the same distances from C and D, \DE 0 C = 50, and E 0 is inside square ABCD. But there is only one such oint with all these roerties, namely, the given oint E. We conclude that oint E 0 is, after all, the same as oint E. To finish off the roblem, we use the fact that ABE 0 is equilateral by construction; i.e., the original ABE is equilateral. Hence its erimeter is equal to AB = = cm. Solution : Our goal will be to confirm that \EBC = 0. For those who know some trigonometry, recall a formula for tan(half-angle): tan a = cos a sin a = ( cosa) ( + cosa) ) tan 5 = cos0 + cos0 = + With the hel of a little bit of algebra, we rationalize the denominator and obtain: ( )( ) ( + = ( ) )( ) = ( ) q ) tan5 =+ ( ) = = + (> 0).

8 ... BAMO 0 Problems and Solutions March, 07 Now back to our roblem. Dro a erendicular from oint E to side BC of the square and mark by H the foot of this erendicular. Since E is at the same distances from C and D, it easily follows that E is half-way between arallel lines AD and BC; i.e., EH = AB = 7 cm. Just as above, from isosceles DEC we know that \ECD = 5. This means that \CEH = 5 (EH DC, CE is a transversal, and \ECD and \CEH are alternating interior angles). Thus, from right EHC we can calculate CH = 7tan5. This imlies that BH = BC CH = 7tan5. We are ready to calculate cot\ebc from right EHB: cot\ebc = cot\ebh = BH EH = 7tan5 7 = tan5 = ( )=. But cot0 =, so the acute \EBC must be 0. We roved what we aimed for! To finish off the roblem, note that \ABE = 0 \EBC = 0 0 = 0. By symmetry, \ABE = 0 and ABE is equilateral. Thus, its erimeter is AB = cm. Solution : It is not hard to comute the ratios of a right triangle without any trig, rovided one knows the ratios of the triangle. Just divide the 75-degree angle into 5- degree and 0-degree angles, dissecting the original triangle into a and Quite elementary from there. The minimum is achieved by staying on the erimeter of the grid, and the maximum by staying as close to the main diagonal as ossible. To see why this is true, tilt the grid 5 degrees:.... Now every ath must include exactly one number from each row. The n th row consists of the numbers k(n + k) for k =,,...,n. As a quadratic function of k, this exression is greatest when k =(n+)/, and gets smaller as k gets farther from (n+)/ in either direction. Thus the smallest number in row n of our tilted grid is the number on either end of the row, which is n, and the largest number in row n is the number closest to the center, which is b n+ c dn+ e. By staying on the erimeter of the grid, we may select the smallest entry from every row, and so accumulate the smallest ossible total. By staying in the middle, we may select the largest entry from every row and accumulate the largest ossible total. All that remains is to calculate these totals in terms of n. The minimum total is ( (n )) + (n + n + n + + n )=( + + +(n )) + n( n) = n(n ) + n = n(n + n ). n(n + )

9 BAMO 0 Problems and Solutions March, 07 The maximum total is (n ) n + n =( n )+( + + +(n ) n) ale n(n + )(n + ) n = n(n + )(n + ) n + = + n(n + )(n + ) (n )(n)(n + ) = + n(n + )(n ) =, and we are finished. For every oint P in the interior of P, let f (P) be the minimum distance from P to a oint on the erimeter of P. Let r denote the maximum value of f (P), which necessarily exists, and let this maximum be achieved at some (not necessarily unique) oint P = I. (We remark that in the case where P is a triangle, I is the incenter.) We claim that if we choose t = r, then the rectangles described in the roblem meet the desired requirements. First, we show that the rectangles cover P. Let Q be any oint in the interior of P, and let R be any closest oint to Q on the erimeter of P; observe that QR = f (Q) ale r. Let R be on side A i A i+ of P. Then we show by contradiction that segment QR is erendicular to A i A i+. For if this is not the case, then let R 0 be the foot of a erendicular from Q to the line A i A i+.we have QR 0 ale QR, so either R 0 = R or R 0 lies outside P on an extension of side A i A i+. But if R 0 lies outside P, then the segment QR 0 must cross the erimeter of P at some oint R 00, in which case QR 00 < QR 0 ale QR, contradicting our choice of R. Thus we have shown that QR is erendicular to A i A i+, and that QR ale r. It follows that Q lies in or on the rectangle A i A i+ B i C i. Hence all oints in the interior of P are covered by the rectangles. Finally, we show that the total area of the rectangles is at most twice the area of P. To see this, draw the segments IA, IA,...,IA n, dissecting P into n triangles: A A I A A 5 A Denoting the area of a olygon by brackets, we have [P]=[A IA ]+[A IA ]+ +[A n IA ]. Let the i th triangle ( ale i ale n) have base b i = A i A i+ and corresonding altitude h i, which is the erendicular distance from I to line A i A i+. The foot of the altitude is either on the erimeter of P or outside P, so we have h i f (I)=r. Hence [P]= b h + b h + + b nh n (b + b + + b n )(r).

10 BAMO 0 Problems and Solutions March, 07 0 The total area of the rectangles A i A i+ B i C i is (b + b + + b n )(r). The above work shows that this is less than or equal to [P], so we are finished. 5 (a) Suose T is ersistent. Observe that for any u satisfying u, the equation x + /x = u has real solutions (because it can be written in the form x ux + = 0, which has discriminant u ). Choose u large enough so that both u and v = T u have absolute value greater than. Then there exist real x,y, not equal to 0 or, such that x + /x = u and y + /y = v. Let a = x, b = /x, c = y, d = /y. Then a + b + c + d = /a + /b + /c + /d = u + v = T, while T = a + b + c + d = x + /x + y + /y = x + x x + y + y y = x x + y y =, and we are finished. (b) Now we will show that is ersistent. Suose a + b + c + d = and a + b + c + d with roots a, b, c, d. Let f (x) have exansion =. Consider the monic olynomial f (x) By Vieta s formulas, we have and f (x)=x e x + e x e x + e. e = a + b + c + d = e bcd + cda + dab + abc = = e abcd a + b + c + d =. Thus, f (x) has the form x x + rx sx + s. Next, we consider the olynomial g(x)= f ( x), which is the monic olynomial with roots a, b, c, d. We have By Vieta s formulas again, we have Therefore, is ersistent. g(x)=( x) ( x) + r( x) s( x)+s = x x + rx +( r + s)x ( r + s). a + b + c + ( r + s) = d ( r + s) =.