USA Mathematics Talent Search

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1 (a) Since x and y are 3-digit integers, we begin by noting that the condition 6(x y) = (y x) is equivalent to 6(1, 000x + y) = 1, 000y + x = 5, 999x = 994y, which (after factoring out a 7 by long division) can be written as x y = 994 5, 999 = = Clearly, x = 142 and y = 857 satisfy this equation, and since they are both 3-digit integers, we are done for this case. (b) Since x and y are 9-digit integers, we begin by noting that the condition 6(x y) = (y x) is equivalent to 6(1, 000, 000, 000x + y) = 1, 000, 000, 000y + x = 5, 999, 999, 999x = 999, 999, 994y, which (after factoring out a 7 by long division) can be written as x y 999, 999, , 857, , 857, 142 = = = 5, 999, 999, , 142, , 142, 857 Clearly, x = 142, 857, 142 and y = 857, 142, 857 satisfy this equation, and since they are both 9-digit integers, we are done in this case as well. Page 1 of Problem 1

2 Let ABC be the triangle, with AB = AC, and let h a denote the length of the altitude from A (which is of course perpendicular to BC). Additionally, let x denote the ratio of area to perimeter. Given the task to find only three such triangles, we begin our investigation with a few observations. If we are able to find an isosceles triangle such that AB, BC, CA, h a are all integers, and c x = 6, for some integer c, then there will be an isosceles triangle with sides c AB, c BC, c CA satisfying the given conditions. Indeed, increasing every side of ABC by a factor of c will increase the area by a factor of c 2, the perimeter by a factor c, and hence x by a factor c 2 /c = c, giving us c x = 6, and consequently an isosceles triangle with integer sides. We carry out our task by using the simplest and right triangles. Fortunately, the three following attemps all produce a desired triangle. For our first triangle, we set AB = AC = 5 and h a = 3, so that we have BC = 2 4 = 8. Thus the triangle has an area of 12, perimeter of 18, and the ratio x = 2/3. Now we note that c = 9 gives us c x = 6, so from our previous observations, we get one triangle with sides AB = AC = 9 5 = 45 and BC = 9 8 = 72. For our second triangle, we set AB = AC = 5 and h a = 4, so that we have BC = 2 3 = 6. Thus the triangle has an area of 12, perimeter of 16, and the ratio x = 3/4. Now we note that c = 8 gives us c x = 6, so from our previous observations, we get another triangle with sides AB = AC = 8 5 = 40 and BC = 8 6 = 48. For our third triangle, we set AB = AC = 13 and h a = 5, so that we have BC = 2 12 = 24. Thus the triangle has an area of 60, perimeter of 50, and the ratio x = 6/5. Now we note that c = 5 gives us c x = 6, so from our previous observations, we get our final triangle with sides AB = AC = 5 13 = 65 and BC = 5 24 = 120. Page 1 of Problem 2

3 We begin by investigating the last two digits of s n to search for a pattern. We compute the first 21 values (by hand) using the recursion s n+1 = 3s n + 1, as shown below: Last two digits of s n n s n From the table, we observe s 21 and s 1 have the same last two digits. More generally, we notice that the last two digits of s n are periodic, with period 20. This can be justified by using the initial observation, along with the recursion to obtain s n+20 s n (mod 100). Since s n ends in at least two identical digits if and only if n 18, 19, or 20 (mod 20), we must prove that the amount of identical digits is the same for each of these three consecutive values of n. Because no other values of n allow s n to end in at least two identical digits, it suffices to prove the following: If s n ends in k fours (k 2), then s n+1 ends in k threes, and s n+2 ends in k zeros. Let s n = p 10 k i, where p is a positive integer that does not end in 4. Thus, we have s n+1 = 3s n + 1 = 3p 10 k i + 1 = 3p 10 k i i + 1 = 3p 10 k + k i=1 10 i i + 1 = (3p + 1) 10 k i Now we note that if 3p (mod 10), then 3p 2 12 (mod 10) p 4 (mod 10), since 3 and 10 are relatively prime. However, this is false by hypothesis, so it follows that s n+1 ends in k threes if s n ends in k fours, with k 2. Similarly, we prove that s n+1 ending in k threes implies that s n+2 ends in k zeros. Let s n+1 = q 10 k i, where q is a positive integer that does not end in 3. We proceed as follows: Page 1 of Problem 3

4 s n+2 = 3s n = 3q 10 k i + 1 = 3q 10 k i = 3q 10 k + 10 i + 1 k 10 i 10 i + 1 i=1 = (3q + 1) 10 k As before, we note that if 3q (mod 10), then 3q 1 9 (mod 10) q 3 (mod 10), since 3 and 10 are relatively prime. However, this is false by hypothesis, so it follows that s n+2 ends in k zeros, as desired. This completes our (not-so-elegant) proof. Page 2 of Problem 3

5 Let K and L be the intersection of BC with IJ, and DE with IJ, respectively. Let M be the foot of the perpendicular from K to BR. Additionally, let a side of one of the unit squares be s, and denote P H = x, so HS = 26 x. Q B M R K C L D A E J G F I P H S From right triangle AF E, and the Pythagorean Theorem, we get AE 2 = 25s 2 + s 2 = 26s 2. From right triangles AGH and HDE, we get AH 2 = HE 2 = 4s 2 + 9s 2 = 13s 2. Now we note that AH 2 + HE 2 = 26s 2 = AE 2, so AHE is an isosceles right triangle. Similarly, we can use right triangles ABK and KLE to get AK 2 = KE 2 = 13s 2, so AKE = AHE, and it follows that AKEH is a square. Since AKEH is a square, we note that the distance from K to P S is the same as the distance from A to RS, which is P S = 26. Thus, we deduce that MK = = 2. Other observation are that AP = HS, P H = SE, and QA = 28 AP = 28 (26 x) = x + 2. We also have MBK = 90 QBA = QAB, so KBM BAQ. Because AB/BK = 3/2, it follows that QB = 3. And now, we can set up an equation to solve for x by using right triangles QAB, AGH, and AP H as follows: (26 x) 2 + x (x + 2) 2 = AP 2 + P H 2 QB 2 + QA 2 = AH2 AB 2 = 9s2 + 4s 2 9s 2 = 13 9 = 9(2x 2 52x + 676) = 13(x 2 + 4x + 13) = 5(x 2 104x ) = 0 = 5(x 91)(x 13) = 0 And since x < 26, we conclude that x = 13, so H is (remarkably) the midpoint of P S. Finally, we note that the area of ABCDEF GHIJ is equal to 13s 2 (which is in turn to equal to AH 2, the area of square AKEH!). Thus, 13s 2 = AH 2 = AP 2 + P H 2 = = 338. Page 1 of Problem 4

6 Construct a line through F, parallel to BC, and let it intersect plane BDA at point F. Since BC DE, F lies on the plane DEF. Now, let us focus on plane BDA and employ methods of two-dimensional geometry. Let D be the foot of the perpendicular from D to AB, and denote the midpoint of AB by X. Additionally, let the foot of the perpendicular from D to BF be Y. Note that F D = F B are tangents to the circle, since F lies in the planes BF C and DEF. This, along with BX = DX Create PDF with GO2PDF for free, if you wish to remove this line, click here to buy Virtual PDF Printer implies that F XB = F XD. We note that F X = BX 2 + F B 2 = = 5 5, since F B is a radius of the semicircle with diameter BC = Letting θ = F XB = F XD, we have, 2 cos θ = = cos 2θ = 2 cos 2 θ 1 = = cos (180 2θ) = 1 5 and since DXD = 180 2θ, it follows that XD = 2, so DD = 50 2 = 4 3. In particular, we observe that BXD is obtuse, so D lies on segment DA. Next, we note that BY DD is a rectangle (by construction) so BY = DD and BD = Y D. We will use these results later on. Note that DEF is isosceles (by symmetry of ABC), and since AD = 4 2, by symmetry and similar isosceles triangles (projecting DE onto ABC), we deduce that DE = 4 3. Now, let h denote the altitude of DEF. By using the perpendicular bisector of DE (parallel to plane ABC) and a perpendicular from F to BC, we can calculate h by using the Pythagorean Theorem. The right triangle has one leg of length 1 DE BC = 3 5 and another of length F Y = = 3. Thus, we get h = = 4 3, and consequently the area of DEF is 1 2 h DE = = 24. Page 1 of Problem 5