Vermont Talent Search April, 0 School Year 00-0 Test 4 Solutions Problem. Find the area of a triangle whose medians have lengths of 39, 4 and 45. Let M be the center of gravity or centroid of the triangle. Medians intersect at a point 3 the distance from the vertices. Let D bisect AM and E bisect CM. In ACM, PD and PE are midlines, i.e. join midpoints of sides, thus PD = 4 and PE = 3. DMP and EMP are 3,4,5 triangles and their areas are 84. (you can verify using Hero s formula) Since D is a midpoint, the areas of ADP and DMP are equal. Thus each of the 4 triangles in ACM has area 84. Similar results occur in BCM and ABM. Therefore, area of ABC = 3 4 84 = 008. Problem. 3 In quadrilateral ABCD, cot A = 4, cot B =, cot C = 5. Find all possible values for cot D. Let a = tan A, b = tan B, c = tan C and d = tan D Then a =, b = and so on. 4 3 We have A B C D π tan A + B + C + D = 0. Writing A B C D + + + = and hence ( ) + + + as ( A + B) + ( C + D) and applying tan ( x y) yields: ( A B) ( C D) tan x + tan y + = tan x tan y tan + + tan + = 0. Now applying the sum relation to both sums gives: a + b c + d + = 0. Now cross multiply to get: a + b acd bcd + c + d abc abd = 0 ab cd
a + b + c abc Solving for d gives d =. Inserting values for,, ab + ac + bc a b c yields 65 5 d = = 39 3 3 and hence cot D =. 5 Problem 3. In ABC, AB = 4, BC = 5 and AC = 6. Equilateral triangles ABD and CBF are drawn exterior to triangle ABC. CD and AF are drawn. Find the sum of X + Y where X = ACD and Y = CAF DBC = ABF ;( B + 60 ) BD = BA = 4 and BF = BC = 5 BCD ABF by SAS and therefore = and since vertical angles are equal, BFQ PCQ. o Thus FBQ = CPQ = 60 But CPQ is an exterior angle for ACP and X + Y = 60 o Problem 4. 5 + 3 + 5 3 = + 0 0. f n + f n = kf n where k is an integer, find k. Given f ( n) If ( ) ( ) ( ) From the given equation we have 5 + 3 + 5 3 f ( ) = + 0 0 n 5 + 3 + 5 3 f ( ) = + 0 0 5 3 5 5 3 5 kf ( 0) k + k = + 0 0 The equation is true for all n, so let 0 Therefore, factoring out terms we have.. n =, then f ( ) f ( ) kf ( ) n 0 = 0.
5 + 3 + 5 + 5 3 5 k k 0 0 + = 0 For the equation to be zero, the quantities in the brackets must be zero. Therefore + 5 k = 0 + 5 Thus ( )( k ) + 5 = 0 and k = or ( ) k ( ) 6 + 5 4 + 5 = 0 Problem 5. A circle is tangent to leg AC of right ABC and intersects the hypotenuse AB at E and leg BC at F. Point B is on the circles circumference. AC and BC have integral lengths and AC > BC. If AE = 4 and BE = find the radius of the circle. Draw radii OD, OF, and OB. OD AC. Draw DF. Let DOF = BFO = θ Note that AED ADB. Thus ( AD) = AE AB = 4 5 = 00 and AD = 0 The only right 's with hypotenuse of 5 are (7,4,5) and (5,0,5) and (7,4,5) is not possible. Thus AC=0, CD=0 & BC=5. Note also that DCF DCB ; thus 0 CF =. Hence 3 400 0 3 DF = 00 + = 9 3
Now use Law of Cosines in triangle DOF : ( ) 8r 300 DF = r + r r cosθ 5 and cosθ =. But in triangle BOF, which is also isosceles, cosθ =. 8r 6r Equating these relationships and simplifying gives 8r 75r 300 = 0 (3r + 0)(6r 65) = 0 65 and r = 6
Problem 6. Two numbers from the set S = {,,3,...,06} are selected at random and multiplied. What is the probability that the product is a multiple of 5. Any number that is a multiple of 5 will produce a product that is a multiple of 5. There are numbers is S that are multiples of 5 and 85 that are not multiples of 5. Probability first number selected is multiple of 5 is 06. If there is no replacement, the second number is immaterial. 85 Probability first number not multiple of 5, but second one is equals In this case, the probability product is a multiple of 5 equals: 06 05 85 9 + = 06 06 05 53 With the first number replaced, the probability of either number being a multiple of 5 is 06. 85 Equivalently, the probability of neither number being a multiple of 5 is 06. Therefore 85 40 the probability the product is a multiple of 5 is = 06 36 Note: Due to the inexact problem statement, either answer or both receive full credit. Problem 7 Let a, b, c, and d be positive real numbers such that log a b and log = 4 b. Find the numerical value of the product abcd. d a From the given information we can say as follows: c d 3a 4b a = b, b = c, c = d, and d = a Thus the product 4 4 3 4 4 d b b c d a b d d + b+ c + (3 a) + abcd = d a b c = d b b d = b d = a c ( cd + c) ( ab+ 3 a) cd c ab 3a cd ab abcd = a c = a a c c = a b c d cd ab ac = a c so cd = or cd = and ab = so ab = Problem 8. Find the sum S st positive integers such that 0 < s < t and s + t >. = c, log = d, log = 3a, b c abcd = 4 where S is the collection of all ordered pairs (, ) c d s t of relatively prime
The following solution will show by mathematical induction that S st = for any n. Basic Step: Show that is true for the base value n =. Clearly the only term in the st S sum is ( s, t ) = (, ) in which case the sum is. Inductive Step: Assume that S st = for ν = n and prove that S st = for ν = n +. Specifically, the inductive step goes from adding all terms with 0 < s < t n and s + t > n to instead adding all terms with 0 < s < t n + and s + t > n +. In doing so all terms with s + t = n + are removed and all terms with t = n + are added. We claim the net effect of doing this is zero. To see this we need to show that For ν < n + ν = EQ. ν n + ν n + s ( ) s n ( ) gcd( ν, n+ ν ) = gcd(, + ) ν < n+ ν where the LHS of EQ is the sum of the removed terms and the RHS is the sum of the additional terms. Also, observe that ν and n + ν are relatively prime if and only if ν and n + are relatively prime. So we are summing over the same range. Writing the summand on the LHS as = + ν ( n + ν ) n + ν n + ν sums gives the LHS as + ν n + n + n + ν ( ) n ( )( ) gcd( ν, n+ ν ) = gcd( ν, + ν ) = ν < n+ ν ν < n+ ν and splitting into two Now interchange ν and n + ν in the second sum which gives just which is the missing half of the sum on the other side. ( n + ) ν gcd( ν, n+ ) = ν > n+ ν So the net effect of going from n to n + is zero as claimed. The following spreadsheet shows this in the case of 0 < s < t and s + t >.
N= N=3 S T S+T (ST) - S T S+T (ST) - 3 0.083333 3 4 0.07693 3 0.045455 3 5 0.03846 6 3 4 0.030303 33 3 3 6 0.0564 39 3 3 0 3 0.033333 30 3 4 0.030303 33 4 4 5 0.077 44 4 3 7 0.093 5 5 4 9 3 0.07778 36 4 5 0.077 44 6 5 7 0.06667 60 5 3 8 0.05385 65 5 7 0.06667 60 7 5 6 0.088 55 5 6 0.088 55 8 5 9 4 0.0 45 5 9 4 0.0 45 9 5 8 3 0.05 40 6 3 9 0.08 78 0 6 7 0.055 66 6 7 0.055 66 6 7 3 0.038 4 7 3 0 0.00989 9 7 9 0.0905 84 7 9 0.0905 84 3 7 8 0.0987 77 7 8 0.0987 77 4 7 0 7 0.0486 70 7 0 7 0.0486 70 5 7 9 6 0.05873 63 7 9 6 0.05873 63 6 7 8 5 0.07857 56 7 8 5 0.07857 56 7 8 9 0.0364 88 8 3 0.00965 04 8 8 9 7 0.03889 7 8 9 0.0364 88 9 9 0 0.000 99 8 9 7 0.03889 7 0 9 0 9 0.0 90 9 3 0.008547 7 0 0.00909 0 9 0 0.000 99 3 0.007576 3 9 0 9 0.0 90 3 0 3 3 0.00769 30 4 sum= 0.5 0 0.00909 0 5 3 4 0.006993 43 6 3 0.007576 3 7 3 5 0.0064 56 sum= 0.5 removed terms= 0.38709 added terms= 0.38709 Note: Correct solutions supported with sound mathematical reasoning received full credit. It was not necessary to provide the solution by mathematical induction as shown above.