BC Exam Solutions Texas A&M High School Math Contest October 24, p(1) = b + 2 = 3 = b = 5.

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1 C Exam Solutions Texas &M High School Math Contest October 4, 01 ll answers must be simplified, and If units are involved, be sure to include them. 1. p(x) = x + ax + bx + c has three roots, λ i, with λ 1 + λ + λ = and λ 1 λ λ =. Suppose that p(1) =. What is p()? Solution: We know that p(x) = (x λ 1 )(x λ )(x λ ) from which we see that a = and c =. Thus, p(1) = 1 + b + = = b =. Thus, p() = 8 (4) + () + = = 0.. Triangle C is inscribed in a semicircle with side C being a diameter of the circle. Line D is perpendicular to C with D= and DC =. What is the length of D? See the sketch below. D C Solution: Triangles D and CD are similar. Thus, D = D. and we have D = 10.. circle centered at O is inscribed in the equilateral triangle C. Line segment DE is tangent to the circle, is perpendicular to, and intersects and C at points D and E respectively. (See the sketch below) If D = 1 inch, what is the length of one of the triangle s sides? D O E C Solution: Let G be the point on the circle at which DE is tangent. Draw a straight line from O to F (the point where the circle is tangent to ) and from O to G. The quadrilateral GOF D is a square, with side length equal to the radius of the circle. Since triangle C is equilateral, the radius of the circle, r and the triangle s side length, s satisfy the equation r = s. Moreover the point F bisects. Thus. s = F = (r + 1) = ( ) s + 1. Solving this equation for s we have s = 1 = + inches.

2 4. What does (1 1 ) (1 1 ) (1 14 ) ( 1 1 ) 01 equal? Solution: The Product equals = ( ) ( 1 ) ( 1 01 ) ( ) ( ) ( ) = 01 = ( ) ( ) = p(x) = x + ax + bx + c is odd about the point x =, and p(1) = 1. What does c equal? Solution: Since p is odd about x =, p() = 0, and p(1) = 1 implies that p() = 1. Thus we have the equations: 8 + 4a + b + c 0, 1 + a + b + c = 1, 7 + 9a + b + c = 1. The solution to this system is: a = 6, b = 10, c = How many different values of the natural number n are there for which n 440 is a perfect square? Solution: Suppose k is an integer such that n 440 = k. That means n k = (n k)(n + k) = Since (n k)(n + k) is even, the two terms must have the same parity. In this case the parity must be even. So 4 must divide one of the terms n k or n + k, and must divide the other one. The table below lists the possibilities. Thus, there are 4 possible values of n. n k 4 4 n + k n park is in the shape of a regular hexagon and a side length of kilometers. Starting at a corner a woman walks along the perimeter a distance of km. How far is she from her starting point? Solution: The sketch below shows an equilateral triangle and a regular hexagon, each having side length, and the coordinates of their vertices, assuming the left lower corner is at the origin. (, ) (1, ) ( /, /) (, ) (0,0) (,0) (0,0) (,0) The woman s trek ends at the point ( /, /), and the distance from this point to the origin is ( ) ( Distance = ) + 1/ ( ) 1/ + 7 = = 1 km. 4

3 8. The numbers a 1, a, a 01 are consecutive terms of an arithmetic sequence, and 01 a i = a 1 + a + + a 01 = 01. Which, if any, of the a i can be determined, and what are their values? Solution: The terms of the sequence can be written as: a i = a 1 + (i 1)k for i = 1,, 01. where k is a fixed constant. dding these 01 terms together, we have ( ) = a i = (a 1 + (i 1)k) = 01a 1 + k. This implies that a k = 1. Since a k = a 1008, we have a 1008 = finite sequence of three digit numbers has the property that the tens and units digits of each term are respectively the hundreds and tens digits of the next term. Moreover, the tens and units digits of the last term are respectively the hundreds and tens digits of the first term. One such sequence is Let S be the sum of the numbers in the sequence. (S for the above sequence is 1.) What is the largest prime number that divides S, for all possible S? Solution: Suppose that is one such sequence. If we add the terms, we have x 1 x x x x x 4 x n 1 x n x 1 x n x 1 x S = (x + x x 1 + x ) + 10 (x + x + + x 1 ) (x 1 + x + + x n ) = 111 (x 1 + x + + x n ). Thus, every such S is divisible by 111 = 7. Hence the largest prime factor is How many different 1 digit numbers can be made up from the integers 1,, and, assuming that 1 is not in the first five digits, is not in the second five digits, and is not in the last digits? Solution: This is essentially the same as asking how many 1 digit numbers can be made up from just 1 and. The answer is The graph of the function f(x) is shown below. How many solutions does the equation f(f(x)) = have? y (, ) (1, ) x (, ) Solution: The equation f(y) = has two solutions: y = and y = 1. The equation f(x) = has one solution, which lies between and. The equation f(x) = 1 has 4 solutions: two positive and two negative. Thus, the equation f(f(x)) = has solutions. 1. Triangle C has side lengths =, C = 6, and C = 7. Two bugs starting at the same time from crawl in opposite directions along the sides of the triangle at the same speed. They meet at point D. What is D? Solution: The perimeter of the triangle is 18. Thus, each bug crawls 9 units. D must lie unit from C and 4 units from. That is, D = 4

4 1. The integers 1 through 9 are written on separate pieces of paper, which are then placed in a hat. slip is drawn, the number on it noted, and then the slip is returned to the hat. This is done one more time. Which integer is most likely to be the units digit of the sum of the two numbers recorded? Solution: The number 0 occurs most often as the units digit. It will occur 9 times, while each of the other digits will occur exactly 8 times. 14. Simplify the expression bx ( a x + a y + b y ) + ay ( a x + b x + b y ). Solution: bx ( a x + a y + b y ) + ay ( a x + b x + b y ) = bx ( a x + b y ) + ay ( a x + b y ) + a bxy + ab x y = () ( a x + b y ) + () (abxy) = a x + b y + abxy = (ax + by) 1. In the triangle below =, C = 6, and D =. Moreover line D is perpendicular to line C. What is the area of triangle C? D 6 C Solution: Triangles D and CD are right triangles. Hence, D = 9 = 16 = D = 4, CD = 6 9 = 7 = CD = 7. Thus, rea of triangle C = ( 4 + ) 7 = Find the sum of all positive integers n for which n divides n + 1. Note, negative integers are allowable divisors. Solution: Dividing n + 1 by n, we have n + 1 = (n )(n + ) +. Thus, n divides n + 1 if and only if n divides. Thus, n must equal one of, 1, 1,, which means that n =, 1,, or 7. The sum of those n s, which are positive is 11. 4

5 17. Numbers can be represented in different bases. For example 47 is assumed to be written in base 10 and means , while (47) 8 (base 8) means or, as normally expressed in base 10, 167. Find the base b such that (4) 10 = (111) b. Solution: Convert everything to base 10, which means Since b > 0, we must have b = 6. 4 = b + b + 1 = b + b 4 = 0 = b = 1 ± 169 = 7 or In the following Sudoku puzzle you are to fill in the missing digits so that each row, column, and small squares each contain all of the digit 1 through 9 inclusive. Let a ij denote the digit in the i th row and j th column of the Sudoku puzzle, where a 11 denoted the entry in the top left hand position, and a 1,9 denoted the entry in the top right hand position. What is the value of the following sum: a 1 + a 9? Solution: The entries a 4, a, and a 6 must consist of the numbers, 7, and 8, which means the entries a 1 and a 9 must be 1 and. Thus, their sum is.

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