BC Exam Solutions Texas A&M High School Math Contest November 8, 2014

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1 Exam Solutions Texas &M High School Math ontest November 8, 014 If units are involved, include them in your answer, which needs to be simplified. 1. If an arc of 60 on circle I has the same length as an arc of 45 on circle II, what is the ratio of the area of circle I to the area of circle II? Solution: Let r 1 and r denote the radii of circles I and II respectively. Then we have π r 1 3 = r π 4 = r 1 = 3 r 4 = 1 = πr 1 πr = ( r1 r ) = How many positive integers are there, which are less than 1000 and not divisible by 5 or 7? Solution: There are 199 integers less than 1000 divisible by 5 and 14 less than 1000, which are divisible by 7. There are 8 integers less than 1000, which are divisible by 35. Thus, there are integers less than 1000 not divisible by 5 or = 686, 3. In a circle of radius r centered at O a chord of length r is drawn. From O a perpendicular to meets at M. From M a perpendicular to O meets O at. In terms of r, what is the area of triangle M? M O Solution: Note that O is an equilateral triangle. Thus, M = r /, OM = 3r/, and the area of 3r MO =. 8 M is similar to OM, with a scaling coefficient of 1 /. Thus, the area of M is 1 /4 that of OM. Thus, 3r rea of M = The figure below is a half circle with = 3 inches and = 1 inches. Line is perpendicular to the half circles diameter and intersects the circle at point. How long is line? Solution: Let x denote the length of line. Then 15 = +, 3 + x =, 1 + x = 15 = (3 + x ) + (1 + x ) x = = 7 x = 6

2 5. The first 50 terms of an arithmetic series sum to 00 and the next 50 terms sum to 700. What is the value of the first term? Solution: The k th term of the series is a k = a + (k 1)d, and 00 = a k = (a + (k 1)d) = 50a d 700 = 100 k= a k = a k a k = 50a + ( = 500d = d = 1. Solving either of the first two equations for a we get a = )d 6. The figure below shows two intersecting circles. The circle to the left is centered at the origin and has radius 1. The second circle is centered at (R, 0), and it intersects the unit circle orthogonally, i.e., perpendicularly. The second circle also passes through the point (r, h), which is interior to the unit circle. Find R in terms of r and h. (r, h) (0,0) R x Solution: enote the coordinates of the point of intersection in the upper half plane by (x, y). We have the following relations: x + y = 1, the point (x, y) is on unit circle, (x R) + y = (r R) + h, the points (x, y) and (r, h) are on the circle y x R = x y, since the line from the origin to (x, y) is perpendicular to the line from (R, 0) to (x, y). The first and third equations imply xr = 1 Expanding the second equation and using the fact that xr = 1 solve for R and get R = 1 + r + h r. 7. Suppose the polynomial p(x) = x 014 c 1 x c x c 014 has roots {±1, ±,, ±1007}. What do c 1 and c 014 equal? Solution: c 1 = 0 as c 1 is the sum of the roots, and c 014 = [1007!], as it is the product of the roots.

3 8. Triangle is isosceles with base. Points P and Q are respectively in and such that = P = P Q = Q. What are the number of degrees in angle? Solution: Q x y P Let x equal and y equal. Then, since triangles, P, QP and P Q are all isosceles, we have x + y = π, x + QP = π, y + P = π, x = QP = P Q = y x = P. These equations imply Solving for x we have x = π 7 = degrees. x + y = π, 3y x = π. 9. Thirty one books are arranged from left to right in increasing order of price, and the price of each book differs by $.00 from each adjacent book. Moreover, the price of the most expensive book equals the sum of the prices of the middle book and a book adjacent to the middle book. What is the price of the most expensive book? Solution: Let p i, for i = 1,, 31, denote the price of each book. Then we have p i = (i 1) + p 1. We know that one of the following two equations is true: p 31 = p 16 + p 15, or p 31 = p 16 + p 17 (1) 30 + p 1 = ( 15 + p 1 ) + ( 14 + p 1 ), or 30 + p 1 = ( 15 + p 1 ) + ( 16 + p 1 ) () 60 + p 1 = 58 + p 1, or 60 + p 1 = 6 + p 1 (3) The first of equations (3) implies p 1 =, while the second of equations (3) implies p 1 =. Thus, p 1 =, and the price of the most expensive book is p 31 = 60 + = The function p(x) = ax + bx + c describes a parabola. Suppose the parabola s vertex is located at the point ( 1, ). What does b /a equal? Solution: ompleting the square we have p(x) = a (x + ba ) ) x + c = a (x ba b + x + 4a + ) ( (c b = a x + b 4a a ) ) + (c b 4a The ordinate of the vertex occurs when the quadratic term is zero. Thus, 1 + b a = 0, or b a =. 11. Jane can mow a field in 1 hours, while Jane and ill working together can mow the field in 8 hours. How long will it take ill to mow the field by himself? Solution: In 8 hours Jane will have mowed /3 rd of the field. Thus, ill will have mowed 1 /3 rd of the field, which means that it would take ill 4 hours to mow the field by himself. 3

4 1. The probability that event occurs is 7 /8, and the probability that event occurs is 5 /6. What is the smallest possible value of the probability of? Solution: We know that Pr( ) = Pr() + Pr() Pr( ). Since Pr( ) 1, we have Pr( ) = Pr() + Pr() Pr( ) = Triangle in the figure below has area 10. Points, E, and F all distinct from the vertices of the triangle lie on sides,, and respectively; = and =3. If triangle E and quadrilateral EF have equal areas, what is that area? F E 3 Solution: Notice that triangle E is part of both triangle E and quadrilateral EF. Thus, the area of E = area of E + area of E, and area of EF = area of FE + area of E. Hence triangles E and FE have the same area. Moreover, since they have a common base they must have the same altitudes from that base, which means that points and F are equidistant from the line through and E. Thus, the line through and F is parallel to the line through and E, which implies that triangles and E are similar. Thus, the ratio of their altitudes must equal 5 /3. The altitude of triangle from the base equals 4. Thus, the altitude of triangle E =4 3/5 = 1 /5. Triangles E and E have equal altitudes, which means that the area of triangle E equals = In, a point in on so that =. If = π /6, what does equal? Solution: Since =, angles and are equal. enote the common value by α. Let ψ denote the value of angle, θ = π α the value of angle, and x the value of angle. ψ θ α x α We have the following equations: α + x ψ = π, ψ + (π α) + x = π. 6 dding these two equations we have x + π = π + π /6, which implies that x = π or 15 degrees. 1

5 15. aisy has twenty 3 stamps and twenty 5 stamps. Using one or more of these stamps, how many different amounts of postage can she make? Solution: Twenty 5 stamps has the same monetary value as thirty 3 stamps with two 5 stamps. It is easy to see that the postage amounts for twenty 3 and twenty 5 is the same as that for fifty 3 and two 5. The postage amounts for the latter are 50 using only 3 stamps, 51 using none or more 3 stamps with exactly one 5 stamp, 51 using none or more 3 stamp with exactly two 5 stamps. The total is = In the figure below E F M G is a trapezoid with and parallel. M is a median of, is a diagonal of the trapezoid with the median M meeting the diagonal at F. Line EG passes through F and is parallel to. If EF has area equal to 6 sq. cm., what is the area of F G? Solution: Note that EF is similar to M and that F G is similar to. Moreover, since the three horizontal lines are parallel we have E = G. Since we also have EF M = E = G = GF, and = M. we may conclude that GF = EF. Since the triangles EF and GF have the same altitudes we may conclude that the area of GF is twice that of EF, so the area of GF is 1 sq. cm. 17. Triangle has a right angle at, =, and = 3. The bisector of meets at. Find. Solution: Let E be the altitude of. E 5

6 Then note that is congruent to E, and so E = =. Then = 13. Let = x. Then E = x, E = 13, and = 3 x. pplying the Pythagorean Theorem to E yields x + ( 13 ) = (3 x), from which x = = 3 ( 13 ). 18. wooden rectangular prism has dimensions 4 by 5 by 6. This solid is painted green and then cut into 1 by 1 by 1 cubes. Find the ratio of the number of cubes with exactly two green faces to the number of cubes with exactly three green faces. Solution: The cubes with two green faces are the cubes along the edges, not counting the corner cubes. In each dimension, we lost two cubes to the corners so we then have four edges with 4 cubes, four with 3 cubes and four with cubes. The total number of cubes with paint on two faces is then = 36. The number of cubes that have paint on three sides are the corner cubes of which there are eight. The required ratio is then 36 : 8 or 9 :. 19. The circle shown below is centered at O, has radius equal to 1, and θ = 4. What is the sum of angles and O? θ O Solution: ngle = θ /, and O = (π θ) /. Thus the sum of these two angles is + O = θ + (π θ) = π. 6

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