2012 Mu Alpha Theta National Convention Theta Geometry Solutions ANSWERS (1) DCDCB (6) CDDAE (11) BDABC (16) DCBBA (21) AADBD (26) BCDCD SOLUTIONS

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1 01 Mu Alpha Theta National Convention Theta Geometry Solutions ANSWERS (1) DCDCB (6) CDDAE (11) BDABC (16) DCBBA (1) AADBD (6) BCDCD SOLUTIONS 1. Noting that x = ( x + )( x ), we have circle is π( x + ) = π(x + x + ), D. x x = x +. Hence the area of the. The angles α 1 and α 7 are same-side exterior angles. Hence α 1 +α 7 = 180 α 7 = = 110, C.. There are three possibilites we can have: α 1 + α or equivalent, α 1 + α or equivalent, and α + α or equivalent. The highest of these is α + α = = 0, D.. S > 60 only when α k is acute. Exactly half of the α i are acute; therefore our answer is 1/, C. 5. Consider the following diagram and let AC = x and BC = x. Then if D is chosen some distance a from C, we have CD = a and BD = x a. If we can form a triangle with these three lengths, then by the Triangle Inequality we have two inequalities: x + (x a) > a a < 5x (1) x + a > x a a > x () Since x < CD < 5x, and we can choose D from a total length of 5x x = x, the probability we desire is ( 5x x ) = 1 x, B. 1

01 Mu Alpha Theta National Convention Theta Geometry Solutions 6. Choose point G on the hypotenuse of right triangle ADC such that DG is an altitude of the triangle.

2 01 Mu Alpha Theta National Convention Theta Geometry Solutions 6. Choose point G on the hypotenuse of right triangle ADC such that DG is an altitude of the triangle. Since AC EF, DG is also perpendicular to EF. Thus DG is the height of parallelogram ( ACEF ), (DG)(AC) and we can write [ACEF ] = (DG)(AC). But since [ABCD] = [ADC] = = (DG)(AC), we have [ACEF ] = [ABCD] = 96, C. 7. A unit triangle is an equilateral triangle with an area of 1. Hence if its side length is s, we have s = 1 s = s =. Thus the perimeter is p = s = 6, D. 8. As the number of sides of regular polygon increases, it approaches the shape of a circle. In the case of the unit polygons, the area of this circle is 1. Thus we want to find the circumference of a circle with area 1. We have πr = 1 r = 1 π and the circumference is πr = π( 1 π ) = π, D. 9. When x 0, we have f(x) = x. However, when x < 0, we can let x = k for some k > 0. Then f(x) = x + k = k + k = 0. For 0 x < 1, f(x) is a triangle with a height of (1) = and base of 1 and thus has an area of 1. For 1 < x < 0, f(x) is a straight line along the x-axis and thus has no area. The total area is 1, A. 10. Note that a n is the number of diagonals for a regular polygon when n. Thus our answer is simply a z = 75, E. 11. Cylic quadrilaterals have the property that opposite angles add up to 180. Hence since opposite angles in a rhombi are equal, in this case, they must both be 90. Furthermore since opposite sides in a rhombus are equal, this rhombus must be a square. Its area is simply 1 = 7, B.

3 01 Mu Alpha Theta National Convention Theta Geometry Solutions 1. The trisection points divide the triangle into a hexagon and three triangles. Each of these triangles are similar to T by a ratio of 1. Thus the ratio of the area of each of these triangles to the area of T is ( 1 ) = 1 9. Since there are three triangles, they account for ( 1 9 ) = 1 of the ( ) total area 1 of T. Thus the ratio of the area of the hexagon to the area of the entire triangle is 1 = 9, D. 1. Let the center be (x, y). Thus the distance from the center to the three points is equal; using distance formula, we have (x ) + (y 5) = x + (y + 1) x x + + y 10y + 5 = x + y + y + 1 x + y = 7 (x ) + (y 5) = (x + ) + (y 1) x x + + y 10y + 5 = x + 6x y y x + 8y = 19 Solving the systems of equations, we have x = 1 51 and y =. Our answer is = 7, A. 1. Call the smaller circle ω. Then the area of R can be found by subtracting the area of ω from the circular segment formed by AB and then dividing by two (since it is half of the remaining area). First, it is easy to find that since AB is at a distance of from O, ω has a radius of = 1 and thus has an area of π. To find the area of the circular segment, we first find the area of sector AOB and then subtract the area of triangle AOB. It is easy to verify that AOB = 10. The area of the sector is then one-third the area of the circle, or π = 16π ()() sin 10. It follows that [ AOB] = =. Thus we have [R] = 1 [( ) ] 16π π = 1π 6, B. 15. Let the respective altitudes be a 1, a, and a. Then the area of the triangle can be written as A = a 1 = a = 5a. From this, we have a 1 = a = 5a. The least common multiple of,, and 5 is 0. Thus we can easily verify that the altitudes are in ratio 10 : 5 :. We have a + b + c = = 19, C. 16. The graph of x + y < is the region between the lines y = x and y = x. Thus the area we seek is just the area of this region for 0 < x <. This is a parallelogram with base of length ( ) = and height. The desired area is ()() = 8, D. 17. Since EF is comprised of the two heights of the triangles, it has length equal to h = ( ) =. Furthermore the altitude of CEF to base EF is equal to half the side length of the square, or.

4 01 Mu Alpha Theta National Convention Theta Geometry Solutions Thus we have [CEF ] = ( )() =, C. 18. Let the radius equal r and the height equal r. Then we have 1 π(r )(r) = 1 πr = 1 r = 1, B. π1/ 19. Call the points (1, ), (, 8), and (, 1), A, B, and C, respectively. If AB is a side, then the fourth point is either D = ( +, 1 + 6) = (6, 7) or D = (, 1 6) = (, 5). If BC is a side, then the fourth point is either D = (1 + 1, 7) = (, 5) (which we already found), or D = (1 1, + 7) = (0, 9). Therefore our answer is = 19, B. Remark. The three given points form the medial triangle of the triangle formed by the possible fourth points of the parallelogram. 0. It is clear that CD = 1 and AC =. Thus CE must equal in order to retain the geometric progression. The height of ABE is then ( 1 ) 1 = Thus the area is ( 1 )(1)( 15 1) = 15 1, A. 1. We have V = πr h = π(r )( 1 r ) = π, A.. We can use Stewart s Theorem to compute the length of AM. Letting AM = p, we have ( )() + ( )() = 6p + ()()(6) p = 1. Now, we can apply Power of a Point around M. We have (CM)(MB) = (AM)(MD) MD = 1 = 9 1 7, A.

01 Mu Alpha Theta National Convention Theta Geometry Solutions. Call the square EF GH and call the sector OAB where O is the center, as shown below. Let the side of the square be equal to x.

5 01 Mu Alpha Theta National Convention Theta Geometry Solutions. Call the square EF GH and call the sector OAB where O is the center, as shown below. Let the side of the square be equal to x. If we draw segment OH and drop a perpendicular from O to GH at P, we obtain right triangle OP H. Since OP bisects AOB, AOP = 0 and OEQ is a triangle (where Q is the intersection of OP and EF ). Then clearly OQ = EQ = x. Hence OP = x( + 1), HP = x, and OH = 6. Applying Pythagorean Theorem gives ( x ) + x ( ) ( + 1 = 6 x 7 + x + ) = 6 x = 6 + x = 6( ) Hence the area of the square is 6( ), D.. We have A = rs, or 8 ( ) 1 18 = A = 7, B. 5

Since EDC = HDA = 60 and CDA = 90, EDH = 150. We know that ED = HD = DC =, so applying Law of Cosines gives us c = + ()()() cos 150 = + 16, D. 6. It is well known that F (n) = 60.

6 01 Mu Alpha Theta National Convention Theta Geometry Solutions 5. The smallest square is formed by connecting the outer vertices of the equilateral triangles in consecutive order as shown in the following diagram. Let EH = c. Since EDC = HDA = 60 and CDA = 90, EDH = 150. We know that ED = HD = DC =, so applying Law of Cosines gives us c = + ()()() cos 150 = + 16, D. 6. It is well known that F (n) = 60. Thus we have F () + F () + + F (n) = 60(n + 1) = 60n 70, B. 7. Consider the following diagram and notice that triangle GF C is a right triangle with sides of length,, and 5: Similarly, it is easy to see that triangle AEG is a right triangle as well, with sides of length 6, 8, and 10. Thus we have EG BC and GF AB which implies that EG GF. Hence triangle EGF is a right triangle with lengths of length and 8. Its area is then 1 ()(8) = 1, C. 6

7 01 Mu Alpha Theta National Convention Theta Geometry Solutions 8. Each successive square has exactly half the area of the previous one. Hence the sum is just an infinite geometric sequence with first term 10 = 100 and common ratio 1/. Our answer is 100 = 00, 1 1 D. 9. We simply have AP + CP = BP + DP 5 + = + DP DP =, C. 0. Note that point A suffices the conditions for the British Flag Theorem on square EBDF. We have F A + BA = CA + EA EA F A = 8 6 (EA F A)(EA + F A) = 8, D. 7

Section 5.1. Perimeter and Area

Section 5.1. Perimeter and Area Section 5.1 Perimeter and Area Perimeter and Area The perimeter of a closed plane figure is the distance around the figure. The area of a closed plane figure is the number of non-overlapping squares of