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January Regional Geometry Team Solutions 1) 1; Statement A is false since it is not stated that the shape is a polygon, a circle does not have exterior angles. Statement B is false. Just because Travis will not go to the beach if it rains, it does not necessarily mean that he will go to the beach if it does not rain. Travis could decide not to go to the beach, even if it does not rain. Statement C is false, the slopes of perpendicular lines are not equal. Statement D is true, vertical angles are congruent. Only one statement is true. 2) 180 degrees; The sum of the angles of a star with n points on it is degrees. 3) . Plugging in 5, we get 180 ; A 20-gon has 170 diagonals = A, given by n(n – 3)/2. The measure of an angle in a regular polygon is give by . Setting this expression equal to 156, we get . Solving for n we get 360/24 = 15 sides = B. The perimeter of a rhombus with diagonals of lengths 12 and 14 can be found by splitting the rhombus into four right triangles using the diagonals. The legs of all 4 of these triangles are 7 and 6 because the diagonals of a rhombus perpendicularly bisect each other. This rhombus is depicted in the diagram to the right. Since all sides of a rhombus are congruent, we can use the Pythagorean Theorem to find the length of one side of the rhombus, and then multiply the result by 4. Doing so, we get: = C. There are only 3 polygons with less than 9 diagonals, D = 3. Now, we are supposed to find the value of 2A + B – C + D. This is equal to . 4) 467; The sum of all of the angles formed by 2 intersecting lines is 180 + 180 = 360 = A. The number of diagonals of a polygon with 15 sides is given by the formula , where n is the number of sides of the polygon. If we substitute 15 for n, we get 90 = B. There are only 3 kinds of triangles: equilateral, isosceles, and scalene, so this statement is true, meaning that it has a value of 1 = C. Polygon ABCDEFGHIJKLMNOP has 16 sides, so 16 = D. A + B + C + D = 360 + 90 + 1 + 16 = 467 5) -52 inches; To solve this problem, refer to the diagram at the right. The perimeter of the frame and picture together is, as shown in the diagram, . The opposite inverse reciprocal of the perimeter is . 6) ;A rhombus with diagonals of length 14 and 48 can be divided into 4 right triangles with legs of length of 7 and 24, and hypotenuse of length 25. Thus the side length of the rhombus is 25, making the perimeter 25(4) = 100 = A. To find the length of the diagonal of a square, divide the square into two 45-45-90 triangles with legs of length 225. Then, using the properties of 45-45-90 triangles, we know that the length of the hypotenuse of one of these right triangles is times larger than a leg of the triangle. Since the side length of one of these triangles is 225, the hypotenuse of one of these right triangles is diagonal January Regional Geometry Team Solutions length of square = B. The midpoint of a line segment with endpoints (8, 2) and (3, -4) is and the sum of the coordinates of the midpoint is 9/2 = C. The distance between (5, 7) and (-12, 3) is = D. The value of . 7) ; This problem is depicted in the drawing to the right. As a result of the triangle being isosceles, angle bisector AD splits the isosceles triangle into two congruent right triangles. This means that side BC is split into to equal halves, and since we know 2 sides of both these right triangles, we can use the Pythagorean theorem to find the length of angle bisector AD. ; A triangle with angles (x – 3), (10x + 3), and (3x + 12) has the sum of its angles equal to 180. 14x = 168, x = 12. The smallest angle of this triangle is 12 – 3 = 9. The compliment of 9 is 81 = A. To find the largest possible value for x in a triangle with sides 8, 13, and x, use the triangle inequality theorem. , and . Simplifying this, we get . The largest possible value of x is 20 = B. The smallest possible side of a triangle with side lengths 14 and 18 can be found using this triangle inequality: . The smallest possible side length for this triangle is 5 = C . The length of the longest side a triangle with coordinates (-5, 3), (1, 2), and (3, 1) is , which can be 8) found using the distance formula. . = D. The value of 9) 1058400; The sum of the squares of the angles of a regular nonagon can be found by first finding one of the angles of a regular nonagon. The measure of an angle of a regular nonagon, where n = the number of sides is . The sum of the squares of the angle measures of this nonagon will be: = A. Skew lines intersect at zero points, so B = 0. 6A – B = 6(176400) – 0 = 1058400 10) 1620; A hendecagon has 11 sides, and the sum of the angles of a polygon with n sides is can substitute 11 in for n to get 180(9) to get 1620. . We 11) 0; The first statement is false, the perimeter of a regular heptagon with side length 7 is 49, not 77, meaning that statement 1 is equal to -1. The second statement is true, the contrapositive of a conditional statement is always true, so statement 2 is equal to 3. The third statement is false, line C can be perpendicular to line A without being perpendicular to line B, so statement 3 is equal to -1. The fourth statement is false, the diagonals of a rectangle do not bisect the angles of a rectangle, so statement 4 is equal to -1. The sum of the values of all of the statements is: 12) The number of ways in which you can arrange n points on a circle is equal to (n – 1)!. Plugging in 6, we get 120 = A. The number of planes formed by n noncollinear points is , where C denotes combination. If we have 12 points, the number of planes formed is 220 = B. The number of arcs formed by n points on a circle is n(n – 1). Plugging in 7 for n, we get 42 = C. The ratio of the sum of the angles of the 2 similar triangles is 180/180 = 1 = D. The value of A + B + C + D = = 383. 13) ; Let the length of the longest piece of string equal X. The shortest piece is now equal to X/3. The second to longest piece is equal to (X – 15). The remaining piece is X/2. The sum of the lengths of all of these pieces is 125 cm. Add the lengths of all of the pieces and set the result equal to 125. This will give us January Regional Geometry Team Solutions and finally 14) . The length of the shortest piece is This problem is illustrated in the diagram below, where x = the height of the point at which the two ropes intersect. The value of x is equal to are the heights of the two towers. 15) ASA; The triangles are congruent by ASA because angles are congruent. , which is given by the formula , where a and b , and also due to the fact that vertical