Florida Association of Mu Alpha Theta January 2017 Geometry Team Solutions

Transcription

1 Geometry Team Solutions Florida ssociation of Mu lpha Theta January 017 Regional Team nswer Key Florida ssociation of Mu lpha Theta January 017 Geometry Team Solutions Question arts () () () () Question Question True Question Question Rhombus Question 5 15 cute < < 4 Question Question Question 8 3 Outside Question 9 3x y = 7 5x 6y = 9 3x y = 17 (9, 9) Question Question 11 0 (4, 1) Question Question 13 entroid ircumcenter Incenter Orthocenter Question 14 Rhombus d s Question Obtuse uthor: Sayeed Tasnim, Seminole High School lass of 011 1

2 Geometry Team Solutions Florida ssociation of Mu lpha Theta January 017 Regional Question 1 Geometry Team Solutions () Since is a trisection point, = 1. Since is a midpoint, = so = + + = = 1. () y subtracting, m = 3x, m = x, m = 8x = 144, so x = 1. m = 3x = 54. (54 and 54 are both acceptable.) () points cannot uniquely define a plane. 3 non-coplanar points may uniquely defined a plane. () pplying the formula, we have d = 3( )+4(7) = 30 5 = 6. Question () The contrapositive of the inverse of a statement is the converse of a statement. The converse of statement 1 is statement 3. () The inverse of the converse of a statement is the contrapositive of a statement. The contrapositive of statement is statement 3. () The converse of the converse of a statement is the original statement so statement 3. () If the antecedent is false, regardless of the consequent, the statement is true. false antecedent with a true or false consequent is valid and always true. Question 3 p a b q (6x 3) s (x + 11) r (x + 8) t (5c 11) (3c + 3) d 6 () Noticing at the angles (x+8) and (x+11) add up to (6x 3) by opposite angles and corresponding angles, x x + 8 = 6x 3 so x = 1. a = x + 8 by alternate exterior angles so a = 49. (49 and 49 are both acceptable.) () b is the complement of 6x 3 = 103 from above. Therefore, b = = 77. (77 and 77 are both acceptable.

3 Geometry Team Solutions Florida ssociation of Mu lpha Theta January 017 Regional () Triangles and are similar because splits the sides into the same ratio of 3 : so it can be viewed as a dilation. Therefore, =, 5c 11 = 3c + 3, and c = 17. (17 and 17 are both acceptable.) () d = (5c 11) = = 66 using the result from (3). (66 and 66 are both acceptable.) Question 4 () Let be the intersection of the two diagonals. y SS theorem, =. Therefore, = = 6. () Triangle is an isosceles triangle so m = 65. Thus, m = 115. (115 and 115 are both acceptable.) () onsider quadrilateral where = and =. Without loss of generality, let. rawing diagonal, = by SSS congruence so =. lso, = by parallel lines so = = so triangle is isosceles with =. Similarly, =. Hence, the figure has equal side lengths which makes it a rhombus. The next most specific figure would be a square and it is easy to see that a rhombus which is not a square satisfies the criteria. () Since m = 135, m = 45. Since is isosceles, m = 45 and is a right triangle with leg length of 4. Thus, the height is and = 4. = so 1 h = 1 (5 + 4 ) =

4 Geometry Team Solutions Florida ssociation of Mu lpha Theta January 017 Regional Question 5 () Using the triangle inequality, m = 17 8 = 9 and n = = 5. Therefore, mn = 15. () Using the triangle classification test, check that = 130 > 11 = 11 so the triangle is acute. () Using the triangle side inequality theorem, the angles are in the order of the opposite sides from smallest to largest so < <. () Noting that : is in the same ratio as : or 3 : where + = = 10, = 4. Question 6 () Triangles and are similar so = = 4 5 = =. = 5 by the ythagorean theorem. Therefore, () is a right triangle because it is a triple. This is an application of the median to the hypotenuse of a right triangle theorem. The median from to is 1 = 10. () a M a b N Let M = M = a and N = N = b. y ythagorean theorem on N, M, and, we have the following respective equations b N = (a) + b = 4a + b, M = a + (b) = a + 4b, = (a) + (b) = 4a + 4b. To solve for, add the first two equations together and multiply by 4 5 yielding Hence = 4 5 (N + M ) = 4a + 4b =. 4 5 ( ) =

5 Geometry Team Solutions Florida ssociation of Mu lpha Theta January 017 Regional () ue to the construction, is an isosceles triangle so =. m = 150, m = 30, so is a right triangle. Since = 1, = 6. Question 7 () Let diagonals and intersect at. y properties of a kite, they are perpendicular. lso by properties of a kite, bisects so = 1. pplying ythagorean theorem on right triangles and, = 5 and = 9. Therefore, = + = 14. () Noting that and by symmetry (or other methods), is a parallelogram. In addition, since =, it is more specifically a rhombus. Since m = 135, m = 45 by adjacent supplementary angles of a parallelogram. Therefore, m = 90. Since =, triangle is a right triangle. (This is just one example of many other ways to show this.) Therefore, = 3. F G H 5

6 Geometry Team Solutions Florida ssociation of Mu lpha Theta January 017 Regional () 15. Reflect point about point M to point. Since diagonals and bisect each other, is a parallelogram with = M = 10. Using the parallelogram law, + = +, = 10 +, = 15. M () y symmetry, = 30 and = 40. Triangle has side lengths of 30, 40, and 50 (in the ratio 3-4-5) so it is a right triangle with right angle. Similarly, is also right. Let and Q be the feet of the altitudes from and respectively. Using right triangle similarity ratios, h = = Q = 4 and = Q = 18. = Q = Q = 14. Therefore, 1 h( + ) = 1 4 ( ) = 768. Q Question 8 () This corresponds to a right triangle. The ratio of the smaller leg to the hypotenuse is 1. () This corresponds to a right triangle. The ratio of one of the legs to the hypotenuse is 1 =. () This corresponds to a right triangle. The ratio of the larger leg to the smaller leg is 3. () onsider a general triangle with side length of a, b, and c opposite angles,, and respectively. ssume a b c without loss of generality. Let the angle bisector at intersect at. The feet of the altitudes, and are the feet of the altitudes of triangles and respectively. n altitude is always inside an acute triangle and an altitude is outside an obtuse triangle if it is not from the obtuse angle. xactly one of triangles and will be acute and the other will be obtuse (unless they are both right for which the altitude will neither be inside nor outside but rather on the side length). This results in determining which of and is acute and which is obtuse. m + m + m = m + m + m since the angles of triangles sum to the same 180. Since m = m by angle bisector, m + m = m + m. y the triangle side inequality theorem, m > m since a > c, which implies that m = (m m ) + m > m so triangle is acute and triangle is obtuse. Therefore, the inside altitude is towards the smaller side length and the outside altitude is towards the larger side length. Hence, is outside the triangle. 6

7 Geometry Team Solutions Florida ssociation of Mu lpha Theta January 017 Regional more intuitive way to view is the result is to notice that right triangles and are similar and the altitude with the longer side length is clearly outside the triangle. Sometimes, when drawing geometric figures, it s not always obvious where line segments and figures may go. This question helps to get a solver thinking about some of these configuration issues. Question 9 () First, the slope of Y Z is = 3. Take the negative reciprocal to get the slope of a line perpendicular to Y Z which is 3. Use the point-slope form equation of a line to get the equation of the line that passes 3 through X: = y+1 x+3 3x y = 7. () First, the median of Y Z is ( 4+6, 8+0 ) = (15, 14). Next, the slope between X and the median: 14 ( 1) 15 ( 3) = 5 6. Then with the point slope form of a line passing through X: 5 6 = y+1 x+3 5x 6y = 9. () From (), the median is (15, 14) and the slope of a line perpendicular to Y Z is 3 3 point-slope form of a line: = y 14 x 15 3x y = 17. () Using the formula for the center of mass, the answer is (9, 9). Question 10 from (1). Using () Note that square F G can be viewed as a rotation and a dilation (with a factor of 3 5 ) of square. However, due to this, we have similar triangles F because F = = 1 and = F due to the same angle of rotation for the line segments. Hence, F = 1 so F = 13 = 6. This rotation and dilation is called spiral symmetry or a homothety. G F There are two potential configurations for this problem which can be seen considering the loci of points for point. Fixing square. oint must be one of the points of intersections of a circle of radius 3 centered at and a circle of radius 13 centered at. The solution for this problem works regardless of either configuration. () y parallel lines,. Therefore, Therefore, = = 17. = = 1 5. y ythagorean theorem, = 13. 7

8 Geometry Team Solutions Florida ssociation of Mu lpha Theta January 017 Regional () line that is parallel to the bases of the trapezoid and inside the trapezoid is the weighted average of the bases based on it s distance from each of these bases. Since : = 15 : 10 = 3 :, F = = 30. F To formally prove this, consider an arbitrary trapezoid where = a, = b, F = c, and a < b. xtend and to meet at. F. Suppose = f and = g are known (or at least their ratio) and let = h. (This can be done for the other leg of the trapezoid or even the heights of the trapezoids and triangles.) y similar triangles, h a = h+f c Here is a very important ratio technique: given that w x = y z, then w x = y z = w+y = h+f+g b. x+z. (Try proving this on your own and try to make intuitive sense of this.) Using this on the first equality, we have h a = h+f c = f h+f c a. Using it on the second, we have c = h+f+g b = g b c. Therefore, f c a = g b c so fb fc = gc ga c = g f+g a + f f+g b which is a weighted average of a and b. very similar treatment may be done if F is outside the trapezoid. = w y x z () rop the altitude from to to meet at Q. y parallel lines Q, Q and Q Q. Using corresponding ratios, = Q and Q = Q. dding these two equations, Q ( 1 + ) 1 = Q+Q = = 1. Solving for Q, Q = ( 1 + ) 1 1 = + = = 1. Q 8

9 Geometry Team Solutions Florida ssociation of Mu lpha Theta January 017 Regional Question 11 () The formula for the number of diagonals of an n-gon is n(n 3). Hence, for an octagon, this is 8 5 = 0. To prove this, consider the n vertices of n-gon. There are ( n ) to select two points to make a line segment. However, there are n line segments which are not diagonals but rather sides of the n-gon. The remaining are certainly diagonals. Hence, the number of diagonals is ( ) n n = n(n 3). () Since rs is an even element of the triple, and the only even element of is 8, then rs = 8 or rs = 4. Since r > s and they are both integers, r = 4 and s = 1. lso, to check, r s = 15 and r + s = 17. () onsider the vertices of convex quadrilateral. If and are joined, they form sides of the quadrilateral so they do not intersect. Similarly, if and do not intersect. iagonals and do intersect. Hence, this is one case out of three so the fraction of groupings is 1 3. () The coincident vertices of the octagon can be either a side length of a square or the diagonal of a square. Go through cases based on what kinds of vertices may be chosen based on how far the vertices of the octagon are from each other. (a) The vertices are adjacent such as and is a side of the square. There are 8 possibilities for the adjacent vertices and the square can be either inside or outside the octagon for a total of 8 = 16. (b) The vertices are adjacent such as and is a diagonal of the square. There are 8 possibilities for the adjacent vertices and one possibility for the square for a total of 8. (c) The vertices are two sides apart such as and is a side of the square. This yields a square that has all four of its vertices on the octagon. y inspection, there are only two such squares for a total of. (d) The vertices are two sides apart such as and is a diagonal of the square. There are 8 possibilities for the vertices and one possibility for the square for a total of 8. (e) The vertices are three sides apart such as and is a side of the square. There are 8 possibilities for the vertices and the square can be either towards the inside or towards the outside the octagon for a total of 8 = 16. (f) The vertices are three sides apart such as and is a diagonal of the square. There are 8 possibilities for the vertices and one possibility for the square for a total of 8. (g) The vertices are four sides apart such as and is a side of the square. There are 4 possibilities for the vertices (because there are only 4 of these diagonals in an octagon) and two directions for the square for a total of 4 = 8. (h) The vertices are four sides apart such as and is a diagonal of the square. However, these coincide with the squares in (c) so they have already been counted. ny vertices that are separated by a number of sides greater than four can be viewed by a smaller separation on the minor arc. Totaling these values, there are 66 squares. Question 1 F 9

10 Geometry Team Solutions Florida ssociation of Mu lpha Theta January 017 Regional () Since F = =, triangle F is isosceles. m F = m m F = = 48. Therefore, m F = 66. () m F = m m F = = 4. () m F = 360 m F m F m F = = 168. lternatively, using quadrilateral F, m F = 360 m F m m F = = 168. () y symmetry, F = F so F is an isosceles triangle. Hence, m F = 6. Question 13 () Median. () ircumcenter. () Incenter. () Orthocenter. Question 14 () y symmetry (or other means), and. Therefore, is at least parallelogram. In addition, = so it is also a rhombus. is not a square because m = () Since is a parallelogram, = s. Therefore, = = d s. () m = m = 36 because is an isosceles triangle with vertex angle 108. Similarly, m = m = 36 as well so. Taking ratios, = s d = d s s s = d sd 1 = ( d s ) d s. Then by the quadratic formula or completing the square, d s = () From the isosceles triangles such as, drop an altitude to form a right triangle. One leg is half a diagonal, 1 d by the isosceles triangle and the hypotenuse is s. Taking the cosine of 36 gives is 1 d s =

11 Geometry Team Solutions Florida ssociation of Mu lpha Theta January 017 Regional Question 15 l () is equal to the distance between and along the direction parallel to the line. The distance between and perpendicular to the line is 15 5 = 10. This forms a right triangle with hypotenuse 6, leg of 10, and a leg of. Therefore, = 4 by the ythagorean theorem. () There must exist some point at which the path meets line l, that is point. Given a choice of point on line l, the shortest distance from to and from to are with straight lines. ut how is this point chosen? Reflect point and line segment about line l to point and line segment respectively. + = + so now the shortest distance can be considered as a path from to to line l at point and then to point and is the intersection of this line and line l. However, the shortest distance for this route is the straight line from to. Note that this implies that = by opposite angles and = by reflection. This along with the right angles imply that = and. The distance between and parallel to line l is 4 from (1). The distance between and perpendicular to line l is = 0. Hence, the shortest distance is = () Using the results from () that with a ratio of similarity : = 5 : 15 = 1 : 3, = 6 and = 18. Therefore, tan(m ) = 5 6. () Using the results from the previous parts, = = 61 and = = long with = 6, applying the ythagorean theorem converse, + = 610 < 676 = 6 =. Hence, the triangle is obtuse. lternatively, one may notice that tan(m ) = 5 6 < 1 = tan(45 ), so m < 45 by the monotonicity of the tangent function. m = 180 m m > = 90 so m is obtuse. 11