* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download For all questions, the choice “E) NOTA” denotes “None
Survey
Document related concepts
Dessin d'enfant wikipedia , lookup
Problem of Apollonius wikipedia , lookup
Tessellation wikipedia , lookup
Euler angles wikipedia , lookup
Line (geometry) wikipedia , lookup
History of geometry wikipedia , lookup
Apollonian network wikipedia , lookup
Golden ratio wikipedia , lookup
Rational trigonometry wikipedia , lookup
History of trigonometry wikipedia , lookup
Trigonometric functions wikipedia , lookup
Reuleaux triangle wikipedia , lookup
Euclidean geometry wikipedia , lookup
Incircle and excircles of a triangle wikipedia , lookup
Transcript
March Regional Geometry Individual Solutions 1. If we translate this philosophy as ~ p ~ q , the contrapositive of this is q p and the converse of this is p q , or “If there is a question whose answer is E, I will place individually.” E. 2. Regardless how the shapes are fitted together, their area remains the same. Thus, we can fit these rhombi s2 3 3 2 4 2 s 3 . A to form a regular hexagon whose area with side s is A 6 3. Refer to figure at right. Suppose the radius of the circle is x. Because the triangle is right, the distance from the point of tangency to the radius to the right angle must also be x. From this, the other portions of the sides can be represented by 20 – x and 21 – x, and because tangents that meet at the same external point must be equal, the hypotenuse must be composed of 20 – x and 21 – x, as shown in the figure. So, 20 x 21 x 29 , and x 6 . Therefore, Atriangle Acircle .5 21 20 6 210 36 . B 2 4. SA r 2 rl r 2 r r 2 h2 , and since r h , SA r 2 r 2r 2 r 2 1 2 . C 5. The longest length of DE occurs when the two triangles are on opposite sides of the line segment. Incidentally, the sides of the two triangles form DE ; to prove this, mACD mBCE 60 because of equilateral triangles, and since they both lie on segment AB they become vertical angles and the segments EC and DC must coincide on the same line as DE . Thus, DE = 6. D 52 102 102 225 15 . B 6. 7. I and III are false when the triangle inscribed is a right triangle. The larger circle’s center coincides with the triangle’s circumcenter and not its incenter, so II is false. IV is true since the ratio of the circle’s radii is 2 :1 as proven by the 30-60-90 special triangle rules. Therefore, A 8. Given a set perimeter, the circle maximizes the enclosed area. A 9. For the moment we can divide the trapezoid into two triangles and one rectangle. Since the hypotenuse of the triangle including the 45 angle is 4, each side must be 2 2 , which is also the rectangle’s height. Thus, the area of this triangle is .5 2 2 2 2 4 and the area of the rectangle is 2 2 4 8 2 . The height of 2 2 is across the 30 angle in the other triangle, making the other leg of length 2 6 ; thus the area is .5 2 6 2 2 2 12 4 3 , and the area of the trapezoid is 4 8 2 4 3 . C 10. Because the original triangle is equilateral, its tangent circles are of equal radius length and are equidistant from each other—and thus the triangle formed by their centers is also equilateral. In fact, the original triangle makes up a fourth of the larger triangle’s area, and its dimensions are half of the larger one’s (in the case of an equilateral triangle); because the circles are equidistant, point A bisects the larger triangle’s March Regional Geometry Individual Solutions side, as do its equivalents on the other sides. Therefore, we only need the radius of the circle to determine AB . If we draw a perpendicular from the radius of a circle to its tangency with the triangle, a special 30-6090 triangle is formed, thus, the radius is 3 3 3 3 63 3 . Therefore, AB 3 .A 2 2 2 11. The area of a triangle given side lengths a, b, and c is A s s a s b s c , where s is the semiperimeter of the triangle. So, s 567 9 , and A 9 9 5 9 6 9 7 6 6 . C 2 12. The angle measure in a regular pentagon is 108 . The numbers 106, 107, 108, 109 and 110 are consecutive natural numbers and have the same sum as 5 108 , and can therefore be the angles of a pentagon; the smallest of these numbers is 106. D 13. Given the trapezoid is isosceles and since the midsegment m is parallel to its bases, m Since Atrapezoid 1 b1 b2 12 . 2 1 h b1 b2 , 60 12h and h 5 . B 2 14. Solving for the cone’s radius, 6 1 r 2 8 , 3 3 18 1 2 . Because the right cone’s 2 r , and r 2 8 circular cross sections’ radii are all proportional to the perpendicular distance to the cone’s apex (its height), 3 6 3 8 , and 2 , where x is the radius of the sliced cone; 8x , x 4 4 x 2 2 6 3 21 1 3 3 1 3 6 . Since Vlargecone .D 4 8 , our answer is 3 4 4 3 2 4 4 we can create a proportion: Vsmallcone 15. Since AB AD , AE AE , and BE DE (because of the properties of the kite), we can say that triangles BEA and DEA are congruent by the SSS theorem, and we can also say that the area of triangle BEA is half that of BAD. Therefore, the ratio of the areas of triangle BCE and BEA is 5: 3 . Suppose BE x and 4x ABCE 5 12 4 AE y . Then, 2 , and y . C 5 ABAE 3 xy y 2 16. We can treat the handshakes as the number of diagonals and sides of a regular polygon, and add Werther’s extra handshakes afterwards. By deductive reasoning, since 3 sides yields 3, 4 yields 6, 5 yields 10 and etc, we can describe this pattern as n n 1 15 15 1 15 7 105 . Then, for n 2 . With n 15 , 2 2 to add Werther’s additional 14 handshakes (since he presumably does not give himself a handshake), 105 14 119 . C March Regional Geometry Individual Solutions 17. By the tangent-secant theorem, 7 3 JL , and JL 2 49 40 40 3 . Since C d , C .B 3 3 3 1 18. mNLJ 110 46 32 . B 2 49 . Therefore, JM JL ML 3 19. If we bisect a 60 angle, we end up with angle measures of 30 , and then if we repeat this process we get 15 . Similarly, bisecting a 45-degree angle gives us 22.5 , and repeating the process gives us 11.25 . Because 26.25 15 11.25 , 33.75 22.5 11.25 , 37.5 22.5 15 and 67.5 45 22.5 , our answer is E 20. Triangle ABC is proportional to triangle BDC , which is proportional to both triangle BED and triangle DEC . We can say that 9 8 6 , and thus DC . Using the Pythagorean Theorem to find 2 6 DC 2 3 7 9 the length of BC , we get BC 6 . Finally, setting up one more proportion, we 2 2 2 BC DC have , DC EC 3 7 9 2 2 , and EC 81 2 27 7 . D 9 14 EC 4 3 7 2 21. The intersection of the altitudes and the intersection of the perpendicular bisectors of a triangle, or otherwise the orthocenter and the circumcenter respectively, can exist outside the perimeter of the triangle if the given triangle is obtuse. Only the centroid and the incenter, or the intersection of the medians and angle bisectors respectively, exist solely within the triangle for all cases. Therefore, B 22. Setting up the proportion between the man and the lamppost, from the lamppost), 24 6x 12 and x 2 . B 12 x 2 (where x is how far he is 6 2 23. If the angle is not divisible by 180, then it cannot be the sum of the interior angles in a regular polygon. Since 1080 3420 7380 6940 6, 19 and 41 , but 38 39 , the answer is C 180 180 180 180 24. A triangular prism capped with tetrahedrons has faces composed of three rectangles and six equilateral triangles. These rectangles make up an area of 3 10 4 120 square units. The six equilateral triangles 42 3 have area of 6 24 3 square units, and so the total surface area is 120 24 3 . B 4 25. The shape described is a rectangle, and rotated about the given line gives us a cylinder of radius 2 and height of 6. Therefore, our volume is V r 2 h 2 6 24 . B 2 March Regional Geometry Individual Solutions 26. A is false, since two skew lines do not like in the same plane. C is false, as the vertices of a triangle prove this. D can be false if one line lies on the xy-plane and the other on the xz-plane; both can cross the x-axis at the same place and both can have the same slope, but they are not the same line. B is true, since all sets of three points determine some plane, and thus three points must be coplanar. Therefore, B 27. By corresponding angles and by vertical angles, we can prove that angles 1, 2, and 3 form a 180 angle. Therefore, our answer is 180 30 150 . B 28. Since mCAB m2 , and since the other corresponding angles of the triangles are the same (as a result of parallel lines), the two triangles are proportional by the AAA theorem. Thus, 4 AB 6 2 , and the ratio between the areas is therefore . A 9 AE 9 3 29. Because the lines are parallel we can conclude that m1 mABC 30 . Drawing the height perpendicular to base AB , we can say A .5bh .5 6 h 3 7sin 30 21 . A 2 30. Choice A gives us SAS, choice B gives us SSA, choice C gives us SAA, and choice D gives us ASA. Of these, SSA does not prove congruency. B