HMMT 二月 1999 · 几何 · 第 9 题

英文解析

Geometry Solutions
Harvard-MIT Math Tournament
February 27, 1999
Problem G1 [3]
Two 10 × 24 rectangles are inscribed in a circle as shown. Find the shaded area.
Solution: The rectangles are 10 × 24, so their diagonals, which are diameters of the circle, havelength 26. Therefore the area of the circle is π 13 , and the overlap is a 10 × 10 square, so the 2
2 2
shaded area is π 13 − 2 · 10 · 24 + 10 = 169 π − 380 .
Problem G2 [3]
A semicircle is inscribed in a semicircle of radius 2 as shown. Find the radius of the smallersemicircle.
Solution: Draw a line from the center of the smaller semicircle to the center of the larger one, anda line from the center of the larger semicircle to one of the other points of intersection of the twosemicircles. We now have a right triangle whose legs are both the radius of the smaller semicircle
√
and whose hypotenuse is 2, therefore the radius of the smaller semicircle is 2 .
Problem G3 [4]
In a cube with side length 6, what is the volume of the tetrahedron formed by any vertex and thethree vertices connected to that vertex by edges of the cube?
Solution: We have a tetrahedron whose base is half a face of the cube and whose height is the side
1 1
length of the cube, so its volume is · ( · 6 ) · 6 = 36 .2
3 2
Problem G4 [4]1
A cross-section of a river is a trapezoid with bases 10 and 16 and slanted sides of length 5. At thissection the water is flowing at π mph. A little ways downstream is a dam where the water flowsthrough 4 identical circular holes at 16 mph. What is the radius of the holes?
Solution: The volume of water going through any cross-section of the river in an hour (assumingthe cross-sections are parallel) is the area times the velocity. The trapezoid has height 4, hencearea 52, so the volume of water going through at any hour is 52 π . Let r be the radius of the holes,
√
2 2 213
then the total area is 4 πr , so the volume of water is 64 πr . Therefore 64 πr = 52 π , so r = .
Problem G5 [5]4
In triangle BEN shown below with its altitudes intersecting at X , N A = 7 , EA = 3 , AX = 4, and
N S = 8. Find the area of BEN .
B EH
S AX
Solution: The idea is to try to find a base and height for the triangle so that we can find the area.N
√
By the Pythagorean theorem, EX = 5, N X = 65, and SX = 1. Triangles AXE and BXS aresimilar since they have the same angles. The ratio of their side lengths is 4:1, so BS = 3 / 4 and
1 3
BX = 5 / 4. Now using either N E or N B as a base, we get that the area of BEN is · (8 + ) · 6
2 4
1 5 105
or · (4 + ) · 10, both of which simplify to .
2 4 4
Problem G6 [5]
A sphere of radius 1 is covered in ink and rolling around between concentric spheres of radii 3 and
5. If this process traces a region of area 1 on the larger sphere, what is the area of the region tracedon the smaller sphere?
Solution: The figure drawn on the smaller sphere is just a scaled down version of what was drawnon the larger sphere, so the ratio of the areas is the ratio of the surface area of the spheres. Thisis the same as the ratio of the squares of the radii, which is .9
Problem G7 [5]25
A dart is thrown at a square dartboard of side length 2 so that it hits completely randomly. Whatis the probability that it hits closer to the center than any corner, but within a distance 1 of acorner?
Solution: By symmetry it will suffice to consider one quarter of the dartboard, which is a square 2
of side length 1. Therefore the probability is the area of the desired region in this square. Thedesired region is the part of the circle of radius 1 centered at a corner that is closer to the oppositecorner. The points closer to the opposite corner are those that are on the other side of the diagonalthrough the other two corners, so the desired region is a quarter of a circle of radius 1 minus a rightπ − 2
triangle with legs of length 1. Therefore the area (and hence the probability) is .
Problem G8 [6]4
Squares ABKL, BCM N, CAOP are drawn externally on the sides of a triangle ABC . The line
′ ′ ′ ′ ′ ′
segments KL, M N, OP , when extended, form a triangle A B C . Find the area of A B C if ABCis an equilateral triangle of side length 2.
√
Solution: Triangle ABC has area 3, and each of the three squares has area 4. The three remainingregions are congruent, so just consider the one that includes vertex B . Triangle KBN has two sides
°
of length 2 and an angle of 120 between them, to bisecting that angle we get two halves of an
√
equilateral triangle of side length 2, so the area is again 3. The remaining region is an equilateral
√ √ √ √
2 ′ ′ ′
triangle of side length 2 3, so its area is (2 3) 3 / 4 = 3 3. Therefore the area of A B C is
√ √ √ √
3 + 3 · 4 + 3 · 3 + 3 · 3 3 = 12 + 13 3 .
Note that this problem is still solvable, but much harder, if the first triangle is not equilateral.
Problem G9 [7]
A regular tetrahedron has two vertices on the body diagonal of a cube with side length 12. Theother two vertices lie on one of the face diagonals not intersecting that body diagonal. Find theside length of the tetrahedron.
Solution: Let ABCD be a tetrahedron of side s . We want to find the distance between two ofits opposite sides. Let E be the midpoint of AD , F the midpoint of BC . Then AE = s/ 2,
√
°
AF = s 3 / 2, and angle AEF = 90 . So the distance between the two opposite sides is EF =
√ √
√
2 2 2 2
AF − AE = 3 s / 4 − s / 4 = s/ 2.
Now we find the distance between a body diagonal and a face diagonal of a cube of side a . Let Obe the center of the cube and P be the midpoint of the face diagonal. Then the plane containing
P and the body diagonal is perpendicular to the face diagonal. So the distance between the body
√
a 2
and face diagonals is the distance between P and the body diagonal, which is (the altitude
2 3
√
s a 2
√
from P of right triangle OP Q , where Q is the appropriate vertex of the cube). So now = ,
2 3
√ √ √2
thus s = a/ 3 = 12 / 3 = 4 3 .
Problem G10 [8]
In the figure below, AB = 15 , BD = 18 , AF = 15 , DF = 12 , BE = 24, and CF = 17. Find
BG : F G .
Solution: Our goal is to find the lengths BG and F G . There are several ways to go about doingthis, but we will show only one here. We will make several uses of Stewart’s theorem, which can
B C3
FDAGbe proved using the law of cosines twice. By Stewart’s theorem on triangle ABD and line BF ,E
2 2 2
15 · 12 + 18 · 15 = BF · 27 + 15 · 12 · 27, so BF = 10 and EF = 14. By Stewart’s theorem
√
2 2 2
on triangle ABE and line AF , AE · 10 + 15 · 14 = 15 · 24 + 14 · 10 · 24, so AE = 561. By
2 2
Stewart’s theorem on triangle AED and line EF , ED · 15 + 561 · 12 = 14 · 27 + 12 · 15 · 27,
√ √
2 2
so ED = 2 57. By Stewart’s theorem on triangle CF E and line F D , 14 · CD + 17 · 2 57 =
√ √ √ √ √
12 · ( CD + 2 57) + 2 57 · CD · ( CD + 2 57), so CD = 57 and CE = 3 57. Note that 2
DG = 18 − BG and apply Menelaus’ theorem to triangle BED and the line through C , G , and F
18 − BG 10 135
to get 3 · · = 1, so BG = . Similarly CG = 17 − F G , so applying Menelaus’ theorem to
BG 14 11
24 F G 1 85
triangle CF E and the line through B , G , and D we get · · = 1, so F G = . Therefore
10 17 − F G 2 11
BG : F G = 27 : 17 . 4