FGCU Invitational
Geometry Team Solutions
2014
1)
𝟐
(−𝟏𝟏𝟑, − )
𝟑
2)
E
Each of the regular polygons listed can be constructed with only a compass and
straightedge: square, regular pentagon, regular dodecagon, and regular 17-gon.
3)
𝟓
𝟐
Area of a trapezoid = ∙ ℎ(𝑏1 + 𝑏2 ) =
4)
𝟖 + 𝟖√𝟐 − 𝟖√𝟑
x-coordinate: −171 +
1
(3 −
3
(−171)); y-coordinate: 26 −
1
2
1
2
1
(26 −
3
(−54))
∙ 2(12 + 𝑏2 ) = 27, so 𝑏2 = 15.
3 2
3
25
Half the excess length of 𝑏2 is 2. The desired length is √22 + (2) = √ 4
Add four corner triangles to form a square. A side of the square is 2 + 2√2.
Area of the square, minus area of the four corner triangles, minus the area of
2
1
1
all eight equilateral triangles = (2 + 2√2) − 4 ( ∙ √2 ∙ √2) − 8 ( ∙ 2√3)
2
2
= 4 + 8√2 + 8 − 4 − 8√3.
5)
9 hours
Volume of frustrum = (1/3)h(a2 ab b2) = (1/3)(12)(a62 6(15) 152) = 1404 ft3
fill time = 1404 ft3 /(26 ft3 /10 min) =
4∙9∙39 𝑓𝑡 3 ∙10𝑚𝑖𝑛
2∙13 𝑓𝑡 3
= 2 ∙ 9 ∙ 3 ∙ 10 𝑚𝑖𝑛
or 9(60 min) = 9 hr
6)
x=3
area(□ABCD) = area(∆AEF) + [area(∆ABE) + area(∆ADF)] + area(∆CEF)
25 =
21
1
1
+ 2 [ ∙ 𝐴𝐷 ∙ 𝐹𝐷 ] + ∙ 𝐶𝐸 ∙ 𝐶𝐹
2
2
2
21
1 2
25 =
+ 5(5 − 𝑥) + 𝑥
2
2
0 = 21 − 10𝑥 + 𝑥 2
The solutions to the quadratic are x = 3 and x = 7, but x = 7 makes line segment
CE longer than a side of square ABCD, thus is extraneous.
7)
362
1
For a convex n-gon the number diagonals is 𝐷 = 2 𝑛(𝑛 − 3) and the number
of degrees of its interior angles is 𝑇 = 180(𝑛 − 2). We want the minimum
1
n such that 𝐷 > 𝑇: 𝑛(𝑛 − 3) > 180(𝑛 − 2)
2
𝑛(𝑛 − 3) > 360(𝑛 − 2)
𝑛2 − 3𝑛 > 360𝑛 − 720)
𝑛2 − 363𝑛 + 720 > 0
We can see that the left side is positive for n = 363; it reduces to 720. It just
remains to see if the left side is positive for any slightly smaller value of n.
𝑛 = 362: 3622 − (363)(362) + 720
= 3622 − 3622 − 362 + 720 = 358
𝑛 = 361: 3612 − (363)(361) + 720
= 3612 − 3612 − 2(361) + 720 = −2
Page 1 of 2
FGCU Invitational
8)
Kite
9)
𝟗
√𝟑
𝟐
Geometry Team Solutions
Slide two of the triangles to each of vertices
A, C, and E, as shown. The star’s area can now
be divided into three equal regular hexagons
of side 1, which can further be divided into 18
equilateral triangles of side 1. And that total
1
area is 18 (2 ∙ 1 ∙
10)
B. trapezoid
11)
117 feet
12)
𝟒𝟓 ≤ 𝒉 ≤ 𝟓𝟐. 𝟓
2014
√3
)
2
9
√3
2
=
quadrilateral LJKC: x + 930 + 890 + 580 = 3600  x = 1200
∠CLN = 870, ∠LCN = 600
triangle LCN: y + 870 + 600 = 1800  y = 330
triangle BHG: 3y = 390 + ∠HBG  ∠HBG = ∠IBL = 600
⃡ || 𝐶𝐷
⃡ .
so we now know 𝐴𝐵
5. ∠DMN = 10y – 2x = 3300 – 2400 = 900
6. quadrilateral ABLM: ∠BAM + 1200 + 930 + 900 = 3600  ∠BAM = 570
OR: triangle MDN: ∠MDN + 900 + 330 = 1800  ∠MDN = 570
so we now know ⃡𝐴𝐷 𝑛𝑜𝑡 || ⃡𝐵𝐶 .
1.
2.
3.
4.
Similar triangles:
ℎ
27.3 𝑓𝑡
=
36 𝑖𝑛
8.4 𝑖𝑛
 ℎ=
9(27.3)𝑓𝑡
2.1
=
3(273)
7
𝑓𝑡 = 3(39)𝑓𝑡
The shadow of the hedge extends 4 feet from the face of the hedge farther
from the tree.
𝑠
5 𝑓𝑡
8 𝑖𝑛
= 10 𝑖𝑛  𝑠 =
8(60𝑖𝑛)
10
= 48 𝑖𝑛
Thus the shadow of the tree is minimally 36 ft (to the hedge) and maximally
42 ft (to the end of the hedge’s shadow).
Similar triangles:
and:
13)
14)
𝟏𝟕 ≤ 𝑨 ≤ 𝟔𝟗
3598
Page 2 of 2
ℎ𝑚𝑖𝑛
36 𝑓𝑡
ℎ𝑚𝑎𝑥
42 𝑓𝑡
10 𝑖𝑛
8 𝑖𝑛
10 𝑖𝑛
=
8 𝑖𝑛
=
minimum: 16
6x1 = 6
1x1 = 1
1x9 = 9
 ℎ𝑚𝑖𝑛 =
 ℎ𝑚𝑎𝑥 =
360 𝑓𝑡
= 45 𝑓𝑡
8
420 𝑓𝑡
= 52.5 𝑓𝑡
8
maximum: 69
7 x 10 = 70
less 1 x 1 = 1
𝑥 2 + 4𝑛 > 1202
9 + 4𝑛 > 14400
4𝑛 > 14391
𝑛 > 3597.75
min 𝑛 = 3598