Download Geometry Team Solutions Vero Beach Invitational January16, 2010

Geometry Team Solutions
#1
297
#2
98
#6
#3
10  8 3
#7 11
Vero Beach Invitational
#5 32
#4 43
#8
#1 297
A: 20
#9
4
151
15
7
#10 12
January16, 2010
#13 9
#14
12c  4d  24b  9
12
#11 235
#12
6
2
B: 18
C: 2880
D: 135
360
 18 , total of exterior angles, one at each vertex,
20
n(n  3) 18  15
divided by one exterior angle; C = 160  18  2880 ; D =

 135 ;
2
2
C
2880
BD
 18  135 = 297
A
20
A : 180  160  20 ; B =
#2 98
A: 140
B: 42
A: Let mOCA  x, mOCB  40  x , then
mOBA  40  x, mOBC  x, making mBOC  140
B : 15x  180, x  12, m3  48, measure of the complement is 42.
A  B  140 – 42 = 98
#3 10  8 3
A: 4  2 3
B: 6  6 3
C: 4 3
A: Extend FE so that it is perpendicular to AB and call the point where it intersects AB, G.
The length of FG is the same as a side of the square which is 4. AE  EB  4 because the
triangle is equilateral and AB = 4 because it is a side of the square, making EG  2 3 since
is a 30-60-90 triangle and EG is opposite the 60 degree angle.
Since FG  4, EG  2 3, EF  4  2 3 .
B: The longer leg is 6, side opposite 30 is
6
 2 3 , hypotenuse is 4 3 making the perimeter 6  6 3
3
C: One side is 2 5 . Diagonals bisect each other and are perpendicular, so one small triangle formed
by the diagonals intersection has a hypotenuse of 2 5 , one leg of 2 2 (1/2 diagonal) and using
Pythagorean theorem the other side of the triangle is 2 3 making the 2nd diagonal 4 3 .
R  S  T  4  2 3  6  6 3  4 3  10  8 3 .
Geometry Team Solutions
#4 43
Vero Beach Invitational
January16, 2010
A: 60
B: 117
C: 100
A: Let the upper base and the legs each have a value of x. The lower base is 2x . Drawing the altitudes
forms 2 triangles and a rectangle. The base of the rectangle is equal to the upper base. Making
1
the other two sides of the lower base equal to x . Since this is ½ the leg (hypotenuse of the
2
triangles formed), the triangles are 30-60-90. The angles of the trapezoid are 60 and 120 with
60 being the smallest angle measure.
B: Angles Q and T are supplementary since they are same side interior angles of the parallel bases.
3x  6  2 x  11, x  37, mQ  117, mT  63 , largest angle is 117.
C: Same side interior angles are supplementary so mABC  70, mCBD  30 so the
mABD  mBAD  40 . mADB  180   mBAD  mDBA  180  80  100 .
C  A  B = 100 + 60 – 117 = 43
#5 32
R: 22
S: 70
T: 60
R: mABO  mOBC  90 so we have 2 x  y  6 x  8  90,8 x  y  82.
mAOB  mBOC  180 so we have 4 x  23 y  86 . Solve this system of equations
to get x  10, y  2, mABO  22.
S: Let x be the measure of the angle. 180  x  180  (90  x)   60, x  20. The angle has
measure 20 and the complement has a measure of 70.
T: 2  ACB since the legs are congruent. mACB  m3  2 x  12. m2  180  m1  180  5 x .
2 x  12  180  5x, x  24 . The m1  120, m  60 .
R  S  T = 22 + 70 – 60 = 32.
#6 4
A: 5
B: 36
C: 45
A: A short stack has a height of 3.8 mm. Dividing 55 by 3.8 we get a whole number answer of 14,
which means there are 14 short stacks before we get to 55 mm. Since 3.8 x 14 = 53.2, we know that
we will start a new stack at this point. Then: 14 stacks + 1 quarter = 54.5 mm.
14 stacks + 1 quarter + 1 nickel = 56.0. thus the coin at the top is a nickel which is 5 cents.
B: There are 8 rectangles singly. Using two at a time, there are 7, using 3 at a time there are 6, see a
pattern? For a total of 36.
x
2
C: Let AB  x, BP  75  x ,
 , x  30  AB, BP  45.
75  x 3
A
5
B   36   4
C
45
#7 11
With the given information, ABC  ADC by SAS. Using this, makes the corresponding
parts congruent and makes ABD isosceles. Therefore: 1) T, 2) F, 3) T, 4) F, 5) T, 6) F, 7) T.
4 true statements are worth 8, 3 false statements are worth 3, 8 + 3 = 11.
Geometry Team Solutions
#8 151
Vero Beach Invitational
January16, 2010
20
7
A: Let the angles be x, y, z . x  30  y, x  z  15, x  y  z  180. Substituting for y and z
A: 35
B: 51
C:
x  x  30  x  15  180, x  65. Making y  35, z  80. The smallest angle is 35.
B: Since RT is the base, XT  XR,2 x  8  5x  7, x  5 . RT  15, RX  18, TX  18
and the perimeter is 51.
C: Let the longer part of the hypotenuse be x and the shorter part would be 5  x .
The angle bisector bisects the third side into lengths proportional to the sides of the
5 x x
20
triangle, so we have
.
 ,x 
3
4
7
20
A  C  B  35   51  151 .
7
#9
15
7
R:
3
S:
45
7
R: Diagonals of a rectangle bisect each other making BE  4 3  6 . BEC is a 30-60-90
right triangle. BT  2 3  3 since it’s opposite the 30  angle. ET  6  3 3 . We are looking for
the length of AT which is AE  ET  4 3  6  6  3 3  3 .
S: mDAB  mBAF  45 . Now using BDA, mADB  90, mDBA  45 . Since DB  9, AD  9 .
The sides of the rectangle are now 12 and 9 making the diagonal DF  15 .
15  x 12
45
DBC ~ FAC so letting DC  x, DF  15  x we have
.
 ,x 
x
9
7
45
S
15
The value of 2  7  .
R
3
7
#10 12
The three triangles are similar so for lengths of sides either use proportions, geometric mean formulas,
or Pythagorean Theorem. And be sure when listing triangles or angles, that corresponding parts must
correspond. 1) F, 2) F, 3) F, 4) T, 5) F, 6) T, 7) F, 8) T, 9) F . 3 true statements X 2 = 6. 6 false statements
6+6 = 12.
Geometry Team Solutions
#11 235
V: 14
Vero Beach Invitational
E: 39
R:
January16, 2010
17
3
V: To solve this, draw the triangle formed by connecting the midpoints of each side. Then you
will have similar triangles, so proportions can be set up to solve for the lengths of the
segments drawn by connecting the points of trisection. As is true in the formula for connecting
midpoints that each side is ½ triangle formed is, we can conclude that the sides of the hexagon
7 8 7 8
are 1/ 3 of each side of the triangle. The sides of the hexagon are 2, , ,2, , making the
3 3 3 3
perimeter 14.
2x  3 4x  6
5 3
3
E: Set up the proportion:

, x  , . Reject the as it makes 6 x  6 a negative number.
8x  2 6 x  6
2 5
5
5
The perimeter of 20 x  11 and substituting makes the perimeter 39.
2
R: Since TA=15, AM=8, using Pythag, TM=17. The centroid divided the median into two parts
17
with a ratio of 2:1. with QM being the 1 part. So QM =
.
3
V  E  R  14  39 
6
2
#12
17
 235
3
Letting the side of the triangle equal x, makes the perimeter of the triangle 3x . A side of the
square with a perimeter of 3x is
#13 9
3
x.
4
3
3 2
2 x
x.
4
4
A: The diagonal of the square is
2 times the side length so
B. The altitude of an equilateral is
1
x
3.
3 times the hypotenuse so the altitude has length
2
2
3
x 2
A 4
3 2
6


The exact value of 
.
x
B
2
3 2 3
2
It can be proved that MQS  MNS by SAS. Be sure when listing similarity and
congruency that corresponding parts match. 1) T, 2) F, 3) F,4) T, 5) F, 6) T.
3 True statements x 2 = 6, 3 false statements, 6+3=9.
#14
12c  4d  24b  9
12
The sum of the remote interior angles equals the exterior angle.
Therefore, mZ  mX  mY .
d
3
c   2b  is what is wanted. The common denominator is 12.
3
4