Solutions - FloridaMAO

February Regional
Geometry Individual Solutions
1. B – 25! − 24! = 7. 7(24)/2 = 84.
2. D –
5−2
!
+ 18 − 4
!
=
205. x = 205. Midpoint: (
!!! !"!!
!
,
!
) = (7/2, 11). 7/2 + 11 = y. x
+ y = 205 + 7/2 + 11 = 439/2
3. A – 2x + 40 + 3x + 10 + 5x + 30 = 180. x = 10. Z is the largest angle. 5(10) + 30 = 80. The
complement of 80 is 10.
4. D – (5x + 20) = (3x + 44). x = 12. 5(12) + 20 = 80. Consecutive angles in a parallelogram are
supplementary. So y = 180 – 80 = 100.
5. B – AF bisects CD so ED = EC = 84. Triangle ADE is a right triangle with hypotenuse 85 and leg
84. Using Pythagorean theorem, AE is equal to 13. EF = AF – AE = 85 – 13 = 72.
6. E – AE = AD = AC = 8 since these are all radii of the circle. Angle EAD = ADC = 30 since AE and
CD are parallel. Since triangle ADC is isosceles with AD = AC, angle ADC = angle ACD = 30.
Draw a segment from point A, bisecting CD. Triangle CAD is now split into two 30-60-90
triangles. Using these ratios, the height of triangle ADC = 4 and the base = 8 3. bh/2 = 16 3
7. D – Angle EAD = 30 and angle CAD = 120, so angle CAE = 150. 360 – 150 = 210. The radius of
𝟏𝟏𝟐!
the circle is 8. (210/360)(𝜋)(8! ) =
𝟑
8. A – One side length “s” measures 27/3 = 9. Area = (𝑠 ! 3)/4 =
!" !
!
9. D – cos(BAC) = adjacent/hypotenuse = 8/17. So 8 is the length of one side of the triangle and
17 is the length of the hypotenuse. Using Pythagorean theorem, the third side is 15. Since
tangent(BAC) = opposite/adjacent = 15/8.
10. C – If the original statement is true, so is the contrapositive.
11. B – The contrapositive of the contrapositive = the original statement. The inverse of the inverse
= the original statement. The converse of the converse = the original statement. The original
statement = “If I like to eat potatoes, then I am cool”
12. B – Connect points A to O and O to B. Draw a circle with diameter AB, circumscribed around
quadrilateral ACBO. Label the radius of the circle as R. Hypotenuse AB of triangle AOB forms the
diameter, 2R, and angle AOB is a right angle. AOandBOwouldbothberadiiofthesquare.Thus
anglesOABandOBAwouldeachmeasure45degreesandAOandBOwouldeachmeasure𝑅 2. Via
Ptolemy: (CO)(2R) = (𝑅 2)(x) +(𝑅 2)(y). R can be divided through by both sides; factor
out 2, and divide through by 2 to determine (CO) = [( 2)(x + y)]/2
13. B – [n(n – 3 )]/2 = 9. n = 6, hexagon. Area = (6𝑠 ! 3)/4 and s = 4. So the area = 24 3.
14. C – CG = CF, FB = AB, AD = GD. Perimeter = 2(2) + 2(5) + 3(5) = 20.
15. C – Inradius = Area of triangle/semi-perimeter of triangle. Using Heron’s formula, the area of
the triangle with side lengths 5, 7, 8 is 10 3. The semi-perimeter is 10. So the radius of circle E
= 3. The area is 3𝜋.
16. A – Definition
17. E – tan 30 =
!"# !"
!"# !"
. So (cos 30)
!
!"# !"
tan 30 tan 45 = tan 45. tan 45 =
!"# !"
!"# !"
= 1.
18. D – Using the Angle Bisector Theorem, AB/AD = BC/DC. 4/3 = 8/DC. AC = AD + DC = 3 +6 = 9.
19. B – Triangle inequality states that the sum of the two shortest sides must be greater than the
longest side. 16 + x > 23. 16 + 23 > x. 7 < x < 39. x can be 8, 9, 10, … , 38 = 31 values.
20. D – Using shoelace theorem, starting with (3, 1) and going in counter clockwise order: 3(13) + 5(4) + 7(1) = x. 1(-5) + 13(7) + 4(3) = y. |x – y | divided by 2 = 36.
February Regional
Geometry Individual Solutions
21. A – GH is a median. (Base 1) – (Base 2) divided by 2 = length of IJ =
22. A – 𝐸𝐹 =
23.
24.
25.
26.
27.
28.
29.
30.
!(!"#$ !)(!"#$ !)
!"#$ ! !(!"#$ !)
=
!(!)(!)
!!!
=
!"
!!
!
!
OR Extend the non-parallel sides of the trapezoid to
intersect and form a triangle. Let L be the height of the triangle (where a is the base of the
triangle and L does not include the height of the trapezoid.)Let h be the height of the trapezoid
with bases a and c. Using similar triangles: L/(L + h) = 3/8. 8L = 3L + 3h. 5L = 3h. The
distance from point K to base a is 3h/11 and the distance from point K to base c is 8h/11. Using
similar triangles, 3/EF = L/(L + 3h/11). So EF = 3(L + 3h/11)/L = 3 + 9h/11L = 3 + 15L/11L
(from 5L = 3h) = 3 + 15/11 = 48/11.
C – (12)(GD) = (9)(18 – 9). GD = 81/12. CD = CG + GD = 12 + 81/12 = 75/4.
B – [(x + 40) + (2x + 90)]/2 = 80. x = 10. Minor arc CF = 10 + 40 = 50.
B – BC is the shortest side of triangle ABC. So there is a 12:4 or 3:1 ratio between the side
lengths of triangle DEF to ABC. The perimeter of triangle DEF = 13(3) + 4(3) + 10(3) = 81.
E – Triangle AFE is a right triangle, so angle AEF = 180 – 90 – 54 = 36. Angle DEF = 180 – 36 =
144.
D – Angle EDC = angle BCD = 90 degrees. Figure FEDCX forms a pentagon. The sum of the
interior angles of a pentagon add up to 540 degrees. From #26, we already know that angle DEF
= 144. 540 – 90 – 90 – 144 – 66 = 150 degrees.
C – a + b + c = 60. 𝑎 ! + 𝑏 ! = 𝑐 ! . 𝑎 ! + 𝑏 ! + 𝑐 ! = 1352. 2𝑐 ! = 1352, c = 26. a + b = 34,
𝑎 ! + 2𝑎𝑏 + 𝑏 ! = 1156. a = 34 – b and ab = 240. a(34 – b) = 240. (a – 24)(a – 10). Length of the
legs = 24 and 10. 4 and 0 are the units digits, so the sum = 4.
E – AB corresponds to DE. The ratio of the areas = (DE)^2: (AB)^2 = 64: 4 = 16:1.
C – AB = DC = 7 and BC = AD = 10. Triangle ABD has side length AB = 7 and AD = 10 and angle
A = 60 degrees. Using law of cosines: 7! + 10! − 2 7 10 𝑐𝑜𝑠60 = 𝐵𝐷 ! = 79.