Download 2013 Student Delegates Statewide ______ Solutions

2013 Student Delegates Statewide _______ Solutions - Geometry Individual Test
1. (B) Although C, D, E, and F are not on the same face, they are coplanar.
2. (E: 25) Using the distance formula, the distance is
−7 − 8
!
+ 12 − −8
!
= 625 = 25.
Alternatively, use the 15-20-25 Pythagorean triple.
3. (C) The altitude makes two congruent right triangles (congruent by Hypotenuse-Leg) with
hypotenuse 12 and one leg 12/2 = 6. (In general, due to HL and CPCTC, the altitude to the base
of an isosceles triangle is also an angle bisector and a median.) The altitude is the other leg
whose length, using the Pythagorean formula, is 12! − 6! = 6 3. 4. (D): All four are true, well-known facts. Note that being “on” a circle does not mean being
“inside” a circle.
5. (D) For each of the 7 vertices, there is one diagonal from it to a vertex that is not itself or a
neighboring vertex (a side is not a diagonal). Thus there are 7-3=4 diagonals emanating from
each vertex, and 7 vertices, making 28 diagonals. But we have counted each diagonal twice, once
for each of its endpoints. Thus there are 28/2 = 14 diagonals. In general, a polygon with n sides
has n(n-3)/2 diagonals.
6. (C) Choose an arbitrary vertex and draw all the diagonals which come from it. This splits the
interior of the octagon into those of 6 triangles whose interior angles comprise those of the
octagon. The sum of the interior angles of the octagon is thus 180(6) = 1080.
7. (A) Let AE and BF be heights. By AAS (base angles of an isosceles trapezoid are congruent),
∆BCF and ∆ADE are congruent. Thus DE = CF = (16-10)/2 = 3. Pythagorean Theorem on ∆BCF
or ∆ADE gives a height of 8! − 3! = 55.
8. (C) Angles AMN and ABC are corresponding angles, so ∆AMN is similar to ∆ABC by AngleAngle. Thus AM/AB = AN/NC, so 6/(6+10) = 8/(8+NC). Solving gives NC = 40/3.
9. (C) See diagram. Let MA = 3x and MB = x. Also let NA = 3y and NB = y. Then AB = 3y-y =
2y. But also AB = 3x+x = 4x, so we have 2y = 4x. We also have MN = x + y = 18. Solving this
system of equations gives x = 6 and y = 12, so AB = 4x = (4)(6) = 24. Remark: In this kind of
setup, M and N are said to create a harmonic division of the line segment AB.
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10. (C) Draw a chord passing through O and P. The point P splits this chord into segments of length
r + OP and r – OP. By the Power of a Point theorem for intersecting chords (or using similar
triangles), we have (AP)(PB) = (r + OP)(r – OP) = r2 – OP2, so (6)(4) = r2 – 22, so r2 = 28. Thus
the area of the circle is 28π.
11. (C) They are not collinear because no two lengths add to give the third. If ∆ABC is a right
triangle, then CA! would be 11! + 30! = 1021 < 32! = 1024. Since the larger the side, the
larger the angle opposite it, angle ABC is greater than 90 degrees and so ∆ABC is obtuse. We
can make this argument more rigorous with the law of cosines.
12. (D) The angle immediately above angle θ is also 42 degrees because the two angles are
corresponding. It and angle θ form a straight angle of 180 degrees, so angle θ is 180-42=138.
13. (B) Let the isosceles triangle be ∆ABC with vertex A, and let the median be BD. Drop
perpendiculars AE and DF onto the base BC. Note that EC = 16/2 = 8 by congruent triangles.
Triangles ∆AEC and ∆DFC are similar with coefficient 2. Then FC = 8/2 = 4 and BF = BC–FC
= 16–4 = 12. Also, since the altitude AE = 6 by the Pythagorean Theorem, DF = 6/2 = 3. By the
Pythagorean Theorem on right triangle ∆BFD, BD = 153 = 3 17.
14. (B) Let the triangle be ∆ABC with AB = 13, BC = 21, AC = 20. Let AD be an altitude with
length h, and let BD = x. Then by Pythagorean Theorem on right triangles ∆ABD and ∆ACD, we
have the following system of equations:
x2 + h2 = 132
(21-x)2 + h2 = 202
Solving for h gives h = 12, so the area is bh/2 = BC*h/2 = 21*12/2 = 126.
Remark: The answer can also be found with Heron’s Formula, which can be derived in the exact
same way.
15. (B) Opposite interior angles are congruent, so ∆ADB is similar to ∆EDC by Angle-Angle. Then
BA/BD = CE/CD. But note that ∆ACE is isosceles: because ∠BCE = ∠B by opposite interior
angles, so ∠AEC = 180-(∠A/2)- ∠B-∠C = ∠A/2 = ∠EAC. So CE = CA. Thus BA/BD = CA/CD.
This is the angle bisector theorem.
By the angle bisector theorem, 24/8 = 15/CD, so CD = 5.
16. (D) Use the well-known fact that the medians of a triangle split its interior into those of 6
triangles with equal areas. Let the midpoint of AC be M. Then the area of triangle ∆AMG is 72/6
= 12. Using the common formula for area, we have 12 = bh/2 = (AM)(h)/2 = 15h/2, where h is
the desired distance. Solving, we get h = 8/5.
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17. (B) Let the rectangular prism’s dimensions be l, w, and h. Then we have the following system of
three equations: lw = 8, wh = 9, and lh = 12. Multiplying the three equations together gives
(lwh)2 = (8)(9)(12) = (25)(33). Then the volume is the square root of this, lwh = 12 6.
18. (C) This is equal to the geometric mean of 6 and 24, 6 ∗ 24 = 12. The proof of this fact uses
the fact that the altitude cuts the triangles into two smaller right triangles which are similar to the
large right triangle and each other.
19. (B) Use the Triangle Inequality: the sum of two side lengths of a triangle is always strictly
greater than the third. Here we have 7+m > 14, so m > 7. Also, 7+14 > m, so m < 21. Putting
these together gives 7 < m < 21.
20. (A) In this solution, [P] denotes the area of polygon P. Let S be on AB such that NS || CM. Since
triangles ∆CMB and ∆NSB are similar with coefficient 2, SB = 1, so MS = 1, so AS = 2.
Triangles ∆ANS and ∆APM are also similar with coefficient 2; the height of ∆ANS is 2, so the
height of ∆APM is 1. Then [APM] = (1)(1)/2 = 1/2. [ACM] = (1)(4)/2 = 2, so [APC] = [ACM] –
[APM] = 2 – 1/2 = 3/2.
21. (D): (i) is true because it can be seen by SAS that there are congruent triangles, which by
CPCTC make equal opposite interior angles, proving the opposite sides parallel; (ii) is true, also
by using equal opposite interior angles; (iii) is true; it is the contrapositive of (ii), which is true,
and the contrapositive of a statement is logically equivalent to the original statement; (iv) is true
since it is logically equivalent to (i).
22. (C) It is a 30-60-90 right triangle with the side of length 3 opposite the 60-degree angle. Thus the
angle bisector makes a smaller 30-60-90 triangle with the side of length 3 opposite the 60degree angle, so the side length opposite the 30-degree angle is 1. Thus the length of the
hypotenuse of this triangle, which is the angle bisector of the original triangle, is 2.
23. (E: 4) We denote BC, CA, AB as a, b, c respectively. Since two tangents from the same point are
of equal length, let the lengths of the tangent segments from A, B, C be x, y, z respectively. Thus
we can write a = y + z, b = x + z, c = x + y (Duality Principle). Let s = x+y+z = (a+b+c)/2 be the
semiperimeter; then we have a = s – x, b = s – y, c = s – z. This becomes x = s – a, y = s – b, z = s
– c. Here, we want BD = y = s – b = 31 – 27 = 4.
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24. (A): (i) is true: this is SAS congruence, considered axiomatic in many systems; (ii) is false: the
first triangle has 2 squared = 4 times the area of the second. If two triangles are similar, their
corresponding altitudes are in the same ratio. Thus the ratio of the areas (bh/2) is the square of
the coefficient of similarity. This result holds for any two similar closed figures, not just
triangles. (iii) is false: counterexamples abound, such as the 3-4-5 and 5-12-13; (iv) is false; one
quadrilateral could be convex and the other could be concave (two adjacent sides ‘caving in’).
25. (D): The large triangle is a 30-60-90, so AC = 50 3. BDC = 90-40 = 50, so tan(50) = BC/50, so
BC = 50tan(50). Thus AB = AC-BC = 50 3 − 50 tan 50°. (Note that tan(50) = 1/tan(40). In
general, tan(x) = 1/tan(90-x).)
26. (B) Solution 1: The
incenter is the intersection
point of all three angle
bisectors. The triangle is
isosceles, so the angle
bisector to the base is also
an altitude and a median.
Thus the altitude is
25! − 15! = 20. The
two small right triangles at
the lower left (shaded-in
on the diagram) are congruent by AAS, so by CPCTC the lower part of the leg is also 15. Then
the upper part of the leg is 25-15=10. The length of the altitude above the incenter is 20-r. We
now have a right triangle with legs 10 and r and hypotenuse 20-r. By the Pythagorean theorem,
10! + 𝑟 ! = 20 − 𝑟 ! , which gives r = 15/2.
Solution 2: Use the formula A = rs, where A is the area and s is the semiperimeter.
27. (B) We denote the feet of the altitudes from A, B, C as A1, B1, C1 respectively. Since ∆ABA1 is a
right triangle, ∠BAA1 = 90 – ∠B. Since ∆AHC1 is a right triangle, ∠AHC1 = 90 – (90 – ∠B) =
∠B, so ∠CHA1 = ∠B because ∠AHC1 and ∠CHA1 are vertical angles. Similarly, ∠BHA1 = ∠C.
So ∠BHC = B + C = 180 – A = 146. (Note that we did not need the measure of angle B or C.)
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28. (A) We denote the intersection of the diagonals as P. Then by Pythagorean Theorem, we have
ZP2 + PA2 = 32
CP2 + PA2 = 42
CP2 + PH2 = 52
ZP2 + PH2 = HZ2
(1)
(2)
(3)
(4)
(1) – (2) + (3) gives ZP2 + PH2 = 32 – 42 + 52 = 18.
Substituting into (4), we get 18 = HZ2 so HZ = 18 = 3 2.
29. (C) Let the octagon be ABCDEFGH. We can group the diagonals into categories such that all
diagonals in a category are parallel and no two diagonals in different categories are parallel.
There are 8 such categories, with a unique category containing AC, AD, AE, AF, AG, BF, BG,
BH. (There are no other categories because AC || DH. If P is the foot of the perpendicular from C
to AB, we can show that DHC is similar to CAP, so DHC = CAP. Since AB || CH, AC || DH by
corresponding angles.) Thus if n <= 8, it is possible that Daniel drew at most one diagonal in
each category. But if n = 9, he must have drawn two diagonals in the same category (Pigeonhole
Principle). The answer is 9.
30. (B) Since the sides of an equilateral
triangle have equal lengths, and same for a
square, ∆EAB is congruent to ∆FAD by
Hypotenuse-Leg (a.k.a. SAS).
Let EB = FD = x. Then CE = CF = 1 – x.
Then EF ! = 2 1 − x ! by the Pythagorean
Theorem on ∆FCE.
We also have EA! = x ! + 1 by the
Pythagorean Theorem on ∆ABE or ∆ADF.
Since EF ! = EA! , 2 1 − x ! = x ! + 1.
Solving gives x = 2 ± 3, but since x < 1
(the side length), x = 2 − 3.
Then the side length of the equilateral
triangle is EF = 1 − x 2
= 3 − 1 2 = 6 − 2.
Diagram for #1 from http://commons.wikimedia.org/wiki/File:Hexahedron.jpg,
edited with IrfanView (www.irfanview.com)
All other diagrams created with GeoGebra (www.geogebra.org)
Test and solutions written by Diego Hernandez, Rickards ‘13
Checked by Alex Yu, Rickards ’14
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