ARML Practice
October 18, 2015
Problems from the 2014 Lehigh University Mathematics Contest.
1. What is the least common multiple of 4004 and 26026?
2. What is the minimal number of positive divisors of x if x > 1 and x, x5 6 , and x7 8 are all
integers?
3. In parallelogram ABCD, CF  AB and CE  AD . If CF = 2, CE = 4 and FB is one-fifth
of AB, what is the ratio of the area of quadrilateral AFCE to that of parallelogram ABCD?
4. In octagon ABCDEFGH, sides AB, BC, CD, DE, and EF are part of a regular decagon,
while sides FG, GH, and HA are part of a regular hexagon. What is the degree measure
of CAH ?
5. Points A, C, and D lie on a circle. Point B lies outside the circle such that B, D, and C are
collinear with D between B and C, and BA is tangent to the circle. If AB = 2, AC = 3, and
BD = 1, what is the area of triangle ABC?
6. A right circular cone has height equal to radius. What is the ratio of its volume to that of
a cube inscribed in it, with the base of the cube lying on the base of the cone? Be sure to
simplify your answer, using the Binomial Theorem if relevant.
7. Rectangle ABCD has side AB of length 8, and side AD of length 3. A point particle is
ejected into the rectangle from A at a 45 degree angle, and reflects off sides with angle of
reflection equal to angle of incidence. Write the ordered pair (V, n), where V is the letter
of the vertex that it hits next, and n in the number of sides that it hits before hitting the
vertex. (The vertex from which it is ejected does not count in the total.)
8. ABCD is a rectangle in which the length AB minus the length AD equals 10. Inside
ABCD is a square WXYZ with sides parallel to those of the rectangle, and W closest to A,
and X closest to B. The total of the areas of the trapezoids XBCY and AWZD is 1000,
while the total area of the trapezoids ABXW and ZYCD is 400. What is the area of the
square WXYZ?
9. Let S denote the locus of all points P in the plane such that as a point Q moves along the
line segment from (0, 0) to P, the distance from Q to (2, 0) decreases throughout the
segment. What is the area of S?
10. Let BE be a median of triangle ABC, and let D be a point on AB such that BD/DA = 3/7.
What is the ratio of the area of triangle BED to that of triangle ABC?
11. An equilateral triangle in the first quadrant has vertices at the points (0, 0), (x1, 4), and
(x2, 11). What is the ordered pair (x1, x2)?
12. The diagonals of a parallelogram partition it into four triangles. Let G be the centroid of
one of the triangles, and let T be a triangle formed by G and two vertices of the
parallelogram. What is the largest possible ratio of the area of T to that of the
parallelogram?
13. In triangle ABC, P lies on AB with AP  1 , and N lies on AC with AN  3 . Let M be the
PB
2
NC
midpoint of BC, and G the intersection of lines AM and PN. What is the ratio of AG ?
GM
Hints:
n
n
1. If an integer P can be factored into prime factors as follows: A  p1 1  p2 2  p3 3  pmm , the
n
number of factors of P is given by:
2.
n
 n1  1 n2  1 n3  1   nm  1 .
One way to find the area of a triangle ABC, with sides a, b, and c is: Area 
1
ab sin C .
2
3. Another way to find the area of a triangle that has sides lengths of a, b, and c and a semiperimeter
of s is by using Heron’s Formula: A  s  s  a  s  b  s  c 
4. In the complex plane (which looks like the Cartesian coordinate system), a rotation of  degrees
or radians can be performed on a + bi by performing the multiplication:
5. Binomial Theorem (abridged version):
 a  b
n
 cis  a  bi  .
n
n
n
n
   a n    a n1b    a n2b 2  ...    b n .
0
1
 2
 n
And, you may want to get the binomial coefficients from Pascal’s Triangle.
ARML Practice – SOLUTIONS
October 18, 2015
Problems from the 2014 Lehigh University Mathematics Contest.
1. 52052. Factor each number. The 1001 is obvious.
2. 25. Since x must be a 6th power and an 8th power to make this work, x is a 24th power.
The smallest number of factors would be when x  p 24 for some prime number p.
3. 1/2. Let BF = s, so AF = 4s. Triangle DEC has sides twice those of triangle BFC. Our
desired ratio is:
[ ABCD]  [ FBC ]  [CED]
s  4s 1
 1

[ ABCD]
10s
2
4. 114. The desired angle equals CAF  FAH . FAH is half an interior angle of a
regular hexagon, hence equal to 60°. Angle BAF is half an interior angle of a regular
decagon, hence equal to 72°. Triangle ABC is isosceles with vertex angle of 144°. Thus
angle BAC is 18°, and CAF = BAF – BAC = 54.
5.
3
15 . Using Power of a Point, BC  BD  AB 2 , and that gives us BC = 4. Using
4
Heron’s Formula, the desired area is
6.

9 1 3 5
   .
2 2 2 2

π
10 + 7 2 . Let h denote the height of the cone, and s the side of the square. The top
12
of the cube will meet the cone at a level where the radius of the cone is s
. Thus, by
2
similar triangles,
3


s 2 2
hs
, and the ratio of the volumes is:
 1 . Then h 
s
2
2
 h 3 
3



2 2 
20  14 2 
10  7 2
3
s
24


7. (B, 9). Count carefully!
24


12


8. 3600. The trapezoids contain equal triangles from two of the corners, as indicated below.
Then 1000 – 400 equals the sum of the areas of the rectangles adjacent to the square,
inside the original rectangle, and this is s(AB – AD), where s denotes the side of the
square. Thus s = 600/10.
9. . If Q moves along a line from (0, 0), the distance from (2, 0) to Q decreases until Q
gets to a point where a perpendicular from (2, 0) to the line intersects the line. This
extreme point lies on the circle whose diameter is the segment from (0, 0) to (2, 0). Thus
S consists of all points in a circle of radius 1, and hence the area is .
10. 3/20. We have BD/AB = 3/10, which equals the ratio of the indicated altitude of triangle
BDE to that of triangle BAE. Since their bases are the same, that is also the ratio of their
areas. Since BE is a median, the areas of triangles BAE and BCE are equal. Thus the
area of BDE is 3/20 times that of ABC.


11. 6 3, 3 . It helps to view this equilateral triangle as being located on the complex
plane. Then (a, b) would be a + bi. The complex number x2  11i is the result of x1  4i
1
3 
being rotated 60°. So,  cis60    x1  4i    
i    x1  4i   x2  11i . This leads to
2
2


the equations
1
3
x1  2 3  x2 and 2 
x1  11 , which has the claimed solution.
2
2
12. 5/12. In the diagram below, M and N are midpoints, and HH’ is an altitude. We have
2
EM  EM
GAD
GH  GE  EN 3
5



 .
ABCD 2 HH 
2MN
4 EM
12
It is easily checked that this is the largest triangle.
13. 6/7. See below.