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SW-MO-ARML
ARML/NYSML Practice Problems
Solutions
1. (1993 ARML I2) Square ABCD is positioned on the axes as shown. Compute the
area of the square.
B
C (13,8)
A
Drop a perpendicular from C to the x-axis. You can
argue that the triangle formed by that segment and C and
D is congruent to the right triangle with one vertex at the
origin. The you get legs of those right triangles to be 5
and 8, giving the length of the side of the square as
52 + 82 = 89 and the area as 89 u2.
D
2. (1994 ARML R2-2) In triangle ABC, median BM = 29. A perpendicular from C
to ray BM meets BM at R. If CR = 4, compute the area of triangle ABC.
B
29
C
M
A
4
The area of the shaded triangle is ½ the area of triangle
ABC. With CR as the altitude and BM as the base, the
area of triangle BMC is ½ x 29 x 4. So the total area
is 29 x 4 = 116 u2.
R
3. (1989 NYSML I1) Points P and Q are on circle O, and chord PQ is drawn. A
second circle is drawn with diameter OP , crossing the chord at point S. If
OP = 7 and PQ = 12, compute PS.
P
PQ=12
S
7
<PSO is a right angle, so OS ⊥ PQ , so
Q
O
OS must bisect PQ . PS = 6
4. (1989 NYSML I3) A quadrilateral is circumscribed about a circle. If three sides
of the quadrilateral are 17, 18 and 21, not necessarily in that order, compute the
smallest possible value for the fourth side.
a
a
17
b
b
d
18
d
Tangent segments are congruent! Set up a
system of equations for a, b, c, and d. Find a +
d. Then change the locations of the 17, 18, and
21 and repeat the process. Try all three
arrangements to determine the smallest, which
is 14
c
21
c
5. (1990 NYSML T9) In convex quadrilateral ABCD, AB = 15, BC = 7, CD = 24,
AD = 20, and m∠A = 90° . Compute AC.
D
24
C
20
A
7
90
15
B
Since triangle ABD is right, compute BD. It is
25. That makes triangle BCD a right triangle by
the converse of the Pythagorean Theorem. Then
ABCD is a cyclic quadrilateral and Ptolemy’s
Theorem applies, so
AC ⋅ BD = AB ⋅ CD + BC ⋅ AD . This gives
AC=20
6. (1990 NYSML R1-2) In the diagram, minor arcs AB and CB are each ¼ of a circle
of radius 2, and are tangent at B. Arc ADC is a semicircle of radius 2. Compute
the area of the enclosed region.
D
C
A
B
Find the area of the shaded rectangle!!
It is 8u2
7. (1991 NYSML T8) Two circles have radii 1 and 7, and one of their common
internal tangents is perpendicular to a common external tangent. Compute the
distance between their centers.
Since the tangents are perpendicular, you could form
squares by drawing radii to the points of tangency.
This means that the distance between the centers is
82 + 6 2 = 10 .
E
F
8. (1991 NYSML I4) In a convex polygon of n sides, one interior angle contains x°,
while each of the remaining n-1 interior angles contains 133°. Compute all four
possible values for x.
Obviously, this cannot be a triangle or quadrilateral. Start with a pentagon.
The sum of the angles of a pentagon is 540°. Subtracting 4 angles of 133°,
leaves an angle of 8°. Do this for a hexagon, heptagon , and octagon and the
angles are 55°, 102°, and 149°. A nonagon will not work.