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Title2010_ARML
TagsCircle Quadratic Equation Logarithm Tetrahedron Curvature
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Total Pages45
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Page 1

ARML Competition 2010

Paul J. Karafiol, Head Writer
Paul Dreyer

Edward Early
Zuming Feng
Benji Fisher

Zachary Franco
Chris Jeuell

Andy Niedermaier
Leo Schneider
Andy Soffer
Eric Wepsic

June 4-5, 2010

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1 Individual Problems

Problem 1. Compute the number of positive integers less than 25 that cannot be written as the difference of two
squares of integers.

Problem 2. For digits A,B, and C, (AB)2 + (AC)2 = 1313. Compute A+B + C.

Problem 3. Points P,Q,R, and S lie in the interior of square ABCD such that triangles ABP , BCQ, CDR, and
DAS are equilateral. If AB = 1, compute the area of quadrilateral PQRS.

Problem 4. For real numbers α, B, and C, the zeros of T (x) = x3 + x2 +Bx+C are sin2 α, cos2 α, and −csc2α.
Compute T (5).

Problem 5. Let R denote the circular region bounded by x2 + y2 = 36. The lines x = 4 and y = 3 partition
R into four regions R1,R2,R3, and R4. [Ri] denotes the area of region Ri. If [R1] > [R2] > [R3] > [R4],
compute [R1]− [R2]− [R3] + [R4].

Problem 6. Let x be a real number in the interval [0, 360] such that the four expressions sinx◦, cosx◦, tanx◦,
cotx◦ take on exactly three distinct (finite) real values. Compute the sum of all possible values of x.

Problem 7. Let a1, a2, a3, . . . be an arithmetic sequence, and let b1, b2, b3, . . . be a geometric sequence. The
sequence c1, c2, c3, . . . has cn = an + bn for each positive integer n. If c1 = 1, c2 = 4, c3 = 15, and c4 = 2,
compute c5.

Problem 8. In square ABCD with diagonal 1, E is on AB and F is on BC with m\ BCE = m\ BAF = 30◦. If
CE and AF intersect at G, compute the distance between the incenters of triangles AGE and CGF .

Problem 9. Let a, b,m, n be positive integers with am = bn = 120 and a 6= b. In the coordinate plane, let
A = (a,m), B = (b, n), and O = (0, 0). If X is a point in the plane such that AOBX is a parallelogram,
compute the minimum area of AOBX.

Problem 10. Let S be the set of integers from 0 to 9999 inclusive whose base-2 and base-5 representations end
in the same four digits. (Leading zeros are allowed, so 1 = 00012 = 00015 is one such number.) Compute the
remainder when the sum of the elements of S is divided by 10,000.

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where (as before) a is taken to be negative if B,C,D are internally tangent to A, and correspondingly for b,c,
or d. This Power Question does not involve the proof of Descartes’ Circle Formula, but the formula may be
used for all problems below.

3a. Two unit circles and a circle of radius 2
3

are mutually externally tangent. Compute all possible values of r such
that a circle of radius r is tangent to all three circles. [2]

3b. Given three mutually tangent circles with curvatures a, b, c > 0, suppose that (a, b, c, 0) does not satisfy
Descartes’ Circle Formula. Show that there are two distinct values of r such that there is a circle of radius r
tangent to the given circles. [3]

3c. Algebraically, it is possible for a quadruple (a, b, c, 0) to satisfy Descartes’ Circle Formula, as occurs when
a = b = 1 and c = 4. Find a geometric interpretation for this situation. [2]

4. Let φ = 1+


5
2

, and let ρ = φ+

φ.

a. Prove that ρ4 = 2ρ3 + 2ρ2 + 2ρ− 1. [2]
b. Show that four pairwise externally tangent circles with nonequal radii in geometric progression must have

common ratio ρ. [2]

As shown in problem 3, given A,B,C,D as above with s = a + b + c + d, there is a second circle A′ with
curvature a′ also tangent to B,C, and D. We can describe A and A′ as conjugate circles.

5. Use Descartes’ Circle Formula to show that a′ = 2s− 3a and therefore s′ = a′ + b+ c+ d = 3s− 4a. [4]

In the context of this problem, a circle configuration is a quadruple of real numbers (a, b, c, d) representing
curvatures of mutually tangent circles A,B,C,D. In other words, a circle configuration is a quadruple (a, b, c, d)
of real numbers satisfying Descartes’ Circle Formula.

The result in problem 5 allows us to compute the curvatures of the six plots removed on day 2. In this case,
(a, b, c, d) = (−1, 2, 2, 3), and s = 6. For example, one such plot is tangent to both of the circles C and C ′
centered at

(
± 1

2
, 0
)
, and to one of the circles of radius 1

3
removed on day 1; it is conjugate to the unit circle (the

boundary of the original kingdom U). If this new plot’s curvature is a, we can write (−1, 2, 2, 3) ` (a, 2, 2, 3),
and we say that the first circle configuration yields the second.

6a. Use the result of problem 5 to compute the curvatures of all circles removed on day 2, and the corresponding
values of s′. [2]

6b. Show that by area, 12% of the kingdom is sold on day 2. [1]

6c. Find the areas of the circles removed on day 3. [2]

6d. Show that the plots sold on day 3 have mean curvature of 23. [1]

7. Prove that the curvature of each circular plot is an integer. [3]

Descartes’ Circle Formula can be extended by interpreting the coordinates of points on the plane as complex
numbers in the usual way: the point (x, y) represents the complex number x + yi. On the complex plane, let
zA , zB , zC , zD be the centers of circles A,B,C,D respectively; as before, a, b, c, d are the curvatures of their
respective circles. Then Descartes’ Extended Circle Formula states

(a · zA + b · zB + c · zC + d · zD )
2 = 2

(
a2z2A + b

2z2B + c
2z2C + d

2z2D
)
.

8a. Suppose that A′ is a circle conjugate to A with center zA 0 and curvature a′, and ŝ = a·zA +b·zB +c·zC +d·zD . Use
Descartes’ Extended Circle Formula to show that a′ ·zA 0 = 2ŝ−3a·zA and therefore a′ ·zA 0+b·zB +c·zC +d·zD =
3ŝ− 4a · zA . [2]

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8b. Prove that the center of each circular plot has coordinates
(
u
c
, v
c

)
where u and v are integers, and c is the

curvature of the plot. [2]

Given a c-triangle T , let a, b, and c be the curvatures of the three arcs bounding T , with a ≤ b ≤ c, and let d
be the curvature of the incircle of T . Define the circle configuration associated with T to be C(T ) = (a, b, c, d).
Define the c-triangle T to be proper if c ≤ d. For example, circles of curvatures −1, 2, and 3 determine two
c-triangles. The incircle of one has curvature 6, so it is proper; the incircle of the other has curvature 2, so it
is not proper.

Let P and Q be two c-triangles, with associated configurations C(P ) = (a, b, c, d) and C(Q) = (w, x, y, z). We
say that P dominates Q if a ≤ w, b ≤ x, c ≤ y, and d ≤ z. (The term “dominates” refers to the fact that the
radii of the arcs defining Q cannot be larger than the radii of the arcs defining P .)

Removing the incircle from T gives three c-triangles, T (1), T (2), T (3), each bounded by the incircle of T and
two of the arcs that bound T . These triangles have associated configurations

C(T (1)) = (b, c, d, a′),
C(T (2)) = (a, c, d, b′),
C(T (3)) = (a, b, d, c′),

where a′, b′, and c′ are determined by the formula in problem 5.

9. Let P and Q be two proper c-triangles such that P dominates Q. Let C(P ) = (a, b, c, d) and C(Q) = (w, x, y, z).

a. Show that P (3) dominates P (2) and that P (2) dominates P (1). [2]

b. Prove that P (1) dominates Q(1). [2]

c. Prove that P (3) dominates Q(3). [2]

10a. Prove that the largest plot sold by the king on day n has curvature n2 + 2. [3]

10b. If ρ = φ +

φ, as in problem 4, prove that the curvature of the smallest plot sold by the king on day n does

not exceed 2ρn. [3]

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17 Tiebreaker Solutions

Problem 1. Let set S = {1, 2, 3, 4, 5, 6}, and let set T be the set of all subsets of S (including the empty set
and S itself). Let t1, t2, t3 be elements of T , not necessarily distinct. The ordered triple (t1, t2, t3) is called
satisfactory if either

(a) both t1 ⊆ t3 and t2 ⊆ t3, or
(b) t3 ⊆ t1 and t3 ⊆ t2.

Compute the number of satisfactory ordered triples (t1, t2, t3).

Solution 1. Let T1 = {(t1, t2, t3) | t1 ⊆ t3 and t2 ⊆ t3} and let T2 = {(t1, t2, t3) | t3 ⊆ t1 and t3 ⊆ t2}. Notice
that if (t1, t2, t3) ∈ T1, then (S \ t1, S \ t2, S \ t3) ∈ T2, so that |T1| = |T2|. To count T1, note that if t1 ⊆ t3
and t2 ⊆ t3, then t1 ∪ t2 ⊆ t3. Now each set t3 has 2|t3| subsets; t1 and t2 could be any of these, for a total of(
2|t3|

)2
= 4|t3| possibilities given a particular subset t3. For n = 0, 1, . . . , 6, if |t3| = n, there are

(
6
n

)
choices for

the elements of t3. So the total number of elements in T1 is

|T1| =
6∑
k=0

(
6
k

)
4k

= (4 + 1)6 = 15625,

by the Binomial Theorem. However, T1 ∩ T2 6= ∅, because if t1 = t2 = t3, the triple (t1, t2, t3) satisfies both
conditions and is in both sets. Therefore there are 64 triples that are counted in both sets. So |T1 ∪ T2| =
2 · 15625− 64 = 31186.

Alternate Solution: Let T1 and T2 be defined as above. Then count |T1| based on the number n of elements
in t1 ∪ t2. There are

(
6
n

)
ways to choose those n elements. For each element a in t1 ∪ t2, there are three

possibilities: a ∈ t1 but not t2, or a ∈ t2 but not t1, or a ∈ t1 ∩ t2. Then for each element b in S \ (t1 ∪ t2),
there are two possibilities: either b ∈ t3, or b /∈ t3. Combine these observations in the table below:

|t1 ∪ t2| Choices for Ways of dividing |S \ (t1 ∪ t2)| Choices for t3 Total
t1 ∪ t2 between t1 and t2

0 1 1 6 26 64
1 6 3 5 25 576
2 15 32 4 24 2160
3 20 33 3 23 4320
4 15 34 2 22 4860
5 6 35 1 21 2916
6 1 36 0 20 729

The total is 15625, so |T1| = |T2| = 15625. As noted in the first solution, there are 64 triples that are counted
in both T1 and T2, so |T1 ∪ T2| = 2 · 15625− 64 = 31186.

Problem 2. Let ABCD be a parallelogram with \ ABC obtuse. Let BE be the altitude to side AD of 4ABD.
Let X be the point of intersection of AC and BE, and let F be the point of intersection of AB and

←→
DX. If

BC = 30, CD = 13, and BE = 12, compute the ratio AC
AF

.

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Solution 2. Extend AD to a point M such that CM ||BE as shown below.

30

13

Because CD = AB = 13 and BE = 12 = CM , AE = DM = 5. Then AC =


352 + 122 =


1369 = 37.
Because EX||CM , XE/CM = AE/AM = 1

7
. Thus EX = 12

7
and XB = 72

7
, from which EX/XB = 1

6
. Apply

Menelaus’s Theorem to 4AEB and Menelaus line FD:

AD

ED
·
EX

XB
·
BF

FA
= 1

30
25
·

1
6
·

13− FA
FA

= 1

13− FA
FA

= 5.

Thus FA = 13
6

. The desired ratio is:

37
13/6

=
222
13

.

Alternate Solution: After calculating AC as above, draw BD, intersecting AC at Y . Because the diagonals
of a parallelogram bisect each other, DY = Y B. Then apply Ceva’s Theorem to4ABD and concurrent cevians
AY ,BE,DF :

AE

ED
·
DY

Y B
·
BF

FA
= 1

5
25
· 1 ·

13− FA
FA

= 1.

Thus FA = 13
6

, and the desired ratio is 222
13

.

Problem 3. Compute the number of two-digit factors of 232 − 1.

Solution 3. Using the difference of squares, 232−1 =
(
216 − 1

) (
216 + 1

)
. The second factor, 216+1, is the Fermat

prime 65537, so continue with the first factor:

216 − 1 =
(
28 + 1

) (
28 − 1

)
28 − 1 =

(
24 + 1

) (
24 − 1

)
24 − 1 = 15 = 3 · 5.

Because the problem does not specify that the two-digit factors must be prime, the possible two-digit factors
are 17, 3 · 17 = 51, 5 · 17 = 85 and 3 · 5 = 15, for a sum of 17 + 51 + 85 + 15 = 168.

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