Solutions of Selected
2001 Texas A& M
Math Contest Problems
Dr. Richard Newcomb
Cistercian Preparatory School
CD 13
Let r be the radius of the inscribed circle. Draw the three
radii to the points of contact of the circle and triangle. Then AB = a + b,
BC = b + r and CA = a + r. Apply the Pythagorean theorem and simplify
to obtain r 2 + (a + b)r = ab. On the other hand it is well known that the
area A of a triangle is rs where s is the semiperimeter. But s = a + b + r so
A = rs = r(a + b + r) = r 2 + (a + b)r = ab. Hence the area is ab.
EF 19
To me perhaps the most interesting problem from the subject
exams this year. Here is my heuristic approach. Assume the limit in question
exists and is positive. Call its value L. Then what that means is that the
sequence an resembles more and more an arithmetic sequence with common
difference L as n gets large. Now from the given recursion this new one
readily follows: a2n = a2n−1 + an−1 . We can rewrite this as a2n − a2n−1 = an−1 .
Then factor the difference of squares as (an + an−1 )(an − an−1 ) = an−1 . Now
divide both sides by n − 1 in this way
an + an−1
an−1
(an − an−1 ) =
n−1
n−1
The first factor on the left hand side approaches 2L while the second factor
approaches L (since the sequence is becoming nearly arithmetic with common
difference L). The right hand side approaches L of course. Thus 2L2 = L.
Hence either L = 1/2 or L = 0. As of this writing I’m not sure how to
rule out the case L = 0 if a1 = 1 (probably one can prove that for n ≥ 4
that the sequence an /n is monotone increasing). So going by the principle
of the more interesting answer, we take L = 1/2. A double check using
Mathematica confirms this as likely.
1
12
22
32
42
+
+
+
+ · · · . The sum S may be dis21
22
23
24
sected into multiples of geometric series and resummed as follows (legitimate
because the sum converges absolutely)
1 1 1
+ + +···
S=1
2 4 8
1
1 1
+ +
+···
+3
4 8 16
1
1
1
+5
+
+
+··· +···
8 16 32
3 5 7
9
=1+ + + +
+···
2 4 8 16
EF 20
Let S =
In turn this new sum may be redone as
1 1 1
S =1+2
+ + +···
2 4 8
1 3 5
+
+ + +···
2 4 8
S
S =1+2+
2
This last simple linear equation in S may be solved to give S = 6.
1
= 1 − x2 +
1 + x2
x4 − x6 + · · · where we used the formula for the sum of a geometric series.
By Taylor’s Theorem, the coefficient of x4 in the Taylor series is the fourth
derivative of the function at 0 divided by 4!. But this coefficient is clearly 1,
so f (4) (0) = 4! = 24.
EF 21
Let x = 1/n. Then we may take f (x) =
EF 22
The sum in question is a Riemann sum for the integral
2n 1
dx = ln(2n) − ln(n) = ln(2n/n) = ln 2. As n increases the Riemann
n
x
sum more closely approximates the integral (handwaving but heuristics are
OK on non-USAMO type competitions) so the answer is ln 2.
Best 7
Since f is piecewise linear and continuous, the minimum
must occur at an endpoint or corner. So all you have to do is plug in
2
x = 0, 1/3, 1/2, 3/4, 1 and see what is the least value you get from those.
That occcurs at x = 3/4 and that least value is −1/4.
Best 9
A nice way to do this is: fix a particular letter, say a. It
occurs in exactly 16 subsets, so it contributes exactly 16 to the sum. Likewise
for the other letters. So all in all that’s 5 times 16 or 80.
Best 13
Replace h by h2 so the limit is the more familiar looking
difference quotient for the second derivative (check by applying L’Hospital’s
rule twice). Answer: f (a).
Best 14
Factor x7 − 1 as (x − 1)(x6 + x5 + · · · + x + 1). Then w is a
7th root of unity, i.e., w 7 = 1. So then w 14 = 1.
Best 16
The question is asking how to spread out five points in the
unit square is as uniform way as possible. The intuitive answer is to place
four points
on the corner and one in the center. The intuitive answer is right.
√
So 1/ 2 is the answer. In general this problem is I believe unsolved for most
cases larger than 12 points.
Best 19
that
Multiply by the conjugate on top and bottom to determine
√
√
a−b
x→∞
2
where a and b are constants. So our answer is (3 − (−3))/(9 − (−9)) = 1/3.
lim
x2 + ax −
x2 + bx =
Best 21
Rewrite as x4 − x2 + (y 4 − y 2 ) = 0 and view as a quadratic
in x2 with y as fixed. Then by the quadratic formula
1
+
1 + 4y 2 − 4y 4
x2 =
2
What is the largest discriminant possible if y is real? Let u = y 2, so the
problem is equivalent to maximizing the parabola 4u + 1 − 4u2 . By locating
the vertex we see that the maximum height of the parabola is 2. Substitute
this for the discriminant
and then take the square root of both sides.
above
√
1+ 2
.
We get the answer of
2
3