2010 AIME 1 – Solutions
Problem 1
. Thus there are
divisors, of which are squares (the exponent of each
prime factor must either be or ). Therefore the probability is
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Problem 2
Note that
arithmetic). That is a total of
,
, ,
(see modular
integers, so all those integers multiplied out are congruent to
. Thus, the entire expression is congruent to
.
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Problem 3
We solve in general using instead of
. Substituting
, we have:
Dividing by
, we get
.
Taking the th root,
In the case
, or
,
.
,
,
, yielding an answer of
.
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Problem 4
We perform casework based upon the number of heads that are flipped.

Case 1: No heads.
The only possibility is TTT (the third coin being the unfair coin). The probability for this to happen to Jackie is
Thus the probability for this to happen to both players is

Case 2: One head.
We can have either HTT, THT, or TTH. The first two happen to Jackie with the same
chance, but the third
happens of the time, since the unfair coin is heads instead of tails. With 3 possibilities for Jackie and 3 for
Phil, there are a total of 9 ways for them both to have 1 head.
Multiplying and adding up all 9 ways, we have a
overall chance for this case.

Case 3: Two heads.
With HHT

, HTH
, and THH
possible, we proceed as in Case 2, obtaining
Case 4: Three heads.
Similar to Case 1, we can only have HHH, which has
chance. Then in this case we get
Finally, we take the sum:
, so our answer is
.
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Problem 5
Solution 1
Using the difference of squares,
equality must hold so
the answer is
and
. Then we see
is maximal and
, where
is minimal, so
.
Solution 2
Since
quantity
must be greater than
, it follows that the only possible value for
is (otherwise the
would be greater than
). Therefore the only possible ordered pairs for
are
,
, ... ,
, so has
possible values.
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Problem 6
Solution 1
Let
and
,
, so it follows that
Also,
. Completing the square, we have
for all (by the Trivial Inequality).
, so
must be of the form
, and
obtains its minimum at the point
for some constant ; substituting
,
. Then
yields
. Finally,
.
Solution 2
It can be seen that the function
must be in the form
for some real and . This
is because the derivative of
is
, and a global minimum occurs only at
(in addition,
because of this derivative, the vertex of any quadratic polynomial occurs at
we obtain two equations:
, and
Solving, we get
and
, so
). Substituting
and
.
. Therefore,
.
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Problem 7
Let each pair of two sets have one element in common. Label the common elements as , , . Set will
have elements and , set will have and , and set will have and . There are
ways to
choose values of , and . There are unpicked numbers, and each number can either go in the first set,
second set, third set, or none of them. Since we have choices for each of numbers, that gives us
.
Finally,
, so the answer is
.
Problem 8
The desired region consists of 12 boxes, whose lower-left corners are integers solutions of
, namely
Since the points themselves are symmetric about
symmetric about
. The distance from
, the boxes are
to the furthest point on an axis-box, for instance
The distance from
, is
to the furthest point on a quadrant-box, for instance
, is
The latter is the larger, and is
, giving an answer of
.
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Problem 9
Add the three equations to get
. Now, let
, so
.
,
. Now cube both sides; the
Solve the remaining quadratic to get
. To maximize
giving
and
terms cancel out.
choose
and so the sum is
.
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Problem 10
If we choose and such that
equality hold. So is just the number of combinations of
anything from to
. If
then
or
there is a unique choice of
and we can pick. If
or
. Thus
and
that makes the
we can let be
.
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Problem 11
The inequalities are equivalent to
points of intersection,
. We can set them equal to find the two
. This implies that one of
, from which we find that
triangle, as shown above. When revolved about the line
right cones that share the same base and axis.
Let
. The region is a
, the resulting solid is the union of two
denote the height of the left and right cones, respectively (so
common radius. The volume of a cone is given by
), and let denote their
; since both cones share the same base, then the
desired volume is
. The distance from the point
given by
. The distance between
to the line
and
is
is given by
. Thus, the answer is
.
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Problem 12
We claim that
is the minimal value of . Let the two partitioned sets be and ; we will try to partition
and
such that the
condition is not satisfied. Without loss of generality, we place in .
Then must be placed in , so must be placed in , and must be placed in . Then
cannot be
placed in any set, so we know is less than or equal to
.
For
, we can partition into
in neither set are there values where
and
(since
and
, and
). Thus
.
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Problem 13
Solution 1
The center of the semicircle is also the midpoint of
. Let this point be O. Let be the length of
Rescale everything by 42, so
Since
Let
is a radius of the semicircle,
,
, and be the areas of triangle
To find we have to find the length of
and
are similar. Thus:
. Then
. Thus
, sector
. Project
so
.
is an equilateral triangle.
, and trapezoid
and
.
onto
to get points
respectively.
and
. Notice that
.
Then
. So:
Let be the area of the side of line containing regions
. Then
Obviously, the is greater than the area on the other side of line . This other area is equal to the total area
minus . Thus:
.
Now just solve for .
Don't forget to un-rescale at the end to get
Finally, the answer is
.
.
Solution 2
Let be the center of the semicircle. It follows that
equilateral.
Let
be the foot of the altitude from
Finally, denote
, and
. It then follows that
, such that
. Extend
. Since
, so triangle
and
to point so that is on
and
are similar,
is
.
and
is perpendicular to
Given that line
divides
into a ratio of
, we can also say that
where the first term is the area of trapezoid
, the second and third terms denote the areas of a full
circle, and the area of
, respectively, and the fourth term on the right side of the equation is equal to .
Cancelling out the
on both sides, we obtain
By adding and collecting like terms,
.
Since
,
, so the answer is
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Problem 14
Solution 1
Observe that is strictly increasing in . We realize that we need
need some sequence of s, s, and then s.
It follows that
is
.
Solution 2
. Manually checking shows that
terms to add up to around
and
, so we
. Thus, our answer
Because we want the value for which
, the average value of the 100 terms of the sequence should
be around . For the value of
to be ,
. We want kn to be around the middle
of that range, and for k to be in the middle of 0 and 100, let
, so
, and
yields
, so our answer is still
.
, so we want to lower . Testing
.
Solution 3
For any where the sum is close to
, all the terms in the sum must be equal to , or . Let be the
number of terms less than or equal to and be the number of terms equal to (also counted in ). With
this definition of and the total will be
, from which
. Now
is the smallest integer for which
If is not a divisor of
assumption,
also not a divisor of
, we have
and so
and with
or
, thus
Similarly,
and
. Dividing
we get
Under the same
by
we see that
) and the sum is obviously monotone in , the largest value is
. Since
(thus a sum of
.
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Problem 15
Solution 1
is
Let
, then
. Also let
Clearly,
each area by the rs formula. Then
yields
. We can also express
. Equating and cross-multiplying
or
Note that for d to be positive, we must have
.
By Stewart's Theorem, we have
or
Brute forcing by plugging in our previous result for , we have
Clearing the fraction and gathering like terms, we get
Aside: Since must be rational in order for our answer to be in the desired form, we can use the Rational
Root Theorem to reveal that
is an integer. The only such in the above-stated range is
.
Legitimately solving that quartic, note that
and
should clearly be solutions, corresponding to
the sides of the triangle and thus degenerate cevians. Factoring those out, we get
The only solution in the desired range
is thus
. Then
, and our desired ratio
, giving us an answer of
.
Solution 2
Let
and
so
respectively, and their equal inradii be . From
Let the incircle of
is a rectangle. Also,
respectively and
notice that the altitude of
and adding these we have
. Let the incenters of
and
be
and
, we find that
meet
at and the incircle of
meet
at . Then note that
is right because
and
are the angle bisectors of
and
. By properties of tangents to circles
and
. Now
to
is of length , so by similar triangles we find that
(3). Equating (3) with (1) and (2) separately yields