Number 1 There are 7 different characters that can be picked, with 0 being the only number that can be repeated twice.

  • If appears 0 or 1 times amongst the sequence, there are sequences possible.
  • If appears twice in the sequence, there are places to place the s. There are ways to place the remaining three characters. Totally, that gives us .

Thus, , and .

Number 2 Denote and . The last condition reduces to . Therefore, is equal to one of the 9 factors of .

Subtracting the one, we see that . There are exactly ways to find pairs of if . Thus, there are solutions of .

Alternatively, note that the sum of the divisors of is (notice that after distributing, every divisor is accounted for). This evaluates to . Subtract for reasons noted above to get . Finally, this changes , so we have to add one to account for that. We get .

Number 3

Denote and . The last condition reduces to . Therefore, is equal to one of the 9 factors of .

Subtracting the one, we see that . There are exactly ways to find pairs of if . Thus, there are solutions of .

Alternatively, note that the sum of the divisors of is (notice that after distributing, every divisor is accounted for). This evaluates to . Subtract for reasons noted above to get . Finally, this changes , so we have to add one to account for that. We get . Extend and to their points of intersection. Since and are both right triangles, we can come to the conclusion that the two new triangles are also congruent to these two (use ASA, as we know all the sides are and the angles are mostly complementary). Thus, we create a square with sides .

is the diagonal of the square, with length ; the answer is .

A slightly more analytic/brute-force approach:

File:AIME II prob10 bruteforce.PNG

Drop perpendiculars from and to and , respectively; construct right triangle with right angle at K and . Since , we have . Similarly, . Since , we have .

Now, we see that . Also, . By the Pythagorean Theorem, we have . Therefore, .

Number 4 Suppose that it takes hours for one worker to create one widget, and hours for one worker to create one whoosit.

Therefore, we can write that (note that two hours is similar to having twice the number of workers, and so on):

Solve the system of equations with the first two equations to find that . Substitute this into the third equation to find that , so .


Number 5

Solution 1

edit

There are squares in total formed by the rectangle with edges on the x and y axes and with vertices on the intercepts of the equation, since the intercepts of the lines are .

Count the number of squares that the diagonal of the rectangle passes through. Since the two diagonals of a rectangle are congruent, we can consider instead the diagonal . This passes through 8 horizontal lines () and 222 vertical lines (). At every time we cross a line, we enter a new square. Since 9 and 223 are relatively prime, we don’t have to worry about crossing an intersection of a horizontal and vertical line at one time. We must also account for the first square. This means that it passes through squares.

The number of non-diagonal squares is . Divide this in 2 to get the number of squares in one of the triangles, with the answer being .

Solution 2

edit

Count the number of each squares in each row of the triangle. The intercepts of the line are .

In the top row, there clearly are no squares that can be formed. In the second row, we see that the line gives a value of , which means that unit squares can fit in that row. In general, there are

triangles. Since , we see that there are more than triangles. Now, count the fractional parts. . Adding them up, we get .

Solution 3

edit

From Pick's Theorem, . In other words, and I is .

Number 6


Let's set up a table of values. Notice that 0 and 9 both cannot appear as any of because of the given conditions. A clear pattern emerges.

For example, for in the second column, we note that is less than , but greater than , so there are four possible places to align as the second digit.

Number   1st   2nd   3rd   4th
0 0 0 0 64
1 1 4 16 64
2 1 4 16 64
3 1 4 16 64
4 1 4 16 64
5 1 4 16 64
6 1 4 16 64
7 1 4 16 64
8 1 4 16 64
9 0 0 0 64

For any number from 1-8, there are exactly 4 numbers from 1-8 that are odd and less than the number or that are even and greater than the number (the same will happen for 0 and 9 in the last column). Thus, the answer is .

Number 7

For , we see that all work, giving 7 integers. For , we see that in , all of the even numbers work, giving 10 integers. For , we get 13, and so on. We can predict that at we get 70.

To prove this, note that all of the numbers from divisible by work. Thus, (the one to be inclusive) integers will fit the conditions. .

The maximum value of . Therefore, the solution is .


Number 8

Solution 1

edit

Denote the number of horizontal lines as , and the number of vertical lines as . The number of basic rectangles is . . Substituting, we find that .

FOIL this to get a quadratic, . Use to find the maximum possible value of the quadratic: . However, this gives a non-integral answer for . The closest two values that work are and .

We see that . The solution is .

Solution 2

edit

We realize that drawing vertical lines and horizontal lines, the number of basic rectangles we have is . The easiest possible case to see is vertical and horizontal lines, as . Now, for every 4 vertical lines you take away, you can add 5 horizontal lines, so you basically have the equation maximize.

Expanded, this gives . From you get that the vertex is at . This is not an integer though, so you see that when , you have and that when x=6, you have . , so the maximum integral value for x occurs when . Now you just evaluate which is .


Number 9 Several Pythagorean triples exist amongst the numbers given. . Also, the length of .

Use the Two Tangent theorem on . Since both circles are inscribed in congruent triangles, they are congruent; therefore, . By the Two Tangent theorem, note that , making . Also, . .

Finally, . Also, . Equating, we see that , so .

Solution 2

edit

By the Two Tangent theorem, we have that . Solve for . Also, , so . Since , this can become . Substituting in their values, the answer is .

Number 10


  • has 6 elements:
    • Probability:
    • must have either 0 or 6 elements, probability: .
  • has 5 elements:
    • Probability:
    • must have either 0, 6, or 1, 5 elements. The total probability is .
  • has 4 elements:
    • Probability:
    • must have either 0, 6; 1, 5; or 2,4 elements. If there are 1 or 5 elements, the set which contains 5 elements must have four emcompassing and a fifth element out of the remaining numbers. The total probability is .

We could just continue our casework. In general, the probability of picking B with elements is . Since the sum of the elements in the th row of Pascal's Triangle is , the probability of obtaining or which encompasses is . In addition, we must count for when is the empty set (probability: ), of which all sets of will work (probability: ).

Thus, the solution we are looking for is .

The answer is .


Number 11

File:2007 AIME II-11.png

If it weren’t for the small tube, the larger tube would travel . Consider the distance from which the larger tube first contacts the smaller tube, until when it completely loses contact with the smaller tube.

Drawing the radii as shown in the diagram, notice that the hypotenuse of the right triangle in the diagram has a length of . The horizontal line divides the radius of the larger circle into on the top half, which indicates that the right triangle has leg of 48 and hypotenuse of 96, a .

Find the length of the purple arc in the diagram (the distance the tube rolled, but not the horizontal distance). The sixty degree central angle indicates to take of the circumference of the larger circle (twice), while the central angle in the smaller circle indicates to take of the circumference. This adds up to .

The actual horizontal distance it takes can be found by using the s. The missing leg is equal in length to . Thus, the total horizontal distance covered is .

Thus, we get , and our answer is .

Number 12

Suppose that , and that the common ratio between the terms is .


The first conditions tells us that . Using the rules of logarithms, we can simplify that to . Thus, . Since all of the terms of the geometric sequence are integral powers of , we know that both and must be powers of 3. Denote and . We find that . The possible positive integral pairs of are .


The second condition tells us that . Using the sum formula for a geometric series and substituting and , this simplifies to . The fractional part . Thus, we need . Checking the pairs above, only is close.


Our solution is therefore .

Number 13 Label each of the bottom squares as .

Through induction, we can find that the top square is equal to .

Examine the equation . All of the coefficients from will be multiples of (since the numerator will have a ). Thus, the expression boils down to . Reduce to find that . Out of , either all are equal to , or three of them are equal to . This gives possible combinations of numbers that work.

The seven terms from can assume either or , giving us possibilities. The answer is therefore .

Number 14

Note:The following solution(s) are non-rigorous.

Substitute the values . We find that , and that . This means that . This suggests that are roots of the polynomial, and so will be a root of the polynomial.

The polynomial is likely in the form of ; appears to satisfy the same relation as , so it also probably has the same roots. Thus, is the solution. Guessing values for , try . Checking a couple of values shows that works, and so the solution is .


Number 15

Solution

edit

File:2007 AIME II-15.png

Solution 1

edit

First, apply Heron's formula to find that the area is  . Also the semiperimeter is  . So the inradius is  .

Now consider the incenter I. Let the radius of one of the small circles be  . Let the centers of the three little circles tangent to the sides of   be  ,  , and  . Let the centre of the circle tangent to those three circles be P. A homothety centered at   takes   to   with factor  . The same homothety takes   to the circumcentre of  , so  , where   is the circumradius of  . The circumradius of   can be easily computed by  , so doing that reveals  . Then  , so the answer is  .

Solution 2

edit

File:2007 AIME II-15b.gif

Consider a 13-14-15 triangle.   [By Heron's Formula or by 5-12-13 and 9-12-15 right triangles.]

The inradius is  , where   is the semiperimeter. Scale the triangle with the inradius by a linear scale factor,  

The circumradius is   where     and   are the side-lengths. Scale the triangle with the circumradius by a linear scale factor,  .

Cut and combine the triangles, as shown. Then solve for 4u:

 
 
 
 
 
 

The solution is  .