La AMC – 10 participa elevi din clasele a IX-a si a X-a, iar la AMC – 12, elevi

din clasele a XI-a si a XII-a. Concursul s-a desfasurat in 9 februarie, la

Universitatea de Nord Baia Mare, in organizarea Centrului de Pregatire a

Concursurilor :colare de la Departamentul de Matematica si Informatica, in

colaborare cu Inspectoratul Scolar Judetean (inspector de specialitate

profesor Gheorge Maiorescu).

In acest an au fost admisi, pe baza rezultatelor la concursurile de matematica

din anul precedent si a rezultatelor din anii anteriori la concursurile AMC, 50

de elevi din clasele VIII-XII, provenind de la mai multe licee din judet.

Profesor dr. Vasile Berinde a reamintit faptul ca fiecare dintre competitii cere

rezolvarea in 75 minute a 25 de probleme tip grila, formulate in limba

engleza. La fiecare problema elevii trebuie sa aleaga raspunsul corect din 5

variante posibile. Pentru fiecare raspuns corect se acorda cate 6 puncte,

pentru fiecare problema la care nu a fost ales un raspuns se acorda 2,5

puncte, iar pentru un raspuns gresit se acorda 0 puncte. Un elev poate obtine

asadar cel mult 150 puncte.

2010 AMC 10A ProblemsProblem 1

Mary’s top book shelf holds five books with the following widths, in centimeters:  ,  , 

,  , and  . What is the average book width, in centimeters?

SolutionTo find the average, we add up the widths  ,  ,  ,  , and  , to get a total sum of  .

Since there are   books, the average book width is   The answer is  .

Problem 2

Four identical squares and one rectangle are placed together to form one large

square as shown. The length of the rectangle is how many times as large as its width?

Solution

Let the length of the small square be  , intuitively, the length of the big square is  .

It can be seen that the width of the rectangle is  . Thus, the length of the rectangle is   times large as the width. The answer is  .

Problem 3

Tyrone had   marbles and Eric had   marbles. Tyrone then gave some of his marbles

to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles

did Tyrone give to Eric?

Solution

Let   be the number of marbles Tyrone gave to Eric. Then,  . Solving for   yields  and  . The answer is  .

Problem 4

A book that is to be recorded onto compact discs takes   minutes to read aloud.

Each disc can hold up to  minutes of reading. Assume that the smallest possible

number of discs is used and that each disc contains the same length of reading. How

many minutes of reading will each disc contain?

Solution

Assuming that there were fractions of compact discs, it would take   CDs

to have equal reading time. However, since the number of discs can only be a whole number, there are at least 8 CDs, in which case it would have   minutes on each of the 8 discs. The answer is 

Problem 5

The area of a circle whose circumference is   is  . What is the value of  ?

Solution

If the circumference of a circle is  , the radius would be  . Since the area of a circle is  , the area is  . The answer is  .

Problem 6

For positive numbers   and   the operation   is defined as

What is  ?

Solution. Then,   is   The answer is 

Problem 7

Crystal has a running course marked out for her daily run. She starts this run by

heading due north for one mile. She then runs northeast for one mile, then southeast

for one mile. The last portion of her run takes her on a straight line back to where she

started. How far, in miles is this last portion of her run?

Solution

Crystal first runs North for one mile. Changing directions, she runs Northeast for

another mile. The angle difference between North and Northeast is 45 degrees. She

then switches directions to Southeast, meaning a 90 degree angle change. The distance now from travelling North for one mile, and her current destination is 

miles, because it is the hypotenuse of a 45-45-90 triangle with side length one (mile). Therefore, Crystal's distance from her starting position, x, is equal to  ,

which is equal to  . The answer is 

Problem 8

Tony works   hours a day and is paid $  per hour for each full year of his age.

During a six month period Tony worked   days and earned $ . How old was Tony at

the end of the six month period?

Solution

Tony worked   hours a day and is paid   dollars per hour for each full year of his

age. This basically says that he gets a dollar for each year of his age. So if he is   

years old, he gets   dollars a day. We also know that he worked   days and

earned   dollars. If he was   years old at the beginning of his working period, he

would have earned   dollars. If he was   years old at the beginning of his

working period, he would have earned   dollars. Because he earned   

dollars, we know that he was   for some period of time, but not the whole time,

because then the money earned would be greater than or equal to  . This is why he

was   when he began, but turned   sometime in the middle and earned   dollars in

total. So the answer is  .The answer is  . We could find out for how long he was   

and  .  . Then   is   and we know that he was   for   days,

and   for   days. Thus, the answer is  .

Problem 9

A palindrome, such as  , is a number that remains the same when its digits are

reversed. The numbers  and   are three-digit and four-digit palindromes,

respectively. What is the sum of the digits of  ?

Solution

 is at most  , so   is at most  . The minimum value of   is  .

However, the only palindrome between   and   is  , which means that   

must be  .

It follows that   is  , so the sum of the digits is  .

See also

(E) 2017

There are 365 days in a non-leap year. There are 7 days in a week. Since 365 = 52 * 7

+ 1 (or 365 is congruent to 1 mod 7), the same date (after February) moves "forward"

one day in the subsequent year, if that year is not a leap year.

For example: 5/27/08 Tue 5/27/09 Wed

However, a leap year has 366 days, and 366 = 52 * 7 + 2. So the same date (after

February) moves "forward"two days in the subsequent year, if that year is a leap

year.

For example: 5/27/11 Fri 5/27/12 Sun

You can keep count forward to find that the first time this date falls on a Saturday is in

2017:

-------------------------------------------------------------------------------

5/27/13 Mon 5/27/14 Tue 5/27/15 Wed 5/27/16 Fri 5/27/17 Sat

Since we are given the range of the solutions, we must re-write the inequalities so

that we have   in terms of  and  .

Subtract   from all of the quantities:

Divide all of the quantities by  .

Since we have the range of the solutions, we can make them equal to  .

Multiply both sides by 2.

Re-write without using parentheses.

Simplify.

We need to find   for the problem, so the answer is 

Problem 12

Logan is constructing a scaled model of his town. The city's water tower stands 40

meters high, and the top portion is a sphere that holds 100,000 liters of water.

Logan's miniature water tower holds 0.1 liters. How tall, in meters, should Logan

make his tower?

SolutionThe water tower holds   times more water than Logan's miniature.

Therefore, Logan should make his tower   times shorter than the actual tower. This is   meters high, or choice  .

See also

Problem13

Angelina drove at an average rate of   kph and then stopped   minutes for gas.

After the stop, she drove at an average rate of   kph. Altogether she drove   km in

a total trip time of   hours including the stop. Which equation could be used to solve

for the time   in hours that she drove before her stop?

The answer is   because she drove at   kmh for   hours (the amount of time before the stop), and 100 kmh for   because she wasn't driving for   minutes, or   hours.

Multiplying by   gives the total distance, which is   kms. Therefore, the answer is 

Problem 14

Triangle   has  . Let   and   be on   and  , respectively, such

that  . Let   be the intersection of segments   and  , and suppose

that   is equilateral. What is  ?

Solution

Let  .

Since  , 

See also

Problem15

In a magical swamp there are two species of talking amphibians: toads, whose

statements are always true, and frogs, whose statements are always false. Four

amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make

the following statements.

Brian: "Mike and I are different species."

Chris: "LeRoy is a frog."

LeRoy: "Chris is a frog."

Mike: "Of the four of us, at least two are toads."

How many of these amphibians are frogs?

Solution

Solution 1

We can begin by first looking at Chris and LeRoy.

Suppose Chris and LeRoy are the same species. If Chris is a toad, then what he says is

true, so LeRoy is a frog. However, if LeRoy is a frog, then he is lying, but clearly Chris

is not a frog, and we have a contradiction. The same applies if Chris is a frog.

Clearly, Chris and LeRoy are different species, and so we have at least   frog out of

the two of them.

Now suppose Mike is a toad. Then what he says is true because we already have   

toads. However, if Brian is a frog, then he is lying, yet his statement is true, a

contradiction. If Brian is a toad, then what he says is true, but once again it conflicts

with his statement, resulting in contradiction.

Therefore, Mike must be a frog. His statement must be false, which means that there

is at most   toad. Since either Chris or LeRoy is already a toad, Brain must be a frog.

We can also verify that his statement is indeed false.

Both Mike and Brian are frogs, and one of either Chris or LeRoy is a frog, so we have   frogs total. 

Solution 2

Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a

frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.

As Mike is a frog, his statement is false, hence there is at most one toad.

As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the

other one tells the truth, and therefore is a toad.

Hence we must have one toad and three frogs.

See also

Problem 16

Nondegenerate   has integer side lengths,   is an angle bisector,  ,

and  . What is the smallest possible value of the perimeter?

SolutionBy the Angle Bisector Theorem, we know that  . If we use the lowest possible

integer values for AB and BC (the measures of AD and DC, respectively),

then  , contradicting theTriangle Inequality. If we use the

next lowest values (  and  ), the Triangle Inequality is satisfied. Therefore, our answer is  , or choice 

Problem 17

A solid cube has side length 3 inches. A 2-inch by 2-inch square hole is cut into the

center of each face. The edges of each cut are parallel to the edges of the cube, and

each hole goes all the way through the cube. What is the volume, in cubic inches, of

the remaining solid?

Solution

Solution 1

Imagine making the cuts one at a time. The first cut removes a box  . The

second cut removes two boxes, each of dimensions  , and the third cut does

the same as the second cut, on the last two faces. Hence the total volume of all cuts

is  .

Therefore the volume of the rest of the cube is  .

Solution 2

We can use Principle of Inclusion-Exclusion to find the final volume of the cube.

There are 3 "cuts" through the cube that go from one end to the other. Each of these

"cuts" has   cubic inches. However, we can not just sum their volumes, as

the central   cube is included in each of these three cuts. To get the correct

result, we can take the sum of the volumes of the three cuts, and subtract the volume

of the central cube twice.

Hence the total volume of the cuts is  .

Therefore the volume of the rest of the cube is  .

Solution 3

We can visualize the final figure and see a cubic frame. We can find the volume of the

figure by adding up the volumes of the edges and corners.

Each edge can be seen as a   box, and each corner can be seen as

a   box.

.

Problem18

Bernardo randomly picks 3 distinct numbers from the set   and

arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set   and also arranges them in descending

order to form a 3-digit number. What is the probability that Bernardo's number is

larger than Silvia's number?

Solution

We can solve this by breaking the problem down into   cases and adding up the

probabilities.

Case  : Bernardo picks  . If Bernardo picks a   then it is guaranteed that his number will be larger than Silvia's. The probability that he will pick a   is  .

Case  : Bernardo does not pick  . Since the chance of Bernardo picking   is  , the

probability of not picking   is  .

If Bernardo does not pick 9, then he can pick any number from   to  . Since Bernardo

is picking from the same set of numbers as Silvia, the probability that Bernardo's

number is larger is equal to the probability that Silvia's number is larger.

Ignoring the   for now, the probability that they will pick the same number is the

number of ways to pick Bernardo's 3 numbers divided by the number of ways to pick

any 3 numbers.

We get this probability to be 

Probability of Bernardo's number being greater is

Factoring the fact that Bernardo could've picked a   but didn't:

Adding up the two cases we get 

Problem19

Equiangular hexagon   has side lengths   

and  . The area of   is   of the area of the hexagon. What is

the sum of all possible values of  ?

Solution

It is clear that   is an equilateral triangle. From the Law of Cosines, we get that  . Therefore, the area of   

is  .

If we extend  ,   and   so that   and   meet at  ,   and   meet at  ,

and   and  meet at  , we find that hexagon   is formed by taking

equilateral triangle   of side length  and removing three equilateral

triangles,  ,   and  , of side length  . The area of   is therefore

.

Based on the initial conditions,

Simplifying this gives us  . By Vieta's Formulas we know that the sum of the possible value of  is  .

See also

Problem20

A fly trapped inside a cubical box with side length   meter decides to relieve its

boredom by visiting each corner of the box. It will begin and end in the same corner

and visit each of the other corners exactly once. To get from a corner to any other

corner, it will either fly or crawl in a straight line. What is the maximum possible

length, in meters, of its path?

Problem21

The polynomial   has three positive integer zeros. What is the

smallest possible value of  ?

Solution

By Vieta's Formulas, we know that   is the sum of the three roots of the polynomial  . Also, 2010 factors into  . But, since there

are only three roots to the polynomial, two of the four prime factors must be

multiplied so that we are left with three roots. To minimize  ,   and   should be multiplied, which means   will be   and the answer is  .

Problem22

Eight points are chosen on a circle, and chords are drawn connecting every pair of

points. No three chords intersect in a single point inside the circle. How many

triangles with all three vertices in the interior of the circle are created?

Solution

To choose a chord, we know that two points must be chosen. This implies that for

three chords to create a triangle and not intersect at a single point, six points need to be chosen. Therefore, the answer is   which is equivalent to 28, 

Problem23

Each of   boxes in a line contains a single red marble, and for  , the box

in the   position also contains   white marbles. Isabella begins at the first box and

successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let   be the probability that Isabella stops after drawing exactly   marbles. What is the smallest value of   for which  ?

SolutionThe probability of drawing a white marble from box   is  . The probability of

drawing a red marble from box   is  .

The probability of drawing a red marble at box   is therefore

It is then easy to see that the lowest integer value of   that satisfies the inequality is  .

See also

Problem24

The number obtained from the last two nonzero digits of   is equal to  . What is  ?

Solution

We will use the fact that for any integer  ,

First, we find that the number of factors of   in   is equal to  . Let  . The   we want is therefore the last two

digits of  , or  . Since there is clearly an excess of factors of 2, we know

that  , so it remains to find  .

If we divide   by   by taking out all the factors of   in  , we can write   as   

where where every multiple of 5 is replaced by

the number with all its factors of 5 removed. Specifically, every number in the form   

is replaced by  , and every number in the form   is replaced by  .

The number   can be grouped as follows:

Using the identity at the beginning of the solution, we can reduce   to

Using the fact that   (or simply the fact that   if you

have your powers of 2 memorized), we can deduce that  . Therefore  .

Finally, combining with the fact that   yields  .

See also