AMC 12A Problem and Solution

2017 AMC 12A

Therefore is at

and is at

.

Let the radius of circle be .

Using Distance Formula, we get the following system of three equations:

By simplifying, we get

By subtracting the first equation from the second and third equations, we get

which simplifies to

When we add these two equations, we get

So

Problem17

There are different complex numbers such that real number?

. For how many of these is

a

Solution 1

Note that these such that

This is real if even

are for integer . So

is even. Thus, the answer is the number of

which is

.

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2017 AMC 12A

Solution 2

By Euler's identity,

, where is an integer.

Using De Moivre's Theorem, we have where that produce unique results. Using De Moivre's Theorem again, we have For

to be real,

,

has to equal to negate the imaginary component. This occurs

even

whenever is an integer multiple of , requiring that is even. There are exactly

, so the answer is

.

values of on the interval

Solution 3

From the start, recall from the Fundamental Theorem of Algebra that have solutions (and these must be distinct since the equation factors into

must

), or notice that the question is simply referring

. Notice that

,

to the 24th roots of unity, of which we know there must be

so for any solution , will be one of the 4th roots of unity (, , , or ). Then solutions will satisfy , will satisfy (and this is further justified by knowledge of the 6th roots of unity), so there must be

such .

Problem18

Let

equal the sum of the digits of positive integer . For example,

. Which of the following could be the value

. For a

particular positive integer , of

?

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2017 AMC 12A

Solution 1

Note that since

answer choices

is , so

, we have that

.

. So,

. The only one of the

Solution 2

One possible value of

would be

, but this is not any of the choices. Therefore, we

know that ends in , and after adding , the last digit carries over, turning the last digit into . If the next digit is also a , this process repeats until we get to a non- digit. By the end, the sum of digits would decrease by multiplied by the number of carry-overs but increase by as a result of the final carrying over. Therefore, the result must be less than original value of condition is

, , since

, where is a positive integer. The only choice that satisfies this

. The answer is

.

Solution 3

Another way to solve this is to realize that if you continuously add the digits of the number

, we get . Adding one to that, we get . So,

satisfies the .

if we assess each option to see which one attains , we would discover that requirement, because

. . The answer is

Problem19

A square with side length is inscribed in a right triangle with sides of length , , and so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length is inscribed in another right triangle with sides of length , , and so that one side of the square lies on the hypotenuse of the triangle. What is ?

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2017 AMC 12A

Solution 1

Analyze the first right triangle.

Note that as

and . Solving,

are similar, so .

. This can be written

Now we analyze the second triangle.

Similarly, Thus, Thus,

and are similar, so , and .

.

. Solving for , we get

.

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