Problem 5 - JEJAK1000PENA – PENGAYAAN MATEMATIKA Web viewLet the group of people who all know each other be , and let the group of people who know no one be . Handshakes occur between

2017, AMC 12 AProblem 1At a gathering of people, there are people who all know each other and people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur?

SolutionLet the group of people who all know each other be , and let the group of people who know no one be . Handshakes occur between each pair such that and , and between each pair of members in . Thus, the answer is

Solution - Complementary CountingThe number of handshakes will be equivalent to the difference between the number of total

interactions and the number of hugs, which are and , respectively. Thus, the total

amount of handshakes is

Problem 2

Chloé chooses a real number uniformly at random from the interval . Independently, Laurent chooses a real number uniformly at random from the interval . What is the probability that Laurent's number is greater than Chloe's number?

Solution

Suppose Laurent's number is in the interval . Then, by symmetry, the probability of

Laurent's number being greater is . Next, suppose Laurent's number is in the interval . Then Laurent's number will be greater with probability . Since each case is

1 | Husein Tampomas, Solved Problems AMC 12A-12B 2017, 2017

equally likely, the probability of Laurent's number being greater is , so the answer is C.

Alternate Solution: Geometric ProbabilityLet be the number chosen randomly by Chloé. Because it is given that the number Chloé choose is interval , . Next, let be the number chosen randomly by Laurent. Because it is given that the number Chloé choose is interval ,

. Since we are looking for when Laurent's number is Chloé's we write the equation . When these three inequalities are graphed the area captured by

and represents all the possibilities, forming a rectangle 2017 in width and 4034 in height. Thus making its area . The area captured by

, , and represents the possibilities of Laurent winning, forming a trapezoid with a height 2017 in length and bases 4034 and 2017 length, thus

making an area . The simplified quotient of these two areas is the

probability Laurent's number is larger than Chloé's, which is .

Problem 3In the figure below, semicircles with centers at and and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter . The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered at is drawn externally tangent to the two smaller semicircles and internally tangent to the largest semicircle. What is the radius of the circle centered at ?

Solution 1Connect the centers of the tangent circles! (call the center of the large circle )


Notice that we don't even need the circles anymore; thus, draw triangle with cevian :

and use Stewart's Theorem:

From what we learned from the tangent circles, we have , , , , , and , where is the radius of the circle centered

at that we seek.

Thus:

Solution 2

Like the solution above, connecting the centers of the circles results in triangle with cevian . The two triangles and share angle , which means we can use Law of Cosines to set up a system of 2 equations that solve for respectively:


(notice that the diameter of the largest semicircle is 6, so its radius is 3 and is 3 - r)

We can eliminate the extra variable of angle by multiplying the first equation by 3 and subtracting the second from it. Then, expand to find :

, so =

Solution 3

Let be the center of the largest semicircle and be the radius of . We know that , , , , and . Notice that

and are bounded by the same two parallel lines, so these triangles have the same heights. Because the bases of these two triangles (that have the same heights) differ by a factor of 2, the area of must be twice that of , since the area of a triangle is

.

Again, we don't need to look at the circle and the semicircles anymore; just focus on the triangles.


Let equal to the area of and equal to the area of . Heron's Formula

states that the area of an triangle with sides and is where ,

or the semiperimeter, is

The semiperimeter of is Use Heron's Formula to obtain

Using Heron's Formula again, find the area of with sides , , and .

Now,

Solution 4

Let , the center of the large semicircle, to be at , and to be at .

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

Problem 4A 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 ?

Solution 1Analyze the first right triangle.

Note that and are similar, so . This can be written as

. Solving, .

Now we analyze the second triangle.


Similarly, and are similar, so , and . Thus,

. Solving for , we get . Thus,

.

Solution 2 (Shortcut using answer choices)

As stated in Solution 1, the square in the first triangle has a side length of . Then we look at the answers. The only answer with a factor of in the denominator is .

Problem 5

How many ordered pairs such that is a positive real number and is an integer between and , inclusive, satisfy the equation

SolutionBy the properties of logarithms, we can rearrange the equation to read with

. If , we may divide by it and get , which implies . Hence, we have possible values , namely

Since is equivalent to , each possible value yields exactly solutions , as we can assign to each . In total, we have

solutions.


Problem 6

A set is constructed as follows. To begin, . Repeatedly, as long as possible, if is an integer root of some polynomial for some

, all of whose coefficients are elements of , then is put into . When no more elements can be added to , how many elements does have?

Solution 1

At first, .

At this point, no more elements can be added to . To see this, let

with each in . is a factor of , and is in , so has to be a factor of some element in . There are no such integers left, so there can be no more additional elements.

has elements

Problem 7

For certain real numbers , , and , the polynomial has three distinct roots, and each root of is also a root of the polynomial

What is ?

Solution

Let and be the roots of . Let be the additional root of . Then from Vieta's formulas on the quadratic term of and the cubic term of , we obtain the following:


Thus .

Now applying Vieta's formulas on the constant term of , the linear term of , and the linear term of , we obtain:

Substituting for in the bottom equation and factoring the remainder of the expression, we obtain:

It follows that . But so

Now we can factor in terms of as

Then and

Hence .

Solution 2

Since all of the roots of are distinct and are roots of , and the degree of is one more than the degree of , we have that

for some number . By comparing coefficients, we see that . Thus,

Expanding and equating coefficients we get that

The third equation yields , and the first equation yields . So we have that


Problem 8Quadrilateral is inscribed in circle and has side lengths

, and . Let and be points on

such that and . Let be the intersection of line and the line through parallel to . Let be the intersection of line and the line through parallel to . Let be the point on circle other than that lies on line . What is

?

Solution

It is easy to see that First we note that

with a ratio of Then with a ratio of ,

so Now we find the length of . Because the quadrilateral is cyclic, we can simply use the Law of Cosines.

By Power of a Point,

. Thus

-solution by FRaelya

Problem 9Call a positive integer if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, ,

, and are monotonous, but , , and are not. How many monotonous positive integers are there?

Solution 1Case 1: monotonous numbers with digits in ascending order


There are ways to choose n digits from the digits 1 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in ascending order. Note that 0 is not included since it will always be a leading digit and that is not

allowed. The sum is equivalent to

Case 2: monotonous numbers with digits in descending order

There are ways to choose n digits from the digits 0 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in descending order. Note that 0 is included since we are allowed to end numbers with zeros. The sum is

equivalent to We discard the number 0 since it is not positive. Thus there are here.

Since the 1-digit numbers 1 to 9 satisfy both case 1 and case 2, we have overcounted by 9.

Thus there are monotonous numbers.

Solution 2Like Solution 1, divide the problem into an increasing and decreasing case:

Case 1: Monotonous numbers with digits in ascending order.

Arrange the digits 1 through 9 in increasing order, and exclude 0 because a positive integer cannot begin with 0.

To get a monotonous number, we can either include or exclude each of the remaining 9 digits, and there are ways to do this. However, we cannot exclude every digit at once, so we subtract 1 to get monotonous numbers for this case.

Case 2: Monotonous numbers with digits in descending order.

This time, we arrange all 10 digits in decreasing order and repeat the process to find ways to include or exclude each digit. We cannot exclude every digit at once,

and we cannot include only 0, so we subtract 2 to get monotonous numbers for this case.

At this point, we have counted all of the single-digit monotonous numbers twice, so we must subtract 9 from our total.

Thus our final answer is .


Problem 10

The number has over positive integer divisors. One of them is chosen at random. What is the probability that it is odd?

SolutionIf a factor of is odd, that means it contains no factors of . We can find the number of factors of two in by counting the number multiples of , , , and that are less than or equal to .After some quick counting we find that this number is . If the prime factorization of has factors of , there are choices for each divisor for how many factors of should be included ( to inclusive). The probability that a randomly

chosen factor is odd is the same as if the number of factors of is which is .

Solution by: vedadehhc

Solution 2We can write as its prime factorization:

Each exponent of these prime numbers are one less than the number of factors at play here. This makes sense; is going to have factors: , and the other exponents will behave identically.

In other words, has factors.

We are looking for the probability that a randomly chosen factor of will be odd--numbers that do not contain multiples of as factors.

From our earlier observation, the only factors of that are even are ones with at least one multiplier of , so our probability of finding an odd factor becomes the following:

Solution submitted by David Kim

Problem 11

A coin is biased in such a way that on each toss the probability of heads is and the

probability of tails is . The outcomes of the tosses are independent. A player has the choice


of playing Game A or Game B. In Game A she tosses the coin three times and wins if all three outcomes are the same. In Game B she tosses the coin four times and wins if both the outcomes of the first and second tosses are the same and the outcomes of the third and fourth tosses are the same. How do the chances of winning Game A compare to the chances of winning Game B?

The probability of winning Game A is less than the probability of winning Game B.

The probability of winning Game A is less than the probability of winning Game B.

The probabilities are the same.

The probability of winning Game A is greater than the probability of winning Game B.

The probability of winning Game A is greater than the probability of winning Game B.

SolutionThe probability of winning Game A is the sum of the probabilities of getting three tails and

getting three heads which is . The probability of winning Game B is the sum of the probabilities of getting two heads and getting two tails

squared. This gives us . The probability of winning

Game A is and the probability of winning Game B is , so the answer is


Problem 12The diameter of a circle of radius is extended to a point outside the circle so that

. Point is chosen so that and line is perpendicular to line . Segment intersects the circle at a point between and . What is the area of ?


Solution 1

Let be the center of the circle. Note that

. However, by Power of a Point,

,

so . Now

. Since .

Solution 2: Similar triangles with Pythagoreanis the diameter of the circle, so is a right angle, and therefore by AA similarity,

.

Because of this, , so .

Likewise, , so .

Thus the area of .


Solution 3: Similar triangles without PythagoreanOr, use similar triangles all the way, dispense with Pythagorean, and go for minimal calculation:

Draw with on . .

.

. ( ratio applied twice)

.

Problem 13Let be the -digit number that is formed by writing the integers from to in order, one after the other. What is the remainder when is divided by

?

SolutionWe will consider this number and . By looking at the last digit, it is obvious that the number is . To calculate the number , note that

so it is equivalent to

Thus it is and , so it is

Problem 14Real numbers and are chosen independently and uniformly at random from the interval

. What is the probability that , where denotes the greatest integer less than or equal to the real number ?


Solution

First let us take the case that . In this case, both and lie in the

interval . The probability of this is . Similarly, in the case that , and lie in the interval , and the probability is

. It is easy to see that the probabilities for for

are the infinite geometric series that starts at and with common ratio . Using the formula for the sum of an infinite geometric series, we get that the probability is

.


Problem 15

The graph of , where is a polynomial of degree , contains points , , and . Lines , , and intersect the graph again at points , ,

and , respectively, and the sum of the -coordinates of , , and is 24. What is ?

Solution

First, we can define , which contains points , , and . Now we find that lines , , and are defined by the equations ,

, and respectively. Since we want to find the -coordinates of the intersections of these lines and , we set each of them to , and synthetically divide by the solutions we already know exist (eg. if we were looking at line , we would synthetically divide by the solutions and , because we already know intersects the graph at and , which have -coordinates of and ). After completing this

process on all three lines, we get that the -coordinates of , , and are , ,

and respectively. Adding these together, we get which gives us

. Substituting this back into the original equation, we get


, and

Solution by: vedadehhc and tdeng

Problem 16Quadrilateral has right angles at and , , and . There is a point in the interior of such that and the area of

is times the area of . What is ?

Solution 1

Let , , and . Note that . By the Pythagorean Theorem, . Since , the ratios of side

lengths must be equal. Since , and . Let F be a point on such that is an altitude of triangle . Note that

. Therefore, and . Since and form altitudes of triangles and , respectively, the areas of these

triangles can be calculated. Additionally, the area of triangle can be calculated, as it is a right triangle. Solving for each of these yields:

Therefore, the answer is

Solution 2

Draw line through , with on and on , . WLOG let ,

, . By weighted average .

Meanwhile, .

. We obtain , namely .


The rest is the same as Solution 1.

Problem 17A set of people participate in an online video basketball tournament. Each person may be a member of any number of -player teams, but no two teams may have exactly the same members. The site statistics show a curious fact: The average, over all subsets of size of the set of participants, of the number of complete teams whose members are among those people is equal to the reciprocal of the average, over all subsets of size of the set of participants, of the number of complete teams whose members are among those people. How many values , , can be the number of participants?

SolutionSolution by Pieater314159

Let there be teams. For each team, there are different subsets of players including that full team, so the total number of team-(group of 9) pairs is

Thus, the expected value of the number of full teams in a random set of players is

Similarly, the expected value of the number of full teams in a random set of players is

The condition is thus equivalent to the existence of a positive integer such that


Note that this is always less than , so as long as is integral, is a possibility. Thus, we have that this is equivalent to

It is obvious that divides the RHS, and that does iff . Also, divides it iff . One can also bash out that divides it in out of the possible residues .

Using all numbers from to , inclusive, it is clear that each possible residue is reached an equal number of times, so the total number of working in that

range is . However, we must subtract the number of "working" ,

which is . Thus, the answer is .


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Problem 5 - JEJAK1000PENA – PENGAYAAN MATEMATIKA Web viewLet the group of people who all know each other be , and let the group of people who know no one be . Handshakes occur between