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from the equation. Then, all coordinates that satisfy

intersect the circle. Squaring, we find After multiplying through by

find

and rearranging, we

and consider

. We see this is a quadratic in

taking the determinant, which tells us that solutions are real when, after factoring:

We plot this inequality on the number line to find it is

satisfied for all values except: extraneous as both there are a total of

and values.

. We then eliminate 0 because it is

are points which coincide. Therefore,

Problem 14

Nondegenerate and

has integer side lengths,

is an angle bisector,

,

. What is the smallest possible value of the perimeter?

Solution

By the Angle Bisector Theorem, we know that then

use the next lowest values (satisfied. Therefore, our answer is

. If we use the lowest

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

, contradicting the Triangle Inequality. If we and

), the Triangle Inequality is

, or choice

.

Problem 15

A coin is altered so that the probability that it lands on heads is less than and when the coin is flipped four times, the probaiblity of an equal number of heads and tails is . What is the probability that the coin lands on heads?

Solution

Let be the probability of flipping heads. It follows that the probability of flipping tails is

.

The probability of flipping heads and tails is equal to the number of ways to flip it times the product of the probability of flipping each coin.

As for the desired probability both and consider the positive root, hence

are nonnegative, we only need to

Applying the quadratic formula we get that the roots of this equation are .

As the probability of heads is less than , we get that the answer is .

Problem 16

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 number is larger than Silvia's number?

and also arranges them in

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

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

Problem 17

Equiangular hexagon

has side lengths

. The area of

is

and

of the area of the hexagon.

What is the sum of all possible values of ?

Solution

It is clear that that

.

If we extend at

, and

, and ,

and

so that

and

meet at

,

and

meet

is an equilateral triangle. From the Law of Cosines, we get

. Therefore, the area of

is

meet at and

, we find that hexagon of side length

, of side length . The area of

is formed by

is

taking equilateral triangle triangles, therefore

and removing three equilateral

.

Based on the initial conditions,

Simplifying this gives us sum of the possible value of is

. By Vieta's Formulas we know that the .

Problem 18

A 16-step path is to go from boundary of the square

, to

with each step increasing either the

at each step?

-coordinate or the -coordinate by 1. How many such paths stay outside or on the

Solution

Brute Force Solution

The number of ways to reach any point ways to reach

on the grid is equal to the number of

. Using this

plus the number of ways to reach

recursion, we can draw the diagram and label each point with the number of ways to reach it and go up until we reach the end. Luckily, the figure is not so big that this is too time-consuming or difficult to do. For example:

etc.

We soon reach

Combinatorial Solution 1

By symmetry we only need to count the paths that go through the second quadrant (

,

).

be the first point when it reaches

.

. Clearly

For each of these paths, let

and the previous point on such path has to be

Fix the value of . There are ways how the path can go from to

, and then there are ways how the path can go from to .