#1**+1 **

This is how I would personally approach this problem.

#1: List all the integers from 1 to 100.

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 |

21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 |

31 | 32 | 33 | 34 | 35 | 36 | 37 | 38 | 39 | 40 |

41 | 42 | 43 | 44 | 45 | 46 | 47 | 48 | 49 | 50 |

51 | 52 | 53 | 54 | 55 | 56 | 57 | 58 | 59 | 60 |

61 | 62 | 63 | 64 | 65 | 66 | 67 | 68 | 69 | 70 |

71 | 72 | 73 | 74 | 75 | 76 | 77 | 78 | 79 | 80 |

81 | 82 | 83 | 84 | 85 | 86 | 87 | 88 | 89 | 90 |

91 | 92 | 93 | 94 | 95 | 96 | 97 | 98 | 99 | 100 |

#2: Eliminate all even numbers and 1 and 100 from the list since we are only interested in "square-free" numbers bounded between 1 and 100.

3 | 5 | 7 | 9 | ||||||

11 | 13 | 15 | 17 | 19 | |||||

21 | 23 | 25 | 27 | 29 | |||||

31 | 33 | 35 | 37 | 39 | |||||

41 | 43 | 45 | 47 | 49 | |||||

51 | 53 | 55 | 57 | 59 | |||||

61 | 63 | 65 | 67 | 69 | |||||

71 | 73 | 75 | 77 | 79 | |||||

81 | 83 | 85 | 87 | 89 | |||||

91 | 93 | 95 | 97 | 99 |

#3: Eliminate multiples of the first perfect square I will test, 3^2, or 9. I am not testing 2^2, or 4 because I have already eliminate the even numbers from the list, so no multiples of 4 remain in the list right now.

3 | 5 | 7 | |||||||

11 | 13 | 15 | 17 | 19 | |||||

21 | 23 | 25 | 29 | ||||||

31 | 33 | 35 | 37 | 39 | |||||

41 | 43 | 47 | 49 | ||||||

51 | 53 | 55 | 57 | 59 | |||||

61 | 65 | 67 | 69 | ||||||

71 | 73 | 75 | 77 | 79 | |||||

83 | 85 | 87 | 89 | ||||||

91 | 93 | 95 | 97 |

#4: Eliminate the multiples of the second perfect square I will test, 5^2, or 25. Again, I am not testing 4^2, or 16, because even numbers have already been filtered out.

3 | 5 | 7 | |||||||

11 | 13 | 15 | 17 | 19 | |||||

21 | 23 | 29 | |||||||

31 | 33 | 35 | 37 | 39 | |||||

41 | 43 | 47 | 49 | ||||||

51 | 53 | 55 | 57 | 59 | |||||

61 | 65 | 67 | 69 | ||||||

71 | 73 | 77 | 79 | ||||||

83 | 85 | 87 | 89 | ||||||

91 | 93 | 95 | 97 |

#5: Eliminate the multiples of 7^2, or 49. Only one number gets eliminated.

3 | 5 | 7 | |||||||

11 | 13 | 15 | 17 | 19 | |||||

21 | 23 | 29 | |||||||

31 | 33 | 35 | 37 | 39 | |||||

41 | 43 | 47 | |||||||

51 | 53 | 55 | 57 | 59 | |||||

61 | 65 | 67 | 69 | ||||||

71 | 73 | 77 | 79 | ||||||

83 | 85 | 87 | 89 | ||||||

91 | 93 | 95 | 97 |

#6: Normally, I would eliminate the multiples of 9^2, but I already sifted out the multiples of nine, so all the multiples of 9^2 have already been eliminated from the list.

#7: There is no need to eliminate the multiples of 11^2, or 121, since that multiple is larger than the largest number we are testing, so we can stop the process here. The numbers that remain are "square-free."

In a list, they are 3,5,7,11,13,15,17,19,21,23,29,31,33,35,37,39,41,43,47,51,53,55,57,59,61,65,67,69,71,73,77,79,83,85,87,89,91,93,95,97.

TheXSquaredFactor03.01.2019

#2**0 **

Since I was in the process of making this solution, I will post it. It is slightly different to Cphill's solution, but the concepts used are mostly the same.

Generating a diagram for problems for geometry is generally a useful strategy so that one can visualize the problem and make observations. Here is the diagram that I will reference throughout the solving process:

I have included all the given information and labeled extra points and lines because I plan to use them while solving.

\(\triangle ABC\) has a 30- and a 60-degree angle. The remaining angle of the triangle, \(m\angle CAB=90^{\circ}\) . The triangle is therefore classified as a 30-60-90 triangle.

30-60-90 triangles have a special property: their side lengths are always at a constant ratio of \(1:\sqrt{3}:2\) . Since I know the length of \(\overline{AB}\) , 1, I can determine the lengths of the others by setting up a proportion.

\(\frac{AB}{BC}=\frac{1}{2}\) | This is the proportion that I mentioned earlier. |

\(BC=2AB\) | I am solving for BC to make the proportion into a simple equation. Substitute 1 for AB. |

\(BC=2\) | |

Now we know that BC=2. In order to go further with this solution, you must know what a centroid is. A centroid is a point of concurrency where all three medians of a triangle intersect in the interior of the triangle. This means you must know what a median is. A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side of that vertex. This information will help us.

In the diagram above, \(\overline{AE}\) is a median. Therefore, point E is a midpoint of \(\overline{BC}\) , so point E bisects the segment. Therefore, \(BE=\frac{BC}{2}=\frac{2}{2}=1\) . This means that we have identified two side lengths and the measure of one angle of \(\triangle ABE\) . You may already recognize it based on the sides and the angles that the triangle is equilateral, but you can use the law of cosines to find \(AE\).

\(c^2=a^2+b^2-2ab\cos C\) | c is the length that we are trying to find. a and b are the side lengths that we know already. |

\(AE^2=AB^2+BE^2-2*AB*BE*\cos60^{\circ}\) | Substitute all the known values and solve for the remaining side length. |

\(AE^2=1^2+1^2-2*1*1*0.5\) | Simplify the right-hand side. |

\(AE^2=2-1=1\) | Take the square root of both sides. |

\(AE=1\) | Of course, in the context of geometry, the negative answer is nonsensical, so we reject the answer. This is the length of the median. |

The centroid is located at point G, and there is one property about centroids that we must know in order to crack this one. It is that it divides the median into a ratio of 2:1. \(\overline{AG}\) is contained within \(\overline{AE}\), so we can generate a system of equations.

\(\boxed{1}\frac{AG}{GE}=\frac{2}{1}\\ \boxed{2}AG+GE=AE=1\) | This is a system of equations that we can solve for AG. |

\(AG=2GE\\ GE=\frac{AG}{2}\) | I manipulated the first equation and wrote in terms of GE. I will substitute this into the second equation to solve for AG. |

\(AG+\frac{AG}{2}=1\) | Multiply by 2 on both sides to eliminate the fraction. |

\(2AG+AG=2\) | Combine the like terms on the left-hand side of the equation. |

\(3AG=2\) | Divide by 3 on both sides of the equation. |

\(AG=\frac{2}{3}\) | |

... And finally, we are done!

TheXSquaredFactor08.12.2018

#1**+2 **

I think the best thing to do here is to construct a diagram. The problem gives you vertices with coordinates from the Cartesian plane, and it even gives you an inequality. Here is the construction I created. I will reference it throughout the solving process:

I first located the given coordinates and connected them with vertical and horizontal lines. I then graphed the inequality\(x>7y\), which can be rewritten as \(y<\frac{x}{7}\) . I added a few points, as they will become relevant when I solve.

One way to solve this problem is to figure out the ratio of the area created by the inequality \(y<\frac{x}{7}\) and the sides of the rectangle to the area of the entire rectangle. I will do this by first finding the coordinates of the point E. Point E is located on \(\overline{BC}\) , so its x-coordinate is also 2009. If you know the x-coordinate, then the y-coordinate of any point on \(y=\frac{x}{7}\) is \(\frac{x}{7}\) . Therefore, the y-coordinate is \(\frac{2009}{7}\) .

The formula for the area of the triangle is \(\frac{1}{2}bh\) . \(\triangle ABE\) is a right triangle, so the base and height are the side lengths that are not the hypotenuse. We know the length of the base, \(AB\) , because it is in the diagram. The height is the length of \(BE\) , or \(\frac{2009}{7}\)

\(A_{\triangle}=\frac{1}{2}*AB*BE=\frac{1}{2}*2009*\frac{2009}{7}=\frac{2009^2}{14}\)

We can also calculate the area of the rectangle

\(A_{\text{rect.}}=bh=AB*BC=2009*2010\)

As aforementioned, the probability of (x,y) such that x>7y is just the ratio of these areas.

\(P(x,y|x>7y)=\frac{A_{\triangle}}{A_{\text{rect.}}}\) | As aforementioned, the probability of (x,y) such that x>7y is just the ratio of these areas. Substitute in the values and simplify. |

\(P(x,y|x>7y)=\frac{\frac{2009^2}{14}}{2009*2010}\) | Use algebraic manipulation to simplify this. |

\(P(x,y|x>7y)=\frac{2009^2}{14*2009*2010}\) | |

\(P(x,y|x>7y)=\frac{2009}{14*2010}\) | At this point, you must use a calculator. |

\(\)\(P(x,y|x>7y)=0.07139\) | |

TheXSquaredFactor08.12.2018

#1**+1 **

Let's first think of the domain of this particular function.

The only portion of this function that can affect the state of the domain is the denominator, so let's investigate it.

If there was a value for x such that \(|x-3|+|x+1|=0\), then a restriction for the domain of this function would exist. Let's see if there are any values for x such that this is true:

\(|x-3|+|x+1|=0\) | I think the best way to solve this is to think about it logically. You could compute the sign changes, but I think there is a better way.
The absolute value signs always output a nonnegative result. In this equation, we are adding two nonnegative outputs together. There is only one way that the sum of two nonnegative numbers can equal 0. That way is 0+0=0. |

\(|x-3|=0\\ x-3=0\\ x=3\) | I have determined that the only x-value that makes the first addend, \(|x-3|\), equal to 0 is when x=3. |

\(|3+1|\stackrel{?}=0\\ 4\stackrel{?}=0\\ \text{false}\) | x=3 is the only candidate that can make this equation equal to 0. The second addend, |x+1|, evaluates to 4 when x=3. However, we determined earlier that |x+1| has to evaluate to 0 in order for solutions to exist. Therefore, there are no solutions, for the real number set at least. |

\(\text{Domain}: x \in {\rm I\!R}\) | The conclusion from this investigation is that there are no restrictions in the domain. |

The nature of this function is generally one of a cubic function because there are no vertical asymptotes. Dividing by the addition of a sum of nonnegatives numbers does not really affect the range.

\(\text{Range}: x \in {\rm I\!R}\)

.TheXSquaredFactor02.12.2018

#1**+4 **

I think that this question is fairly difficult. I have added a few extra lines and points to the diagram where I saw fit:

Here are my initial observations in regard to this current diagram:

- The x-axis, which is blue, is tangent to both circles.
- The green line is also tangent to both circles.
- The green and blue lines intersect at one common point on the x-axis.

These initial observations are enough to get me started with this problem. I constructed a parallel line to the x-axis, \(\overleftrightarrow{AK}\). \(\angle BKA\) is, therefore, a right angle. Point K is located at (8,1) because it acquires the y-coordinate of Point A and the x-coordinate of Point B. \(\triangle BKA\) is a right triangle, and it is possible to find \(\tan(m\angle KAB)\).

\(\tan m\angle KAB=\frac{9-1}{8-2}=\frac{8}{6}=\frac{4}{3}\). Notice how 4/3 is also the slope of \(\overleftrightarrow{AB}\). The actual measure of the angle is not really important.

Because \(\overleftrightarrow{AK}||\overleftrightarrow{JH}\), \(\angle HJB\cong\angle KAB\). By extension, \(m\angle HJB=m\angle KAB\). This also means that \(\tan m\angle HJB=\tan m\angle KAB=\frac{4}{3}\).

Because two tangent lines of a circle intersect at an external point, J, \(\angle IJB\cong\angle HJB\). Using the same reasoning as before, \(\tan m\angle IJB=\tan m\angle HJB=\frac{4}{3}\). \(m\angle IJH=m\angle IJB+m\angle HJB=m\angle HJB+m\angle HJB=2m\angle HJB\)

Both tangent lines meet at an angle of \(2m\angle HJB\). Just like finding \(\tan m\angle KAB\) yielded the slope, \(\tan 2m\angle HJB\) will yield the slope of the desired line, \(\overleftrightarrow{IJ}\).

Luckily, we can utilize the Double-Angles Identities, specifically the \(\tan 2m\angle HJB=\frac{2\tan m\angle HJB}{1-\tan^2 m\angle HJB}\).

\(\tan 2m\angle HJB=\frac{2\tan m\angle HJB}{1-\tan^2 m\angle HJB};\\ \tan m\angle HJB=\frac{4}{3}\) | This is the Double-Angle Identity for the tangent function. Replace all instances of \(\tan m\angle HJB\) with 4/3. |

\(\tan 2m\angle HJB=\frac{2*\frac{4}{3}}{1-\left(\frac{4}{3}\right)^2}\\ \tan 2m\angle HJB=\frac{\frac{8}{3}}{\frac{-7}{9}}\\ \tan 2m\angle HJB=\frac{-24}{7} \) | Simplify. The result is the slope of the line \(\overleftrightarrow{IJ}\) . |

\(y=\frac{-24}{7}x+b\) | This is the equation of the line \(\overleftrightarrow{IJ}\). |

The equation of the circle with the center A is \((x-2)^2+(y-1)^2=1\). Since \(y=\frac{-24}{7}x+b\), substitute that into the equation for the circle:

\((x-2)^2+(y-1)^2=1;\\ y=\frac{-24}{7}x+b \) | These are the equations we are grappling with. |

\((x-2)^2+(\frac{-24}{7}x+[b-1])^2=1\) | It is time to expand this monstrosity. |

\(x^2-4x+4+\left(\frac{-24}{7}x\right)^2+2*\frac{-24}{7}x*(b-1)+(b-1)^2=1\) | |

\(x^2-\frac{28}{7}x+4+\frac{576}{49}x^2-\frac{48b-48}{7}x+b^2-2b+4=0\) | |

\(\textcolor{red}{\frac{625}{49}}x^2\textcolor{blue}{-\left(\frac{48b-20}{7}\right)}x+\textcolor{green}{(b^2-2b+4)}=0\\ \textcolor{red}{a}x^2+\textcolor{blue}{b}x+\textcolor{green}{c}=0\) | Notice the parallelism between a quadratic and the current equation. Tangent lines always intersect a circle at only one point, so let's set the discriminant equal to zero to ensure only one solution is possible. |

\(\Delta=b^2-4ac;\\ \Delta=\left(-\frac{48b-20}{7}\right)^2-4*\frac{625}{49}*(b^2-2b+4)\\ \Delta=\frac{-196b^2+3080b-9600}{49}\) | Set the discriminant equal to zero and solve. |

\(\frac{-196b^2+3080b-9600}{49}=0\Rightarrow-196b^2+3080b-9600=0\) | Use the quadratic formula here. |

\(b_1=\frac{30}{7}\text{ or }b_2=\frac{80}{7}\) | Reject b_2 because that is on the other side of the circle. |

TheXSquaredFactor16.11.2018