Problem number 11. The square ABCD lies in a plane and its edge measures 1. Consider all squares that share at least two vertices with the square ABCD. What is the area of the region covered by all such squares, not including ABCD? So to start, let's go ahead and write out our original square ABCD. Now, four squares that obviously share two vertices each with this square are the four that are either adjacent or adjacent to it either vertically or horizontally, like so. However, there are also four squares that share two diagonals, two diagonal vertices with the square. So there are two squares that share these two vertices, and there are two squares that share these two vertices. And if we include those, well, based on the side length, the side length would be square root of 2, which is the diagonal of the square. So it would also happen to go like this. So one side would go from one diagonal to the next. Another side of that same square would go from this diagonal to this one, and then from this diagonal to this one, and then from this diagonal to this one, since all of them have the same diagonal length, which is square root of 2. So filling all of those in, we have this sort of a shape where also notice that while we're not counting the original square, the entire original square is covered by other squares that share the diagonal vertices with it. So in fact, we are covering a 9 by 9 area missing four half squares with a side length of 1. See, we're missing half of a square of side length 1, half a square of side length 1, half here, and half here. So it would be a 3 times 3 equals 9 area, but we're missing four halves, which is equal to 2, and 9 minus 2 is equal to 7. So we have a total surface area of 7, and thus our final answer is C. Problem number 12. Angle B measures 25% less than angle C and 50% more than angle A. The measure of angle C is what? So instead of talking about this in terms of percentages, let's talk about it in terms of fractions, since that'll make it a little bit easier. If angle B measures 25% less than angle C, then that means angle B is equal to 75% of angle C, or three-fourths of angle C, and if it's 50% more than angle A, then that means that it's three halves of angle A. So the measure of B is equal to three-halves of A, which is equal to three-quarters of C. Now the measure of angle C, as we see here, three-quarters of C is equal to three-halves of A. So what we can do with this equality right here is we can multiply both sides by four-thirds. That'll just give us C here, and that'll give us 4 over 2, which equals 2 times A here. So C is equal to 2 times A, which is the same as saying that C is 100% greater than angle A. So our final answer is D. Problem number 13. If x and y are integers such that 2 to the x plus 1 plus 2 to the x equals 3 to the y plus 2 minus 3 to the y, then the value of x is what? So let's start by rewriting both sides of this equation in a way that's a little more convenient. So 2 to the x plus 1 is just another way of saying 2 to the x times 2 plus 2 to the x from this 2 to the x term that we have here. Now the reason why this is nicer is because now we have 2 times this number here and another copy of this number here, so that's going to turn to 2 to the x times 3. Similarly here, 3 to the y plus 2 is just 3 to the y times 3 squared minus 3 to the y, and so that's going to be, since 3 squared is equal to 9, we're subtracting a 3 to the y. That's going to give us 3 to the y times 8. So now if we divide both sides by 3 to the y and by 3, we get 2 to the x over 3 to the y equals 8 thirds. Now the numerator here is clearly a power of 2 and the denominator is clearly a power of 3, and since there's no way that we could multiply, there's no way to multiply the numerator and denominator by the same number or to simplify the fraction in some way to get a different answer than this. There's only one way that we could fill in x and y so that we can get it to equal 8 thirds, and instead of putting it into words, I would say just go ahead and try playing with it and you'll see that the only way that this equation can hold is if x is equal to 3 and y is equal to 1, and since x is equal to 3, we know that our final answer is b. Problem number 14. What is the sum of cosine 1 degree plus cosine 2 degrees adding all of the cosines up till 359 degrees? Now a good thing to notice here is that if we're adding all of the cosines together except for cosine 360 and cosine 0, well the cosine 1 degree for instance, since you can think of it as the x value on the unit circle, we can look over on the mirror image of it along the vertical axis from 1 degree over to 179 degrees, and we know that the cosine of 1 degree will be equal to the negative of 179 degrees, and similarly cosine 2 degrees is equal to negative cosine 178, and then for instance cosine 181 degrees is equal to negative cosine 359 degrees. These are all the angles that are mirror images of each other along the vertical axis of the unit circle. Now because all of these mirror images cancel each other out, most of the terms here cancel each other out, with the exception of one term, and that's cosine 180 degrees, since its mirror image would be either cosine 0 or cosine 360, but neither of those are being considered here. So cosine 180 degrees is as far left on the unit circle as you can go, so that's equal to negative 1. All of the other terms cancel out because of this mirror imaging that they have, and so our final answer is E. Problem number 15. The figure shows two semicircles. The chord CD of the larger semicircle is parallel to the segment AB and tangent to the smaller semicircle. What is the area of the shaded region if CD equals 4? So ideally what we would like to do is we would like to find the area of the large semicircle and just subtract the area of the small semicircle, if that would somehow be possible. Now notice the point from the center of the smaller semicircle all the way to the chord, that's going to be equal to the radius of this smaller semicircle, and it's also going to be equal to, well it's also going to be equal to the side of a triangle. We can just go ahead and shift this radius over a little bit, and it's equal to one of the sides of a triangle that we can form here, and the point here we can say is at the center of the larger semicircle, and so this hypotenuse here we can call the radius of the larger semicircle, and now since this is the center point of the larger semicircle, then this length here from here to D is exactly half the length of the chord, which is 4, so this length is 2. So here we have the little radius, here we have the big radius, and here we have 2. Now what we can gather from this is that big R squared, the radius of the big semicircle, is equal to little R squared plus 2 squared, which equals 4, and that's just using the Pythagorean theorem on this right triangle here, where again this is the little radius, this is the big radius, and this is a length of 2, since this point here is the center of AB. Now how is this helpful? Well we can go ahead and recall the area formula for a semicircle, and remember we want to subtract the area of the smaller semicircle from the area of the larger semicircle, or 1 half pi big R squared minus 1 half pi little r squared, which is equal to 1 half pi big R squared minus little r squared, but from this equation we could tell that that's going to be equal to 4. So in fact we have 4 in the parentheses here, which means it's 1 half pi times 4, which is equal to 2 pi, and so our final answer is C. Problem number 16. In the following diagram, the length of each edge is an integer. What is the value of x? So what we know about the value of x is that it's the side length of two different triangles, one with side lengths of 5 and 9, and one with side lengths of 5 and 17. Now one important fact about triangles in general is that if you add up the lengths of two sides, then the length of the third side will always be smaller, and that's a well-established fact that you definitely learned in your geometry class at some point, and so we're going to use that fact here. Now we know that for instance x should be less than 9 plus 5, right, because these are the other two side lengths, so when we add them together x should be less than that, and also 5 plus x should be less than 17, or greater than 17, sorry. So just from the fact that one side length must be less than the sum of the other two side lengths, we can gather these two inequalities, but then manipulating these inequalities a little bit, we get x is less than 14, and x is greater than 12. So if x falls between 12 and 14, and it's an integer, we know that it must be 13, so our final answer is b. Problem number 17. Tom was born on his mother's 20th birthday, and both celebrate their birthdays on the same day. How many times will Tom's age divide his mother's age? So the nice thing about this problem is that, well, there's a very clever solution to it. Really you don't need to consider, you don't need to like add one to Tom's age, and one to his mom's age, and then two to Tom's age, and two to his mother's age, and keep on adding like that, and checking if they divide. That's completely unnecessary. In fact, all you have to do is consider all of the factors of 20, and here they are. 1, 2, 4, 5, 10, and 20. Now after each of these numbers of years, Tom's age will divide his mother's age. So for instance, when Tom's mother is 21, and Tom is 1, their ages will divide. When Tom's mom is 22, Tom will be 2. The ages will divide. When Tom's mom is 24, he will be 4, and again the ages will divide. So the idea is this. All of the numbers that divide 20 already divide the first 20 years of his mom's age, and then the additional however many years will just be the same copy of that number. So for instance, we know that 5 divides 20, so after 5 years, 5 will also divide 25. That's sort of the reasoning behind it. So go ahead and try to let that sink in for a bit. I think it's a nice clever sort of quick way to confirm that these are the only ages that will divide Tom's mom's age, but once you realize this trick, you realize that it's only the factors of 20 that make sense. So we know that there are six such numbers, and thus our final answer is C. Problem number 18. A certain island is inhabited by knights and liars. Every liar always lies, and every knight always tells the truth. An islander named Avers, when asked about his neighbor Barrows, answered, at least one of us is a liar. Which of the following statements is true? So we're going to try to use process of elimination here. Now let's suppose they are both knights, just to pick one of the answers. If they are both knights, then that means Avers is a knight, and that means he's telling the truth. But Avers says that at least one of them is a liar. But if they're both knights, then neither of them can be liars, so this would be a lie. And so we have a logical contradiction here. So we can eliminate answer choice D. Now if they are both liars, then that means Avers is a liar, but he's saying at least one of us is a liar. And that would happen to be true in this case, because they're both liars, and we can't have Avers telling the truth. So again, this is a contradiction. Now what about answer choice C? Avers is a liar, and Barrows is a knight. Now if Avers is a liar, and Barrows is a knight, that means that one person is a liar. But Avers is saying at least one of us is a liar, so Avers is telling the truth here. So Avers cannot be a liar. So again, we can eliminate this answer choice. Now Avers is a knight, and Barrows is a liar. Well let's think about this one. If Avers is telling the truth, and saying at least one of us is a liar, and Barrows is a liar, then there's no logical contradiction here. It totally makes sense. There's no, there's nothing that sets off any sort of red flags. And so by process of elimination, we see that our final answer is E. Problem number 19. Consider a sphere with a radius of 3 that is centered at the origin of a Cartesian coordinate system. How many points on the surface of this sphere have integer coordinates? So let's recall the formula for a sphere of radius 3. It's x squared plus y squared plus z squared equals 3 squared. So now let's go ahead and consider how we could set it up so that there are only integer coordinates. Well there's basically two kinds of points we can have. We can have a point with two zeros and a three, or we could have two twos and a one. Now in particular, let's look at the case where we have two zeros and a three. It doesn't matter where the three is. It can be in the x value, it can be the y value, or it can be the z value, because it doesn't change the outcome of the equation. Also the three can be positive or negative. So we have two choices here, two choices here, and two choices here. So a total of six different points of the form where there's a three or minus three and two zeros. So there are six points in this category. Notice that plus or minus zero is the same, so we don't consider whether zero is positive or negative. But there are six possibilities here. Now in the case where we have two twos and a one, the pluses and minuses do matter. So first of all, we can have the one here, the one here, or the one here. That's three different possibilities. And then we can have either the first term being a plus or minus term, the second term being a plus or minus term, or the third term being a plus or minus term. So that's two times two times two times three, because we don't know where the one is. We have three different choices for where the one is. So that's equal to eight times three, or 24 different possibilities for points of this form. So we have 24 possibilities and six possibilities. We add them together for a total of 30 different points that exist on the sphere that satisfies this equation. So our final answer is A. Problem number 20. The figure shows a graph of a function y equals ax cubed plus bx squared plus cx plus d. What is the value of b? So looking at this equation, we know that there are three points that we can try working with. When x is equal to one or negative one, we have y equals zero. And when x equals zero, we have y equals two. So that's three points. Now, just right away, we can figure out that if x equals zero, all of these terms disappear. And so we're just left with d. And that's supposed to be equal to two, since that's this point right here. So from that, we can conclude that d is supposed to be equal to two. Now, this might not even be that important for actually solving the problem. It's necessary at the end, but it's a relatively quick step. More important is realizing that y is equal to zero for both of these points at x equals one and negative one. So we could set up an equation where we plug in one and negative one for x, and we get a plus b plus c plus two is equal to negative a plus b minus c plus two, since x cubed is equal to negative one, or since one negative one cubed is equal to negative one. Now, from this equation, we can subtract b plus two from both sides, and we get that a plus c is equal to negative a minus c, which is the additive inverse of a plus c. So the only case where a number equals its additive inverse, or its negative, however you want to phrase it, is when the number is equal to zero. That's the only number where the positive and negative are equal to each other. So we know that a plus c and negative a minus c are both equal to zero. And that means that if we go back to either of these equations, let's say a plus b plus c plus two is equal to zero, then we can just go ahead and erase the a plus c, since we know that's also equal to zero, and we know that b plus two must then be equal to zero. But then b is clearly equal to negative two, so our final answer is b.

logical deduction

geometric reasoning

exponential equation

cosine symmetry

Cartesian plane