Problem number 1 states, the sum of the ages of Tom and John is 23, the sum of the ages of John and Alex is 24, and the sum of the ages of Tom and Alex is 25. What is the age of the oldest person? So first the problem tells us that the sum of the ages of Tom and John is 23, we'll represent Tom's age with a T and John's age with a J, so T plus J is equal to 23. Then the problem tells us that the sum of the ages of John and Alex is 24, we'll represent Alex's age with A, so in this case you would have J plus A equals 24. And lastly it tells us that the sum of the ages of Tom and Alex is 25, so T plus A equals 25. So we have 3 equations and 3 unknowns, so we can solve. So first let's add up all of our equations together to get 2T plus 2J plus 2A is equal to 72. Now let's divide everything by 2 to get T plus J plus A equals 36. So now if we want to solve for example for Alex's age, we can subtract the equation T plus J equals 23 from the first one to get that A, Alex's age, is 13. And we get that J plus 13 equals 24, so J is equal to 11. Now we can plug A into this equation to get T plus 13 equals 25, or that T equals 12. So the oldest of these 3 ages is Alex, with an age of 13. So the question asks us, what is the age of the oldest person? The answer is 13, letter D. Problem number two states, the sum of one-tenth plus one-hundredth plus one-thousandth is. So in order to add up all of these fractions, we need to have a common denominator. So for 1 over 10, let's multiply by 100 over 100. And for 1 over 100, let's multiply by 10 over 10. So it'll give us a denominator of 1000 in all three terms. So for 1 over 10, that's transformed into 100 over 1000. For 1 over 100, we get 10 over 1000. And then 1 over 1000 already had the denominator of 1000, so we didn't have to do anything with it. So now we can simply add up all the numerators, so 100 plus 10 plus 1 is 111. So the final sum is 111 over 1000. So the question asked us, the sum of 1 over 10 plus 1 over 100 plus 1 over 1000 is? 111 over 1000. Letter C. Problem number 3 states, Maria wants to build a bridge across a river and knows that the shortest possible bridge from each point on one shore is always of the same length. Which of these pictures cannot be a picture of her river? So first let's take a look at the first river. We can see that it is just two parallel lines. The distance between two parallel lines is always going to be the same. So this one is a possibility for her river. Next let's take a look at the half circle one. In this case the distance between the edge of a circle and the center of the circle is always going to be the same because the radius is always consistent for a circle. So this one also works. Next if we take a look at the saw one, we can see that for example this length is going to be the same as this length. But it's not going to be the same as this length because this length is going to be the hypotenuse of the triangle which has to be longer than the side length. The side lengths in this case are the two yellow lines. So this one is actually not possible. But let's look at the last two. The last two are kind of the same. They're constructed from concentric circles and we know the distance between two concentric circles is always going to be consistent assuming that their centers are in the same spot. So this one works and then this next one also works for the same reason. So the only one that doesn't work is the saw jagged one. So the question asked us, which of these pictures cannot be a picture of her river? And it's this one. Letter B. Problem number four states, how many integers are greater than 2015 times 2017, but less than 2016 times 2016? So in order to solve this problem, we could literally just multiply 2016 by 2016, and 2015 by 2017, and then see how many numbers there are between them. But multiplying four digit numbers takes a long time and isn't very fun, so instead let's tackle it in a different way. First let's say that x equals 2016. This means that x minus 1 is going to be equal to 2015, and x plus 1 is going to be 2017. Now we know that we have x squared, which is 2016 squared, and we know that we have x minus 1, 2015, times x plus 1, 2017, and we don't want to know how many numbers there are between these. So on the right side let's FOIL it out, and we get x squared minus 1, which means that the difference between 2015 times 2017 and 2016 times 2016 is 1. So now how many numbers are between these two numbers? Well the answer is zero, since they're consecutive numbers. And if you actually want to do all the multiplication, this is how it ends up. You get 4,064,256 and 255, which are two consecutive numbers, and there are zero numbers between them. So the question asked us, how many integers are greater than 2015 times 2017, but less than 2016 times 2016? The answer is zero. Letter A. Problem number 5 states, a set of points forms a picture of a kangaroo in the xy plane as shown. For each point, the x and y coordinates are swapped. What is the result? So first let's take a look at a very basic example of this swapping behavior. Let's take this red point on the x-axis. If we swap its x and y coordinate, well its y coordinate is 0 right now, so its x coordinate will be 0 after the swap. And its x coordinate is right now positive, so its y coordinate will also be positive. So it will end up right here. If we take a number on the y-axis, and then swap the y and the x values, we'll end up with a value on the x-axis. If we take a number in between, it will end up getting reflected over the line y equals x. This is actually what it means. When you swap the y and the x values, you're effectively taking the inverse of a function, or mirroring it over y equals x. Now our kangaroo is not a function, but it's a set of points, so we can apply the same principle. So we're going to be mirroring it over this line. So let's take some key points of the kangaroo. For example, the end of its foot. We know the end of its foot is going to end up in the top left corner. So it cannot be this one, since its snout is in the top left corner in this one. Now let's take a look at the snout. The snout really doesn't change much, and still ends up in the top right corner. Three of our remaining four answers do not have the snout in the top right corner, so they cannot be the answer. Therefore, this is the kangaroo with points representing the x and the y coordinates swapped. So the question asked us, what is the result? The answer is this kangaroo, letter A. Problem number six states, what is the smallest number of planes that are needed to enclose a bounded part in three-dimensional space? So in order to solve this problem, let's figure out first what happens in one dimension, and then in two, and then we can figure out what happens in three dimensions. So first let's start out with one dimension. One dimension is just a number line. If we have one point on the number line, that effectively splits our number line in half, but both of those are still unbounded since they go to infinity. But as soon as we put down another point that isn't at the same spot as the first point, we can see that we have a region that is bounded. This line in one dimension is bounded by two points. Next let's take a look at two dimensions. For two dimensions we can draw a line, and this will split our space up into two regions. But both of these regions are still infinite, so let's draw another line. Assuming this line is not parallel, it will break up our space into four regions, but each of these regions is also infinite. But if we make a third line, we can see that we can make a triangle. So we have a bounded region in two dimensions using three lines. So now if we analyze our data, in one dimension it took us two points, in two dimensions it took us three lines. So in three dimensions it's going to take us four planes. We can further verify this by thinking of the simplest shape in three dimensions. The simplest shape in three dimensions is a tetrahedron, and it consists of four faces. It is simply four triangles glued together. So the question asks us, what is the smallest number of planes that are needed to enclose a bounded part in three-dimensional space? The answer is 4, letter B. Problem number seven states, Diana wants to write nine integers in the circles on the diagram so that for the eight small triangles whose vertices are joined by segments, the sums of the numbers in the vertices are identical. What is the largest number of different integers she can use? So first let's label each of the values inside each of the circles. We'll label them a through i. So now if we take a look at this triangle for example we know that its sum is going to be a plus b plus e and for this triangle it's going to be a plus d plus e and we can do that for all of the triangles and we get eight equations since there are eight triangles. So now let's consider these two triangles a plus d plus e and g plus d plus e. Of course we know that their sums are equal to one another so we can set these two values equal to each other. We have a common d and a common e in each of them so we can subtract out d and e and get that a equals g. So now we know what g is. g is exactly the same as a and as a matter of fact this is mirrored. This happens in four different spots so all of the corners of this shape have a value of a. So now these are how our equations look and as you can see they're mostly the same except for this middle term and if all of them are equal to one another that means that this middle term must also be equal to each other. So b must be equal to d which must be equal to h which must be equal to f. So we'll say that each of these values is b. So this means that she can choose a b and e and she can choose any integer for these and it would work but she has three integers to choose. So the question asked us what is the largest number of different integers she can use? The answer is three. Letter c. Problem number 8 states the rectangles S1 and S2 in the picture have the same area. Determine the ratio X over Y. So first let's figure out the areas of each of these rectangles. So we know that the area of a rectangle is its base times its height. So in the case of S1 the red region would be its base and the yellow region would be its height. So in this case the base has a value of X and the height has a value of Y less than 5 since the height starts at Y. For S of 2 the base is in red and the height is in yellow. So the base is X less than 8 and the height is Y. So now we have our two expressions for area and we know that their areas are equal to one another so we can set them equal to each other. Now let's distribute the X and the Y from both sides to get 5 X minus X Y equals 8 Y minus X Y. Let's add X Y to both sides and we get 5 X equals 8 Y and the problem asks us for the ratio X over Y so let's divide both sides by Y and now let's divide both sides by 5 to get the ratio X over Y is equal to 8 over 5. So the question asked us determine the ratio X over Y. The answer is 8 over 5. Letter E. Problem number 9 states, if x squared minus 4x plus 2 is equal to 0, then x plus 2 over x equals... So when you first think about solving this problem, you might be tempted to solve for x. However, x is not that easy to solve for in this case. It's not an integer. So instead, let's look at some algebraic manipulation that we can do. So we know that the value we're looking for is x plus 2 over x. So first, let's add 4x to both sides on the left side to get x squared plus 2 is equal to 4x. Now, if we divide both sides by x, we get x plus 2 over x is equal to 4. Well, there we go. We have the value of x plus 2 over x. It's 4. So the question asked us, then x plus 2 over x equals... 4. Letter E. Problem number 10 states, the lengths of arc AP and arc BP in the figure are 20 and 16 respectively. Then the measure of the angle AXP equals, so first let's give the angle AXP a name. We'll call it theta, and since we're dealing with arc length, it's going to be much easier if we say that theta is in radians. Just how there are 360 degrees in a circle, there are also 2 pi radians in a circle, so it's just a different way of measuring the angle. What's so special about radians is if you multiply the arc angle in radians by the radius of the circle that is swept, you will get the arc lengths, and we'll see that used later. So first let's define this angle BOP as alpha. Then let's draw a line going from point O to point C, and we'll say that OC is parallel to PX. Since we know that angle COP is equal to 90 degrees, or pi over 2 radians, and angle COB is equal to theta, because they're alternating interior angles, we know that angle alpha is equal to pi over 2 minus theta, or theta less than pi over 2. Next let's define this angle beta, and let's extend our line segment to go through point D, and we'll say that DC is parallel to PX. So again we know that angle POD is going to be pi over 2, or 90 degrees, and then we know that angle DOA is going to be equal to theta because of corresponding angles. So we know ultimately that beta is going to be equal to pi over 2 plus theta. Let's define the radius of the circle as R, and now let's take a look at the arc lengths. This orange arc length from P to B we know is equal to 16, but in terms of the angle it's going to be equal to alpha times R. This is that magic of radians that I was talking about earlier. And this arc length right here in magenta is going to be 20, and that's going to be equal to R times beta. So to recap, the arc length swept out by some angle in radians is going to be equal to that value times the radius of the circle. So now let's try and solve for theta. First let's substitute in our alpha and beta in terms of theta into our arc length equations. To get 20 is equal to R quantity pi over 2 minus theta, and 16 is equal to R quantity pi over 2 plus theta. Now let's add these two equations together to get 36 is equal to R times pi, or that the radius of the circle is equal to 36 divided by pi. Now let's subtract the two equations. We get 20 minus 16 is equal to R pi over 2 minus R theta minus R pi over 2 minus R theta. So the positive R pi over 2 and the negative R pi over 2 cancel out, and we get 4 is equal to 2 R theta. We know the value of R, it's 36 over pi, so let's replace that. So we get 4 is equal to 2 times 36 over pi theta. Let's divide both sides by 2 and by 36 over pi, and now we get that theta is equal to 2 pi over 36, or pi over 18. So now let's just convert this into degrees. In order to do this, let's multiply that value by 360 over 2 pi, since there are 360 degrees in 2 pi radians. Doing this, the 360 divided by 2 gets us 180, and the pi's cancel out, so we get 1 over 18 times 180 equals theta. 1 over 18 times 180 is 10 degrees, so theta equals 10 degrees. So the question asked us, then the measure of the angle AXP equals 10 degrees. Letter E. Problem number 11 states, a, b, c, d are positive integers satisfying a plus 2 equals b minus 2 equals c times 2 equals d divided by 2, which is the largest of the four numbers a, b, c, and d. So here we have our four equations. And first let's consider the two multiplication ones, or I guess the one multiplication one and the division one. So c times 2 equals x and d divided by 2 equals x. They all equal x because they all equal each other, and this way we can split them up easier. So since these two equations both equal x, we can set them equal to one another. So c times 2 equals d divided by 2. We can multiply both sides by 2 to see that 4c is equal to d. Now since we know that c has to be a positive integer, we know that it has to be greater than or equal to 1. This means that 4c must be greater than c, since 4 times any positive integer is going to be greater than that positive integer. For example, if you pick 10, 10 is going to be less than 40, or 40 is going to be greater than 10. So this tells us that c is less than d. So now we've shown that d is bigger than c, so let's continue trying to prove that. So let's consider this equation, a plus 2 equals x, and let's set it equal to c times 2 equals x, to get c times 2 equals a plus 2. Subtracting 2 from both sides, we get 2c minus 2 equals a. Well we know that 2c minus 2 is going to be less than 2c, and we already know that c is going to be greater than or equal to 1, so we know that 2c is going to be less than 4c, and we know that 4c is d. So therefore we know that a is less than d. Now let's consider our last equation, and let's hopefully try and prove that b is less than d, or maybe that d is less than b. So b minus 2 equals x. Let's set that equal to c times 2 equals x. Adding 2 to both sides, we get 2c plus 2 equals b. So now we have a problem because we have some addition. Are we sure that 2c plus 2 is less than 4c? Well if c is 1, then it would be equal to each other. But we can actually figure this out by taking a look at a restriction on c. If c is equal to 1, then a, which is equal to 2c minus 2, would be equal to 2 times 1 minus 2, which means that a would be 0. And that's impossible since the problem told us that a, b, c, and d were all positive integers. So therefore c cannot be 1, and c has to be greater than 1. So what does this tell us? Well this tells us now that the minimum value for c would be 2. So if we plug in 2, we would get 4 plus 2 is less than 8, which is 6 is less than 8, which is true. And as we keep on adding more numbers, the 4 times c term is going to grow faster than the 2c plus 2 term, so that's going to be true for all values of c. So therefore b is less than d. So now we've proven that a is less than d, b is less than d, and c is less than d, which means that d must be the greatest number. So the question asked us, which is the largest of the four numbers a, b, c, and d? The answer is d, letter d. Problem number 12 states, in the pyramid of numbers shown to the right, each upper field is the product of the two fields directly underneath. Which of the following numbers cannot appear in the top field if the three bottom fields only contain natural numbers bigger than one? So first let's populate our bottom row with natural numbers. We'll just call them a, b, and c. So now we know that the block above two other blocks is going to be the product of those bottom two blocks. So for the left we're going to have a times b, and for the right we're going to have b times c. And then the one above that's going to be the product of those two, which is going to be a times b squared times c. And these are our five possible answers. Let's break each of these up into their prime factorization and see if we can put their prime factors into a, b, and c. So for example let's start out with 56. 56's prime factorization is 2, 2, 2, and 7. So what we could say is that a equals 2, b also equals 2, and c equals 7. Nowhere in the problem did it say that the natural numbers couldn't be the same, so we can say that a is equal to b, which is equal to 2. And we can see that if we say 2 times 2 squared times 7, that's going to be equal to 56. So 56 works. Breaking up 84 into its prime factorization, we get 2, 2, 3, and 7. Here it's a little bit easier. We can see that a can be 3, b is 2, and then c is 7. We can obviously swap a and c, but that doesn't matter. And if we multiply them all together, you can see that we get 84. So 84 works. Now for 90, 90 breaks up into 2, 3, 3, and 5. So again it's kind of easy. You can see that a is 2, b is 3, and c is 5. So we get 2 times 3 squared times 5, which is equal to 90. So now let's go into 105, and 105's prime factorization is 3, 5, and 7. But we don't have any numbers that repeat, so there is no number that we can put for b. So 105 is impossible. Next for 220, 220's prime factorization would be 2, 2, 5, and 11. So a can be 5, b will be 2, and c will be 11, which will get us 5 times 2 squared times 11, which will get us 220. So that one also works. So the only one that we cannot get is 105. So the question asked us, which of the following numbers cannot appear in the top field if the three bottom fields only contain natural numbers bigger than 1? The answer is 105, letter d. Problem number 13 states, what is x sub 4 if x sub 1 is equal to 2? And x sub n plus 1 is equal to x sub n raised to the x sub n power, for n is greater than or equal to 1. So first, we know the first term in the sequence, we know that x sub 1 is equal to 2, and then we know how to find every other term in the sequence, or every next one. We know that the next term is going to be the previous term raised to the previous term's power. So for example, x sub 2 is going to be x sub 1 raised to the x sub 1 power, or 2 to the second power. For x sub 3, it's going to be x sub 2 to the x sub 2 power, and we know x sub 2, it's 2 squared, so we can plug that in. In the outer power, we have 2 squared, which is 4, and then we're going to have a power to a power, so we have to multiply those two together. So we would have 2 to the power of 2 times 4, which would be 2 to the 8th power. So x sub 3 is 2 to the 8th. Now it's time for x sub 4. x sub 4 is going to be x sub 3 to the x sub 3 power, so it would be 2 to the 8th to the power of 2 to the 8th. Same thing we did last time, 2 to the 8th is going to be 256, and then we have to multiply that by 8 since it's a power to a power. So 2 to the power of 8 times 256 is 2 to the power of 2048. So we found x sub 4, but sadly it doesn't match up with any of our answers, because our answers are in a different form. They're in the form 2 to the power of 2 to the power of something. So let's try and figure out what that something is. In order to do this, let's figure out 2048 is equal to 2 to the what power. You could do a log base 2 by hand, but that's very complicated, so I think it's better just to guess and check in this case. We could have 3, 4, 11, 16, or 768, and 11 is going to work. 2 raised to the 11th power is 2048. So we know it's this one. So the question asked us, what is x sub 4 of x sub 1 is equal to 2, and x sub n plus 1 is equal to x sub n to the x sub n power for n is greater than or equal to 1? The answer is 2 to the power of 2 to the power of 11. Letter c. Problem number 14 states, in rectangle ABCD the length of the side BC is half the length of the diagonal AC. Let M be a point on CD such that length AM is equal to the length of MC. What is the size of angle CAM? And so now for this question they don't give us a diagram so we got to draw it by ourselves. So first let's draw a rectangle ABCD and then we'll call the length of the edge BC X and we know the length of the diagonal is twice that or actually we know that the length of BC is half the length of the diagonal but that's the same thing so we'll say that the diagonal is 2x. Next we know there is some point M on DC such that the length of AD is equal to the length of MC. So we also know that triangle AMC is isosceles since two of its sides AM and MC are equal to one another. The problem told us that. So therefore we know that angle CAM and angle ACM are equal to one another. That's a property of isosceles triangles. So if we solve for angle ACM we'll figure out what angle CAM is and I think it's easier here to solve for angle ACM since it's right angle is only broken up into two parts while the other one is broken up into three parts. So let's try solving for it. We'll call this angle ACB gamma. Now we know that the cosine of gamma is going to be equal to the adjacent side over the hypotenuse side and we know both of those the adjacent side is going to be X and the hypotenuse is going to be 2x. The X's cancel out and we get cosine of gamma is equal to 1 half. If you remember the unit circle cosine is equal to 1 half when gamma is equal to 60 degrees. So now we know that angle 60 degrees. Let's say that this angle right here is going to be alpha and we of course know that our rectangle is made of right angles so we know that BCD is going to be a right angle. So alpha plus gamma is going to be equal to 90 degrees. We know gamma so it's really easy to solve for alpha. Simply substitute in 60 for gamma subtract 60 from both sides and we get that alpha is equal to 30 degrees. Since we know that angle ACM is equal to angle CAM we also know that angle CAM is equal to 30 degrees and we were looking for angle CAM. So the question asked us what is the size of angle CAM? The answer is some other angle. Letter E. Problem number 15 states, Diana cut up a rectangle of area 2016 into 56 equal squares. The lengths of the sides of the rectangle and of the squares are integers. What is the number of different rectangles for which Diana can do this? So here we have the rectangle that she cuts up into 56 squares. We'll say that this rectangle consists of n high squares and m wide squares. So we know that n times n is going to be 56, since she cuts it up into 56 equal squares. So now the question is, how many ways can we make up 56 from a product? For example, we could say that n is equal to 1 and m is equal to 56, which would be a very long rectangle. We could say that n equals 2 and m equals 28, n equals 4, m equals 14, or n equals 7 and m equals 8. Now we could also flip the n's and the m's around and get four more possible values. However, these rectangles are just going to be rotated versions of the other one. And a rotated version of one of them is not a different rectangle, so we don't have to account for these. So there are only four ways that we can make this rectangle. So the question asked us, what is the number of different rectangles for which Diana can do this? The answer is 4, letter B. Problem number 16 states, on the island of knights and knaves, every citizen is either a knight who always speaks the truth or a knave who always lies. During your travels on the island, you meet seven people sitting around a bonfire. They all tell you, I'm sitting between two knaves. How many knaves are there? So we have seven people sitting around the bonfire. And first, let's figure out what possible combinations we can have. If the person says I'm sitting between two knaves and is a knight, then they're telling the truth and they are actually sitting between two knaves. Here we'll represent knights in green and knaves in red. However, what could also be happening is it could be a knave, in which case they could be sitting between two knights. Or perhaps they're sitting between one knight and one knave. This would also be lying since they're saying they're sitting between two knaves. So these are our three possible combinations. A knave surrounded by two knights, a knight surrounded by two knaves, or a knave surrounded by a knight and another knave. So first, let's start out at one of the extremes. Let's assume that everybody's a knight. And we can see that none of these individual people satisfy any of the conditions, so all of them are problematic. However, if you switch one of them to a knave, in this case we'll just say this one, then this one no longer becomes problematic because it's possible to have a knight, then a knave, and then another knight. However, it's not possible to have a knight, a knight, and a knave. Since the two people sitting next to that knave cannot be knaves, then let's move on to the next people. So these two. Now five out of the seven are satisfied, but still two are not. In order to resolve this, let's set one of those knights to a knave. And now we can see this final combination with four knaves and three knights. We can see here that everybody is satisfied since they all follow the patterns that we discovered on the left. So we're good. And it doesn't matter how you started, you'll end up with four knaves and three knights anyway. You might just get this structure rotated a little bit. So the question asked us, how many knaves are there? The answer is four. Letter. B. Problem number 17 states the equations x squared plus ax plus b equals zero and x squared plus bx plus a equals zero both have real roots it is known that the sum of squares of the roots of the first equation is equal to the sum of squares of the roots of the second one and a is not equal to b then a plus b equals so before we solve this problem you need to be familiar with vieta's formula and vieta's formula essentially states if we have a polynomial with coefficients an an minus one all the way through a1 and then the constant one a sub zero and it has roots which we know it has n roots we'll call them r1 r2 all the way through rn minus 1 rn then the sum of all of these roots is going to be equal to minus an minus one over an so the xn minus one coefficient and the xn coefficient and then that the product of all the roots is going to be equal to minus one to the n power times a sub zero the constant term divided by the first coefficient of x of n now this might not be very clear so let's look through a specific example for example in the case of x squared plus 5x plus 6 its roots are minus 2 and minus 3 so the first thing vieta's formula tells us is that the sum of the roots minus 2 plus minus 3 is going to be equal to the second highest coefficient over the highest coefficient and in this case not highest in value but highest in associated power so we'll have the linear coefficient over the quadratic coefficient the linear coefficient here is 5 and the quadratic coefficient is 1 so we will have minus 5 over 1 so now if we multiply the two roots together negative 2 times negative 3 what we'll get is we'll get the quotient of the lowest coefficient divided by the highest coefficient again lowest and highest here doesn't mean value but rather with their associated power so 6 is the constant coefficient and 1 is the quadratic coefficient so negative 2 times negative 3 is positive 6 which is equal to 6 over 1 so that checks out so you need to know this to solve the problem so now on to solving the problem first let's review what it told us it told us that x squared plus ax plus b equals zero and x squared plus bx plus a equals zero and both of these equations have two roots we'll call the roots of the first one alpha and beta and the roots of the second one gamma and delta and then it told us that the sum of the squares of the roots in each equation are equal to one another so alpha squared plus beta squared is equal to gamma squared plus delta squared so now using vieta's method we know that alpha plus beta is equal to minus a over one and using the other form we know that alpha times beta is equal to b over one we're simply dividing by one so we can drop those and on the other side we have a similar thing where we know that gamma plus delta is equal to minus b over one and gamma times delta is going to be equal to a over one we have divided by one here so we can just cancel those out and so now we have five formulas relating alpha beta gamma delta a b so first let's take this equation alpha plus beta is equal to minus a and square it so squaring both sides we get alpha squared plus two alpha beta plus beta squared is equal to a squared now ideal what we would want here is we would want alpha squared plus beta squared so we can then later substitute it into our top equation in order to isolate alpha squared plus beta squared let's subtract two times this equation from the original equation so subtracting that we would get minus two alpha beta and on the other side minus two b and the positive two alpha beta and the negative two alpha beta cancel out and we get alpha squared plus beta squared is equal to a squared minus two b let's do that for the other equation again exactly the same as before in order to cancel out the two gamma delta we'll subtract gamma delta equals a twice and then two gamma delta cancels out with minus two gamma delta and we get gamma squared plus delta squared is equal to b squared minus 2a so now we can see that we can actually substitute this into our top equation rather than substituting the left sides though let's substitute the right sides and this is what we get a squared minus 2b is equal to b squared minus 2a so let's subtract b squared from both sides and add 2b and now on the left side you can see we have a difference of squares that's going to be equal to a plus b times a minus b so now let's try and get an a minus b term on the right side so we can divide it out so we can factor it out like this to get minus two quantity a minus b quantity a minus b and now can we divide by a minus b well yes because the problem told us that a is not equal to b so we won't be dividing by zero so that's okay so we end up with a plus b is equal to minus two so the question asked us then a plus b is equal to minus two letter b Problem number 18 states, if the perimeter of the square in the figure to the right equals 4, then the perimeter of the equilateral triangle equals... So first we know the perimeter of the square is 4, so we know that each of its side lengths is going to be 1. So the square consists of four equidistant sides. Next let's call the side length of a triangle T, and we can see that this little length right here is going to be 1 less than T, so T minus 1. Now upon closer inspection, we know this is an equilateral triangle, so we know that this angle is 60 degrees. We also know that this is a right angle here, since we know that the square shares an edge with the triangle, and if we have 90 degrees and 60 degrees, that means that the last angle is 30 degrees. So we have a 30-60-90 triangle. Well what does that mean? Well that means that we know all the side lengths relative to another side length. So if we know that the bottom side length is T minus 1, we know the side length on the right is going to be the square root of 3 times T minus 1, and we also know that the hypotenuse is going to be 2 times T minus 1, but in this case that's not important. Only this one's important, because we know that this length is also 1. So we can set these two values equal to one another, and now we can solve for T, the side length of the triangle. So let's divide by square root 3, and now let's add 1, and now we need to find a common denominator, so let's multiply by root 3 over root 3, and now we can add them up to get 1 plus root 3 over root 3 equals T. In order to clean it up and avoid the roots in the denominator, let's multiply everything by root 3 to get root 3 plus 3 over 3 equals T. Now if you remember, T is only the side length of the triangle, and a triangle has three side lengths, and we know it's an equilateral triangle, so we know all of its side lengths are the same. So in this case, the perimeter is going to be equal to 3 times T, so if we multiply T by 3, we get 3 times square root 3 plus 3 over 3. The 3 in the numerator and the denominator cancel out, and we get the square root of 3 plus 3. So the question asked us, then the perimeter of the equilateral triangle equals 3 plus square root 3, letter B. Problem number 19 states, each of 10 points in the figure on the left is marked with 0, 1, or 2. We know the sum of the numbers in the vertices of any white triangle is divisible by 3, while the sum of the numbers in the vertices of any black triangle is not divisible by 3. Three of the points are marked as shown in the figure. What number can be used to mark the central point? So we know that if we add up all of the vertices of one of the white triangles, it should be divisible by 3. So for example, for this one highlighted in green, we know that it has to be divisible by 3, and the only way we can do that is if the last one is 1, because 0 plus 1 plus 2 is 3, so therefore it's divisible by 3. And for the black one, we want to do the opposite thing. We want to make sure that it's not divisible by 3. So 0 plus 1 plus the mystery number cannot be divisible by 3. However, if the mystery number is 2, therefore it would be divisible by 3. So we know that the mystery number cannot be 2. So we know that the mystery number is either 0 or 1. So let's take a look at both possibilities. First, let's take a look at if it's 1. If we take a look at these two, we can figure out the two vertices at the top. Since we know that 0 plus 1 plus something has to be divisible by 3, we know that that one is 2. We know that 1 plus 2 plus something has to be divisible by 3, so we know that that is 0. And we have a problem. Since we know that this triangle right here cannot be divisible by 3, but it is, so therefore the center point cannot be 1. That means that the center point must be 0. And we can confirm this by solving for the rest of the triangle. We can solve for the top one. So on the left one we're going to have 0 plus 0 plus 0, which would be divisible by 3. And for the right one, 2 plus 1 is going to be divisible by 3. So in the top triangle there would be a 2. And now considering the one at the bottom, we would have 1 plus 0 plus 2 to be divisible by 3. And in the bottom right one we would need to have 2 plus 2 plus 2 to be divisible by 3. We'd get 6. So with 0 it works. So the mystery number is no longer a mystery number. We know it's 0. So the question asked us, What numbers can be used to mark the central point? The answer is only 0. Letter A. Problem number 20 states, Patina draws five points A, B, C, D, and E on a circle as well as a tangent to the circle at A, such that all five angles marked with X are equal. Note that the drawing is not to scale. What is the measure of the angle A, B, D? So first, let's consider this line. We know that the sum of all of the angles that it consists of has to be equal to 180 degrees, and the only angles it consists of are X. As a matter of fact, five copies of X. So 180 is equal to 5X. We can divide that by 5 to solve for X, and we get that X is equal to 36 degrees. So now we solved for X. Next, let's draw a line between A and F. We'll make point F right here. And we'll make it so that AF bisects angle CAD. So what does that mean? Well, that means that line segment AC and AD are going to be equal to one another. So therefore, triangle ACD will be isosceles, which means that the two angles that it consists of that aren't X are going to be equal to one another. We'll call that Y. So we know that 180 is equal to X plus Y plus Y, and we know X is 36, so we can substitute that in. Subtract 36 from both sides to get 144 degrees is equal to 2Y, and now divide by 2 to get Y is equal to 72 degrees. So next, let's consider these two angles, angle ABD and ACD. We know that both of those angles are subtended by arc AED. So what does this imply? Well, this implies that the two angles are actually equal to one another. So now we've solved the mystery of what angle ABD is. It's the same as ACD, which we solved to be 72 degrees. So the question asked us, what is the measure of the angle ABD? The answer is 72 degrees. Letter C. Problem number 21 states, how many different real solutions are there to the x squared minus 4x plus 5 all raised to the power of x squared plus x minus 30 equals 1 equation. So here we have the equation that we need to find all the x values so that when we evaluate it we get 1. And we know that this is some number raised to the power of another number. And the question is, when can this number be equal to 1? Well there are actually two possibilities. The first one is if a is equal to 1 because 1 raised to any power will yield a 1. And another one is if b equals 0 because any number raised to the zero power will be equal to 1. So we need to find all of the values of x such that the base term x squared minus 4x plus 5 equals 1 and that the power expression x squared plus x minus 30 equals 0. So first let's start out with the base and let's try and make it equal to 1. So in this case we would have x squared minus 4x plus 5 equals 1. We can break this up into x minus 2 squared plus 1 equals 1. And of course if we want this to be equal to 1 we simply have to make the x minus 2 quantity squared term disappear. So with x equals 2 we can do this because then we would have 2 minus 2 which would be 0 squared so we would get 1. So one possible solution is if x equals 2. So one possible solution is if x equals 2. And this is the only solution to x squared minus 4x plus 5 equals 1 since we see we have a double multiplicity. So we have two roots of x equals 2 but only one answer. So now let's take a look at the power x squared plus x minus 30 and we want this to be equal to 0 because again any number raised to the zero power is going to be equal to 1. And we can simply factor this out. I mean it's not too difficult. We would get x minus 5 and x plus 6. So in order for this to be equal to 0 x could equal 5 or x could equal minus 6. So in total we have three options x could be equal to 5 or minus 6 to make the power of 0 or x could be equal to 2 to make the base 1. So we have three possible solutions. So the question asked us how many different real solutions are there to the equation? The answer is 3. Letter C. Problem number 22 states a quadrilateral contains an inscribed circle, i.e. a circle tangent to the four sides of the quadrilateral. The ratio of the perimeter of the quadrilateral to that of the circle is 4 to 3. Then the ratio of the area of the quadrilateral to that of the circle is... so we know what we're looking for. We're looking for the ratio of the area of the quadrilateral to the area of the circle. And we're given the perimeter of the quadrilateral to the circumference of the circle, and we know that's 4 to 3. So now let's write down some equations for the perimeter of the quadrilateral and for the circumference of the circle. For the perimeter of the quadrilateral we'll just call that P, and for the circumference of the circle we know that's going to be 2 pi r. And now let's do the same thing for the area equations. So for the circle it's going to be pi r squared, and then for the area of the quadrilateral we're going to have half of the perimeter of the quadrilateral times the radius of the inscribed circle. And what you actually might notice here is if we simplify the equation at the top we get P over 2 pi r, which is exactly the same equation that we have at the bottom, P over 2 pi r. That means the ratio of the area of the quadrilateral to the area of the circle is exactly the same as the ratio of the perimeter of the quadrilateral to the circumference of the circle. And we know that this is 4 over 3. So the question asked us, then the ratio of the area of the quadrilateral to that of the circle is 4 to 3. Letter E. Problem number 23 states how many quadratic functions in X have a graph passing through at least three of the marked points. So first let's discuss the fact that if we have three points that aren't collinear then they can define a unique parabola. And some of you may already know this but for those of you who don't we can quickly prove this. For example let's say that we have three points p1 q1 p2 q2 p3 and q3 and we'll assume that p1 p2 and p3 are not the same because then that would fail the vertical line test and we couldn't have a function anyway. So essentially just three points scattered along the x-axis. And then we have our quadratic in its typical form ax squared plus bx plus c. And now we can make three equations plugging in the p-coordinate and expecting out the q-coordinate which is essentially what a function is. So for example for p1 q1 we would have a p1 squared plus a b p1 plus c equals q1. So effectively we're replacing y with the q's and x with the p's. So now we have three equations here and actually if we take the first one and subtract it from the second one and if we take the third one and subtract it from the first one and then simplify it what we end up having is three equations with three unknown variables a b and c. And you might be like well p1 p2 p3 q1 q2 and q3 are also variables but those are points and we're assuming that you would obviously know those. So three equations three unknowns mean that it's trivial to solve and there is a parabola that goes through those three points. So now that we're done with that let's go back to the graph. Let's say that we pick three points that are on top of one another. This would fail the vertical line test therefore this function could not be a function. If we have a shape like this well then it also fails the vertical line test so it also couldn't be a function. The only way that we can get a function is if we get one point from each of the three columns. We'll name them column 1 column 2 and column 3. If we have one from each of them we can create a parabola out of those three. And let's say that first we're picking from the first column we would have three choices and from those three choices we can move on to three more within the second column and then from those we can pick three more in the third column. So in the end there are 27 ways that we can pick three points one in each column. However some of these ways cannot make a parabola. For example these three. These three would be horizontal lines so therefore the a component of the parabola would be zero and it wouldn't be a parabola at all it would be a line. So for example these three ways are invalid. Another two invalid ways could come if the lines are selected diagonally. This would also form a line and the a coefficient would be equal to zero which bumps up to five invalid ways. And those are all the invalid ways so we know we have 27 ways and five of them are invalid so therefore that means that we have 22 valid ways. So the question asked us how many quadratic functions in X have a graph passing through at least three of the marked points? The answer is 22. Letter D. Problem number 24 states, in a right triangle ABC, right angle at A, the bisectors of the acute angles intersect at point P. If the distance from P to the hypotenuse is a root 8, what is the distance from P to A? So in this problem, they don't give us a figure, so we need to draw it out. So here we have our right triangle, we know it is ABC and we know that the right angle is at A. Next thing the problem tells us is that we have two lines going from the acute angles, which are bisectors. So therefore, these angles all would be congruent. And then we know where they meet. There is a point, which is P. And the distance from P to the hypotenuse is the square root of 8. Now there is something very special about this point P. And that is that point P is constructed from the bisectors of the acute angles of the triangle. And this means that point P is actually an incenter. And this is the point that is furthest away from any edge. But it's also equidistant from all the edges. Which means that we can draw a circle around this point, and this circle will touch each of the edges. And the problem even gives us the radius of the circle. That's going to be the length from point P to the hypotenuse, or square root 8. So with this knowledge, since all the radii of the circle are the same, we know that this length is square root 8. And we also know that this length is square root 8. So that means that we can solve for this length right here. Which is what we're looking for, the distance between point P and point A. So using Pythagoras' theorem, A squared plus B squared equals C squared, where C is our hypotenuse and A and B are the side lengths, we can plug in square root 8 for both A and B. To get 8 plus 8 equals C squared, or 16 equals C squared. And we get that C equals 4. The distance from point P to point A is 4. So the question asked us, what is the distance from P to A? The answer is 4. Letter E. Problem number 25 states, three three-digit numbers are formed using the digits from 1 to 9. Each digit is used exactly once. Which of the following numbers cannot be equal to the sum of these three numbers? So we know we have the numbers 1 through 9 and we know they have to take the positions of digits in three three-digit numbers. So represent the three-digit numbers as ABCD, DEF, and GHI. So the values of each of these numbers for ABC would be a hundred A plus ten B plus C. Since A is in the hundred spot B is in the ten spot and C is in the one spot. For DEF it would be a hundred D plus ten E plus F for the same reasons and for the last one would be a hundred G plus ten H plus I again for the same reasons. So now the problem asks about the sum of all these so let's add all these up and we get something like this. So now let's combine all the like terms and factor out a hundred, a ten, and then a one. So we'll have a hundred times A plus D plus G plus ten times B plus E plus H plus one times C plus F plus I. So now you might think that we're stuck here but really we're not. We can take out one term from the hundred quantity A plus D plus G and one term from the ten quantity B plus E plus H and then at the very end what we end up having is 99 times A plus D plus G plus 9 times B plus E plus H plus and then all of the terms. So A, B, C, D, E, F, G, H, and I. And we actually know the value of A plus B plus C plus D plus E plus F plus G plus H plus I. Since we know their values we don't know which one's which but we know that one of them has to be one, one of them has to be two, and so on. So therefore that sum is equal to 1 plus 2 plus 3 plus 4 plus 5 plus 6 plus 7 plus 8 plus 9 or 45. So we can actually substitute that in. And now this is what we get. You can see that all of our terms are multiples of 9 so we can factor out the 9. So what does this tell us? Well we know all the numbers A, D, G, B, E, and H are integers and we know that multiplying and adding integers doesn't change their identity as integers and 9 times any integer is going to be a multiple of 9. So we know that our number is a multiple of 9. So now let's take a look at our answers and see if any of them aren't divisible by 9. So here are our answers and let's divide them by 9 and 1500 is not divisible by 9. So 1500 cannot be a possible sum. So the question asks us which of the following numbers cannot be equal to the sum of these three numbers? The answer is 1500. Letter A. Problem number 26 states, a cube is dissected into six pyramids by connecting a given point in the interior of the cube with each vertex of the cube. The volumes of the five of these pyramids are 2, 5, 10, 11, and 14. What is the volume of the sixth pyramid? So here we have our six pyramids, and then we know that they all have volumes 2, 5, 10, 14, and 11, and then there's a sixth one that we don't know the volume of. That's what we're trying to find out. So now we know that the equation for the volume of a pyramid is one-third the base times the height. The base of the pyramid in this case is the same for all of them, it is a square. Since a cube is made up of six squares, the base is all going to be the same. The same square, we'll call the side length of this cube s, so the area of the square would be s squared. So the volume of this pyramid would be one-third s squared times h, where h is the height of the pyramid. And the only thing that's going to vary here is the height. The side length of the cube is going to stay constant. So we'll label our heights h1, h2, h3, h4, h5, and then the mystery one, h6. So now let's take a look at some relations. Here we have a side view of our cube, and then we have two of our pyramids cut in half. And we can see that for the top one we'll call the height of that pyramid hy, and for the bottom one we'll call it hx. And we can see that the sum of the two heights of the pyramids are going to be equal to s. On the other side we can see a very similar thing. We'll call the left pyramid's height ha, and the right pyramid's height hb. So ha plus hb is also going to be equal to s. So now we've tackled the heights of four of the pyramids, but then we still have two. But these pyramids would be coming in and out of the screen. They would look kind of something like this. And we'll say that their heights are hr and ht. And again, hr plus ht is equal to s. So now we know the volumes of the pyramids. Now obviously this isn't completely correct, but it's fine since we're only solving the answers relative to one another. So let's try and plug in these volumes, which now we know are proportional to their heights, and try and make three of the same side lengths. So we know that 2 has to be one of the heights, so let's try and find a corresponding one such that the remaining ones also make the same sum. And we'll see that shortly. So first let's try 2 and 5. 2 and 5 would be equal to 7. And there's no other way of making 7 out of our remaining numbers, so 2 is probably not paired with 5. Let's try and pair it with 10. With 10 it would make 12. And again, there's no real way of making another 12. Well, we can say that the mystery number is 1, and then we would have 11 plus 1 would be 12, but then we'd have 5 plus 14, so that doesn't work. So let's try 2 paired with 11. 2 paired with 11 would give us 13. And we can make 13 with maybe 10 plus 3 or 5 plus 8, but then we can't make another 13, so this one's also not going to work. If we pair 2 with 14, we'll make 16, and we can see that we can actually pair up 5 and 11 to make another pair of 16. And then we can say that our mystery number is 6, and have 10 plus 6 equals 16. So our mystery number here would be 6. So the question asked us, what is the volume of the sixth pyramid? The answer is 6. Letter C. Problem number 27 states, a rectangular strip ABCD is 5 centimeters wide and 50 centimeters long and is made out of paper that is yellow on one side and green on the other. Folding the strip, Christina makes vertex B coincide with the midpoint M on the side CD. Folding again, she makes vertex D coincide with the midpoint N on side AB. What is the area in centimeters squared of the visible yellow part of the strip in the last picture? So first let's take a look at our original piece of paper, and here we can see that the region marked in yellow is the region exposed in the picture all the way on the right, and then the red region is folded up and we only see the other side of it, the green side. So this is still our original piece of paper though, so we can label it ABCD. Problem also tells us that it's 5 centimeters high and 50 centimeters wide, so we can mark those two lengths. Next let's mark all the points in the middle. We know that M and N are midpoints of A, B, and D, C respectively, and we'll just mark F and E. Now we know that the area of the parallelogram in the middle is going to be the base times its height, and we actually know its height, it was given to us, it's 5 centimeters, and then the base is going to be the length of line segment MF, which means in order to find this area we just need to find the length of line segment MF. And analyzing how we folded this, we can see that the line segment FB and FN, after they're folded, are actually going to be the same length, since that's how the folding is going to work. And same thing for DE and EM, they're also going to be the same length. So now we know that N is the midpoint between D and C, so we can say that NC is going to be 25 centimeters. And if we call the distance that we're looking for, which is MF, we'll call that X, we can say that FB is X less than 25, or 25 minus X. Since we know that FB and FN are the same, we can say that FN is also 25 minus X. And now if we look at this right triangle, we can set up a relation using Pythagoras' theorem and then solve for X. So we're going to have A squared plus B squared equals C squared. In this case, one of the side lengths is going to be X and the other one is going to be 5, so we're going to get X squared plus 5 squared, and it's going to be equal to the hypotenuse squared, which is 25 minus X, and then that squared. So if we FOIL out the right side, we get X squared minus 50 X plus 625. Subtracting X squared from both sides, we get 50 X is equal to 600, which means that X is equal to 12 centimeters. Of course, X represents the line segment MF, so MF is 12 centimeters. And now all we have to do is calculate the area, and we figured out that equation before. We know that the area of the yellow parallelogram is equal to the length of MF times 5. Again, base times height, the height is known, since we know the strip is 5 centimeters tall. So 5 times 12 is going to give us 60 centimeters squared. So the question asked us, what is the area in centimeters squared of the visible yellow part of the strip in the last picture? The answer is 60. Letter B Problem number 28 states, Anne chooses a positive integer n and wrote down the sum of all the positive integers from 1 to n. A prime number p divides the sum, but not any of the sumands. Which of the following could be n plus p? So we know that she picks a number n and then we know that she adds them all up. So we'll have 1 plus 2 plus 3 plus 4 plus all the way up to n. The sum of the elements from 1 to n is equivalent to n quantity n plus 1 divided by 2. And if you're ever not sure about an equation you can always just plug in a number like 7. You can see that 7 times 7 plus 1 which is 8 is going to be 56 divided by 2 will be 28. And if we add up the numbers 1 through 7 that does in fact equal 28. So there we go. Now we verify that our equation works. So this is going to be the sum of numbers from 1 through n. So now we know that this number is going to be divisible by p. This also tells us that n quantity n plus 1 is also going to be divisible by p. Now we know that the numbers 1, 2, 3, 4 all the way through n are not going to be divisible by p. And I mean the problem tells us this. So therefore we know that p is greater than n. Well how do we know this? Because if p is less than or equal to n then some number from 1 through n would have to be divisible by p. Because at some point that number would be equal to p. So therefore we know that p has to be greater than n. So pooling together all of our knowledge we know that n quantity n plus 1 is divisible by p, n is not divisible by p, and p is greater than n. So therefore p must be equal to n plus 1. So now let's try and convert this into the p plus n form which is a form in which the answers are given. So let's subtract 1 from both sides and then let's add p to both sides. And we get n plus p is equal to 2p minus 1. So now let's say that this number, this sum, is g. Therefore we know that g plus 1 divided by 2 must be prime. And all of the answers that we were given are g. So let's just add 1 and divide by 2 and see if they're going to be prime. So here are our values of g. Let's add 1 and divide by 2 for all of them. And the only one of them that is prime is 109. So that one must be our answer. g must be 217. So the question asked us, which of the following could be n plus p? The answer is 217, letter a. Problem number 29 states. Consider a five by five square divided into 25 cells. Initially, all its cells are white. In each move, we can change the color of any three consecutive cells in a row or in a column to the opposite color, i.e. white cells become black and black ones become white. What is the smallest possible number of moves needed to obtain this chess board coloring shown in the figure? So for this one, I think it's easier just to try and do this without doing any math, but we can do the math here as well. So first we have these six spots right here, and we know that all of them need to be black at the end, which means that we need to transform them at least once. None of these are in line or within three blocks from each other, so therefore they all need separate moves to turn these into black. So right now we're gonna do this So right now we're gonna need six moves to turn all of these to black. But now you can see these two red ones. These two red ones are adjacent to the edge and are surrounded by three other whites. Therefore, they need to be converted one more time for the adjacent ones to be turned to white. So therefore, we're gonna need an additional two moves to turn those squares to white. So in total, we're gonna need eight moves. So now let's show how this can be done. For the first move, let's change these ones. For the second one, let's move these three. Then for the third one, the top row, and then for the fourth one, the bottom row. Now let's turn those adjacent ones to white so we can do a move like this and another one like this. So now we're up to six moves. And now in two more moves, flipping these three, and then these three, we can achieve the checkerboard pattern in just eight moves. So the question asked us, what is the smallest possible number of moves needed to obtain the chessboard coloring shown in the figure? The answer is less than 10. Letter A. Problem number 30 states, the positive integer n has exactly 6 distinct positive divisors including 1 and n. The product of 5 of these is 648. Which one of the following is the 6th divisor of n? So we know that n has 6 divisors including 1 and n. So let's take a look at the prime factorization of n. First we'll say that there's only one prime factor. We'll call it p. Therefore n would be divisible by 1 and p. In this case n is p so it is also divisible by n but that is p. So that's not enough divisors. We need 6 and we only have 2. So let's add another prime factor. We'll call it q. So in this case n would be divisible by 1, p, q, and p times q. This is only 4 divisors and we need 6. So let's add another prime factor. We'll call this one r. And if n's prime factorization is pqr then n would be divisible by 1, p, q, r, pq, pr, qr, and pqr. This is 8 divisors which is too much. So we need to somehow get one in the middle. And the way that we do this is by saying that the prime factorization is pq squared. This is possible of course. For example let's say that p is 5 and q is 7. You can have a prime factorization of 5, 7, and 7. So in this case n would be divisible by 1, p, q, pq, q squared, and pq squared. So we know n is divisible by 1 and n so it is divisible by 1 and it is divisible by n which is pq squared. And if we multiply all these together we have three terms of p and six terms of q. So we would get p cubed times q to the sixth. So now we know the product of five of these divisors is 648. And if we break it up into its prime factorization we get 2 cubed times 3 to the fourth. Now we can clearly see that 2 is p and 3 is q. So which divisor didn't we divide? Well that's easy. If we have p cubed times q to the fourth times that last divisor equals p to the third, q to the sixth, that means the one that we're missing is q squared. And we know what q is. We know that q is 3. That means that q squared is 9. So that's the missing divisor. So the question asked us, which of the following is the sixth divisor of n? The answer is 9. Letter c.

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