Hello and welcome to the Math Kangaroo 2007 level 11 and 12 solutions video series. First of all, I'd like to congratulate you for even participating in the Math Kangaroo. When I was in elementary school and in high school I would participate in every single year and I'm definitely familiar with how big of a mathematical challenge it is. It goes way beyond what your schools will typically expect of you, so you should really be patting yourself on the back for taking on this challenge. Now the purpose of this video series is to show you a logical path to get to the correct answer of every single question on this exam. Keep in mind the methods which you see in these videos are not necessarily the only methods to solve the problems. In fact, you might use a completely different method and still get the correct answer, and in that case your method might be equally good or it might be better or worse in some situations. If you have any questions about this sort of thing, please feel free to email me at thomas.mathkangaroo.org, or if you have any other questions in general about the exam, please don't hesitate to email me. Now before moving on to the videos, I'd like to offer two quick pieces of advice that I offer with every video series, and that's first of all, before you watch any of the videos, please at least try to solve the problem for which video you're going to be watching, because if you just watch the video without trying to solve the problem, it's going to be like you're spoon-feeding yourself information, and you'll hear everything that the video says, and you'll see all of the work, and you'll think, oh, okay, that makes sense, and you'll move on to the next one, but you won't have actually learned anything. You won't have any idea of how to generalize the approach or how to apply it to other problems, because you just had everything given to you. If you really want to learn, you need to struggle with the problem first. Then you, when you watch the video, you're going to see where you went wrong, where you could have improved, where you could have been more efficient, and overall your problem-solving skills will improve, but the key is to struggle with the problem first. The second piece of advice I have to offer, and this one I can't stress enough, please, please write down as much of your work as possible. Whether you're doing practice problems now, or whether you're doing the actual work on the exam, write out as much of your work as possible, because all too often people who are good at math try keeping too much of it in their head just for the sake of speeding through everything. They think it's a little bit faster to keep it in their head instead of writing it out, but the problem is you're keeping so much stuff in your head that you're liable to forget something, or maybe make a silly mistake while you're remembering it, so just to avoid any sort of silly, completely unnecessary mistakes, you can prevent all of that just by writing out your work. And also, if you write out your work for practice problems, then you'll be able to more easily go back to what you were thinking about during those problems when you're watching the videos, so you'll be able to find whatever mistake there is in your math much more easily. So there's only benefits to writing down your work. Now with these two pieces of advice in mind, I hope that you find the video series extremely useful, and that you learn a lot from it. So good luck, have fun, and enjoy! Problem number one. A billiard ball always bounces off the side of a billiard table at an angle of 45 degrees, as shown. If it continues on the path shown, which pocket will the ball fall into? So if a ball is bouncing off at 45 degrees, then as you can see on the grid that's laid out on the table, it forms a line of slope either 1 or negative 1, depending on which way it's going. So all we have to do is draw lines of slope 1, and follow the path that the ball makes. And that path will look like this, going from here, to here, to here, to here, all the way into pocket C. And we know that it's slope 1 because it's moving from one diagonal of the square to the next diagonal. This is how you know it will always be slope 1, from one diagonal to the next. So following this path, we see it's in pocket C, and that is our final answer. Problem number two. How many zeros does the number 12 squared times 15 cubed end with? So probably the easiest way to do this problem is to try breaking it up into smaller factors with the same exponents. So we could rewrite 12 as 3 squared times 4 squared. 12 squared is 3 squared times 4 squared. And we could rewrite 15 as 3 cubed times 5 cubed. And in particular, we have a 3 squared and 3 cubed, so we could pair those up, and that'll give us 3 to the 5th times 4 squared times 5 cubed. And now we know what each of these are. These are easier to evaluate than 15 cubed, well, or 12 squared. Although if you want to, you can just multiply these out on your own, and then multiply out their products, and see how many zeros there are at the end. But this way is just a little bit quicker, I think. We know 3 to the 5th is equal to 243, 4 squared is 16, and 5 cubed is 125. And in particular, if we multiply 125 by 243, we know that since there's a 3 in the ones digit, and a 5 in the ones digit here, that we're going to have a 5 in the ones digit of their product. So that won't necessarily be helpful. But if we multiply the 5, the one with the 5 in its ones digit, by the number with the 6 in its ones digit, then we know that we're going to get a 0 in the final digit. Think of it as, you know, 5 times 3 gets us 15, and 5 times 6 gets us 30. Well, 30 has a 0 in the ones digit. So when we multiply these two, we get 2,000 times 243. And so we know that when we take the product of these two numbers, it's going to be these three zeros at the end of it. So our final answer is D. Problem number 3. Adam, Barry, and Carl have a total of 30 marbles. When Barry gave Carl 5 marbles, Carl gave Adam 4 marbles, and Adam gave Barry 2 marbles. Each of them had the same number of marbles. How many marbles did Adam have to begin with? So let's go ahead and call A the number of marbles that Adam originally had, B the number for Barry, and C the number for Carl, before they did any sort of trading. Now, we know that A plus B plus C should be equal to 30, since that's the total number of marbles they have. We can also write out how many marbles each person has in the new situation. So for instance, we know that Barry originally had B marbles, but he gave away 5 to Carl, and then he received 2 from Adam. So it's B minus 5 plus 2. This is the new amount that Barry has. And by similar logic, we find that the new amount for Carl is C plus 5 minus 4, and we have A plus 4 minus 2 for Adam. And each of these should be equal to 10, as stated by the problem, since each of them had the same number of marbles after the trades. Simplifying each of these expressions, we have B minus 3, C plus 1, and A plus 2. And we know that A plus 2 is supposed to be equal to 10, which means A is equal to 8. And in particular, we can double-check this, because we could solve for all three of these letters, A, B, and C, and then add up what we get. So A will be equal to 8, C will be equal to 9, and B will be equal to 13. And 8 plus 9 plus 13 does equal 30. So just as a double-check, to be extra safe, we see that the numbers make sense. So again, A should be equal to 8, since 8 plus 2 equals 10. And so our final answer is E. Problem number 4. Triangle ABC is inscribed inside the circle centered at O. The shaded triangle has an area of square root of 3. What is the area of triangle ABC? So looking at this problem, we know that the area of this shaded triangle is square root of 3. And in particular, we also know the formula for the area of a triangle is 1 half base times height. So the base in this case is AO, and the height is going from B down to this diameter here that is going horizontally across. And in fact, if we go ahead and look at this unshaded triangle here, since triangle ABO has a base that goes from the edge of the circle right to the center, and OC is also going from the center of the circle right to the edge, we know that AO and OC must be of the same length. And that length is equal to the radius of the circle, incidentally. But that means that both of these triangles have the same base. Furthermore, we know that in both cases, there's the edge of the triangle going from O to B. And although this isn't the height of the triangle, they both have the same base and they both have the same top point, which means that both of these triangles actually have the same height as well. So they have the same base and the same height, which means they must have the same area. So if this area is square root of 3, then this area must also be square root of 3. And so the area of the total triangle is just the sum of these two areas, which is 2 times square root of 3. So our final answer is A. Problem number 5, what is the value of the sine of 1 degree over the cosine of 89 degrees? So at this point, it would be worth noting that when we're talking about the first 90 degrees on a unit circle, it's very easy to convert the sine of one angle to its equivalent cosine. As an example, consider the very well-known sine 30 degrees is equal to cosine 60 degrees, or sine 60 degrees is equal to cosine 30 degrees. In fact, the sine of any angle is equal to the cosine of its complement. If you look at the unit circle, you'll see that this holds true because the sine is effectively the y-value along the unit circle, and the cosine is effectively the x-value along the unit circle. And just based on how that first quadrant, that first 90 degrees looks, you know it's always going to be, you change whether it's sine or cosine, and then you change it to the complementary angle. So the reason why this is useful is because we know that the complement of one degree is 89 degrees, so sine of one degree is just equal to cosine 89 degrees. So these two things are actually equal to each other, and since they're equal to each other, we know that the fraction must be equal to 1. So our final answer is E. Problem number six. Michael correctly answered 80% of the test questions and left the remaining five questions unanswered. How many questions were there on the test? So we know that Michael answered 80% of the questions, and he left all of the remaining questions unanswered, which means that five questions make up the remaining 20% of the exam. There's no questions that he got wrong. So if it's asking how many questions there were on the test, and we know that five questions corresponds to 20%, well all we have to do is multiply 20% by 5 to get 100%, and that's the same as multiplying 5 by 5 to get 25. So we know that 25 is the total number of questions, since that corresponds to 100%. So our final answer is B. Problem number seven. The points B, C, and D divide the segment AE into four equal parts. The three arcs are semicircles with diameters AE, AD, and DE. What is the ratio of the length of arc AE to the sum of the lengths of the arcs AD and DE? So we're looking at the ratio between this arc length and the sum of these two arc lengths. Now let's start by naming one of these segments. They're all of equal length, so it doesn't matter which one we pick, but let's go ahead and call one of them, let's say D. All right, we'll go ahead and name the length of this little d. Now the length of this arc, since it's a semicircle, and the semicircle has a radius of 2D, since it's going from A to C, this is the radius of it, we know that that the circumference of any circle is pi times the diameter, or pi times 2 times the radius. And since this is half a circle, it's half the circumference, so the diameter would be 4D, and we want half of that, so it's 1 half pi times 4D, and this gives us the the arc length of the big semicircle. And incidentally, we can rewrite this as 1 half pi times 3 plus 1 times D, since 3 plus 1 is just equal to 4. And in particular, we could then break up this 3 plus 1 into two separate terms, 1 half pi 3D plus 1 half pi 1D. But in fact, 1 half pi 3D, well the diameter of this circle with the diameter AD is just equal to 3D, so this portion right here is just equal to this arc length, and similarly this portion right here with the diameter of 1D is equal to this arc length. So we just showed algebraically that this arc length is exactly equal to the sum of these two arc lengths, just by breaking up this 3 plus 1 here. So since these two arc lengths are equal to each other, their ratio must be 1 to 1, so our final answer is E. Problem number 8. The sum of five consecutive integers is equal to the sum of the next three consecutive integers. What is the largest of these eight integers? So before starting on this problem, I'd just like to point out that in all of my work with the kangaroo videos, it seems like every single exam has at least one problem like this at this grade level. So this is a generally good thing to learn about, and it's not super tricky either. You just have to go ahead and name the consecutive numbers all with respect to one variable. So to explain what I mean, let's suppose that the smallest of all of the integers is called X. In that case, the next integer, the next consecutive integer, will be X plus 1, and then the next one will be X plus 2, and then X plus 3, and so on, all the way up to X plus 7. Those will be the eight integers, and based on what the problem says, we know that X plus all of the numbers up to X plus 4, the first five consecutive integers, are equal to the next three, X plus 5, X plus 6, and X plus 7. So if we add up both sides of the equation, we get 5X plus 10 is equal to 3X plus 18. Subtract 3X from both sides and subtract 10 from both sides, and we get 2X equals 8, or X equals 4. And remember, we don't want the smallest integer, which is X, we want the largest integer, which is X plus 7. So X equals 4, which means X plus 7 equals 11, and so our final answer is D. Problem number nine, which of the following are the same as the given figure? So to solve this question, there's really no sort of special trick. It all comes down to your visualization. In particular, you have to be able to imagine rotating around this shape and keeping track of where this centerpiece will be. Will it be down here, or will it be up here? Will this edge be pointing out or in? Will this edge be pointing out or in, left or right? Keeping track of all of this, there's no special trick to it. It's just something that you have to practice. So in particular, if you don't have the ability to visualize this shape rotating, try to cut out a bunch of pieces of maybe index cards or some other firm sheets of paper, and then paste or tape them together, or maybe tape together some dice, for example. You could go ahead and take a bunch of dice and line them up and tape them together at the appropriate points to make the shape, and try rotating it around, and this should help you improve your visualization. Basically, the idea is if you can't visualize this on your own, then you should try making a 3D model for it, so that way you could practice visualizing the thing by working with a physical copy of it. So to solve this question, well, there are two figures that match the one that are given in the problem. The first one is figure W. It's just a rotation. It's a slight rotation going clockwise 90 degrees, if you're looking at it from the top. So that's the rotation that W is making. And then we also have figure Y being the other one. It's as if you're taking the original figure and just sort of pushing it, tilting it backwards. That's how I like to think about it. The other two are not rotations of it. It involves flipping a portion of it, so that it's not possible to achieve just by rotating it around. And if you go ahead and use a 3D model, this will be more visible. But W and Y are the ones that actually result from somehow moving around this figure. So our final answer is A. Problem number 10. Some historians claim that the ancient Egyptians used a string with two knots to construct right triangles. If the length of the string is 12 yards and one of the knots is at a point X, three yards from one end, at what distance from the other end should the second knot be tied to obtain a right angle at X? So we're trying to obtain a right triangle where essentially this piece of rope and this piece of rope will form the legs of the triangle, and this longer piece of rope will form the hypotenuse of the triangle. We also know that the total length is 12, and that the shortest length is equal to 3. Well the nice thing to realize here is that 3 plus 4 plus 5 is equal to 12. And if you remember from your high school geometry class, there are these things called Pythagorean triples, which are basically very nice right triangles, which pair up like so that you know that 3 squared plus 4 squared equals 5 squared. There are a few other well-known Pythagorean triples. If you don't remember them, go ahead and look them up. But basically the idea is if you see two of the three numbers from the triple, you don't have to do any extra math. You know right away what the third number should be. So in this case we see that since we have 3 for one end and the three lengths have to add up to 12, we could put in 4 and 5 for the other two string lengths, and we'll have a Pythagorean triple, which means we will definitely have a right triangle. No extra math required. And so the length of this part of the string, which is what the problem is asking for, is equal to 5. So our final answer is C.

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