Hello, and welcome to the Math Kangaroo Media Library. You are about to view interactive solutions to levels 11 and 12 of the 2015 competition. You will likely notice that some of the solutions are slightly different from the suggested solutions you may have already reviewed. So as you follow along, compare your own solutions and the suggested solutions with the current presentation and make sure you understand the differences. If you have any questions or comments, feel free to contact me at the address provided. My name is Luke, and I'm a past Math Kangaroo participant, and I hope you will find this presentation useful in preparing for your next Math Kangaroo competition. Question number one. Andrea was born in 1997 and her younger sister Charlotte in 2001. The age difference of the two sisters at any time is therefore which one of the following. And we have here some time considerations, so let's draw a timeline. And I'll have here on one end Andrea's birthday sometime in 1997, so I will need two years here, 1997, 8, 9, 2000, 2001, and 2010, too. Let me just draw that in. So this, open that here, will represent the year 1997 on our timeline, and then 1998, 1999, the year 2000 is next. And then starting with 2001, we can look for Charlotte's birthday, and we'll need one more year, 2002. So we don't know exactly the date of Andrea's birthday, but she was born sometime in 1997, so I'll mark all of this region here as the possible location of Andrea's birthday on our timeline. Likewise, we don't know Charlotte's birthday for sure, but any place in here between 2001 and 2002 is a good place for placing Charlotte's birthday. Like so. And now, in the worst possible scenario for this problem, the closest time span between the two birthdays would be right here after 1998, if Andrea is born as late as 1997 as possible, and we're as close to this open dot on the timeline as possible on one end, and at the other end, Charlotte is born as close to the beginning of 2001 as possible, then the shortest possible time span between their birthdays would be here, indicated in blue. Well, let me mark that as the shortest time between birthdays. And that is just under three years if we're not actually extending to the open circles on our timeline. And so, looking at the answers, the best answer here is answer E. The difference in ages of the two sisters at any time cannot be less than three years. Question number two. The expression quantity A minus B raised to the fifth power plus quantity B minus A raised to the fifth power is equal to which of the following five expressions? Let's rewrite our quantity that we have to simplify. A minus B raised to the fifth power plus B minus A raised to the fifth power, and quickly we notice that we have the same letters here, A minus B, but subtracted in different order. Subtraction is not commutative, so we cannot add these, these are not like terms, but we can do something about the order here. We can rearrange the order and the second term and the second set of parentheses by factoring out minus inside that set of parentheses. Let's do that, and then we'll see what happens. We have the first quantity unchanged, and then here I'll write, instead of B minus A, I'll write A minus B. I'll have to multiply by a negative one out front to make these equivalent, and then, of course, everything is still raised to the fifth power, like so. So the expressions here, I'll just underline them quickly. What I wrote on the right hand side of equal sign, and what we started with on the left hand side of equal sign are equal, and then we can distribute the exponent. We have the first quantity here changed again, and then I have plus, there's going to be A minus one to the fifth power, multiplied with quantity A minus B to the fifth power, of course, raising negative one to the fifth power results in a negative one here as a factor, so we're subtracting from quantity A minus B to the fifth power, the same quantity, and that gives us, in the end, just zero. So the answer here, question number two, what this expression simplifies to is zero, and that would be choice A. Question number three, how many solutions does the equation two raised to the power two X is equal to four raised to the power X plus one have? And let's take a closer look at our equation here. We are not going to actually try to solve it. We are going to decide how many solutions exist. Type of equation we're looking at is an exponential equation because we're looking to isolate the variables that are present in the exponents. The best way usually to approach such a problem is to work with the same base on both sides. So we have base two on the left-hand side and the base four on the right-hand side, and the exponent, we can rewrite our exponents, so on the left-hand side, we have two and then two times X. Applying that to our knowledge of rules of exponents, we can multiply two times X like so. A power raised to a power results in the total exponent that's the product of those exponents, so that's equivalent to the left-hand side. On the right-hand side, I'll reverse the addition rule of exponents, and I have here four to the power one times four to the power X. With the same base, the exponents add, and so that expression is equivalent to what I had above on the right-hand side. And let's keep going. Here inside the parentheses, we have now another name for four, so that's four to the power X. On the right-hand side, I have four times four to the power X. And we have, we know this a common factor in the expressions on either side of the equal sign. So since four to the power X is always greater than zero for any value of X, we can divide both sides by four to the power X and not worry about dividing by zero. And when we do that, we have a one on the left-hand side, and we have a four on the right-hand side when we factor out four X and divide by it. And now what we notice is that our original equation here simplifies to this equation. So this last line is equivalent to our first line, and I'll copy that here again. But there is no value of X for which one is equal to four. But one is not equal to four, irrespective of the value of X. Okay, so this is an equation with no solution. There is no value of X. I can choose to make one equal to four, therefore what we have is a contradiction. The correct answer here to the problem is to choose zero or choice A, there are no solutions. Question number four, Diana drew a bar chart representing the quantity of the four tree species registered during a biology excursion. And here on the right, we have that chart. Jasper thinks that a pie chart would better represent the ratios of the different tree species. What does the respective pie chart look like? So our task in this problem is to transform the data represented here in this bar chart into an equivalent pie chart. And looking at the bars, we focus our attention at the shortest, smallest bar, which is in black over here. So on the corresponding pie chart, the black slice of the pie chart has to be the smallest. And we can compare that against other slices. And it is easy to see that the choice here in D cannot be the answer because black is not the smallest slice. But it seems like in the other four choices, it is. So let's continue. And focusing again on the black bar over here, we see that together, the black and the white look very much like all of the bar shaded here in light gray. So I can show this to you by moving that black bar in position above the white bar. Let me do that. So I can select it and move it up like so. I'll just move it right above here so you can see how they line up. And they stack up pretty well. And I can even draw a line here up above to see that these indeed look like together they are equal. Together, I mean the white with the black represent the same quantity of trees as the lightly shaded gray bar over here. So looking at our choices, choice E here doesn't look very good because the white slice here together with the black slice that is much less than the lightly shaded slice. And choice C, it actually looks like white together with black would be accounting for about half the pie, which is more than the lightly shaded slice. Here in choice B, they kind of look equal, the white with the black, and also in choice A. So then we have to finally pay attention to the last color we haven't really looked at, the bar that's to the very right. I'll say that it's a darker shade of gray, not quite completely black. And that bar here is less than the white bar in quantity, the shorter, so it represents less trees of that particular species. And we see that in choice B over here, those two colors are essentially equal. They take up half the pie. They are not equal, however, so we have to eliminate also choice B. And so finally, by the process of elimination, the best pie chart to represent the bar chart is here, the very first one in choice A. Question number five, how many of the figures shown on the right can be drawn with one continuous stroke without drawing any of the segments twice? And here we have the four figures made up of concentric circles, or sometimes just one circle. And it looks like there is a full ear segment drawn along the diameter, extending a little bit out of the circle. And here the diameter is omitted in picture B. And our task here is to decide if I were to draw this or if I were to trace along it, can I actually complete that task of tracing the figure without tracing any segment twice or without drawing over any segment twice? I'll just make a larger version of this diagram so we can better see what's going on. So let's scale this up a little bit. This here is two percent. And I'll show how to trace some of these figures in figure A, we have that circle with the diameter and little notches on the outside. What I would do is I would start here on the left and go towards the circle, like so, along the segment, and then follow the circle on the outside, making a complete trace, like so, all the way around, until I get to this point that I have seen. Again, now I cannot continue along the circle because that would be retracing my steps, so I take a right turn over here, pass along the diameter, and once I get to the next intersection I cannot go left or right because I've been there already, so I just continue out of the circle. And that way I have not traced anything more than once. So certainly figure A here is possible. And now let's look at figure C because figure C already contains this figure A. So what I can do is try that same method again, now I'll just do it a little faster without drawing arrows. So I start over here on the outside, I follow my circle all the way around, like I did before, and once I get to this intersection I have to go right. And now what happens is I'm essentially looking at the same figure again, just shrunken, sitting inside what I just traced. So let's do it again. Same plan, follow the circle along like that clockwise until I get to the intersection where I have to turn right, and then we turn right and keep going past this intersection, past the next intersection, and out of the circle. So I have traced figure C without overdrawing any of the places I've already visited. So that is possible to do, and figure D is also possible to trace the same way. So follow the line in, all the way outside, and then clockwise along the circle, move in, clockwise along the next inner circle, move in, clockwise along that circle, and then all the way out. So hopefully that is clear how to do that. We have three figures here that are possible to be traced without drawing over any of the segments twice. What about figure B? And we actually have four. Well, let's use a different color here, I'll do it blue, follow the same method. I get to start on the outside, trace the circle clockwise all the way out. At the first intersection, if I were to turn and make that segment, I would have to double back on it, and therefore that's not allowed, so I keep going, and I get over here, and now at this intersection, I have to stop. So I cannot go left, because that would continue over where I have already been, I cannot go up because I've been there already, and so here there is a little piece that has not been traced over. So maybe I just got unlucky, let's try it another way. I started at the end here, let's start over here, if I go left, then I would have to back up, that's not allowed, so I can go up or down, let's just continue clockwise, and at this intersection over here, I go follow that little short piece, I have to double back, so that's not possible, and once I go past it, there's really no way to go back over there. So we see that it is really not possible to trace figure B in a way such that we're not drawing over any segment twice, the other three figures that we managed to trace out in such a way, so the answer here is D, that three of these four figures we can continuously trace without drawing over any segment twice. Question number six, we add the 31 integers from 2001 to 2031, and divide the sum by 31, what result do we get? So what I'm looking at is a sum consisting of 31 terms, and it begins with 2001, then 2002 is the next term, all the way, continuing with consecutive integers up to 31, 2031. So I can rearrange the sum here, I can split up each term, each of the 31 terms into a pair consisting of the number 2000, and whatever is left over, so 2000 plus 1 would be the first pair, 2002 would be the second pair, and then continue like so until the last pair is 2000 plus 31. And there are exactly 31 pairs like this, so the number 2000 appears 31 times, and after I have collected those, I have 1 plus 2, all the way up to 31, so the sum of the first consecutive 31 positive integers is what's here in parentheses. There are various ways of evaluating that, other than just adding these up. One way is to use a formula that goes something like this, the first n positive consecutive integers added up yield the product here, n with n plus 1, and then all of that is divided by 2, and here n is equal to 31. So we can write that down we have here 31 times 2000 and then following the formula 31 times 32 divided by 2. Okay so we notice that 31 here is a common factor in both terms let's just factor it out and what I have then is 2000 plus 16 since 32 divided by 2 is the 16 over here. Okay so that is our expression and we factored it with a factor of 31 on purpose because we recall the directions here ask us to divide this sum by 31. So if I divide by 31 I can essentially go and drop this factor of 31 and then what's remaining is 2000 plus 16. So for number six the result of performing all these calculations is 2016 or in other words answer D. Question number seven a square piece of paper is folded along the dashed lines as shown on the right one after another in any order or direction from the resulting square one corner is cut off. Now the paper is unfolded how many holes are there in the paper? Here's our diagram I'll make a larger version of it I'll move it over scale it up and then we can better see what's going on and what we have here is the sheet of paper that is a square and we're going to create three columns and three rows of it and fold along the dashed lines in any way or order we like as long as in the end we have a square sheet that represents exactly one ninth of the original area. So we can immediately say that this sheet of paper over here that square is nine pieces thick because we have folded over nine of those squares together on top of one another. So cutting a corner off makes exactly nine cuts in the original sheet of paper. So let's imagine we're doing that I'm going to make a cut here in the corner let's do the upper right hand corner like that cut that off so that's my cut line and then I'll just erase and then I'll just erase here what is being put away so that corner that I removed actually corresponds to nine corners being removed in the original sheet of paper and now let's say that we have some unfolding going on and by symmetry here let's say that I'm imagining the unfolding is being done along this red line so I'll copy that red line over here and if I unfold this now like that I should be making a cut that is symmetrical about this red line so let me see if I can sort of sketch what the end result would look like I should have a piece of paper exactly mirror image about this line something roughly like this and I really need to focus on this feature over here and what I have is two of the nine corners after one unfolding so I can look for that in my original diagram and that should be somewhere let's say over here I need to make this v and then erase of course what has been removed so that little piece and that little piece and there we go that has been cut away now unfolding unfolding this again now like so down and then down again would create a pair of cuts right over here let me draw them that be this v upside down and then by symmetry I would have another v like that and so when four of these corners line up in just the right way and I remove them well what I'm creating is a hole so this is this is that that hole that we're looking for okay so this is hole number one okay and now if we were to fold this over again we should have a completely unfolded sheet of paper and with the symmetry we're considering here there would be cuts over here and removing those gives me sort of a edge here that's being a wedge here that's being cut off that's not quite a hole and then one more cut is necessary for the total of nine and that would be that would be this corner that's being cut up over here okay so now I have a total of nine corners removed and that guarantees that one hole exists in my sheet of paper so I can right away eliminate answer a and then I can think about whether two holes are possible so that's the question can we have a second hole and we have another hole here and the answer is no why not because if I were to try to place another hole over here um let me let me do that in one of the places where it's uh possible to make a hole where actually maybe let me do something like this I'll kind of draw a uh an outline of a hole and then I'll try to place it over uh one of the possible intersections of these cross line lines and we see that if I place it over here what I'm now creating is a second hole or it's possible to have one over here or to have one over here but no matter where I place it let me just put it here for clarity what I'm going to create is at least one square with two corners removed okay so I'll have at least one square with two corners removed if there is a second hole and that means in the original I would have to make two cuts because being folded back together into a stack of nine and having two holes means that there is some square with two corners missing and that is only possible if I cut off a second corner here in the folded position so we cannot have a second hole we did discover that at least one is guaranteed so the answer here has to be b there's going to be exactly one hole in the paper after unfold question number eight a drinking glass has the shape of a truncated cone as shown here in the figure to the right the outside of the glass without the base should now be covered with colored paper what shape does the paper need to be when laid flat in order to completely cover the whole glass without overlap so what we have here is this container and we're going to see which shape of a sheet of paper is necessary out of the five shapes given here so that once we have this sheet of paper cut in such a way we can simply wrap it around this container and cover it completely on the outside not including the base of course and so maybe let's think backwards if we start with the rectangle and try to coil it around and glue together some of these edges that I'll identify over here so if I glue in the rectangle this red edge together with this red edge by curling them around and what I'm going to get is a cylinder and clearly this shape here is not a cylinder it has slanted sides now in the trapezoid if I try to do the obvious thing and glue together the blue edges that I'm outlining over here what's going to happen is I won't be able to actually do that these these don't match so here we don't really we don't really get anything as the edges do not align now in the circle sector in C that is possible to glue together so I can in orange here outline the edges I'm interested in and if I glue them together then actually what we end up looking at is a circular cone so it looks like we're on the right track a circular cone is what we're talking about here in the problem and then we need to truncate it how are we going to truncate it we are going to cut away an arc over here of that of that cone like so we try that in orange we're going to truncate it by making a circular cut over here like that okay and that piece that's removed here is exactly the piece that is necessary if I were to continue this like so to make my drinking glass shape into a into a cone so let's try this a little further something like that that's exactly what I need and we can see that this shape that is outlined over here matches exactly the shape that has to be removed in order to make this glass shape that we're talking about in the problem so it's not quite a circle sector that gives us a circular cone but it's a part of a circle sector that I need and that is the shape here in in E it's a circle sector with a smaller sector there removed from from the bottom question number nine three semicircles have diameters which are the sides of the right triangle their areas are x y and z as shown here in the diagram to the right which of the following is necessarily true okay so we have some expressions over here one inequality and four equations relating the areas of the semicircles here labeled with x y and z I'll just make the diagram slightly larger so let's move it over let's enlarge it and then and label the diameters because that is an important part of this question so the semicircle have diameters which are the sides of a right triangle so let's label them in red I'll have let's work alphabetically so the circle here the semicircle with area label x let's say that the diameter is going to be a capital a the next one let's say in blue alphabetically y going to have a diameter of capital b and then z let's make it green I'll call that diameter capital c and almost immediately we're tempted to use the Pythagorean theorem so we have c squared is going to be equal to and then we have a squared plus b squared and that is the Pythagorean theorem and somehow we have to relate the capital a b and c to the capital x y and z well those would be the areas of these circles or semicircles rather so x is the area of the semicircle with diameter a so that would be one half the area of a circle so that is i times the radius squared and the radius here is exactly half of the diameter like that now I can do the same thing for y that's one half the area of a circle with diameter b so the radius is half of b squared and for z finally I have one half i times the radius so half the diameter c over 2 squared like that and I'll simplify a little bit I have here for x one half times pi and then a is squared and then I have an extra one half squared so that's one fourth times a squared and likewise here for y I'll have a one fourth b squared and then for z I'll have a one fourth c squared okay so finally simplifying this all the way I can just multiply the fractions and I have one eighth pi a squared is equal to x one eighth pi b squared is equal to y and one eighth pi c squared is equal to z now we know that c squared is equal to a squared plus b squared so we can start replacing here what we know and let's do that so now I'll use orange let's say and we have c is going to be equal to one eighth pi c squared and then from the Pythagorean theorem that's equal to one eighth pi c squared is equal to a squared plus b squared like so and I can distribute so let's have one eighth pi together with the a squared plus one eighth pi together with the b squared and lo and behold we've seen those already those are names for x and names for y after replacing over here the expressions that we have already seen initially okay so I made that replacement and I add here as a conclusion the statement from the very left hand side I read off as z and that z is equal to x plus y and looking at our choices here the answer to question nine is c x plus y is equal to z question number 10 which of the following is the complete list of the number of acute angles a convex quadrilateral can have we have here a mention of a convex quadrilateral a quadrilateral is simply a shape with four sides such as a rectangle over here and let me draw another four-sided figure and I'll use that to demonstrate this idea of convexity any any line that I draw between any two points on the boundary of a of a rectangle like so or like so will remain completely inside of my rectangle this other shape I have over here sometimes has this property for example these lines remain completely inside that shape but if I try to make a connection here like that or maybe like this then my line starts to pass outside of the boundary and that is not allowed so here we would have on the top convex as an example the convex shape and this one here is not convex okay well looking at my rectangle again I have an angle sum here I can sort of play with I have four 90 degree angles usually denoted like this and so that is a total of 360 4 times 90 and we note that here we have zero acute angles okay so that allows us to discount answer e right away it's possible to have zero acute angles and then we can imagine that I'm somehow distorting this shape for example if I were to if I were to push on the bottom right hand corner and kind of keep the upper left corner intact I'm affecting the the angle measures here and I'm creating probably an acute angle here looks like and then it looks like this angle over here will also be acute so how would I do that numerically well I don't have to work too hard I would just make two of my angles less than 90 degrees and then adjust the other two angles so that the angle sum still comes out to 360 degrees okay so here I have an example with two acute angles okay let's see if we can make one acute angle I won't mess up my picture anymore but numerically you can have a acute angle one of them and then an obtuse angle to make up for having this acute angle and then two angles with measures of 90 degrees like that so here I have one acute angle and then let's follow the same thinking to look at three acute angles and then I'll need an obtuse angle so the angle sum here is again 360 degrees okay and so now I have 0, 1, 2, and 3 on the list so that discounts choice D and also choice A now the remaining choices here differ in one respect we have to choose C if it's possible to have actually four acute angles but that is not possible as we will note here we cannot have an angle sum of less than 360 degrees which does happen with four angles that are less than nine with four acute angles angles okay at the border case borderline situation is at the very top over here where we have exactly four right angles and if we decrease all of them just ever so slightly so they become acute well we have to decrease the angle sum here past 360 okay so it's not possible to have four acute angles and we eliminate choice C and the answer here to question number 10 is the complete list is represented in B we can have 0, 1, 2, or 3 acute angles in the convex quadrilateral

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