Hello and welcome to the Math Kangaroo Media Library. You are about to view interactive solutions to levels 9 and 10 of the 2012 competition. You will likely notice that some of the solutions are slightly different from the suggested solutions you may have already reviewed. 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. In problem number 1, we are given the following isosceles triangle. We are told that by means of two line segments, its area has been divided into two regions. Now, three of those regions are displayed and the areas are indicated. We are asked to compute the area of the remaining region indicated here by the question mark. The first thing that I will do is name the isosceles triangle, triangle ABC, by labeling the vertices as follows. And immediately we can note that the area of this triangle ABC can be computed by adding together the areas of the two smaller triangles, which have names ABN and ANC. The next thing to observe is that the triangles ABN and ANC, they share a side here, and this line segment AN emanating from the vertex A joining the midpoint of the opposite side is by definition a median of the triangle ABC. It therefore divides the area of triangle ABC into two equal pieces, and those pieces are exactly triangle ABN and triangle ANC. They therefore have equal areas, and we can compute the area of ABN easily by adding the areas of those two smaller regions. So we have 3 plus 6 is equal to the area of triangle ABN, which is now equal to the area of triangle ANC. And that triangle consists of the shaded region with area 3 and the region with area that we are trying to compute indicated by the question mark. And that gives us our final calculation here, 9 is equal to the area of the region indicated by the question mark plus 3, and that gives us the area of the region indicated by the question mark. And so the answer to the problem is D, 6. In problem number 2, we are asked to perform the calculation 11 and 11 hundredths minus 1 and 111 thousandths. We begin by rewriting the problem like so, 10 times 1 and 111 thousandths is the same thing as 11 and 11 hundredths, so that will slightly simplify our calculation and make it easier to keep track of the decimal. And so now we write 10 times 1 and 111 thousandths minus 1 and 111 thousandths. We factor out the common factor, and that leaves us with 9 times 1 and 111 thousandths, which is equal to 9 and 999 thousandths. So the answer to the question is D. In problem number 3, we are looking at a rectangular prism that we are told consists of 4 pieces, each made up of 4 cubes of a single color. So we have here 3 pieces in shades of gray, and the piece that is least visible is in white. We are asked to determine the shape of that piece. The first thing to note is that the front face of this prism does not contain any white cubes. Therefore, the shape of the white piece has to be flat. In other words, there is no protruding cube like in C or B or A. So now we have narrowed down our choices to D and E. Now E cannot be the answer, because looking at the back corner here, that piece, the cube that would be directly below it, is not of the same color as these 3. And it cannot be the lightest shade of gray, because we can see all 4 of them. And it cannot be the darkest shade of gray, because then it wouldn't be connected to this piece. Therefore, the cube that is directly below that corner has to be white. And so, we are looking at a piece of this shape. So the answer is D. In problem number 4, we are told that Alice and Bob communicate by means of coded messages, which Alice constructs as follows. She takes a letter in the alphabet and assigns to it a number that corresponds to the letter's position. So A is first and it gets 1, B is 2, C is 3, and so forth. She takes that number and multiplies it by 2 and then adds 9 to the result. Bob receives a sequence of numbers, and because he is familiar with the encoding scheme, he is able to recover the position of the original letters, and hence the message that Alice is sending him. So if he receives 25, 19, 45, and 38, what is the original message? Here are the possible answers, and we notice that because the words differ in the last letter, we should begin by deciding what letter corresponds to the code 38. So to undo Alice's calculation, we would say that 38 is equal to 2 times a number corresponding to the position of the letter in the alphabet, plus 9. And so 29 is 2 times that number, and so 29 over 2 is equal to a number corresponding to the position of the letter in the alphabet. However, this is not an integer, and so we must conclude that Alice made a mistake. And the answer to the problem is E, there is a mistake. In problem number 5, we are looking at a square ABCD of side length 4 cm, a triangle ECD that shares its base with the square, the line segment CD, and we are also looking at a line G that is the extension of the line segment AB. The problem is asking us to determine the distance between the vertex E and the line G. So in other words, the distance of the line connecting those two that is perpendicular to the line G. So let me sketch in here a perpendicular line, and the problem will be to determine the length of that line. So let's also label those line segments with their known lengths, CD, DA, CB, and AB all have a length equal to 4. Now comparing the areas, we know that the area of triangle ECD is equal to the area of the square, which is 4 cm squared, or 16 cm squared like that. We also know that the area of triangle ECD can be calculated by taking 1 half times its base, the line segment CD with length 4 cm, times its height, which is right here, that's the height. So there is a point right above C, and the distance from E to that point is the height of triangle ECD. We can solve for H, because we have the following relationship, 16 cm squared is equal to 1 half times 4 cm times H, and this tells us that H is equal to 8 cm. Therefore, the length of the line connecting E to the line G is 8 cm, and this is 4 cm, so the answer to the problem is C, 12 cm. Question number 6. The sum of the digits of a 7-digit number is 6. What is the product of these digits? So we have 1, 2, 3, 4, 5, 6, 7 digits, and let's say that all of them are 1. We have 7 1's, and so the sum of the digits is 7, but we cannot have that, we must have the sum equal to 6, therefore one of these has to be decreased, and so we now have a 0 as a digit. The product of any number with 0 is equal to 0, therefore our digits have a product equal to 0, and the answer is A. Question number 7. We are asked to consider a right triangle. I sketched that triangle. Here we have ABC, a right triangle, with legs of length 6 and 8 cm. The points KL and M are midpoints of the sides, and we connect them with lines to obtain the triangle KLM in red. The question is asking us to determine the perimeter of triangle KLM. So we make a very important observation that because the line segment MK is parallel to the leg CB, and the line segment KL is parallel to the leg AC, the triangle KLM is similar to the triangle ABC, therefore their perimeters are constant multiples of one another. And we have perimeter of triangle ABC is equal to some constant multiple of the perimeter of the triangle KLM. So let's compute the perimeter of the triangle ABC and then figure out what the constant multiple is. We have lengths of two of the legs, and we do not know the length of the hypotenuse. However, by the Pythagorean theorem, since this is a right triangle, we can compute the hypotenuse, which I will call lowercase c. c squared is equal to 6 squared plus 8 squared, which comes out to 100. Therefore, c is equal to 10 centimeters. That tells us that the perimeter of the triangle ABC is equal to 24 centimeters. Now, to determine the constant K, we note that M is the midpoint of the side AC, therefore the length of the segment MC is 3 centimeters. The length of the segment AM is also 3 centimeters. Similarly, CL has length 4 centimeters because L is the midpoint of that side. And we computed the hypotenuse, so we have AK is 5 centimeters and so is KB. We now note that the side KL is exactly half the side AC, the side MK is exactly half the side CB, and also the hypotenuse ML of the red triangle is exactly half the hypotenuse AB. So the scaling factor here, K, is equal to 1 half. And that allows us to compute the perimeter of triangle KLM as 1 half of the perimeter of the triangle ABC, and so we have the answer equal to B, 12 centimeters. In question number 8, we are asked to determine which of the following expressions does not have the property that replacing 8 by any other positive number throughout does not yield the same answer. Let's start to simplify these expressions and see what we can come up with. In A, inside the parentheses, we have 8 divided by 8, and so that always gives us 1, no matter if we use a different positive number than 8. In B, we have 8 plus 1 minus 8, and that always yields 1. And it doesn't matter if we were to use a number other than 8. In C, we end up with 8 divided by 3 times 8, and so after division, we end up with 3, irrespective of whether we were to replace 8 by another positive number. Now in D, we have 8 minus 1 plus 8, and notice that unlike in question B, we have a different order of operations, and so 8 minus 1 plus 8 leaves us with 16 minus 1, or if we were to replace 8 with some positive number x, we would have 2x minus 1 as our answer, and for various choices of positive numbers x, we obtain different quantities. Therefore, the answer to this question is D. The quantity 8 minus the quantity 8 divided by 8 plus 1 is not invariant under the replacement of 8 by different positive numbers. In question number 9, we are asked to compute the perimeter of a quadrilateral. A picture is not provided, so I sketched in a four-sided figure. We are told in the problem that two of the sides have lengths 1 and 4, and also that a diagonal of length 2 cuts the quadrilateral into two isosceles triangles. So let's give lengths to sides according to the information in the problem. One of the sides has length 4, so let's say it's this long side, and in order to create an isosceles triangle, we have to assign length 2 or 4 to the remaining side, but in order for this to become a triangle, we must choose 4 rather than 2, otherwise the triangle inequality fails for a triangle with sides of length 2, 2, and 4. Similarly, we have several choices here for the second isosceles triangle. We must choose a 2 for one of the sides for this to be an isosceles triangle, and then the remaining side can have length 1 and 4, so we must choose 1 because it's stipulated in the problem that two sides have lengths of 1 and 4, and we check that this is indeed an isosceles triangle. And now the perimeter is 1 plus 2 plus 4 plus 4, which gives us 11, and so the answer to this question is D, the perimeter is equal to 11. Question number 10. Each of the numbers 144 and 220 is divided by the positive integer n, giving a remainder of 11 in each case. What is the value of n? So what the question is asking us to do is, first of all, to observe that the number n divides the numbers 144 minus 11 and also 220 minus 11. In other words, n divides the numbers 133 and 209. We obtain a factorization of 133, which is 7 times 19, and also we factor 209 to obtain 11 times 19. Since n is supposed to divide both of these numbers, we must choose n to be 19 because according to our prime factorization, that is the only common factor between 133 and 209. Therefore, the answer to question number 10 is D, 19.

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