Session # 1 SOLUTIONS

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09/23/2017 Session # 1 SOLUTIONS Problems marked with a star ( ) are from Lectures and Problems, A Gift to Young Mathematicians, by V.I. Arnold, c 2015 Mathematical Sciences Research Institute. 1. Mary and Murray want to buy their mother a cake for her birthday.

1 09/23/2017 Session # 1 SOLUTIONS Problems marked with a star ( ) are from Lectures and Problems, A Gift to Young Mathematicians, by V.I. Arnold, c 2015 Mathematical Sciences Research Institute. 1. Mary and Murray want to buy their mother a cake for her birthday. Murray was seven dollars short of the price and Mary was one dollar short. They combined their money to buy the cake, but even then they did not have enough. How much did the cake cost? (The price of the cake is in dollars, no cents.) The cake cost $7. Murray had 0 money; if Murray had had as much as a dollar, they could have bought the cake. 2. A bottle with a cork costs $1.10 while the bottle alone costs 10 cents more than the cork. How much does the cork cost? The cork costs 50 cents. Given that the bottle costs 10 cents more than the cork, the bottl plus cork is equal (in cost) to two corks plus 10 cents. That equals $1.10, so two corks equal $ Two people left at dawn, at the exact same time, one traveling from A to B, the other one from B to A. They travel at a constant speed, without stopping. They meet at noon. The first one arrives at B at 4 p.m., the second one arrives at A at 9 p.m. At what time was dawn that day? Let s draw a picture. C marks the spot where they meet. We can use trial and error, assume dawn happened at (say) 4, see if it works. If not,try 5,then 6, etc.,until it works. Doing one trial might not be a bad idea. Also, since the question does not depend on the actual value of the distance from A to B, we could give it a value. Say 10 miles. Let s see if dawn at 4 a.m. works. There are then 8 hours to 12 noon, meaning that in the end the first person traveled a total of = 12 hours, the second one a total of = 17hours. So the first person s speed is 10/12 = 5/6 mph, the second person s speed is 10/17 mph. At this speed, in 8 hours, the first person covers 8 5/6 = 20/3 miles and is at C, the second person 8 10/17 = 80/17, and is at C. Now AC + CB = 10 so we should have 20/3 + 80/17 equals 10. Unfortunately, it equals 580/51 = To solve the problem, let us call x the time it took for the two people to meet; that means that dawn happened

2 MCFAU/2017/09/23 2 at x before 12 noon. The first person went from A to C in x hours, her speed is AC/x. The second person did this same trajectory (in reverse) after 12 noon, taking 9 hours, so the second person s speed is AC/9. The ratio of speeds is: speed first p. speed second p. = AC x AC/9 = 9 x. Now let s start at B. The first person takes 4hours to get from C to B, the second person takes x hours to get from B to C. Thus CB speed first p. speed second p. = 4 CB/x = x 4. This means that we must have x/4 = 9/x, or x 2 = 36, thus x = 6. Dawn happened at 12 6 = 6 a.m. 4. What is the maximum area that a right triangle with a hypotenuse that is 10 inches long can have? 25 square inches. If you use the hypotenuse of a right triangle as the diameter of a circle. the other vertex must be on the circle. This means that the altitude has to be at most equal to the radius of the circle, so 5 inches in this case. 5. P Q is the diameter of a circle with center O; R is a point on the circumference. If P O = OQ = QR = 1, find P R. The triangle must have a right angle at R so that by the Theorem of Pythagoras P R = (P O + OQ) 2 QR 2 = Two vertical poles 20m and 80m high stand apart on a horizontal plane. Find the height in meters of the point of intersection of the lines joining the top of each pole to the foot of the other.

3 MCFAU/2017/09/23 3 We labeled some points. We have AP Q ACD so that thus Also P CQ ACB so that implying P Q 80 = AP AC, P Q = (80 AP )/AC. P Q 20 = P C AC, P Q = (20 P C)/AC.

MCFAU/2017/09/23 4 Equating the two expressions for P Q and canceling AC we see that P C = 4AP so that AC = AP +P C = 5AP. Using this in the first expression for P Q, The answer is 16 meters.

4 MCFAU/2017/09/23 4 Equating the two expressions for P Q and canceling AC we see that P C = 4AP so that AC = AP +P C = 5AP. Using this in the first expression for P Q, The answer is 16 meters. P Q = (80AP )/(5AP ) = Something easy now. An equilateral triangle has sides of length 6. What is its area. There are at least two ways. The altitude of the triangle can easily be found by the theorem of Pythagoras; it is one leg of a right triangle whose other leg has length 3, the hypotenuse has length 6, so the altitude has length = 27 = 3 3. The area is A = = 9 3. Another way is to use Heron s formula according to which the area of a triangle of sides of lengths a, b, c is A = s(s a)(s b)(s c) where s = (a + b + c)/2. For our triangle s = 9, s a s b = s c = 3 so that as before. A = = ABCD is a quadrilateral, AD = CD, AB = 8, BC = 6, ADC = 60 and the angle at B is a right angle. Find the area. We draw the segment AC.

5 MCFAU/2017/09/23 5 Triangle ABC is a right triangle with legs of length 6 and 8; by the theorem of Pythagoras its hypotenus AC has length 10. Now ACD is an isosceles triangle with an angle of 60 ; it must be equilateral. We now have Area of ABC = 1 (6 8) = 24, 2 Area of ACD = 25 3, Area of ABCD = And now for something completely different. For this problem you need to know the the volume of a ball (sphere) of radius r is V = 4 3 πr3. A plastic ball of radius 12 inches has been hollowed out and filled with lead: So we have a ball of lead covered by plastic. This ball when placed in water sinks so that is even with the surface of water. If the specific gravity of lead is 10 times that of water, of the plastic is 1/36 that of water, what is the radius of the lead ball? (The specific gravity of water can be taken as 1).

$MCFAU/2017/09/23 6 The answer should be the root (square root?, cubic root?) of a fraction times an integer. Let r be the radius of the lead ball.$

6 MCFAU/2017/09/23 6 The answer should be the root (square root?, cubic root?) of a fraction times an integer. Let r be the radius of the lead ball. By the principle of Archimedes, the weight of the volume of water displaced by the ball must equal the weight of the ball. The weight of a ball of water of radius 12 is (given the specific density of 1) 4 3 π(12)3. The weight of the lead ball is 4 3 πr3 10 and of the plastic shell ( 4 3 π(12)3 4 ) 3 πr3 1. We should have π(10)r π 1 ( 12 3 r 3) = π123. We can solve this for r 3 to get r 3 = so that the answer is r =

4. Solve for x: 5. Use the FOIL pattern to multiply (4x 2)(x + 3). 6. Simplify using exponent rules: (6x 3 )(2x) 3

4. Solve for x: 5. Use the FOIL pattern to multiply (4x 2)(x + 3). 6. Simplify using exponent rules: (6x 3 )(2x) 3 SUMMER REVIEW FOR STUDENTS COMPLETING ALGEBRA I WEEK 1 1. Write the slope-intercept form of an equation of a. Write a definition of slope. 7 line with a slope of, and a y-intercept of 3. 11 3. You want