Geometry Kaleidoscope. the sum of the distances from each point to AB and to CD equals to the sum of the distances from the same point to BC and AD.

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Geometry Kaleidoscope October 25, 2014 Problem set 1: squares and circles 1. Given a square ABCD, find the set of all points on the plane such that the sum of the distances from each point to AB and to CD equals to the sum of the distances from the same point to BC and AD. The set consists of the inside of the square (together with the boundary!) and the two diagonals of the square extended indefinitely in all directions. A complete to the problem must have two parts: Show that all the points in the set satisfy the required property; this is straightforward; Show that no other point does (need to use triangle inequality) 2. A coin rolls around another coin and makes a full circle. How many times does the coin rotate? First, make a guess. Then, make an experiment. Finally, find an explanation. When a coin of radius r rolls along a straight line and travels a distance of 2πr, it makes one full rotation. This almost obvious fact might make it seem that the coin would make just one rotation when it rolls around another coin of the same radius. It is worth it to give the following hint: suppose that a coin rolls for the distance of πr along a straight segment, and then continues along a segment of length πr which is at an angle α with respsect to the first one. What is the total angle by which the coin turns? During the first 1

half of motion, the coin turns an angle of π radians. It then has to turn an angle of 2π α to start rolling along the second segment. During the second half of the motion, the coin turns another π radians. Thus, the total rotation is 2π +(2π α). The next step is to understand what happnes when you roll a coin around a polygon of perimeter 2πr. The total rotation of the coin will be equal to 2π + (Sum of external angles of the polygon). Since the sum of external angles equals to 2π, a coin rotates 2π +2π = 4π degrees when rolling around a polygon. This corresponds to two full rotations! Finally, consider a circle of radius r as the limit of polygons of perimeter 2πr as you increase the number of sides. Since the coin rotates twice when it rolls around any polygon, it will rotatied twice when it rolls around th circle. 3. A ladder standing next to a wall starts sliding down and eventually falls. What is the trajectory of a kitten sitting in the middle of the ladder? (Assume the kitten is very calm and does not jump when the ladder starts moving). A good way to start is to just draw several positions of the ladder and mark positions of the midpoint. Notice that the ladder is vertical before it starts moving. Let O be the point in the corner and A,B be the end points of the ladder (with A being the upper end point). Let M be the midpoint where the kitten sits. Then OM is the median of OAB. Extend OM to a point P such that OM = MP and OAPB is a rectangle. Then, M is also a midpoint of OP. Since diagonals of a rectangle are equal, OM = AM = MB. After this, one can conjecture that the set of points is a (portion of) a circle centered at O with radius half the length of the ladder. The 2

above shows that the midpoint of the ladder lies on an arc of the circle. We also need to show that any point on the arc is a midpoint of the ladder at some point. To do this, take a point M on the arc. Extend OM to OP so that OM = MP. Let A be the projection of P onto the wall and B be the projection of P onto the floor. Then AB is the position of the ladder for which M is the midpoint. 4. A circle of radius R rolls inside of a circle of radius 2R. What is the trajectory of a point on the smaller circle? Let P be a point on the smaller circle which at some point was touching the bigger circle at a point A. Let O be the center of the bigger circle. Let T be the point at which the two circles are tangent to each other at the given moment. Because of the motion, the arcs TA and TP are equal to each other. Since the radius of the smaller circle is half of the radius of the bigger one, the angle subtended by the arc KT is twice that of the angle subtended by the arc AT. Therefore, AOT = POT. Thus, P lies on the segment AO. The trajectory in question is a straight line. 5. Given a triangle, inscibed a square into it in the following sense: draw a square inside the triangle such that two of its vertices are on the base of the triangle and two vertices are on the other two sides of the triangle. 3

Problem set 2: Lighthouse and other problems 1. How far can you see from a lighthouse of heighthassuming the Earth s radius is R? Given that R 6,400km and h 16m for the Kileau lighthouse (located on the North shore of the island of Kauai), estimate the maximal distance at which you can see a ship from the top of this lighthouse. Consider the triangle formed by the center of the Earth (point O), the top point on the lighthouse (point L), and one of the farthest points P you can see from L (i.e., a point on the horizon, from the point of view of an observer at the top of the lighthouse). This is a right angle triangle as OP LP. By Pythagorian theorem, we have d 2 +R 2 = (R+h) 2, where d = LP is the distance in question. Therefore, d 2 = (R+h) 2 R 2 = h (2R+h), where we used the formula for the difference of squares. Since h R, we have For R = 6,400km, we get d 2 h 2R = 2Rh d 2Rh. d = 80 2hkm. For h = 16m = 16 10 3 km, we get approximately 14km. Extension: ask them, how far we can see from our height? Also, how much further can you see when you increase the height of a lighthouse by a factor of 2? 2. What is the biggest number of squares on a 6 6 checkerboard that you can color in such a way that none of them are touching each other (even at a vertex)? 4

The difficulty here is not only to find a solution (which is not very hard), but to prove that this it is impossible to color a bigger number of squares. The trick to prove this is to look at squares of size 2 2 on the checkerboard. Each of those can contain at most one painted square. 3. Cut a rectangle into two pieces which can be put together to make a triangle. Connect midpoints of two adjacent sides. Cut out the resulting triangle and flip it to the other side of the rectangle. 4. Cut a triangle into four pieces which can be put together to make a rectangle. Drop an altitude (that lies inside of the triangle) and draw the line connecting the two midpoints on the remaining two sides of the triangle. These two lines divide the triangle into four parts that can be reassembled into a rectangle. 5

Problem set 3: examples and computations 1. What is the biggest number of sides that a polygon obtained as an intersection of a triangle and a quadrilateral can have? A nonconvex quadrilateral can intersect a triangle so that the resulting shape is an 8-gon. 2. Give an example of a triangle cut into (a) three congruent triangles; (b) four congruent triangles; (c) five congruent triangles; 3. A sail has the shape of a quadrilateralabcd with angles A = B = D = 45. Given that AC = 4 meters, find the area of the sail. Let H be the point of intersection of the extention of BC with AD. Introduce notation for AH and DH. Then, use the formula for the area of a triangle for ABH and CDH. Use Pythagorean theorem to find a relation between AH and DH from ACH. 4. Suppose that a square of side length a is inscribed into a circle. Take one of the four segments between the circle and the square and inscribe a new square into this segment. What is the side length of this square? Let O be the center of the circle, B be a vertex of the smaller square lying on the circle, and A be the midpoint of the side of the smaller square that connects the two vertices lying on the circle. Then AOB is a right angle triangle with the sides AB = x 2, OB = r = a 2 2, and AO = x+ a 2, where a is the side of the larger square and x is the side of the smaller square. Using Pythagorean theorem for this triangle, we get x = a 5. 6

Problem set 4 (Polygons) 1. What types of polygons can you get when you intersect a cube by a plane? Consider all possible cases. A point, a segment, a triangle, a quadrilateral (discuss all possible types!), a pentagon and a hexagon. You can not get more sides as a cube has only 6 faces. 2. What is the sum of external angles of an n-gon (polygon with n sides)? How does it change with the number of sides? Find a good explanation of your result. Here is an instructive argument: To count external angles, travel around the polygon as follows: start with a side, travel to the next angle, turn to continue along the next side, reach the next vertex, turn again to continue to the next side, etc. Eventually, you make a full rotation. Thus, the total angle is 2π. You can also do a straightforward angle count. 3. What kind of quadrilateral do you get if you connect the four midpoints of the sides of an arbitrary quadrilateral? Parallelogram. This result is known as Varignon s theorem. The two opposite sides of the quadrilateral in question are both parallel to the diagonal that lies between them. 4. What is the biggest number of parts you can get when drawing n straight lines on the plane? What is the biggest number of parts you can get for n = 2014? They will not necessarily get to a general formula, but working it out for small ns and figuring out a pattern can be very instructive. 7

Problem set 5: Circles 1. Suppose you are walking around the Earth s equator. What is the difference between the distance travelled by your head and your feet? (First, make a guess. The, make a computation. You can use the height of 1.6m (which is nearly 5 3 ) in the computation. Your answer should be a number. The distance your feet walk is 2πR. The distance your head walks is 2π(R+h). The difference is 2πh, or about 10 meters. 2. You tie a belt around the middle of a (perfectly round) orange. Then you increase the length of the belt by one meter and pull it away from the orange so that it is at the same distance from the orange on all sides. Now you do the same thing with the (perfectly round) Earth: tie a belt around the equator; increase its length by one meter, and pull it away from the surface to be at the same distance in all directions. Will the belt be closer to the Earth or to the orange? For example, would a mouse be able slip under the belt tied around the Earth? (First make a guess, then make a computation). This is very similar to the first problem. If you added the same distance, the circumferences increased by the same amount. Since C = 2π r, the radius changes by the same amount. 3. If you cool down steel by 1 1 C, it shrinks by a factor 100,000 (i.e., loses this fraction of its length). A steel wire is tied around the Earth s equator. If you cool it by 1 C, the wire will shrink and sink into the Earth to some depth. How far will it sink? Make a guess, then make a computation. 8

Problem set 6: Tying Goats 1. (Warm-up) What section of a pasture is consumed by a goat if the goat is tied to a single stake planted in the pasture? The inside of a circle. 2. You take a walk on a field hodling a goat on a 1-meter-long rope. You are walking along the path which is a rectangle with dimensions 3 meters 5 meters. What is the section of the field that goat has accesses to while on the walk? The rectangle with round corners and a hole in the middle. 3. How can you constraint a goat to an eye shaped field: In other words, how can the goat be tied using ropes and stakes so that it can eat only the grass in the eye-shaped field above? Tie the goat to two stakes which are at the centers of these two circles, using ropes whose length is equal to the radius. 4. A rope has been stretched between two stakes in a field. A goat is tied to this rope with another rope that is free to slide aong the first rope. What is the shape of the portion of the field the goat can eat? A rectangle with two semicircles on both sides. 5. Suppose that you know how to constraint the goat to a shape F and how to contstraint the goat to a shape G. Explain how to constraint 9

the goat to the intersection of F and G. Make both ties. 6. Use the previous problems to constraint the goat to a field in the shape of a semicircle. Use problems 1 and 4. 7. Use the previous problems to constraint the goat to a filed in the shape of a square. Two fields as in problem 4 perpendicular to each other. 10

Problem set 7: Bridges and Roads 1. As you are standing near a river (but not at the river bank!), you see a fire at a house on the same side of the river. You want to put out the fire as fast as possible, so you run to the river to get water and then to the house. At what point on the river should you get the water so that the distance you cover before you get to the burning house is minimized? 2. You are standing on a path connecting you with a tall tree. You want to place a mirror at some point on the path in such a way that when you look into the mirror you see the top of the tree. Where should you place the mirror? 3. Two cities are located on opposite sides of a river with parallel straight sides. You need to build a road between the cities. The road includes a bridge that needs to be perpendicular to the river. Where should the bridge be located so that the total length of the road is the shortest possible? 4. To get to school from your house, you need to cross two streets which intersect perpendicularly. You can walk to a street in any way you like, but have to cross a street perpendicularly to the street. What is the shortest path from your house to school? Make a picture and consider all possible cases. 5. Two towns are separated by two rivers. Each river has parallel banks. However, the two rivers are not parallel to each other. Where should bridges that cross the rivers perpendicularly to the banks be built so that the distance between the two towns is as short as possible. 11