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2002-2003 Problems Maine Mathematics Science and Engineering Talent Search Round 2 Grades 9-12 SOLUTIONS 1. Connect the midpoints of two opposite sides of a convex quadrilateral to the corners as shown. Showe that the area of the blue region is equal to the area of the red. Solution . Let ABCD the convex quadrilateral, M and m the midpoints , Dh and BH the distance between sides AB and CD as shown. The diagonal BD (as shown in blue color separates the area of the quadrilateral for the area of triangles ABD and BCD. H C M D A B h m 1 1 Area of (ABCD) = Area of (ABD) + Area of (BCD) = ( AB hD) (CD HB) 2 2 1 1 1 1 ( Am hD) (mB hD) ( MC HB) ( MD HB) = 2 2 2 2 1 1 1 1 Blue region{ (mB hD) + ( MD HB) } +Red region { ( Am hD) + ( MC HB) } 2 2 2 2 As m and M are the midpoints of the sides AB and BD respectively, we know that Am = mB and CM = MD : substituting these terms in the equation, 1 1 1 1 Blue region{ (mB hD) + ( MD HB) }+Red region { (mB hD) + ( MD HB) } 2 2 2 2 The terms multiplied together are identical, wherefore the areas of blue region and red region are equal and ½- ½ of the area of the convex quadrilateral. 2.Given three spheres that pairwise touch each other and also touch the plane P, determine the radius of the sphere that has its center on the plane P and that touches all the three spheres. Solution First, we have to impose some conditions on the spheres. Draw the three spheres and connect the respective centers and also connect the projections of the centers on the plane. The centers are C1 , C 2 , C 3 and projections P1 , P2 , P3 . Since C1C2 connects the spheres of two tangent spheres, C1C2 r1 r2 , likewise C2C3 r2 r3 . and C1C3 r1 r3 (1) If the spheres are tangent to the plane, the triangle P P2 P3 must exist. 1 Using the Pythagorean theorem (C1C3 )2 (r1 r3 )2 P P P P3 (r1 r3 ) 2 (r1 r3 ) 2 = 2r1r3 (2r1r3 ) 2 r1r3 2 1 3 1 Equivalently, P P2 2 r1r2 1 and P2 P3 2 r2 r3 (2) So, since the triangle must exist, its sides must obey the triangle inequality. 2 r1r2 2 r2 r3 2 r1r3 r1r2 r2 r3 r1r3 Likewise r3r1 r2 r3 r1r2 and r1r3 r1r2 r2 r3 These conditions are necessary for the spheres to be tangent to each other and the plane. There are two half-spheres with centers on the plane and tangent to the three given spheres. The First sphere The small half-sphere shaded with a darker tone of gray is the first sphere. Let P the center of the small half-sphere sphere and x be the radius of the sphere. PC1 x r1 PP1 ( x r1 ) 2 r1 x 2r1 x 2 2 PC 2 x r2 PP2 x 2r2 x 2 PC3 x r3 PP3 x 2r3 x 2 Drawing the triangle P1P2P3, the area of the triangle equals the sum of the areas of P1PP2 and P1PP3 and P2PP3 together. Using the Heron formula for the areas the radios x can be expressed in terms of r1, r2 and r3. P2 P1 P P3 Second sphere: the second sphere is covering the three spheres Let P be the center of the half-sphere and x be the radius. PC1 x r1 PC2 x r2 PC3 x r3 PP ( x r1 ) 2 r1 x 2 2r1 x 2 1 PP x 2 2r2 x 2 PP x 2 2r3 x 3 By a similar method, using the area of the respective triangles that are formed, we can determine x as a function of r1, r2 and r3. (Solution of: Cristina Domnisoru, Philip Exeter Academy, grade 12) . 3. Prove that the following numbers are all composite. 1000...001 (with 1991 0's) 111...1111 (with 1989 1's) 111...1111 (with 1990 1's) Solution First, we prove that the number: 1000...001 (with 1991 0's) is a composite number. We’ll use the fact that a n 1 (a 1)( a n 1 a n 2 a n 3 ..... a 2 a 1) if n odd 1000 ...... 01 101990 1 (1010 )199 1 (1010 1)(...) the last expression is the product of two numbers, therefore the original number is composite. Next, show that 111...1111 (with 1989 1's) is composite number. The sum of the digits (that is, the sum of the 1989 1’s) equal 27. As 27 is divisible by 3, therefore the number written as 1989 1’s is divisible by 3. Therefore it is composite number. Finally to show that 11……..1 (with 1990 1's) is composite number. As 1990 is an even number, the number 11…….1 is divisible by 11. More precisely, 11……..1 = 111010..... therefore, divisible by 11 and so, it is a 01 composite number. 4. Connect any point in the interior of a parallelogram with the four corners. Show that the area of the blue region is equal to the area of the red. Solution Let P the point as described, inside the parallelogram. The perpendicular distance of AB and CD is DH2. The parallel sides AB = CD and AD = BC. Draw perpendiculars from P to all four sides. The area of the blue triangles: 1 1 AD PH3 ' BC PH3 the ½ of the area of the parallelogram 2 2 1 1 The area of the red triangles AB PH1 DC PH1 ' the ½ of the area of the 2 2 parallelogram 5. We have twenty coins of the same size and external appearance some are real gold, while others are counterfeit. The counterfeit coins are lighter. Using at most eleven weighings on a pan balance, how can we determine how many coins are counterfeit? Solution First we make a “weight trial” by placing one coin on each pan. There are three possible outcomes: 1. One of the pans is heavier. Then one of the coins is gold, the other is the lighter counterfeit coin. Place both coins in one pan and weigh them against pairs of remaining blocks (those being divided into nine pairs arbitrarily). Any pair of coins which outweigh the first pair must consist of two gold coins. If the first pair is the heavier then both coins in the second pair are counterfeit. If both pair balance, the second pair has one gold and one counterfeit. 2. On the first trial weigh both pans balance. In this case both coins are gold or both coins are counterfeit. Just as before, place both coins in one pan and weigh them against pairs of the remaining coins. Assume that the first k pairs of blocks from the nine blocks have the same weight as the test blocks and that the (k+1)st pair tested have a different weight. If k=9then all the coins weigh the same and so there are no counterfeit coins. Assume, without loss of generality that the ( k+1 )st pair tested is lighter. Then the original two blocks as well as all those of the first k pairs tested must be gold coins. Therefore in the 1+(k+1) = k+2 weight trials already made we have found (k+1) gold coins. Now we compare the two blocks of the (K+1)st (heavier) pair. (This is the (k=3)rd trial). If both blocks are the same weight, they both must be gold. If they are not of the same weight, one is gold and one is counterfeit. In either event we are able after k+3 weight trial to display a pair of blocks of which one is counterfeit and the other gold. 3. By using this pair we can determine in, at most, 8-k weight trials how many gold coins remain among the 20-2(k+2)=16-2k unweighed ones. Using the techniques employed in the first event. The number of weight trials used in the second event is then equal to k+3+(8-k) = 11 6. We have a faulty clock, that is uniformly fast, and is ahead every day less than an hour. On January 1st 2002 it showed the correct time, and the same day at 1:05 p.m. the two hands of the clock were in perfect overlap. When will this faulty clock show the correct time again? This problem will be on Round 4 again. 7. A volleyball net has the form of a rectangular lattice with dimensions of 50 x 600. What is the maximum number of unit strings you can cut before the net falls apart into more than one piece? Solution We consider the volleyball net as a graph, its nodes as vertices, and the strings as edges. A cycle is any closed path in a graph which does not pass through the same vertex of the graph twice . A connected graph without cycles is a tree. Our objective is to erase as many edges as possible, while keeping the graph connected. We delete the edges one by one as long as we can.. Notice, that if the graph has a cycle , then we can delete any of the edges in this cycle. But a connected graph without a cycle is a tree, thus, when we have obtained a tree, we cannot delete any more of the graph’s edges. Let us calculate the number of edges in our graph at this final moment. The number of vertices is the same as originally. , that is, equals 51 601 30651. On the other hand a tree with this many vertices must have 30651-1 = 30650 edges. At the very beginning we had 601 50 600 51 60650 edges. Thus we can delete no more than 30000 edges - and it is easy to see that we actually can do this. 8. Two knights of each color sit on an irregular fragment of a chessboard as shown. They move as in chess (without the requirement that alternate moves be by alternate colors) and must remain on the fragment. Can they be moved so that the white and black knights exchange positions?(In my idea folder) Solution. Number the squares as shown in the figure and build a graph by connecting two nodes if they are associated with squares that a knight can traverse in a single move. The graph, annotated to show the initial positions of the knights, can be seen here. Node 3 provides convenient "off-street parking" for any knight the player wishes to place in the next available upper position; that is, the knights can be shuttled into the lower nodes while dropping off any desired knight at node 3. That knight can then be moved upward and the process repeated. The puzzle is so flexible that more complicated variations can also be solved. If the knights were individually distinguished, say by the letters A through D, you could create any of the 4!=24 possible permutations of their initial positions -- even if squares 4 and 5 were removed from the board fragment! 9. There are seven lakes in Lakeland. They are connected by 10 canals so that one can swim through the canals from any lake to any other. How many islands are there in Lakeland ? Mathematical Circles Page 145 (Note: to solve this problem, you need to know Euler’s formula for planar graphs. You may find in our links, or by using any search engine. If you need help to find it, please e-mail us) If a planar graph is depicted properly, then it divides the plane into several regions called faces. Let us denote the number of faces by F, the number of vertices by V and the number of edges of the graph by E. Euler’s formula for planar graphs: for a properly depicted connected planar graph the equality V-E +F = 2 Solution. We have a planar graph consisting of 7 lakes (these are the vertices), 10 canals (these are the edges) and we want to determine how many islands (faces) we have. If the graph is connected, we have V-E + F = 2 7 – 10 + F = 2 f=5 that is 4 islands and the land outside If any edge is disconnected, the total number of the connected portion is V-1 and the total number of edges is E-1 . The number of faces doesn’t change it is always 4, as long as the canals don’t cross. Answer: Always 4 islands. 10. Find all natural numbers which becomes ten times greater if you insert a zero between their units digit and their tens digit. Solution. All numbers ending in 0 have this property. No other natural number has this property. Note: this problem is the prelude” of a somewhat more challenging problem coming up in Round 4. *