The length of CE is 2, so CDE has the same length base as ABC. Since D is the midpoint of AC, the perpendicular distance from D to BC is half that of the perpendicular distance from A to BC. Thus the height of CDE is half that of ABC.
CDE thus has the same base as ABC and half the height of ABC. Its area is thus half that of ABC.
The area of an equilateral triangle of side length s is (s2√3)/4. So the area of ABC is √3, and the area of CDE is (√3)/2.
Sunday, July 10, 2011
Saturday, July 9, 2011
Summer Problem Solving Marathon Solution #30
Diagonal AC is the hypotenuse of right triangle ABC, with legs 18 and 21. So AC is √(182 + 212) = &radic(324 + 441) = √765.
AD is then the hypotenuse of right triangle ACD, with legs 14 and √765. AD is thus √(142 + 765) = √(196 + 765) = √961 = 31.
The perimeter of ABCD is thus 18 + 21 + 14 + 31 = 84.
AD is then the hypotenuse of right triangle ACD, with legs 14 and √765. AD is thus √(142 + 765) = √(196 + 765) = √961 = 31.
The perimeter of ABCD is thus 18 + 21 + 14 + 31 = 84.
Summer Problem Solving Marathon Solution #29
First we determine the length of the longest possible growing path. We do this by enumerating the distances between points. Two points in the grid can be separated by 0 to 3 units horizontally and 0 to 3 units vertically. Using the Pythagorean Theorem, that gives us the following possible distances:
√1: 0,1 and 1,0 separation
√2: 1,1 separation
√4: 0,2 and 2,0 separation
√5: 1,2 and 2,1 separation
√8: 2,2 separation
√9: 0,3 and 3,0 separation
√10: 1,3 and 3,1 separation
√13: 2,3 and 3,2 separation
√18: 3,3 separation
That's a total of 9 different distances, so the longest possible growing path would be of length 10.
Next we check to see if a length 10 path is possible, and (if it is possible) how many such paths there are. Since the movement options are more constrained when the distances are long than when they are short, we construct the path in reverse.
First we need a length √18 gap. This is possible only by moving from one corner dot to the opposite corner dot. For convenience, we label the dots (x,y), where x and y both range from 1 to 4. Then our path can start:
(1,1), (4,4)
(4,4), (1,1)
(1,4), (4,1)
(4,1), (1,4)
So there are four choices for the first pair of elements in the path.
Since the diagram is symmetric, these four choices will proceed in the same way. For convenience, we focus on the case (1,1), (4,4). From (4,4) we need a gap of √13. This requires moving 3 units in one direction and 2 in the other. There are two points that meet this requirement: (2,1) and (1,2). Symmetry is still preserved, so we focus on the case (2,1).
Next we need a gap of √10. This requires a movement of 3 units in one direction and 1 in the other. There are two points meeting this condition: (1,4) and (3,4).
However, (1,4) will not work. We next need a gap of distance √9, which requires a separation of 3 in one direction and 1 in the other. The points meeting this constraint are (1,1) and (4,4), but both of these are already in the path.
So we must proceed with (3,4). The only point of distance √9 from (3,4) is (3,1), so this move is forced. From here, we need a gap of distance √8, which requires a separation of 2 units in both directions. The only point meeting this constraint is (1,3), so this move is also forced.
Next we need a distance of √5, with separations of 2 and 1. (2,1), (3,2), and (3,4) all meet this constraint, but only (3,2) is unused, so this move is also forced.
Next we need a distance of &rdic;4, with separations of 2 and 0. (3,4) and (1,2) both meet this constraint, but only (1,2) is unused, so this move is also forced.
Next we need a distance of √2, with a separation of 1 unit in both directions. (2,1) and (2,3) meet this constraint, but only (2,3) is unused, so this move is also forced.
Finally we need a distance of √1, which requires a move of a single unit horizontally or vertically. There are four points meeting this constraint, but (3,1) is already in the path. The other three are available.
Looking back, we had 4 choices for the first move, 2 for the second, and 3 for the final. All other moves were forced. So there are a total of 4 x 2 x 3 = 24 paths of length 10.
The desired answer is then 10 x 24 = 240.
√1: 0,1 and 1,0 separation
√2: 1,1 separation
√4: 0,2 and 2,0 separation
√5: 1,2 and 2,1 separation
√8: 2,2 separation
√9: 0,3 and 3,0 separation
√10: 1,3 and 3,1 separation
√13: 2,3 and 3,2 separation
√18: 3,3 separation
That's a total of 9 different distances, so the longest possible growing path would be of length 10.
Next we check to see if a length 10 path is possible, and (if it is possible) how many such paths there are. Since the movement options are more constrained when the distances are long than when they are short, we construct the path in reverse.
First we need a length √18 gap. This is possible only by moving from one corner dot to the opposite corner dot. For convenience, we label the dots (x,y), where x and y both range from 1 to 4. Then our path can start:
(1,1), (4,4)
(4,4), (1,1)
(1,4), (4,1)
(4,1), (1,4)
So there are four choices for the first pair of elements in the path.
Since the diagram is symmetric, these four choices will proceed in the same way. For convenience, we focus on the case (1,1), (4,4). From (4,4) we need a gap of √13. This requires moving 3 units in one direction and 2 in the other. There are two points that meet this requirement: (2,1) and (1,2). Symmetry is still preserved, so we focus on the case (2,1).
Next we need a gap of √10. This requires a movement of 3 units in one direction and 1 in the other. There are two points meeting this condition: (1,4) and (3,4).
However, (1,4) will not work. We next need a gap of distance √9, which requires a separation of 3 in one direction and 1 in the other. The points meeting this constraint are (1,1) and (4,4), but both of these are already in the path.
So we must proceed with (3,4). The only point of distance √9 from (3,4) is (3,1), so this move is forced. From here, we need a gap of distance √8, which requires a separation of 2 units in both directions. The only point meeting this constraint is (1,3), so this move is also forced.
Next we need a distance of √5, with separations of 2 and 1. (2,1), (3,2), and (3,4) all meet this constraint, but only (3,2) is unused, so this move is also forced.
Next we need a distance of &rdic;4, with separations of 2 and 0. (3,4) and (1,2) both meet this constraint, but only (1,2) is unused, so this move is also forced.
Next we need a distance of √2, with a separation of 1 unit in both directions. (2,1) and (2,3) meet this constraint, but only (2,3) is unused, so this move is also forced.
Finally we need a distance of √1, which requires a move of a single unit horizontally or vertically. There are four points meeting this constraint, but (3,1) is already in the path. The other three are available.
Looking back, we had 4 choices for the first move, 2 for the second, and 3 for the final. All other moves were forced. So there are a total of 4 x 2 x 3 = 24 paths of length 10.
The desired answer is then 10 x 24 = 240.
Friday, July 8, 2011
Summer Problem Solving Marathon Question #31
[Value = 4 points]
ABC is an equilateral triangle with side length 2. D is the midpoint of AC. Point E is chosen so that C is the midpoint of BE. What is the area of the triangle CDE?
ABC is an equilateral triangle with side length 2. D is the midpoint of AC. Point E is chosen so that C is the midpoint of BE. What is the area of the triangle CDE?
Thursday, July 7, 2011
Summer Problem Solving Marathon Solution #28
Working from right to left, we find that the first place with an obvious error is 4 + 2 = 1 in the 10,000 place. Assuming the previous parts are correct, we should have carried 1 from the previous place. So we have two possibilities:
(1) The 1 is correct, and the sum should be 11. Then either 4 needs to be replaced by 8, or 2 needs to be replaced by 6. However, replacing 4 with 8 creates an error in the hundreds place, so this possibility can be ignored.
(2) The 1 is incorrect. Then the 4 and 2 are correct, so 1 needs to be replaced with 7. But replacing 1 with 7 creates an error in the tens place.
So the only option is that 2 is replaced with 6. Making this substitution throughout, we find that the result is a correct addition. So the answer is 2 + 6 - 8.
(1) The 1 is correct, and the sum should be 11. Then either 4 needs to be replaced by 8, or 2 needs to be replaced by 6. However, replacing 4 with 8 creates an error in the hundreds place, so this possibility can be ignored.
(2) The 1 is incorrect. Then the 4 and 2 are correct, so 1 needs to be replaced with 7. But replacing 1 with 7 creates an error in the tens place.
So the only option is that 2 is replaced with 6. Making this substitution throughout, we find that the result is a correct addition. So the answer is 2 + 6 - 8.
Summer Problem Solving Maraton Question #30
[Value = 4 points]
Let ABCD be a quadrilateral in which AB is of length 18, BC is of length 21, and CD is of length 14. Let angle ABC be right, and let the diagonal AC be perpendicular to CD. What is the perimeter of ABCD?
Let ABCD be a quadrilateral in which AB is of length 18, BC is of length 21, and CD is of length 14. Let angle ABC be right, and let the diagonal AC be perpendicular to CD. What is the perimeter of ABCD?
Wednesday, July 6, 2011
Summer Problem Solving Marathon Solution #27
There are eight triangles of minimal size in the diagram. For convenience, number these 1 through 8, reading left to right on the top half of the square and then left to right on the bottom half of the square.
There are four triangles each made up of two of the smallest triangles -- one made from triangles 2 and 3, one made from 4 and 8, one made from 6 and 7, and one made from 1 and 5.
There are no triangles of size 3. There are four triangles of size 4 -- one made of triangles -- one made from triangles 1, 2, 3, and 5; one made from triangles 2, 3, 4, and 8; one made from triangles 4, 6, 7, and 8; and one made from triangles 1, 5, 6, and 7.
There are no triangles of any other size. Thus there are a total of 8 + 4 + 4 = 16 triangles in the diagram.
There are four triangles each made up of two of the smallest triangles -- one made from triangles 2 and 3, one made from 4 and 8, one made from 6 and 7, and one made from 1 and 5.
There are no triangles of size 3. There are four triangles of size 4 -- one made of triangles -- one made from triangles 1, 2, 3, and 5; one made from triangles 2, 3, 4, and 8; one made from triangles 4, 6, 7, and 8; and one made from triangles 1, 5, 6, and 7.
There are no triangles of any other size. Thus there are a total of 8 + 4 + 4 = 16 triangles in the diagram.
Subscribe to:
Posts (Atom)