Instructions
Please answer the following questions to the best of your ability. Show your work where appropriate, as the method of arriving at a solution is as important as the solution itself. You may require a separate sheet of paper for your calculations. Calculators are discouraged for the first four problems to better exercise your problem-solving skills.
Section 1: The Pythagorean Journey
A 17-foot ladder is placed against a vertical wall such that its base is 8 feet from the foot of the wall. The top of the ladder then slips down the wall by 4 feet. How much farther from the wall is the base of the ladder now? Express your answer in simplest radical form.
Section 2: The Cosmic Clock
Two comets, Comet Halley and Comet Austen, were observed from Earth in the year 1835. Comet Halley has an orbital period of exactly 75 years, while the lesser-known Comet Austen has a period of exactly 90 years. Assuming their periods remain constant, what is the very next year that both comets will be observable from Earth together?
Section 3: A Matter of Ratios
The ratio of apples to oranges to pears in a fruit basket is 3:5:2. After 4 apples are added to the basket, the ratio of apples to oranges becomes 1:1. How many pieces of fruit were in the basket originally?
Section 4: The Expanding Square
The side length of a perfect square is increased by 20%. By what percentage does the area of the square increase?
Section 5: The Triangle Challenge
A right triangle is known to have integer side lengths. One of its sides has a length of 15. This side could be one of the legs or the hypotenuse. Find all possible perimeters for such a triangle, and then calculate the sum of these perimeters.
Answer Key
Section 1: The Pythagorean Journey
Answer: The base of the ladder is now (2√42 - 8) feet farther from the wall.
- Step 1: Find the initial height. Let the height be h. Using the Pythagorean theorem (a² + b² = c²), we have h² + 8² = 17². This simplifies to h² + 64 = 289, so h² = 225, and h = 15 feet.
- Step 2: Find the new height. The ladder slips down 4 feet, so the new height is 15 - 4 = 11 feet.
- Step 3: Find the new base distance. Let the new base distance be b. Now, 11² + b² = 17². This gives 121 + b² = 289, so b² = 168. Therefore, b = √168.
- Step 4: Simplify and find the difference. To simplify √168, we find the largest perfect square factor: √168 = √(4 × 42) = 2√42 feet. The distance it slid is the new distance minus the old distance: 2√42 - 8 feet.
Section 2: The Cosmic Clock
Answer: The next year they will appear together is 2285.
- This problem requires finding the Least Common Multiple (LCM) of 75 and 90.
- Step 1: Prime Factorization.
75 = 3 × 25 = 3 × 5²
90 = 9 × 10 = 2 × 3² × 5 - Step 2: Calculate LCM. The LCM is found by taking the highest power of each prime factor present in either number: 2¹ × 3² × 5² = 2 × 9 × 25 = 450.
- Step 3: Find the year. They will appear together again in 450 years. The next year is 1835 + 450 = 2285.
Section 3: A Matter of Ratios
Answer: There were 20 pieces of fruit in the basket originally.
- Step 1: Set up variables. Let the original number of apples, oranges, and pears be 3x, 5x, and 2x, respectively. The total fruit is 10x.
- Step 2: Form the new ratio. When 4 apples are added, the number of apples becomes 3x + 4. The number of oranges remains 5x. The new ratio of apples to oranges is (3x + 4) : 5x, which is given as 1:1.
- Step 3: Solve the equation. We set up the proportion: (3x + 4) / 5x = 1 / 1. Cross-multiplying gives 3x + 4 = 5x. Solving for x, we get 4 = 2x, so x = 2.
- Step 4: Calculate the original total. The original total number of fruits was 10x. Substituting x = 2, we get 10(2) = 20.
Section 4: The Expanding Square
Answer: The area increases by 44%.
- Step 1: Define the original square. Let the original side length be s. The original area is A₁ = s².
- Step 2: Define the new square. Increasing the side by 20% means the new side length is s + 0.20s = 1.2s.
- Step 3: Calculate the new area. The new area is A₂ = (1.2s)² = 1.44s².
- Step 4: Find the percentage increase. The increase in area is A₂ - A₁ = 1.44s² - s² = 0.44s². The percentage increase is (increase / original) × 100 = (0.44s² / s²) × 100 = 0.44 × 100 = 44%.
Section 5: The Triangle Challenge
Answer: The sum of all possible perimeters is 466.
There are two main scenarios to consider:
Scenario A: 15 is a leg of the triangle.
Let the sides be 15, b, and c (hypotenuse). Then 15² + b² = c², which means 225 = c² - b² = (c - b)(c + b). We need to find factor pairs of 225. Since c and b are integers, (c-b) and (c+b) must have the same parity (both even or both odd). Since their product (225) is odd, both must be odd. The factor pairs of 225 are (1, 225), (3, 75), (5, 45), and (9, 25).
- Pair (1, 225): c-b=1, c+b=225 → c=113, b=112. Perimeter = 15+112+113 = 240.
- Pair (3, 75): c-b=3, c+b=75 → c=39, b=36. Perimeter = 15+36+39 = 90.
- Pair (5, 45): c-b=5, c+b=45 → c=25, b=20. Perimeter = 15+20+25 = 60.
- Pair (9, 25): c-b=9, c+b=25 → c=17, b=8. Perimeter = 15+8+17 = 40.
Scenario B: 15 is the hypotenuse.
Let the sides be a, b, and 15. Then a² + b² = 15², so a² + b² = 225. We need to find integer pairs (a, b) that satisfy this. By checking integers for 'a' less than 15, we find a solution.
- If a = 9, then 9² + b² = 225 → 81 + b² = 225 → b² = 144 → b = 12.
- This gives the sides (9, 12, 15). Perimeter = 9+12+15 = 36.
Final Calculation:
The possible perimeters are 240, 90, 60, 40, and 36. The sum is 240 + 90 + 60 + 40 + 36 = 466.